TEST BANK for Molecular Biology of the Cell 6th Edition by Bruce Alberts-1-1 by ACADEMIAMILL

1. Scientists discover more than ten thousand new species of living organisms every year. What is shared between all of these organisms? A. They are made of cells, whose nuclei enclose their DNA. B. They obtain their energy from sunlight. C. They produce and use adenosine triphosphate (ATP). D. Their genome contains at least 1000 genes. E. All of the above. 2.

All cells … A. have membrane transport proteins. B. synthesize proteins on the ribosome. C. replicate their genome by DNA polymerization. D. transcribe their genetic information by RNA polymerization. E. All of the above.

3. Imagine a segment of DNA (within a gene) encoding a certain amount of information in its nucleotide sequence. When this segment is fully transcribed into mRNA and then translated into protein, in general, … A. the protein sequence would carry more information compared to the DNA and mRNA sequences, because its alphabet has 20 letters. B. the protein sequence would carry less information compared to the DNA and mRNA sequences, because several codons can correspond to one amino acid. C. the amount of information in the mRNA sequence is lower, because the mRNA has been transcribed using only one of the DNA strands as the template. D. the amount of information in the mRNA sequence is higher, because several mRNA molecules can be transcribed from one DNA molecule. 4. Which of the following processes that happens inside a cell DOES NOT normally require consumption of free energy by the cell? A. Replication of the genetic material

B. Import of nutrients from the environment C. Diffusion of small molecules within the cell D. Regulation of gene expression E. Synthesis of enzymes that catalyze cellular reactions 5.

Which of the following would you NOT expect to find in a bacterial cell? A. Swimming using flagella B. Having a cell wall around the plasma membrane C. ATP production in mitochondria D. Protein production on the ribosome E. Sexual exchange of DNA with other bacteria

6. To trace family relationships between distantly related organisms such as humans, algae, bacteria, and archaea, one should compare their genomes in regions … A. that evolve rapidly. B. that have a higher mutation rate. C. that code for proteins. D. where mutations are hardly tolerated. E. where most mutations are selectively neutral. 7. Laboratory strains of the model organism Escherichia coli that are resistant to antibiotics are very often used in research laboratories as well as in the biotechnology industry. If cultures of such bacteria were allowed to contaminate the environment uncontrollably, it is possible that at some point, pathogenic bacteria such as Neisseria meningitidis (which causes meningitis and can cause death, especially in children) could acquire the same antibiotic-resistance gene, causing a meningitis outbreak that is difficult to treat. In this scenario, which of the following mechanisms is a more likely source of the antibiotic-resistance gene in N. meningitidis? A. Random new gene generation B. Intragenic mutation C. Gene duplication D. DNA segment shuffling E. Horizontal gene transfer 8.

A virus … A. is a type of cell. B. has genetic material made of proteins.

C. can only infect a single host species. D. can act as a vector for gene transfer. E. cannot persist in its host for more than one cell generation. 9.

Which of the following does NOT typically involve horizontal gene transfer? A. Sexual reproduction in humans B. Bacteriophage infection of bacteria C. The evolutionary history of the eukaryotic cell D. The accidental duplication of a small region of a bacterial chromosome followed by cell division E. Introduction of plasmids into bacteria in a laboratory

10. Gene duplication can give rise to homologous genes that are part of gene families. For example, there are six actin genes in the genome of most mammalian species. In humans, the ACTB gene, which encodes a cytoskeletal actin, is expressed ubiquitously, while ACTC1 is expressed mainly in cardiac cells. Although bacteria lack the eukaryotic cytoskeletal organization, the bacterial MreB gene bears recognizable sequence similarity to mammalian actin genes and codes for a protein that is similar to actin in structure and function. Which of the following statements is true about these genes? A. ACTB is homologous to ACTC1 but not to MreB. B. ACTB is orthologous to ACTC1 but not to MreB. C. ACTB is paralogous to ACTC1 but not to MreB. D. MreB is orthologous to ACTB but not to ACTC1. E. ACTB is paralogous to both ACTC1 and MreB. 11. Out of nearly 5000 protein-coding gene families, there is a set of nearly 300 conserved gene families that are found in species from all domains of life. When one looks at the general functions assigned to these gene families, it is found that … A. the majority of them function in cell-to-cell signaling. B. the majority of them are poorly characterized. C. more than one-third of them are involved in translation or amino acid transport and metabolism. D. more than one-half of the shared families are involved in DNA replication and transcription. E. Nearly all of them are involved in energy production and carbohydrate metabolism.

12.

Which of the following is true regarding Escherichia coli? A. Most of our understanding about mitosis comes from studies on this model organism. B. It is a rod-shaped bacterium that can only grow in the gut of humans and other vertebrates. C. Two strains of E. coli can differ by up to 0.1% in their genomes. D. E. coli strain K-12 encodes about 4300 proteins. E. The E. coli (strain K-12) genome is about 430 million nucleotide pairs long.

13.

Which of the following is NOT true regarding the tree of life? A. Most bacteria and archaea have 1000 to 6000 genes in their genomes. B. Eukaryotes are more similar to archaea than to bacteria with respect to the proteins that act on their DNA. C. Most bacteria and archaea have genome sizes between one and ten million nucleotide pairs, whereas eukaryotic genomes can be millions of times larger. D. Archaeal species were thought to belong to the eukaryotic world before sequence analysis placed them in a separate domain of life. E. Photosynthetic bacteria are thought to be the ancestors of the eukaryotic chloroplasts.

14.

A mutation in the cdc28 gene in the budding yeast Saccharomyces cerevisiae causes cell-

cycle arrest, giving rise to unbudded cells that look like “dumbbells.” Treatment of wild-type cells with nocodazole, a drug that destabilizes some cytoskeletal polymers, leads to a similar phenotype. Based only on these observations, which statement is true regarding cdc28? A. cdc28 codes for a master regulatory kinase that phosphorylates other proteins. B. Nocodazole binds to the protein coded by the cdc28 gene. C. The product of the cdc28 gene is responsible for resistance to nocodazole. D. The product of the cdc28 gene is involved in cell cycle regulation. E. The product of cdc28 destabilizes the same cytoskeletal polymers that nocodazole also destabilizes. 15.

Which of the following structures is exclusively found in eukaryotic cells? A. Plasma membrane B. Cell wall C. Chromosome D. Ribosome E. Lysosome

16. Mitochondria and chloroplasts are thought to have evolved from free-living aerobic bacteria that were engulfed by an ancestral anaerobic cell and established a mutually beneficial (symbiotic) relationship with it. Which of the following statements is NOT true about these organelles? A. They are similar in size to small bacteria. B. They have their own circular genomic DNA. C. They have their own ribosomes. D. They have their own transfer RNAs. E. They are found in all eukaryotes. 17. In terms of cellular lifestyle, different kingdoms of life can be likened to hunters, farmers, and scavengers. Which of the following is true in this scheme? A. The ancestral eukaryotic cell was a farmer, but it turned into a hunter once it acquired mitochondria. B. Plant cells are considered scavengers, because their cell wall does not allow them to move. C. Most protozoa are hunters, whereas animal cells are farmers. D. The ancestral eukaryotic cell was a hunter, but upon acquiring chloroplasts it made the transition into farming. E. Fungi are scavengers without mitochondria. 18. Comparing the genomes of present-day mitochondria or chloroplasts with the genomes of their corresponding bacteria reveals that these organelles do not have many of the genes that are essential for their function. For instance, they lack the many genes that are required for DNA replication. What has happened to these genes? A. They have been lost during evolution, since the organelles no longer rely on DNA replication. B. The required genes are kept in the nucleus, but many have evolved by gene transfer from the organelle. C. These genes have undergone mutations and have changed beyond recognition, but are still present in the organelle. D. The organelles do not replicate their DNA; they import new DNA from the nucleus. E. The required genes are on plasmids that are separate from the organelle’s genome.

19. Based on the variation of genome size and gene number in the organisms presented in the following graph, which organism has the highest number of genes per unit length of their genome? (Note the logarithmic scale.)

A. H. sapiens B. M. musculus C. A. thaliana D. C. elegans E. E. coli 20. Which of the following groups of living organisms has the highest variation in haploid genome size? A. Mammals B. Fish C. Fungi D. Protozoa E. Prokaryotes 21. All cells in a multicellular organism have normally developed from a single cell and share the same genome, but can nevertheless be wildly different in their shape and function. What in the eukaryotic genome is responsible for this cell-type diversity? A. The genes that encode transcription regulatory proteins B. The regulatory sequences that control the expression of genes

C. The genes that code for molecules involved in receiving cellular signals D. The genes that code for molecules involved in sending cellular signals to other cells E. All of the above 22. Didinium nasutum is a single-celled eukaryote that can hunt and feed on other living cells. It has an elaborate anatomy with beating cilia, a “mouth opening,” an “anal aperture,” and a set of contractile bundles; it can also shoot “darts” to paralyze its prey. What group of living cells does D. nasutum represent? A. Protozoa B. Yeasts C. Algae D. Animals E. It can belong to any of the above 23. It is a model organism used to study various cell processes such as regulation of the eukaryotic cell cycle. Mutants are available for every gene in its exceptionally small genome. It can live indefinitely in either a haploid or a diploid state. Which of the following describes this organism? A. It can reproduce only asexually. B. It is a fungus. C. It lacks a cell wall. D. Its cell cycle is typically much slower than that of human cells. E. All of the above 24. It is a model organism used to study various eukaryotic cell and developmental processes such as cell division and cell death. Its hermaphrodite adult is composed of exactly 959 somatic (non-germ) cells, the lineage of each of which has been worked out with great precision. It is approximately 1 mm long. Which of the following describes this organism? A. It is a vertebrate. B. It is a plant pathogen that destroys many crops. C. Its genome codes for a few thousand genes. D. It can fly. E. It can be frozen indefinitely in a state of suspended animation. 25. It is a model organism used to study various cell and developmental processes such as the growth of developing body parts in the right place and with the correct shape. It develops from a

fertilized egg to an adult in a little over a week, and has been a favorite of geneticists for almost a century. Some of its cells have giant chromosomes whose banding patterns have been extremely helpful in classical genetic studies. Which of the following describes this organism? A. It is a vertebrate. B. Its genome is only 10 million nucleotide pairs long. C. There are many more duplicate genes in this organism compared to humans. D. Although useful for genetic studies, the molecular mechanisms governing its development are irrelevant to human development. E. It normally only reproduces sexually. 26.

This model organism is particularly well suited for studying developmental processes in

higher animals. It develops from a fertilized egg to an adult in only two to three months, and its body is transparent for the first two weeks, making it easy to observe cell behavior during development. Which of the following describes this organism? A. It is a vertebrate. B. It is well suited for genetic analysis. C. Its early stages of development occur outside of the mother’s body. D. Its genome size is almost half that of humans. E. All of the above. 27. Judged by the average number of nucleotide-pair differences per 1000 nucleotide pairs, which of the following pairs show the highest difference? A. The genomes of S. cerevisiae and M. musculus B. The genomes of two different E. coli strains C. The ribosomal RNA genes from human and E. coli D. The transfer RNA genes in a human and M. musculus E. The genomes of two humans 28. Indicate if each of the following descriptions matches RNA (R) or DNA (D). Your answer would be a five-letter string composed of letters R and D only, e.g. RDDRR. ( ) It is mainly found as a long, double-stranded molecule. ( ) It contains the sugar ribose. ( ) It normally contains the bases thymine, cytosine, adenine, and guanine. ( ) It can normally adopt distinctive folded shapes. ( ) It can be used as the template for protein synthesis.

29. Indicate if each of the following descriptions matches messenger RNAs (M), ribosomal RNAs (R), or transfer RNAs (T). Your answer would be a four-letter string composed of letters M, R, and T only, e.g. RRRT. ( ) They contain codons. ( ) They contain anticodons. ( ) They are (covalently) attached to amino acids. ( ) They are at the core of a complex that carries out protein synthesis. 30. Indicate if each of the following descriptions matches lithotrophic (L), organotrophic (O), or phototrophic (P) organisms. Your answer would be a four-letter string composed of letters L, O, and P only, e.g. LLPP. ( ) They feed on other living organisms or their organic products. ( ) They are responsible for the current oxygen-rich atmosphere of the Earth. ( ) They are all known to be prokaryotic. ( ) They are the primary energy converters in hydrothermal vents in the ocean floor.

31. Indicate if each of the following statements is true (T) or false (F). Your answer would be a five-letter string composed of letters T and F only, e.g. FFTFF. ( ) Animals ultimately depend on bacteria for fixation of the atmospheric nitrogen. ( ) If one finds animals in an isolated ecosystem, there should be photosynthetic organisms in that ecosystem as well. ( ) Carbon fixation can be carried out by bacteria only. ( ) All eukaryotes are organotrophs. ( ) Compared to eukaryotic cells, prokaryotes show greater biochemical diversity. 32. In the following paragraph, fill in the blanks (indicated by numbers) with either Archaea (A), Bacteria (B), or Eukaryotes (E). Your answer would be a three-letter string composed of letters A, B, and E only, e.g. ABE. “Methanococcus jannaschii is an anaerobic thermophilic microbe that belongs to the domain ... (1) and is found in extreme environments such as hydrothermal vents. It lives at temperatures near the boiling point of water and at pressures over 200 times higher than at sea level. Its genome sequence, identified in 1996, revealed that most of its metabolic processes are similar to those in ... (2), while its genetic machinery is more similar to that of ... (3).”

33. Imagine two spherical cells, one of which is 5000 times larger in volume than the other. The smaller is a prokaryote, and the larger cell is a eukaryote with 20% of its volume confined in a spherical nucleus. If the diameter of the prokaryotic cell is 0.7 micrometers, what is the diameter of the nucleus in the eukaryotic cell in micrometers? Write down your answer as a number only. 34. What is the order of the following evolutionary landmarks (A to D), from the oldest to the most recent? Your answer would be a four-letter string composed of letters A to D, with the oldest event on the left. A. Divergence of human and bird lineages B. Divergence of human and chimpanzee lineages C. Divergence of A. thaliana lineage from the conifers lineage D. Divergence of fish and insect lineages

Answers 1. Answer: C Difficulty: 2 Section: The Universal Features of Cells on Earth Feedback: All living organisms are made of cells. However, only a minority of them are eukaryotes with defined cell nuclei. Not all cells obtain their energy directly from sunlight. However, regardless of the energy source, the main energy currency used by all known living cells is ATP. Some prokaryotic genomes contain only a few hundred genes. 2. Answer: E Difficulty: 1 Section: The Universal Features of Cells on Earth Feedback: All living cells replicate their DNA by the same overall mechanism of templated polymerization. They also transcribe the genetic information into RNA form using the same mechanism. Translation of this information into protein form is mediated by the ribosome in all cells. Additionally, all cellular membranes have membrane transport proteins that control the flow of material through the membranes. 3. Answer: B Difficulty: 3 Section: The Universal Features of Cells on Earth Feedback: There are 64 (that is, 43) nucleotide triplets or codons using four types of nucleotide monomers, whereas each codon encodes one out of only 20 amino acids. Therefore, some sequence information is lost when translating from mRNA to protein sequences, such that it is not usually possible to deduce the original mRNA sequence given a translated polypeptide sequence. 4. Answer: C Difficulty: 2 Section: The Universal Features of Cells on Earth Feedback: Diffusion is a spontaneous process driven by the thermal energy of randomly moving molecules, and normally does not require energy expenditure by the cell. For the other mentioned processes, free energy is required. If a cell “dies,” these processes would stop or get out of control, but small molecules would still diffuse. 5. Answer: C Difficulty: 2 Section: The Diversity of Genomes and the Tree of Life

Feedback: Bacterial cells do not have membrane-enclosed organelles such as mitochondria. 6. Answer: D Difficulty: 2 Section: The Diversity of Genomes and the Tree of Life Feedback: Highly conserved genes that code for optimized, essential, protein or RNA molecules should be studied to trace family relationships across distant taxa. Mutations in these regions (such as the genes that code for ribosomal RNAs) are usually eliminated, making the sequence recognizable even after billions of years of evolutionary history. 7. Answer: E Difficulty: 2 Section: The Diversity of Genomes and the Tree of Life Feedback: Horizontal gene transfer is common in bacteria and can result in the acquisition of new antibiotic-resistance genes by pathogenic bacteria. 8. Answer: D Difficulty: 1 Section: The Diversity of Genomes and the Tree of Life Feedback: Viruses are not living cells, but use the cellular machinery of their hosts to replicate their genetic material, which is made of DNA or RNA. They can act as vectors for horizontal gene transfer between cells of the same or different species. Infection is followed by either lysis (killing of the host cell) or the persistence of the viral genome in the host cell (which can last for many cell generations). 9. Answer: D Difficulty: 2 Section: The Diversity of Genomes and the Tree of Life Feedback: Introduction of genes into bacteria or eukaryotes by viral infection or artificially in the laboratory are examples of horizontal gene transfer, as is sexual reproduction between members of the same eukaryotic species. A great deal of horizontal gene transfer has occurred during the evolution of eukaryotic cells, such as in the process that led to the development of mitochondria (see Figure 12–3). 10. Answer: C Difficulty: 2 Section: The Diversity of Genomes and the Tree of Life Feedback: Gene duplication within a single genome results in paralogous genes (such as human ACTB and ACTC1 genes), whereas other homologous genes that share direct

ancestry are called orthologs (in this case, some ancestor of MreB is presumably a distant actin ortholog). These genes all belong to the actin family. 11. Answer: C Difficulty: 2 Section: The Diversity of Genomes and the Tree of Life Feedback: A significant fraction of the ubiquitous gene families across the three domains of life is involved in translation and amino acid metabolism and transport. Other metabolic processes such as carbohydrate and coenzyme metabolism and transport also constitute a large subset. This is only a rough sketch of the core ancestral gene set, but is nevertheless informative. Please refer to Table 1–1 for details. 12. Answer: D Difficulty: 1 Section: The Diversity of Genomes and the Tree of Life Feedback: E. coli is a prokaryotic model organism; although the discovery of many fundamental molecular mechanisms were aided by studies using E. coli, the eukaryotic process of mitosis is best studied in simple eukaryotic cells. E. coli can be easily cultured in the lab in nutrient media. Its genome of 4.6 million nucleotide pairs codes for about 4300 different proteins. But this number is reported for the K-12 strain. Other strains can be different in up to 50% of their genes, a significant diversity which has been explained by the process of horizontal gene transfer. 13. Answer: D Difficulty: 1 Section: The Diversity of Genomes and the Tree of Life Feedback: Most prokaryotes have small genomes (106 to 107 nucleotide pairs) and code for between 1000 and 6000 genes. Archaea are more similar to bacteria in their metabolism, but more similar to eukaryotes with respect to their DNA replication, DNA repair, and DNA packaging proteins (archaeal histones). Before the comparison of DNA sequences of highly conserved genes placed them in a separate domain, archaeal organisms were classified as bacteria. 14. Answer: D Difficulty: 3 Section: The Diversity of Genomes and the Tree of Life Feedback: Genetic analysis of mutations can give clues about gene functions. For example, if a mutation causes cell-cycle arrest, the affected gene is interpreted to have a role in cell-cycle progression. However, this does not necessarily mean that its product

physically interacts with (binds to) another cellular component that is also involved in this process. 15. Answer: E Difficulty: 2 Section: Genetic Information in Eukaryotes Feedback: Membrane-bound organelles such as lysosomes are normally not found in prokaryotes. Please note that using the same names for some structures in both prokaryotes and eukaryotes (e.g., cell wall) does not necessarily mean that these structures share the same molecular structure. 16. Answer: E Difficulty: 1 Section: Genetic Information in Eukaryotes Feedback: Many eukaryotes lack chloroplasts, and some have lost their mitochondria. 17. Answer: D Difficulty: 2 Section: Genetic Information in Eukaryotes Feedback: In this lifestyle analogy, acquiring a symbiont chloroplast allowed cells to transition from hunting to farming. Other cells such as fungi evolved into scavengers instead, while protozoa and animal cells maintained their hunting habits. Fungi contain mitochondria. 18. Answer: B Difficulty: 2 Section: Genetic Information in Eukaryotes Feedback: During the evolution of mitochondria and chloroplasts, many essential organelle genes are thought to have been transferred to the nuclear genome. 19. Answer: E Difficulty: 2 Section: Genetic Information in Eukaryotes Feedback: E. coli has the highest ratio of gene number to genome size among the organisms shown in the graph. The lowest ratio belongs to H. sapiens. The ratio for each organism corresponds to the linear difference in the length of the red and blue bars for that organism. If the graph was not drawn in logarithmic scale, this comparison might not have been as straightforward. 20. Answer: D Difficulty: 2 Section: Genetic Information in Eukaryotes

Feedback: The smallest and largest genomes in eukaryotes are found in protozoans. Please refer to Figure 1–32. 21. Answer: E Difficulty: 1 Section: Genetic Information in Eukaryotes Feedback: The eukaryotic genome codes for not only various proteins and RNA molecules, but also regulatory molecules that control the expression of other genes. Noncoding regulatory sequences are used to tune the expression level of the cellular genes. The cell actively exchanges signals with other cells in the organism, which help it decide which genes to express. 22. Answer: A Difficulty: 1 Section: Genetic Information in Eukaryotes Feedback: Protozoa are generally quite complicated in their structure and behavior despite being single-celled, as exemplified by D. nasutum. 23. Answer: B Difficulty: 2 Section: Genetic Information in Eukaryotes Feedback: It is the yeast Saccharomyces cerevisiae. 24. Answer: E Difficulty: 2 Section: Genetic Information in Eukaryotes Feedback: It is the nematode worm Caenorhabditis elegans. 25. Answer: E Difficulty: 2 Section: Genetic Information in Eukaryotes Feedback: It is the fruit fly Drosophila melanogaster. 26. Answer: E Difficulty: 2 Section: Genetic Information in Eukaryotes Feedback: It is the zebrafish Danio rerio. 27. Answer: A Difficulty: 2 Section: Genetic Information in Eukaryotes Feedback: Mostly due to horizontal gene transfers, different strains of bacteria can vary wildly in their genomic sequences; however, the difference between the whole genomes

of two species as distantly related as the budding yeast and the mouse is by far greater. In comparison, two human individuals differ in a few nucleotide pairs per 1000 nucleotide pairs of their genome. Genes such as ribosomal RNA genes are highly conserved and show a significant identity even between H. sapiens and E. coli. 28. Answer: DRDRR Difficulty: 1 Section: The Universal Features of Cells on Earth Feedback: Unlike DNA, cellular RNA is mainly polymerized as single-stranded molecules, allowing for folding into distinctive structures. Messenger RNAs (mRNAs) carry the information for protein synthesis. RNA contains the sugar ribose. Thymine is found in DNA. 29. Answer: MTTR Difficulty: 1 Section: The Universal Features of Cells on Earth Feedback: Ribosomal RNA forms the core of the ribosome, an RNA–protein complex. Each transfer RNA is covalently attached to a specific amino acid at one end, and displays at its other end the anticodon sequence that can pair with a codon in a messenger RNA. 30. Answer: OPLL Difficulty: 1 Section: The Diversity of Genomes and the Tree of Life Feedback: While organotrophs (such as animals) feed on organic material produced by other organisms, the primary energy converters obtain their energy from the nonliving world. Phototrophs (such as plants) use the energy of sunlight. Molecular oxygen, a byproduct of phototrophs, has changed the Earth’s atmosphere and the entire biosphere. In hydrothermal vents, anaerobic lithotrophic microbes live in conditions similar to those in the early days of life on Earth. 31. Answer: TFFFT Difficulty: 2 Section: The Diversity of Genomes and the Tree of Life Feedback: We rely on plants as our source of organic carbon and nitrogen. Plants in turn rely on bacteria for organic nitrogen, but can fix carbon by themselves. Hydrothermal vents provide an example of a secluded ecosystem that lacks photosynthetic organisms. 32. Answer: ABE Difficulty: 2 Section: The Diversity of Genomes and the Tree of Life

Feedback: Archaea usually live in extreme environments, and resemble eukaryotes more in their genetic machinery, and bacteria more in their metabolic machinery. 33. Answer: 7 Difficulty: 3 Section: Genetic Information in Eukaryotes Feedback: The volume of the nucleus is 1000 times that of the prokaryotic cell in this example (that is, 20% × 5000), which means its diameter should be 10 times larger. 34. Answer: DACB Difficulty: 3 Section: Genetic Information in Eukaryotes Feedback: The vertebrates and insects separated about 700 million years ago. Humans and birds separated about 300 million years ago. All flowering plants share a common ancestor that lived about 150 million years ago. Humans and chimpanzees are separated by only a few million years.

Which of the following elements is not normally found in cells? A. Copper B. Iron C. Silver D. Cobalt E. Zinc

A hydrophobic molecule is typically … A. able to form hydrogen bonds with itself but not with water. B. able to form hydrogen bonds with water. C. charged. D. hard to dissolve in a solvent. E. incapable of interacting favorably with water.

3. For each of the following pairs, indicate whether they interact via hydrogen bonds (H) or ionic bonds (I), or do not favorably interact (N). Your answer would be a four-letter string composed of letters H, I, and N only, e.g. HNNI. ( ) ATP and Mg2+ ( ) Urea and water ( ) Glucose and the enzyme hexokinase (which uses glucose as a substrate) ( ) A phospholipid tail and inorganic phosphate 4.

Which of the following chemicals do you NOT expect to be readily dissolved in water? A. Uric acid B. Hexane C. Glycerol D. Ethanol

E. Potassium chloride 5. Weak noncovalent attractions in the cell can be very strong in a nonaqueous environment. Some of these attractions are as strong as covalent interactions in a vacuum (their bond energy is approximately 340 kJ/mole), but become more than twenty-five times weaker (their bond energy becomes approximately 13 kJ/mole) in water. What type of attraction shows this phenomenon? A. Electrostatic attractions B. Hydrogen bonds C. van der Waals attractions D. Hydrophobic force E. All of the above 6. The bond energies associated with noncovalent attractions in the cell are too weak to resist disruption by thermal motion. However, cellular macromolecules can interact specifically AND strongly with each other (or fold by themselves) merely via such interactions. How is this possible? A. The bond energies increase radically when two interacting molecules approach each other. B. The interacting molecules also fortify their binding via covalent bonds to keep them from dissociation. C. Many weak bonds together in a complementary geometry can afford a strong binding. D. The cell lowers its internal temperature to reduce thermal motion of molecules and enhance the weak attractions. 7. What is the pH of a 10–8 M solution of hydrochloric acid? Round the pH value to the nearest integer, e.g. 10. A. 8 B. 7 C. 6 D. 5 E. 4 8. The cell can change the pH of its internal compartments using membrane transport proteins that pump protons into or out of a compartment. How many protons should be pumped into an endocytic vesicle that is 10–15 liters in volume and has a neutral pH in order to change the

pH to 5? Avogadro’s number is 6 × 1023. Omit complications such as the membrane potential, buffers, and other cellular components. A. 6000 B. 60,000 C. 120,000 D. 600,000 E. 6,000,000 9.

Which of the following is true regarding a fatty acid molecule in water? A. It is positively charged at physiological pH, but can become neutral when the pH is high enough. B. It is positively charged at physiological pH, but can become neutral when the pH is low enough. C. It is negatively charged at physiological pH, but can become neutral when the pH is high enough. D. It is negatively charged at physiological pH, but can become neutral when the pH is low enough. E. It is not charged at physiological pH.

10. The amino acid serine has an amino group, a carboxyl group, and a hydroxyl group. Which of the following better represents the structure of this amino acid at neutral pH? A H3N+

B O–

H3N+

CH3

C H3N+

NH2

H3N+

O–

CH2 OH

H2N

11. The three families of cellular macromolecules are polymerized and depolymerized by a general mechanism involving water. Each of them has a set of monomers whose polymerization

changes the total free energy of the system. Which of the following statements is true regarding these macromolecules? A. Each polymerization step requires free-energy input and proceeds by the consumption of one water molecule. B. Each depolymerization step requires free-energy input and proceeds by the consumption of one water molecule. C. Each polymerization step requires free-energy input and proceeds by the release of one water molecule. D. Each depolymerization step requires free-energy input and proceeds by the release of one water molecule. 12. Sort the following from a low to a high contribution to the total mass of an E. coli bacterium. Your answer would be a four-letter string composed of letters A to D only, e.g. DCBA. (A) Water (B) Sugars (C) Proteins (D) Nucleic acids 13.

Which of the following statements is true regarding cellular metabolism? A. A living organism decreases the entropy in its surroundings. B. During catabolism, heat is generated, and the cell uses this heat to perform work during anabolism. C. The heat released by an animal cell as part of its metabolic processes comes from the bond energies in the foodstuffs that are consumed by the animal. D. Living organisms defy the second law of thermodynamics, but still obey the first law.

14.

The folding of proteins can be considered a simple conversion from the unfolded to the

natively folded state. At about 27°C (or 300 K), the free-energy change of folding for a particular protein is measured to be –40 kJ/mole. If the enthalpy change (ΔH) of folding is –640 kJ/mole, what is the entropy change (ΔS) of folding for this protein? Write down your answer with the appropriate sign (+ or –) and in kJ/mole/K, e.g. –1000 kJ/mole/K. 15.

Which of the following correctly summarizes the overall process of photosynthesis? A. CO2 + O2 → H2O + sugars B. CH2O + CO2 + O2 → H2O + sugars

C. CO2 + H2O → H2 + CO2 D. CO2 + H2O → O2 + sugars 16. Which of the following statements is true regarding reactions involving oxidation and reduction? A. The carbon atom is more oxidized in formaldehyde (CH2O) than in methanol (CH3OH). B. Oxidation of food in all organisms requires oxygen. C. A molecule is oxidized if it gains an electron (plus a proton) in a reaction. D. A dehydrogenation reaction is a reduction. E. In an organic molecule, the number of C–H bonds increases as a result of oxidation. 17. Enzymes are the cell’s catalyst crew. They make the life of the cell possible by carrying out various reactions with astounding performance. Which of the following is NOT true regarding cellular enzymes? A. Enzymes lower the activation energy of the reactions that they catalyze. B. Enzymes can specifically drive substrate along certain reaction pathways. C. Enzymes can push energetically unfavorable reactions forward by coupling them to energetically favorable reactions. D. Enzymes are proteins, but RNA catalysts, called ribozymes, also exist. E. Enzymes can change the equilibrium point for reactions that they catalyze. 18. In the following diagram showing the reaction pathway for a simple single-substrate enzymatic reaction, which of the quantities corresponds to the activation energy of the forward reaction?

Total energy

b c Reaction pathway

A. (a – b)

B. (a + b) C. (a – c) D. (a + c) E. (b – c)

Number of molecules

19. In the following diagram showing the distribution of thermal energy in a population of substrate molecules, the energy thresholds indicated by numbers represent ...

Energy per molecule

A. the activation energy at high and low temperature. B. the reaction rate at high and low pH. C. the activation energy with and without an enzyme. D. the reaction rate at high and low substrate concentrations. E. the activation energy at high and low substrate concentrations. 20. A cellular enzyme catalyzes the catabolic reaction shown below. Its coenzyme is shown in the box. Which of the following is correct regarding this reaction?

State 1

State 2

A. The substrate is reduced in this reaction and the coenzyme is converted from state 1 to state 2. B. The substrate is oxidized in this reaction and the coenzyme is converted from state 1 to state 2. C. The substrate is reduced in this reaction and the coenzyme is converted from state 2 to state 1. D. The substrate is oxidized in this reaction and the coenzyme is converted from state 2 to state 1. 21. The molecules inside the cell constantly collide with other molecules and diffuse through the cytoplasm in a random walk. The average net distance traveled by such a molecule after a certain time period t is proportional to the square root of t, i.e. (t)0.5, as well as to its diffusion coefficient. If, on average, it takes a molecule 100 milliseconds to travel a net distance of 0.5 µm from its starting point, how long would it normally take for the same molecule to travel a net distance of 5 µm from the same starting point? A. 0.2 second

B. 0.3 second C. 1 second D. 10 seconds E. 0.32 seconds 22. Sort the following molecules from a low to high rate of diffusion inside the cytosol. Your answer would be a four-letter string composed of letters A to D only, e.g. ADCB. (A) Myoglobin (a protein) (B) Glycine (an amino acid) (C) Ribosome (a protein–RNA complex) (D) CO2 23. The equilibrium constant for the reaction that breaks down each molecule of substrate A to one molecule of B and one molecule of C is equal to 0.5. Starting with a mixture containing only molecules A at 1 M concentration, what will be the concentration of molecule A after reaching equilibrium under these conditions? A. 0.5 M B. 0.25 M C. 0.125 M D. 0.333 M E. 0.667 M 24. The free-energy change (ΔG) for a simple reaction, A → B, is 0 kJ/mole at 37°C when the concentrations of A and B are 10 M and 0.1 M, respectively. What is the free-energy change for the reaction when the concentrations of A and B are instead 0.01 M and 1 M, respectively? Recall that ΔG° = –5.9 × log(Keq). Write down your answer as a number with the appropriate sign (+ or –) and in kJ/mole, e.g. +11.8 kJ/mole.

25. Imagine the reaction A → B with a negative ΔG value under experimental conditions. Which of the following statements is true about this reaction? A. The reaction is energetically unfavorable. B. The reaction proceeds spontaneously and rapidly under these conditions. C. Increasing the concentration of B molecules would increase the ΔG value (toward more positive values). D. The reaction would result in a net decrease in the entropy (disorder) of the universe.

E. The reaction cannot proceed unless it is coupled to another reaction with a positive value of ΔG. 26. In the first reaction of the glycolytic pathway, the enzyme hexokinase uses ATP to catalyze the phosphorylation of glucose, yielding glucose 6-phosphate and ADP. The ΔG° value for this reaction is –17 kJ/mole. The enzyme glucose 6-phosphatase catalyzes a “reverse” reaction, in which glucose 6-phosphate is converted back to glucose, and a phosphate is released. The ΔG° value for this reaction is –14 kJ/mole. What is the ΔG° value for the following reaction? ATP + H2O → ADP + Pi A. –3 kJ/mole B. +3 kJ/mole C. –31 kJ/mole D. +31 kJ/mole E. –15.5 kJ/mole 27. The enzyme phosphoglucose isomerase converts glucose 6-phosphate to its isomer fructose 6-phosphate in the second step of glycolysis. The equilibrium constant for the reaction is 0.36. Evaluating the ΔG° of the reaction (ΔG° = –5.9 × log Keq), decide which of the following conclusions is true. A. The ΔG° is negative, therefore the reaction proceeds in the forward direction. B. The ΔG° is negative, but whether or not the reaction proceeds would depend on ΔG, not ΔG°. C. The ΔG° is positive, but in a cell that is active in glycolysis, the reaction can still proceed in the forward direction. D. The ΔG° is positive, therefore the reaction proceeds in the reverse direction. 28.

Which of the following represents an “activated” carrier molecule? A. AMP B. NADH C. NAD+ D. NADP+ E. CoA

29. ATP is the main energy currency in cells, and it can especially be used to drive condensation reactions that produce macromolecular polymers. How does ATP normally catalyze the condensation reaction, which by itself is energetically unfavorable? A. It transfers its terminal phosphate to an enzyme and is released as ADP. B. It transfers its two terminal phosphates to an enzyme, and is released as AMP. C. It covalently attaches to both of the substrates. D. It transfers either one or two terminal phosphate(s) to one of the substrates and is released as either ADP or AMP. E. It covalently attaches to the enzyme, forming an enzyme–AMP adduct. 30.

Despite their overall similarity, NADH and NADPH are not used indiscriminately by the

cell. What are the distinctive features of these two carrier molecules? A. NADPH has an extra phosphate near its nicotinamide ring, giving it distinct electrontransfer properties. B. In the cell, NADH is usually in excess over NAD+, but NADP+ is usually in excess over NADPH. C. NADH is normally involved in anabolic reactions, whereas NADPH is normally involved in catabolism. D. Both NADPH and NADH are recognized by the same enzymes with similar affinities, since the extra phosphate group in NADPH is not involved in such recognition. E. In the cell, NADH is found mostly in the form that acts as an oxidizing agent, whereas NADPH is found mostly in the form that acts as a reducing agent. 31. In an enzymatic reaction involving NADH or NADPH, reduction of a substrate accompanies the oxidation of these carrier molecules to NAD+ or NADP+, respectively. What else typically happens in such a reaction? A. A molecule of water is released to the solution upon completion of the reaction. B. A proton is released during the oxidation of the carriers. C. A proton is taken up by the substrate that is being reduced. D. A proton is taken up by the carrier molecule that is being oxidized. E. A phosphate group is transferred to the substrate. 32. What is the reaction performed on the molecule labeled as substrate in the following diagram? What is the name of the activated carrier?

Substrate

Product

A. This is a methylation reaction and the activated carrier is ATP. B. This is a methylation reaction and the activated carrier is S-adenosylmethionine. C. This is a carboxylation reaction and the activated carrier is ATP. D. This is a carboxylation reaction and the activated carrier is carboxylated biotin. E. This is an acetylation reaction and the activated carrier is acetyl CoA. 33. Under anaerobic conditions, glycolysis provides most of the ATP that the cell needs. In animal cells, pyruvate, the end product of glycolysis, is converted to lactic acid by lactate dehydrogenase, as shown below: CH3(CO)COO– + X →CH3(CHOH)COO– + Y What is the correct carrier pair (in place of X and Y) in this reaction? A. X is (ADP + Pi), and Y is (ATP) B. X is (NADP+), and Y is (NADPH + H+) C. X is (NAD+), and Y is (NADH + H+) D. X is (NADH + H+), and Y is (NAD+) E. X is (NADP++ H+), and Y is (NADPH) 34. Macromolecules in the cell can be made from their monomers using one of two polymerization schemes. One is called head polymerization, in which the reactive bond required for polymerization is carried on the end of the growing polymer. In contrast, in tail

polymerization, the reactive bond is carried by each monomer for its own incorporation. In the figure, indicate the polymerization scheme and the type of macromolecule.

A. Head polymerization of a protein B. Tail polymerization of a protein C. Head polymerization of a polysaccharide D. Head polymerization of a nucleic acid E. Tail polymerization of a nucleic acid 35. What is the end product of glycolysis in the cytoplasm of eukaryotic cells? How many carbon atoms does the molecule have? A. Acetyl CoA; it has two carbon atoms attached to coenzyme A B. Phosphoenolpyruvate; it has three carbon atoms C. Glucose 6-phosphate; it has six carbon atoms D. Pyruvate; it has three carbon atoms E. Glyceraldehyde 3-phosphate; it has three carbon atoms 36. The substrate for the glycolytic enzyme glyceraldehyde 3-phosphate dehydrogenase is glyceraldehyde 3-phosphate (with one phosphate group) while its product is 1,3bisphosphoglycerate (with two phosphate groups). Where does the extra phosphate group come from? A. From combining two molecules of the substrate

B. ATP C. Fructose 1,6-bisphosphate D. Pi E. NADH 37. Steps 6 and 7 of glycolysis are catalyzed by the enzymes glyceraldehyde 3-phosphate dehydrogenase and phosphoglycerate kinase, respectively. Together, they ... A. result in the oxidation of an aldehyde to a carboxylic acid. B. produce both ATP and NADH. C. couple the oxidation of a C–H bond to the activation of carrier molecules. D. catalyze the only glycolytic reactions that create a high-energy phosphate linkage directly from inorganic phosphate. E. All of the above. 38. Arsenate is a toxic ion that can interfere with both glycolysis and oxidative phosphorylation. Arsenate resembles Pi (inorganic phosphate) and can replace it in many enzymatic reactions. One such reaction is catalyzed by glyceraldehyde 3-phosphate dehydrogenase in step 6 of glycolysis. Upon completion of the reaction, instead of the normal product, 1,3-bisphosphoglycerate, the mixed anhydride 1-arsenato-3-phosphoglycerate is formed; this undergoes rapid spontaneous hydrolysis into arsenate plus 3-phosphoglycerate, the latter being a normal product of step 7 in glycolysis. What would be the effect of arsenate poisoning in glycolysis? A. It results in more ATP and NADH molecules generated for every glucose molecule. B. It results in fewer ATP molecules generated per glucose molecule, but NADH generation is not directly affected. C. It brings glycolysis to an abrupt stop. D. It results in fewer ATP and NADH molecules generated per glucose molecule. E. It does not affect the number of ATP or NADH molecules generated per glucose molecule. 39. Which of the following is true regarding energy production and storage in plants and animals? A. Plant and animal cells make starch for long-term energy storage. B. Most of the ATP in a plant cell has been generated in the chloroplast and transported to other parts of the cell.

C. Oxidation of one gram of starch releases more energy than oxidation of fat, but since starch absorbs a lot of water, it is not as efficient as fat in energy storage. D. Animals, but not plants, can store fats in the form of triacylglycerol (triglyceride). E. Plant seeds often contain large amounts of fats and starch. 40. What are the molecules that normally supply carbon and oxygen atoms, respectively, for the citric acid cycle? A. Oxaloacetate, oxaloacetate B. Acetyl CoA, O2 C. Oxaloacetate, O2 D. Acetyl CoA, H2O E. Pyruvate, pyruvate 41. Indicate if each of the following descriptions matches lipids (1), nucleic acids (2), polysaccharides (3), or proteins (4). Your answer would be a four-digit number composed of digits 1 to 4 only, e.g. 1332. ( ) Their monomers contain phosphorus and nitrogen. ( ) They constitute almost half of the cell’s dry mass. ( ) They are the main constituent of all cellular membranes. ( ) They are largely hydrophobic and can store energy. 42. Sort the following molecules (A to E) based on the oxidation of the carbon atom, from higher to lower oxidation states. Your answer would be a five-letter string composed of letters A to E only, e.g. ADCBE. Put the letter corresponding to the highest oxidation level on the left.

43. Indicate true (T) and false (F) statements below regarding glycolysis. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTT. ( ) Molecular oxygen is used in glycolysis to oxidize glucose. ( ) Along the glycolytic pathway, ATP is both consumed and generated.

( ) In the course of glycolysis, one molecule of NADH is formed per molecule of glucose. ( ) Following the production of one molecule of fructose 1,6-bisphosphate, the rest of the glycolytic pathway generates four molecules of ATP. 44.

Fill in the blank in the following paragraph. “During intense ‘anaerobic’ physical exercise, the high energy demand in the muscle cells leads to an accumulation of lactic acid in these cells and their surrounding tissues. Similarly, the yeast Saccharomyces cerevisiae can produce ethanol when grown anaerobically. The lactate or ethanol production takes place in a process called ...”

45. Sort the following molecules based on the amount of energy that is released when their phosphate bond is hydrolyzed as indicated. Your answer would be a four-letter string composed of letters A to D only, e.g. ADCB. Put the molecule with the highest amount of hydrolysis energy on the left. (A) ATP when hydrolyzed to ADP (B) Glucose 6-phosphate when hydrolyzed to glucose (C) 1,3-bisphosphoglycerate when hydrolyzed to 3-phosphoglycerate (D) Phosphoenolpyruvate when hydrolyzed to pyruvate 46. Indicate true (T) and false (F) statements below regarding fatty acid metabolism. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTT. ( ) Most animals derive their energy from fatty acids between meals. ( ) Fatty acids are converted to acetyl CoA in the cytosol, which is then transported into mitochondria for further oxidation. ( ) Fatty acids are stored in fat droplets in the form of triacylglycerols. ( ) The breakdown of fatty acids into each acetyl CoA unit requires the hydrolysis of two ATP molecules. 47. Indicate whether each of the following descriptions matches glycolysis (G) or the Krebs cycle (K). Your answer would be a four-letter string composed of letters G and K only, e.g. GGGK. ( ) It oxidizes acetyl CoA to CO2. ( ) In eukaryotic cells, it is carried out in the cytosol.

( ) It produces FADH2. ( ) α-Ketoglutarate, one of its intermediates, is used to synthesize the amino acid glutamic acid. 48. Indicate whether each of the following molecules is an intermediate in glycolysis (G) or in the tricarboxylic acid cycle (T). Your answer would be a four-letter string composed of letters G and T only, e.g. GGTT. ( ) Fumarate ( ) Malate ( ) Phosphoenolpyruvate ( ) Succinate

The Citric Acid Cycle: Questions 49-52 The citric acid cycle is summarized in the following figure. Answer the following question(s) about this cycle.

49. In step 1 of the citric acid cycle drawn above, what is the molecule indicated with a question mark? A. O2 B. ATP C. H2O D. H+ E. Pyruvate 50.

In the citric acid cycle shown above, which steps produce CO2 as a by-product? List all

such steps by their number, from the smallest number to the largest. Your answer would be a number composed of digits 1 to 8 only, e.g. 258. 51. In the citric acid cycle shown above, which steps produce either NADH or FADH2? List all such steps by their number, from the smallest number to the largest. Your answer would be a number composed of digits 1 to 8 only, e.g. 258. 52. Aconitase catalyzes an isomerization reaction in the citric acid cycle shown above, in which H2O is first removed and then added back to the substrate. Which step is catalyzed by this enzyme? Write down the step number as your answer, e.g. 5.

53. The electron carriers NADH and FADH2 donate their electrons to the electron-transport chain in the inner mitochondrial membrane, leading to ATP synthesis powered by an H+ gradient across the membrane. If, on average, the oxidation of each NADH or FADH2 molecule in this pathway results in the production of 2.5 and 1.5 molecules of ATP, respectively, how many ATP (and GTP) molecules are produced on average as a result of the complete oxidation of one molecule of acetyl CoA in the mitochondrion? Consider only the citric acid cycle and oxidative phosphorylation. A. 10 B. 12 C. 13.5 D. 14.5 E. 15

54. Indicate true (T) and false (F) statements below regarding the cellular metabolism of nucleotides and amino acids. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTT. ( ) Nitrogen fixation occurs in the mitochondria in most animal cells to generate amino acids. ( ) All 20 natural amino acids must be provided in our diet and are therefore “essential.” ( ) There are NO essential nucleotides that must be provided in the diet. ( ) Catabolism of amino acids in our body leads to the production of urea which is excreted.

Answers 1. Answer: C Difficulty: 1 Section: The Chemical Components of a Cell Feedback: Metal ions such as copper, iron, cobalt, and zinc are used as cofactors that are necessary for the function of some enzymes. 2. Answer: E Difficulty: 1 Section: The Chemical Components of a Cell Feedback: Hydrophobic molecules usually have no charge and form no or few hydrogen bonds, and are therefore not favored by the network of hydrogen bonds in liquid water. They do dissolve in nonpolar organic solvents. 3. Answer: IHHN Difficulty: 2 Section: The Chemical Components of a Cell Feedback: ATP is negatively charged and can form ionic bonds with magnesium ions. Urea is highly soluble in water due to its hydrogen-bonding capacity. Similarly, interaction of a polar molecule like glucose with the active site of an enzyme can be mediated by hydrogen bonds and other noncovalent (or even covalent) bonds. In contrast, the fatty acid tails in phospholipids are hydrophobic and do not favorably interact with negatively charged phosphate molecules. 4. Answer: B Difficulty: 3 Section: The Chemical Components of a Cell Feedback: Hexane is an alkane hydrocarbon incapable of hydrogen-bonding with water molecules, which results in an entropically unfavorable state when the two interact. All the other mentioned chemicals can be readily dissolved in water because they have polar bonds or can dissociate into ions. 5. Answer: A Difficulty: 2 Section: The Chemical Components of a Cell Feedback: The probing of the charged ions by water molecules greatly reduces the bond energy of ionic bonds (electrostatic interactions) in aqueous solutions. Hydrogen bonds are also weakened in water, but they are not that strong in a vacuum to begin with. 6. Answer: C Difficulty: 1

Section: The Chemical Components of a Cell Feedback: Although each noncovalent bond is weak, when many of them are formed simultaneously (in a complementary interface), their energies can sum to produce a tight binding. 7. Answer: B Difficulty: 3 Section: The Chemical Components of a Cell Feedback: The concentration of hydronium ions would be the sum of those obtained from the dissociation of water and the acid: [H3O+] = 10–7 + 10–8 = 1.1 × 10–7. The pH value will then be calculated as: pH = –log [H3O+] = –log [1.1 × 10–7] = 7 – log (1.1) = 6.96. This is very close to neutral pH. 8. Answer: A Difficulty: 3 Section: The Chemical Components of a CellFeedback: The initial number of hydronium ions would be: [H3O+]1 = 10–15 L × 10–7 mole/L = 10–22 mole. The final number at pH 5 would be: [H3O+]2 = 10–15 L × 10–5 mole/L = 10–20 mole. The difference is: [H3O+]2 – [H3O+]1 = 9.9 × 10–21 mole. This is equivalent to approximately 6000 protons that need to be pumped in. 9. Answer: D Difficulty: 3 Section: The Chemical Components of a Cell Feedback: Due to the presence of the carboxyl group, a fatty acid molecule carries a negative charge at neutral pH. However, lowering the pH can reverse the ionization of this group to the neutral (protonated) state. 10. Answer: C Difficulty: 2 Section: The Chemical Components of a Cell Feedback: The amino and carboxyl groups are common to all amino acids. The serine side chain contains a hydroxyl group. 11. Answer: C Difficulty: 2 Section: The Chemical Components of a Cell Feedback: The polymerization reaction generally requires a free-energy input. Also, the addition of each monomer to the growing polymer is a condensation reaction that is accompanied by the release of one water molecule. The opposite reaction (depolymerization) involves hydrolysis and consumes one water molecule.

12. Answer: BDCA Difficulty: 2 Section: The Chemical Components of a Cell Feedback: Water accounts for about 70% of the total mass in a typical cell. In the remaining “dry mass,” proteins constitute about half, the nucleic acids RNA and DNA are next, and polysaccharides (and their sugar monomers) are still less abundant. 13. Answer: C Difficulty: 2 Section: Catalysis and the Use of Energy by Cells Feedback: For a living animal cell, heat-generating reactions from burning of foodstuffs are “coupled” to other reactions that increase order inside the cell. Concomitantly, there is an increase in the overall entropy of the universe (cell plus its environment), with no violation of the laws of thermodynamics for a spontaneous process. 14. Answer: –2 kJ/mole/K Difficulty: 3 Section: Catalysis and the Use of Energy by Cells Feedback: For the folding reaction, the free-energy change can be written as: ΔG = ΔH – TΔS Therefore: ΔS = (ΔH – ΔG)/T = (–640 kJ/mole + 40 kJ/mole) / (300 K) = –2 kJ/mole/K The negative value of ΔS means a decrease in entropy. This is not unexpected since folding results in the formation of a single conformation (or a limited set of conformations) out of an enormous number of possible coils. 15. Answer: D Difficulty: 2 Section: Catalysis and the Use of Energy by Cells Feedback: Photosynthesis consumes water and atmospheric CO2 to make simple sugars and the by-product oxygen. 16. Answer: A Difficulty: 3 Section: Catalysis and the Use of Energy by Cells Feedback: Oxidation involves the full or partial removal of electrons from an atom, and does not necessarily involve oxygen. In the cell, organic molecules usually release a proton to their surrounding when oxidized in a dehydrogenation reaction, decreasing the number of C–H bonds in the molecule. 17. Answer: E

Difficulty: 2 Section: Catalysis and the Use of Energy by Cells Feedback: Enzymes catalyze most cellular reactions by lowering the activation energy, but they cannot change the equilibrium constant of the reactions that they catalyze; that is, both forward and reverse reactions are sped up by the same factor. However, they can selectively drive substrates along one of various cellular metabolic pathways, and can also couple unfavorable reactions to spontaneous heat-generating reactions. 18. Answer: A Difficulty: 2 Section: Catalysis and the Use of Energy by Cells Feedback: The activation energy corresponds to the height of the energy barrier between the reactant and the product, and is the minimum amount of energy that should be provided in order for the reaction to proceed. 19. Answer: C Difficulty: 2 Section: Catalysis and the Use of Energy by Cells Feedback: In the presence of an enzyme (line 1), the fraction of substrate molecules that have enough thermal energy to proceed through the reaction is increased compared to that in the uncatalyzed reaction (line 2). 20. Answer: B Difficulty: 3 Section: Catalysis and the Use of Energy by Cells Feedback: This is the reaction catalyzed by the enzyme succinate dehydrogenase in the citric acid cycle. Succinate is oxidized to fumarate, and the FAD carrier is reduced to FADH2. By subsequently donating its two electrons to the electron-transport chain, FADH2 will be converted back to FAD for another round of the reaction. 21. Answer: D Difficulty: 3 Section: Catalysis and the Use of Energy by Cells Feedback: The net distance of 5 µm is 10 times higher than 0.5 µm, and would on average take 10 seconds (i.e. 102 × 100 milliseconds) to reach. 22. Answer: CABD Difficulty: 2 Section: Catalysis and the Use of Energy by Cells

Feedback: In general, larger molecules diffuse more slowly compared to smaller molecules. Interaction with other molecules (including the solvent) and the shape of the molecule will also affect the diffusion coefficient. 23. Answer: A Difficulty: 3 Section: Catalysis and the Use of Energy by Cells Feedback: The equilibrium constant for this reaction is calculated as: Keq = 0.5 = [B]eq [C]eq / [A]eq Since the initial mixture contains only molecule A, it follows that: [B]eq = [C]eq = 1 M – [A]eq Combining these equations and solving for [A]eq, we will have: [A]eq = 0.5 M which means the molecules B and C will also be present at 0.5 M at equilibrium. 24. Answer: +23.6 kJ/mole Difficulty: 4 Section: Catalysis and the Use of Energy by Cells Feedback: The free-energy change can be written as: ΔG = ΔG° + RT ln([B]/[A]) When ΔG is equal to zero, the system is at chemical equilibrium, and ΔG° = – RT ln([B]eq/[A]eq) = – RT ln(10–2) = –5.9 × log (10–2) = +11.8 kJ/mole When the concentrations are changed, we have: ΔG = ΔG° + RT ln([B]/[A]) = ΔG° + RT ln(102) = ΔG° – RT ln(10–2) = 2 ΔG° = +23.6 kJ/mole 25. Answer: C Difficulty: 3 Section: Catalysis and the Use of Energy by Cells Feedback: The negative ΔG value indicates that the reaction is favorable under these conditions and would increase the entropy of the universe. However, unless we know the steps of the reaction, the ΔG value cannot predict the reaction rate, because the latter depends on the activation-energy barrier. Finally, the ΔG value changes as the concentrations of reactants and products change. As the products accumulate, the reaction will eventually reach an equilibrium, where ΔG is equal to zero. 26. Answer: C Difficulty: 3 Section: Catalysis and the Use of Energy by Cells Feedback: The two reactions described in the question can be written as:

ATP + glucose → ADP + glucose 6-phosphate ΔG°= –17 kJ/mole glucose 6-phosphate + H2O → glucose + Pi ΔG°= –14 kJ/mole Combining these reactions yields the ATP hydrolysis reaction presented in the question. Since the free-energy changes are additive, the ΔG° value for this combined reaction is the sum of the ΔG° values for the two reactions above: (–17 kJ/mol) + (–14 kJ/mole) = –31 kJ/mole. 27. Answer: C Difficulty: 3 Section: Catalysis and the Use of Energy by Cells Feedback: Since the equilibrium constant is less than 1, the log Keq term is negative, making the ΔG° positive, which means the reaction is unfavorable under standard conditions. But inside a cell performing glycolysis, a lower concentration of fructose 6phosphate than of glucose 6-phosphate can drop ΔG to a negative value. The reaction thus proceeds in the forward direction, providing a continuous supply of substrate for the next step in the pathway. Note that due to “coupling” of the reactions in the glycolytic pathway, and even though some steps can have positive ΔG° values, the overall negative ΔG° of the pathway can drive the entire chain of reactions in the forward direction, even under the standard conditions. 28. Answer: B Difficulty: 1 Section: Catalysis and the Use of Energy by Cells Feedback: The activated carrier molecules carry chemical groups in high-energy linkages and can deliver the group or the energy (or both) to metabolic reactions when necessary. They then need to be activated again. NADH is an activated carrier, while its oxidized form NAD+ is not. 29. Answer: D Difficulty: 1 Section: Catalysis and the Use of Energy by Cells Feedback: By transferring either a phosphate group or a pyrophosphate group to a hydroxyl group on one of the monomers involved in the polymerization, ATP “activates” the monomer, making the overall reaction favorable. 30. Answer: E Difficulty: 3 Section: Catalysis and the Use of Energy by Cells Feedback: The extra phosphate in NADPH does not affect its electron-transfer properties, but makes it different enough to be recognized by a different set of enzymes. NADPH

operates chiefly with enzymes that catalyze anabolic reactions, which normally need reducing power. Accordingly, NADPH is found mostly in its reduced form (i.e. in excess over NADP+) in the cell. The opposite is true for NADH, which is normally involved in catabolic reactions. 31. Answer: C Difficulty: 2 Section: Catalysis and the Use of Energy by Cells Feedback: In a reduction reaction, NADPH (or NADH) is oxidized, donating a hydride ion to the substrate. A substrate can also capture a proton from the surroundings, creating two C–H bonds. The carrier is converted to its oxidized form (NADP+ or NAD+). 32. Answer: D Difficulty: 2 Section: Catalysis and the Use of Energy by Cells Feedback: The reaction shown is catalyzed by the enzyme pyruvate carboxylase, which uses a covalently bound carboxylated biotin to carboxylate pyruvate and produce oxaloacetate. 33. Answer: D Difficulty: 3 Section: Catalysis and the Use of Energy by Cells Feedback: Under anaerobic conditions, NAD+ can be recycled in this reaction (in which pyruvate is reduced), so that glycolysis can continue in the absence of oxidative phosphorylation. 34. Answer: E Difficulty: 2 Section: Catalysis and the Use of Energy by Cells Feedback: Each nucleotide monomer is activated—at the expense of hydrolysis of two ATP molecules—into an intermediate carrying a reactive phosphoanhydride bond. 35. Answer: D Difficulty: 1 Section: How Cells Obtain Energy from Food Feedback: In glycolysis, two pyruvate molecules (each with three carbon atoms) are produced from each molecule of glucose (with six carbon atoms). 36. Answer: D Difficulty: 2 Section: How Cells Obtain Energy from Food Feedback: In step 6 of glycolysis, this enzyme couples the oxidation of the substrate with the production of NADH, as well as incorporation of inorganic phosphate. The P i is then transferred to ADP to generate ATP in step 7.

37. Answer: E Difficulty: 2 Section: How Cells Obtain Energy from Food Feedback: Please refer to Figure 2–48. 38. Answer: B Difficulty: 3 Section: How Cells Obtain Energy from Food Feedback: Since step 7 is bypassed, the ATP molecules that are naturally generated in that step are no longer produced; however, NADH is still made as before in the first half of step 6. Arsenate also has other effects on cell metabolism that collectively make it a toxic compound. Please refer to Figure 2–48. 39. Answer: E Difficulty: 1 Section: How Cells Obtain Energy from Food Feedback: Compared to the polysaccharides glycogen (in animals) and starch (in plants), fat is more efficient as a long-term energy storage both per gram and per volume. It can be stored as triglycerides in both plants and animals, although the types of fatty acids vary. In plant cells, chloroplasts generate sugars that can be oxidized by the mitochondria to generate ATP for the cell. The ATP produced in the chloroplast by photosynthesis cannot be transported out of that organelle. 40. Answer: D Difficulty: 2 Section: How Cells Obtain Energy from Food Feedback: One molecule of acetyl CoA enters the cycle by combining with oxaloacetate. 41. Answer: 2411 Difficulty: 2 Section: The Chemical Components of a Cell Feedback: The nucleotides contain one to three phosphate groups and a nitrogencontaining base, and are polymerized to form long nucleic acid molecules such as DNA. Proteins are made of amino acids and make up half of the dry mass of the cell, i.e. approximately 15% of the total cell weight. Lipids have large hydrophobic fatty acid chains and, in addition to forming bilayer membranes, can store food energy and release it when necessary. 42. Answer: DAECB Difficulty: 3 Section: Catalysis and the Use of Energy by Cells

Feedback: As a general rule, in organic molecules, a lower number of C–H bonds corresponds to more oxidized carbon atoms. 43. Answer: FTFT Difficulty: 2 Section: How Cells Obtain Energy from Food Feedback: During the stepwise oxidation of glucose in the course of glycolysis, two molecules of ATP are used to make fructose 1,6-bisphosphate, which is then cleaved and eventually converted to two molecules of pyruvate, generating four molecules of ATP and two molecules of NADH. 44. Answer: fermentation Difficulty: 1 Section: How Cells Obtain Energy from Food Feedback: Fermentation is an energy-yielding pathway and is often anaerobic. 45. Answer: DCAB Difficulty: 2 Section: How Cells Obtain Energy from Food Feedback: Hydrolysis of phosphoenolpyruvate to pyruvate is the most exergonic (releases the highest amount of energy). ATP can be generated from ADP upon the hydrolysis of 1,3-bisphosphoglycerate to 3-phosphoglycerate. ATP hydrolysis can be used to drive the phosphorylation of glucose. 46. Answer: TFTF Difficulty: 2 Section: How Cells Obtain Energy from Food Feedback: Between meals, fatty acids stored in the fat droplets in adipocytes in the form of triacylglycerol are released by hydrolysis and enter the bloodstream. Upon entry into other cells, they are transported to the mitochondria where they are mostly converted to acetyl CoA in a step-by-step manner, each step producing one FADH2 and one NADH molecule. 47. Answer: KGKK Difficulty: 2 Section: How Cells Obtain Energy from Food Feedback: In the citric acid cycle (Krebs cycle), which takes place in the mitochondria, the carbon atoms of acetyl CoA are oxidized and released as CO2, while NADH, FADH2, and GTP are generated in the process. Many intermediates of the citric acid cycle and glycolysis are precursors for the biosynthesis of important small molecules in the cell. 48. Answer: TTGT

Difficulty: 2 Section: How Cells Obtain Energy from Food Feedback: Phosphoenolpyruvate is converted to pyruvate in the last step of glycolysis. Succinate, fumarate, and malate are three consecutive citric acid cycle intermediates leading to the regeneration of oxaloacetate. 49. Answer: C Difficulty: 3 Feedback: In the first step of the citric acid cycle, CoA is hydrolyzed by water after the formation of a citryl CoA intermediate. 50. Answer: 34 Difficulty: 3 Feedback: Steps 3 and 4 are catalyzed by isocitrate dehydrogenase and the αketoglutarate dehydrogenase complex, respectively, and involve decarboxylation of the substrates and the release of carbon dioxide. 51. Answer: 3468 Difficulty: 3 Refer to: The Citric Acid Cycle Section: How Cells Obtain Energy from Food Feedback: Steps 3, 4, and 8 produce NADH, while step 6 produces FADH2. 52. Answer: 2 Difficulty: 2 Feedback: Aconitase converts citrate to isocitrate through an aconitate intermediate created by dehydration of the substrate. 53. Answer: A Difficulty: 3 Section: How Cells Obtain Energy from Food Feedback: The complete oxidation of a molecule of acetyl CoA results in the production of three NADH molecules plus one FADH2 and one GTP (or ATP) molecule. Therefore, as a result of oxidative phosphorylation, a total of 10 molecules are generated: (3 × 2.5) + (1 × 1.5) + 1 = 10 54. Answer: FFTT Difficulty: 2 Section: How Cells Obtain Energy from Food Feedback: All known nitrogen-fixing cells are prokaryotic microorganisms. Animals rely on their dietary intake of protein and nucleic acids as sources of useful nitrogen. However, only 9 of the 20 amino acids that make up proteins and none of the nucleotides that make up nucleic acids are essential; the remainder can be synthesized from other

ingredients in the diet. When amino acids in our body are degraded, their nitrogen atoms eventually appear in urea molecules which are excreted.

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION CHAPTER 3: PROTEINS Copyright © 2015 by W.W. Norton & Company, Inc. 1. At physiological pH, what is the approximate net charge of a hexapeptide with the following amino acid sequence? Asp-Val-Ile-Glu-Arg-Ser A. –2 B. –1 C. 0 D. +1 E. +2 2. Which of the following pairs of amino acid residues would you expect to form ionic bonds? A. Glutamic acid and glutamine B. Arginine and lysine C. Lysine and glutamic acid D. Tryptophan and tyrosine E. Tyrosine and glutamine 3. Which of the following stretches of amino acid residues would you expect to find in the interior of protein molecules? A. Ala-Asp-Asp-Tyr-Arg B. Gly-Lys-Ser-Pro-Thr C. Phe-Glu-Gln-Glu-Asn D. Ala-Val-Leu-Ile-Trp E. Gly-Tyr-His-Arg-His 4. Which of the following is NOT the role of molecular chaperones in the folding of cellular proteins? A. They assist proteins in folding into their correct conformations. B. They help prevent formation of protein aggregates. C. They specify the final three-dimensional shape of proteins. D. They catalyze the folding of proteins in the crowded environment of the cell.

E. They make the protein-folding process in the cell more reliable. 5. The observation that proteins often renature into their original conformations after they have been unfolded by denaturing solvents implies that … A. the information needed to specify the three-dimensional shape of a protein is encoded in its amino acid sequence. B. the cell does not need molecular chaperones for survival. C. the final folded structure of a protein is usually NOT the one with the lowest free energy. D. each protein folds into several different conformations inside the cell. E. All of the above. 6. Imagine a cellular protein composed of 3000 amino acid residues in one continuous polypeptide chain. This protein is almost certainly … A. extracellular. B. globular. C. multidomain. D. composed of mostly α-helical regions. E. intrinsically disordered. 7. For each of the following cartoon representations from left to right, indicate whether the protein structure is composed of only α helices (1), only parallel β sheets (2), only antiparallel β sheets (3), α helices plus parallel β sheets (4), or α helices plus antiparallel β sheets (5) as repetitive secondary structure elements. Your answer would be a four-digit number composed of digits 1 to 5 only, e.g. 5513.

Images created in PyMOL, from PDB entries 1STU, 2LPC, 1SA8, and 3BOB.

8. Protein secondary structure elements such as α helices and β sheets constitute the major regular folding patterns in proteins. With regard to these elements, … A. hydrogen-bonding between the amino acid side chains defines the type of secondary structure. B. a certain short amino acid sequence always adopts the same secondary structure. C. only a few specific amino acid sequences can adopt these repetitive structures. D. the folding patterns result from hydrogen-bonding between the N–H and C=O groups in the polypeptide backbone. E. All of the above. 9.

Which of the following is NOT true regarding the members of a protein family in

general? A. They have similar three-dimensional conformations. B. They share an ancestry; i.e. they are homologs. C. They can functionally replace each other. D. Their gene sequence is less well conserved than their structure. E. Over evolutionary time scales, the family has expanded mainly through gene duplication events. 10. The number of ways in which protein domains fold in nature is limited. Which of the following is a better estimate of this number? A. 200 B. 2000 C. 200,000 D. 2,000,000 E. 20 million 11.

The human estrogen receptor is a symmetrical dimeric nuclear protein that can regulate

gene expression by binding to a DNA sequence called an estrogen response element (ERE) near the promoter of its target genes. Each subunit of the receptor binds to about six base pairs of DNA. Which of the following sequences is a likely candidate for the ERE? The sequences are written in the 5′-to-3′ direction. The letter N represents any of the four DNA bases. A. AGGTCANNNTGACCT B. AGGTCANNNAGGTCA C. AGGTCANNNACTGGA D. AGGTCANNNATATAT

E. AGGCCTNNNTCATGA 12.

Which of the following can be a function for intrinsically disordered protein sequences? A. High-specificity binding to other proteins B. Cell signaling through covalent modification of the protein sequence C. Tethering to hold interacting proteins in close proximity D. Formation of a diffusion barrier from a dense network of such sequences E. All of the above

13.

You have purified a multisubunit extracellular protein that has several interchain

disulfide bonds. Which of the following chemicals would you add to your purified protein mixture if you wanted to eliminate the disulfide bonds? A. NaCl, a salt B. SDS, an ionic detergent and denaturing agent C. H2O2, an oxidizing reagent D. Tris, a buffering agent E. DTT, a reducing agent 14.

Many viruses have large capsids in the form of a hollow sphere, made of hundreds of

identical protein subunits. What are the advantages of having coats made of several copies of only a few subunits? A. Assembly can be readily regulated. B. Disassembly can be readily regulated. C. It requires a smaller amount of genetic information. D. The effect of mistakes in protein synthesis on the overall assembly is minimized. E. All of the above. 15.

Stable β-sheet aggregates can form from many proteins, forming intertwined cross-beta

strands that have the potential to kill cells or damage tissues. Which of the following is NOT true regarding these aggregates? A. They form almost exclusively in the cells of the nervous system. B. Different types of such aggregates can form from the same protein. C. Their formation is associated with conditions such as Parkinson’s disease and Kuru. D. They can form spontaneously, but also can be triggered to form by an infection with the same aggregate. E. Some healthy cells form these aggregates to store their secretory proteins.

16. An amino acid residue that is not part of the active site of an enzyme and does not interact with the ligand is nevertheless critical for ligand binding and is highly conserved. How can this be explained? A. This residue is critical for the correct folding and the placement of ligand-binding site residues. B. The residue helps with restricting the access of water to the ligand-binding site. C. This residue can affect the chemical properties of the residues in the ligand-binding site. D. All of the above. 17. Evolutionary tracing has identified clusters of the most invariant amino acids in the SH2 domain family. Which of the following is true regarding these sites? A. Mutation in these sites generally keeps the protein functional. B. The invariant amino acids generally have negatively charged side chains. C. These sites correspond to the binding site for peptides containing phosphorylated tyrosine. D. The amino acids at these sites are generally hydrophobic. E. These sites have evolved fast because they have critical functions. 18. Protein A can bind to each of the proteins B or C. The association rate constants are the same for forming the AB and the AC complexes. However, the dissociation rate constant for AB is 100 times higher than that for AC. Given that every tenfold increase in the equilibrium constant (of the association reaction) corresponds to about –5.9 kJ/mole difference in the standard free-energy change for the reaction (ΔG°), what is the value of (ΔG°AB – ΔG°AC) in kJ/mole? A. –11.8 B. –5.9 C. 0 D. +5.9 E. +11.8 19. In interactions between proteins, each hydrogen bond contributes to the free energy of binding by about –4 kJ/mole. If two proteins bind to each other through nine hydrogen bonds, six of which are eliminated when one of these proteins is mutated, how much would you expect the

equilibrium constant for their binding to change as a result of the mutation? (ΔG° = –5.9 × log Keq) A. Increase sixfold B. Decrease fourfold C. Decrease by six orders of magnitude D. Increase 1 million-fold E. Decrease by four orders of magnitude 20. The turnover number for an enzyme is equivalent to the number of substrate molecules processed per second per enzyme molecule. To a test tube containing a 100 mM concentration of its substrate, you have added an enzyme at a final concentration of 10 µM, and have measured the rate of the reaction to be approximately 500 µM/sec. If the Km for the binding of the enzyme to this substrate is about 100 mM, what is the turnover number? A. 100 B. 500 C. 1000 D. 5000 E. 10,000

21. Enzymes can catalyze cellular reactions through various mechanisms. Which of the following statements is NOT true regarding enzymes? A. They can provide the chemical groups necessary for simultaneous acid and base catalysis. B. They have a higher affinity for the transition state of the substrate than for its stable form. C. They can form covalent bonds with the substrate during catalysis. D. They can strain a substrate to force it toward a specific transition state. E. They accelerate a cellular reaction by destabilizing the transition state. 22. An enzyme acts on a tyrosine residue in a target protein to create a binding site for the SH2 domain. This enzyme is most specifically … A. an isomerase. B. a nuclease. C. a phosphatase. D. a kinase.

E. a synthase. 23. The enzyme lysozyme catalyzes the cutting of a polysaccharide chain through hydrolysis. Which of the following is NOT true regarding the catalytic cycle for this enzyme? A. It involves acid catalysis. B. It involves base catalysis. C. It involves strain catalysis. D. It involves covalent catalysis. E. It involves metal ion catalysis. 24.

How do multienzyme complexes in the cell, such as the fatty acid synthase, enhance

reaction rates? A. By increasing the diffusion rate of their substrate in the cell B. By allowing the channeling of pathway intermediates from one enzyme to the next C. By limiting the diffusion of the substrates in membrane-bound compartments D. By increasing the intrinsic rate of catalysis for individual enzymes E. All of the above Answer: B

Difficulty: 2

Section: Protein Function

25. Which of the regulatory interactions 1 to 5 depicted in the following diagram is NOT an example of a negative feedback regulation?

3 P

A. 1 B. 2 C. 3 D. 4

4 R

S 5

E. 5 26. Two ligands, A and B, bind to two different conformations of the enzyme X. The ligand A is the enzyme’s substrate, whereas ligand B binds to a remote allosteric site. Which of the following is a consequence of this arrangement? A. Binding of A to X does not affect the affinity of X for binding to B. B. Binding of B to X does not affect the rate of reaction catalyzed by X. C. Binding of A to X increases the affinity of X for B. D. Binding of B to X decreases the affinity of X for A. E. Binding of B to X has a large effect on the binding of A to X, but binding of A to X has a small effect on X–B binding. 27. The phosphofructokinase (PFK) enzyme is one of the key players in the glycolytic pathway in which glucose eventually breaks down into pyruvate, some ATP is generated, and some NAD+ is reduced. PFK catalyzes the committed step in the pathway and is under extensive regulation. Which of the following compounds would you expect to activate PFK? A. ADP B. NADH C. Pyruvate D. ATP E. All of the above 28.

Phosphorylation of a protein by a protein kinase … A. adds two positive charges to the protein. B. activates the protein. C. deactivates the protein. D. can create a binding site for other proteins. E. requires the hydrolysis of two molecules of ATP per phosphorylated residue.

29. In the following schematic diagram of a simple signaling pathway, protein Z regulates the activity of protein X, which is an upstream protein kinase, through a negative feedback loop. Which of the following better describes protein Z?

P X

X P Y

Y P Z

A. It is a protein kinase that is activated by phosphorylation. B. It is a protein kinase that is inactivated by phosphorylation. C. It is a protein phosphatase that is activated by phosphorylation. D. It is a protein phosphatase that is inactivated by phosphorylation. 30. Most GTPases are present inside the cell at a much higher concentration than their upstream GAP and GEF proteins. Imagine a mutation in a certain GTPase, such as a Rab protein, resulting in an extremely tight binding between the GTPase and its GEF, and a very slow dissociation. What would you expect to happen in the cell as a result? A. Rab proteins will be activated, because the tightly bound GEFs will be unavailable. B. Rab proteins will be inactivated, because the tightly bound GEFs will be unavailable. C. Rab proteins will be activated, because the Rab-GAPs will become activated. D. Rab proteins will be inactivated, because their GAPs will become activated. E. Rab proteins will be activated, because there are fewer GAPs available. 31. A viral version of the Src kinase called v-Src is found in some retroviruses. Unlike the cellular Src, the v-Src kinase is constitutively active, and can drive the cell into uncontrolled growth and tumor formation. Which of the following molecular differences between the two versions of Src is more likely to be responsible for this? A. v-Src lacks the active-site tyrosine residue. B. v-Src lacks a lobe of the kinase domain. C. v-Src has multiple inhibitory tyrosine phosphorylation sites in the kinase domain.

D. v-Src lacks the C-terminal tail that can bind to Src’s SH2 domain. E. v-Src has a longer C-terminal tail. 32. Switch proteins that bind and hydrolyze GTP are ubiquitous cell regulators in a wide variety of molecular processes. Which of the following statements is NOT true regarding these proteins? A. EF-Tu is an example of such proteins. B. GTP hydrolysis by the GTPase generally renders it inactive. C. Activation of the GTPase involves addition of a phosphate group to its bound guanine nucleotide. D. Guanine nucleotide exchange factors generally activate the GTPases. E. Guanine nucleotide exchange factors accelerate the release of bound GDP from the GTPases. 33. A polyubiquitin chain has been attached to a protein. The ubiquitin molecules are linked together via isopeptide bonds between Lys48 of one molecule and the carboxyl end of the next one. This protein is expected to … A. be a part of chromatin. B. undergo proteasomal degradation. C. be involved in DNA repair. D. be targeted to endocytic vesicles. E. None of the above. 34.

The marking of a protein by polyubiquitylation to signify degradation … A. requires the hydrolysis of one ATP molecule to ADP per polyubiquitin chain. B. involves covalent attachment of the target protein to the E1 enzyme. C. is carried out by the proteasome complex. D. is typically done on an arginine residue in the target protein. E. involves the recognition of the target protein by an E2–E3 ligase.

35. The SCF ubiquitin ligase can recognize and mark various target proteins at different stages of the cell cycle. In this complex, … A. different F-box subunits recognize different target proteins. B. the F-box subunit and the E2 ubiquitin-conjugating enzyme are at the opposite ends of the C-shaped molecule.

C. a scaffold protein arranges the other subunits such that the two ends of the complex are separated by a gap. D. the use of interchangeable parts such as the F-box subunits makes economical use of the genetic information and allows for rapid evolution of new functions. E. All of the above. 36.

In terms of molecular function, what do Ras and myosin have in common? A. They normally bind to GTP. B. They couple the hydrolysis of a bound nucleoside triphosphate to protein movements. C. They switch between two distinct conformations controlled by protein phosphorylation and dephosphorylation. D. They are multisubunit proteins and contain a scaffold subunit. E. All of the above.

37. The compound GDPNP is a GTP analog that can bind to GTPases in the same way as GTP. However, unlike GTP, it cannot undergo the hydrolysis reaction that normally releases Pi. Therefore, … A. GDPNP and GDP bind to the same conformation of a GTPase. B. a GDPNP-bound Ras is constitutively active. C. a GDPNP-bound EF-Tu cannot bind to tRNA molecules. D. binding of GDPNP to EF-Tu allows multiple cycles of tRNA binding and release without the use of free energy. E. None of the above. 38.

Consider the proteins Ras, Src, kinesin, and the ATP synthase pump. All of them … A. are allosteric proteins. B. can have two or more distinct conformations. C. can bind to ATP (or GTP). D. can hydrolyze ATP to ADP (or GTP to GDP). E. All of the above.

39. Many macromolecular complexes in the cell contain scaffold proteins. What do these proteins do that benefits the cell? A. They can enhance the rate of critical cellular reactions. B. They can hold the many subunits of a large complex together.

C. They can confine and concentrate a specific set of interacting proteins to a particular cellular location. D. They can provide a large macromolecular complex with either flexibility or rigidity. E. All of the above. 40. Imagine a protein that can be independently phosphorylated on any of its 10 tyrosine residues and acetylated on either of its 2 lysine residues. In principle, how many different combinations of these modifications are possible for this protein? A. 210 B. 220 C. 1010 D. 1220 E. 212 41. A simple protein interaction map is shown below for human cytochrome c (Cyt), a hemecontaining protein that is normally found inside the mitochondria and is associated with the electron-transport chain (ETC) of the inner mitochondrial membrane. Under special emergency conditions, this protein can also moonlight as a signal transducer for the onset of a pathway leading to programmed cell death (PCD). As shown in the map, it also interacts with a group of phosphoprotein phosphatases (PPP) involved in cell signaling pathways. Assuming that the functions of the proteins labeled as X, Y, and Z are unknown, which of the following points can be reasonably argued from this interaction map?

ETC X

Cyt c

PPP

PCD Y Z

A. The protein labeled X probably functions in PCD as well as in the ETC, because it interacts with cytochrome c, which is known to be closely involved in PCD, and it also interacts with several proteins from the ETC group. B. The protein labeled Y probably has a role in PCD, especially if its orthologs in other organisms have such a role and show a similar pattern of interactions. C. If the proteins in the PCD group are known to form a large complex, then the protein labeled Z is likely to be the scaffold protein for that complex. D. The protein labeled Y probably has a phosphatase function, because it interacts with the proteins of the PPP group. E. The protein labeled Z is probably not an essential protein because it only interacts with one other protein in this map. Reference: The 20 Amino Acid Residues: Questions 43-47 The structural formulas for the 20 naturally occurring amino acid residues are shown in the panel below. Answer the following question(s) with the help of this panel.

42. Which of the amino acids shown above has the most limited combinations of phi (φ) and psi (ψ) angles in its Ramachandran plot? Write down the one-letter abbreviation for it, e.g. A. 43. Which amino acid residues shown above have acidic side chains? Write down the oneletter abbreviations for them, in alphabetical order, e.g. ACGNV. 44. What is the amino acid sequence of the peptide depicted below? Write down the sequence from the N- to the C-terminus, and use only the one-letter abbreviations, e.g. AGCNT.

45. What is the amino acid sequence of the peptide depicted below? Write down the sequence from the N- to the C-terminus, and use only the one-letter abbreviations, e.g. AGCNT.

46. What is the amino acid sequence of the peptide depicted below? Write down the sequence from the N- to the C-terminus, and use only the one-letter abbreviations, e.g. AGCNT.

47. In an α helix, each amino acid residue is rotated by about +100° and raised by about 0.15 nm relative to the previous residue. This means that each full turn of the helix rises by … nm parallel to the helix’s axis. 48. There are 3.6 residues per full turn of a (right-handed) α helix. In a typical coiled-coil motif called a leucine zipper, each of the two participating α helices has a leucine residue at every seventh position in the sequence. This places all of these leucine residues roughly on one side of each helix, providing an interaction interface with the leucine side chains of the other helix. Determine the handedness of such a leucine zipper coiled-coil; i.e. do the helices twist around each other in a right-handed (R) or left-handed (L) fashion? Write down R or L as your answer. 49. Imagine that 1 L of a solution containing each of the 20 naturally occurring amino acids at 50 mM concentration each (total concentration of 1 M) is allowed to polymerize in a perfectly stepwise fashion such that at each step, a random amino acid can be incorporated into a growing

polypeptide. The steps are repeated, until eventually the solution is only composed of 40-mers (and virtually all of the monomers have been used). What fraction of all 40-mers can possibly be present in this solution? Round your number to four decimal places. Avogadro’s number is 6 × 1023. 50. Some protein domains are found in many different proteins and have been especially mobile during evolution. A domain of this kind is called a “protein …” 51. For a simple enzymatic reaction that involves only one substrate and follows Michaelis– Menten kinetics, the changes in the concentrations of substrate, product, free enzyme, and the enzyme–substrate complex over the course of the reaction are depicted by solid curves in the

concentrations

graph below. Which curve (1 to 4) corresponds to [ES]? Write down the number as your answer.

4 time

52. A protein is drawn in the following simplified diagram undergoing a set of covalent modifications including the addition of a chain of ubiquitin protein monomers (U) to one of its lysine side chains, a phosphate moiety (P) to its tyrosine side chain, and a lipid farnesyl (F) to its cysteine side chain. Indicate the order of these modifications (1 to 3) in the diagram. Your answer would be a three-letter string composed of letters F, U, and P only, e.g. PFU.

Unmodified protein

The protein is associated with the plasma membrane

The protein is recognized by an active E2– E3 ligase

The protein is specifically degraded

Answers 1. Answer: B Difficulty: 2 Section: The Shape and Structure of Proteins Feedback: Each of the acidic amino acids Asp (aspartic acid) and Glu (glutamic acid) carries one negative charge on its side chain at neutral pH, while Arg (arginine) carries one positive charge. The side chains of the other residues shown are not ionized at neutral pH. Additionally, the terminal amine group (one positive charge) and carboxyl group (one negative charge) neutralize each other. 2. Answer: C Difficulty: 2 Section: The Shape and Structure of Proteins Feedback: Electrostatic attractions can hold together amino acid residues of opposite charge, i.e. an acidic and a basic residue. 3. Answer: D Difficulty: 2 Section: The Shape and Structure of Proteins Feedback: Hydrophobic side chains in a protein—belonging to amino acids such as phenylalanine, leucine, isoleucine, valine, and tryptophan—tend to be buried in the interior of the molecule to avoid the unfavorable contact with water. 4. Answer: C Difficulty: 1 Section: The Shape and Structure of Proteins Feedback: The molecular chaperones assist a protein in adopting its correct conformation, which is specified by the protein’s amino acid sequence. 5. Answer: A Difficulty: 2 Section: The Shape and Structure of Proteins Feedback: This observation indicates that the amino acid sequence of a protein contains all the information needed for specifying its three-dimensional shape. 6. Answer: C Difficulty: 3 Section: The Shape and Structure of Proteins Feedback: This large size is about ten times that of an average protein and about thirty times that of an average protein domain. Such a protein is extremely unlikely to constitute a single domain or have an overall globular shape. Although proteins have disordered regions, it is not likely that such a large protein is entirely disordered either. Many large multidomain proteins are mostly made up of β sheets. 7. Answer: 5134 Difficulty: 3 Section: The Shape and Structure of Proteins Feedback: The first protein (on the left) is composed of α helices and antiparallel β sheet. The second protein is made almost entirely from α helices. The third protein, in contrast, is composed of antiparallel β sheets only. The protein on the right contains both α helices and parallel β sheets.

8. Answer: D Difficulty: 1 Section: The Shape and Structure of Proteins Feedback: It is the repeating hydrogen-bond pattern between backbone atoms that defines each of the secondary structure elements. Many different sequences can adopt each of these structures and, depending on the context, the same sequence may be found in different folding patterns. 9. Answer: C Difficulty: 2 Section: The Shape and Structure of Proteins Feedback: Although members of the same family share a similar sequence, ancestry, and threedimensional conformation, they have nevertheless evolved to have their own specialized functions. 10. Answer: B Difficulty: 2 Section: The Shape and Structure of Proteins Feedback: We know the three-dimensional shapes of over 100,000 proteins. However, most of them fold up into conformations that are not entirely novel, and for which known representative proteins already exist. The number of different protein folds is probably not more than about 2000. 11. Answer: A Difficulty: 3 Section: The Shape and Structure of Proteins Feedback: Such a symmetrical dimer is expected to recognize a corresponding symmetrical DNA sequence with inverted repeats, such that both strands have the same sequence when read in the 5′-to-3′ direction. Each subunit recognizes the sequence AGGTCA. 12. Answer: E Difficulty: 1 Section: The Shape and Structure of Proteins Feedback: Intrinsically disordered protein sequences are frequent in nature, and many of them are involved in one or more of these four important functions. 13. Answer: E Difficulty: 2 Section: The Shape and Structure of Proteins Feedback: A reducing agent such as dithiothreitol (DTT) converts S–S bonds back to SH groups. 14. Answer: E Difficulty: 1 Section: The Shape and Structure of Proteins Feedback: Especially for some viruses, small genome sizes confer high evolutionary fitness. 15. Answer: A Difficulty: 2 Section: The Shape and Structure of Proteins Feedback: The amyloid fibrils can occur in many tissues, but the nervous system is the most sensitive to their formation and shows the most symptoms when affected. Studies in yeast cells have shown that different types of infectious prions can form from the same type of protein.

Although these aggregates are associated with some diseases in humans, healthy cells are known to use them as well, for storage of secretory proteins for example. 16. Answer: D Difficulty: 2 Section: Protein Function Feedback: Residues that are far from the binding site can still influence ligand binding by affecting global or local folding of the protein, by controlling the access of water to the binding site, or by changing the chemical properties of other residues. 17. Answer: C Difficulty: 1 Section: Protein Function Feedback: The invariant amino acids generally correspond to the conserved binding site for the binding partners of the SH2 proteins. 18. Answer: E Difficulty: 2 Section: Protein Function Feedback: The equilibrium constant for complex formation is equal to the ratio of association and dissociation rate constants; thus, it is 100 times higher for AC compared to AB. This makes ΔG°AB more positive than ΔG°AC by a 11.8 kJ/mole difference. 19. Answer: E Difficulty: 3 Section: Protein Function Feedback: Elimination of six hydrogen bonds in the interaction increases the ΔG° value by about 24 kJ/mole, which corresponds to a 10,000-fold decrease in the binding equilibrium constant. 20. Answer: A Difficulty: 3 Section: Protein Function Feedback: Under these conditions, we have: [S] = Km = 100 mM. This means the rate of reaction is: V = ½ × Vmax. Thus, Vmax = 2 × 500 µM/sec = 1000 µM/sec. Finally, turnover number is equal to this rate divided by the enzyme concentration; that is: kcat = (1000 µM/sec) / (10 µM) = 100 per second. 21. Answer: E Difficulty: 3 Section: Protein Function Feedback: Transition-state stabilization is a general strategy for enzymes to lower the activation energy and accelerate a reaction. 22. Answer: D Difficulty: 2 Section: Protein Function Feedback: This enzyme is a tyrosine kinase. 23. Answer: E Difficulty: 2 Section: Protein Function

Feedback: Lysozyme uses several mechanisms to accelerate the reaction by several orders of magnitude. The process involves using amino acid side chains that act as acids (donate protons) or bases (accept protons) during the catalytic cycle, straining the substrate to expose the target bonds, and even forming a temporary covalent linkage with one half of the substrate. However, metal ions are not used by this enzyme for catalysis. 24. Answer: B Difficulty: 2 Section: Protein Function Feedback: When the diffusion of substrates is limiting the overall rate of an enzymatic reaction or pathway, the use of multienzyme complexes enhances the rate by bringing the various enzymes together and passing the pathway intermediates between them. 25. Answer: E Difficulty: 2 Section: Protein Function Feedback: In 5, T is not inhibiting its own production via the regulation of an enzyme that acts earlier in the pathway. 26. Answer: D Difficulty: 3 Section: Protein Function Feedback: Due to the negative “linkage” between the two binding events, binding of each of the two ligands to the enzyme reduces the affinity of the enzyme for the other ligand, and this effect is quantitatively reciprocal as well. 27. Answer: A Difficulty: 3 Section: Protein Function Feedback: NADH, pyruvate, and ATP are all the products of the pathway and are not expected to activate PFK. On the other hand, a low energy charge in the cell (with a high ADP/ATP ratio) should trigger the activation of the glycolytic and oxidative respiratory pathways to generate more ATP. 28. Answer: D Difficulty: 1 Section: Protein Function Feedback: Phosphorylation of a protein (typically coupled to one ATP hydrolysis per reaction) adds negative charges to the protein, potentially changing its surface chemistry and recruiting binding partners that recognize the phosphorylated protein. Some proteins are activated by this modification, while some are not, or are even inactivated. 29. Answer: C Difficulty: 2 Section: Protein Function Feedback: Protein Z, when phosphorylated by the protein kinase Y, catalyzes the removal of the phosphate from the protein kinase X, rendering it inactive in a feedback inhibition mechanism. 30. Answer: B Difficulty: 3 Section: Protein Function

Feedback: The tight binding to the mutant titrates out the available GEFs (guanine nucleotide exchange factors), which are not as abundant as the GTPase itself. Therefore, not enough GEF will be available to activate the rest of the GTPase molecules; thus, compared to the normal situation, the GTPase molecules will become inactivated, with GDP remaining bound. 31. Answer: D Difficulty: 2 Section: Protein Function Feedback: By lacking the C-terminal tail that includes the inhibitory tyrosine phosphorylation site, v-Src bypasses the autoinhibitory mechanism mediated by the SH2 domain. Thus it no longer requires the dephosphorylation event to become active. 32. Answer: C Difficulty: 2 Section: Protein Function Feedback: Activation of the GTPases involves exchange of a bound GDP for GTP, rather than direct phosphorylation of the bound GDP. 33. Answer: B Difficulty: 2 Section: Protein Function Feedback: A polyubiquitin chain with Lys48 linkages generally targets the attached protein to the proteasome for degradation. However, other forms of modification by ubiquitin or related proteins can be involved in a variety of different processes. 34. Answer: E Difficulty: 2 Section: Protein Function Feedback: In this pathway, each ubiquitin molecule is activated by linkage to an E1 enzyme via a thioester bond, at the expense of two ATP equivalents. The ubiquitin is then transferred onto an E2 enzyme that is part of an E2–E3 ubiquitin ligase complex. The latter recognizes the target protein and transfers the ubiquitin to a lysine residue in the target protein or to Lys48 of the previous ubiquitin in the chain. A protein marked with this polyubiquitin chain is then recognized and degraded by the proteasome. 35. Answer: E Difficulty: 1 Section: Protein Function Feedback: In the SCF complex, the F-box subunit is mainly responsible for substrate specificity, and is separated from the E2 end of the complex by a gap that accommodates the substrate. The use of interchangeable parts in such a complex is obviously advantageous for the cell. 36. Answer: B Difficulty: 2 Section: Protein Function Feedback: Through fundamentally similar molecular mechanisms, both myosin and Ras use the free energy of nucleoside triphosphate hydrolysis to perform mechanical work that involves significant conformational changes in a switchlike fashion. 37. Answer: B Difficulty: 2

Section: Protein Function Feedback: GDPNP can trap a GTPase in an active conformation, since hydrolysis is not possible. 38. Answer: E Difficulty: 1 Section: Protein Function Feedback: Although being as functionally diverse as a switch protein, a protein kinase, a motor, and a pump, these proteins share many structural and mechanistic features. 39. Answer: E Difficulty: 1 Section: Protein Function Feedback: Scaffolds can hold the subunits of a large complex together, affecting (enhancing) intersubunit interactions as well as the cellular localization of the whole complex. Scaffolds can confer rigidity or flexibility to large complexes. 40. Answer: E Difficulty: 1 Section: Protein Function Feedback: Each modification can have two different states, which therefore doubles the number of combinations. 41. Answer: B Difficulty: 3 Section: Protein Function Feedback: Just because two proteins closely interact with each other, it does not mean that they have the same enzymatic activity. For example, a scaffold protein is generally expected to interact with many of the subunits in a complex. Proteins that display similar patterns in maps derived from different organisms are likely to have similar functions—almost certainly if they also have homology. Note that such a map only shows the small subset of interactions from the whole organism that are clustered around cytochrome c, and does not reflect all of the interactions for all of the proteins shown. 42. Answer: P Difficulty: 3 Section: The Shape and Structure of Proteins Feedback: Proline, being an imino acid, can adopt only a limited set of configurations within a polypeptide chain. 43. Answer: DE Difficulty: 3 Section: The Shape and Structure of Proteins Feedback: Aspartic acid and glutamic acid residues have acidic side chains, and are usually negatively charged in cellular proteins. 44. Answer: CELL Difficulty: 3 Section: The Shape and Structure of Proteins Feedback: The sequence is cysteine–glutamic acid–leucine–leucine. 45. Answer: SCIENCE Difficulty: 3

Refer to: The 20 Amino Acid Residues Feedback: The sequence is serine–cysteine–isoleucine–glutamic acid–asparagine–cysteine– glutamic acid. 46. Answer: GRPCNQFYC Difficulty: 3 Section: The Shape and Structure of Proteins Feedback: The sequence is that of the mammalian peptide hormone vasopressin: glycine– arginine–proline–cysteine–asparagine–glutamine–phenylalanine–tyrosine–cysteine. Note the disulfide bond between the two cysteines. 47. Answer: 0.54 Difficulty: 3 Section: The Shape and Structure of Proteins Feedback: A full (360°) turn in the α helix contains 3.6 residues (360°/[100° per residue]), and is thus raised by 0.54 nm:

3.6 residues per turn × 0.15 nm rise per residue = 0.54 nm rise per turn. 48. Difficulty: 4 Section: The Shape and Structure of Proteins Feedback: Seven is slightly short of the number of residues in two full turns of the right-handed α helix (7.2), which places each leucine not perfectly on top of the previous leucine, but slightly rotated in the left-handed sense. Thus, to interact with the leucine residues in the other helix, the helix should twist around it in a left-handed fashion. 49. Answer: 0.0000 Difficulty: 3 Section: The Shape and Structure of Proteins Feedback: The number of possible 40-mers is equal to 2040 or approximately 1052. If every 40mer in the solution has a unique sequence, there will be 0.025 moles of different 40-mers made in this solution (since the total amount of monomer was 1 mole), which is equivalent to 1.5 × 1022 molecules. Thus, only about one out of every 1030 possible 40-mer peptides has been sampled in this solution. Even the whole universe does not have enough atoms to sample all possible sequences for a protein with 300 amino acid residues in this way. 50. Answer: module Difficulty: 1 Section: The Shape and Structure of Proteins Feedback: Protein modules are a subset of protein domains that have been especially mobile during evolution and seem to have particularly versatile structures. 51. Answer: 3 Difficulty: 2 Section: Protein Function Feedback: After a brief initial period of pre-steady state, the concentration of the formed ES complexes remains more or less constant in the apparent steady-state phase. This assumption allows the derivation of the Michaelis–Menten equation for reaction rate that involves Vmax, Km, and [S]. 52. Answer: FPU

Difficulty: 2 Section: Protein Function Feedback: Attachment of lipid compounds such as the farnesyl group to a protein causes it to become membrane-associated. At the right time, a specific phosphorylation can result in recognition of the protein by certain partners, which in this case target the protein for degradation by polyubiquitin addition.

1. In a double-stranded DNA molecule, one of the chains has the sequence CCCATTCTA when read from the 5′ to the 3′ end. Indicate true (T) and false (F) statements below regarding this chain. Your answer would be a four-letter string composed of letters T and F only, e.g. TTFT. ( ) The other chain is heavier, i.e. it has a greater mass. ( ) There are no C residues in the other chain. ( ) The 5′-terminal residue of the other chain is G. ( ) The other chain is pyrimidine-rich. 2. Indicate which numbered feature (1 to 5) in the schematic drawing below of the DNA double helix corresponds to each of the following. Your answer would be a five-digit number composed of digits 1 to 5 only, e.g. 52431.

2 4

( ) Hydrogen-bonding ( ) Covalent linkage ( ) Phosphate group ( ) Nitrogen-containing base ( ) Deoxyribose sugar

3. Complete the DNA sequence below such that the final sequence is identical to that of the complementary strand. Your answer would be a seven-letter string composed of letters A, C, T, and G only, e.g. TTCTCAG. 5′-

-3′

4. A DNA nucleotide pair has an average mass of approximately 660 daltons. Knowing the number of nucleotides in the human genome, how many picograms of DNA are there in a diploid human nucleus? Avogadro’s number is 6 × 1023. Write down the picogram amount without decimals (round the number to the closest integer), e.g. 23 pg. 5.

Which of the following features of DNA underlies its simple replication procedure? A. The fact that it is composed of only four different types of bases B. The antiparallel arrangement of the double helix C. The complementary relationship in the double helix D. The fact that there is a major groove and a minor groove in the double helix

Which of the following correlates the best with biological complexity in eukaryotes? A. Number of genes per chromosome B. Number of chromosomes C. Number of genes D. Genome size (number of nucleotide pairs)

7. Indicate true (T) and false (F) statements below about the human genome. Your answer would be a six-letter string composed of letters T and F only, e.g. FTFFFT. ( ) Only about 1.5% of the human genome is highly conserved. ( ) Almost half of our genome is composed of repetitive sequences. ( ) Genes occupy almost a quarter of the genome. ( ) There are roughly as many pseudogenes in the human genome as functional genes. ( ) Transposable elements occupy almost 10% of our genome. ( ) On average, exons comprise 1.5% of our genes.

8. Chromosome 3 contains nearly 200 million nucleotide pairs of our genome. If this DNA molecule could be laid end to end, how long would it be? The distance between neighboring base pairs in DNA is typically around 0.34 nm. A. About 7 mm B. About 7 cm C. About 70 cm D. About 7 m E. None of the above 9.

For the Human Genome Project, cloning of large segments of our genome was first made

possible by the development of yeast artificial chromosomes, which are capable of propagating in the yeast Saccharomyces cerevisiae just like any of the organism’s 16 natural chromosomes. In addition to the cloned human DNA, these artificial vectors were made to contain three elements that are necessary for them to function as a chromosome. What are these elements? Write down the names of the elements in alphabetical order, and separate them with commas, e.g. gene, histone, nucleosome. 10. Indicate whether each of the following descriptions better applies to a centromere (C), a telomere (T), or an origin of replication (O). Your answer would be a seven-letter string composed of letters C, T, and O only, e.g. TTTCCTO. ( ) It contains repeated sequences at the ends of the chromosomes. ( ) It is NOT generally longer in higher organisms compared to yeast. ( ) Each eukaryotic chromosome has many such sequences. ( ) There are normally two such sequences in each eukaryotic chromosomal DNA molecule. ( ) There is normally one such sequence per eukaryotic chromosomal DNA molecule. ( ) It is where DNA duplication starts in S phase. ( ) It attaches the chromosome to the mitotic spindle via the kinetochore structure. 11. The eukaryotic chromosomes are organized inside the nucleus with a huge compaction ratio of several-thousand-fold. What is responsible for such a tight packaging? A. The various chromatin proteins that wrap and fold the DNA B. The nuclear envelope which encapsulates the chromosomes C. The nuclear matrix that provides a firm scaffold D. All of the above

12.

The two chromosomes in each of the 22 homologous pairs in our cells ... A. have the exact same DNA sequence. B. are derived from one of our parents. C. show identical banding patterns after Giemsa staining. D. usually bear different sets of genes. E. All of the above.

13.

Compared to the human genome, the genome of yeast typically has … A. more repetitive DNA. B. longer genes. C. more introns. D. longer chromosomes. E. a higher fraction of coding DNA.

14. Indicate true (T) and false (F) statements below regarding histones. Your answer would be a six-letter string composed of letters T and F only, e.g. TTFFFF. ( ) The histones are highly acidic proteins. ( ) The histone fold consists of three α helices. ( ) The core histones are much more conserved than the H1 histone. ( ) The N-terminal tails of the core histones undergo a variety of reversible posttranslational modifications. ( ) Every nucleosome core is made up of three polypeptide chains. ( ) The H1 histone is absent in the 30-nm fibers. 15. Indicate which feature (1 to 4) in the schematic drawing below of a chromatin fiber corresponds to each of the following. Your answer would be a four-digit number composed of digits 1 to 4 only, e.g. 2431. 2

4 3 1

( ) Nucleosome core particle ( ) Linker DNA ( ) Histone octamer ( ) Non-histone protein 16. In assembling a nucleosome, normally the …(1) histone dimers first combine to form a tetramer, which then further combines with two … (2) histone dimers to form the octamer. A. 1: H1–H3; 2: H2A–H2B B. 1: H3–H4; 2: H2A–H2B C. 1: H2A–H2B; 2: H1–H3 D. 1: H2A–H2B; 2: H3–H4 E. 1: H1–H2; 2: H3–H4 17. The chromatin remodeling complexes play an important role in chromatin regulation in the nucleus. They … A. can slide nucleosomes on DNA. B. have ATPase activity. C. interact with histone chaperones. D. can remove or exchange core histone subunits. E. All of the above. 18.

Which of the following is true regarding heterochromatin in a typical mammalian cell? A. About 1% of the nuclear genome is packaged in heterochromatin. B. The DNA in heterochromatin contains all of the inactive genes in a cell. C. Genes that are packaged in heterochromatin are permanently turned off. D. The different types of heterochromatin share an especially high degree of compaction. E. Heterochromatin is highly concentrated in the centromeres but not the telomeres.

19. The position effect variegation (PEV) phenotype described in this chapter can be used to identify new genes that regulate heterochromatin formation. For instance, strains of Drosophila melanogaster with the White variegation phenotype have been subjected to mutagenesis to screen for dominant mutations (in other genes) that either enhance or suppress PEV, meaning the mutations result in either lower or higher red pigment production, respectively. Which of the following mutations is expected to be an enhancer of variegation?

A. A mutation that results in the loss of function of the fly’s HP1 (heterochromatin protein 1) gene. B. A loss-of-function mutation in a gene encoding a histone deacetylase that deacetylates lysine 9 on histone H3. C. A gain-of-function mutation in a gene encoding a histone methyl transferase that trimethylates lysine 9 on histone H3, resulting in a hyperactive form of the enzyme. D. A gain-of-function mutation in a gene encoding a histone acetyl transferase that normally acetylates lysine 9 on histone H3, resulting in higher expression of the protein. 20.

The acetylation of lysines on the histone tails … A. loosens the chromatin structure because it adds positive charges to the histone. B. recruits the heterochromatin protein HP1, resulting in the establishment of heterochromatin. C. can be performed on methylated lysines only after they are first demethylated. D. is sufficient for the formation of an open chromatin structure. E. is a covalent modification and is thus irreversible.

21.

Nucleosomes that are positioned like beads on a string over a region of DNA can interact

to form higher orders of chromatin structure. Which of the following factors can contribute to the formation of the 30-nm chromatin fiber from these nucleosomes? A. Interactions that involve the histone tails of neighboring nucleosomes B. Interaction of the linker histone H1 with each nucleosome C. Binding of proteins to DNA or the histones D. ATP-dependent function of chromatin remodeling complexes E. All of the above 22. Indicate whether each of the following histone modifications is generally associated with active genes (A) or silenced genes (S). Your answer would be a four-letter string composed of letters A and S only, e.g. SSAS. ( ) H3 lysine 9 acetylation ( ) H3 serine 10 phosphorylation ( ) H3 lysine 4 trimethylation ( ) H3 lysine 9 trimethylation

23. Indicate whether each of the following histone modifications adds a negative charge to the histone (A), removes a positive charge from the histone (B), or does neither of these (C). Your answer would be a four-letter string composed of letters A, B, and C only, e.g. CABA. ( ) H3 lysine 9 acetylation ( ) H3 serine 10 phosphorylation ( ) H3 lysine 4 trimethylation ( ) H3 lysine 9 trimethylation 24. To study the chromatin remodeling complex SWR1, a researcher has prepared arrays of nucleosomes on long DNA strands that have been immobilized on magnetic beads. These nucleosomes are then incubated with an excess of the H2AZ–H2B dimer (which contains the histone variant H2AZ) in the presence or absence of SWR1 with or without ATP. She then separates the bead-bound nucleosomes (bound fraction) from the rest of the mix (unbound fraction) using a magnet, elutes the bound fraction from the beads, and performs SDS-PAGE on the samples. This is followed by a Western blot using an antibody specific to the H2AZ protein used in this experiment. The results are shown below, with the presence (+) or absence (–) of ATP, SWR1, or the H2AZ–H2B dimer in each incubation reaction indicated at the top of the corresponding lane. ATP SWR1 H2AZ–H2B

Bound H2AZ - - - + + + + + - + - + - + - + + - - + +

Unbound H2AZ -

- - + - + - + +

+ + + + - + - + - - + +

H2AZ

Which of the following statements is confirmed by the Western blot shown? A. SWR1 deposits H2AZ histones into the nucleosome arrays. B. SWR1 function is not ATP-dependent. C. The antibody used in this experiment binds to the SWR1 complex. D. All of the above. 25. Indicate whether each of the following descriptions better matches the major histones (M) or the histone variants (V). Your answer would be a six-letter string composed of letters M and V only, e.g. VVMVMV. ( ) They are more highly conserved over long evolutionary time scales.

( ) They are present in much smaller amounts in the cell. ( ) They are synthesized primarily during the S phase of the cell cycle. ( ) Their incorporation often requires histone exchange. ( ) They are often inserted into already-formed chromatin. ( ) They are assembled into nucleosomes just behind the replication fork. 26.

A chromatin “reader complex” … A. is always coupled to a “writer complex” and spreads specific chromatin modifications. B. can recognize any histone code. C. binds tightly to the chromatin only when a specific set of histone marks is present. D. can only bind to a single specific histone mark. E. has at least five protein subunits.

27. The centromeric regions in the fission yeast Schizosaccharomyces pombe are wrapped by nucleosomes containing the CENP-A histone H3 variant, and are flanked by clusters of tRNA genes that separate them from the surrounding pericentric heterochromatin. If the tRNA clusters are removed from this region, the HP1-bound heterochromatin spreads further to cover the centromeric regions. The tRNA genes are transcribed by strong RNA polymerase III promoters, which can associate with transcription factors and recruit chromatin-modifying enzymes. Based on these observations, indicate which blanks (A to E) in the paragraph below correspond to each of the following phrases. Your answer would be a five-letter string composed of letters A to E only, e.g. BCDEA. “The …(A) are not sufficient to prevent heterochromatin expansion to the centromeric regions. Instead, the …(B) are acting as …(C) in S. pombe, similar to the role of the …(D) in the β-globin locus in chickens and humans. Likely candidates for the histone-modifying enzymes recruited by the RNA polymerase III complexes are …(E).” ( ) HS4 element ( ) chromatin boundaries ( ) histone acetyl transferases ( ) tRNA genes ( ) CENP-A-containing histones

28.

In human cells, the alpha satellite DNA repeats … A. have a specific sequence indispensable for the seeding event that leads to chromatin formation. B. can be seen to be packaged into alternating blocks of chromatin, one of which contains the histone H3 variant CENP-A. C. are sufficient to direct centromere formation. D. are necessary for centromere formation. E. All of the above.

29.

It has been shown that inhibition of a key chromatin remodeling complex known as

NuRD, by deleting one of its subunits, can result in a significant increase in the efficiency of reprogramming of somatic cells into pluripotent stem cells. The reprogramming is normally done by the induced expression of a battery of transcription factors in the somatic cells, but is typically not very efficient. Such an observation suggests that the NuRD complex is normally involved in … A. erasing the epigenetic memory in somatic cells. B. maintaining the epigenetic memory in somatic cells. C. preventing DNA replication. D. formation of extended loops from chromosome territories. 30. Imagine a chromosome translocation event that brings a gene encoding a histone acetyl transferase enzyme from its original chromosomal location to a new one near heterochromatin. Which of the following scenarios is definitely NOT going to happen? A. The gene gets silenced due to heterochromatin expansion, leading to the misregulation of gene expression for a number of critical genes. B. The translocation event also brings along a chromatin barrier that can prevent heterochromatin expansion into the gene, and there is no phenotypic anomaly. C. Since the gene encodes a histone acetyl transferase, it resists heterochromatin expansion by acetylating its own histones. D. The level of the gene product decreases due to a position effect, leading to an imbalance in the chromatin state of the cell that results in the activation of programmed cell death. 31.

Lampbrush chromosomes … A. are transcriptionally inactive. B. are readily observed in the oocytes of humans and insects.

C. have thousands of duplicated DNA molecules arranged side by side. D. are mitotic chromosomes with two sister chromatids attached together only at the centromere. E. are thought to have a structure that is relevant to mammalian chromosomes in interphase. 32. Findings from a number of experiments on human chromatin have suggested that the DNA in our chromosomes is organized into loops of various lengths. Approximately how long is a typical loop (in nucleotide pairs of DNA)? A. 50 B. 2000 C. 100,000 D. 10 million E. 50 million 33. You have performed a chromosome conformation capture (3C) experiment to study chromatin looping at a mouse gene cluster that contains genes A, B, and C, as well as a regulatory region R. In this experiment, you performed in situ chemical cross-linking of chromatin, followed by cleavage of DNA in the nuclear extract with a restriction enzyme, intramolecular ligation, and cross-link removal. Finally, a polymerase chain reaction (PCR) was carried out using a forward primer that hybridizes to a region in the active B gene, and one of several reverse primers, each of which hybridizes to a different location in the locus. The amounts of the PCR products were quantified and normalized to represent the relative crosslinking efficiency in each analyzed sample. You have plotted the results in the graph below. The same experiment has been done on two tissue samples: fetal liver (represented by red lines) and fetal muscle (blue lines). Interaction of the A, B, and C genes with the regulatory R region is known to enhance expression of these genes. Indicate true (T) and false (F) statements below based on the results. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF.

Gene cluster

efficiency (arbitrary unit)

Relative cross-linking

100

other genes

4 Liver

Muscle 2 1

–10

Position (kb)

+10

+20

( ) The B gene would be predicted to have higher expression in the fetal liver compared to the fetal muscle tissue. ( ) Interaction between the R region and the B gene involves the A gene looping out. ( ) Interaction between the R region and the B gene involves the C gene looping out. ( ) In fetal muscle, the B gene definitely does not engage in looping interactions with any other elements in the cluster. 34.

Polytene chromosomes are useful for studying chromatin because they … A. are smaller than regular chromosomes and easier to manipulate. B. lack heterochromatin. C. have distinct visible banding patterns. D. can make polyploid cells. E. All of the above.

35. Five major types of chromatin were identified in studies performed on Drosophila melanogaster cells, although much more remains to be learned about chromatin diversity and dynamics. Which of the following is correct regarding these findings in Drosophila? A. These results were obtained using the 3C technique, which determines the positions of loops in the chromatin. B. According to these results, there are four types of heterochromatin and only one type of euchromatin. C. The Polycomb form of chromatin belongs to the euchromatin type.

D. In addition to these five major types of chromatin, there seem to be additional minor types as well. E. The pattern of chromatin types in the chromosomes is constant across the different cell types in a multicellular organism. 36. For each of the following classifications, indicate whether you would expect to find an actively transcribed gene in the first category (1) or the second (2). Your answer would be a sixdigit number composed of digits 1 and 2 only, e.g. 222121. ( ) 1: Heterochromatin, or 2: euchromatin ( ) 1: Chromosome puffs, or 2: condensed chromosome bands ( ) 1: Nuclear periphery, or 2: the center of the nucleus ( ) 1: Within the chromosome territory, or 2: extended out of the territory ( ) 1: Apart from, or 2: close to actively transcribed genes within the nucleus ( ) 1: 11-nm “beads-on-a-string” fibers, or 2: 30-nm fibers 37. A gene that had been turned off in a liver cell has just been induced to be highly expressed as the cell responds to a new metabolic load. What observations do you expect to accompany this change? A. More than 100 proteins would become associated with the gene for its transcription. B. The nuclear position of the gene would change to place it in a “transcription factory.” C. Chromatin modifications associated with the gene would change in favor of higher expression. D. All of the above. 38. Fill in the blank in the following paragraph regarding chromatin organization. Do not use abbreviations. “Our ~6.4-giganucleotide nuclear DNA is organized into 46 chromosomes, each occupying a territory inside the interphase nucleus. In each chromosome, the chromatin is thought to be composed of ‘open’ and ‘closed’ chromatin compartments. At the meganucleotide scale, each compartment is organized into knotfree arrangements called … that allow tight packing and simultaneously avoid entanglement.” 39. What are the consequences for the cell of the “fractal globule” arrangement of chromosome segments?

A. The chromosomes cannot be condensed maximally this way. B. The neighboring regions of DNA are furthest from each other in the threedimensional space. C. The ability of the chromatin fiber to fold and unfold efficiently is maintained. D. Dense packing is permitted, but the ability to easily fold and unfold the chromatin is prohibited. E. None of the above. 40.

Which of the following is NOT true about the nuclear subcompartments? A. Nucleoli, Cajal bodies, and speckles are examples of such subcompartments. B. Each specialized subcompartment has a distinct biochemical environment and a selected set of proteins and/or RNA molecules. C. The subcompartments are constitutively present in a cell except during nuclear divisions. D. They are likely to be organized by a tethered network of macromolecules in gel-like structures.

41. What features do Cajal bodies, interchromatin granule clusters, and nucleoli have in common? A. High permeability to the surrounding nucleoplasm B. A network of macromolecules bound together by covalent linkages C. A lipid bilayer membrane D. The same set of RNA and protein molecules E. The same size 42. As each cell in our body prepares for mitosis, its chromosomes start to look different. What are the changes in chromosome appearance that accompany the entry into M phase? Indicate true (T) and false (F) descriptions below. Your answer would be a five-letter string composed of letters T and F only, e.g. TTFFT. ( ) The chromosomes become readily visible by the naked eye. ( ) The chromosomes coil up further to become about 10 times shorter. ( ) Each chromosome is condensed and then replicated to form two sister chromatids. ( ) The typical diameter of a mitotic chromosome arm is about 70 nm. ( ) The two sister chromatids are disentangled from each other by the time chromosome condensation is complete.

43. The genetic information carried by a cell is passed on, generation after generation, with astonishing fidelity. However, genomes are still altered over evolutionary time scales, and even their overall size can change significantly. Which of the following genome-altering events has increased the size of the mammalian genome the most? A. Transposition B. Point mutation C. Chromosomal deletion D. Chromosomal inversion E. Chromosomal translocation 44.

Fill in the blanks in the following paragraph regarding genome evolution. In your answer,

separate the two missing phrases with a comma, e.g. protein, plasma membrane. Do not use abbreviations. “Conservation of genomic sequences between humans and chickens is mainly due to … selection, whereas the conservation observed between humans and chimpanzees is mostly due to the short time available for mutations to accumulate. Even the DNA sequences at the … position of synonymous codons are nearly identical between humans and chimpanzees.”

Reference: Phylogenetic Tree Questions 45 and 46 Phylogenetic trees based on nucleotide or amino acid sequences can be constructed using various algorithms. One simple algorithm is based on a matrix of pairwise genetic distances (divergences) calculated after multiple alignment of the sequences. Imagine you have aligned a particular gene from different hominids (humans and the great apes), and have estimated the normalized number of nucleotide substitutions that have occurred in this gene in each pair of organisms since their divergence from their last common ancestor. You have obtained the following distance matrix. A

0.40

3.61

3.79

1.36

1.44

3.94

Answer the following question(s) based on this matrix.

45. If species A in the distance matrix represents human, indicate which of the other species (B to D) represents chimpanzee, gorilla, and orangutan, respectively. Your answer would be a three-letter string composed of letters B, C, and D only, e.g. DCB.

46.

The following tree can be constructed from these distances assuming a constant

molecular clock, meaning that the length of each horizontal branch corresponds to evolutionary time as well as to the relative genetic distance from the common ancestor that gave rise to that branch. Indicate which one of the species in the matrix (B to D) corresponds to branches 1 to 3, respectively. Your answer would be a three-letter string composed of letters B, C, and D only, e.g. DCB. Human 1 2 3

47. The regions of synteny between the chromosomes of two species can be visualized in dot plots. In the example shown in the following graph, a chromosome of a hypothetical species A has been aligned with the related chromosome in species B. Each dot in the plot represents the observation of high sequence identity between the two aligned chromosomes in a window located at the two corresponding chromosome positions. A series of close dots can make a continuous line. Choosing a sufficiently large window size allows a “clean” dot plot with solid lines that show only the long stretches of identity, allowing ancient large-scale rearrangements to be identified. Several chromosomal events can be detected in such dot plots. Indicate which feature (a to g) in the dot plot is best explained by each of the following events. Your answer would be a seven-letter string composed of letters a to g only, e.g. cdbagef. Each letter should be used only once.

Position on the chromosome of species B (million nucleotide pairs)

g f

50 e c b 0

100

Position on the chromosome of species A (million nucleotide pairs)

( ) A duplication that exists in both species ( ) A duplication in species A only ( ) A triplication in species B only ( ) An inversion without relocation ( ) An inversion combined with relocation ( ) A deletion in species A ( ) A translocation in species A from a different chromosome

48. In each of the following comparisons, indicate whether the molecular clock is expected to tick faster on average in the first (1) or the second (2) case. Your answer would be a four-digit number composed of digits 1 and 2 only, e.g. 2222. ( ) 1: The exons, or 2: the introns of a gene ( ) 1: The mitochondrial, or 2: the nuclear DNA of vertebrates ( ) 1: The first, or 2: the third position in synonymous codons ( ) 1: A gene, or 2: its pseudogene counterpart 49. Most fish genomes are at least 1 billion nucleotide pairs long. However, the genome of the puffer fish Fugu rubripes is quite small at only about 0.4 billion nucleotide pairs, even though the number of Fugu genes is estimated to be comparable to that of its relatives which have larger genomes. What do you think mainly accounts for the Fugu genome being this small? A. Evolutionary advantage of extremely small exon sizes in the Fugu lineage B. Unusual disappearance of all intronic sequences from the Fugu genome

C. Increased abundance of transposable elements in the Fugu genome D. Increased occurrence of mitotic whole-chromosome loss in the Fugu lineage E. Low relative rate of DNA addition compared to DNA loss in the Fugu lineage 50. The copy number of some human genes, such as the salivary amylase gene AMY1, varies greatly between different individuals. The salivary amylase breaks down some of the dietary starch into smaller sugars. In the case of AMY1, a positive correlation has been observed between the copy number and the amount of amylase in the saliva. Gene copy number per diploid genome can be estimated by performing a quantitative polymerase chain reaction (PCR) using primers specific to the gene of interest. You have performed such PCR experiments on samples from two human populations that have traditional diets with low and high starch levels, respectively, and

Percentage of each population

have plotted the data in the histogram below. Which population (A or B) in the histogram is likely to be the one with traditionally higher dietary starch? Write down A or B as your answer.

40 A B

0 2

AMY1 gene copy number

51.

Indicate true (T) and false (F) statements below regarding a genome and its evolution.

Your answer would be a four-letter string composed of letters T and F only, e.g. FFTF. ( ) The genome of the last common ancestor of mammals can be investigated only if a DNA sample of the ancestor can be obtained. ( ) All of the “ultraconserved” elements found in the human genome have been shown to encode long noncoding RNAs.

( ) If a mouse carrying a homozygous deletion of a highly conserved genomic sequence survives and shows no noticeable defect, the highly conserved sequence has to be functionally unimportant. ( ) The “human accelerated regions” are genomic regions that are found in humans with no homologs in chimpanzees or other animals. 52. The globin gene family in mammals, birds, and reptiles is organized into α- and β-globin gene clusters that are located on two different chromosomes. In most fish and amphibians, however, the globin genes are close to each other on one chromosome. At which point (A to E) in the following simplified phylogenetic tree is a chromosomal translocation likely to have happened that placed the α- and β-globin genes on two separate chromosomes? Chicken E

Lizard

Human

D B

Frog A

Goldfish 400

300

200

100

Million years ago

53. To discover genes that have undergone accelerated evolution in the human lineage, you compared the amino acid sequences of dozens of proteins from orthologous protein-coding genes in humans, chimpanzees, and mice. For each gene, you build an unrooted phylogenetic tree in which the branch lengths (a, b, or c) correspond to the number of amino acid substitutions in that branch, as depicted below. Primates and rodents diverged ~90 million years ago, and humans and chimpanzees diverged ~5.5 million years ago. For each individual gene shared by the three species, you therefore define the “normalized substitution rate” parameters h and k as h = (a/5.5)/[c/([2 × 90] – 5.5)], and k = (b/5.5)/[c/([2 × 90] – 5.5)]. Based on these definitions, which genes are more likely to be responsible for “uniquely human” traits?

Human

Chimpanzee

Mouse A. Genes with very high h and k values B. Genes with very low h and k values C. Genes with very high h values but not very high k values D. Genes with very high k values but not very high h values 54. Which of the following would most reliably suggest that a genomic sequence is functionally important? A. The presence of a long open reading frame in the sequence B. Multispecies conservation of the sequence C. Low copy number variation of the sequence D. The presence of active chromatin marks over the sequence 55. Imagine a human protein containing 33 repeats of a simple domain arranged in tandem. In contrast, a homolog found in bacteria contains only one domain. What is the minimum number of duplication events that can account for the evolution of this protein since our divergence from bacteria? Write down the number as your answer, e.g. 200. 56. Indicate true (T) and false (F) statements below regarding human genetic variations. Your answer would be a four-letter string composed of letters T and F only, e.g. TFTF. ( ) The genomes of two randomly chosen humans are expected to be identical with respect to at least 99.99% of the nucleotides. ( ) Copy number variations can contain genes. ( ) If the frequency of a point mutation in a population is only 0.1%, with no mutation at this site in the rest of the population, then the variation is NOT considered to constitute a single-nucleotide polymorphism. ( ) Most of the common genetic variants in the current human population could have been present in a human ancestral population of only about 10,000 individuals.

57. Assume two isolated human communities with 500 individuals in each. If the same neutral mutation happens at the same time in two individuals, one from each community, what is the probability that it will be eventually fixed in both of the populations? How would the result change if the two communities fully interbreed? Write down the numbers in scientific notation and separate the two answers with a comma, e.g. 10–5, 3 × 10–2.

Answers 1. Answer: TTFF Difficulty: 2 Section: The Structure and Function of DNA Feedback: The sequence of the other chain is 5′-TAGAATGGG-3′, which makes it slightly heavier because it is mainly composed of the bulkier purine bases. 2. Answer: 43215 Difficulty: 1 Section: The Structure and Function of DNA Feedback: In the double-stranded DNA, the sugar-phosphate backbones form two covalently continuous chains, while the nitrogenous bases from one chain are hydrogenbonded to those of the other chain according to the Watson–Crick model. 3. Answer: TCTAAAG Difficulty: 2 Section: The Structure and Function of DNA Feedback: The final sequence should be 5′-CTTTAGATCTAAAG-3′. The complementary strand would then have the exact same sequence. This is an example of a palindromic sequence. 4. Answer: 7 pg Difficulty: 3 Section: The Structure and Function of DNA Feedback: There are about 6.4 × 109 nucleotide pairs (np) in a diploid nucleus. The average mass of 660 daltons is equivalent to 660 grams per mole (6 × 1023) of the nucleotide pairs. Thus, the total mass of nuclear DNA is: (6.4 × 109 np) × (660 g/mole) / (6 × 1023 np/mole) = ~7 × 10–12 g = ~7 pg 5. Answer: C Difficulty: 1 Section: The Structure and Function of DNA Feedback: In principle, replication would have been conceptually as simple without any of the other features. 6. Answer: C Difficulty: 1 Section: Chromosomal DNA and its Packaging in the Chromatin Fiber Feedback: Biological complexity correlates better with the number of genes than it does with genome size, number of chromosomes, or number of genes per chromosome. 7. Answer: FTTTFF Difficulty: 2 Section: Chromosomal DNA and its Packaging in the Chromatin Fiber Feedback: Only about 5% of our genome is highly conserved. Our relatively long genes (including their exons, introns, and some regulatory sequences) cover almost a quarter of

the genome, but only about 1.5% of our genome is composed of exonic sequences. These exons constitute roughly 6% of our genes. In contrast, a whopping 50% of our genome is made of various repeated sequences, most notably the transposable DNA elements. Please refer to Table 4–1 for the data. 8. Answer: B Difficulty: 3 Section: Chromosomal DNA and its Packaging in the Chromatin Fiber Feedback: Multiplying the size of the chromosome by the distance between consecutive bases gives the end-to-end distance as follows: (200 × 106 nucleotide pairs [np]) × (0.34 × 10–9 m/np) = ~0.068 m = ~7 cm. 9. Answer: centromere, origin of replication, telomere Difficulty: 2 Section: Chromosomal DNA and its Packaging in the Chromatin Fiber Feedback: These three elements are required in a functional chromosome. 10. Answer: TTOTCOC Difficulty: 1 Section: Chromosomal DNA and its Packaging in the Chromatin Fiber Feedback: There is normally one centromere per chromosomal DNA molecule, and two per mitotic chromosome. The mitotic kinetochore structure forms at the centromere. In contrast, the telomeres are at the two ends of each linear chromosome. There are usually many replication origins per eukaryotic chromosome. Replication origins and centromeres are both generally much longer in higher eukaryotes compared to yeast. 11. Answer: A Difficulty: 1 Section: Chromosomal DNA and its Packaging in the Chromatin Fiber Feedback: The various chromatin proteins, including histone and non-histone proteins, wrap and fold the DNA to achieve an astonishing compaction ratio. The nuclear envelope and the nuclear matrix are dispensable for this effect, as evident by the high compaction seen in mitotic chromosomes when the nucleus is disassembled. 12. Answer: C Difficulty: 1 Section: Chromosomal DNA and its Packaging in the Chromatin Fiber Feedback: The homologs are derived from both parents, normally have the same set of loci, and show the same banding pattern. However, the sequences are not expected to be identical. 13. Answer: E Difficulty: 1 Section: Chromosomal DNA and its Packaging in the Chromatin Fiber Feedback: The yeast has a concise genome with a much higher ratio of coding to noncoding DNA.

14. Answer: FTTTFF Difficulty: 1 Section: Chromosomal DNA and its Packaging in the Chromatin Fiber Feedback: The highly basic histone proteins are made of the histone fold that consists of three α helices connected by two loops. Additionally, each core histone has an N-terminal tail which, along with the rest of the protein, can be modified post-translationally. Each nucleosome core particle contains eight histone proteins, two copies of each type. The less well conserved histone H1 is not part of the nucleosome core, but is present in the 30-nm fibers. 15. Answer: 2431 Difficulty: 1 Section: Chromosomal DNA and its Packaging in the Chromatin Fiber Feedback: The nucleosome core particle is composed of a histone octamer wrapped by 147 nucleotide pairs of DNA, and is connected to its neighbors via a linker DNA of variable length. Non-histone proteins are also abundant in the chromatin. 16. Answer: B Difficulty: 1 Section: Chromosomal DNA and its Packaging in the Chromatin Fiber Feedback: The H3–H4 and the H2A–H2B dimers appear to be stable intermediates in histone assembly and exchange. The H3–H4 tetramers are also stable, and are thought to be assembled (and inherited) mostly as a single unit. 17. Answer: E Difficulty: 1 Section: Chromosomal DNA and its Packaging in the Chromatin Fiber Feedback: The ATP-dependent chromatin remodeling complexes can move the nucleosomes on DNA, interact with histone chaperones, and exchange histones or remove them from DNA. 18. Answer: D Difficulty: 2 Section: Chromatin Structure and Function Feedback: Heterochromatin covers a significant fraction of the mammalian genome including the centromere and telomere neighborhoods, but contains very few, mostly inactive, genes. Even though heterochromatin has a very high degree of compaction, its genes can still become active under appropriate conditions through the remodeling of the chromatin into more open arrangements. There are many genes that are not packaged in heterochromatin but are nevertheless transcriptionally inactive. 19. Answer: C Difficulty: 3 Section: Chromatin Structure and Function

Feedback: An enhancer of variegation would facilitate the spread of heterochromatin into the nearby White gene, and consequently decrease red pigmentation. For example, the production of a hyperactive methyl transferase enzyme would result in elevated levels of H3 lysine 9 methylation, leading to enhanced heterochromatin expansion. The other described mutations bring about suppression of variegation. 20. Answer: C Difficulty: 2 Section: Chromatin Structure and Function Feedback: Histone acetylation on lysine residues is generally an activating mark favoring more open chromatin structures (in contrast to what HP1 does), but this effect is mediated mostly through the recruited non-histone proteins that recognize this mark, not the chemical modification itself. The acetylation eliminates the positive charge on lysine, which helps—but is not sufficient alone—to loosen the nucleosome packaging. All known natural covalent modifications on the histone tails are reversible. 21. Answer: E Difficulty: 2 Section: Chromatin Structure and Function Feedback: Chromatin structure is highly dynamic, undergoing tightly regulated conformational rearrangements. In the formation of the 30-nm fiber, in addition to the interactions of histone tails (such as the H4 tail) with neighboring nucleosomes and the 1to-1 deposition of the linker histone H1, non-histone proteins that bind to the nucleosomes also play important roles. Among these are the chromatin remodeling complexes that can slide the nucleosomes in order to attain optimal positioning on the DNA. 22. Answer: AAAS Difficulty: 1 Section: Chromatin Structure and Function Feedback: Acetylation of lysine 9, phosphorylation of serine 10, and trimethylation of lysine 4 in histone H3 are associated with active genes. Trimethylated lysine 9 in this histone, however, is a mark associated with silenced genes. 23. Answer: BACC Difficulty: 1 Section: Chromatin Structure and Function Feedback: Histone H3 lysine 9 acetylation removes a positive charge from the histone and is associated with actively transcribed genes and open chromatin conformations; similar effects arise from H3 serine 10 phosphorylation, which adds a negative charge to the histone. Trimethylation of H3 lysine 4 is also associated with active genes. It keeps the positive charge on the lysine, as does the H3 lysine 9 trimethylation that is associated with silenced genes and heterochromatin formation. 24. Answer: A

Difficulty: 3 Section: Chromatin Structure and Function Feedback: Remodeling complexes can catalyze the exchange of histone subunits (including the histone variants) in nucleosomes with the help of histone chaperones. In this experiment, addition of the SWR1 complex and ATP in the presence of excess H2AZ–H2B dimers results in the appearance of H2AZ-containing nucleosomes in the arrays, i.e. a band appears for the bound fraction. This indicates that the SWR1 complex transfers the H2AZ histones to the immobilized nucleosomes. There is no reason to believe that the antibody binds to the SWR1 complex. 25. Answer: MVMVVM Difficulty: 2 Section: Chromatin Structure and Function Feedback: Compared to the histone variants, the major histones are more highly conserved, are much more abundant, and become available in a burst of synthesis at S phase in order to associate with the newly replicated, histone-deficient, DNA molecules. 26. Answer: C Difficulty: 2 Section: Chromatin Structure and Function Feedback: There are various chromatin reader complexes in the nucleus, each recognizing a limited set of histone mark combinations, not merely a single specific histone mark. A reader complex is highly specific thanks to its modular design: it binds tightly only if the several histone marks that it recognizes are present. A reader complex may bear a number of recognition modules all linked on one single protein, and does not have to be associated with a writer complex. 27. Answer: DCEBA Difficulty: 2 Section: Chromatin Structure and Function Feedback: In this example, the tRNA genes serve as barrier sequences to block the spread of heterochromatin by recruiting histone-modifying enzymes such as histone acetyl transferase enzymes. This is similar to the role of the HS4 sequence in protecting the βglobin locus in our red blood cells. Heterochromatin can spread over the centromeric regions in the absence of the tRNA genes in the fission yeast, suggesting that the presence of CENP-A-containing histones does not provide a sufficient protection. 28. Answer: B Difficulty: 2 Section: Chromatin Structure and Function Feedback: Human alpha satellite DNA has been observed to be organized in an alternating pattern, implying a folding scheme that positions the CENP-A-containing nucleosomes in proximity to the proteins of the outer kinetochore. Even though it enhances the seeding step in de novo centromere formation, the alpha satellite DNA is

neither necessary nor sufficient for centromere formation: centromeres can form on a DNA molecule that lacks these repeats altogether; furthermore, the same repeat sequences are also found at non-centromeric chromosomal regions. 29. Answer: B Difficulty: 2 Section: Chromatin Structure and Function Feedback: The NuRD chromatin remodeling complex is thought to be normally involved in maintaining the repressive epigenetic state of several developmentally important genes in the somatic cells, thus resisting the pluripotency induction and preventing efficient reprogramming of these cells into stem cells. Inhibition of this complex relieves this repression, allowing a more efficient reprogramming. 30. Answer: C Difficulty: 2 Section: Chromatin Structure and Function Feedback: All the other scenarios can happen with some degree of likelihood. In contrast, the reasoning in (C) is flawed. The fact that a gene encodes a histone acetyl transferase does not mean that it can automatically resist heterochromatin expansion. That would require a mechanism for targeting of the protein to its own genetic locus. 31. Answer: E Difficulty: 2 Section: The Global Structure of Chromosomes Feedback: The gigantic lampbrush chromosomes are meiotically paired chromosomes first found in amphibian eggs and are readily seen by light microscopy. Even though they have not been observed in mammals, studying their structure has provided important insights into the ways the chromosomes (such as those in a mammalian interphase nucleus) fold into loops of different lengths. 32. Answer: C Difficulty: 1 Section: The Global Structure of Chromosomes Feedback: The typical loop size is between 50,000 and 200,000 nucleotide pairs, although significantly longer loops may also exist. A DNA stretch of 2000 nucleotide pairs would have only about 10 nucleosomes. On the other hand, 10 million nucleotide pairs would be comparable to the size of a small chromosome arm. 33. Answer: TTFF Difficulty: 3 Section: The Global Structure of Chromosomes Feedback: The 3C data show high relative cross-linking between the B gene and the R region in the liver (red) but not the muscle (blue) sample. This suggests that, in the liver cells, these two elements spatially interact by looping out the region containing the A gene, and supports the prediction that the B gene is expressed at higher levels in the liver

compared to muscle. Even though no significant looping interaction is detected between the B gene and the R region in the muscle sample, the gene might still interact with other elements in the locus, as suggested by the elevated cross-linking efficiency with the neighboring C gene in this sample. 34. Answer: C Difficulty: 2 Section: The Global Structure of Chromosomes Feedback: The enormous polytene chromosomes have proven very useful in visualizing chromosome organization; they can be readily studied by light microscopy and show distinct banding patterns corresponding to certain genomic positions, providing clues as to how the chromatin is organized into domains at a large scale. 35. Answer: D Difficulty: 2 Section: The Global Structure of Chromosomes Feedback: The five major chromatin types were found by analyzing localization data for tens of proteins and histone marks. Such data can be obtained through techniques such as chromatin immunoprecipitation. Three heterochromatin (compacted) and two euchromatin (open) chromatin types were identified in this way, including the known classical HP1 heterochromatin and the developmentally important Polycomb heterochromatin. Note that the chromatin types are not fixed during development and across different cell types, and that in addition to the five major types, other minor chromatin forms appear to be present. 36. Answer: 212221 Difficulty: 3 Section: The Global Structure of Chromosomes Feedback: More open chromatin is usually associated with higher transcriptional activity. Thus, an actively transcribed gene is likely to be found in euchromatin (which is generally not associated with the nuclear lamina), and more so in the extended 11-nm fibers. In a polytene chromosome, “puffs” represent the sites where active transcription occurs. An active gene can even leave the territory of its parent chromosome and extend out into “factories” of transcription along with other active genes. 37. Answer: D Difficulty: 2 Section: The Global Structure of Chromosomes Feedback: High transcriptional activity is accompanied by the recruitment of a large number of proteins including the RNA polymerase transcription machinery. This may coincide with the migration of the gene from its previous environment in the chromosome territory out toward the foci of active transcription. The chromatin profile of the gene is also expected to change toward a more open state. 38. Answer: fractal globules

Difficulty: 1 Section: The Global Structure of Chromosomes Feedback: Most regions of our chromosomes are folded into a conformation referred to as a fractal globule. This knot-free arrangement facilitates maximally dense packing, but also preserves the ability of the chromatin fiber to unfold and fold. 39. Answer: C Difficulty: 2 Section: The Global Structure of Chromosomes Feedback: The fractal globule model is permissive to maximal packing and, at the same time, allows the chromatin to fold and unfold easily. In this model, compared to a “random coil” arrangement, the spatial distance between two neighboring regions of the DNA is shorter on average. 40. Answer: C Difficulty: 1 Section: The Global Structure of Chromosomes Feedback: Nuclear subcompartments such as nucleoli, interchromatin granule clusters (or speckles), and Cajal bodies carry out specialized functions using distinct sets of macromolecules organized in gel-like structures. These structures are highly permeable and highly selective at the same time. They are formed only when needed and are not constitutively present. 41. Answer: A Difficulty: 1 Section: The Global Structure of Chromosomes Feedback: Each nuclear subcompartment is a non-membrane-bound network of a specific set of RNA and protein molecules linked via noncovalent interactions to produce a distinct biochemical environment, despite their high permeability. They come in various sizes. 42. Answer: FTFFT Difficulty: 2 Section: The Global Structure of Chromosomes Feedback: The cells achieve an extra 10-fold compaction ratio of their DNA by condensing the chromatin into mitotic chromosomes that can be seen using a light microscope. This is done after the DNA replication in S phase, and ends in the complete disentanglement of the sister chromatids. A mammalian mitotic chromatid is normally between 0.5 and 1 µm in diameter. A 70-nm-thick chromosome would be only about twice as wide as a 30-nm fiber. 43. Answer: A Difficulty: 2 Section: How Genomes Evolve

Feedback: Transposition by various “parasitic” mobile DNA elements has changed the mammalian genome so profoundly that almost half of our genome is composed of recognizable products of transposition. In addition, chromosomal duplications can be facilitated by transposable elements. 44. Answer: purifying, third Difficulty: 1 Section: How Genomes Evolve Feedback: Human and chimpanzee genomes are nearly identical, even in the third position of synonymous codons, where there is no functional constraint on the nucleotide sequence. This reflects the short time available for mutations to accumulate in the two closely related lineages. In contrast, the sequence conservation found between the genes of humans and chickens is almost entirely due to purifying selection. 45. Answer: BDC Difficulty: 2 Section: How Genomes Evolve Feedback: The lowest pairwise distance in the matrix (0.40) is between species A and B. Species A and B are each separated from their common ancestor by a distance of 0.20 (= 0.40 / 2). Next, the average distance of species A and B from species D is only 1.40 (= [1.36 + 1.44] / 2), which is lower than the average distance from species C—that is, 3.7 (= [3.61 + 3.79] / 2)—and so on. Based on these distances and the known phylogeny of the hominid family, one would expect species B to be chimpanzee, species C to be orangutan, and species D to be gorilla. 46. Answer: BDC Difficulty: 2 Section: How Genomes Evolve Feedback: Based on the distances and the known phylogeny of the hominid family, one would expect species B (1) to be chimpanzee, species C (3) to be orangutan, and species D (2) to be gorilla. 47. Answer: bcfedag Difficulty: 4 Section: How Genomes Evolve Feedback: A diagonal line in this plot indicates a continuous region of synteny between the two chromosomes. Because of a genomic region that is absent (e.g. deleted) from one end of the chromosome of species A, the long diagonal line is shifted up (represented by a in the plot). A symmetrical set of parallel lines (as at b) on the two sides of a diagonal reflects the existence of a repeat in both chromosomes, whereas the parallel line on one side of the diagonal at c indicates duplication only in species A. A chromosomal inversion shows up as a diagonal with a negative slope, which may or may not be accompanied by translocation of the inverted region (as at d and e, respectively). Two tandem duplications in species B may appear as a triplication (as at f) represented by

three parallel lines. Finally, g represents a region at the end of the chromosome in species A that is absent from the chromosome of species B. 48. Answer: 2122 Difficulty: 3 Section: How Genomes Evolve Feedback: The molecular clock runs faster for sequences that are subject to less purifying selection. 49. Answer: E Difficulty: 2 Section: How Genomes Evolve Feedback: The balance between the rates of DNA addition and DNA loss has been biased toward a net DNA loss in the Fugu lineage, resulting in a “cleansing” of functionally unnecessary sequences including most transposable elements and large parts of the introns. The exons, which are under purifying selection, were not affected as significantly. 50. Answer: B Difficulty: 3 Section: How Genomes Evolve Feedback: Since the correlation of gene copy number and expression level has been established for this gene, it is reasonable to expect the histogram for the population with higher dietary starch to be shifted toward higher gene copy numbers. 51. Answer: FFFF Difficulty: 2 Section: How Genomes Evolve Feedback: By comparing genomic DNA sequences from various existing mammals, as well as non-mammals, much can be inferred about the genome of the last common ancestor that lived about 100 million years ago. Although some of the ultraconserved elements may encode long noncoding RNAs, the function of the majority of these elements is still a mystery. Since even a tiny selective advantage can support the conservation of a particular DNA sequence, lack of an “obvious” defect in a laboratory strain of mouse does not preclude functional importance. The “human accelerated regions” show accelerated genetic changes during recent evolution in the human lineage, but are not entirely novel genetic elements; in fact, they had been highly conserved before the sudden spurt of nucleotide changes. 52. Answer: C Difficulty: 2 Section: How Genomes Evolve Feedback: It appears that a chromosome translocation event happened about 300 million years ago in the lineage leading to mammals and birds after it diverged from the amphibian lineage.

53. Answer: C Difficulty: 3 Section: How Genomes Evolve Feedback: Genes with very high h and k values have shown an accelerated rate of change in both human and chimpanzee lineages compared to the mouse lineage. Even though such genes are interesting candidates to study general hominid evolution, the genes that have evolved uniquely in humans are probably those with high h but low k values. Note that in the normalized substitution rates, a and b are divided by 5.5 because each of them represents amino acid changes during 5.5 million years since the divergence of human and chimpanzee lineages. The parameter c represents the changes that occurred in the mouse lineage since its divergence from the primate lineage (90 million years ago), as well as changes that occurred in the primate lineage before human and chimpanzee lineages diverged (between 90 million and 5.5 million years ago). Therefore, c is normalized by dividing by the combined period (90 + 90 – 5.5). 54. Answer: B Difficulty: 1 Section: How Genomes Evolve Feedback: Multispecies conservation of a genomic sequence is a strong indication of its functional importance. 55. Answer: 6 Difficulty: 3 Section: How Genomes Evolve Feedback: Duplication of DNA segments has been a widespread evolutionary event that can happen over long or short blocks of the genome. It can exponentially expand the block to create tandem repeated arrays. For example, in an ideal case, only five duplications are needed to create 32 (that is, 25) repeats of an original DNA segment that encodes a protein domain. One more is needed to make the total 33 repeats. In reality, however, more than six duplication events might have been involved. 56. Answer: FTTT Difficulty: 2 Section: How Genomes Evolve Feedback: A number of variations—such as scattered rare mutations and abnormalities, single-nucleotide polymorphisms, simple tandem repeat expansions [such as those of the (CA)n repeats], and copy number variations—makes each two human individuals of the same sex different in, at least, about 1 in every 1000 nucleotides of their genomes; therefore, 99.99% identity is an overestimation. Nearly half of the copy number variations contain genes. For a variation at a nucleotide position to be considered a single-nucleotide polymorphism, it must be common enough such that the probability that two individuals have a different nucleotide at this position is at least 1%. It is

estimated that the current pattern of human genetic variation was mostly in place when our ancestral population size was about 10,000. 57. Answer: 10–6, 10–3 Difficulty: 3 Section: How Genomes Evolve Feedback: The probability that the mutation becomes fixed in each isolated population of 500 individuals is approximately 1/(2 × 500) = 10–3. The combined probability for fixation in two independent populations is thus equal to 10–3 × 10–3 = 10–6. If the two populations fully interbreed, they can be considered as a single population of 1000 people with two mutations. The fixation probability is therefore calculated to be 2/(2 × 1000) = 10–3. Hence, it is more likely for the same rare neutral variations to become fixed in the entire population in these communities if they highly interbreed.

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION CHAPTER 5: DNA REPLICATION, REPAIR, AND RECOMBINATION Copyright © 2015 by W.W. Norton & Company, Inc. 1. Which of the following is correct regarding the mutation rate of genomic DNA in different organisms? A. Human cells have a much higher mutation rate compared to bacteria when the rate is normalized to a single round of replication over the same length of DNA. B. Mutation rates limit the number of essential genes in an organism’s genome. C. Mutations in the somatic cells cannot be lethal. D. Even if the mutation rate was 10 times higher than its current value, germ-cell stability in humans would not have been affected. E. All of the above. 2. The mutation rate in bacteria is about 3 nucleotide changes per 10 billion nucleotides per cell generation. Under laboratory conditions, bacteria such as Escherichia coli can divide and double in number about every 40 minutes. If a single Escherichia coli cell is allowed to exponentially divide for 10 hours in this manner, how many mutations would you expect to observe on average in the genome (4.5 million nucleotide pairs) of each of the resulting bacteria compared to the original cell? Assume all mutations are neutral; that is, they do not affect the cell-division time. A. Less than 0.001 B. About 0.02 C. One or two D. About 10 E. About 100 3. On average, errors occur in DNA synthesis only once in every ten billion nucleotides incorporated. Which of the following does NOT contribute to this high fidelity of DNA synthesis? A. Complementary base-pairing between the nucleotides B. “Tightening” of the DNA polymerase enzyme around its active site to ensure correct pairing before monomer incorporation

C. Exonucleolytic proofreading by the 3′-to-5′ exonuclease activity of the enzyme to correct mispairing even after monomer incorporation D. A strand-directed mismatch repair system that detects and resolves mismatches soon after DNA replication E. All of the above mechanisms DO contribute to the fidelity. 4.

The nuclear DNA polymerases in human cells … A. polymerize about 1000 nucleotides per second during DNA replication in vivo. B. are incapable of 3′-to-5′ exonuclease activity. C. are capable of 3′-to-5′ DNA polymerase activity. D. have a single active site that is used for both polymerization and editing. E. are unable to initiate polymerization de novo (i.e. in the absence of a primer).

5. What is the main source of the free energy for the mechanical work performed by DNA helicases during DNA replication in our cells? A. The hydrogen-bonding energy in the DNA double helix B. Thermal energy in the nucleus C. ATP hydrolysis by the helicase D. The energy of SSB binding to single-stranded DNA E. ATP hydrolysis by DNA topoisomerases 6. During DNA replication in the cell, DNA primase makes short primers that are then extended by the replicative DNA polymerases. These primers … A. are made up of DNA. B. generally have a higher number of mutations compared to their neighboring DNA. C. are made more frequently in the leading strand than the lagging strand. D. are joined to the neighboring DNA by DNA ligase. E. provide a 3′-phosphate group for the DNA polymerases to extend. 7. DNA ligases are used in both DNA replication and repair to seal breaks in the DNA. But DNA damage can result in single- or double-strand breaks that are not normal ligase substrates. These need to be processed first before a ligase can act on them. One of the enzymes that is recruited to some of such breaks is called PNK. It has two separate activities on the DNA, both of which can help provide a canonical ligase substrate. Which of the following activities would you expect PNK to have in this context? A. 5′ kinase (phosphorylation of a free 5′-OH group) and 3′ kinase

B. 5′ phosphatase (dephosphorylation to create a free 5′-OH group) and 3′ phosphatase C. 3′ kinase and 3′ phosphatase D. 5′ phosphatase and 3′ kinase E. 5′ kinase and 3′ phosphatase 8. Fill in the gap in the following paragraph using what you know about the activities of the proteins involved in DNA replication. “Mitochondrial DNA replication requires a set of proteins similar to those used for the replication of the nuclear genome. However, mitochondria lack a dedicated DNA ... and use the mitochondrial RNA polymerase instead.” 9.

During DNA replication, the single-strand DNA-binding (SSB) proteins … A. are generally found more on the leading strand than the lagging strand. B. bind cooperatively to single-stranded DNA and cover the bases to prevent basepairing. C. prevent the folding of the single-stranded DNA. D. bind cooperatively to short hairpin helices that readily form in the single-stranded DNA. E. All of the above.

10. This protein is present at every replication fork and prevents DNA polymerase from dissociating, but does not impede the rapid movement of the enzyme. Which of the following is true regarding this protein? A. It self-assembles onto DNA at the replication fork. B. It is assembled on DNA as soon as DNA polymerase runs into a double-strand region of DNA. C. Its assembly normally follows the synthesis of a new primer by the DNA primase. D. It disassembles from DNA as soon as DNA polymerase runs into a double-strand region. E. All of the above. 11. At the replication fork, the template for the lagging strand is thought to loop around. This looping would allow the lagging-strand polymerase to move along with the rest of the replication fork instead of in the opposite direction. The single-strand part of the loop is bound by the singlestrand DNA-binding (SSB) proteins. As each Okazaki fragment is synthesized toward

completion, how does the size of the loop change? What about the size of the SSB-bound part of the loop? A. Increases; increases. B. Increases; decreases. C. Decreases; increases. D. Decreases; decreases. E. Decreases; does not change. 12. The Dam methylase is responsible for methylating the adenine base in GATC sequences in Escherichia coli. Imagine two E. coli strains, one without any active Dam methylase, and the other with a hyperactive version of the enzyme that operates faster than the wild-type enzyme. Which of these strains would you expect to show a “mutator” phenotype? A. Both of the strains B. Neither of them C. Only the first strain D. Only the second strain 13. In the following schematic drawing, two DNA molecules are shown before and after the action of a protein that is also involved in the process of DNA replication. What is this protein called?

A. DNA ligase B. DNA helicase C. DNA polymerase I D. DNA topoisomerase I E. DNA topoisomerase II 14.

What do the enzymes topoisomerase I and topoisomerase II have in common?

A. They both have nuclease activity. B. They both create double-strand DNA breaks. C. They both require ATP hydrolysis for their function. D. They both can create winding (tension) in an initially relaxed DNA molecule. E. All of the above. 15. In Escherichia coli, replication of DNA can occur throughout the cell cycle while the cell is also actively transcribing its genes. This means collisions between replication forks and RNA polymerases are inevitable. Depending on the orientation of the genes, collisions can be rear-end (when both machines are traveling in the same direction) or head-on (when they are traveling in opposite directions). In the following paragraph, match each of the letters (A to D) to one appropriate number below. Do not use a number more than once. Your answer would be a fourdigit number composed of digits 1 to 5 only, e.g. 1253. “Typically, in a rear-end collision, the (A) of RNA polymerase collides with the (B) in the replication fork. In contrast, in a headon collision, the (C) of RNA polymerase hits the (D) in the fork.” 1. front edge (of RNA polymerase) 2. rear edge (of RNA polymerase) 3. DNA helicase 4. leading-strand DNA polymerase 5. lagging-strand DNA polymerase 16. Which of the following features is common between the replication origins in Escherichia coli and Saccharomyces cerevisiae? A. They both normally exist in one copy per genome. B. Both are specified by DNA sequences of tens of thousands of nucleotide pairs. C. Both contain sequences that attract initiator proteins, as well as stretches of DNA rich in A-T base pairs. D. Both contain GATC repeats that are methylated to prevent the inappropriate “firing” of the origin. E. All of the above. 17. You have found a strain of Escherichia coli that has an unusually short doubling time of only 15 minutes, despite the fact that its complete DNA replication should take almost 35 minutes. You also find that there is only one replication origin on its chromosome from which two forks originate, just like the normal process described for E. coli. However, you discover

that the origin of replication in this strain has a significantly shorter “refractory period,” resulting in the reactivation of the origin before the previous round of replication is over. Based on this model, if you examine the chromosomes of this strain (under conditions of fast growth), how many replication forks would you expect to observe per chromosome on average? A. Two, just like the wild-type strain B. Four C. Six D. Eight E. Ten 18.

The telomerase enzyme in human cells … A. has an RNA component. B. extends the telomeres by its RNA polymerase activity. C. polymerizes the telomeric DNA sequences without using any template. D. removes telomeric DNA from the ends of the chromosomes. E. creates the “end-replication” problem.

19. If this protein complex does not function normally, the ends of the eukaryotic chromosomes would activate the cell’s DNA damage response, causing chromosomal fusions and other genomic anomalies. What is this protein complex called? A. Telomerase B. T-loop C. ORC D. Shelterin E. RecA 20. Which of the following schematic drawings better depicts the end of mammalian chromosomal DNA? 5′ 3′ A.

5′ 3′

3′ 5′

5′ 3′

3′ 5′

21. Which of the following spontaneous lesions in DNA occurs most frequently in a mammalian cell? A. Depurination B. Cytosine deamination C. Guanine oxidation D. Guanine alkylation E. Depyrimidination 22. DNA glycosylases constitute an enzyme family found in all three domains of life. They can … A. add sugar moieties to DNA. B. remove sugar moieties from DNA. C. add a purine or pyrimidine base to DNA. D. remove a purine or pyrimidine base from DNA. E. remove a nucleotide from DNA. 23. Upon heavy damage to the cell’s DNA, the normal replicative DNA polymerases may stall when encountering damaged DNA, triggering the use of backup translesion polymerases. These backup polymerases … A. lack 3′-to-5′ exonucleolytic proofreading activity. B. are replaced by the replicative polymerases after adding only a few nucleotides. C. can create mutations even on undamaged DNA.

D. may recognize specific DNA damage and add the appropriate nucleotide to restore the original sequence. E. All of the above. 24.

What are the products of deamination of cytosine and 5-methyl cytosine, respectively? A. Thymine and uracil B. Thymine in both cases C. Uracil and thymine D. Uracil in both cases E. Xanthine and hypoxanthine

25.

Which of the following repair pathways can accurately repair a double-strand break? A. Base excision repair B. Nucleotide excision repair C. Direct chemical reversal D. Homologous recombination E. Nonhomologous end joining

26.

This protein folds into a doughnut shape that can encircle DNA. It can load on the DNA

only when the DNA is broken in both strands, so that the DNA can thread through the hole in the protein. Which of the following proteins do you think matches this description? A. PCNA, the sliding clamp for DNA polymerases at the replication forks B. Ku, the protein that recognizes DNA ends and can initiate nonhomologous end joining C. MCM, the helicase critical for the initiation and elongation of replication D. Topoisomerase II, which can create or relax superhelical tension in DNA E. RecA/Rad51, which carries out strand invasion in homologous recombination 27. In contrast to vertebrates, there is very little DNA methylation in the genomes of invertebrates such as Drosophila melanogaster and Caenorhabditis elegans. Indicate whether you expect each of the following statements to be true (T) or false (F) regarding 5′-CG-3′ dinucleotide sequences in the genome. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTT. ( ) On average, approximately one dinucleotide out of every 16 in the human genome is a CG dinucleotide.

( ) On average, approximately one dinucleotide out of every 256 in the D. melanogaster genome is a CG dinucleotide. ( ) The proportion of CG dinucleotides in the human genome is more than that of C. elegans. ( ) The proportion of CG dinucleotides in the human genome is more than that expected by chance. 28. In which phases of the eukaryotic cell cycle does homologous recombination often occur to repair DNA damage? A. G1 and S phases B. S and G2 phases C. G2 and M phases D. M and G1 phases E. G1 and G2 phases 29. Which of the following is NOT correct regarding homologous recombination and its regulation? A. Loss of heterozygosity can occur if a broken chromosome is repaired using a sister chromatid instead of its homologous chromosome. B. Repair of double-strand breaks by homologous recombination is favored during or soon after DNA replication. C. Homologous recombination can rescue broken or stalled replication forks in S phase. D. Excessive use of homologous recombination by human cells can lead to cancer. E. Low usage of homologous recombination by human cells can lead to cancer. 30. In the following schematic drawing of a Holliday junction that undergoes branch migration, cutting at which combination of the sites a to d would generate a crossover?

a Branch migration

d c

A. a and b B. a and c C. a and d D. b and c E. b and d 31. In meiosis, a crossover in one position is thought to inhibit crossing-over in the neighboring regions. This regulatory mechanism … A. results in a very uneven distribution of crossover points along each chromosome. B. ensures that even small chromosomes undergo at least one crossover. C. controls how the Holliday junctions are resolved. D. All of the above. 32. What group of mobile genetic elements is largely responsible for the resistance of the modern strains of pathogenic bacteria to common antibiotics? A. DNA-only transposons B. Retroviral-like retrotransposons C. Nonretroviral retrotransposons D. Site-specific recombinases 33.

DNA-only transposons … A. can be recognized by the presence of short inverted repeats at each end. B. often encode a transposase that mediates the transposition process. C. leave double-strand breaks in the donor chromosome. D. can move by a cut-and-paste mechanism. E. All of the above.

34.

Which of the following is true regarding retroviral-like retrotransposons? A. They encode both a reverse transcriptase and an RNA polymerase. B. They have directly repeated long terminal repeats at their two ends when integrated into chromosomal DNA. C. Their genomic RNA can be translated to produce viral coat proteins. D. They leave double-strand breaks in the original donor DNA. E. The Alu element in our genome is an example of retroviral-like retrotransposons.

35. Which of the following is NOT correct regarding long and short interspersed nuclear elements? A. Each of them encodes a reverse transcriptase. B. They rely on the cellular transcription machinery to produce their RNA transcript. C. They use one of the strands in the target DNA as a primer to synthesize their DNA. D. Together, they make up about 40% of our genome. E. They can move into new regions of genome that do not have any homology with their DNA. 36. Phase variation helps protect the bacterium Salmonella typhimurium against the immune system of its host by switching the orientation of a certain promoter. This process … A. is carried out through a DNA transposition mechanism. B. is irreversible. C. can often result in the excision of the promoter from the chromosome altogether. D. is mediated by enzymes that form transient covalent bonds with the DNA. 37. A replication fork is shown schematically below. The strand labeled A is called the … strand.

A 3′ 5′

38. The sliding clamp and the DNA helicase that function at the replication fork both have three-dimensional structures resembling a ring with a central hole through which DNA is threaded. Which of these proteins, the clamp (C) or the helicase (H), do you think has a wider hole in its structure? Write down C or H as your answer. 39. Indicate true (T) and false (F) statements below regarding the initiation of replication in human cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) Tens of thousands of replication origins are used each time a cell in our body replicates its DNA. ( ) Different cells in our body use different sets of replication origins. ( ) Both replication forks in a replication bubble are normally active in replication. ( ) Gene expression and chromatin structure can affect the choice of the origins to use as well as the order in which they are activated. 40. What combination of the following events normally prevents the origins of replication in the yeast Saccharomyces cerevisiae from “firing” more than once during the cell cycle? Your answer is a two-letter string composed of letters A to E only, e.g. AB. Order the letters in your answer alphabetically. (A) The helicase-loading proteins Cdc6 and Cdt1 are only active in S phase. (B) The helicase Mcm1 can be delivered to the origin recognition complex (ORC) only in S phase. (C) The ORC can only become active (by dephosphorylation) in G1 phase. (D) The helicase can only become active (by phosphorylation) in S phase. (E) The ORC can bind to the origin only in S phase. 41. Examples of two general types of DNA damage are shown in the following drawing. Which type of damage (1 or 2) is more common in our cells?

42. In cells that are exposed to sunlight, ultraviolet (UV) light can result in covalent linkage between two adjacent DNA bases. If not repaired in time, such dimers can lead to inheritable mutations. Consider the sequence 5′-GGTATGATCATTATAA-3′ in the chromosome of a cell that is exposed to intense sunlight. How many possible dimers can form in the region of DNA double helix corresponding to this sequence? 43. Indicate whether each of the following DNA lesions is typically repaired via the base excision (B) or nucleotide excision (N) repair pathway? Your answer would be a four-letter string composed of letters B and N only, e.g. BBBB. 1. Deaminated cytosines 2. Depurinated residues 3. Thymine dimers 4. Bulky guanine adducts 44. Indicate true (T) and false (F) descriptions below regarding the nucleotide excision repair pathway. Your answer would be a four-letter string composed of letters T and F only, e.g. FFFF. ( ) It involves recognition of distortions in the DNA double helix rather than specific base changes. ( ) It involves endonucleolytic cleavage and helicase-mediated strand removal.

( ) It involves cleavage by the AP endonuclease. ( ) It is coupled to the DNA transcription machinery of the cell. 45.

Fill in the blank in the following paragraph. Do NOT use abbreviations. “In human cells, the predominant pathway to repair double-strand breaks is …, in which the broken ends are simply rejoined with the concomitant loss of a few nucleotides. This leaves scars at the breakage sites. This pathway can potentially create chromosome translocations.”

46.

Sort the following steps in the order that they normally happen during the process of

repairing double-strand breaks by homologous recombination. Your answer would be a six-letter string composed of letters A to F only, e.g. DEFABC. (A) Ligation (B) DNA synthesis using undamaged DNA as the template (C) DNA synthesis using original DNA as the template (D) Release of the invading strand (E) Strand invasion (F) Nuclease digestion (resection) 47. The RecA/Rad51 protein carries out strand exchange during homologous recombination. Indicate true (T) and false (F) statements about this process below. Your answer would be a fourletter string composed of letters T and F only, e.g. FFFF. ( ) RecA hydrolyzes ATP upon binding to the invading DNA strand. ( ) RecA forces the invading strand into a conformation that fully mimics the geometry of a long DNA double helix. ( ) Sampling of the homologous duplex by the invading strand is likely to occur in triplet nucleotide blocks. ( ) RecA-bound, invading, single-stranded DNA binds and destabilizes the homologous duplex to allow the sampling of its sequence by base-pairing. 48. Indicate true (T) and false (F) statements below regarding the use of homologous recombination in meiosis. Your answer would be a four-letter string composed of letters T and F only, e.g. FFFF. ( ) Meiotic recombination starts with a double-strand break caused by errors in DNA replication.

( ) Meiotic recombination occurs preferentially between DNA from maternal and paternal chromosome pairs. ( ) Holliday junctions can form during meiotic recombination, sometimes in pairs. ( ) During meiotic recombination in human cells, the majority of the invading strands are released, leading to no crossover. 49. Indicate whether each of the following mobile elements in Drosophila is a DNA-only transposon (D), a retroviral-like retrotransposon (R), or a nonretroviral retrotransposon (N). Your answer would be a four-letter string composed of letters D, R, and N only, e.g. RRRR. ( ) P elements; these have inverted terminal repeats and move with the help of a transposase enzyme. ( ) Copia elements; these have directly repeated long terminal repeats and move with the help of a reverse transcriptase and an integrase. ( ) F elements; these are long interspersed nuclear elements (LINEs). ( ) Mariner elements; these move via a DNA intermediate. 50. Consider three types of mobile genetic elements that are found in our genome: the DNAonly transposons (D), the retroviral-like retrotransposons (R), and the nonretroviral retrotransposons (N). Which type appears to still be active and move in our genome, accounting for a detectable fraction of human mutations? Write down your answer as D, R, or N. 51. As shown in the following drawing, a researcher has engineered three pairs of LoxP sites (for conservative site-specific recombination) in a region that contains three reporter genes coding for red, yellow, or cyan fluorescent proteins, respectively. Each type of LoxP sequence (shown as a black, gray, or white arrowhead) is specific, meaning it does not recombine with the other types of LoxP sequences. Upon Cre recombinase activation, depending on which recombination event occurs first (which we assume is random), a number of possible combinations of reporters can remain in the final DNA. For each of the following combinations, indicate whether it can (C) or cannot (N) result from this recombination scheme. Do not consider the re-integration of excised DNA, which happens very rarely. Your answer would be a six-letter string composed of letters C and N only, e.g. CCCCNN. Red

Yellow

Cyan

( ) Red and yellow ( ) Red only ( ) Yellow only ( ) Cyan only ( ) Yellow and cyan ( ) Red and cyan

Answers: 1. Answer: B Difficulty: 2 Section: The Maintenance of DNA Sequences Feedback: The mutation rate per nucleotide pair per replication cycle in humans is similar to that of Escherichia coli, and both are extremely low, less than one in a billion. Nevertheless, current mutation rates are thought to limit the number of essential genes in an organism to about 30,000. The stability of the germ-line genome depends on mutation rates, as does somatic-cell stability. Cancers typically arise from somatic mutations. 2. Answer: B Difficulty: 3 Section: The Maintenance of DNA Sequences Feedback: About 15 cell generations pass in 10 hours of growth. Thus, the expected average number of mutations is equal to: (15 generations) × (3 × 10–10 mutations/generation/nucleotide) × (4.5 × 106 nucleotides/genome) = ~2 × 10–2 mutations per genome. This is equivalent to about 98% of the bacterial cells having zero mutations and the remaining 2% having only one mutation each. Please note that the number of mutations in different cells in a given culture will vary widely around the average calculated above. 3. Answer: E Difficulty: 2 Section: DNA Replication Mechanisms Feedback: Base-pairing is the basis of all DNA replication and repair. The other three mechanisms each increase accuracy by about 100-fold. 4. Answer: E Difficulty: 1 Section: DNA Replication Mechanisms Feedback: Eukaryotic DNA polymerases are not as fast as their bacterial counterparts, presumably due to the difficulty in replicating through nucleosomes. All known DNA polymerases polymerize DNA from 5′ to 3′ and require a free 3′ end provided by a primer. The self-correcting polymerases have an additional 3′-to-5′ exonuclease activity that takes place at a distinct site in the enzyme. 5. Answer: C Difficulty: 1 Section: DNA Replication Mechanisms

Feedback: The helicase at the replication fork binds and hydrolyses ATP to move on the DNA in a stepwise fashion. 6. Answer: B Difficulty: 1 Section: DNA Replication Mechanisms Feedback: The DNA primase enzyme syntheses RNA primers (with free 3′-OH groups) that are extended and later replaced by DNA that is more accurately synthesized. This happens more frequently in the lagging strand. 7. Answer: E Difficulty: 3 Section: DNA Replication Mechanisms Feedback: The DNA ligase reaction normally requires a free 3′-OH group and a 5′phosphate group on the two DNA ends to proceed. If a break has created 3′-phosphate and 5′-OH ends instead, for example, an enzyme such as PNK can dephosphorylate the 3′ end in the upstream molecule, and phosphorylate the 5′ end in the downstream molecule, to create the canonical ends. 8. Answer: primase Difficulty: 3 Section: DNA Replication Mechanisms Feedback: Like the RNA polymerases involved in transcription, primases do not require priming and can synthesize RNA molecules that can be used as primers for DNA synthesis by the replicative DNA polymerases. RNA polymerase is used to create RNA primers for mitochondrial DNA replication. 9. Answer: C Difficulty: 2 Section: DNA Replication Mechanisms Feedback: The SSB proteins bind cooperatively to the regions of exposed single-stranded DNA (which are routinely found in the lagging strand) and prevent the formation of hairpin helices without blocking the base-pairing potential. 10. Answer: C Difficulty: 2 Section: DNA Replication Mechanisms Feedback: The sliding clamp is normally loaded (more frequently on the lagging strand) on the primer–template duplex by the clamp loader, a process that requires ATP hydrolysis by the loader. As soon as the polymerase runs into a double-strand region downstream, the clamp releases the polymerase but is not itself immediately disassembled from DNA. 11. Answer: A Difficulty: 3 Section: DNA Replication Mechanisms

Feedback: The double-strand region of the loop represents the growing Okazaki fragment and gradually increases in size. Additionally, as each Okazaki fragment is growing, the fork is also traveling, exposing more single-stranded DNA in the loop. The loop size resets to zero upon initiating the synthesis of the next Okazaki fragment. 12. Answer: A Difficulty: 3 Section: DNA Replication Mechanisms Feedback: An inactive or hyperactive Dam methylase can interfere with strand-directed mismatch repair, as both make it harder to distinguish between the newly replicated and old DNA strands. Both situations are expected to increase the overall mutation rate, giving rise to a mutator phenotype. 13. Answer: E Difficulty: 2 Section: DNA Replication Mechanisms Feedback: This untangling of intertwined DNA is mediated by the enzyme topoisomerase II. This type of reaction helps solve the winding problem during replication. 14. Answer: A Difficulty: 3 Section: DNA Replication Mechanisms Feedback: Unlike topoisomerase I, which can only relieve the tension in DNA through introducing a nick (single-strand break) in one of the DNA strands, topoisomerase II can actively introduce or relieve tension by creating double-strand breaks (that remain tightly associated with the enzyme) using energy from ATP hydrolysis. 15. Answer: 2413 Difficulty: 3 Section: DNA Replication Mechanisms Feedback: The replication fork in E. coli can progress almost 20 times faster than the RNA polymerase, meaning that rear-end collisions are typically between the rear edge of the RNA polymerase and the front edge of the leading-strand DNA polymerase that is traveling (faster) in the same direction on the same template strand. In a head-on collision, however, the front edge of RNA polymerase hits the DNA helicase that is traveling on the same (lagging) strand in the opposite direction. Note that the polymerases translocate with respect to their template in the 3′-to-5′ direction, while the helicase in the fork translocates in the 5′-to-3′ direction. 16. Answer: C Difficulty: 1 Section: The Initiation and Completion of DNA Replication in Chromosomes Feedback: The yeast origins of replication are similar to the bacterial origins in that they are A-T-rich and have particular sequences that attract replication initiator proteins. In higher eukaryotes, in contrast, the determinants of the replication origins are probably

less sequence-specific. E. coli and S. cerevisiae both have relatively short origins compared to higher organisms. 17. Answer: C Difficulty: 3 Section: The Initiation and Completion of DNA Replication in Chromosomes Feedback: Even though each round of replication takes 35 minutes, the “firing” interval can be adjusted independently, and even go out of control. In this strain, possibly due to mutations in the methylation pathway, the origins can become activated every 15 minutes, resulting in an average of six forks (three bubbles) per chromosome, the newest of which have started about 30 minutes after the oldest ones. 18. Answer: A Difficulty: 1 Section: The Initiation and Completion of DNA Replication in Chromosomes Feedback: Telomerase uses its RNA component as a template to polymerize DNA sequences at the chromosome ends to solve the “end-replication” problem. 19. Answer: D Difficulty: 1 Section: The Initiation and Completion of DNA Replication in Chromosomes Feedback: With the help of the t-loop structure, shelterin “hides” the telomere ends from the cell’s double-strand break detector and repair systems. 20. Answer: D Difficulty: 3 Section: The Initiation and Completion of DNA Replication in Chromosomes Feedback: The telomeres have 3′ overhangs and are folded in structures called t-loops, which help protect the chromosome ends. 21. Answer: A Difficulty: 1 Section: DNA Repair Feedback: Depurination (that is, the hydrolytic removal of purine bases from DNA) is by far the most common endogenous DNA lesion in our cells. Please refer to Table 5–3 for details. 22. Answer: D Difficulty: 1 Section: DNA Repair Feedback: These enzymes can recognize altered bases in DNA and catalyze their removal by hydrolyzing the glycosidic bond between the base and the deoxyribose sugar. 23. Answer: E Difficulty: 1 Section: DNA Repair

Feedback: The translesion polymerases come in different flavors and some of them are very efficient in repairing specific damage, but they generally have a low fidelity as they cannot proofread. The cell thus limits their usage to very few polymerization steps. 24. Answer: C Difficulty: 2 Section: DNA Repair Feedback: The use of thymine instead of uracil by DNA has the advantage that deamination of cytosine (to uracil) can be readily detected as an abnormal base in DNA. However, deamination of 5-methyl cytosine (a modification that is found in vertebrate DNA) can produce thymine and lead to mutations at a higher rate. 25. Answer: D Difficulty: 1 Section: DNA Repair Feedback: Compared to the faithful homologous recombination pathway to repair doublestrand breaks, nonhomologous end joining is a “quick and dirty” solution that often results in mutations at the site of repair. 26. Answer: B Difficulty: 3 Section: DNA Repair Feedback: The Ku dimer forms a hole that accommodates a DNA double helix. In addition, its structure reflects the fact that it should only bind to free DNA ends (for example, from double-strand breaks) in a fashion similar to threading a needle. 27. Answer: FFFF Difficulty: 3 Section: DNA Repair Feedback: In a random DNA sequence, one out of every 16 dinucleotides is a CG. However, these dinucleotides are found at a much lower frequency in the human genome, which is probably related to the fact that cytosine methylation at these sites (which has a role in transcription regulation) can be followed by deamination to cause a C-to-T substitution. Consistent with this notion, the CG frequency in D. melanogaster and C. elegans is close to the expected value. 28. Answer: B Difficulty: 2 Section: Homologous Recombination Feedback: Homologous recombination is used to repair DNA damage during and shortly after the S phase of the cell cycle, when a sister chromatid is available for faithful repairs. 29. Answer: A Difficulty: 2 Section: Homologous Recombination

Feedback: If a homologous chromosome is used (instead of a sister chromatid) to repair a double-strand break, one of the parental alleles of the affected gene can be replaced by the other allele, leading to loss of heterozygosity. 30. Answer: E Difficulty: 3 Section: Homologous Recombination Feedback: Cutting at b and d would result in a crossover. Cutting at a and c would only lead to a heteroduplex (no crossover). The other combinations are not normally chosen. 31. Answer: B Difficulty: 2 Section: Homologous Recombination Feedback: The poorly understood regulatory mechanism of crossover control ensures a roughly even distribution of crossover points along chromosomes. It also ensures that each chromosome undergoes at least one crossover every meiosis. For many organisms, roughly two crossovers per chromosome occur, one on each arm. 32. Answer: A Difficulty: 1 Section: Transposition and Conservative Site-Specific Recombination Feedback: DNA-only transposons that carry genes encoding antibiotic-inactivating enzymes are largely responsible for the spread of antibiotic resistance in bacterial strains. 33. Answer: E Difficulty: 1 Section: Transposition and Conservative Site-Specific Recombination Feedback: DNA-only transposons often encode a transposase to move by a cut-and-paste mechanism that results in a broken donor chromosome. Short inverted repeats of DNA sequence are found at the ends of these transposons. 34. Answer: B Difficulty: 1 Section: Transposition and Conservative Site-Specific Recombination Feedback: Retroviral-like retrotransposons resemble retroviruses in their replication, but lack the ability to produce a protein coat. They have directly repeated long terminal repeats at each end and are transcribed by cellular RNA polymerases. The transcript is then used as a template by a reverse transcriptase (encoded by the element) to make a double-stranded DNA copy that is then integrated into a new site on the chromosome using an integrase (also encoded by the element). This mechanism keeps the original copy unchanged. Alu elements are nonretroviral retrotransposons. 35. Answer: A Difficulty: 2 Section: Transposition and Conservative Site-Specific Recombination

Feedback: Most short interspersed nuclear elements (SINEs) do not encode any proteins and spread by pirating enzymes encoded by other transposons, such as long interspersed nuclear elements (LINEs). 36. Answer: D Difficulty: 2 Section: Transposition and Conservative Site-Specific Recombination Feedback: Phase variation in Salmonella is brought about by a conservative site-specific recombinase encoded by the bacterium. The recombinases that catalyze conservative sitespecific recombination form transient, high-energy covalent bonds with the DNA and use this energy to complete the DNA rearrangements, an action reminiscent of topoisomerases. 37. Answer: lagging Difficulty: 2 Section: DNA Replication Mechanisms Feedback: Strand A is extended 5′-to-3′ (from right to left) whereas the fork moves in the opposite direction; therefore, the strand should be polymerized discontinuously. 38. Answer: C Difficulty: 3 Section: DNA Replication Mechanisms Feedback: The central hole in the clamp can accommodate a double-stranded DNA molecule with extra room for smooth sliding. However, the helicase only allows a single strand of DNA to pass through its central hole. 39. Answer: TTTT Difficulty: 2 Section: The Initiation and Completion of DNA Replication in Chromosomes Feedback: Human cells have probably hundreds of thousands of potential origins, out of which about 40,000 are used each time a cell divides. Depending on gene expression and chromatin state, different sets of origins can be used in different cells of the body. Once an origin is activated, the resulting replication bubble expands on both sides of the origin. 40. Answer: CD Difficulty: 3 Section: The Initiation and Completion of DNA Replication in Chromosomes Feedback: The helicase is delivered to the ORC by active Cdc6 and Cdt1 proteins in G1 phase, but is activated by phosphorylation only in S phase. At the same time, phosphorylation of ORC prevents its reactivation in S phase. ORC can be reactivated again later in G1. 41. Answer: 2 Difficulty: 1 Section: DNA Repair

Feedback: Depurination (2) and deamination (1) are both common, although depurination is by far the most common endogenous DNA lesion in our cells. 42. Answer: 5 Difficulty: 3 Section: DNA Repair Feedback: UV irradiation can cause two adjacent pyrimidine bases (T or C) to form these covalent dimers. Please note that the complementary strand should also be taken into account. The possible sites of damage are TC and TT doublets in the strand shown, and TT, TC, and CC doublets in the complementary strand. 43. Answer: BBNN Difficulty: 2 Section: DNA Repair Feedback: Please note that depurinated residues are repaired through only part of the base excision repair pathway. 44. Answer: TTFT Difficulty: 1 Section: DNA Repair Feedback: In nucleotide excision repair, a multienzyme complex recognizes DNA double-helix distortions caused by damage. This pathway involves cutting the strand containing the damage at two positions flanking the damage. A helicase is also recruited and catalyzes strand removal. This pathway is also linked to the transcription machinery. 45. Answer: nonhomologous end joining Difficulty: 1 Section: DNA Repair Feedback: In nonhomologous end joining, the broken ends are simply brought together and rejoined by DNA ligation, often concomitant with the loss of nucleotides at the site of joining. 46. Answer: FEBDCA Difficulty: 2 Section: Homologous Recombination Feedback: This order of events results in the accurate repair of double-strand breaks. 47. Answer: FFTT Difficulty: 2 Section: Homologous Recombination Feedback: RecA binds cooperatively to the invading strand, forcing blocks of three nucleotides into a conformation mimicking that of three nucleotides in a double helix (although, between adjacent triplets, the DNA backbone is untwisted and stretched out). This complex then binds and destabilizes the target homologous duplex, allowing the sampling to take place by base-pairing interactions. RecA hydrolyzes its bound ATP after the strand exchange is completed. 48. Answer: FTTT

Difficulty: 2 Section: Homologous Recombination Feedback: Meiotic recombination requires an initial double-strand break. In contrast to the breaks resulting from DNA damage, this one is provided in a controlled manner by a specialized meiotic protein. 49. Answer: DRND Difficulty: 2 Section: Homologous Recombination Feedback: P elements and mariner elements are DNA-only transposons in different Drosophila species, copia is an abundant retroviral-like retrotransposon, and F elements are nonretroviral retrotransposons. 50. Answer: N Difficulty: 1 Section: Transposition and Conservative Site-Specific Recombination Feedback: Some nonretroviral retrotransposons (such as Alu elements) are still moving in our genome, accounting for perhaps two mutations out of every thousand new human mutations. Although our genome is also littered with relics of retroviral-like retrotransposons, none appear to be active today. 51. Answer: CNNCNC Difficulty: 3 Section: Transposition and Conservative Site-Specific Recombination Feedback: The arrangement of LoxP sites matters. For example, here, it does not allow the final combination (yellow + cyan) because deleting the red reporter requires the use of the white sequence, which also deletes the yellow reporter.

DNA and RNA polymerase differ in all of the following EXCEPT... A. the nucleotide substrates they incorporate. B. their requirement for a primer. C. their error rate. D. the type of chemical reaction they catalyze. E. their processivity.

What enzyme is depicted in the following schematic drawing?

A. DNA polymerase B. RNA polymerase C. Ribosome D. Reverse transcriptase E. Topoisomerase 3. The sequence of a region of DNA around the 5′ end of a gene in Escherichia coli is shown below. The –10 hexamer and the transcription start site are highlighted. What would be the sequence of the first 10 nucleotides of the mRNA transcribed from this gene? Write down the sequence from 5′ to 3′, e.g. CGGAUAAACT. 5′…GCGCTTGGTATAATCGCTGGGGGTCAAAGAT…3′

4. Due to their high transcription rate, active ribosomal RNA (rRNA) genes can be easily distinguished in electron micrographs of chromatin spreads. They have a characteristic “Christmas tree” appearance, where the DNA template is the “trunk” of the tree and the nascent RNA transcripts form closely packed “branches.” At the base of each branch is an RNA polymerase extending that branch, while RNA processing complexes at the tip of the branch form terminal “ornaments.” The top of the tree represents the ... of the rRNA gene, and the “ornaments” are at the ... end of the nascent rRNA molecules. A. end; 3' B. end; 5' C. beginning; either 3' or 5' D. beginning; 3' E. beginning; 5' 5. Which of the following types of noncoding RNA chiefly functions in the processing and chemical modification of ribosomal RNAs (rRNAs)? A. Small nuclear RNAs (snRNAs) B. Small nucleolar RNAs (snoRNAs) C. Small interfering RNAs (siRNAs) D. Transfer RNAs (tRNAs) E. MicroRNAs (miRNAs) 6. For the bacterial transcription machinery, which of the following mRNA sequences would you expect to constitute a potent transcriptional termination signal? Note that the two underlined regions in each sequence are complementary to each other. A. 5'... UGGCCCAGUCGGAAGACUGGGCCUUUUGUUUU...3' B. 5'... UGGCCCAGUCGGAAGACUGGGCCCGCGGAGCU...3' C. 5'... UUUUGUUUUAGGCCCAGUCGGAAGACUGGGCCA...3' D. 5'... CGCGGAGCUAGGCCCAGUCGGAAGACUGGGCCA...3' 7. The transcript for which of the following noncoding RNA in our cells is expected to undergo 5' cap addition after transcription? A. 5S rRNA B. miR-21 (a microRNA) C. tRNAPhe D. 5.8S rRNA

E. 18S rRNA 8. This large and complex general transcription factor has a DNA helicase activity that exposes the template for RNA polymerase II transcription. It also has a kinase activity that phosphorylates the C-terminal domain of the polymerase on Ser5 leading to promoter clearance. It is... A. TFIIB B. TFIID C. TFIIE D. TFIIF E. TFIIH 9. This general transcription factor recognizes the TATA box in RNA polymerase II promoters. It is... A. the only single-subunit general transcription factor. B. able to introduce a rather sharp kink in the double helix upon binding to DNA. C. responsible for the phosphorylation of the RNA polymerase CTD during transcription initiation. D. also responsible for the recognition of the BRE element in the promoter. E. All of the above. 10. Of the following proteins or protein complexes, which one does NOT typically interact with an elongating RNA polymerase II? A. Histone-modifying enzymes B. Capping enzymes C. Chromatin remodeling complexes D. Mediator complex E. Histone chaperones 11. How does a eukaryotic cell deal with the superhelical tension in its genomic DNA resulting from the activity of RNA polymerases? A. DNA gyrase introduces negative supercoils, keeping the DNA under constant tension. B. The RNA polymerases are allowed to rotate freely around their templates during transcription, leading to the relaxation of the tension. C. DNA topoisomerases rapidly remove the superhelical tension caused by transcription.

D. The nucleosomes adjust the tension by binding to positively supercoiled regions behind a moving RNA polymerase. E. All of the above. 12. Comparing mRNA molecules from human and Escherichia coli cells, which of the following is typically NOT true? A. A human mRNA has a special 5' cap, while a bacterial mRNA does not. B. A human mRNA has a poly-A tail, while a bacterial mRNA does not. C. A human mRNA undergoes alternative splicing, while a bacterial mRNA does not. D. A human mRNA contains noncoding sequences, while a bacterial mRNA does not. E. A typical human mRNA encodes one protein, while many bacterial mRNAs encode several different proteins. 13. Eukaryotic pre-mRNAs undergo a number of modifications such as capping at the 5' end. A 5' cap... A. consists of a modified terminal adenine nucleotide. B. has a 3'-to-5' linkage between the terminal nucleotide and the 5' end of the premRNA. C. contains a triphosphate bridge between the terminal base and the 5' end of the premRNA. D. carries a negative charge in the terminal base due to methylation. E. is identical for all mRNAs that are transcribed by RNA polymerase II. 14. After the first and before the second chemical step of RNA splicing, the intron of the premRNA... A. is still covalently connected to the 3' exon and has an internal branch in the shape of a lariat. B. is still covalently connected to the 3' exon and is linear. C. is still covalently connected to the 5' exon and has an internal branch in the shape of a lariat. D. is still covalently connected to the 5' exon and is linear. E. is still covalently connected to both of its flanking exons and is linear.

15. In the following qualitative histogram, which two curves better correspond to human exon and intron length distributions, respectively?

frequency

Relative

1 3 2

0 0

Exon or intron length (nucleotide pairs)

500

A. Curves 1 and 2 B. Curves 2 and 1 C. Curves 2 and 3 D. Curves 3 and 2 E. Curves 3 and 1 16.

To ensure the fidelity of splicing, the spliceosome... A. hydrolyzes ATP to undergo complex rearrangements. B. examines the splicing signals on the pre-mRNA several times. C. assembles on the pre-mRNA co-transcriptionally. D. takes advantage of “exon definition.” E. All of the above.

17. A primary mRNA transcript with three exons is depicted below. Which of the following mature mRNA products of this transcript is a result of exon skipping? 5'

A. B. C. D. E.

18. The enzyme poly-A polymerase is responsible for adding 3' poly-A tails to eukaryotic mRNAs. This enzyme... A. cuts the mRNA after recognition of the cleavage/polyadenylation signal by CstF and CPSF proteins. B. polymerizes the tail using an RNA template that is part of the enzyme. C. is extremely processive. D. normally adds about 1000 A nucleotides to the mRNA. E. uses ATP as a substrate. 19.

Several mechanisms contribute to the diversity of the mRNAs and proteins encoded by a

single gene in our genome. Which of the following is normally NOT one of them? A. Alternative choice of polyadenylation sites B. Alternative choice of translation initiation sites C. Alternative choice of transcription initiation sites D. Alternative choice of the reading frames E. Alternative choice of splice sites 20. Which of the following better describes a typical, actively translated mRNA in its journey from the nucleus to the cytosol? A. Initially in a circular conformation, the mRNA linearizes, enters the cytosol (5' end first), and remains linear. B. Initially in a circular conformation, the mRNA linearizes, enters the cytosol (3' end first), and remains linear. C. Initially linear, the mRNA enters the cytosol (5' end first), and adopts a circular conformation. D. Initially linear, the mRNA enters the cytosol (3' end first), and adopts a circular conformation. E. Initially linear, the mRNA enters the cytosol (3' end first), and remains linear. 21. As an mRNA molecule is processed in the nucleus, it loses some proteins and binds to new ones, some of which are used in mRNA surveillance pathways. The presence of which of the following molecules on an mRNA is a signal that the mRNA is still NOT ready for nuclear export? A. Cap-binding complex B. Exon junction complex C. snRNPs used in splicing

D. poly-A-binding proteins E. SR proteins 22. The nucleolus is a dynamic subcompartment within the nucleus and its size varies depending on the circumstances. In which of the following cells would you NOT expect to see the nucleoli? A. A yeast cell undergoing DNA replication B. A human neuron in a quiescent (G0) state C. A human macrophage active in phagocytosis D. A fruit fly embryonic nucleus in the G2 phase of the cell cycle E. A mouse embryonic cell in the metaphase stage of mitosis 23.

Cajal bodies in the eukaryotic cell nucleus... A. are stockpiles of fully mature snRNPs and other RNA processing components. B. can only be observed by electron microscopy. C. are absolutely required in all cell types. D. speed up the maturation and assembly of snRNPs and snoRNPs. E. are the main sites of pre-mRNA splicing.

24. Which of the following processes takes place in the nucleoli within the eukaryotic nucleus? A. Ribosome assembly B. rRNA gene transcription C. Telomerase assembly D. tRNA processing E. All of the above 25.

Which of the following is true about the molecule shown in the following drawing?

A. Its gene is transcribed by RNA polymerase I. B. There are about 50 genes in our genome encoding this type of molecule. C. This molecule normally undergoes various covalent modifications. D. It is normally composed of about 20 monomers. E. Its gene transcript normally undergoes alternative splicing. 26.

How is tRNA splicing different from mRNA splicing in eukaryotic cells? A. tRNA splicing does not proceed via transesterification reactions. B. tRNA splicing is carried out by proteins only. C. tRNA splicing does not create a lariat intermediate. D. tRNA splicing involves RNA endonuclease and RNA ligase activities. E. All of the above.

27.

The modified nucleotide shown below is normally found in mature...

A. tRNAs. B. ribosomes. C. spliceosomes. D. snoRNAs.

E. All of the above. 28. Each aminoacyl-tRNA synthetase activates a certain amino acid by attaching it to its corresponding tRNA(s) through a high-energy linkage... A. between the amino group of the amino acid and the 3' hydroxyl group of the tRNA. B. between the amino group of the amino acid and the 5' hydroxyl group of the tRNA. C. between the carboxyl group of the amino acid and the 3' or 2' hydroxyl group of the tRNA. D. between the carboxyl group of the amino acid and the 5' hydroxyl group of the tRNA. E. between the amino group of the amino acid and the 3' or 2' hydroxyl group of the tRNA. 29. Activation of an amino acid by an aminoacyl-tRNA synthetase involves the covalent linkage of... A. AMP to the amino acid. B. ADP to the amino acid. C. AMP to the tRNA. D. ADP to the tRNA. E. ADP to the enzyme. 30. On the ribosome, the mRNA is read from ..., and the polypeptide chain is synthesized from... A. 5' to 3'; C- to N-terminus. B. 5' to 3'; N- to C-terminus. C. 3' to 5'; C- to N-terminus. D. 3' to 5'; N- to C-terminus. 31.

An elongating ribosome is bound to appropriate tRNAs in both the A and the P sites and

is ready for peptidyl transfer. What happens next? A. The carboxyl end of the polypeptide chain is released from the P-site tRNA and joined to the free amino group of the amino acid linked to the A-site tRNA. B. The amino end of the polypeptide chain is released from the P-site tRNA and joined to the free carboxyl group of the amino acid linked to the A-site tRNA. C. The carboxyl end of the amino acid is released from the A-site tRNA and joined to the free amino group of the polypeptide chain linked to the P-site tRNA.

D. The amino end of the amino acid is released from the A-site tRNA and joined to the free carboxyl group of the polypeptide chain linked to the P-site tRNA. 32. Which of the following nucleotides is hydrolyzed in both transcription and in translation elongation? A. ATP B. GTP C. TTP D. UTP E. CTP 33.

How fast does a bacterial ribosome move on an mRNA? A. At about 2 nucleotides per second, significantly lower than the speed of the RNA polymerase. B. At about 5 nucleotides per second, comparable to the speed of the RNA polymerase. C. At about 10 nucleotides per second, significantly lower than the speed of the RNA polymerase. D. At about 20 nucleotides per second, comparable to the speed of the RNA polymerase. E. At about 60 nucleotides per second, comparable to the speed of the RNA polymerase.

34. Which of the following features is common between the bacterial and eukaryotic ribosomes in translation initiation? A. They both use an initiator tRNA that carries formylmethionine. B. They both bind to the 5' end of the mRNA and move forward to find the start codon. C. They both recognize the start codon by interacting with the Shine–Dalgarno sequence. D. They both interact with various translation initiation factors. E. All of the above. 35. The following mRNA sequence is taken from the middle of exon 3 in a mature mRNA that has 12 exons. Knowing that this mRNA does not undergo nonsense-mediated decay, which of the reading frames shown is the correct one for this mRNA? Write down 1, 2, or 3 as your answer.

Reading frames:

36.

5’…AGUGAUUCGAUACAGCUAGCGGACAGCUA…3 1 …’ 2… 2 3 3 2… 2 32 2 3 3 3

Polysomes... A. are large cytoplasmic assemblies made of several ribosomes each translating their exclusive mRNA. B. are only found in the eukaryotic cytoplasm. C. can take advantage of the circularization of eukaryotic mRNA (by interactions between the 5' and 3' ends of the mRNA) to further speed up the rate of protein synthesis. D. are mostly translationally inactive and are normally used by the cell to store the ribosomes and their associated mRNAs for future use. E. All of the above.

37. The specific transfer RNA used for the incorporation of selenocysteine in proteins recognizes the UGA codon, which is normally a translation stop codon. What prevents this tRNA from reading through all the other “legitimate” UGA stop codons in the cells and causing a massive, disastrous recoding? A. The cells that have selenocysteine in their proteins do not normally use the UGA stop codon and always use the other two stop codons (UAA and UAG) instead. B. This tRNA is made in such a small amount that its side effects are negligible. C. This tRNA only interacts with the UGA codon in the P site of the ribosome, and therefore does not interfere with the normal function of the codon in translation termination, which takes place in the A site. D. The mRNAs encoding the selenocysteine-containing proteins also contain additional sequences required for the recoding event. 38. In a gene that normally contains three exons, which of the following changes probably will NOT activate the nonsense-mediated mRNA decay pathway? The sizes of exons 1 to 3 are 100, 150, and 200 nucleotide pairs, respectively. A. A nonsense mutation in exon 1. B. A nonsense mutation in exon 2. C. A mutation in the first intron resulting in the inclusion of a large intronic fragment in the mature mRNA.

D. A frameshift mutation in exon 1. E. A mutation in exon 2 leading to its loss through exon skipping. 39. This complex uses ATP and forms an “isolation chamber” for its misfolded protein substrates, allowing them to fold to their native conformation. It is called... A. a proteasome. B. a ribosome. C. Mediator. D. a chaperonin. E. an exosome. 40.

What is the function of the 19S cap in the proteasome complex? A. Recognizing the proteins that will be degraded B. Unfolding the proteins that will be degraded and threading them into the central 20S cylinder for digestion C. Removing the polyubiquitin chain from the proteins that will be degraded D. Hydrolyzing ATP to make the digestion process highly efficient E. All of the above

41. How many ATP molecules must be hydrolyzed by the proteasome for the digestion of a protein that has been tagged for degradation with a polyubiquitin chain? A. None; the digestion does not require ATP hydrolysis and is exergonic B. One C. Two D. Ten E. Many; the number depends on the specific protein 42.

In humans, nearly 80% of proteins are acetylated on their N-terminal residue, a

modification known to be recognized by a specific E3 enzyme, which directs the ubiquitylation of the protein for rapid degradation. Does this mean that all of these proteins are actively degraded at any given time? A. Yes; this high turnover rate ensures that their activity is under tight control. B. Yes; but they are not fully degraded in this way and can still function as protein fragments. C. No; the destruction signal can be buried in the interior of the protein or bound to other proteins.

D. No; the E3 enzyme recognizing this mark is inactive most of the time. 43. In the following representation of a single-stranded RNA molecule, the nucleotide residues are drawn as gray circles, while base-pairing interactions between them are indicated by dashed lines. The structure of this RNA molecule has...

A. one hairpin loop and one pseudoknot. B. one pseudoknot only. C. three hairpin loops and a four-stem junction. D. two hairpin loops and a three-nucleotide bulge. E. three hairpin loops. 44. In the following diagram showing the overall flow of genetic information in all living cells, what is the name of the process indicated with a question mark?

45. Sort the following events in the order that they occur during transcription initiation in Escherichia coli. Your answer would be a four-letter string composed of letters A to D only; e.g. DCAB. (A) Abortive initiation trials (B) σ Factor release from the RNA polymerase holoenzyme (C) Binding of the holoenzyme to the promoter in the “closed” complex (D) Formation of the transcription bubble 46. Your friend who is studying the sequence conservation around the branch-point site on a subset of pre-mRNAs in Macaca mulatto (rhesus macaque) has sent you the following sequence

logo of the region of interest, but has forgotten to indicate the position of the branch-point nucleotide itself within this region. Where do you think it is? Write down the position number (1 to 8) as your answer.

Information content (bit)

0 1

Position

47. You have identified an RNA-binding protein that can bind directly (with varying affinities) to mRNA molecules that bear the following sequences near the 3' splice sites of their exons. Based on these results, what is the five-nucleotide consensus sequence for the binding site of this protein? Your answer would be a five-letter string composed of letters A, U, C, or G only; e.g. UUUAG. 5'...AACGG...3' 5'...AUCGG...3' 5'...GACGC ...3' 5'...AACC C ...3' 5'...AUCGC ...3' 5'...AACAC ...3' 5'...AACC C ...3' 48. In the following schematic drawing, a transcribing RNA polymerase is moving toward the right on the DNA template as indicated. If the polymerase is allowed to rotate freely while the DNA is kept from doing so, would the enzyme rotate in a right-handed (R) or left-handed (L) fashion around the template? Write down R or L as your answer.

Right-handed rotation of the polymerase

Left-handed rotation of the polymerase

49. Indicate true (T) and false (F) statements below regarding mRNA splicing in human cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) Introns must be removed in the order in which they occur along the pre-mRNA. ( ) Nucleosomes tend to be positioned over exons (on average, about one nucleosome per exon). ( ) Exon size tends to be much more variable than intron size. ( ) The exon junction complexes mark the sites on the mRNA where splicing has not been successful. 50. No matter where translation begins, only one polypeptide sequence pattern can be obtained from the translation of an mRNA chain with the sequence (AC)n; that is, an mRNA with the sequence 5'...ACACACACAC...3'. In contrast, an mRNA chain with the sequence (ACG)n can be translated to polypeptide sequences of three distinct patterns, depending on the choice of the reading frame. Using similar judgment, how many different polypeptide sequence patterns can be obtained from an mRNA chain with the sequence (ACGU)n? Write down your answer in digits, e.g. 12. 51. Three consecutive nucleotides in RNA (such as AUC or GUA) constitute a triplet called a ..., which can specify an amino acid or a stop signal for translation. 52.

The anticodon sequence of a phenylalanine-specific tRNA is 5'-GAA-3'. Given the

wobble base-pairing rules presented in the following table, how many different codons can be “read” by this tRNA? Write down your answer in digits, e.g. 12.

Wobble base in the codon

Possible anticodon bases

A, G, or I

G or I

53.

Indicate true (T) and false (F) statements below regarding the accuracy of translation by

the ribosome. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) The difference in the binding affinity of a codon to a correct versus incorrect anticodon CANNOT by itself account for the high accuracy of translation. ( ) GTP hydrolysis by EF-Tu both speeds up translation and increases its accuracy. ( ) The ribosome can detect correct Watson–Crick base-pairing between the codon and anticodon in the A site, and consequently trigger GTP hydrolysis by EF-Tu. ( ) Even after GTP hydrolysis by EF-Tu, the ribosome can prevent wrong amino acid incorporation by kinetic proofreading. 54. Indicate whether each of the following is located (or takes place) in the large (L) or the small (S) ribosomal subunit. Your answer would be a four-letter string composed of letters L and S only, e.g. LLLS. ( ) Peptide bond formation ( ) Codon–anticodon interaction ( ) The path of the mRNA ( ) The polypeptide exit tunnel 55.

Indicate true (T) and false (F) statements below regarding the folding of proteins upon

their synthesis. Your answer would be a five-letter string composed of letters T and F only, e.g. TFFFF. ( ) A globular protein can fold to its native form while still within the ribosome. ( ) A protein can begin to fold and to bind to other proteins while still being synthesized on the ribosome. ( ) A protein may undergo several cycles of hsp70 binding and release while still being synthesized on the ribosome.

( ) A molten globule can form only after the synthesis of the protein is complete. ( ) A protein typically undergoes hsp60-assisted folding while still being synthesized. 56. Fill in the blank. These RNA molecules can catalyze a wide variety of biochemical reactions, including peptide bond formation and RNA splicing. They can be found in nature or selected in vitro for a desired function. Such a molecule is called a(n) ... 57. Indicate true (T) and false (F) statements below regarding the RNA world hypothesis. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) According to this hypothesis, RNA in primitive cells was responsible for storing genetic information, while proteins were responsible for the catalysis of chemical reactions. ( ) The existence of natural ribozymes supports this hypothesis. ( ) In support of this hypothesis, all peptides in present-day cells are made by the ribosome. ( ) All present-day cells use DNA as their hereditary material.

Answers: 1. Answer: D Difficulty: 2 Section: From DNA to RNA Feedback: Despite the lack of homology, the reactions catalyzed by these two polymerases are analogous, even though the enzymes work on different substrates. DNA polymerases require a primer and have a much lower error rate compared to RNA polymerases. They are also not as processive on their own. 2. Answer: B Difficulty: 2 Section: From DNA to RNA Feedback: The enzyme synthesizes RNA using one of the DNA strands as template. 3. Answer: GGUCAAAGAU Difficulty: 3 Section: From DNA to RNA Feedback: The mRNA sequence starts from the transcription start site and is the same as the sequence of the coding strand of the DNA (and complementary to the “template strand”) except that T is replaced with U. 4. Answer: E Difficulty: 3 Section: From DNA to RNA Feedback: The shortest rRNA “branches” at the “top” of the tree emanate from the beginning of the gene since their transcription has just started. All branches extend at their 3′ end at their base while their 5′ end starts to associate with RNA processing complexes. 5. Answer: B Difficulty: 1 Section: From DNA to RNA Feedback: In the nucleolus, the snoRNAs help process and chemically modify the rRNAs in various ways. 6. Answer: A Difficulty: 3 Section: From DNA to RNA Feedback: Many bacterial transcription terminators consist of a stable hairpin followed by a string of U nucleotides (that form weak A-U base pairs with the template), which together help disengage the mRNA transcript from RNA polymerase. 7. Answer: B Difficulty: 2 Section: From DNA to RNA

Feedback: MicroRNAs are mainly transcribed by the RNA polymerase II machinery and, similar to mRNAs, their primary transcripts normally undergo capping and polyadenylation. 8. Answer: E Difficulty: 1 Section: From DNA to RNA Feedback: TFIIH is a large complex with several key functions in transcription initiation. 9. Answer: B Difficulty: 2 Section: From DNA to RNA Feedback: It is the multisubunit TFIID, containing a TATA-binding protein, which, upon binding to the TATA box, creates a sharp kink in the double helix that serves as a landmark to attract other transcription factors. 10. Answer: D Difficulty: 2 Section: From DNA to RNA Feedback: In the initiation phase of transcription, Mediator mediates the interaction of RNA polymerase with the transcriptional activators and is required for the activation of transcription by enhancers. 11. Answer: C Difficulty: 2 Section: From DNA to RNA Feedback: Unlike bacteria, which maintain a fairly constant level of negative supercoiling in their circular DNA, eukaryotic topoisomerases rapidly remove the superhelical tension caused by DNA translocating machines such as the RNA polymerases. 12. Answer: D Difficulty: 2 Section: From DNA to RNA Feedback: Noncoding regions are also found in E. coli mRNAs, although these regions are typically short compared to those in human cells. 13. Answer: C Difficulty: 1 Section: From DNA to RNA Feedback: The cap is a 7-methyl G residue linked to the 5' end of the transcript via a 5'to-5' triphosphate bridge. Methyl transferase enzymes methylate the terminal guanine base (giving it a positive charge) and sometimes other nucleotides, creating different 5' caps. 14. Answer: A Difficulty: 2 Section: From DNA to RNA

Feedback: Due to the first transesterification step, the branch point in the intron is covalently bound to the 5' end of the intron, and the intron is no longer continuous with the upstream (5') exon. 15. Answer: A Difficulty: 3 Section: From DNA to RNA Feedback: Our exons have a narrow distribution of size around the mean value of about 150 nucleotide pairs, whereas introns can vary greatly in size from about 10 to over 100,000 nucleotide pairs. 16. Answer: E Difficulty: 1 Section: From DNA to RNA Feedback: The spliceosome uses ATP hydrolysis to produce a series of rearrangements, allowing the splicing signals to be checked and rechecked to ensure the accuracy of the process. The fact that splicing is coupled to transcription provides an advantage and helps the cell keep track of the exons, as does exon definition by SR proteins and other factors that mark the splice sites around each exon. 17. Answer: B Difficulty: 2 Section: From DNA to RNA Feedback: Splicing “mistakes” include exon skipping and intron retention. Additionally, selection of cryptic splice sites can lead to the partial inclusion of exons or introns. 18. Answer: E Difficulty: 2 Section: From DNA to RNA Feedback: After the mRNA is cleaved, this enzyme adds about 200 A nucleotides (from ATP) one by one without using a template. 19. Answer: D Difficulty: 2 Section: From DNA to RNA Feedback: Eukaryotic cells take advantage of various mechanisms such as alternative splicing and alternative choices of polyadenylation sites, transcription start sites, and translation start sites to create a highly complex pool of gene products in the cell. Normally, however, a cellular gene is interpreted in only one of the possible reading frames. 20. Answer: C Difficulty: 2 Section: From DNA to RNA Feedback: Normally, the export of an mRNA to the cytosol occurs with the 5' cap proceeding first. The mRNA can then circularize through the interaction of the cytosolic

cap-binding proteins with the cytosolic poly-A-binding proteins to facilitate rapid translation. 21. Answer: C Difficulty: 2 Section: From DNA to RNA Feedback: The snRNPs remain associated with the pre-mRNA until it is fully spliced. Their presence indicates that the mRNA is still being processed. 22. Answer: E Difficulty: 2 Section: From DNA to RNA Feedback: During the M phase, the chromosomes condense and the nucleoli disappear. In telophase, the chromosomal regions contributing their rRNA genes to the subcompartment coalesce again and their transcription resumes. 23. Answer: D Difficulty: 1 Section: From DNA to RNA Feedback: In Cajal bodies, snRNPs and snoRNPs undergo maturation and assembly at enhanced rates compared to the rates estimated for cells in which these aggregates are disrupted. However, this enhancement is not crucial for all cells. Pre-mRNA splicing is thought to happen in mRNA production “factories” dispersed throughout the nucleus. 24. Answer: E Difficulty: 1 Section: From DNA to RNA Feedback: The nucleolus is a factory for the processing of various noncoding RNAs and their assembly into larger complexes. 25. Answer: C Difficulty: 2 Section: From RNA to Protein Feedback: It is a transfer RNA (tRNA) molecule, which is normally composed of nearly 80 nucleotides, is heavily modified, and is transcribed from one of the ~500 tRNA genes in our genome by RNA polymerase III. 26. Answer: E Difficulty: 2 Section: From RNA to Protein Feedback: tRNA splicing appears to be simpler compared to the splicing of pre-mRNAs. 27. Answer: E Difficulty: 2 Section: From RNA to Protein Feedback: The modified nucleotide is pseudouridine, one of the most abundant modified nucleotides in cellular RNA, and it is found in tRNAs, rRNAs, snRNAs, and snoRNAs.

The “guide” snoRNAs in the nucleolus are themselves involved in the pseudouridylation of the other RNAs. 28. Answer: C Difficulty: 2 Section: From RNA to Protein Feedback: The amino group of the amino acid is free, and will serve as the attacking group in the formation of the peptide bond on the ribosome. The carboxyl group, on the other hand, forms a high-energy and susceptible ester bond with the tRNA in preparation for that same reaction. 29. Answer: A Difficulty: 1 Section: From RNA to Protein Feedback: The amino acid is initially activated by an ester linkage, through its carboxyl group, to an AMP molecule resulting from the hydrolysis of an ATP molecule. The amino acid is then transferred onto the appropriate tRNA molecule. 30. Answer: B Difficulty: 2 Section: From RNA to Protein Feedback: During translation elongation, the ribosome translocates on the mRNA from the 5′ end to the 3′ end in consecutive sets of three nucleotides. The polypeptide is concomitantly elongated at its C-terminus by addition of individual amino acids. 31. Answer: A Difficulty: 2 Section: From RNA to Protein Feedback: In the fundamental reaction of protein synthesis, a peptide bond is formed between the carboxyl group at the end of a growing polypeptide chain and a free amino group on an incoming aminoacyl-tRNA. 32. Answer: B Difficulty: 1 Section: From RNA to Protein Feedback: In transcription, GTP is a polymerization substrate along with ATP, UTP, and CTP. Additionally, GTP is hydrolyzed by several factors involved in translation, including the eukaryotic elongation factors eEF1 and eEF2. 33. Answer: E Difficulty: 3 Section: From RNA to Protein Feedback: The polypeptide chain grows on the ribosome by about 20 amino acids per second, which is equivalent to an mRNA translocation rate of 60 nucleotides per second. This is comparable to the rate of transcription by the bacterial RNA polymerase (about 50 nucleotides per second). Indeed, a bacterial ribosome can translate mRNAs co-

transcriptionally, by closely following an RNA polymerase and translating the same mRNA molecule that is being produced by the polymerase. 34. Answer: D Difficulty: 2 Section: From RNA to Protein Feedback: The initiator tRNA in eukaryotes carries methionine. Unlike the cap-binding and the 5'-to-3' “scanning” of the mRNA by the eukaryotic translation initiation machinery, bacterial ribosomes find the start codon by directly interacting with the ribosome-binding sites in the mRNA. Initiation factors in both systems assist the ribosomes in initiating with high efficiency and speed. 35. Answer: 1 Difficulty: 3 Section: From DNA to RNA Feedback: Frames 2 and 3 contain a stop codon (UAG at the sixth codon, UGA at the first codon, respectively), which is not consistent with the sequence coding for a protein. 36. Answer: C Difficulty: 2 Section: From RNA to Protein Feedback: Polysomes form as a result of frequent initiation events leading to several ribosomes bound to and translating each mRNA simultaneously. If the mRNA is circularized, the ribosomes that have finished the synthesis of the protein can reinitiate immediately for another round. 37. Answer: D Difficulty: 2 Section: From RNA to Protein Feedback: The presence of the UGA codon is only one of the signals required for the recoding event. Other sequences (that can form stem-loop structures) in the vicinity of the UGA codon on the mRNA collaborate with a specific protein factor to incorporate the amino acid into the growing protein chain only for particular mRNAs and not any mRNA with a UGA codon. . 38. Answer: E Difficulty: 3 Section: From RNA to Protein Feedback: Nonsense mutations in the upstream exons can directly trigger nonsensemediated decay (NMD). Frameshift mutations in these exons are likely to introduce premature termination codons and activate NMD as well. Similarly, retention of intronic sequences carries a high risk of premature stop codons. In some cases, exon skipping can lead to NMD if the size of the skipped exon is not a multiple of 3 and therefore its absence changes the reading frame of the downstream exon.

39. Answer: D Difficulty: 1 Section: From RNA to Protein Feedback: The hsp60-like proteins form a large barrel-shaped complex, sometimes called a chaperonin, that forms an “isolation chamber” for the folding process of a protein after it has been fully synthesized. 40. Answer: E Difficulty: 1 Section: From RNA to Protein Feedback: The 19S cap in the proteasome is a large complex with various functions including substrate recognition, de-ubiquitylation, unfolding and feeding into the 20S core, as well as regulatory roles. 41. Answer: E Difficulty: 1 Section: From RNA to Protein Feedback: The AAA proteins in the cap hydrolyze tens or hundreds of ATP molecules to unfold and thread each protein into the central 20S digestive core. 42. Answer: C Difficulty: 3 Section: From RNA to Protein Feedback: The modification is not necessarily exposed for the E3 enzymes to recognize. Exposure of the modification can occur if, for example, the protein becomes misfolded. 43. Answer: D Difficulty: 3 Section: The RNA World and the Origins of Life Feedback: The RNA molecule folds into a structure (shown below) with two hairpins; one with a five-nucleotide loop, and the other with a four-nucleotide loop. The former hairpin also has a three-nucleotide bulge in its 5' stem. A three-way junction is formed at the base of the two hairpins.

5 3 44. Answer: Translation ' ' Difficulty: 1 Section: From DNA to RNA Feedback: The diagram shows the pathway from DNA to protein.

45. Answer: CDAB Difficulty: 2 Section: From DNA to RNA Feedback: In bacterial transcription initiation, binding of the RNA polymerase holoenzyme to the promoter region in the closed complex is followed by the formation of the open complex involving the formation of the transcription bubble. After a number of abortive initiations, RNA polymerase clears the promoter and σ factor is released. 46. Answer: 7 Difficulty: 3 Section: From DNA to RNA Feedback: The branch-point nucleotide is often an A residue surrounded by other intronic sequences that are required for splicing. 47. Answer: AACGC Difficulty: 3 Section: From DNA to RNA Feedback: The most common nucleotide at each position of the alignment is chosen for the consensus sequence. Note that none of the examined mRNAs has the consensus sequence. 48. Answer: R Difficulty: 3 Section: From DNA to RNA Feedback: When the polymerase is moving to the right, it induces positive supercoiling which hinders helix opening. If the enzyme is allowed to rotate freely (which is normally not the case in vivo), right-handed rotation will eliminate the superhelical tension, as if the polymerase is twisting to follow along the right-handed double helix. 49. Answer: FTFF Difficulty: 2 Section: From DNA to RNA Feedback: Exon size tends to be much more uniform than intron size and averages at about 150 nucleotide pairs, close to the length of DNA wrapped by one nucleosome. The bound nucleosome is suggested to act as a “speed bump” to allow the assembly of splicing proteins involved in exon definition. 50. Answer: 1 Difficulty: 3 Section: From RNA to Protein Feedback: Even though the polypeptide sequence pattern for the (ACGU)n RNA polymer is more complicated than for the other two examples, there is only one such pattern regardless of the choice of the reading frame: (Thr.Tyr.Val.Arg)n. For the (AC)n RNA, also only one polypeptide sequence pattern is possible: (Thr.His)n. 51. Answer: codon

Difficulty: 1 Section: From RNA to Protein Feedback: Three consecutive nucleotides in a protein-coding sequence of mRNA constitute a codon. Each codon specifies either one amino acid or a stop in the translation process. 52. Answer: 2 Difficulty: 3 Section: From RNA to Protein Feedback: The Phe codons UUC and UUU can be recognized by this tRNA, according to the wobble pairing rules. 53. Answer: TTTT Difficulty: 2 Section: From RNA to Protein Feedback: The difference in affinity between correct and incorrect codon–anticodon matches cannot by itself account for the high accuracy of translation. The ribosome further assesses the correctness of the match; in the case of a correct match, EF-Tu hydrolyzes its bound GTP. This contributes to the speed and fidelity of translation. Even after EF-Tu release, there is a second proofreading opportunity for the ribosome to prevent an incorrect amino acid from being added to the growing chain. 54. Answer: LSSL Difficulty: 2 Section: From RNA to Protein Feedback: The RNA components of the ribosome in the large and small subunit are responsible for peptide bond formation and the verification of the codon–anticodon match, respectively. mRNA is threaded through the small subunit, while the nascent protein emerges from the large subunit. 55. Answer: FTTFF Difficulty: 2 Section: From RNA to Protein Feedback: Proteins can start folding as soon as they emerge from the ribosome (or even within the exit tunnel of the ribosome, but only for small helical segments) and one domain can be fully folded and functional before an entire multidomain protein is completely synthesized. However, folding within the hsp60 chaperonin complex typically happens only post-translationally. 56. Answer: ribozyme Difficulty: 1 Section: The RNA World and the Origins of Life Feedback: Ribozymes are RNAs capable of catalyzing many different biochemical reactions. The ribosome provides an example in which the rRNA of the large subunit

bears peptidyl transferase activity. Many other functions not found naturally can be screened for in the laboratory. 57. Answer: FTFT Difficulty: 2 Section: The RNA World and the Origins of Life Feedback: The hereditary material in all present-day cells is DNA, but according to the RNA world hypothesis, RNA preceded DNA (as genetic material) as well as proteins (as catalysts). Some short peptides are synthesized without the ribosome, an observation that provides clues as to how the early transition from an RNA world to a protein-dominated world might have initiated.

1. Indicate true (T) and false (F) statements below regarding gene control. Your answer would be a four-letter string composed of letters T and F only, e.g. TTFT. ( ) Development of multicellular organisms from a fertilized egg only rarely involves DNA rearrangements in specialized cells. ( ) A typical human cell expresses less than 1% of its approximately 30,000 genes at any given time. ( ) Genes that are expressed in all cell types usually vary in their level of expression in different cell types. ( ) Many differentiated plant cells can be fully de-differentiated and give rise to an entire plant. 2. In the following schematic graph of a hypothetical set of RNA-seq data, the number of reads is plotted for a region of chromosome containing two genes, from samples obtained from two different tissues. Which gene (X or Y) do you think is more likely a “housekeeping” gene? Which region (1 or 2) within gene Y most likely corresponds to an exon?

Number of reads

Liver Brain Gene X

Gene Y Position on chromosome

A. Gene X; region 1 B. Gene X; region 2 C. Both genes; region 1 D. Gene Y; region 1 E. Gene Y; region 2

3. Studying the expression of a transcription regulatory protein in two cell types, you have performed experiments showing that the mRNA encoding the protein is present at comparable levels in the cytosol of both cell types. However, based on the expression of its target genes, you suspect that the protein activity might be significantly different in the two cell types. Which of the following steps in expression of the gene encoding this protein is more likely to be differentially controlled in these cell types? A. Transcription B. Translation C. mRNA transport D. mRNA degradation 4. In analysis using two-dimensional gel electrophoresis of the proteins expressed in different cell types, the number of spots that are different in different cells usually exceeds the number of common spots, and even the common spots can still have different intensities. The spots representing which of the following proteins would you expect to be among the common spots when compared across several cell types? A. RPL10 (a ribosomal protein) B. HBA1 (a hemoglobin subunit) C. Insulin (a hormone) D. Tyrosine aminotransferase (a metabolic enzyme) E. All of the above 5.

What determines the time and place that a certain gene is transcribed in the cell? A. The type of cis-regulatory sequences associated with it B. The relative position of cis-regulatory sequences associated with it C. The arrangement of various cis-regulatory sequences associated with it D. The specific combination of transcription regulators present in the nucleus E. All of the above

6. The majority of transcription regulators make sequence-specific contacts with DNA in the major groove. In the two diagrams below, where are the contact surfaces that are exposed in the major groove?

3 1

A. 1, 3 B. 1, 4 C. 2, 3 D. 2, 4 7. Considering the diagrams below that show hydrogen-bond donors and acceptors as well as hydrophobic groups in four DNA base pairs, which of the following do you think is the most difficult to accomplish by DNA-binding proteins?

A. Distinguishing between A-T and T-A in the major groove B. Distinguishing between A-T and C-G in the major groove C. Distinguishing between C-G and G-C in the major groove D. Distinguishing between C-G and G-C in the minor groove E. Distinguishing between C-G and T-A in the minor groove 8.

Protein subunits that interact specifically with DNA sequences … A. typically recognize sequences of two to three nucleotide pairs in length. B. do so via hydrogen bonds, ionic bonds, and hydrophobic interactions. C. typically form about five weak interactions at the protein–DNA interface. D. often bind loosely to DNA. E. All of the above.

Which of the following DNA-binding motifs uses β sheets to recognize DNA bases? A. The helix–turn–helix motif B. The leucine zipper C. The zinc finger motif D. The helix–loop–helix motif E. None of the above

Information content

10. The following sequence logo represents the preferred cis-regulatory sequences of an imaginary transcription regulator that functions as a dimer. Would you expect this sequence to be recognized by a homodimer (M) or a heterodimer (T)? Write down M or T as your answer.

Position 11. A DNA-binding protein recognizes a specific eight-nucelotide sequence in DNA. Assuming that its binding is perfectly specific, how many binding sites are expected to exist for it in human genomic DNA, which is composed of about 6 × 109 nucleotide pairs? What about if the protein forms a homodimer? Note that the target sequence can be oriented either way in the double-stranded DNA. A. About 200,000; about 30,000 B. About 2000; about 300 C. About 200; about 30 D. About 200,000; about 300 E. About 200,000; about 3 12. A DNA-binding protein binds cooperatively to DNA by fairly weak homodimerization. The binding of this protein to DNA … A. shows a sigmoidal binding curve. B. occurs in more of an all-or-none manner, compared to a protein that is a constitutive dimer. C. occurs despite the fact that the protein molecules are predominantly monomers in solution. D. All of the above.

13. Indicate true (T) and false (F) statements below regarding transcription regulators. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) About 10% of the protein-coding genes in most organisms encode transcription regulators. ( ) The packaging of eukaryotic DNA into chromatin dampens the effect of cooperative binding of transcription regulators. ( ) Some DNA-binding proteins can induce bends or kinks in their target DNA. ( ) Dimerization of transcription regulators increases the chance of binding to nonspecific sequences. 14.

The Trp operon in Escherichia coli encodes the components necessary for tryptophan

biosynthesis. In the presence of the amino acid in a bacterium, … A. the tryptophan operator is bound to the tryptophan repressor. B. the tryptophan repressor is bound to bacterial RNA polymerase. C. the expression of the tryptophan repressor is shut off. D. the operon genes are expressed. E. All of the above. 15.

Under which of the following conditions is the Lac operon in Escherichia coli fully

turned on? A. Low glucose and lactose levels B. Low glucose but high lactose levels C. High glucose but low lactose levels D. High glucose and lactose levels E. Low cAMP and high glucose levels 16. Consider a cis-regulatory enhancer sequence in the Escherichia coli chromosome that is located thousands of nucleotide pairs upstream of the gene that it regulates. If the regulatory sequence is mutated to become nonfunctional, the introduction of the wild-type enhancer on a plasmid fails to regulate the gene. This implies that … A. the regulatory sequence encodes a regulatory protein that binds near the promoter of the target gene and controls RNA polymerase binding. B. the regulation of the target gene involves looping out of the intervening DNA, and the promoter of the cis-regulatory sequence must be on the same chromosome. C. the regulatory sequence can bind directly to the RNA polymerase. D. the regulatory sequence cannot bind to a protein.

17. Transcription regulation has similarities and differences in bacteria and in eukaryotes. Which of the following is correct in this regard? A. Most bacterial genes are regulated individually, whereas most eukaryotic genes are regulated in clusters. B. The rate of transcription for a eukaryotic gene can vary in a much wider range than for a bacterial gene (which is, at most, only about 1000-fold). C. DNA looping for gene regulation is the rule in bacteria but the exception in eukaryotes. D. Transcription regulators in both bacteria and eukaryotes usually bind directly to RNA polymerase. E. The default state of both bacterial and eukaryotic genomes is transcriptionally active. 18. Which of the following is directed by transcription activators in eukaryotic cells in order to provide a more accessible DNA for the transcription machinery? A. Nucleosome remodeling B. Histone removal C. Histone replacement D. Histone modifications E. All of the above 19. The following schematic drawing represents the activation of transcription for a eukaryotic gene (gene X). Indicate what component (A to D) in the drawing corresponds to each of the following. Your answer would be a four-letter string composed of letters A to D only, e.g. ADBC.

( ) Mediator ( ) cis-Regulatory sequence ( ) A general transcription factor ( ) “Spacer” DNA that may encode lncRNAs

20. To prevent spurious transcription from a gene, acetylation of histones—which is carried out by histone acetyl transferases ahead of a moving RNA polymerase II—is quickly reversed by histone deacetylases and histone methyl transferases in the wake of the polymerase, leaving a trail of specific methylated histones. Which of the following curves do you think better represents the distribution of this specific histone methylation mark with respect to a gene?

Degree of methylation

B D C

Transcribed region Promoter Genomic location 21. Two transcription activators cooperate to recruit a coactivator to a cis-regulatory sequence and activate transcription of a nearby gene. If each of the activators increases the affinity of the coactivator for the reaction site (and therefore the rate of transcription) by 100fold, how much would you expect the affinity to increase when both activators are bound to DNA compared to when none is bound? A. 10-fold B. 100-fold C. 200-fold D. 10,000-fold 22. In the following schematic diagram of a region of a mammalian genome, genes A to D (triangles) are located in between a number of insulator elements (white squares) and barrier sequences (black squares). If the cis-regulatory sequence (oval) is bound by an abundant repressor protein, which gene would you expect to be expressed at a higher rate in this cell?

A .

Heterochromatin

Reference: Even-skipped Regulatory Module (Questions 23-26) Expression of the Even-skipped (Eve) gene in early Drosophila embryos is under the control of several transcription regulators. In one example, one of the Eve stripes is positioned near the anterior region of the embryo, and its regulatory module contains binding sites for Bicoid and Hunchback (activators) as well as Giant and Krüppel (inhibitors) such that the gene is expressed only in the region where concentrations of the two activators are high and the concentrations of the two inhibitors are low. A reporter gene can be placed under the control of this module, and it can be shown to form a stripe in the same place in the embryo as the corresponding stripe of Eve. Answer the following question(s) based on these findings.

23. In the following schematic diagrams of an early Drosophila embryo, in which region would you expect to find the reporter protein put under the control of the regulatory module mentioned above?

Bicoid

Hunchback

Giant

Krüppel

A B C D E

24. What would you expect to happen to the pattern of reporter expression in flies that lack the gene encoding Giant? A. It would be expressed in all seven stripes. B. It would be expressed in stripe 5 only. C. It would expand to cover a broad anterior region of the embryo. D. It would fail to express efficiently in the stripe 2 region. E. It would be expressed throughout the whole embryo. 25.

What would you expect to happen to the pattern of reporter expression in flies that lack

the gene encoding Bicoid? A. It would be expressed in all seven stripes. B. It would be expressed in stripe 5 only. C. It would expand to cover a broad anterior region of the embryo. D. It would fail to express efficiently in the stripe 2 region. E. It would be expressed throughout the whole embryo. 26.

In the Eve regulatory modules, the binding sites for Giant and Krüppel are usually close

to, or even overlapping with, those of Bicoid and Hunchback. This implies that Eve expression is regulated by … between the activators and regulators. A. Feedback regulation B. Cooperation C. Competition D. Feed-forward regulation

27. In the following schematic drawings showing four different network motifs in gene transcription circuits, indicate whether each of the motifs 1 to 4 corresponds to a feed-forward loop (E), flip-flop device (F), negative feedback loop (N), or positive feedback loop (P). Your answer would be a four-letter string composed of letters E, F, N, or P only, e.g. PNEF.

Answer: PENF Difficulty: 2 Section: Molecular Genetic Mechanisms that Create and Maintain Specialized Cell Types Feedback: The circuits 1 and 3 include indirect feedback loops (with positive and negative feedback, respectively). Circuit 4 has two interconnected flip-flop switches.

Reference: Feed-forward Loops (Questions 28-30) Consider the two feed-forward loops below containing three transcription regulators A, B, and C, where A receives the input signal and C generates the output. In the so-called coherent loop (left), A activates C both directly and indirectly, whereas in an incoherent loop (right), A activates C via one route and inactivates it via the other. Answer the following question(s) based on these network motifs.

Coherent

Incoherent

28. Considering the two particular feed-forward designs shown, would you expect the coherent (C) or the incoherent (I) loop to have a stable output and respond best to input signals that are above a certain threshold? Write down C or I as your answer. 29. Considering coherent and incoherent feed-forward loops, eight different designs are possible using three components and involving activation (positive) and inhibition (negative) regulation only. In all such loops, A would regulate both B and C, and B would regulate only C. How many of these designs constitute a coherent loop? Write down your answer as a number, e.g. 7. 30. Would you expect a coherent (C) or an incoherent (I) loop to generate the following response pattern? In this example, A is stimulated by the input, and the transcription of C is measured as the output. Write down C or I as your answer.

Signal

Input Output

0 Time

31. Indicate true (T) and false (F) statements below regarding DNA methylation in humans. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) Methylation of adenines is the most common DNA methylation in humans. ( ) Methylated cytosine can be accidentally deaminated to produce thymine, leading to a C-to-T transition. ( ) Cytosine methylation often occurs within a 5′-CG-3′ sequence. ( ) Shortly after fertilization, a genome-wide wave of demethylation takes place. 32.

Indicate true (T) and false (F) statements below regarding DNA methylation in humans.

Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) DNA methylation makes repression of gene expression “leakier.” ( ) DNA methylation at the promoter region is usually an indication of active transcription. ( ) CG islands are nonpermissive for RNA polymerase assembly. ( ) The dinucleotide 5′-GC-3′ is much more common than 5′-CG-3′ in the human genome. 33.

Indicate which of the following enzymatic activities corresponds to A, B, or C in the

following schematic drawing of DNA methylation events in the human genome. Your answer would be a three-letter string composed of letters A, B, and C only, e.g. CBA.

Genomic DNA after replication

Methyl group

A C

( ) DNA demethylating enzyme ( ) Maintenance methyl transferase enzyme ( ) De novo methyl transferase enzyme

34. Beckwith-Wiedemann syndrome in humans is characterized by “overgrowth” in childhood, sometimes leading to unusually large parts of the body. An imprinted gene cluster on chromosome 11 is associated with this disease. The cluster contains several genes including Igf2 and H19. Igf2 encodes an insulin-like growth factor that is maternally imprinted, i.e. the maternal copy is not expressed. However, the DNA methylation pattern of this locus is not different between the two homologous chromosomes. On the other hand, H19 is also imprinted and its methylation pattern does differ between the two parental chromosomes. H19 is transcribed into a noncoding RNA that appears to silence the transcription of the Igf2 gene in cis. Would you expect the H19 locus to be hypermethylated in the maternally inherited chromosome (M) or paternally inherited chromosome (P)? Write down M or P as your answer.

35. In the following pedigree, females and males are indicated by circles and squares, respectively, and the presence of a rare disease caused by a loss-of-function mutation in an imprinted autosomal gene is indicated by black color. Is the gene maternally imprinted (M; i.e.

not expressed from the maternally inherited chromosome) or paternally imprinted (P)? Write down M or P as your answer.

36.

Which of the following is true regarding genomic imprinting? A. It is an epigenetic phenomenon. B. It occurs in most animals. C. It always involves inactivation of genes through direct DNA methylation. D. It can “unmask” recessive alleles but cannot “mask” dominant ones. E. All of the above.

37. You have engineered the X chromosomes in female mice such that one X chromosome expresses green fluorescent protein when active, while the other expresses red fluorescent protein. You have used these mice to study cancer in females. You know that each tumor is a clonal cellular proliferation, meaning all of its proliferating cells are descendants of a single original cancer-causing cell. It follows that, unless X-chromosome inactivation is perturbed in tumors, … A. all tumor cells in one mouse should express the same fluorescent protein (either red or green), but tumor cells from different mice can show either red or green fluorescence. B. the cells in any tumor should all express the same fluorescent protein (either red or green), but independently derived tumors in the same mouse can show either green or red fluorescence. C. different cells within each tumor can express different fluorescent proteins, and the tumors would therefore show yellow fluorescence, but each cell shows either red or green fluorescence.

D. each cell can express both fluorescent proteins and would therefore emit yellow fluorescence, and the tumors would glow in yellow as well. E. different tumors would show red, yellow, or green fluorescence. 38. Fill in the blank in the following paragraph regarding X-chromosome inactivation. Do not use abbreviations. “The presence of two copies of the X chromosome in females but only one in males poses a potential problem for regulation of gene expression: everything else being equal, the higher copy number of the X-linked genes can lead to an imbalanced overproduction of their products in females compared to males. Organisms have evolved different … mechanisms to solve this problem. In mammals, for example, the almost complete inactivation of one of the X chromosomes provides the solution to the dosage problem.” 39. Two copies of the gene A exist in a diploid mammal. However, only one copy is expressed in every somatic cell. Different somatic cells in the body of each organism have inactivated one or the other allele in a seemingly random fashion, but when they divide, the daughter cells inherit the choice of the inactive allele faithfully. This is an example of … A. X-inactivation B. Genomic imprinting C. Loss of heterozygosity D. Monoallelic gene expression E. Alternative gene splicing 40. Indicate true (T) and false (F) statements below regarding riboswitches. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) A riboswitch permits transcription elongation only when bound to its small-molecule ligand. ( ) Riboswitches appear to be a modern evolutionary invention. ( ) Riboswitches are often located near the 3′ end of mRNAs, and therefore fold after the rest of the mRNA. ( ) Riboswitches are exclusively found in prokaryotes. 41. In the following schematic diagram of five pairs of alternative RNA splicing patterns, which pair depicts exon skipping only? In exon skipping, an exon is spliced out of the RNA

transcript along with its flanking introns. In the diagram, exons are colored dark blue and introns are colored yellow. Light blue indicates possible exons.

42. Imagine a gene encoding a pre-mRNA with twelve exons, ten of which can be alternatively spliced in vivo. Assuming that alternative splicing for this RNA only occurs through exon skipping, how many different proteins can possibly be made from this pre-mRNA as a result of alternative splicing? A. About 10 B. About 20 C. About 100 D. About 200 E. About 1000 43. The alternative splicing of a certain transcript can result in the production of two mRNA isoforms, one predominant in muscle cells and the other in neurons. The gene contains an exon that is skipped in muscle cells but retained in neurons. You create a mutant version of the gene in which the 3′ splice site near this exon is deleted. However, when you introduce this into a culture of neural cells, an even longer pre-mRNA is produced, consistent with the activation of a

secondary splice site located near the deleted one. Is the secondary splice site within the exon (E) or its neighboring intron (I)? Write down E or I as your answer. 44. Stimulated B lymphocytes switch from the synthesis of membrane-bound to secreted antibody molecules by increasing the concentration of a subunit of the trimeric CstF complex that cleaves and polyadenylates mRNAs. How does this up-regulation of CstF bring about the production of soluble antibodies? A. It favors a strong polyadenylation site in the immunoglobulin primary transcript, creating a longer antibody molecule that is secreted. B. It activates a weak polyadenylation site in the immunoglobulin primary transcript and prevents splicing, creating a shorter molecule that is secreted. C. It favors a strong polyadenylation site in the immunoglobulin primary transcript, creating a shorter antibody molecule that is secreted. D. It activates a weak polyadenylation site in the immunoglobulin primary transcript, creating a longer antibody molecule that is secreted. E. It favors a strong polyadenylation site in the immunoglobulin primary transcript, aborting translation and creating a shorter antibody molecule that is secreted. 45.

Indicate true (T) and false (F) statements below regarding RNA editing. Your answer

would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) RNA editing in animal cells mainly takes place through adenine or cytosine deamination. ( ) RNA editing can change the pattern of pre-mRNA splicing. ( ) RNA editing in coding regions can result in changes in the protein sequence. ( ) Retroviruses such as HIV encode RNA-editing components to confront the host defense mechanisms. 46.

You have transfected HeLa cells with a gene encoding the human immunodeficiency

virus (HIV) protein Rev and have induced the expression of the protein. You incubate the cell culture in the presence or absence of leptomycin B and later measure the localization of Rev inside the cells by immunofluorescence microscopy. Leptomycin B specifically inhibits the cellular Crm1 protein, a nuclear transport receptor that is essential for the normal function of Rev in the HIV life cycle. Your results are tabulated below. Which condition (1 or 2) do you think corresponds to the presence of leptomycin B?

Condition

Cytoplasmic Rev (%)

Nuclear Rev (%)

1 2

30 3

70 97

47. When unfolded proteins accumulate in the lumen of the endoplasmic reticulum due to inefficient folding conditions, a membrane-bound protein kinase is activated and phosphorylates a subunit of the translation initiation factor eIF2. Indicate which of the following events would (Y) or would not (N) occur as a result. Your answer would be a four-letter string composed of letters Y and N only, e.g. YYYY. ( ) eIF2 binds tightly to the ribosome. ( ) Translation initiation is stimulated. ( ) The GDP-bound form of eIF2 accumulates. ( ) eIF2B is activated. 48. A researcher studying the human immunodeficiency virus (HIV) has infected human T cells with wild-type or Rev-deficient viruses. She extracts cytoplasmic RNA from the cells and separates each isolated RNA mixture by agarose-gel electrophoresis. She then uses northern analysis using HIV-specific probes to identify those bands that contain viral RNA. The results are presented in the following schematic drawing. Which lane (1 or 2) corresponds to cells infected with the Rev-deficient virus? What is the size of the full-length HIV genome?

Apparent length (×1000 nt) 9 4 2

A. Lane 1; about 9000 nucleotides B. Lane 1; about 4000 nucleotides C. Lane 1; about 2000 nucleotides D. Lane 2; about 9000 nucleotides E. Lane 2; about 2000 nucleotides

49. Some viruses encode a protease that cleaves the translation initiation factor eIF4G, rendering it unable to bind to eIF4E. What is the consequence of this cleavage? A. It shuts down most of the cellular translation machinery, which causes the release of virus particles. B. It favors viral protein synthesis because IRES-dependent translation initiation is inhibited. C. It shuts down most host-cell protein synthesis and diverts the translation machinery to IRES-dependent initiation, thus favoring viral protein synthesis. D. It favors viral protein synthesis by shutting down translation from uORFs. E. It shuts down IRES-dependent translation, forcing the virus into latency. 50. Gcn4 is a protein kinase that can phosphorylate and inactivate the initiation factor eIF2. The Gcn4 mRNA contains several short upstream open reading frames (uORFs) that negatively regulate its translation initiation. The phosphorylation of eIF2 by Gcn4, therefore, can control Gcn4 expression … A. in a negative feedback loop. B. in a positive feedback loop. C. in a feed-forward loop. 51. Indicate true (T) and false (F) statements below regarding mRNA stability and degradation. Your answer would be a four-letter string composed of letters T and F only, e.g. TTFT. ( ) Any factor that affects the translation efficiency of an mRNA tends to have the opposite effect on its degradation. ( ) Decapping of mRNA normally occurs after poly-A tail shortening removes the polyA tail and starts chewing into the 3′-UTR. ( ) Most mRNA decay in the cells generally proceeds via endonucleolytic cleavage. ( ) As a general rule, eukaryotic mRNAs have a shorter half-life compared to bacterial mRNAs. 52.

Under iron-starvation conditions in the cell, A. cytosolic aconitase binds to the 3′ UTR of the mRNA encoding the transferrin receptor. B. cytosolic aconitase binds to the 3′ UTR of the mRNA encoding ferritin.

C. cytosolic aconitase blocks translation of the mRNA encoding the transferrin receptor. D. the mRNA encoding the transferrin receptor is cleaved by an endonuclease. E. All of the above. 53. Indicate true (T) and false (F) statements below regarding P-bodies. Your answer would be a four-letter string composed of letters T and F only, e.g. TTFT. ( ) P-bodies are membrane-enclosed organelles scattered throughout the cytoplasm. ( ) mRNAs that are to be degraded are transferred from P-bodies to stress granules containing decapping and exonuclease enzymes. ( ) When translation initiation is blocked, stress granules typically shrink. ( ) mRNAs associated with stress granules are primed for translation. 54. Which of the following classes of noncoding RNAs is NOT directly involved in RNA interference? A. miRNA B. snoRNA C. piRNA D. siRNA 55. Indicate whether each of the following descriptions better matches miRNAs (M) or siRNAs (S) in animal cells. Your answer would be a four-letter string composed of letters M and S only, e.g. MMMM. ( ) They usually cleave their target RNAs using the slicing activity of the Argonautes. ( ) They typically show perfect complementarity with their target RNA. ( ) They can bind to the RITS as well as the RISC complex. ( ) They seem to be the more ancient form of small noncoding RNAs. 56.

Which strand in an miRNA precursor will serve as the guide strand in the RNA-induced

silencing complex (RISC)? Analysis of miRNA sequences has revealed an asymmetry in the stability of the double-stranded RNA precursor. The strand showing lower thermodynamic stability near its 5′ end (nucleotides 2 to 6 in the mature guide strand) is normally selected as the guide strand, and the other strand is usually degraded. In the following miRNA precursor, which strand (1 or 2) do you think will be incorporated into the active RISC? Write down 1 or 2 as your answer.

1 2

5′ CAAAGUGCUUACAGCGCAGGUAG 3′ 3′ GAUGUUCACGGAAGUGGCGUC-A 5′

57. The schematic diagram below shows the processing of a class of small RNAs that are involved in RNA interference (RNAi). Which of the following is true regarding these RNAs?

58.

What is the function of RNA-dependent RNA polymerases in RNAi? A. They prevent the spread of the RNAi pathway by replicating the target RNAs. B. They help amplify the RNAi response by replicating the target RNAs. C. They produce additional copies of the siRNAs to ensure that the RNAi response is sustained and spread. D. They are viral proteins that prevent the spread of RNAi by preferentially replicating siRNA sponges.

59. A certain region of a mammalian genome is transcribed at low but sustained levels using two flanking promoters. The RNA products are annealed together, processed, and used to recruit the RITS complex to this region, as well as chromatin modifying enzymes, RNA-dependent RNA polymerase, and a Dicer enzyme. Indicate whether this pathway is more closely related to that seen in miRNA (M), piRNA (P), or siRNA (S) interference pathways. Write down M, P, or S as your answer.

60. In many animals, siRNAs can only function within the cell in which the siRNA is introduced. In contrast, the worm Caenorhabditis elegans can be fed with bacteria that synthesize double-stranded RNAs, and the RNAi spreads throughout the animal. A genetic screen to identify genes involved in systemic RNAi led to the discovery of Sid-1, a transmembrane protein expressed in most tissues in the adult worm. Loss-of-function mutations in the gene encoding Sid-1 restrict the RNAi activity to the cells surrounding the digestive tract after siRNA feeding. Ectopic expression of Sid-1 in Drosophila melanogaster cells that normally lack systemic RNAi and are unable to take up siRNAs from the medium, enables these cells to passively take up siRNA . These findings suggest that … A. Sid-1 is necessary and sufficient for systemic RNAi and siRNA uptake, respectively. B. Sid-1 is necessary but not sufficient for systemic RNAi and siRNA uptake, respectively. C. Sid-1 is not necessary but is sufficient for systemic RNAi and siRNA uptake, respectively. D. Sid-1 is not necessary and not sufficient for systemic RNAi and siRNA uptake, respectively. 61.

Comparing the bacterial CRISPR system and the eukaryotic RNAi mediated by siRNAs,

indicate whether each of the following CRISPR components is most analogous to Argonaute (A), siRNA (S), or target mRNA (T) in the RNAi pathway. Your answer would be a three-letter string composed of letter A, S, and T only. ( ) Cas ( ) Viral DNA ( ) crRNA 62. Consider the descriptions below for three different lncRNAs, and determine whether they function in cis (C) or in trans (T). Your answer would be a three-letter string composed of letters C and T only, e.g. CCT. ( ) The tsiX RNA is transcribed from the X-inactivation center in the X chromosome that will not be inactivated. It is antisense with respect to Xist, repressing its transcription and function. ( ) The ANRIL lncRNA is an antisense transcript in a gene cluster expressing inhibitors of cell-cycle proteins. It silences the entire cluster by recruiting heterochromatin proteins of the Polycomb group.

( ) The PTENP1 lncRNA is similar to an mRNA that encodes the protein PTEN, and contains many of the miRNA response elements present in PTEN mRNA; it therefore protects the PTEN mRNA from miRNA-mediated silencing by acting as a decoy.

Answers: 1. Answer: TFTT Difficulty: 1 Section: An Overview of Gene Control Feedback: The changes in gene expression that underlie the development of multicellular organisms do not generally involve changes in the DNA sequence of the genome. At any one time, a typical human cell expresses 30–60% of its approximately 30,000 genes at some level. But even those genes that are expressed in all cell types usually vary in their level of expression from one cell type to the next. Under the right conditions, differentiated plant cells can regenerate an entire adult plant. 2. Answer: D Difficulty: 3 Section: An Overview of Gene Control Feedback: A housekeeping gene is expected to be expressed in nearly all cells, including brain and liver cells. This is the case for gene Y, for which the mRNA levels are significant in both cell types, according to the RNA-seq data. The regions with higher numbers of reads in the RNA-seq results correspond to exons. 3. Answer: B Difficulty: 2 Section: An Overview of Gene Control Feedback: Since the mRNA levels are similar in both cell types, difference in protein activity can be attributed to differences in translation or in post-translational control. 4. Answer: A Difficulty: 2 Section: An Overview of Gene Control Feedback: Ubiquitously expressed gene products include ribosomal proteins, DNA and RNA polymerases, and DNA repair enzymes, among others. 5. Answer: E Difficulty: 2 Section: Control of Transcription by Sequence-specific DNA-binding Proteins Feedback: The expression of a gene can be determined by its cis-regulatory sequences and transcription regulators that function in trans. 6. Answer: B Difficulty: 2 Section: Control of Transcription by Sequence-specific DNA-binding Proteins Feedback: Nearly all transcription regulators make the majority of their contacts with the major groove (1 and 4 in the drawings) because the major groove is wider and displays more molecular features than does the minor groove (2 and 3). 7. Answer: D Difficulty: 2

Section: Control of Transcription by Sequence-specific DNA-binding Proteins Feedback: In the minor groove, precise recognition of A-T versus T-A (or C-G versus GC) is not readily feasible. However, an A-T can still be distinguished from C-G, for example. 8. Answer: B Difficulty: 1 Section: Control of Transcription by Sequence-specific DNA-binding Proteins Feedback: DNA-binding proteins typically recognize specific sequences that are 5 to 10 nucleotide pairs in length. This involves forming 20 or so weak interactions at the interface, including hydrogen-bonding, ionic bonding, and hydrophobic interactions. The sum of all protein–DNA interactions can result in high affinity and specificity. 9. Answer: E Difficulty: 1 Section: Control of Transcription by Sequence-specific DNA-binding Proteins Feedback: All of the examples presented in A to D use α helices to interact with DNA. A limited number of DNA-binding motifs use β sheets to recognize DNA bases. 10. Answer: T Difficulty: 3 Section: Control of Transcription by Sequence-specific DNA-binding Proteins Feedback: Homodimeric transcription regulators typically have a symmetrical arrangement which is reflected in their target DNA sequence in the form of an inverted repeat. The sequence logo shown is not in agreement with an inverted repeat, or even a direct repeat; the DNA is therefore expected to be recognized by a heterodimer. 11. Answer: E Difficulty: 3 Section: Control of Transcription by Sequence-specific DNA-binding Proteins Feedback: The expected number of target 8-mer sequences is calculated as: (6 × 109 8-mer/genome) × (2 orientations per double strand) / (48 8-mers/target sequence) = ~200,000 target sequences/genome. Similarly, the expected number of target 16-mer sequences is calculated as: (6 × 109 16-mer/genome) × (2 orientations per double strand) / (416 16-mers/target sequence) = ~3 target sequences/genome. 12. Answer: D Difficulty: 1 Section: Control of Transcription by Sequence-specific DNA-binding Proteins Feedback: Cooperative binding creates a sigmoidal binding curve with two main states: at low protein concentrations, the protein is mostly a monomer and the binding site on DNA is mostly unoccupied; at high protein concentrations, the great majority of the DNA sites are occupied by the protein dimer. 13. Answer: TFTF

Difficulty: 1 Section: Control of Transcription by Sequence-specific DNA-binding Proteins Feedback: Nucleosome packaging of DNA and dimerization of transcription regulators promote cooperative DNA binding by these proteins. Some proteins that bind to DNA can significantly bend the DNA molecule. About 10% of protein-coding genes in most cells encode transcription regulators, making them one of the largest classes of proteins in the cell. 14. Answer: A Difficulty: 1 Section: Transcription Regulators Switch Genes On and Off Feedback: When tryptophan is present, it binds to the repressor which then binds to the operator to turn off the operon. 15. Answer: B Difficulty: 1 Section: Transcription Regulators Switch Genes On and Off Feedback: The Lac operon is fully turned on when glucose is absent AND lactose is present. 16. Answer: B Difficulty: 2 Section: Transcription Regulators Switch Genes On and Off Feedback: DNA looping can occur during bacterial gene regulation, where the intervening DNA acts as a tether to enormously increase the probability that the proteins bound near the promoter and those bound to the cis-regulatory sequences will collide with each other. This effect requires that both sequences be on the same DNA molecule, or be somehow linked physically. 17. Answer: B Difficulty: 2 Section: Transcription Regulators Switch Genes On and Off Feedback: Variation in transcription rates across the genome is much greater in eukaryotic cells (about one million-fold) compared to prokaryotic cells (about one thousand-fold). In eukaryotes, genes are often transcribed and regulated individually, and DNA looping occurs in the regulation of nearly every gene. In these cells, transcription regulators often assemble in groups and do not directly contact RNA polymerases. The default state of most eukaryotic DNA packaged into nucleosomes is off. 18. Answer: E Difficulty: 1 Section: Transcription Regulators Switch Genes On and Off Feedback: Eukaryotic transcription activator proteins can direct local alterations in chromatin structure. This can be achieved through covalent histone modifications, nucleosome remodeling or removal, and histone replacement.

19. Answer: DBAC Difficulty: 2 Section: Transcription Regulators Switch Genes On and Off Feedback: Please refer to Figure 7–17. 20. Answer: B Difficulty: 3 Section: Transcription Regulators Switch Genes On and Off Feedback: The methylation mark would be found in the body of actively transcribed genes (but not near their promoter, which requires a more open nucleosome packaging). 21. Answer: D Difficulty: 2 Section: Transcription Regulators Switch Genes On and Off Feedback: The combined effect of two transcription activators binding to the same regulatory sequence in DNA is synergistic, not simply additive. 22. Answer: C Difficulty: 3 Section: Transcription Regulators Switch Genes On and Off Feedback: Gene D is not protected from heterochromatin expansion by any barrier sequence, while genes A and B are repressed by the cis-regulatory element bound to a repressor. Gene C, however, is not under these repressive effects, thanks to its flanking insulator elements and barrier sequences. 23. Answer: B Difficulty: 1 Refer to: Even-skipped Regulatory Module Section: Molecular Genetic Mechanisms that Create and Maintain Specialized Cell Types Feedback: This module is normally activated in the second stripe, where inhibition is weak and activation is strong. 24. Answer: C Difficulty: 2 Section: Molecular Genetic Mechanisms that Create and Maintain Specialized Cell Types Feedback: As expected, lack of Giant or Krüppel results in a broad expression pattern for the second stripe. 25. Answer: D Difficulty: 2 Section: Molecular Genetic Mechanisms that Create and Maintain Specialized Cell Types Feedback: As expected, lack of Bicoid or Hunchback results in an inefficient expression of the second stripe. 26. Answer: C Difficulty: 2 Section: Molecular Genetic Mechanisms that Create and Maintain Specialized Cell Types

Feedback: This helps sharpen the boundaries flanking the Eve stripes. 27. Answer: PENF Difficulty: 2 Section: Molecular Genetic Mechanisms that Create and Maintain Specialized Cell Types Feedback: The circuits 1 and 3 include indirect feedback loops (with positive and negative feedback, respectively). Circuit 4 has two interconnected flip-flop switches. 28. Answer: C Difficulty: 3 Section: Molecular Genetic Mechanisms that Create and Maintain Specialized Cell Types Feedback: This also depends on the particular parameters of the system. However, coherent feed-forward motifs are generally more stable and more robust to perturbations. 29. Answer: 4 Difficulty: 2 Section: Molecular Genetic Mechanisms that Create and Maintain Specialized Cell Types Feedback: Half of the possible designs yield coherent loops. 30. Answer: I Difficulty: 3 Section: Molecular Genetic Mechanisms that Create and Maintain Specialized Cell Types Feedback: The two-phase response reflects the incoherent nature of this feed-forward motif. 31. Answer: FTTT Difficulty: 1 Section: Mechanisms that Reinforce Cell Memory in Plants and Animals Feedback: The most common DNA methylation in vertebrate DNA occurs on cytosine (C) nucleotides largely in the sequence 5′-CG-3′. The accidental deamination of the methylated cytosine can then potentially lead to a mutation. Shortly after fertilization in mammals, a genome-wide wave of cytosine demethylation takes place and the vast majority of methyl groups are lost from the DNA. 32. Answer: FFFT Difficulty: 2 Section: Mechanisms that Reinforce Cell Memory in Plants and Animals Feedback: DNA methylation helps to make repression of gene expression less leaky. RNA polymerase is often found bound to promoters within unmethylated CG islands, even when the associated gene is not being actively transcribed. Except in CG islands, 5′CG-3′ dinucleotides are rare in vertebrate genomes. 33. Answer: CAB Difficulty: 2 Section: Mechanisms that Reinforce Cell Memory in Plants and Animals

Feedback: Arrow A represents the maintenance of DNA methylation after DNA replication, while arrows B and C represent de novo DNA methylation and DNA demethylation, respectively. 34. Answer: P Difficulty: 3 Section: Mechanisms that Reinforce Cell Memory in Plants and Animals Feedback: On the maternally inherited chromosome, H19 is transcribed and silences Igf2. This does not occur on the paternal chromosome. As a result, Igf2 is maternally imprinted with no maternal-specific methylation. 35. Answer: M Difficulty: 3 Section: Mechanisms that Reinforce Cell Memory in Plants and Animals Feedback: If a disease-causing gene is maternally imprinted, carrier females silently pass it to the next generation; their sons may have affected children, but their daughters would not. 36. Answer: A Difficulty: 2 Section: Mechanisms that Reinforce Cell Memory in Plants and Animals Feedback: Genomic imprinting, observed in placental mammals and some flowering plants, is an epigenetic form of inheritance. It is mediated through DNA methylation in mammals, but it does not always involve gene inactivation by direct methylation. It can unmask recessive alleles or mask dominant alleles. 37. Answer: B Difficulty: 2 Section: Mechanisms that Reinforce Cell Memory in Plants and Animals Feedback: Since different cells in the female body have randomly inactivated one or the other X chromosome, tumors derived from them will inherit the inactivation pattern and can therefore express one or the other fluorescent protein. However, all cells within each tumor are expected to express the same fluorescent protein. 38. Answer: dosage compensation Difficulty: 2 Section: Mechanisms that Reinforce Cell Memory in Plants and Animals Feedback: Dosage compensation allows mammals to equalize the dosage of Xchromosome gene products between males and females. 39. Answer: D Difficulty: 1 Section: Mechanisms that Reinforce Cell Memory in Plants and Animals Feedback: Monoallelic expression can result from genomic imprinting, but not all examples of monoallelic expression are due to imprinting. 40. Answer: FFFF

Difficulty: 2 Section: Post-transcriptional Controls Feedback: Riboswitches are short RNA sequences that can change conformation upon ligand binding and regulate transcription. These ancient regulators are common in bacteria as well as in eukaryotes. They are often found near the 5′ end of mRNAs and fold co-transcriptionally. 41. Answer: A Difficulty: 2 Section: Post-transcriptional Controls Feedback: The alternative splicing event shown in A is exon skipping: the third exon is missing (together with its flanking introns) in one of the two mRNA variants. Shown in B is the retention of a single intron (together with its flanking exons) in one of the mRNA variants. 42. Answer: E Difficulty: 2 Section: Post-transcriptional Controls Feedback: Each alternative exon is either skipped or retained, possibly creating 210 (= ~1000) different mature mRNAs. 43. Answer: I Difficulty: 2 Section: Post-transcriptional Controls Feedback: The cryptic splice site resides within the upstream intron, resulting in the production of a longer mRNA when this splice site is chosen. 44. Answer: B Difficulty: 1 Section: Post-transcriptional Controls Feedback: The suboptimal cleavage/poly-A addition site in the transcript is encountered first but is usually skipped in unstimulated B lymphocytes, leading to production of a longer antibody molecule that is anchored in the plasma membrane. When activated to produce antibodies, the B lymphocyte increases its CstF concentration; as a result, cleavage now occurs at the suboptimal site, and the shorter transcript is produced. 45. Answer: TTTF Difficulty: 1 Section: Post-transcriptional Controls Feedback: Two principal types of mRNA editing in animals are the deamination of adenine to produce inosine and, less frequently, the deamination of cytosine to produce uracil. Both types can have profound effects on the meaning of the RNA message, including changing the pattern of pre-mRNA splicing or changing the amino acid sequence of the protein coded by the mRNA. RNA from some retroviruses, including the

human immunodeficiency virus (HIV), is extensively edited by the host cell, presumably as part of a defense mechanism to hold the virus in check. 46. Answer: 2 Difficulty: 2 Section: Post-transcriptional Controls Feedback: By inhibiting Crm1, leptomycin B inhibits the nuclear export of Rev, resulting in its nuclear accumulation. 47. Answer: NNYN Difficulty: 1 Section: Post-transcriptional Controls Feedback: Under stress conditions, eIF2 phosphorylation prevents its release from the guanine nucleotide exchange factor eIF2B, sequestering the latter in a nonfunctional form, and consequently preventing further eIF2 guanine nucleotide exchange and activation. As a result, translation initiation is largely inhibited. 48. Answer: D Difficulty: 1 Section: Post-transcriptional Controls Feedback: In the absence of a functional Rev protein (lane 2), only the fully spliced viral RNAs are exported from the nucleus to the cytosol. Rev is required for the export of the full-length viral RNA. 49. Answer: C Difficulty: 1 Section: Post-transcriptional Controls Feedback: Translation of internal ribosome entry site (IRES)-containing mRNAs does not rely on the common cap-dependent initiation mechanism. 50. Answer: B Difficulty: 1 Section: Post-transcriptional Controls Feedback: Since phosphorylated eIF2 promotes the skipping of the uORFs in favor of the translation of the main Gcn4 ORF, the kinase activity of Gcn4 provides a possible positive feedback control mechanism. 51. Answer: TFFF Difficulty: 1 Section: Post-transcriptional Controls Feedback: Eukaryotic mRNAs are generally more stable than bacterial mRNA. Eukaryotic mRNA decay is mediated by 5′ exonucleases (after decapping) or 3′ exonucleases, usually after poly-A tail shortening leaves only about 25 nucleotides of the tail. Since the poly-A shortening and the translation machinery compete for the same components, factors that affect one of them also affect the other one in an opposite direction. Some mRNAs are destroyed by specific endonucleases.

52. Answer: A Difficulty: 1 Section: Post-transcriptional Controls Feedback: The iron-sensitive RNA-binding protein aconitase can bind to the 3′ untranslated region (UTR) of the transferrin receptor mRNA and increase production of the receptor by blocking endonucleolytic cleavage of the mRNA. The addition of iron to cells results in the release of aconitase from the mRNA and thus decreases the stability of the mRNA. 53. Answer: FFFT Difficulty: 1 Section: Post-transcriptional Controls Feedback: Even though they somehow function as “organelles,” P-bodies are not membrane-enclosed. Many mRNAs are degraded in P-bodies, but some mRNAs move from P-bodies to stress granules, which contain mRNA-binding proteins that prime the mRNAs for translation. Translation itself, however, does not occur in stress granules. When translation initiation is blocked, stress granules enlarge as more and more nontranslated mRNAs are stored in them. 54. Answer: B Difficulty: 1 Section: Regulation of Gene Expression by Noncoding RNAs Feedback: The small nucleolar RNAs do not function in RNA interference. 55. Answer: SSSS Difficulty: 2 Section: Regulation of Gene Expression by Noncoding RNAs Feedback: The small interfering RNAs (siRNAs) appear to be the most ancient form of RNA interference. Having an exact match to their target RNA molecules, they direct the cleavage of the target. siRNAs can also control transcription of target genes by binding to the RNA-induced transcriptional silencing (RITS) complex. 56. Answer: 1 Difficulty: 2 Section: Regulation of Gene Expression by Noncoding RNAs Feedback: Note the presence of mismatches and even a bulge along the duplex. Strand 1 contains mostly A-U base pairs, as well as a mismatch, near its 5′ end (left). Strand 2 contains mostly C-G pairs near its 5′ end (right). 57. Answer: C Difficulty: 2 Section: Regulation of Gene Expression by Noncoding RNAs Feedback: These are miRNAs 58. Answer: C Difficulty: 1

Section: Regulation of Gene Expression by Noncoding RNAs Feedback: The RNA interference (RNAi) response can be amplified in some organisms using RNA-dependent RNA polymerases. In these cases, RNA-dependent RNA polymerases use siRNAs as primers to produce additional copies of double-stranded RNAs that are then cleaved into siRNAs. 59. Answer: S Difficulty: 1 Section: Regulation of Gene Expression by Noncoding RNAs Feedback: RNA interference can direct heterochromatin formation, altering the packaging of nuclear chromatin and expression of genes. 60. Answer: A Difficulty: 2 Section: Regulation of Gene Expression by Noncoding RNAs Feedback: According to these results, Sid-1 is both necessary for systemic RNAi in the nematode and sufficient for siRNA uptake in fly cells: without it, systemic RNAi in C. elegans is lost; additionally, when ectopically expressed, it confers the ability to take up siRNA in D. melanogaster cells. 61. Answer: ATS Difficulty: 1 Section: Regulation of Gene Expression by Noncoding RNAs Feedback: Despite lacking homology, these two pathways share similarities: Cas can be guided by crRNAs to target invading viral DNA; similarly, an Argonaute or Piwi protein uses siRNAs to seek and cleave target mRNAs. 62. Answer: CCT Difficulty: 2 Section: Regulation of Gene Expression by Noncoding RNAs Feedback: The first two lncRNAs act in cis; i.e. they affect only the chromosome from which they are transcribed. The third example, however, acts in trans and regulates the function of other genes after it diffuses from its site of synthesis.

1. You have used fluorescence-activated cell sorting to separate blood T lymphocytes from a healthy person and a patient infected with a virus. To sort the cells, you used antibodies against two cell-surface receptors, CD4 and CD8, each conjugated to a different fluorescent dye. Based on the results below, which cell type do you think is depleted by the viral infection: CD4+ helper T cells (H) or CD8+ cytotoxic T cells (C)? In these dot plots, each cell is represented by a dot whose coordinates are the intensities of CD4 and CD8 signals. Write down H or C as your answer. Virus-infected patient

CD8 signal

Healthy control

CD4 signal

2. Fill in the blank in the following paragraph regarding plant cell cultures. Do not use abbreviations. “Plant cells can be easily rendered totipotent to give rise to all plant tissues on their own. This can be done by culturing a plant tissue in sterile nutrient media to produce an amorphous cell mass called a …”

3. Indicate whether each of the following descriptions applies to transformed cell lines (T), nontransformed cell lines (N), or both (B). Your answer would be a four-letter string composed of letters T, N, and B only, e.g. BBNN. ( ) They can be thawed and recultured after long-term storage in liquid nitrogen. ( ) They often grow only when attached to a surface, such as that of a culture dish. ( ) They proliferate to very high densities. ( ) They usually cause tumors when injected into immunocompromised animals. 4.

What are the advantages of monoclonal antibodies over antisera? A. They can be produced in higher quantities. B. They can be produced with higher purity. C. They have a more uniform specificity. D. They can be made against molecules that constitute only a minor component of a complex mixture. E. All of the above.

5. How are antibody-producing hybridoma cell lines immortalized to provide an unlimited source of monoclonal antibodies? A. By ectopic expression of telomerase B. By transformation with a retrovirus C. By cell fusion D. By overproduction of introduced oncogenes E. By irradiation 6. To create cellular factories for monoclonal antibody production, cells derived from an immortalized B lymphocyte cell line (M) are cultured together with antibody-secreting B lymphocytes (B) on a so-called HAT medium. The medium contains an inhibitor of DHFR, an enzyme essential for de novo nucleotide biosynthesis. Additionally, it contains substrates for the only alternative nucleotide biosynthetic pathway called the salvage pathway. The enzyme HGPRT is essential for the salvage pathway. One of the cells mentioned above lacks the gene encoding one of the enzymes mentioned. Considering the role of the HAT medium in selecting for hybrid cells that are both immortalized and secrete antibodies, which cell (B or M) do you think is missing an enzyme? Which enzyme is missing in this cell? A. M – DHFR B. M – HGPRT C. B – DHFR

D. B – HGPRT 7. Sort the following cellular components to reflect the centrifugal force required to sediment them by preparative ultracentrifugation, from low to high force required. Your answer would be a four-letter string composed of letters R, M, L, and N, e.g. LRNM. (L) Lysosomes (M) Microsomes (N) Nuclei (R) Ribosomes (free cytosolic) Answer: NLMR

Difficulty: 2

Section: Purifying Proteins

( ) It uses a shallow sucrose gradient to stabilize the sedimentation bands against convection. ( ) In this method, subcellular components may either move up or down to form discrete bands. 10. For the two following graphs, the same purified protein has been subjected to velocity sedimentation and gel filtration, as indicated. Two absorption peaks are visible in each graph, corresponding to the monomeric and dimeric forms of the protein. Which peaks correspond to the protein dimer? Velocity sedimentation

A280 (relative protein amount)

Top

Bottom

Fractions

Gel filtration 3

A. 1 and 3 B. 1 and 4 C. 2 and 3 D. 2 and 4

Fractions

100

11. In purifying proteins by column chromatography, elution can be performed in different ways, depending on the type of matrix. For example, bound proteins can be eluted by gradually increasing or decreasing salt concentration in the solution applied to the column. For which of the following matrices do you think increasing and decreasing salt concentration, respectively, are used for elution? A. Ion-exchange; hydrophobic B. Affinity; ion-exchange C. Hydrophobic; affinity D. Hydrophobic; ion-exchange E. Ion-exchange; gel-filtration 12. Which of the following chromatography methods resembles immunoprecipitation in the purification of proteins from mixtures? A. Ion-exchange chromatography B. Gel-filtration chromatography C. Hydrophobic interaction chromatography D. Affinity chromatography 13.

How is tandem affinity purification distinct from other purification schemes such as

simple affinity or ion-exchange chromatography? A. In its use of antibodies B. In its use of proteases C. In its use of glutathione S-transferase D. In its use of viral peptides E. In its use of limited washing steps 14.

In SDS-PAGE of proteins, … A. the proteins are unfolded while being separated. B. the proteins are separated by size, mostly independent of their native charge. C. large proteins move more slowly. D. an ionic detergent is used. E. All of the above.

15. Migration of proteins during separation by SDS polyacrylamide-gel electrophoresis (SDS-PAGE) is generally related to the logarithm of their molecular weight, as depicted in the qualitative graph below. If a protein mixture containing proteins of evenly distributed molecular

Relative migration rate

weights (e.g. 10, 20, 30, 40, etc.) is separated by SDS-PAGE, which of the following staining patterns would you expect to observe?

Log (MW)

16. In isoelectric focusing, is the low-pH end close to the anode (A; positive electrode) or cathode (C)? Would you use octylglucoside (O; a nonionic detergent) or SDS (S) for separation of proteins by this technique? Would a highly basic protein be near the high-pH (H) or low-pH (L) end of the strip at the end of the experiment? Write down your answer as a three-letter string using the letters in the parentheses above, e.g. ASH. 17. Three subunits of a protein complex have been resolved in a two-dimensional polyacrylamide gel, as shown in the schematic drawing below. Indicate whether each of the

following better describes the a, b, or c subunit. Your answer would be a three-letter string composed of letters a, b, and c only, e.g. cca.

SDS migration

Low pH

High pH c

pH gradient

( ) It has the smallest apparent molecular weight. ( ) It has the highest negative charge. ( ) It has the lowest isoelectric point. 18. You have discovered a mutant protein that forms promiscuous disulfide bonds with many other cellular proteins in the endoplasmic reticulum. Using an antibody against the protein, you immunoprecipitate proteins that interact with it. You then separate these proteins using a special type of two-dimensional polyacrylamide-gel electrophoresis (PAGE): in this method, separation in the first dimension is carried out by SDS-PAGE in the absence of β-mercaptoethanol, while separation in the second dimension is carried out by SDS-PAGE in the presence of βmercaptoethanol. The result is shown in the following schematic drawing. Apparent molecular weights are indicated on the left. What do you think is the molecular weight of the mutant protein? What is the molecular weight represented by the question mark?

First dimension

15K

Second dimension

150K

A. 15K, 45K B. 30K, 45K C. 15K, 30K D. 30K, 30K E. 30K, 60K 19.

Where do individual peptides get fragmented in a tandem mass spectrometry (MS/MS)

set-up? A. Electrospray ion source B. Mass filter C. Inert gas chamber D. Mass analyzer E. Detector 20. Which mass spectrometry instrument is most suitable for determining all the proteins present in an organelle? A. MALDI-TOF B. MS/MS C. LC-MS/MS D. GC-MS 21. You performed a co-immunoprecipitation experiment to identify cellular proteins associated with protein X. For the experiment, you used antibodies against protein X that were

immobilized on beads. You incubated these beads with cell lysates from either wild-type cells or cells lacking the gene encoding protein Y. After washing and eluting the bound protein off the beads, you separated each bound and unbound fraction by SDS-PAGE. Finally, you performed immunoblotting using antibodies against either protein Y or protein Z. The results of the Western blots are presented in the following schematic drawings. According to these results, do you think protein Y enhances (E) or inhibits (I) binding of protein X to protein Z? Write down E or I as your answer.

Unbound

Bound

Y-null

Bound

Wild type

Bound

Y-null

Unbound

Fraction:

Protein Z

Wild type

Bound

Lysate:

Protein Y

Unbound

Western antibody:

22. The binding of protein A to two other proteins, B and C, has been studied with fluorescence anisotropy, and the results of the measurements have been plotted in the following schematic diagram. Given equal kon rate constants, which complex do you think has a longer half-life? What is the value of Ka (the association constant) for this complex?

Anisotropy (mP)

200 A-B A-C 150 -

100 0

Concentration of A (µM)

A. A-B; 105 L/mol B. A-B; 10–5 L/mol C. A-C; 1.5 × 105 L/mol D. A-C; 6.7 × 104 L/mol E. A-C; 6.7 × 10–4 L/mol 23. Indicate whether each of the following descriptions better applies to the determination of atomic structure of proteins by x-ray diffraction (X) or by nuclear magnetic resonance (NMR) spectroscopy (N). Your answer would be a five-letter string composed of letters X and N only, e.g. NNXNN. ( ) It is more suitable for complexes larger than about 100 kD. ( ) It requires samples to crystallize. ( ) It can better reveal dynamics of molecular motions. ( ) It involves the calculation of electron-density maps. ( ) It is used to determine the solution structure of proteins. 24. In an unfolded (random coil) protein, amino acid residues are exposed to the solvent and share more or less the same environment, whereas each residue in a folded protein has its own unique neighborhood. This fact can be exploited in using NMR to study protein folding. The two schematic diagrams below represent two-dimensional NMR spectra for the same protein in either its folded (native) or unfolded state. The chemical shifts, which depend on the local neighborhood of each atom, are plotted in these diagrams. Which diagram (A or B) do you think corresponds to the folded protein? Write down A or B as your answer.

25. You have carried out a genetic screen in a species of fish, and have identified a new gene in this organism. You have cloned and sequenced the gene, but do not know anything about the biochemical function of the gene product. Which of the following is the most sensible next step in order to get clues about the function? A. Purifying the protein product and determining its structure by x-ray diffraction B. Screening for small molecules that perturb gene function C. Using the BLAST algorithm to scan sequence databases for similar sequences D. Using MS/MS to sequence the gene product without the need for purification E. Mutating the gene and performing reverse genetics 26. In sequence alignments such as those generated by a BLAST search, the significance of the alignment can be presented as an E-value, which specifies how likely it is for a match this good to be found in a database of a given size by chance. You have scanned a large sequence database with a sequence query, hoping to find a significant match. Which one would you look at: (A) a match with an E-value of 1, or (B) one with an E-value of 0.001? Write down A or B as your answer. 27. Indicate whether each of the following manipulations or procedures used in current recombinant DNA technology commonly rely on an enzyme (E) or are done nonenzymatically (N). Your answer would be a six-letter string composed of letters E and N only, e.g. EEEEEE. ( ) Cleavage of DNA at specific sites

( ) DNA ligation ( ) DNA cloning ( ) Nucleic acid hybridization ( ) Synthesis of DNA of any desired specific sequence ( ) DNA sequencing 28. The EcoRI restriction enzyme recognizes and cuts at the sequence GAATTC. Such an enzyme is expected to cut a typical double-stranded DNA at 1 in every … sites. A. about 400 B. about 4000 C. about 40,000 D. about 400,000 E. about 4 million 29. The AgeI restriction enzyme recognizes the sequence ACCGGT and cuts each strand between the first (A) and the second (C) nucleotide. This can be represented as A|CCGGT, with the vertical bar indicating the cleavage position. If a DNA fragment is cut with AgeI at both ends, cutting a plasmid with which of the following enzymes would then allow insertion of the fragment into the plasmid by simple ligation of sticky ends? A. FseI, with the recognition sequence GGCCGG|CC B. AatI, with the recognition sequence AGG|CCT C. BstZI, with the recognition sequence C|GGCCG D. MroI, with the recognition sequence T|CCGGA E. None of the above 30. Indicate whether ordinary agarose-gel electrophoresis (A), polyacrylamide-gel electrophoresis (P), or pulsed-field gel electrophoresis (F) is used to better separate each of the following DNA molecules. Your answer would be a three-letter string composed of letters A, P, and F only, e.g. APF. ( ) Small DNA oligonucleotides such as hybridization probes and PCR primers ( ) Plasmid DNA ( ) Bacterial chromosomes 31. Your friend has cut a 3000-nucleotide-pair circular plasmid with either or both restriction enzymes A and B, and has separated the digestion products by electrophoresis. The results are shown in the schematic drawing below, with the approximate size of the fragments indicated

3000 3000

Cut with A and B

Cut with B

Cut with A

Uncut

above each band. Based on these results, what is the closest distance (in nucleotide pairs) between an A restriction site and a B restriction site on this plasmid?

2600 1600 1000 400 400

A. 400 B. 600 C. 1000 D. 1200 E. 1600 32. A researcher has labeled the 5′ end of a DNA primer with 32P radioisotopes. The primer can hybridize to a complementary sequence near the 3′ end of a long, spliced mRNA molecule composed of three exons. The mRNA molecule can interact with two RNA-binding proteins: one of them (protein 1) binds to the first exon and the other (protein 2) binds to the second exon. The researcher uses reverse transcriptase to extend the primer in the presence of the mRNA and either of these proteins, then separates the cDNA products by gel electrophoresis, and finally visualizes the labeled DNA bands by autoradiography. The results are presented schematically in the following drawing. Assuming that the reverse transcriptase falls off the template when confronted by bound proteins, which lane (A or B) in the gel corresponds to the reaction in the presence of protein 1? Write down A or B as your answer.

33. You have devised genetic screens to identify gain- and loss-of-function mutations in the Escherichia coli DNA ligase gene. One of the screens (1) seeks mutant cells that are infected more readily with a ligase-deficient strain of T4 phage. The other screen (2) seeks mutant cells that have increased radiation sensitivity at high temperatures. Indicate whether each of the following descriptions better applies to screen 1 (1) or screen 2 (2). Your answer would be a three-digit number composed of digits 1 and 2 only, e.g. 122. ( ) It would identify gain-of-function DNA ligase mutants. ( ) It would identify loss-of-function DNA ligase mutants. ( ) It would identify conditional mutants. 34.

Bacterial artificial chromosomes (BACs) … A. are derived from the E. coli chromosome. B. are present in high copy numbers per cell. C. can maintain DNA sequences of hundreds of thousands of nucleotide pairs. D. are generally not stable. E. All of the above.

35. Indicate whether each of the following descriptions better applies to cDNA libraries (C) or genomic DNA libraries (G). Your answer would be a five-letter string composed of letters C and G only, e.g. CGCGC. ( ) Their production requires use of reverse transcriptase. ( ) They are enriched in protein-coding genes. ( ) Different sequences are represented in them based on their transcription level. ( ) They contain intronic sequences.

( ) They are cell-type specific. 36. You have used a variation of fluorescence in situ hybridization (FISH) called flow-FISH. In this technique, DNA probes are labeled with fluorescence and hybridized to their targets inside the cells in a tissue. The cells are then sorted using a fluorescence-activated cell sorter based on their fluorescent signal intensity. The probes in your experiment are designed to hybridize to human telomeric sequences at the ends of the chromosomes. You perform flowFISH using these probes on two human cell types, an early primary fibroblast culture and a late secondary culture that is showing the signs of replicative cell senescence. The sorting results are presented in the following histograms. Which histogram (1 or 2) would you expect to correspond to the primary culture?

Percentage of cells

2 1

Fluorescent signal intensity per cell 37. The specificity of nucleic acid hybridization is remarkable. Even a single mismatch between a probe and a target sequence can be detected depending on hybridization conditions. This sensitivity can be employed in applications such as SNP haplotyping. Consider a probe that is complementary to a 20-nucleotide-long genomic region including a single polymorphic nucleotide. The probe is perfectly complementary to the more common allele, but has one mismatch in the case of the minor allele. The temperature-dependent hybridization of the probe to the single-strand genomic fragment is schematically presented in the following graph. To detect hybridization, a fluorescent dye is added to the reaction. The dye fluoresces strongly only when intercalated between the bases of double-stranded DNA, and is therefore used to quantify double-strand formation or dissociation. Which “melting curve” (1 or 2) in the graph do you think corresponds to the more common SNP allele? Which curve shows a higher melting temperature?

Fluorescence

2 1

Temperature A. 1; 1 B. 1; 2 C. 2; 1 D. 2; 2 38. The following schematic graph shows temperature change over time in a cycle of a PCR assay. Which steps (A to C) in the cycle correspond to hybridization (annealing), strand separation (denaturation), and DNA synthesis (extension), respectively? Your answer would be a three-letter string composed of letters A to C, e.g. ACB.

Temperature

A C B Time 39. Each cycle of PCR doubles the amount of DNA synthesized in the previous cycle. A pair of primers is used to direct the specific synthesis of a desired DNA segment. The following diagram shows the products of PCR amplification after three cycles. The DNA strands of the desired segment are indicated in yellow boxes in the diagram. As shown, at the end of the third cycle, 8 out of 16 strands have the desired length and are boxed. After four more cycles (i.e. at the end of the seventh cycle), how many strands of DNA are expected to be synthesized? How many of them have the desired length? Write down the two numbers and separate them with a comma, e.g. 16,8.

Desired segment

BEFORE PCR

END OF FIRST CYCLE

END OF SECOND CYCLE

END OF THIRD CYCLE

40. In a forensic lab, DNA samples from a mother (M) and her child (C) have been analyzed at three short tandem repeat (STR) loci using PCR and gel electrophoresis. The same analysis has been performed for five men (1 to 5). According to the results below, which men can be eliminated as possible candidates for the biological father of the child? Your answer would be a number composed of digits 1 to 5 only, e.g. 145.

41. You would like to clone a gene by PCR. The sequence of the two ends of the coding strand for the gene is provided below. Which following pair of primers would direct the amplification of this gene? All sequences are written from 5′ to 3′. ATGAAATCTACGTTTCAC……CCCCCAGTACCCCCCTTA A. ATGAAATCTACGTTTCAC ; CCCCCAGTACCCCCCTTA B. ATGAAATCTACGTTTCAC ; TAAGGGGGGTACTGGGGG C. TACTTTAGATGCAAAGTG; ATTCCCCCCATGACCCCC D. TACTTTAGATGCAAAGTG ; CCCCCAGTACCCCCCTTA E. GTGAAACGTAGATTTCAT; ATTCCCCCCATGACCCCC 42. A piece of DNA has been sequenced by automated dideoxy sequencing, in which each dideoxyribonucleoside triphosphate was labeled with a fluorescent tag of a different color. The data corresponding to a small segment with the sequence 5′-TGCCACA-3′ is shown in the following diagram. Knowing that ddGTP has been labeled with green fluorescence, what color do you think was the fluorescent tag used to label ddTTP in the sequencing experiment?

Direction of electrophoresis

A. Blue B. Green C. Red D. Yellow 43. In Sanger sequencing, does each labeled DNA molecule have a fluorescently labeled dideoxyribonucleotide at its 5′ end or its 3′ end? Do the labeled nucleotides lack the 2′ or 3′ hydroxyl group? A. 3′ end; 2′ hydroxyl group only B. 3′ end; 3′ hydroxyl group only C. 5′ end; 2′ hydroxyl group only D. 5′ end; 3′ hydroxyl group only E. 3′ end; both hydroxyl groups 44. Indicate whether each of the following descriptions better applies to Illumina® sequencing (I), Ion Torrent™ sequencing (T), or both sequencing technologies (B). Your answer would be a four-letter string composed of letters B, I, and T only, e.g. IBBB. ( ) It uses fluorescently labeled nucleotides. ( ) It uses PCR-generated copies of DNA. ( ) It is a second-generation sequencing technology and employs a cyclic wash-andmeasure paradigm. ( ) It relies on the fidelity of DNA polymerase for its accuracy. 45. A piece of DNA has been sequenced by Ion Torrent™ sequencing. The result corresponding to a six-nucleotide segment of this DNA is presented in the graph below, in which pH change is recorded after the addition of each of the four nucleotide triphosphates in cycles. What is the sequence of this segment? All sequences are written from 5′ to 3′.

Number of protons released

2 1 T A C G T A C G T A C G T A C G Sequence of nucleotide added

A. TACGTA B. CGTACG C. CCGTAA D. CGGTTA E. ATTGGC

Reading frames

46. There are six possible reading frames for any double-stranded DNA. In the following schematic drawing of a genomic region in Escherichia coli, stop codons in each of these frames are indicated by black bars. In which reading frame do you think the open reading frame (ORF) is translated in this region? Write down the frame number as your answer, e.g. 2. 3 2 1 –1 –2 –3 Position on chromosome 47. A portion of RNA-seq data obtained from two tissue samples is plotted in the following schematic diagram. Which region (A to E) corresponds to an alternative exon?

Number of reads

Liver

Muscle

Position on chromosome 48. A protein made from an expression vector can be so plentiful as to account for about 10% of total cell protein. You would like to produce a protein of interest that can be expressed at such a level in Escherichia coli. You plan to induce the expression and collect the cells when cell density in the culture medium reaches about 109 cells/mL. Proteins make up about 15% of total cell weight, and each E. coli cell typically has a mass of about 1 picogram (i.e. 10–12 g). Assuming that almost 50% of the protein is lost during purification, how many liters of culture should you prepare in order to obtain 15 milligrams of pure protein? Write down your answer in liters, with two decimal places and without units, e.g. 15.00. 49. Two loci on a human chromosome are about 1,500,000 nucleotide pairs apart and are found to recombine in about 2% of gametes during gametogenesis. Consistently, each genetic map unit (which corresponds to 1% recombination) in humans is equivalent to a physical distance of about … nucleotide pairs. A. 30 million B. 7.5 million C. 3 million D. 0.75 million E. 0.3 50. The p53 gene is an important tumor suppressor gene in our genome. It codes for a transcription regulatory protein that can assemble into tetramers. A number of mutations in p53 render the protein inactive. These mutant protein molecules form mixed tetramers with wild-type p53 proteins and, in this case, the tetramer is inactive. The mutations can therefore be categorized as … A. null.

B. gain-of-function. C. recessive lethal. D. dominant-negative. E. suppressor. 51.

For a complementation test to work, the mutations under study must be… A. recessive. B. dominant. C. gain-of-function. D. null. E. conditional.

52. You have isolated five mutations (1 to 5) in the yeast Saccharomyces cerevisiae that make the haploid cells unable to grow in the absence of histidine. Each haploid mutant can be mated with any of the other ones, forming diploid cells that either can (+) or cannot (–) grow in the absence of histidine, as indicated in the following complementation table. How many complementation groups do these mutations represent? Each complementation group typically corresponds to a separate gene. 1

–

A. 1 B. 2 C. 3 D. 4 E. 5

53. The wild-type nematode Caenorhabditis elegans has a bent body with the shape of a wave. In the presence of increasing concentrations of ethanol, however, the body bend is

progressively diminished. Loss-of-function mutations in SLO-1, a gene that codes for an ion channel, confer ethanol resistance and allow the worm to maintain its body bend up to higher ethanol concentrations. The channel function is defined based on the ability of the channel to conduct ions across the membrane. According to these observations, would you expect the channel to open or close in the presence of ethanol in wild-type worms? Would a gain-offunction mutation in SLO-1 tend to increase or decrease body bend? A. Open; increase B. Open; decrease C. Close; increase D. Close; decrease 54. You have studied mutations in a transcription regulatory protein from Bacillus subtilis by an in vivo reporter gene assay carried out in Escherichia coli. The cis-regulatory sequence that the protein normally recognizes was inserted upstream of a gene encoding an enzyme that can produce a visible blue product. E. coli cells were then transformed with the reporter construct along with the gene encoding one of the various mutant forms of the transcription regulatory protein. Since the transcription regulatory protein is an inhibitor, the colonies turn blue only when the mutant protein is nonfunctional and unable to inhibit gene expression. Interestingly, you have identified a number of “cold-sensitive” loss-of-function mutations that show the mutant phenotype at 22°C but not at 37°C. Which of the following correctly describes the phenotype of these cold-sensitive mutants? A. Their colonies turn blue at permissive (low) temperatures. B. Their colonies turn blue at permissive (high) temperatures. C. Their colonies turn blue at nonpermissive (low) temperatures. D. Their colonies turn blue at nonpermissive (high) temperatures. 55. In a bacterium that is normally cylindrical in shape, loss of either of two genes A and B results in round cells. Additionally, gain-of-function mutations in gene A can rescue the phenotype of gene B loss-of-function mutants. What phenomenon best describes this observation? The product of which gene is more likely to function upstream of the other one in determining cell shape in this bacterium? A. Suppression; gene A B. Suppression; gene B C. Complementation; gene A D. Complementation; gene B

56. Consider a biochemical pathway in petal cells of a plant, in which a white precursor is turned into a final red pigment through a yellow and an orange intermediate by the activity of three enzymes encoded by genes A, B, and C. The petal color in plants harboring functional (capital letter) or nonfunctional (lowercase letter) alleles of these genes is presented in the following table. What do you think is the order in which the enzymes act in the pathway? Your answer would be a three-letter string composed of letters A to C, e.g. BCA. Genotype

Petal color

ABC ABc

Red Orange

AbC Abc aBC aBc abC Abc

White White Yellow Yellow White White

57. Loss-of-function mutations in either imp or yfgL genes in Escherichia coli result in an increased permeability of the outer membrane to small molecules and therefore confer antibiotic sensitivity, as the antibiotics can enter the cell more readily. Interestingly, double mutants, in which both genes are defective, are resistant to antibiotics. Similar to the wild-type strain, but unlike the single mutants, they can grow on media containing antibiotics. This is an example of … A. synthetic lethality. B. complementation. C. synergy. D. a recessive mutation. E. None of the above. 58. Susceptibility to drug and alcohol addiction is partially genetic. For example, the human dopamine receptor DRD2 is associated with cocaine and alcohol abuse; individuals carrying an A nucleotide, instead of the more common C nucleotide, at a certain position within the sixth intron of the DRD2 gene are at a significantly increased risk of abuse. This variation in sequence is an example of … A. a SNP.

B. a CNV. C. an indel. D. epistasis. E. None of the above. 59. If two genomic polymorphic sites were randomly and independently associated with each other, one would expect the frequency of observing a combination of their allelic forms in an individual to be equal to the multiplication product of the probability of the alleles occurring separately. In many cases, however, this is not true. The degree of deviation from random association can be quantified and summarized in diagrams such as the one below. In this diagram, eight SNPs in a genomic region are shown and their pairwise deviation from random association is color-coded by different shades of red, such that the darkest red shade indicates the furthest deviation from random association. In contrast, those pairs that associate randomly are colored white. For example, SNPs A and F seem to associate randomly as indicated by the corresponding box marked with black borders. From this diagram, two haplotype blocks can be readily detected. What subset of the shown SNPs is not part of any of the two blocks? Write down the letter(s) as your answer, in alphabetical order, e.g. CEH. Position on chromosome

60. Indicate whether each of the following descriptions better applies to forward (F) or reverse (R) genetics. Your answer would be a four-letter string composed of letters F and R only, e.g. RFFF. ( ) It involves new gene discovery by genetic screens.

( ) It may also be called genome editing. ( ) In this approach, mutations are introduced deterministically in a known gene. ( ) It is involved in creating knockout mice lacking a particular gene. 61. You have created transgenic mice that carry the gene encoding green fluorescent protein (GFP) under the control of a ubiquitous promoter. The GFP-encoding gene, however, is inactive unless recombination by Cre removes a blocking sequence between the promoter and the gene. These mice have been mated with those harboring the Cre recombinase under the control of the insulin promoter. The Cre recombinase is fused to the estrogen receptor, which keeps the fusion in the cytosol unless it binds to tamoxifen, in which case it allows nuclear entry. After mating, you select adult mice harboring both of these transgenes and inject them with either tamoxifen or water. Which of the following would you expect to observe later as a result when you examine brain and pancreas tissue samples under a fluorescence microscope? A. GFP signal is observed in the brain in the presence of tamoxifen only. B. GFP signal is observed in the brain in the absence of tamoxifen only. C. GFP signal is observed in the pancreas in the presence of tamoxifen only. D. GFP signal is observed in the pancreas in the absence of tamoxifen only. E. GFP signal is observed in the brain in either the presence or the absence of tamoxifen. 62. You mated male and female heterozygous Oct4 knockout mice to study the function of this important gene. Among the progeny, one-third were homozygous wild type and the rest were heterozygous. You then repeated the cross, but this time you genotyped several embryos at different stages of early development and grouped them as either homozygous wild type (+/+), heterozygous (+/–), or homozygous null (–/–). You have summarized your results in the following table, in which the number of embryos in different groups and at different developmental stages are presented. According to these results, complete loss of Oct4 is … Embryonic stage

Genotype

8-cell embryos

64-cell embryos

Implanted embryos

+/+

+/–

–/–

A. NOT lethal. B. synthetic lethal.

C. zygotic lethal. D. embryonic lethal. 63.

Which of the following is NOT correct regarding both Cas9 and EcoRI? A. They are both endonucleases. B. They both create double-strand breaks in DNA. C. They both recognize their target sequences with the help of guide RNAs. D. They are both part of bacterial defense mechanisms against foreign DNA. E. They are both greatly useful in manipulating DNA and studying gene expression and function.

64. A pool of five barcoded yeast mutants, each deleted for a different gene (genes A to E), has been grown under three different conditions (1 to 3). After the growth phase was completed, genomic DNA was isolated from the entire pool and the relative amount of each mutant was determined by quantifying the amount of DNA containing each barcode. The results are presented in the following graph. Which gene (A to E) is essential under condition 1 but not under conditions 2 or 3?

Relative DNA abundance

Conditions: 1 2 3

Deletion mutant 65. In which of the following organisms is systemic RNAi possible through feeding the animal with the RNA? A. Caenorhabditis elegans B. Drosophila melanogaster C. Mus musculus

D. Homo sapiens E. All of the above 66. Which of the following can limit the use of RNA interference in studying gene expression and function? A. RNAi does not efficiently inactivate all genes. B. Certain tissues are resistant to the action of RNAi. C. Off-target effects are common in RNAi treatments. D. All of the above. 67.

You have used a reporter gene system to investigate the contribution to gene expression

of three cis-regulatory DNA sequences (A to C) from a key developmental gene. Reporter gene expression from constructs containing all three or only a subset of the regulatory sequences is summarized in the diagram below. According to these findings, indicate whether each of the following conclusions is (Y) or is not (N) acceptable. Your answer would be a four-letter string composed of letters Y and N only, e.g. NNNN. cis-regulatory DNA sequences A

Relative expression level

C 100% 99% 0% 85% 0% 190%

( ) Region B is likely to normally bind an activator and turn the gene on. ( ) Region C is sufficient for activation of gene expression. ( ) Region A is necessary for activation of gene expression. ( ) Region C is necessary for activation of gene expression.

Fluorescence

68. The following schematic graph shows the result of two quantitative RT-PCR experiments. Total mRNA from two tissue samples (1 and 2) was isolated and subjected to RTPCR using primers designed to amplify a tissue-specific gene. Added to the reaction was a fluorescent dye that fluoresces only when bound to double-stranded DNA. According to the graph, which tissue has a higher level of this mRNA? By how much?

Number of PCR cycles A. About 2-fold higher in tissue 1 B. About 1000-fold higher in tissue 1 C. About 2-fold higher in tissue 2 D. About 1000-fold higher in tissue 2 69. Which of the following methods provides the most sensitive and accurate way of measuring and comparing mRNA levels of a set of genes in two different tissues. The genes are expressed at relatively low levels in these tissues, and do not necessarily code for protein. A. RNA-seq B. RT-PCR C. DNA microarray analysis D. ChIP-seq E. Ribosome profiling 70. You have grown cultures of the yeast Saccharomyces cerevisiae in a complex medium alone or supplemented with either glucose or ethanol. In the presence of glucose, available energy is abundant and the cells can therefore grow by fermenting glucose. In the presence of ethanol, however, the cells switch to a mostly aerobic respiration to meet their energy demands. They also shut down many of their anabolic pathways and activate catabolic pathways to produce more energy to better survive under these conditions. After the growth phase, you extract total mRNA from the cells, convert them to cDNA, and label them with a fluorescent

molecule. You then hybridize the cDNAs to a yeast DNA microarray. Part of the results obtained by cluster analysis after fluorescent measurements is summarized in the following simplified diagram. Here, red and green indicate increase and decrease, respectively, in expression of a gene compared to that in the nonsupplemented medium. Which row (1 or 2) do you think corresponds to the glucose-supplemented medium?

Ribosome biogenesis

Electrontransport chain

1 2

71. Consider a protein composed of only 21 amino acids encoded by a short ORF that is located between two known genes in a mammalian genome. Due to its length, the ORF had not been annotated as a gene. Which of the following methods can demonstrate that it is indeed a gene that is actively translated to protein? A. RNA-seq B. RT-PCR C. DNA microarray analysis D. ChIP-seq E. Ribosome profiling 72. The results of two sets of chromatin immunoprecipitation experiments on two cell types (1 and 2) are shown in the following simplified graphs for the same genomic region containing the first two exons of a gene. Antibodies against a transcription regulatory protein (X) or against a subunit of RNA polymerase II were used in the ChIP-seq experiments, as indicated. Do you think the protein is a transcriptional activator or repressor? The results from a parallel RNA-seq experiment are provided in the third graph. Which profile (a or b) in the graph would you expect to correspond to cell type 1?

Number of sequence reads

ChIP-seq Protein X

1 2

RNA Pol II 2

1 Exon 1 Genomic position

Exon 2

Number of sequence reads

Total RNA-seq

a b Genomic position

A. Activator; profile a B. Repressor; profile a C. Activator; profile b D. Repressor; profile b 73. RNA-seq and ribosome profiling experiments have been carried out on the same cells. The following simplified graphs show the results for the same genomic region containing a gene with three exons. Which graph (1 or 2) do you think corresponds to the RNA-seq results? What feature is represented by the arrow?

Number of sequence reads

Genomic position A. Graph 1; stop codon B. Graph 1; polyadenylation site C. Graph 2; stop codon D. Graph 2; polyadenylation site 74. Sort the following steps in the common procedure to create transgenic plants. Your answer would be a four-letter string composed of letters A to D, e.g. DABC. (A) Callus growth (B) Agrobacterium infection (C) Shoot/root induction (D) Removal of leaf tissue from a plant 75. Consider the binding of a transcription regulatory protein (A) to a promoter DNA (p) according to the following scheme. A + p → A:p At steady state, which of the following formulae correctly yields the bound promoter fraction, i.e. [A:p]/([A:p] + [p])? The association constant is designated as K. A. [A]/(K + [A]) B. K[A]/(1 + K[A]) C. K[A]/(K + [A]) D. [A]/(1 + K[A]) 76. When a promoter p is not saturated with bound regulatory protein A, doubling the total concentration of A would … A. increase steady-state [A:p] exactly twofold. B. increase steady-state [A:p] over twofold. C. increase steady-state [A:p] less than twofold.

D. decrease steady-state [A:p] approximately twofold. E. decrease steady-state [A:p] over twofold. 77. In the following schematic graph, the fraction of promoters (px) bound to the regulatory protein A is plotted at various concentrations of A. Which of the following quantities represents X in the graph?

Fraction bound

100%

50%

0% X

[A]

A. [Total px] B. [px] C. kon / koff D. koff / kon E. kon[px] / koff 78. Consider a transcription regulatory protein (A) that can bind to a promoter (p). After reaching an initial steady state, the concentration of A is suddenly increased tenfold. The following graph shows the formation of A:p over time after this change for two systems, one of which is described by higher koff and kon values compared to the other. Which curve (1 or 2) corresponds to this system? Write down 1 or 2 as your answer.

[A:p] (relative value)

5 1 2

1 Time

[A:p] (relative value)

79. Consider a transcription regulatory protein (A) that can bind to the promoter (p) of a target gene. After reaching an initial steady state, the concentration of A is suddenly increased tenfold. The following graph shows the formation of A:p over time after this change. What is the slope of the curve at time zero? 5

1 0

Time

A. 2 koff B. 2 kon C. 5 koff D. 5 kon E. 9 koff 80. Everything else being equal, if the mean lifetime of a protein is doubled, its concentration at steady state would … A. increase approximately twofold. B. increase less than twofold. C. increase over twofold. D. decrease approximately twofold. E. decrease less than twofold.

81. Consider two proteins A and B with equal transcription and translation rates. The mean lifetime of A, however, is twice as long as that of B. Starting with no protein, the time it takes for a protein to reach 99% of its steady-state concentration is … A. the same for proteins A and B. B. two times longer for A compared to B. C. two times longer for B compared to A. D. less than twofold longer for A compared to B. E. less than twofold longer for B compared to A. 82.

You have set up a system of differential equations to describe the concentration of a

protein as a function of time based on transcription, translation, and degradation rates. You simulate two conditions (1 and 2) that are identical except that the transcription rate for the protein is set tenfold higher in one of them (1) compared to the other (2). Starting from a concentration of zero, the time required to reach 50% of the final steady-state concentration is … A. the same for conditions 1 and 2. B. tenfold longer for 1 compared to 2. C. tenfold longer for 2 compared to 1. D. less than tenfold longer for 1 compared to 2. E. less than tenfold longer for 2 compared to 1. 83. Which equation better describes the steady-state concentration of a protein (X) that is regulated by the simple binding of a transcription repressor (R) to its cis-regulatory DNA sequences? β is the transcription rate, m is the translation rate, K is the association constant in the binding of R to the regulatory sequences, and τX is the mean lifetime of the protein X. A. βmτX / (1 + K[R]) B. βmτX / (1 + [R]) C. βmτX K / (1 + K[R]) D. βmτX [R] / (1 + K[R]) E. βmτX [R] / (K + [R]) 84. The following graph shows the change in the concentration of a protein (A) over time as a fraction of the final steady-state level. The two curves (1 and 2) in the graph are simulations with similar parameters except that one of them involves a negative feedback loop in which A activates the expression of a repressor protein which in turn represses the expression of A. Which

[A] (relative value)

curve (1 or 2) corresponds to the simulation with the negative feedback loop? Write down 1 or 2 as your answer.

1 1 2

0 Time 85. Consider a genetic network consisting of gene A whose product activates the transcription of gene R. Imagine that the concentration of protein A is suddenly changed by a significant amount. The following graphs show how the concentration of protein R would then change over time to reach steady state again. The two curves (1 and 2) in each graph correspond to systems that have different transcription rates for A. For one of the two graphs (A or B) a negative feedback loop is present, in which protein R represses the transcription of gene A. Which conclusion is correct according to these results?

1 2

B 1 2

A. The negative feedback loop makes the system slower in reaching the steady state. B. With negative feedback, the steady-state concentration is less sensitive to fluctuations in kinetic rate constants. C. Higher transcription rates for gene A result in lower steady-state levels of protein R both with and without negative feedback.

D. All of the above. 86. Consider the transcriptional circuits depicted in the following schematic drawings. These circuits share identical kinetic parameters but have a different number of stages. Which circuit (A or B) creates more robust oscillations? Write down A or B as your answer.

87. Consider a promoter sequence with n binding sites for an activator protein. Which Hill coefficient (h) describes noncooperative binding, i.e. independent, noninteracting binding sites? A. h = n B. h = 1 C. h = 0 D. h = n – 1 E. h < 0 88. Bistability in a biological system can be accompanied by hysteresis. Consider switchlike cellular responses to a signal molecule, as shown in the following graphs. In each graph, the solid curve corresponds to the change in cellular response when the signal concentration is being increased over time, whereas the dashed curve corresponds to the change when the signal concentration is being decreased. Which graph do you think corresponds to a system with a greater degree of hysteresis? What mechanism is responsible for this behavior?

Cellular response

Signal strength

A. Graph A; negative feedback B. Graph A; positive feedback C. Graph B; negative feedback D. Graph B; positive feedback

Concentration of Y

89. The schematic graph below represents the nullclines (solid curves) and a few selected trajectories (dotted curves) for a system composed of two proteins (X and Y) with a complex regulatory behavior. According to the graph, indicate whether each of the following statements is (Y) or is not (N) acceptable. Your answer would be a four-letter string composed of letters Y and N only, e.g. YYYY.

Concentration of X ( ) The system is bistable.

( ) There are three steady states in the system. ( ) The system contains positive feedback. ( ) The system contains delayed negative feedback. 90. The quantitative output of a gene depends in part on the relative affinities of transcription regulators. In the following graphs, the red color indicates high gene expression and blue color indicates low gene expression. The expression level is plotted as a function of the concentrations of two transcription regulatory proteins, an activator (A) and a repressor (R). Every other parameter being identical, which graph corresponds to a system where the affinity of protein A for the cis-regulatory sequences is higher than that of protein R? Which graph corresponds to a system that shows a “leakier” expression? 1

A. Graph 1; Graph 1 B. Graph 1; Graph 2 C. Graph 2; Graph 1 D. Graph 2; Graph 2 91. In the schematic graphs below, the red color indicates high gene expression and blue color indicates low gene expression. The expression level is plotted as a function of the concentrations of two transcription regulatory proteins, A1 and A2. Which graph (1 or 2) do you think corresponds to an “AND NOT” combinatorial logic? Write down 1 or 2 as your answer.

92. Some feed-forward motifs are capable of generating a brief pulse of gene activation in response to a sustained input signal: in such motifs, the signal activates an activator (A) and a repressor (R), both of which control the expression of the same gene. The quantitative output of the gene depends in part on the relative affinities of A and R for binding to the cis-regulatory sequences of the target gene. Under which of the following conditions is a brief pulse of gene activation observed in response to a sustained input signal? KA and KR are the association constants for A and R, respectively. A. KA = KR B. KA << KR C. KA >> KR D. KAKR > 1 93. Indicate which of the following network motifs represent negative feedback, positive feedback, incoherent feed-forward, and coherent feed-forward, respectively. Your answer would be a four-letter string composed of letters A to D only, e.g. BACD. A

94. Indicate whether each of the following better applies to a coherent (C) or incoherent (I) feed-forward motif. Your answer would be a three-letter string composed of letters C and I only, e.g. CCC. ( ) It can provide cellular responses with robustness despite perturbations in the input signal. ( ) A cellular pathway with such a motif generally requires a sustained (long) input signal to generate a full response. ( ) A cellular pathway with such a motif can generate a brief pulse of response in the presence of a sustained input signal. 95.

Imagine a bistable system composed of two repressor proteins X and Y, each of which

binds to the cis-regulatory sequences of the gene encoding the other protein, i.e. X represses the expression of Y, and Y represses the expression of X. A researcher engineers the genes such that they encode protein X fused to green fluorescent protein (GFP) and protein Y fused to red fluorescent protein (RFP). Cells containing the fusion constructs are then grown and sorted by fluorescence-activated cell sorting (FACS) based on their GFP and RFP signals. The results are presented in the following dot plot. Each cell is represented by a dot whose coordinates correspond to GFP and RFP fluorescence in the cell. According to these results, which protein (X or Y) do you think binds to its target DNA sequence with a higher degree of cooperativity?

GFP

Write down X or Y as your answer.

RFP

Answers: 1. Answer: H Difficulty: 3 Section: Isolating Cells and Growing them in Culture Feedback: The cells with a high CD4 signal (that is, CD4+) are depleted in the patient. 2. Answer: callus Difficulty: 1 Section: Isolating Cells and Growing them in Culture Feedback: A single totipotent cell from such a callus can be grown into a small clump of cells from which a whole plant can be regenerated. 3. Answer: BNTT Difficulty: 2 Section: Isolating Cells and Growing them in Culture Feedback: Cell lines can be stored in liquid nitrogen at –196°C for an indefinite period and retain their viability when thawed. Unlike nontransformed cell lines, transformed cell lines often grow without attaching to a surface and can proliferate to a much higher density in a culture dish. They can usually cause tumors if injected into a susceptible animal. 4. Answer: E Difficulty: 2 Section: Isolating Cells and Growing them in Culture Feedback: Monoclonal antibodies have several advantages over the polyclonal antibodies present in antisera. 5. Answer: C Difficulty: 2 Section: Isolating Cells and Growing them in Culture Feedback: Fusion of B cell clones with immortalized myeloma cells (a transformed B cell line) creates hybrid cells that are also capable of indefinite proliferation. 6. Answer: B Difficulty: 3 Section: Isolating Cells and Growing them in Culture Feedback: The transformed myeloma (M) cells are immortalized but are not selected for in HAT media. This is because they lack HGPRT and their DHFR-dependent pathway is also inhibited. Non-fused B cells can still make their nucleotides by the salvage pathway but can only proliferate for a few generations. The hybrid cells, on the other hand, are both immortalized and capable of carrying out the salvage pathway. 7. Answer: NLMR Difficulty: 2 Section: Purifying Proteins

Feedback: During cell fractionation by ultracentrifugation, progressively higher speeds are used to fractionate different components. Nuclei sediment in the early steps whereas macromolecular complexes such as ribosomes come out in later steps. 8. Answer: E Difficulty: 3 Section: Purifying Proteins Feedback: The terminal velocity for the 50S particle is calculated as: vt = S × ac = (50S × 10–13 s/S) × (50,000 × 10 m/s2) = 2.5 × 10–6 m/s = 0.9 cm/h It would therefore take about 10 hours for the 50S particle to travel about 9 cm. 9. Answer: EEVE Difficulty: 2 Section: Purifying Proteins Feedback: In equilibrium sedimentation, each subcellular component moves up or down the steep sucrose or cesium chloride gradient to reach a position corresponding to its density. It is sensitive enough to separate DNA molecules containing different nitrogen isotopes. Velocity sedimentation also uses gradients, but these are shallow and are used to limit convective mixing. 10. Answer: C Difficulty: 2 Section: Purifying Proteins Feedback: The larger dimer would sediment faster and would appear near the bottom of the ultracentrifuge tube (peak 2). In gel filtration, the dimer would come off the column in earlier fractions (peak 3). 11. Answer: A Difficulty: 3 Section: Purifying Proteins Feedback: Increasing salt concentration would weaken the charge–charge interactions between bound proteins and an ion-exchange matrix. Decreasing salt concentration, on the other hand, weakens the hydrophobic interactions between bound proteins and a hydrophobic matrix. 12. Answer: D Difficulty: 1 Section: Purifying Proteins Feedback: Immunoprecipitation is a variation on the theme of affinity chromatography. 13. Answer: B Difficulty: 2 Section: Purifying Proteins Feedback: Since tandem affinity purification (TAP)-tagging relies on proteolytic cleavage for elution of the protein from the first column, the protein can be washed

extensively at this step to remove most contaminants. Additionally, it can be purified further due to the presence of another, non-cleaved, tag. 14. Answer: E Difficulty: 1 Section: Analyzing Proteins Feedback: SDS (sodium dodecyl sulfate) is a negatively charged detergent that can release proteins from their binding partners and unfold them into extended polypeptide chains. It also masks each protein’s intrinsic charge. Consequently, during SDS polyacrylamide-gel electrophoresis (SDS-PAGE), proteins are separated mainly based on their size, and not their shape or intrinsic charge; larger proteins move more slowly because the gel matrix hinders their migration to a higher extent compared to smaller proteins. 15. Answer: C Difficulty: 2 Section: Analyzing Proteins Feedback: Within the linear range shown in the graph, smaller proteins are better resolved (separated) in SDS-PAGE compared to larger proteins. 16. Answer: AOH Difficulty: 2 Section: Analyzing Proteins Feedback: If the pH is lower than the isoelectric point of a protein, the protein is protonated (positively charged) and should therefore be repulsed by the nearby anode toward regions with higher pH. Ionic detergents are not suitable for this charge-sensitive technique. A basic protein has a high (basic) isoelectric point. 17. Answer: baa Difficulty: 2 Section: Analyzing Proteins Feedback: The b subunit seems to be the smallest. The a subunit has the lowest isoelectric point and is thus the most acidic (has the highest negative charge). 18. Answer: C Difficulty: 3 Section: Analyzing Proteins Feedback: All protein complexes separated on this gel presumably contain the mutant protein. Those on the diagonal are associated noncovalently with the mutant protein, and therefore run identically in the two dimensions and end up on the diagonal. The proteins that had formed disulfide bonds with the mutant protein, however, are dissociated from it in the second dimension, resulting in two spots for each such complex, one of which is the common 15K mutant protein. The protein also appears to have dimerized with itself (large spot at the intersection), corresponding to a molecular weight of 30K, indicated by the question mark.

19. Answer: C Difficulty: 2 Section: Analyzing Proteins Feedback: In MS/MS, a collision chamber containing a high-energy inert gas connects the two mass analyzers (the first one acting as a mass filter) and is the place where fragmentation of precursor ions takes place. 20. Answer: C Difficulty: 2 Section: Analyzing Proteins Feedback: In this set-up, a chromatographic separation (liquid chromatography; LC) precedes the feeding of different peptides or proteins derived from the organelle into the tandem mass spectrometer (MS) system. 21. Answer: E Difficulty: 2 Section: Analyzing Proteins Feedback: Protein Z is found in the bound fraction of wild-type cell lysates, but mostly in the unbound fraction of lysates lacking protein Y. Thus, protein Y, which itself associates with protein X, appears to promote binding of protein X to protein Z. 22. Answer: A Difficulty: 2 Section: Analyzing Proteins Feedback: The A-B complex formation has a lower Kd (10 µM) as shown by the curves. Accordingly, it should have a lower koff (given identical kon values), and therefore a longer half-life. The Ka is the inverse of Kd. 23. Answer: XXNXN Difficulty: 2 Section: Analyzing Proteins Feedback: Each method has advantages and disadvantages. For instance, NMR has the advantage of providing information about the conformation of proteins in solution and the dynamic changes in their structure. It also requires the preparation of relatively small quantities of the protein of interest. However, for technical reasons, its use is currently limited to molecules smaller than 100,000 daltons. 24. Answer: B Difficulty: 3 Section: Analyzing Proteins Feedback: In the unfolded, random-coil state, variability in the environment surrounding the nuclei (e.g. hydrogens) is decreased significantly. This is reflected in the distribution of peaks in the nuclear magnetic resonance (NMR) spectra, which is broader for the folded state (B). 25. Answer: C

Difficulty: 1 Section: Analyzing Proteins Feedback: Searching a collection of known gene and protein sequences is typically done over the Internet using fast alignment algorithms such as BLAST. The clues generated from such searches typically help the investigator in future, direct experimental studies. 26. Answer: B Difficulty: 2 Section: Analyzing Proteins Feedback: A smaller E-value represents a more significant match. 27. Answer: EEENNE Difficulty: 2 Section: Analyzing and Manipulating DNA Feedback: Hybridization is mostly done nonenzymatically, as is chemical DNA synthesis. Current major DNA sequencing methods rely on enzymes (e.g. polymerases), but some nonenzymatic methods are also in development. 28. Answer: B Difficulty: 2 Section: Analyzing and Manipulating DNA Feedback: Such a restriction site is expected to occur by chance in 1 in 46 random hexamers. 29. Answer: D Difficulty: 2 Section: Analyzing and Manipulating DNA Feedback: Both AgeI and MroI generate the same sticky end (with the sequence CCGG) as a 5′ overhang after cleavage. 30. Answer: PAF Difficulty: 2 Section: Analyzing and Manipulating DNA Feedback: Whereas plasmid DNA of different sizes can be readily separated by ordinary agarose-gel electrophoresis, bacterial chromosomes can be best separated by a variation of this technique called pulsed-field gel electrophoresis. On the other hand, much smaller DNA molecules are best resolved by polyacrylamide-gel electrophoresis. 31. Answer: C Difficulty: 2 Section: Analyzing and Manipulating DNA Feedback: One of the B sites is 1000 nucleotide pairs away from the A site, and the second B site is another 400 nucleotide pairs further, making it 1400 nucleotide pairs away in one direction and 1600 nucleotide pairs away in the other direction around the circular plasmid. 32. Answer: B

Difficulty: 3 Section: Analyzing and Manipulating DNA Feedback: Binding of protein 1 to a region of mRNA that is further upstream allows the reverse transcriptase enzyme to further extend the labeled primer (create a longer DNA) before reaching a bound protein. 33. Answer: 122 Difficulty: 2 Section: Analyzing and Manipulating DNA Feedback: The ligase-deficient phages would readily infect cells that have increased ligase activity. Since ligase activity is required for DNA repair, loss-of-function DNA ligase mutants show radiation sensitivity. 34. Answer: C Difficulty: 2 Section: Analyzing and Manipulating DNA Feedback: BACs are derived from the F plasmid, present in only one or two copies per E. coli cell, and can stably maintain large DNA fragments of up to 1 million nucleotide pairs. 35. Answer: CCCGC Difficulty: 2 Section: Analyzing and Manipulating DNA Feedback: cDNA libraries are enriched in protein-coding genes and lack intronic sequences. They are cell-type specific; how much each gene sequence is represented in such a library depends on its level of transcription in the cell type from which the library was obtained. The cDNA is made by reverse transcription of mRNA extracted from the cell. 36. Answer: 2 Difficulty: 2 Section: Analyzing and Manipulating DNA Feedback: During replicative cell senescence, the telomeres become progressively shorter, resulting in weaker fluorescence in this experiment. 37. Answer: D Difficulty: 3 Section: Analyzing and Manipulating DNA Feedback: The probe hybridized to a perfectly complementary sequence (in curve 2) shows a higher stability (and higher melting point). 38. Answer: BAC Difficulty: 2 Section: Analyzing and Manipulating DNA Feedback: Steps A, B, and C correspond to strand separation, hybridization, and synthesis, respectively.

39. Answer: 256,240 Difficulty: 2 Section: Analyzing and Manipulating DNA Feedback: The total number of DNA strands grows exponentially, whereas the number of longer-than-desired strands grows linearly. After seven cycles, there would be a total of 256 strands (i.e. 2 × 27) among which 16 strands are longer than desired (i.e. 2 original strands plus 2 × 7 longer primer extensions) and the remaining 240 have the desired length. After several more cycles, essentially all of the DNA strands will be the desired length. 40. Answer: 2345 Difficulty: 3 Section: Analyzing and Manipulating DNA Feedback: Those bands in lane C that are not shared with lane M can be accounted for only by lane 1. The other four men can be eliminated as candidates for the biological father. 41. Answer: B Difficulty: 3 Section: Analyzing and Manipulating DNA Feedback: The first (forward) primer would hybridize to the strand that is complementary to that shown (i.e. coding strand) and would be extended toward the right, while the second (reverse) primer would hybridize to the coding strand and would be extended toward the left. 42. Answer: C Difficulty: 2 Section: Analyzing and Manipulating DNA Feedback: The labeled ddTTP (red) is incorporated twice, opposite to the A nucleotides in the sequence. 43. Answer: E Difficulty: 1 Section: Analyzing and Manipulating DNA Feedback: The dideoxyribonucleotides terminate the chains because they lack the 3′ hydroxyl group required for further polymerization. Additionally, all DNA nucleotides lack the 2′ hydroxyl group. 44. Answer: IBBB Difficulty: 1 Section: Analyzing and Manipulating DNA Feedback: Both of them are second-generation sequencing technologies and require DNA amplification before sequencing-by-synthesis, which involves cycles of wash and measurement. Illumina sequencing uses fluorescent tags, but Ion Torrent sequencing measures pH changes.

45. Answer: D Difficulty: 2 Section: Analyzing and Manipulating DNA Feedback: When each nucleotide is incorporated into DNA in the well that contains this sequence, the release of an H+ ion changes the pH, which is registered by a semiconductor chip. If the next position in the sequence has the same nucleotide, the nucleotide is incorporated twice, and twice as many protons are released, as is the case for G and T nucleotides in the example shown. 46. Answer: –1 Difficulty: 1 Section: Analyzing and Manipulating DNA Feedback: An open reading frame is a stretch of DNA with no stop codons. Long open reading frames likely correspond to protein-coding genes in genomes that have uninterrupted genes. 47. Answer: D Difficulty: 2 Section: Analyzing and Manipulating DNA Feedback: As shown in this example, alternative exons (D) can be identified by total RNA-seq. 48. Answer: 2.00 Difficulty: 2 Section: Analyzing and Manipulating DNA Feedback: The amount of purified protein obtained per milliliter of culture is calculated as: 109 cell/mL × 10–12 g/cell × 15% × 10% × 50%= 7.5 × 10–6 g/mL To obtain 15 mg of the protein, therefore, one would need to start with about 2000 milliliters (= 2 L) of culture. 49. Answer: D Difficulty: 2 Section: Studying Gene Expression and Function Feedback: If two genes are reassorted in x% of gametes, they are said to be separated on a chromosome by a genetic map distance of x map units (or x centimorgans). The greater the distance between two loci on a single chromosome, the greater is the chance that they will be separated by crossing-over occurring at a site between them. 50. Answer: D Difficulty: 2 Section: Studying Gene Expression and Function Feedback: These p53 mutations are dominant-negative because they cause a loss-offunction phenotype even in the presence of a normal copy of the gene. This phenomenon

occurs when the mutant gene product interferes with the function of the normal gene product. 51. Answer: A Difficulty: 2 Section: Studying Gene Expression and Function Feedback: If the mutations are recessive, a complementation test can be used to ascertain whether the mutations fall in the same gene or in different genes. Note that most recessive mutations are partial or complete loss-of-function (that is, null) mutations. 52. Answer: C Difficulty: 3 Section: Studying Gene Expression and Function Feedback: One group includes mutants 1 and 3, another includes mutants 2 and 4, and the last one only includes mutant 5. Mutations within a complementation group cannot complement each other, whereas those belonging to different groups can. 53. Answer: B Difficulty: 3 Section: Studying Gene Expression and Function Feedback: The loss-of-function mutations in SLO-1 partially close the channel, an effect opposite to that of ethanol. 54. Answer: C Difficulty: 3 Section: Studying Gene Expression and Function Feedback: As long as the permissive conditions prevail, conditional mutants function normally. In this example, only under nonpermissive conditions (i.e. at low temperatures) do the mutations abort the inhibitory function of the protein, resulting in blue colonies. 55. Answer: B Difficulty: 3 Section: Studying Gene Expression and Function Feedback: The gain-of-function mutations in gene A suppress the loss-of-function mutations in gene B, suggesting that the product of gene A functions further downstream. 56. Answer: BAC Difficulty: 3 Section: Studying Gene Expression and Function Feedback: With the b allele, the petals are white regardless of whether the other gene products are functional. Hence, B should be acting first. The genotype at the C locus makes a difference only when the active A allele is present, placing C at the last step. 57. Answer: E Difficulty: 3 Section: Studying Gene Expression and Function Feedback: This is an example of a synthetic phenotype, but it is not lethal.

58. Answer: A Difficulty: 1 Section: Studying Gene Expression and Function Feedback: This is an example of a single-nucleotide polymorphism (SNP), a polymorphism due to substitution of a single nucleotide. 59. Answer: D Difficulty: 2 Section: Studying Gene Expression and Function Feedback: SNPs A, B, and C appear to constitute one haplotype block, while E, F, G, and H form another one. Haplotype blocks are often interspersed by recombination hotspots, containing SNP D in this example. 60. Answer: FRRR Difficulty: 1 Section: Studying Gene Expression and Function Feedback: Reverse genetics begins with a known gene and mutates it to determine which cell processes require its function. Because the genome of the organism is deliberately altered in a particular way (in gene knockouts, for example), this approach is also called genome engineering or genome editing. New gene discovery through genetic screens and identifying the mutations and the genes responsible for a certain phenotype are examples of the more traditional (forward) approach of classical genetics. 61. Answer: C Difficulty: 2 Section: Studying Gene Expression and Function Feedback: Cre is expressed only in insulin-secreting cells of the pancreas. And it can enter the nucleus only in the presence of tamoxifen. It can then activate the GFP gene by removing the blocking sequence by recombination. 62. Answer: D Difficulty: 2 Section: Studying Gene Expression and Function Feedback: The homozygous null embryos survive up to the 64-cell stage, but die before implantation. 63. Answer: C Difficulty: 2 Section: Studying Gene Expression and Function Feedback: Unlike Cas9, restriction enzymes such as EcoRI recognize DNA sequences directly through protein–DNA interactions. 64. Answer: D Difficulty: 2 Section: Studying Gene Expression and Function

Feedback: Mutants lacking gene D are absent from the pool grown under condition 1, meaning that gene D is required for survival under this condition. 65. Answer: A Difficulty: 1 Section: Studying Gene Expression and Function Feedback: To test gene function by RNA interference (RNAi), double-stranded RNA (dsRNA) can be introduced into C. elegans either by feeding the worms E. coli that express the dsRNA or by injecting the dsRNA directly into the animal’s gut. 66. Answer: D Difficulty: 1 Section: Studying Gene Expression and Function Feedback: RNA interference (RNAi) has made reverse genetics simple and efficient in many organisms, but it has several potential limitations that should be considered when designing RNAi experiments and interpreting their results. 67. Answer: NNYN Difficulty: 3 Section: Studying Gene Expression and Function Feedback: Only region A appears to be necessary for activation of reporter gene expression, as eliminating either B or C (or both) does not shut down gene expression. In the absence of region B, expression is increased almost twofold, suggesting that this region is inhibitory. The presence of region C increases expression, but only when region A is also present. 68. Answer: B Difficulty: 3 Section: Studying Gene Expression and Function Feedback: The sample with a higher concentration of mRNA requires fewer PCR cycles to reach the same half-maximal DNA concentration. A difference of about 10 cycles corresponds to an approximately 1000-fold difference in mRNA level (210 = ~1000). 69. Answer: B Difficulty: 1 Section: Studying Gene Expression and Function Feedback: Please note that RNA-seq can also be used, but it is not as quantitative as RTPCR. 70. Answer: 1 Difficulty: 2 Section: Studying Gene Expression and Function Feedback: In glucose medium, compared to the nonsupplemented medium, the genes involved in ribosome biogenesis would be up-regulated (red) and those involved in the respiratory chain would be down-regulated (green). The opposite would occur in the presence of ethanol.

71. Answer: E Difficulty: 1 Section: Studying Gene Expression and Function Feedback: As revealed by ribosome profiling, many open reading frames (ORFs) that are too short to be normally annotated as genes are actively translated and probably encode functional, albeit very small, proteins. 72. Answer: D Difficulty: 2 Section: Studying Gene Expression and Function Feedback: Protein X is enriched at the cis-regulatory sequences in cell type 1, which corresponds to the lower RNA polymerase II occupancy over the gene, suggesting it is associated with repressed genes. Accordingly, RNA-seq results do not show significant peaks over exons in cell type 1 (profile b). 73. Answer: D Difficulty: 2 Section: Studying Gene Expression and Function Feedback: Peaks in the RNA-seq result also cover the 5′ and 3′ untranslated regions of the mRNAs. 74. Answer: DBAC Difficulty: 2 Section: Studying Gene Expression and Function Feedback: Transgenic plants can be created from plant cells transfected with DNA in culture. Normally, a disc is cut out of a leaf and incubated in a culture of Agrobacterium that carries the desired gene as well as a selectable marker on a plasmid. Only cells transfected with the plasmid are then able to proliferate and form a callus on the selection medium. With careful manipulation of the nutrients and growth regulators, a whole new plant can then be generated from the callus. 75. Answer: B Difficulty: 2 Section: Mathematical Analysis of Cell Functions Feedback: At steady state, [A:p]/[p] = K[A]. Therefore, [A:p]/([A:p] + [p]) = K[A]/(1 + K[A]). Please refer to the equations in Figure 8–72. 76. Answer: C Difficulty: 2 Section: Mathematical Analysis of Cell Functions Feedback: At steady state, [A:p] = K[A][pT]/(1 + K[A]). Therefore, doubling [A] increases [A:p], but not by twofold. Please refer to the equations in Figure 8–72. 77. Answer: D Difficulty: 2 Section: Mathematical Analysis of Cell Functions

Feedback: The bound fraction reaches its half-maximum value when [A] = koff/ kon = Kd. Please refer to the equations in Figure 8–72. 78. Answer: 1 Difficulty: 2 Section: Mathematical Analysis of Cell Functions Feedback: The system with higher koff and kon values reaches the new steady state more rapidly. Please refer to the equations in Figure 8–73. 79. Answer: E Difficulty: 3 Section: Mathematical Analysis of Cell Functions Feedback: The slope of the curve (i.e. d[A:p]/dt) at any time is equal to the rate of complex formation minus the rate of complex dissociation. At the initial steady state (before the concentration jump), the two terms are equal. The rate of complex formation is kon [A][p] and is suddenly increased tenfold at t = 0. The rate of dissociation is koff [A:p] (or just koff since [A:p]t=0 is assigned a value of 1) and is unchanged at t = 0. Therefore, the slope increases from 0 to 9 koff at time zero. Please refer to the equations in Figure 8–73. 80. Answer: A Difficulty: 2 Section: Mathematical Analysis of Cell Functions Feedback: Concentration of a protein at steady state is equal to its production rate multiplied by its lifetime. Please refer to the equations in Figure 8–74. 81. Answer: B Difficulty: 2 Section: Mathematical Analysis of Cell Functions Feedback: The time it takes for a protein to reach any desired fraction of its steady-state concentration (i.e. response time) is a linear function of its lifetime. Please refer to the equations in Figure 8–74. 82. Answer: A Difficulty: 2 Section: Mathematical Analysis of Cell Functions Feedback: It is not intuitive, but the time it takes for a protein to reach a desired fraction of its steady-state concentration (i.e. response time) is independent of its production rate. The steady-state concentration itself, however, does depend on all rates. Please refer to the equations in Figure 8–74. 83. Answer: A Difficulty: 2 Section: Mathematical Analysis of Cell Functions Feedback: Please refer to the equations in Figure 8–75. 84. Answer: 1

Difficulty: 2 Section: Mathematical Analysis of Cell Functions Feedback: Negative feedback can cause oscillations and, as seen in this example, overshoots in approaching the steady state. This allows the system to reach steady state faster. Please refer to Figure 8–76. 85. Answer: B Difficulty: 2 Section: Mathematical Analysis of Cell Functions Feedback: Negative feedback can protect the cell from perturbations that continually arise in its environment. Such perturbations can change the kinetic parameters. With negative feedback, the steady-state concentration in a system changes to a lesser extent upon fluctuations in these parameters. Please refer to Figure 8–77. 86. Answer: B Difficulty: 2 Section: Mathematical Analysis of Cell Functions Feedback: The circuit with more components (B) is more likely to oscillate compared to the similar circuit with fewer components. This is because an increase in the number of components in a negative feedback loop leads to delays in the amount of time required for the cycle of signals to be completed. The system with the longer loop also tends to exhibit stable oscillations within a much broader range of parameters, indicating that this system provides a more robust oscillator. 87. Answer: B Difficulty: 2 Section: Mathematical Analysis of Cell Functions Feedback: When h is 1, there is no cooperativity. 88. Answer: D Difficulty: 2 Section: Mathematical Analysis of Cell Functions Feedback: In hysteresis, the current state of the system depends on the route it has taken in the past. In the presence of hysteresis in a bistable system, the transition points for switching on (solid curve) and switching off (dashed curve) are different. 89. Answer: YYYY Difficulty: 3 Section: Mathematical Analysis of Cell Functions Feedback: The spiral trajectories represent oscillations (resulting from delayed negative feedback) as the system approaches one of the two stable nodes. The steady state in the middle is unstable. 90. Answer: A Difficulty: 2 Section: Mathematical Analysis of Cell Functions

Feedback: Graph 1 corresponds to a system that shows a leakier expression: since KA > KR, even a small concentration of A is capable of overcoming repression by R. Please refer to Figure 8–84. 91. Answer: 1 Difficulty: 2 Section: Mathematical Analysis of Cell Functions Feedback: Graph 1 corresponds to an AND NOT logic function, whereas graph 2 corresponds to an AND logic function. 92. Answer: B Difficulty: 2 Section: Mathematical Analysis of Cell Functions Feedback: With a high binding affinity, the slow repressor eventually represses gene expression even in the presence of a sustained input signal. 93. Answer: DBCA Difficulty: 2 Section: Mathematical Analysis of Cell Functions Feedback: Motifs A to D are coherent feed-forward, positive feedback, incoherent feedforward, and negative feedback, respectively. 94. Answer: CCI Difficulty: 2 Section: Mathematical Analysis of Cell Functions Feedback: A coherent feed-forward interaction can detect persistent inputs and ignore brief fluctuations. An incoherent feed-forward interaction can generate pulses. 95. Answer: Y Difficulty: 3 Section: Mathematical Analysis of Cell Functions Feedback: As expected, most cells express either high levels of X-GFP or high levels of Y-RFP but not both. The switchlike behavior is sharper for X-GFP: there are almost no cells with intermediate GFP signal intensity, whereas many cells show an intermediate RFP signal. This suggests that binding of Y to its gene may occur with a relatively higher cooperativity.

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION CHAPTER 9: VISUALIZING CELLS Copyright © 2015 by W.W. Norton & Company, Inc. 1. If an average globular protein was of the size of a tennis ball, a typical animal cell would be as large as … A. a cubicle. B. a room. C. a tennis court. D. a stadium. E. a city. 2. In the diagram below, a logarithmic scale of sizes is shown. Indicate which of the sizes indicated (A to H) better corresponds to the dimensions of each of the following. Your answer would be a three-letter string composed of letters A to H only, e.g. HCG.

( ) A bacterium ( ) An animal cell ( ) A globular protein 3. Indicate true (T) and false (F) statements below regarding light and light microscopy. Your answer would be a four-letter string composed of letters T and F only, e.g. FFFF. ( ) Two light waves of the same amplitude and frequency will completely cancel each other out if not perfectly in phase.

( ) If the refractive index of a medium is 1.1, light travels in a vacuum 1.1 times faster than it does in the medium. ( ) The limit of resolution for conventional light microscopy is approximately 0.4 µm, corresponding to the wavelength of violet light. ( ) A light-emitting particle can be detected with a light microscope even if it is several times smaller than the resolution limit of the microscope. 4. The following schematic diagram shows the path of light rays passing through in a light microscope. If the angular width of the cone of rays collected by the objective lens (2θ) is increased, would the resolution improve (I; i.e. the resolution limit decreases) or deteriorate (D)? Write down I or D as your answer.

5. In which of the following microscopy techniques are oblique rays of light focused on the specimen? A. Bright-field microscopy B. Dark-field microscopy C. Phase-contrast microscopy D. Differential-interference-contrast microscopy

6. The light used to excite a fluorescent molecule carries … energy and has a … wavelength compared to the light that is then emitted from the molecule. A. greater; longer B. greater; shorter C. the same amount of; shorter D. less; longer E. less; shorter 7. Indicate whether you would use a fluorescent organic molecule (O), in situ hybridization (H), or a coupled fluorescent protein (P) to visualize the cells and their molecules in each of the following cases. Your answer would be a five-letter string composed of letters O, H, and P only, e.g. OHOOO. ( ) You would like to see where in the early Drosophila embryo the mRNA encoding a certain transcription regulator is located. ( ) You would like to see the nuclei and count them in an early mouse embryo. ( ) You would like to visualize chromosome 3 in a human cell culture derived from a patient’s tissue, based on specific sequences present on this chromosome. ( ) You would like to observe the oscillations in Ca2+ ions inside a fertilized frog egg. ( ) You would like to compare the localization of two transcription regulatory proteins in cultured human T cells. 8. What is the advantage of using quantum dots as an alternative to organic fluorochromes such as Cy3 and Alexa dyes? A. They are brighter. B. Their emission spectra can be precisely fine-tuned. C. They have a longer lifetime and bleach more slowly. D. All of the above. 9. Two approaches have been devised to deal with the problem of blurring in light microscopy with thicker samples. Indicate whether each of the following descriptions better applies to confocal design (C) or image deconvolution (D). Your answer would be a three-letter string composed of letters C and D only, e.g. CCD. ( ) It is normally faster. ( ) It requires a higher degree of sample illumination. ( ) It can be used to obtain images from relatively deeper parts of the specimen.

10. Which microscopy set-up uses a longer wavelength of light than usually excites a particular fluorophore? Which one allows researchers to peek deeper into biological samples? A. Single-photon; single-photon B. Single-photon; two-photon C. Two photon; single photon D. Two-photon; two-photon 11. Two segments (S1 and S2) in a viral protein are suspected to be responsible for the nuclear localization of the protein in infected human cells. You have engineered a plasmid to encode the green fluorescent protein (GFP) fused to either or both of these peptide segments, and have introduced the plasmid into the cells. After the expression of the fusion protein is induced, you visualize the cells using a fluorescence microscope equipped with filters appropriate for detection of GFP. Your results are presented in the following schematic drawings in which the GFP signal is represented in green. Which of the following is more consistent with these observations? GFP only

GFP-S1

GFP-S2

GFP-S1-S2

A. Both S1 and S2 are required for nuclear localization. B. S1 is required and sufficient for nuclear localization. C. S2 is required and sufficient for nuclear localization. D. S1 is required but not sufficient for nuclear localization.

E. S2 is required but not sufficient for nuclear localization. 12. When the gene encoding a certain cytoskeleton protein is deleted, the resulting mutant cells round up and do not form their normal appendages. These mutants can be rescued when a gene encoding an N-terminal green fluorescent protein (GFP) fusion of the protein is expressed, but not when a gene encoding a C-terminal GFP fusion is expressed. Which fusion protein (N or C) is appropriate to use in studying cellular localization and activity? Write down N or C as your answer. 13.

Given the absorption and emission spectra of three fluorescent dyes in the simplified

diagrams below, which pair of dyes is better suited for a fluorescence resonance energy transfer (FRET) study? Write down AB, BC, or AC as your answer. A

Excitation

B C

Emission

Wavelength (nm)

14. Imagine a transcription regulatory protein (X) that is known to shuttle back and forth between nucleus and cytosol in an oscillatory pattern. Protein Y is a nuclear protein that can bind to X to create a dimer that binds to DNA. You have fused protein X to green fluorescent protein (GFP) and protein Y to blue fluorescent protein (BFP), and have measured fluorescence resonance energy transfer (FRET) and non-FRET signals in the nucleus at different time points, as indicated in the following simplified plot. At which time period (1 or 2) do you think protein

X is in the nucleus? BFP can be excited at 440 nm, and emits maximally at 470 nm. GFP is excited at 470 nm and emits maximally at 500 nm. Excitation at 440 nm, emission at 500 nm Excitation at 440 nm, emission at 470 nm 1

Fluorescence

Time (h) 15. Tubulin labeled with caged fluorescein can be introduced into dividing cells by microinjection. Various small regions in the mitotic spindle (made up of tubulin subunits) are briefly irradiated with laser light that uncages the fluorescent tubulin. Five minutes after irradiation, the highest spindle fluorescence is observed when the irradiated region is close to the chromosomes near the cell equator, and the lowest fluorescence is observed when regions near the spindle poles are irradiated. Based on this observation, do you think tubulin subunits are incorporated into the spindle mostly near the poles (P) or near the equator (E)? Write down P or E as your answer. 16. A certain GTP-binding protein can exist in two main states. When bound to GDP, it is mostly cytosolic. In its GTP-bound form, however, it associates with the cytosolic face of the endoplasmic reticulum (ER) membrane, where it hydrolyzes the bound GTP after a short delay and is released again into the cytosol. You have created and expressed green fluorescent protein (GFP) fusions of the wild-type protein, as well as that of a mutant protein that does not bind GTP as readily as the wild type. You then perform a fluorescence recovery after photobleaching (FRAP) experiment by photobleaching a small area of the ER membrane and measuring GFP fluorescence recovery over time. According to the results below, which curve (1 or 2) do you think corresponds to the wild-type fusion protein? Write down 1 or 2 as your answer.

1 GFP fluorescence in the bleached area

Time Photobleaching 17.

Which of the following is correct regarding aequorin? A. It is a small molecule used to detect calcium ions in vivo. B. It is a fluorescent dye. C. It emits blue light in the presence of calcium ions. D. It can be used in animal cells but not plant cells. E. All of the above.

18.

Consider an engineered chimeric protein made from fusion of three proteins: a blue

fluorescent protein (BFP), a calmodulin-binding peptide, and a green fluorescent protein (GFP). Calmodulin is an abundant calcium-binding protein in eukaryotes. Once bound to calcium ions, it can recognize the calmodulin-binding peptide in the fusion protein, change conformation, wrap around the peptide, and bring the BFP and GFP components in close proximity. This results in fluorescence resonance energy transfer (FRET) between BFP and GFP. Accordingly, the fusion protein … A. is a luminescent ion-sensitive indicator that red-shifts its emission wavelength in the presence of calcium. B. is a luminescent ion-sensitive indicator that increases its emission in the presence of calcium. C. is a genetically encoded calcium indicator that red-shifts its emission wavelength in the presence of calcium. D. is a genetically encoded calcium indicator that increases its emission in the presence of calcium.

19. Single-molecule detection by fluorescence microscopy is limited by the presence of an excess of out-of-focus fluorescent molecules. How does a TIRF microscope uniquely overcome this limitation? A. By removal of the out-of-focus molecules by selective destruction. B. By using a deconvolution algorithm that reverses the convolution of signals due to the out-of-focus molecules. C. By a confocal set-up that eliminates out-of-focus signals. D. By exciting only in-focus molecules via an evanescent field. E. By using an objective lens with extremely high numerical aperture.

20–21 Atomic Force Microscopy Atomic force microscopy (AFM) is used in an experiment to unfold a multidomain protein by applying mechanical force. The protein contains several copies of an immunoglobulin domain that are unfolded one by one as the two ends of the molecule (one attached to a cover slip, and the other to the AFM tip) are being pulled apart, resulting in the “sawtooth” force–extension curves shown below. The same experiment is done twice, once in the presence and once in the absence of a chaperone protein that stabilizes the immunoglobulin domains. Answer the following questions based on this graph. 1000

Force (pN) 1 2

0 0

200 Extension (nm)

20.

According to the force–extension graph, how many immunoglobulin domains are

unfolded in each of these experiments? Write down the number as your answer, e.g. 9.

21. According to the force–extension graph, which curve (1 or 2) would you expect to correspond to the reaction in the presence of the chaperone protein? Write down 1 or 2 as your answer.

22. Indicate true (T) and false (F) statements below regarding superresolution fluorescence microscopy. Your answer would be a four-letter string composed of letters T and F only, e.g. FFFF. ( ) Resolutions of 5 nm or better can be readily achieved by superresolution fluorescence techniques. ( ) SIM overcomes the limit imposed by the diffraction of light by a computational analysis on images obtained from interference patterns. ( ) PALM and STORM techniques reduce the width (or “spread”) of the point spread function. ( ) Success of STED depends on fluorescence probes that are reversibly switched off and on. 23.

Indicate whether each of the following descriptions better applies to SIM (S), STED (T),

or STORM/PALM (P) superresolution techniques. Your answer would be a four-letter string composed of letters S, T, and P only, e.g. PPTS. ( ) It switches on and off individual fluorophores at random over time to accurately determine their position. ( ) It creates a moiré pattern from the interference of the illuminating pattern and the sample features. ( ) It doubles the resolution of conventional fluorescence microscopy. ( ) It limits excitation to the fluorophores that are located at the center of the focal point by using a doughnut-shaped beam in addition to the excitation beam. 24. Indicate true (T) and false (F) statements below regarding electron microscopy. Your answer would be a four-letter string composed of letters T and F only, e.g. FFFF. ( ) Depending on acceleration voltage, the resolution limit of an electron microscope can be as small as 0.05 nm. ( ) The emission gun and the magnetic coils in an electron microscope are analogous to the light source and the glass lenses in a light microscope, respectively.

( ) Contrast in specimens for electron microscopy can be achieved using electron-dense material. ( ) For biological samples, the effective resolution of electron microscopy is about 1 nm. 25. Electron microscopy samples are often chemically fixed before dehydration, resinembedding, and sectioning. But they can also be “fixed” by rapid freezing, in a way that precludes ice-crystal formation, to ensure minimal damage to the original cell structures. How can this be done? A. High-pressure cooling B. Plunging into liquid nitrogen C. Spraying with a jet of liquid propane D. Contact with a copper block cooled by liquid helium E. All of the above 26. The presence of which of the following provides the sample with the lowest electron density? A. Osmium B. Lead C. Uranium D. Carbon E. Gold 27. Indicate whether each of the following descriptions better applies to the scanning (S) or transmission (T) electron microscopy techniques. Your answer would be a four-letter string composed of letters S and T only, e.g. TSTS. ( ) It generally has a greater depth of field. ( ) It is usually smaller, cheaper, and simpler. ( ) It detects electrons that are scattered or emitted from the specimen. ( ) It is used to create electron-microscope tomograms. 28. You have generated strains of Drosophila melanogaster that are expected to show interesting developmental phenotypes such as misplaced organs in the adult fly. However, some of these phenotypes are not readily seen with light microscopy. You therefore fix each mutant fly, dry it, coat it with a thin layer of gold, and place the entire fly into an electron microscope for viewing. What type of microscope are you using? Write down SEM or TEM as your answer.

29.

Which of the following is NOT correct regarding cryoelectron microscopy? A. It does not require shadowing or negative staining. B. It involves obtaining images from many—sometimes tens of thousands of— individual molecules that may or may not all be in the same orientation. C. Subnanometer resolutions (e.g. 0.5 nm or better) can be achieved by this technique. D. It can only be used to see the exterior surface of molecules and complexes. E. Atomic models obtained from x-ray crystallography can be fitted into the density envelope obtained from this technique.

Answers 1. Answer: D Difficulty: 2 Section: Looking at Cells in the Light Microscope Feedback: The diameter of a typical globular protein is on the order of nanometers, about ten million times smaller than a tennis ball, which has a diameter on the order of centimeters. Ten million multiplied by the diameter of a typical animal cell (10 to 20 micrometers) yields 100 to 200 meters, comparable to the dimensions of a stadium. It should be noted that a plant cell can be 200 µm across, corresponding to the size of a town in our analogy; on the other hand, a bacterium at 2 µm would be only as large as a tennis court. 2. Answer: EFC Difficulty: 1 Section: Looking at Cells in the Light Microscope Feedback: A typical animal cell (F) is between 10 to 20 micrometers in diameter. This is an order of magnitude larger than a typical bacterium (E). A typical globular protein (C) is much smaller, only a few nanometers in diameter. 3. Answer: FTFT Difficulty: 1 Section: Looking at Cells in the Light Microscope Feedback: Total destructive interference (complete canceling) only occurs when the two waves are out of phase by exactly 180 degrees. The formal resolution limit of light microscopy is about 0.2 µm. 4. Answer: I Difficulty: 2 Section: Looking at Cells in the Light Microscope Feedback: A higher numerical aperture and therefore a better resolution can be achieved with higher values of θ. 5. Answer: B Difficulty: 1 Section: Looking at Cells in the Light Microscope Feedback: In dark-field microscopy, oblique rays of light do not enter the objective directly. Instead, some of the scattered rays from objects (such as cells and their components) enter the objective to create a bright image against a dark background. 6. Answer: B Difficulty: 1 Section: Looking at Cells in the Light Microscope

Feedback: The photon emitted by a fluorescent molecule is necessarily of lower energy and longer wavelength than the absorbed photon. Therefore the excitation light has greater energy and a shorter wavelength than the emitted light. 7. Answer: HOHOP Difficulty: 2 Section: Looking at Cells in the Light Microscope Feedback: Sequence-specific staining of nucleic acids can be achieved by in situ hybridization of probes that are attached to fluorescent dyes. If general DNA staining is desired, a fluorescent organic molecule such as DAPI can be used. Fluorescent or luminescent ion-sensitive indicators (many of which are small molecules) can be used to measure cellular ionic concentrations in real time. A protein can be genetically engineered as a fluorescent fusion protein, and then imaged in living cells by fluorescence microscopy. 8. Answer: D Difficulty: 1 Section: Looking at Cells in the Light Microscope Feedback: Quantum dots have several advantages over organic fluorescent dyes, including higher brightness, stability, and emission tunability. 9. Answer: CCC Difficulty: 1 Section: Looking at Cells in the Light Microscope Feedback: Confocal microscopy provides a rapid way of eliminating blurring, but requires brighter illumination. 10. Answer: D Difficulty: 1 Section: Looking at Cells in the Light Microscope Feedback: In the two-photon effect, which is employed by multiphoton microscopes, fluorescent molecules are excited almost simultaneously by two (or more) photons of longer wavelength compared to one-photon excitation. The longer wavelength provides advantages including the ability to image deep within live tissues. 11. Answer: C Difficulty: 3 Section: Looking at Cells in the Light Microscope Feedback: Based on these results, S1 is neither necessary nor sufficient for nuclear import. In contrast, S2 is necessary (because nuclear localization is lost without it) and sufficient (because it causes nuclear localization by itself without the need for other segments). 12. Answer: N

Difficulty: 2 Section: Looking at Cells in the Light Microscope Feedback: If the gene encoding the fusion cannot rescue a loss-of-function mutation of the original gene, the presence of the fused protein might be interfering with gene function. 13. Answer: AB Difficulty: 2 Section: Looking at Cells in the Light Microscope Feedback: Good FRET pairs do not have extensive overlapping between their emission spectra or between their excitation spectra. Additionally, the emission spectrum of the donor should overlap with the excitation spectrum of the acceptor, such that Förster resonance energy transfer can be significant. 14. Answer: 2 Difficulty: 3 Section: Looking at Cells in the Light Microscope Feedback: The FRET signal in the green curve (negatively correlated with donor fluorescence in the blue curve) is expected to be high when the two proteins can interact in the same compartment. 15. Answer: E Difficulty: 3 Section: Looking at Cells in the Light Microscope Feedback: The pulse of laser light photoactivates the incorporated tubulins in the region of the spindle irradiated. After about 5 minutes, most of the activated tubulin molecules near the pole will have quickly moved toward the poles and left the spindle, which no longer fluoresces, while those activated near the equator will have moved along the spindle microtubules toward the poles and they still fluoresce. 16. Answer: 1 Difficulty: 3 Section: Looking at Cells in the Light Microscope Feedback: Since the mutant protein cannot efficiently bind GTP, it takes longer to recover the fluorescence in the ER membrane after photobleaching. 17. Answer: C Difficulty: 3 Section: Looking at Cells in the Light Microscope Feedback: Aequorin is a calcium-sensitive luminescent protein that emits blue light in the presence of the ion. It can be expressed in various cells to monitor changes in calcium concentration. 18. Answer: C

Difficulty: 2 Section: Looking at Cells in the Light Microscope Feedback: On excitation with violet light, such a protein would shift its emission peak toward longer wavelengths, from blue to green, upon a calcium concentration increase. 19. Answer: D Difficulty: 1 Section: Looking at Cells in the Light Microscope Feedback: In a total internal reflection fluorescence (TIRF) microscope, an evanescent wave excites only those fluorescent molecules that lie within an extremely thin plane adjacent to the cover slip. 20. Answer: 4 Difficulty: 3 Section: Looking at Cells in the Light Microscope Feedback: Each peak normally represents one unfolding event. The last peak corresponds to the rupture of the polypeptide and loss of tension at high extensions. 21. Answer: 1 Difficulty: 3 Section: Looking at Cells in the Light Microscope Feedback: The force required to unfold the domains is expected to be higher in the presence of the chaperone protein. 22. Answer: FTFT Difficulty: 2 Section: Looking at Cells in the Light Microscope Feedback: The highest resolution achieved by superresolution techniques has been about 20 nm, an order of magnitude past the limit set by light diffraction. Different methods overcome the limit in various ways. In structured illumination microscopy (SIM), this is done by using a patterned light source that is translated and rotated with respect to the sample to collect a series of interference images that are then processed. Stimulated emission depletion microscopy (STED) overcomes the limit by changing the shape of the point spread function, whereas localization methods approximate the center of the point spread function by stochastic sampling. 23. Answer: PSST Difficulty: 2 Section: Looking at Cells in the Light Microscope Feedback: Different superresolution fluorescence techniques overcome diffractionlimited resolution in different ways. For example, through illumination by a grid, structured illumination microscopy (SIM) can improve the resolution by about a factor of two. Stimulated emission depletion microscopy (STED) uses a bright torus-shaped lased beam to switch the fluorescent molecules off at the periphery of the point spread

function. Single-molecule localization methods such as photoactivated localization microscopy (PALM) or stochastic optical reconstruction microscopy (STORM) use lasers to sequentially switch on a sparse subset of fluorescent molecules in a specimen. 24. Answer: TTTT Difficulty: 1 Section: Looking at Cells and Molecules in the Electron Microscope Feedback: The practical resolving power of an electron microscope can be about 0.05 nm. However, problems in specimen preparation, contrast, and radiation damage can limit the effective resolution to 1 nm for biological samples. Emission gun in electron microscopy is analogous in function to a light source in light microscopy; the magnetic coils focus the electron beams, just as glass lenses focus the light; and the use of electrondense material to achieve contrast in electron microscopy is similar to the use of lightabsorbing dyes in light microscopy. 25. Answer: E Difficulty: 1 Section: Looking at Cells and Molecules in the Electron Microscope Feedback: All of these methods can be used for flash-freezing of the samples. 26. Answer: D Difficulty: 1 Section: Looking at Cells and Molecules in the Electron Microscope Feedback: Contrast in the electron microscope depends on the electron-scattering power of the atoms in the specimen, which in turn depends on their atomic number. Thus all the atoms here have high atomic numbers, and high electron contrast, except carbon, which has an atomic number of 6 and is a major element in biological specimens to start with. 27. Answer: SSST Difficulty: 1 Section: Looking at Cells and Molecules in the Electron Microscope Feedback: Scanning electron microscopy is generally simpler and has a greater depth of field. It is used to see the surface of specimens, which can be entire plant parts or small animals. 28. Answer: SEM Difficulty: 1 Section: Looking at Cells and Molecules in the Electron Microscope Feedback: Often an entire plant part or small animal can be put into the microscope with very little preparation. 29. Answer: D Difficulty: 1 Section: Looking at Cells and Molecules in the Electron Microscope

Feedback: Cryoelectron microscopy reveals the electron density of entire particles, not just their surface.

What is the typical thickness of a lipid bilayer such as the plasma membrane of our cells? A. 0.5 nm B. 5 nm C. 50 nm D. 100 nm E. 500 nm

2. In the following schematic drawing of an abundant plasma membrane phosphoglyceride, which part is positively charged?

A B C

A. A B. B C. C D. D E. E

3. Which of the following is NOT correct regarding the molecule whose structural formula is shown below?

A. It is an amphiphilic molecule. B. It is a sterol. C. It makes the membrane less permeable to small hydrophilic molecules. D. It is found in membranes of virtually all living cells. E. It affects the fluidity of the lipid bilayer. 4.

Which of the following is normally NOT found in a eukaryotic membrane? A. Cholesterol B. Phosphatidylinositol C. Sphingomyelin D. Ganglioside GM1 E. Octylglucoside

5. The motion of lipid molecules in a synthetic bilayer can be studied by various techniques. Which of the following has been observed in these systems? A. Phospholipids diffuse rapidly within and between the two leaflets of a bilayer. B. An average lipid molecule can diffuse the length of about 2 micrometers in a fraction of a millisecond. C. The flip-flops are very rare for phospholipids but cholesterol molecules flip-flop more often. D. Within a bilayer, lipid molecules rarely rotate about their long axis, but diffuse laterally at very high rates. E. All of the above.

6. Why do liposomes not fuse with one another spontaneously when suspended in an aqueous environment? A. Because fusion requires a large number of flip-flops, which are very rare. B. Because the hydration shell of the polar head groups of the lipids needs to be removed. C. Because fusion requires micelle formation. D. Because of the rapid lateral diffusion and rotation of the lipid molecules. 7. Which of the following changes would you expect to increase the phase transition temperature of a synthetic bilayer composed of phosphatidylserine? A. Incorporation of phospholipids with longer fatty acid chains. B. Introduction of double bonds in the fatty acids. C. Addition of cholesterol. D. Removal of serine from the head group. E. None of the above. 8. Which of the following is correct regarding the composition of various biological membranes? A. Bacterial plasma membranes are often composed of one main type of phospholipid and lack cholesterol. B. Cholesterol in the eukaryotic plasma membrane induces phase transition to the gel state. C. Inositol phospholipids are the most abundant lipids in the endoplasmic reticulum membrane. D. The mitochondrial and bacterial membranes are rich in glycolipids. E. Yeast cells synthesize more fatty acids with cis-double bonds when the temperature in the environment rises. 9.

The two monolayers of the plasma membrane in a human red blood cell … A. have different overall electrical charges, with negatively charged phospholipids (e.g. phosphatidylserine) normally enriched in the inner monolayer. B. have the same abundance of phosphatidylinositol.

C. exchange phospholipids only through spontaneous flip-flops. D. both contain glycolipids. E. both contain gangliosides. 10.

Many cells store lipids in droplets of varying sizes. These droplets … A. are enclosed by a phospholipid monolayer (instead of a bilayer). B. mostly store cholesterol and phospholipids. C. are produced by and released from the Golgi apparatus. D. have mostly protein-free bilayer membranes. E. are composed primarily of charged amphiphilic lipids.

11. Which of the following lipids do you expect to be a canonical scramblase substrate in the plasma membrane? A. Ganglioside GM1 B. Cholesterol C. Glycosylphosphatidylinositol D. Phosphatidylethanolamine E. Galactocerebroside 12.

Glycolipids such as gangliosides … A. may contain oligosaccharide chains with negatively charged residues. B. are found to constitute about 10% of the total lipid mass in the plasma membrane of neurons. C. are found in the extracellular leaflet (facing away from the cytosol) in the cellular membranes. D. affect the electrical environment of the membrane. E. All of the above

13.

Transmembrane proteins … A. are typically exposed only to one side of the membrane. B. can be released from the membrane by a gentle extraction procedure such as salt treatment. C. are often further attached to the membrane via a GPI anchor. D. are sometimes covalently attached to a fatty acid chain that inserts into the membrane. E. cannot contain β sheets in the part of their structure that interacts with the membrane interior.

14. Integrins are single-pass integral membrane proteins in the plasma membrane of animal cells and are involved in the interaction of the cell with the surrounding extracellular matrix. Which of the following descriptions do you think matches the transmembrane part of an integrin molecule? A. It forms a β barrel. B. It is an α helix that is bent in the middle. C. It is about 10 amino acids long, with every other amino acid side chain being hydrophobic. D. It folds in a conformation with maximal intrachain hydrogen-bonding. E. It is about 100 amino acids long. 15. The nicotinic acetylcholine receptor is a cation channel in the plasma membrane of some neurons. It is composed of five subunits, each of which is a transmembrane protein. The hydropathy plot for each mature subunit is qualitatively represented in the following diagram. How many membrane-spanning alpha helices do you expect to exist in each subunit?

+25 Hydropathy index

–25 1

100

200

300

400

Residue number 16.

What do all β-barrel transmembrane proteins have in common? A. The number of β strands. B. The diameter of the barrel. C. The number of negative peaks in their hydropathy plots. D. The general function, i.e. membrane transport. E. The structural rigidity compared to α-helical transmembrane proteins.

17. For each membrane protein in the following schematic drawings, indicate whether the cytoplasmic side of the membrane is more likely to be on the left (L) or on the right (R). Your answer would be a four-letter string composed of letters L and R only, e.g. RRRR.

S H

= sugar residue

= fatty acid

= GPI anchor

18. Under certain conditions, about 20 molecules, on average, are predicted to exist in each β-octylglucoside detergent micelle. If the critical micelle concentration for this detergent is about 20 mM under these conditions, at what total β-octylglucoside concentration are about half of the detergent molecules found in micelles? A. 1 mM B. 10 mM C. 20 mM D. 21 mM E. 40 mM 19. On the plasma membrane of the archaeon Halobacterium salinarum, bacteriorhodopsin is found in high density in patches of purple membrane. This protein … A. is a light-gated anion channel. B. is a single-pass transmembrane protein. C. uses the energy stored in the proton gradient to transport small molecules. D. changes conformation in response to light. E. All of the above. 20. Which of the following proteins is NOT in the same superfamily as the other four? The other four share an overall architecture composed of seven closely-packed transmembrane helices. A. Bacteriorhodopsin B. Channelrhodopsin C. G-protein-coupled receptor

D. Aquaporin E. Rhodopsin 21. While examining the crystal structure of a membrane protein, you find several phospholipid molecules bound to the protein. You know that these lipids … A. are thought to help stabilize many membrane proteins. B. may enhance the crystallization of the bound membrane proteins. C. interact specifically with the protein. D. can have head groups of various sizes and charges depending on the protein. E. All of the above.

Relative fluorescence intensity

22. The lateral diffusion of a plasma membrane protein has been studied using the fluorescence recovery after photobleaching (FRAP) technique in a cell under normal conditions, and the following plot has been obtained. Under a different condition (e.g. when a signaling pathway is turned on), the protein binds to a few other proteins and protein complexes in the plasma membrane. As a result, its lateral diffusion coefficient is reduced and a significant fraction of the molecules of this protein are immobilized in large aggregates. How would you expect the FRAP curve to change under these new conditions? The dotted line in each graph is the original curve shown for comparison.

100 Time (s)

E. 23. In intestinal epithelial cells, the different plasma membrane domains are separated from each other by special barriers. Indicate true (T) and false (F) statements below regarding this separation. Your answer would be a four-letter string composed of letters T and F only, e.g. FFFF. ( ) These barriers are set up by nanoscale lipid raft domains. ( ) Proteins normally cannot pass through these barriers, whereas any membrane lipid can do so freely. ( ) The asymmetric distribution of membrane proteins resulting from these barriers is functionally important. ( ) These barriers are a result of proteins of the apical surface forming a large aggregate together that excludes proteins of the other domains. 24. Which of the following changes is more likely to further confine a membrane protein within a corral established by the cortical cytoskeletal network? A. Increased temperature B. Protease cleavage of the extracellular domain of the protein C. Binding to an extracellular ligand D. Protease cleavage of the cytosolic domain of the protein E. Binding of the protein to other transmembrane proteins 25. In the following schematic drawing of the cytosolic side of human red blood cell plasma membrane, which of the labeled proteins is spectrin?

A B C D

26.

0.1 µm

In the schematic drawing below, a protein is present on one side of a membrane in a high-

enough quantity to induce curvature in this region of the membrane. In each of the following scenarios, indicate whether the protein would induce positive (P) or negative (N) curvature in the membrane. Your answer would be a three-letter string composed of letters P and N only, e.g. PPP.

Positive curvature

Negative curvature

( ) The protein inserts a hydrophobic domain into the membrane leaflet that is facing it. ( ) The protein is a phospholipase that cleaves the head group in specific phospholipids. ( ) The protein binds cooperatively to multiple phospholipids with very large head groups.

Answers 1. Answer: B Difficulty: 2 Section: The Lipid Bilayer Feedback: The thickness of cellular lipid bilayers depends on various factors, such as the lipid content, the length and the degree of saturation of the fatty acids, and temperature, but it is normally between 4 and 5 nanometers. 2. Answer: A Difficulty: 3 Section: The Lipid Bilayer Feedback: In the drawing, A can be one of many possible choices (including the positively charged choline or ethanolamine), B is the phosphate moiety, C is the glycerol, and D and E are fatty acids. 3. Answer: D Difficulty: 2 Section: The Lipid Bilayer Feedback: The structure shown is that of cholesterol, an amphiphilic sterol that affects the permeability and fluidity of the membranes in eukaryotic cells. Most prokaryotic membranes lack cholesterol. 4. Answer: E Difficulty: 1 Section: The Lipid Bilayer Feedback: Octylglucoside is a nonionic detergent. Each of the other molecules mentioned can be highly abundant in a eukaryotic cell. 5. Answer: C Difficulty: 2 Section: The Lipid Bilayer Feedback: In contrast to high lateral diffusion and rotation rates, flip-flop rates are very low for phospholipids, although not so low for cholesterol. Due to the high lateral diffusion coefficients, an average lipid molecule can travel the length of a bacterial cell (approximately 2 µm) in about a second. 6. Answer: B Difficulty: 1 Section: The Lipid Bilayer Feedback: For fusion to take place, the water layer that binds the head groups of the lipid molecules should be removed to allow the two membranes to meet. 7. Answer: A Difficulty: 3

Section: The Lipid Bilayer Feedback: With longer hydrocarbon chains, the phase transition temperature increases, i.e. the membrane can freeze more readily. In contrast, introduction of double bonds in the fatty acids chains creates kinks in the chains and makes it harder to pack them together. At physiological pH, removal of serine adds a negative charge to the head group, also tending to limit tight packing. The effect of cholesterol is more complicated, but at the high concentrations found in eukaryotic plasma membranes, cholesterol also prevents the hydrocarbon chains from crystallizing. 8. Answer: A Difficulty: 1 Section: The Lipid Bilayer Feedback: Often one main type of phospholipid makes up the membrane in a bacterium, and these membranes lack cholesterol. Glycolipids are also rare in the mitochondrial and bacterial membranes, as are the functionally important phosphoinositides in eukaryotic membranes. The role of cholesterol in membrane fluidity and phase transition is a complicated one and depends on many variables, but at concentrations found in eukaryotic membranes, cholesterol prevents the packing of phospholipids into the gel state. Unsaturated fatty acids enhance the fluidity of the bilayer, as does the rise in temperature. 9. Answer: A Difficulty: 2 Section: The Lipid Bilayer Feedback: While glycolipids such as gangliosides are limited to the outer monolayer, phosphatidylinositol and the negatively charged phosphatidylserine are mostly found in the inner monolayer. Membrane proteins actively maintain the asymmetry of the bilayer by regulated transport of the lipids between the monolayers. 10. Answer: A Difficulty: 2 Section: The Lipid Bilayer Feedback: Unlike vesicles, lipid droplets are surrounded by a lipid monolayer as their interior (mainly composed of cholesterol esters and triacylglycerides) is hydrophobic. 11. Answer: D Difficulty: 3 Section: The Lipid Bilayer Feedback: Scramblases normally scramble phospholipids (such as phosphatidylethanolamine) across the membrane bilayer. 12. Answer: E Difficulty: 3 Section: The Lipid Bilayer

Feedback: Exclusively found in the extracellular leaflet of the plasma membrane, glycolipids can be negatively charged and can have an effect on the electrical environment in the membrane. They are significantly abundant in neuronal membranes. 13. Answer: D Difficulty: 2 Section: Membrane Proteins Feedback: Many transmembrane proteins are modified with the covalent attachment of lipids that insert into the membrane. 14. Answer: D Difficulty: 2 Section: Membrane Proteins Feedback: Folding of a transmembrane segment into an α -helical conformation maximizes hydrogen-bonding in the absence of water. The hydrogen-bonding requirement is also satisfied in the multipass β barrel structures. 15. Answer: 4 Difficulty: 3 Section: Membrane Proteins Feedback: As suggested by the presence of four positive hydropathy peaks in the plot (three between residues 200 and 300, and one at the C-terminus), there are four transmembrane helices in each subunit. 16. Answer: E Difficulty: 2 Section: Membrane Proteins Feedback: The β-barrel arrangement results in higher structural rigidity compared to α helices. The β-barrel proteins vary in their size and the number of strands, and only some of them form membrane-spanning pores for transport. 17. Answer: LLRL Difficulty: 3 Section: Membrane Proteins Feedback: Disulfide bonds are normally not found in the reducing environment of the cytosol; neither are glycosylated protein regions or glycosylphosphatidylinositol (GPI)anchored proteins. Many cytosolic proteins interact with the membrane via a covalently attached fatty acid chain and/or prenyl groups. 18. Answer: E Difficulty: 3 Section: Membrane Proteins Feedback: Concentration of the free detergent remains constant above the critical concentration of 20 mM. Thus, at 40 mM total concentration, half of the detergent molecules are free and the other half are in micelles. 19. Answer: D

Difficulty: 1 Section: Membrane Proteins Feedback: These bacteriorhodopsins are light-driven proton pumps and, similar to the rhodopsins in our retina, have seven transmembrane helices in their structure. 20. Answer: D Difficulty: 1 Section: Membrane Proteins Feedback: Although functionally diverse, rhodopsins, bacteriorhodopsin, channelrhodopsin, and G-protein-coupled receptors share an overall architecture. 21. Answer: E Difficulty: 1 Section: Membrane Proteins Feedback: Interactions with specific lipids of various types are thought to help stabilize many membrane proteins and enhance their crystallization. 22. Answer: C Difficulty: 3 Section: Membrane Proteins Feedback: The fraction of the molecules that are practically mobile is represented by the final fluorescence recovery. The diffusion rate can also be deduced from the curve and is related to the rate of recovery. 23. Answer: FFTF Difficulty: 2 Section: Membrane Proteins Feedback: The barriers are set up by special cell–cell junctions known as tight junctions, which limit the diffusion of proteins (and mostly the lipids in the extracellular monolayer) of the plasma membrane between the apical and the basolateral domains. 24. Answer: E Difficulty: 2 Section: Membrane Proteins Feedback: The size of the cytoplasmic domain and, more importantly, the association with other proteins affect the extent to which a transmembrane protein is confined within a corral. 25. Answer: A Difficulty: 2 Section: Membrane Proteins Feedback: In the spectrin-based filamentous cytoskeleton on the cytosolic side of the plasma membrane in red blood cells, the spectrin tetramers create a meshwork in association with several other proteins. The spectrin network is important for the normal functioning of the red blood cells and other cells. 26. Answer: PNP

Difficulty: 3 Section: Membrane Proteins Feedback: Proteins can use various strategies to create curvature in cellular membranes: by inserting a hydrophobic domain (or attached lipid anchor), by acting as rigid scaffolds that directly deform the membrane, or by biasing the shape of phospholipids present in a particular region of the membrane (e.g. by clustering phospholipids that have large head groups).

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION CHAPTER 11 MEMBRANE TRANSPORT OF SMALL MOLECULES AND THE ELECTRICAL PROPERTIES OF MEMBRANES Copyright © 2015 by W.W. Norton & Company, Inc. 1. The permeability of a protein-free lipid bilayer to various molecules depends on their properties. Sort the following in order of low to high permeability from left to right. Your answer would be a six-digit number composed of digits 1 to 6, e.g. 123456. (A) O2 (B) ATP (C) RNA oligonucleotide (D) Na+ (E) Glucose (F) Urea 2.

Why do cells not have membrane transport proteins for O2? A. Because they need to keep the oxygen concentration low inside the reducing environment of the cell. B. Because oxygen can dissolve in water and leak in via water channels. C. Because oxygen can dissolve in the lipid bilayer and diffuse in and out rapidly without the need for a transporter. D. Because oxygen is transported in and out of the cell in special oxygen-carrying proteins such as hemoglobin. E. Because oxygen transport across a membrane is energetically unfavorable.

3. For which of the following ions is the intracellular concentration typically higher than the extracellular concentration? A. Sodium B. Calcium C. Magnesium D. Potassium E. Chloride

4. Imagine a small synthetic vesicle made from pure phospholipids enclosing an interior lumen containing 1 mM glucose and 1 mM sodium chloride. If the vesicle is placed in pure water, which of the following happens faster? A. Na+ diffuses out. B. Cl– diffuses out. C. H2O diffuses in. D. Glucose diffuses out. E. Sodium chloride diffuses out. 5.

In contrast to transporters, the channel proteins in cellular membranes … A. interact strongly with the solute(s) that they transport. B. undergo a conformational change every time they transport a solute. C. can only mediate passive transport. D. form pores that are always open.

6. After release into the synaptic cleft, the neurotransmitter dopamine is actively taken up by the cells via specific dopamine transporters. The drug cocaine interferes with this process and is therefore called a reuptake inhibitor. The inhibition of the transporter at a certain cocaine concentration is qualitatively represented in the following graph. Based on this graph, which of

Relative rate of dopamine reuptake

the following describes the effect of cocaine on the kinetics of dopamine reuptake by the transporter?

1 Without cocaine

With cocaine

Dopamine concentration (µM)

100

A. In the presence of cocaine, the maximal rate of transport (Vmax) is unaffected, but the apparent affinity of the transporter for dopamine is reduced. B. In the presence of cocaine, the maximal rate of transport (Vmax) is reduced, but the apparent affinity of the transporter for dopamine is unaffected.

C. In the presence of cocaine, the maximal rate of transport (Vmax) is enhanced, but the apparent affinity of the transporter for dopamine is unaffected. D. In the presence of cocaine, both the maximal rate of transport (Vmax) and the apparent affinity of the transporter for dopamine are enhanced. E. In the presence of cocaine, both the maximal rate of transport (Vmax) and the apparent affinity of the transporter for dopamine are unaffected, but the transporter is nevertheless inhibited. 7. Which of the following transporters mediates primary active transport when transporting the solutes in the directions indicated? The concentration of both solutes (shown as small circles and squares) is higher outside of the cell.

E OUTSIDE

INSIDE

8. The lactose permease in Escherichia coli is an H+–lactose symporter that mediates the inward active transport of lactose if this sugar is present in the environment instead of glucose. Which of the following is true about this transporter? A. It has a twofold pseudosymmetrical structure. B. Lactose and H+ bind to two different conformations of the transporter. C. The transporter goes through an intermediate state in which the bound lactose is open to both sides of the membrane. D. If lactose and H+ concentrations are changed sufficiently, the transporter can act as an H+–lactose antiporter. E. All of the above. 9.

Which of the following normally functions to lower the pH of the cytosol? A. A Na+–H+ exchanger in the plasma membrane. B. A Na+-driven Cl––HCO3– exchanger in the plasma membrane. C. A Na+-independent Cl––HCO3–exchanger in the plasma membrane.

D. A V-type ATPase in the lysosomal membrane. E. Both answers A and B above. 10. Many amino acids in our diet are absorbed via the transcellular transport pathway by the intestinal epithelial cells. This process requires ATP hydrolysis by … A. the Na+–amino acid symporters in the apical domain of the plasma membrane. B. the Na+–amino acid antiporters in the apical domain of the plasma membrane. C. the Na+-K+ pumps in the basal and lateral domains of the plasma membrane. D. the amino acid carriers in the basal domain of the plasma membrane. E. F-type ATPases in the apical domain of the plasma membrane. 11. This family of ATPases is structurally related to the turbine-like pumps that acidify lysosomes and vesicles; however, they usually function in reverse, generating ATP from ADP and Pi using proton gradients across membranes. What are they called? A. P-type pumps B. ABC transporters C. V-type pumps D. F-type pumps E. Permeases 12. Indicate whether each of the following descriptions matches an ABC transporter (A), a Ptype pump (P), or a V-type pump (V). Your answer would be a five-letter string composed of letters A, P, and V only, e.g. PPVVV. ( ) The pumps in this family are phosphorylated at a key Asp residue in each transport cycle. ( ) This family is the largest among membrane transport proteins and includes some channels as well as pumps. ( ) The pumps in this family are responsible for the acidification of synaptic vesicles. ( ) The sodium-potassium pump is a member of this family. ( ) The multidrug resistance protein is a member of this family. 13.

Which of the following transporters is NOT electrogenic? A. Na+–glucose symporter B. Ca2+-pump C. Na+-K+ pump D. Na+-independent Cl––HCO3– exchanger

E. Bacteriorhodopsin 14. Most eukaryotic ABC transporters are involved in exporting small molecules from the cytosol. In this subset of the ABC transporters, does the small molecule bind better to the ATPbound (B) or the ATP-free (F) transporter? Write down your answer as B or F. Hint: The ATPbound transporter “faces outward.” 15. Which of the following is a pump that hydrolyzes two ATP molecules per transport cycle? A. The cystic fibrosis transmembrane conductance regulator protein B. The multidrug resistance protein C. The Na+-K+ pump D. The Ca2+-pump E. The V-type ATPase 16.

An ion channel … A. always mediates passive transport. B. is ion-selective. C. is typically several orders of magnitude faster than a transporter. D. is usually gated. E. All of the above.

17. The Nernst equilibrium potential for an ion that is 10 times more concentrated in the cytosol compared to the extracellular fluid is about –60 mV. How much would the potential be if the extracellular concentration decreases 100-fold with no change in the intracellular concentration? A. –6000 mV B. –6 mV C. –600 mV D. –20 mV E. –180 mV 18. The membrane potential in a particular hyperpolarized cell is measured to be –70 mV. For each of the following channels in this cell, the calculated Nernst equilibrium potential for the corresponding ion is presented. Indicate whether each channel is driven to pass the ion into (I) or

out of (O) the cell. Your answer would be a four-letter string composed of letters I and O only, e.g. IIII. ( ) K+ channel; the K+ equilibrium potential is –60 mV. ( ) Na+ channel; the Na+ equilibrium potential is +70 mV. ( ) Ca2+ channel; the Ca2+ equilibrium potential is +130 mV. ( ) Cl– channel; the Cl– equilibrium potential is –90 mV. 19. Sanshool is a natural compound found in Sichuan pepper, which is commonly used in making spicy Asian food, and creates a numbing sensation in the mouth. It is known to inhibit a subset of potassium leak channels involved in maintaining the resting membrane potential in sensory neurons. How do you think sanshool affects these sensory neurons? A. It elevates the resting potential (to less negative values) and makes it easier to excite the neuron. B. It elevates the resting potential and makes it harder to excite the neuron. C. It lowers the resting potential and makes it easier to excite the neuron. D. It lowers the resting potential and makes it harder to excite the neuron. E. It causes hyperpolarization and decreases the spontaneous firing of the neuron. 20.

Which of the following is correct regarding the electrical properties of cellular

membranes? A. Setting up (or changing) the membrane potential requires changing the bulk concentrations of the ions on the two sides of the membrane. B. The resting potential in most animal cells is between 20 mV and 120 mV (positive inside). C. If the membrane is impermeable to an ion, the membrane potential approaches the equilibrium potential for that ion. D. The resting potential decays immediately following the inhibition of the Na+-K+ pump by a drug. E. None of the above.

21. A membrane potential is plotted over time in the following graph. The Nernst equilibrium potentials for four ions (A to D) that affect the membrane potential in this system are indicated in the graph. What happened in the time period indicated by a question mark?

Relative potential

B C

5 Time (ms)

A. Channels conducting ion A opened. B. Channels conducting ion B opened. C. Channels conducting ion C opened. D. Channels conducting ion D opened. E. Channels conducting ion A and C opened. 22.

A potassium channel conducts K+ ions several orders of magnitude better than Na+ ions,

because … A. the Na+ ion is too large to pass through the channel pore. B. the hydrated Na+ ion occupies a larger volume compared to the hydrated K+ ion, and is too large to pass through the channel pore. C. the Na+ ion is too small to interact with the channel in a way that facilitates the loss of water from the ion. D. the Na+ ion cannot bind to the high-affinity K+-binding sites in the channel pore. 23.

Which of the following is the reason why an aquaporin does not pass ions (such as Na+)

through? A. The Na+ ions are too small to pass through the pore rapidly enough. B. The Na+ ions are too large to pass through the narrow pore. C. The Na+ ions do not favorably interact with the wall of the pore and therefore cannot be dehydrated. D. Two Asn residues in the channel prevent the passage of Na+ ions. E. A Na+ ion must become fully hydrated to pass through.

24. Aquaporin has a pair of key asparagine residues located on the wall almost halfway through its pore. These residues simultaneously bind to the oxygen atom of a passing water molecule, rendering it unavailable for hydrogen-bonding. As a result, … A. the channel can conduct negatively charged (but not positively charged) ions. B. the channel can conduct glycerol as well as water. C. the channel cannot conduct glycerol. D. the channel cannot conduct protons. E. the channel cannot conduct chloride ions. 25.

MscS and MscL are mechanosensitive channels that ... A. constitute the majority of channel proteins in the plasma membrane of auditory hair cells in the human cochlea. B. open in response to membrane tension. C. open at the same osmotic pressure, but differ in their ion selectivity. D. are highly ion selective. E. All of the above.

26.

A neuron’s repetitive firing rate is limited by an absolute refractory period, during which

a new action potential cannot be generated. Which event is chiefly responsible for this limit? A. Opening of K+ leak channels. B. Opening of voltage-gated Na+ channels. C. Opening of transmitter-gated cation channels. D. Inactivation of voltage-gated K+ channels. E. Inactivation of voltage-gated Na+ channels. 27.

If the extracellular concentration of sodium ions is artificially decreased for a neuron, … A. both the resting potential and the peak of the action potential rise significantly. B. the resting potential increases but the action potential peak does not. C. the resting potential hardly changes but the action potential peak is raised. D. the resting potential decreases but the action potential peak is raised. E. the resting potential hardly changes but the action potential peak is lowered.

28. Which of the following situations in a neuron results in action potentials that have extended depolarization phases? A. Inhibiting the voltage-gated Na+ channels.

B. Stimulating the voltage-gated K+ channels. C. Compromising the inactivation mechanism in voltage-gated Na+ channels. D. Compromising the inactivation mechanism in voltage-gated K+ channels. E. None of the above. 29.

Myelination of axons in the peripheral nervous system … A. is carried out by Schwann cells. B. insulates the axons to reduce current leakage. C. increases the speed of action potential propagation through saltatory conduction. D. conserves energy because the active excitations are restricted to the nodes of Ranvier. E. All of the above.

30. Once expressed in a neuron in the mouse brain and subjected to a flash of light, how does channelrhodopsin affect the membrane potential? A. It causes neurotransmitter release, leading to an action potential in the postsynaptic neuron. B. It increases the membrane permeability to cations and makes the membrane potential more negative. C. It depolarizes the membrane, leading to the opening of voltage-gated Na+ channels. D. It hyperpolarizes the membrane, leading to the inactivation of voltage-gated K+ channels. E. It changes the membrane potential such that voltage-gated K+ channels but not voltage-gated Na+ channels are activated. 31. In a voltage-clamp experiment on a squid giant axon, the membrane potential is clamped at +20 mV, as shown below. Within the same time frame, which of the curves A to D better show qualitatively how the total membrane current changes? A positive value of the current represents an outward flow of cations.

mA/cm2

+2 0 –2

32.

A B C

Total current D

Voltage-gated cation channels in the plasma membrane of a neuron open when the

membrane potential passes a threshold value that is about 20 mV above the resting potential. This corresponds to a change of about 4 × 106 newtons per coulomb in the magnitude of the electric field across the membrane compared to the resting state. The tetrameric channels have a voltage-sensing helix S4 in each of their four subunits. Most S4 helices contain seven residues with positively charged side chains. The charge carried by each such residue is about 1.6 × 10–19 coulombs. By reaching threshold depolarization, what are the magnitude and the direction of the collective force that the four S4 helices in the channel experience as a result of the change in the electric field? The force that particles of charge q in an electric field E experience is calculated as F = q.E. A. 18 pN, toward the cell interior B. 18 pN, toward the cell exterior C. 90 pN, toward the cell interior D. 90 pN, toward the cell exterior E. 45 pN, toward the cell interior

33. Which of the following graphs better represents the change in the relative conductance of Na+ and K+ channels during an action potential in a neuronal membrane? Na+

mS/cm2

50 25

K+ 0 0

Na+

mS/cm2

4 2 Time (ms)

K+

25 0 0

4 2 Time (ms)

Na+

mS/cm2

50 25

K+

0 0

4 2 Time (ms)

Na+

mS/cm2

K+ 0 0

Na+

mS/cm2

4 2 Time (ms)

K+

25 0 0

4 2 Time (ms)

34. Indicate true (T) and false (F) statements below regarding ion channels in cellular membranes. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTT. ( ) With patch-clamp recordings, it is possible to study ion transport through a single molecule of a channel protein. ( ) Cells that are not electrically excitable often lack gated ion channels in their plasma membrane. ( ) The aggregate current crossing a patch of plasma membrane represents the degree to which each individual channel is open at that time. ( ) Patch-clamp recordings have shown that any voltage-gated ion channel opens as soon as the membrane potential is altered. 35. In neurons, the synaptotagmin-1 protein is localized to presynaptic axon terminals. When activated, it facilitates the fusion of neurotransmitter-containing synaptic vesicles to the plasma membrane, leading to neurotransmitter release into the synaptic cleft. Which of the following would you expect to directly activate synaptotagmin-1 for this function? A. Binding to Mg2+ B. Binding to Ca2+ C. Binding to phosphatidylcholine D. Binding to K+ E. Binding to the K+ leak channels 36. A certain neuron in a mouse brain is firing about 300 times per second, whereas almost all of its neighboring neurons are firing at much lower rates of only a few times per second. This implies that ... A. this neuron expresses more depolarizing (Na+ and Ca2+) than hyperpolarizing (K+) channels compared to most of its neighbors. B. the mouse is experiencing an epileptic seizure. C. the refractory period of its voltage-gated Na+ channels is at least 300 ms. D. it releases excitatory neurotransmitters at its axon terminals. E. it is probably of a different type compared to most of its neighbors. 37. From left to right, indicate whether the transmitter-gated channels for (1) calcium, (2) chloride, (3) potassium, and (4) sodium at a chemical synapse usually open in response to excitatory (E) or inhibitory (I) neurotransmitters. Your answer would be a four-letter string composed of letters E and I only, e.g. EEIE.

38. Astroglial cells in the brain can actively import the amino acid glutamate from the synaptic cleft, convert it to the amino acid glutamine, and release the glutamine into the cleft to be subsequently taken up by the presynaptic neurons. The neurons then convert it back to glutamate. Why is this function of glial cells important? A. Because it competes with and replaces the direct uptake of glutamate by the neurons and thus regulates neuronal function. B. Because neurons cannot produce the amino acids glutamine and glutamate and rely on the glial cells for their supply. C. Because glutamate cannot be taken up directly by the neurons. D. Because glutamate is an excitatory neurotransmitter and should be cleared from the synaptic cleft rapidly after its release. E. Because glutamate import through glutamate–Na+ symporters in the glial cells is required for the maintenance of Na+ concentration in the synaptic cleft. 39. The nicotinic acetylcholine receptor is an ionotropic receptor that can be activated by nicotine, a stimulant drug found in cigarettes. Additionally, α-bungarotoxin, a neurotoxin found in the venom of some snakes, binds preferentially and tightly to the closed conformation of the receptors. Another drug, donepezil, is a potent acetylcholinesterase inhibitor used to treat patients with Alzheimer’s disease. If applied at a neuromuscular junction, indicate whether each of the compounds (1) acetylcholine, (2) nicotine, (3) α-bungarotoxin, and (4) donepezil would enhance (E) or suppress (S) the Ca2+ influx in the muscle cells. Your answer would be a fourletter string composed of letters E and S only, e.g. EESS. 40. Neuromuscular transmission involves the sequential activation of several ion channels. Sort the following events to reflect the sequence leading to muscle contraction. Your answer would be a five-digit number composed of numbers 1 to 5 only, e.g. 12543. (A) Opening of voltage-gated Na+ channels in the muscle cell plasma membrane. (B) Opening of Ca2+ release channels in the sarcoplasmic reticulum membrane. (C) Opening of ionotropic acetylcholine receptors on the muscle cell plasma membrane. (D) Opening of voltage-gated Ca2+ channels in the T tubules. (E) Opening of voltage-gated Ca2+ channels in the motor neuron. 41. Dopamine is a neurotransmitter involved in the reward pathways in the brain, and its decreased activity has been associated with diseases such as Parkinson’s disease and attention-

deficit/hyperactivity disorder (ADHD). Which of the following drugs is NOT a likely candidate to treat these diseases? A. Methylphenidate, a dopamine reuptake inhibitor B. Amphetamine, a dopamine-releasing agent that triggers the release of dopamine into the synaptic cleft C. Carbidopa, which enhances the availability of the dopamine synthesis precursors in the brain D. Forskolin, a sensitizer of dopamine receptors E. Chlorpromazine, a dopamine antagonist that binds to and inhibits dopamine receptors 42.

Each single neuron acts as a computational device. Typically, the magnitude of the

combined postsynaptic potential is encoded in the ... of firing of action potentials in the postsynaptic neuron. 43. Which of the following better represents the process of adaptation mediated by the voltage-gated Ca2+ channels and Ca2+-activated K+ channels? The vertical axis in each graph is the membrane potential in the postsynaptic neuron (in millivolts), while the horizontal axis is the time of constant stimulation (in milliseconds). +30

–70

100

200

100

200

+30

–70

+30

–70

100

200

100

200

100

200

+30

–70

D. +30

–70

44. The inactivation rate of voltage-gated Na+ and K+ channels can regulate the firing frequency of neurons. Which of the following combinations results in the highest firing frequency? A. Rapid inactivation of both Na+ and K+ channels B. Rapid inactivation of Na+ channels, slow inactivation of K+ channels C. Slow inactivation of Na+ channels, rapid inactivation of K+ channels D. Slow inactivation of both Na+ and K+ channels 45. Which of the following is NOT common between long-term potentiation and long-term depression? A. They are both important for synaptic plasticity in normal individuals. B. They both require the activation of the glutamate-gated NMDA receptors. C. They both require a rise in the intracellular concentration of Ca2+. D. They both result in an enhanced postsynaptic response. E. They can both last for a long time.

46. Sort the following events in a sequence that would lead to long-term potentiation (LTP). Your answer would be a five-digit number composed of numbers 1 to 5 only, e.g. 13245. (A) Activation of AMPA receptors (B) Glutamate release into the synaptic clefts (C) Activation of calcium-dependent protein kinases in the postsynaptic cell (D) Postsynaptic membrane depolarization (E) Activation of NMDA receptors

Answers 1. Answer: CBDEFA Difficulty: 2 Section: Principles of Membrane Transport Feedback: The permeability of the membrane to a molecule depends on various factors. The relative hydrophobicity of a molecule and its size are two important factors: the smaller and more hydrophobic a molecule, the higher the permeability. 2. Answer: C Difficulty: 1 Section: Principles of Membrane Transport Feedback: The diffusion of oxygen through the bilayer is already extremely fast and therefore there is no need for membrane transport proteins. 3. Answer: D Difficulty: 2 Section: Principles of Membrane Transport Feedback: The concentration of potassium inside a typical mammalian cell is more than 20 times higher than that outside. The other ions are usually more concentrated outside the cell. 4. Answer: C Difficulty: 2 Section: Principles of Membrane Transport Feedback: If the solute concentration inside the vesicle is too high, the rapid influx of water can even burst the vesicles. 5. Answer: C Difficulty: 2 Section: Principles of Membrane Transport Feedback: Channels interact weakly with the solutes, resulting in faster transport rates compared to transporters. They do not change their conformation in every transport cycle, but they do have mechanisms for the opening and closing of their pores. Transport through channels is always passive. 6. Answer: A Difficulty: 3 Section: Transporters and Active Membrane Transport Feedback: Transporters resemble enzymes in their mechanism and kinetics. Cocaine in this example is called a competitive inhibitor of the transporter. 7. Answer: E Difficulty: 3 Section: Transporters and Active Membrane Transport Feedback: In secondary active transport, a transporter can use the energy from the transport of one solute down its electrochemical gradient to drive the transport of another

solute against its electrochemical gradient. However, pumping all of the solutes against their electrochemical gradients requires an external energy source and is known as primary active transport. 8. Answer: A Difficulty: 2 Section: Transporters and Active Membrane Transport Feedback: Transporters share structural features that can explain their function as well as their evolution. Permeases such as lactose permease have a twofold pseudosymmetrical structure. In the case of lactose permease, which is a symporter, binding to both solutes happens in one conformation (outward-facing), and transition to the other conformation occurs without exposing the solutes simultaneously to both sides of the membrane. 9. Answer: C Difficulty: 2 Section: Transporters and Active Membrane Transport Feedback: The Na+–H+ exchanger and the Na+-driven Cl––HCO3– exchanger in the plasma membrane help pump out protons from the cytosol (increase the cytosolic pH), and the latter exchanger also neutralizes cytosolic protons by bringing in bicarbonate ions. On the other hand, the Na+-independent Cl––HCO3–exchanger, which is activated at high pH, helps lower the pH by transporting bicarbonate ions out of the cell down their electrochemical gradient. 10. Answer: C Difficulty: 2 Section: Transporters and Active Membrane Transport Feedback: The Na+-K+ pumps in the basal and lateral domains of the plasma membrane of the epithelial cells establish a gradient of sodium ions that drives the unidirectional transport of amino acids and sugars from the intestinal lumen into the bloodstream. 11. Answer: D Difficulty: 2 Section: Transporters and Active Membrane Transport Feedback: The F-type pumps usually function in reverse, and are responsible for the majority of ATP production in our cells. 12. Answer: PAVPA Difficulty: 2 Section: Transporters and Active Membrane Transport Feedback: P-type pumps (e.g. the Na+-K+ pump) phosphorylate a key aspartic acid residue of their phosphorylation domain in each transport cycle. The ABC superfamily is the largest among membrane transport protein families and is of great clinical importance, as exemplified by the multidrug resistance (MDR) protein. Vesicle acidification is normally performed by V-type ATPases. 13. Answer: D

Difficulty: 2 Section: Transporters and Active Membrane Transport Feedback: The Na+-independent Cl––HCO3– exchanger transports one negative charge in and one out in each round of transport, and is therefore not electrogenic. 14. Answer: F Difficulty: 3 Section: Transporters and Active Membrane Transport Feedback: In most eukaryotic ABC transporters, the substrate binds to the inward-facing ATP-free state of the transporter, which then binds ATP to face outward and release the substrate. Thus, most eukaryotic ABC transporters function as exporters. 15. Answer: B Difficulty: 2 Section: Transporters and Active Membrane Transport Feedback: The multidrug resistance protein is an ABC transporter that binds to and hydrolyzes two ATP molecules for the transport of each substrate. The cystic fibrosis transmembrane conductance regulator protein is also an ABC superfamily member, but it is a channel and hydrolyzes ATP just for gating. 16. Answer: E Difficulty: 2 Section: Transporters and Active Membrane Transport Feedback: Ion channels are ion-selective and can fluctuate between open and closed states. They can only mediate passive transport of ions, and do it at remarkable rates, typically several orders of magnitude higher than the rate of transport through transporters. 17. Answer: E Difficulty: 3 Section: Channels and the Electrical Properties of Membranes Feedback: According to the Nernst equation, V = (RT/zF) × ln(Co/Ci). We do not need to numerically evaluate this equation for the purpose of this question. If the potential was – 60 mV when the concentration ratio was 0.1 (which means the ion is positively charged), now that the ratio is 0.001, the potential should be –180 mV, i.e. V = –60 mV × [ln(0.001)/ln(0.1)]. 18. Answer: IIII Difficulty: 3 Section: Channels and the Electrical Properties of Membranes Feedback: The driving force for the transport of these ions comes from the electrochemical gradient across the membrane; that is, the membrane potential minus the Nernst potential for each ion—the ion is transported in the direction that would dissipate this gradient. For example, in this particular cell, the positively charged potassium ions

tend to enter the cytoplasm to change the membrane potential from –70 mV to the more positive value of –60 mV. 19. Answer: A Difficulty: 3 Section: Channels and the Electrical Properties of Membranes Feedback: Inhibition of the potassium leak channels by sanshool leads to depolarization of the membrane in these sensory neurons and sensitizes them. 20. Answer: E Difficulty: 2 Section: Channels and the Electrical Properties of Membranes Feedback: Changing the membrane potential can be achieved by the transport of a small number of ions across the membrane and does not require concentration changes. The resting membrane potential in most cells is close to the K+ equilibrium potential (and is negative inside) because the plasma membrane is most permeable to this ion. 21. Answer: D Difficulty: 3 Section: Channels and the Electrical Properties of Membranes Feedback: Opening of the channels conducting ion D brings the membrane potential down near the equilibrium potential for this ion. 22. Answer: C Difficulty: 1 Section: Channels and the Electrical Properties of Membranes Feedback: The selectivity filter in the potassium channel discriminates between the two ions very efficiently. The sodium ions cannot get rid of their water shell because they are too small to interact favorably with the pore. 23. Answer: C Difficulty: 1 Section: Channels and the Electrical Properties of Membranes Feedback: The pore in the aquaporin channel is not large enough to allow hydrated ions to pass. On the other hand, the hydrophobic wall of the pore makes it too unfavorable for an ion to dehydrate to enter the pore. 24. Answer: D Difficulty: 1 Section: Channels and the Electrical Properties of Membranes Feedback: The simultaneous hydrogen-bonding of the two Asn residues with the water molecule prevents the water molecule from participating in a proton “relay” that can result in the net transport of protons across the membrane. 25. Answer: B Difficulty: 1 Section: Channels and the Electrical Properties of Membranes

Feedback: The mechanosensitive channels of small and large conductance (MscS and MscL) in bacteria open to form relatively nonselective pores of different sizes when the osmotic pressure increases to moderate and high levels, respectively. 26. Answer: E Difficulty: 1 Section: Channels and the Electrical Properties of Membranes Feedback: The inactivation of the voltage-gated Na+ channels delays the firing of the next action potential. The slow inactivation of the voltage-gated K+ channels also contributes to the refractory period. 27. Answer: E Difficulty: 3 Section: Channels and the Electrical Properties of Membranes Feedback: The resting membrane potential mostly depends on the electrochemical potential of the potassium ion, whereas the action potential is more dependent on the sodium ion, and its peak is expected to be lowered if the extracellular sodium concentration is decreased. 28. Answer: C Difficulty: 3 Section: Channels and the Electrical Properties of Membranes Feedback: If the inactivation mechanism in voltage-gated Na+ channels is compromised, the delayed closing of these channels would lead to an extended depolarization phase. A similar effect can be obtained by inhibiting the voltage-gated K+ channels. 29. Answer: E Difficulty: 1 Section: Channels and the Electrical Properties of Membranes Feedback: Myelin sheath is made by Schwann cells around axons in peripheral nerves; it insulates these axons, massively improves their conduction, and confines the “expensive” excitation mechanisms to interspersed nodes along the axon. 30. Answer: C Difficulty: 2 Section: Channels and the Electrical Properties of Membranes Feedback: Channelrhodopsins are light-responsive cation channels that can be engineered to be expressed in specific neural cell types. Opening of these channels in response to light allows sodium ions to leak into the cells, raising the membrane potential to the point when voltage-gated Na+ channels open and further depolarize the membrane in an action potential. 31. Answer: C Difficulty: 3 Section: Channels and the Electrical Properties of Membranes

Feedback: The rapid opening of the sodium channels leads to sodium influx, i.e. a negative current. The slow opening of the potassium channels leads to potassium efflux, i.e. a positive current. The latter current persists as long as the voltage-gated K+ channels stay open. 32. Answer: B Difficulty: 3 Section: Channels and the Electrical Properties of Membranes Feedback: This is an oversimplified picture of the yet-to-be-elucidated gating process in these channels. There are 4 × 7 (= 28) positively charged residues in the four S4 helices combined, whose total charge would be 28 × 1.6 × 10–19 coulombs. As a result of the change in the electric field (by 4 × 106 newtons per coulomb), the change in the force would be equal to: (28 × 1.6 × 10–19 C) × (4 × 106 N/C) = 1.8 × 10–11 N = 18 pN Since the membrane potential has become slightly depolarized, the change in the force is directed outward, tending to push the positively charged S4 helices more toward the extracellular leaflet. Note that the membrane is still polarized (negative inside) at the threshold potential and the electric field has not reversed yet. 33. Answer: B Difficulty: 3 Section: Channels and the Electrical Properties of Membranes Feedback: The voltage-gated Na+ channels open quickly during an action potential and are then rapidly inactivated. The voltage-gated K+ channels open and close more slowly and repolarize the membrane potential. 34. Answer: TFFF Difficulty: 3 Section: Channels and the Electrical Properties of Membranes Feedback: The powerful patch-clamp technique can be used to record ion currents through a single ion channel. Using this technique, individual gated channels are observed to randomly flip between open and closed in an all-or-none fashion to generate the aggregate membrane current. Such experiments have also shown that even cells that are not electrically excitable have a variety of gated ion channels. 35. Answer: B Difficulty: 3 Section: Channels and the Electrical Properties of Membranes Feedback: Synaptotagmin-1 is activated by binding to calcium which flows into the cell through voltage-gated Ca2+ channels upon the arrival of an action potential. This activation drives neurotransmitter release. 36. Answer: E Difficulty: 2 Section: Channels and the Electrical Properties of Membranes

Feedback: Based in part on their firing pattern, neurons can be categorized into functionally different types. The firing properties of each neuron type are largely determined by the dynamic levels of depolarizing (Na+ and Ca2+) and hyperpolarizing (K+) channels, and are under homeostatic control. 37. Answer: EIIE Difficulty: 1 Section: Channels and the Electrical Properties of Membranes Feedback: Excitatory neurotransmitters open cation channels leading to sodium or calcium influx and postsynaptic membrane depolarization, whereas inhibitory neurotransmitters open potassium or chloride channels, suppressing firing by making depolarization harder. 38. Answer: D Difficulty: 3 Section: Channels and the Electrical Properties of Membranes Feedback: The astroglial cells help deplete the neurotransmitter glutamate from synaptic clefts after its release from presynaptic axon terminals. These cells convert the neurotransmitter to the inert amino acid glutamine, which can then be released and taken up by the presynaptic cell to be converted back to glutamate and used during later synaptic transmissions. 39. Answer: EESE Difficulty: 3 Section: Channels and the Electrical Properties of Membranes Feedback: The Ca2+ influx in the muscle cell is stimulated by acetylcholine or nicotine through the opening of ionotropic acetylcholine receptors. Acetylcholinesterase lowers the concentration of acetylcholine at the neuromuscular junction, and its inhibitor would therefore also enhance Ca2+ influx mediated by acetylcholine. In contrast, by favoring the closed state of the channels, α-bungarotoxin suppresses the influx and can cause paralysis. 40. Answer: ECADB Difficulty: 2 Section: Channels and the Electrical Properties of Membranes Feedback: This sequence of events connects the arrival of the action potential at the axon terminal at a neuromuscular junction to the calcium-dependent contraction of muscle fibers. 41. Answer: E Difficulty: 3 Section: Channels and the Electrical Properties of Membranes Feedback: All of these drugs except for the dopamine antagonist have been used with some success for the treatment of the mentioned diseases. 42. Answer: frequency

Difficulty: 1 Section: Channels and the Electrical Properties of Membranes Feedback: The combined magnitude of an excitatory postsynaptic potential is typically encoded into the frequency of firing of action potentials: greater stimulation leads to higher frequency of action potentials. 43. Answer: D Difficulty: 3 Section: Channels and the Electrical Properties of Membranes Feedback: Under constant stimulation, the opening of Ca2+-activated K+ channels increases K+ permeability, making the membrane harder to polarize, increasing the delay between action potential, and decreasing the cellular response. 44. Answer: A Difficulty: 2 Section: Channels and the Electrical Properties of Membranes Feedback: To support high firing rates, both channels should rapidly inactivate. 45. Answer: D Difficulty: 1 Section: Channels and the Electrical Properties of Membranes Feedback: Long-term potentiation and long-term depression share a similar scheme, but result in opposing effects largely due to the different levels of Ca2+ that they bring about. 46. Answer: BADEC Difficulty: 2 Section: Channels and the Electrical Properties of Membranes Feedback: The excitatory neurotransmitter glutamate released by an activated presynaptic axon terminal opens AMPA-receptor channels allowing Na+ influx. This depolarizes the postsynaptic membrane, removing Mg2+ block from NDMA-receptor channels which, when bound to glutamate, allow Ca2+ to enter the cell and activate downstream pathways.

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION CHAPTER 12 INTRACELLULAR COMPARTMENTS AND PROTEIN SORTING Copyright © 2015 by W.W. Norton & Company, Inc. 1. Imagine a protein that has been engineered to contain a nuclear localization signal, a nuclear export signal, a C-terminal peroxisomal targeting sequence, and a canonical endoplasmic reticulum (ER) signal sequence. With all of these signals, where would you expect to find the protein after its synthesis? A. Cytosol B. Nucleus C. Shuttling between the cytosol and the nucleus D. Peroxisomes E. Endoplasmic reticulum 2. Consider a human cell such as a hepatocyte. Which of the following compartments occupies a larger volume in the cell? A. Cytosol B. Nucleus C. Endoplasmic reticulum D. Mitochondria E. Peroxisomes 3. Consider a human liver hepatocyte. Among the following membranes, which one has the largest total area? A. Plasma membrane B. Nuclear inner membrane C. Mitochondrial outer membrane D. Rough ER membrane E. Smooth ER membrane 4. Consider two cells, A and B. Both are approximately spherical, but cell A is a bacterium with a diameter of only about 1 µm, while the diameter of the eukaryotic cell B is about 10 µm. If the plasma membrane in the eukaryotic cell comprises only about 2% of the total cell

membrane, which cell has a higher ratio of total cell membrane to volume? Write down A or B as your answer. 5. Indicate whether the location of each of the following is topologically equivalent to the cytosol (C) or the extracellular space (E). Your answer would be a five-letter string composed of letters C and E only, e.g. CCECE. ( ) Ribosomes ( ) Chromatin ( ) Lysosomal hydrolases ( ) Calcium ions in the ER ( ) Peroxisomal catalase 6. Indicate whether each of the following transport processes occurs via the mechanisms described as gated transport (G), transmembrane transport (T), or vesicular transport (V). Your answer would be a five-letter string composed of letters G, T, and V only, e.g. VTTTG. ( ) Import into nucleus ( ) Export from nucleus ( ) Import into mitochondria ( ) Return from Golgi to ER ( ) Return from ER to cytosol 7. Indicate true (T) and false (F) statements below regarding the compartmentalization of cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) Almost all eukaryotic cells have plastids, but only plant cells have chloroplasts capable of photosynthesis. ( ) The perinuclear space is topologically equivalent to the extracellular space. ( ) The mitochondria and chloroplasts are thought to have evolved by invagination and pinching off from the plasma membrane of the ancient eukaryotic cell. ( ) All organelles in a eukaryotic cell can be constructed de novo, which means that the information to construct them is encoded in the genome. 8.

Fill in the blank in the following paragraph. DO NOT use abbreviations. “A significant fraction of total membrane area in a eukaryotic cell encloses the lumen of the ..., which forms an extended netlike labyrinth of tubules and sacs. Secretory proteins are normally

synthesized by ribosomes bound to a special type of this compartment.” 9. A geneticist has devised a strategy to study protein translocation into the endoplasmic reticulum (ER) in yeast cells. She is interested in two different signal sequences that are thought to operate via slightly different translocation mechanisms. Using genetic engineering, she has fused the first signal sequence to a protein whose cytosolic expression is absolutely necessary for cell survival in the selective medium, but is inactive when in the ER. In the same cell, she has also fused the second signal sequence to a toxic protein whose cytosolic expression leads to cell lysis but is harmless when in the ER. Whereas wild-type cells undergo lysis upon the expression of these fusion proteins, she has been able to identify viable mutants, each of which has a lossof-function mutation in a gene encoding a protein involved in membrane translocation. The products of these genes are probably ... A. involved in the general transport of proteins into the ER, regardless of the type of signal sequence. B. involved in the transport of proteins with the first signal sequence but not the second one. C. involved in the transport of proteins with the second signal sequence but not the first one. D. involved in the transport of proteins with a novel signal sequence (i.e. neither the first signal sequence nor the second one). 10. Indicate true (T) and false (F) statements below regarding the nucleus and nuclear protein transport. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) The inner and outer nuclear membranes are continuous with each other, yet maintain distinct protein compositions. ( ) The outer nuclear membrane is studded with ribosomes engaged in protein synthesis. ( ) The endoplasmic reticulum lumen is continuous with the nuclear interior. ( ) Ribosomal proteins pass through the nuclear pore complexes twice; they are imported into the nucleus after synthesis, and are exported from the nucleus after assembly with ribosomal RNA. 11. In the following schematic diagram of a nuclear pore complex, on which side is the cytosol located? What is the approximate diameter of the pore, as indicated by the vertical bar?

Perinuclear space

A. Right; 1 µm B. Right; 0.1 µm C. Left; 1 µm D. Left; 0.1 µm 12. In the following graph, the magnitude of concentration difference across the nuclear pore complexes (NPCs) is plotted for four molecules (A to D) as a function of time, starting from an

the nuclear envelope (nM)

Concentration difference across

arbitrarily chosen initial concentration difference. Indicate which curve corresponds to each of the following molecules. Your answer would be a four-letter string composed of letters A to D only, e.g. ABCD.

2 A B

1 C D 0 0

10 Time (min)

( ) A large protein that is being actively transported across the NPC ( ) A small water-soluble molecule ( ) A small protein composed of a few dozen residues ( ) A large protein that is NOT actively transported into or out of the nucleus 13.

Which of the following scenarios does NOT normally occur on a nuclear pore complex? A. A protein complex is imported into the nucleus, with ONLY one of its subunits containing a nuclear localization signal (NLS). B. In a single pore, an NLS-containing protein is imported, while at the same time a nuclear export signal (NES)-containing protein is exported. C. A nuclear import receptor is exported from the nucleus through the pore. D. A protein is imported on its own through the pore, without the need for a separate import receptor. E. All of the above are possible scenarios and occur naturally.

14. Consider a transcription regulatory protein that has both a nuclear localization and a nuclear export signal and is normally found both in the nucleus and in the cytosol at comparable concentrations. This protein has a high-affinity binding partner in the nucleus. Upon activation of a certain signaling pathway, the binding protein is ubiquitylated and degraded. As a result of this, ... A. the transcription regulatory protein accumulates in the nucleus. B. the transcription regulatory protein accumulates in the cytosol. C. the distribution of the transcription regulatory protein does not change, but the expression of its target genes may be altered. D. the distribution of the transcription regulatory protein does not change, and the expression of its target genes is not necessarily changed as a result of the degradation event. 15. According to the model for nuclear transport described in this chapter, what do you think would happen if you could artificially limit all Ran-GAP activity to the nucleus and all Ran-GEF activity to the cytosol? A. Proteins containing an NLS would be actively exported from the nucleus, while NEScontaining proteins would be actively imported. B. Both import and export of nuclear proteins would be stalled, as they lose their directionality. C. Protein import into the nucleus would be reversed, but export would be unaffected.

D. Protein import into the nucleus would be stalled, but export would be unaffected. E. Nothing would change; this is the normal Ran-GAP and Ran-GEF distribution. 16. Indicate true (T) and false (F) statements below regarding the nuclear transport of proteins. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) Many nuclear import and export receptors are members of the same protein family. ( ) Most nuclear import receptors contain unstructured domains with FG-repeats. ( ) Adaptor proteins that simultaneously bind to nuclear localization signals and to importins are required for the import of some nuclear cargo proteins. ( ) Ran is mostly found in its GTP-bound form in the nucleus. 17. The formation of a stable ternary complex involving Ran GTPase, a nuclear transport receptor, and a cargo protein occurs in ... A. both nuclear import and export. B. nuclear import , but not export. C. nuclear export, but not import. D. neither nuclear import nor export. 18.

Cyclin B1, a key cell cycle regulatory protein in vertebrates, is mostly cytosolic before

mitosis. Early in mitosis, however, the protein is phosphorylated by certain protein kinases and consequently accumulates in the nucleus. How can phosphorylation bring about nuclear accumulation of this protein? A. Phosphorylation within the nuclear localization signal inhibits the function of the signal. B. Phosphorylation within the nuclear export signal enhances the function of the signal. C. Phosphorylation within the nuclear export signal interferes with the function of the signal. D. Phosphorylation elsewhere on the protein enhances binding to cytosolic proteins. 19. Indicate true (T) and false (F) statements below regarding the mitochondrial protein import system. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) Mitochondrial proteins should fold natively twice: once in the cytosol and once inside the organelle. ( ) β-barrel proteins that are abundant in the mitochondrial outer membrane are imported from the cytosol independently of the TOM complex.

( ) Signal sequences that target precursor proteins to the mitochondrial matrix form an αhelix in which positively charged residues cluster near its N-terminus, while uncharged or hydrophobic residues cluster near the other side. ( ) At least two signal sequences are required to direct proteins to the mitochondrial matrix. 20. Which subset of the following is directly involved in driving protein import into the mitochondrial matrix space? Choose all correct sources. Your answer would be a string composed of letters A to G only, in alphabetical order, e.g. AE. A. ATP hydrolysis inside mitochondria B. ATP hydrolysis outside mitochondria C. GTP hydrolysis inside mitochondria D. GTP hydrolysis outside mitochondria E. Light F. Membrane potential across the inner membrane G. Membrane potential across the outer membrane 21. Tom40 is a nuclear-encoded essential subunit of the TOM complex in the outer mitochondrial membrane. It is a β-barrel protein that forms the pore through which precursor proteins enter the intermembrane space from the cytosol. Indicate true (T) and false (F) statements below regarding the Tom40 protein. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) Incorporation of new Tom40 in the outer membrane requires preexisting Tom40 in that membrane. ( ) Formation of new TOM complexes is dependent on the SAM complex. ( ) Tom40 is partially translocated through the outer membrane and is then transferred in the plane of the membrane to fold into its native conformation. ( ) Tom40 is translocated through the inner membrane as a precursor. 22. Certain cysteine-containing proteins of the mitochondrial intermembrane space are imported from the cytosol with the help of the Mia40 protein via a disulfide relay system. What drives the unidirectional import of these proteins? Are these proteins reduced or oxidized at their cysteine residues upon import? A. ATP hydrolysis; oxidized B. The electron-transport chain; oxidized C. ATP hydrolysis; neither oxidized nor reduced

D. ATP hydrolysis; reduced E. The electron-transport chain; reduced 23. Which of the following proteins or protein complexes is directly required for the targeting of mitochondrial inner membrane multipass proteins, such as metabolite transporters, whose signal sequence is normally not cleaved after import? A. TIM22 B. TIM23 C. OXA D. Mia40 E. SAM 24. Indicate whether each of the following descriptions refers to protein import into mitochondria (M), chloroplasts (C), or both (B). Your answer would be a five-letter string composed of letters M, C, and B only, e.g. BBBMC. ( ) ATP and GTP hydrolysis drive translocation into the organelle. ( ) The organelle has an extra compartment that requires extra signal sequences for protein targeting. ( ) Transport through the double membrane is driven in part by an H+ gradient across the inner membrane. ( ) Imported precursor proteins have amphiphilic N-terminal signal sequences that are usually removed after use. ( ) Hsp70 family chaperones inside the organelle assist in protein translocation during import. 25. Indicate true (T) and false (F) statements below regarding peroxisomal proteins. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) All peroxisomal proteins are encoded in the nucleus. ( ) Some peroxisomal proteins are synthesized by ribosomes attached to the rough endoplasmic reticulum. ( ) All peroxisomal proteins reach the organelle after their synthesis is completed. ( ) Peroxisomal proteins have to be unfolded before import. 26.

Which reaction is normally catalyzed by the enzyme catalase in the peroxisome? A. 2O2– + 2H+ → H2O2 + O2 B. FMNH2 + O2 → FMN + H2O2

C. C2H5OH (ethanol) + NAD+ → C2H4O (acetaldehyde) + NADH + H+ D. 2H2O2 → 2H2O + O2 E. Both C and D above 27. Sort the following events as they occur during the peroxisomal protein import cycle, starting with the release of cargo from Pex5. Your answer would be a five-letter string composed of letters A to E only, e.g. ABEDC. (A) Pex5 deubiquitylation (B) Pex5 ubiquitylation (C) Docking and translocation of the cargo protein along with Pex5 (D) Pex5 export from the peroxisome with the help of ATPases Pex1 and Pex6 (E) Pex5 binding to a cargo protein containing a C-terminal peroxisomal targeting sequence 28. Indicate whether each of the following descriptions better matches the rough (R) or the smooth (S) endoplasmic reticulum (ER). Your answer would be a five-letter string composed of letters R and S only, e.g. SRSRR. ( ) It mostly has a tubular appearance. ( ) It contains the transitional ER. ( ) It is coated by ribosomes. ( ) It can be specialized for functions such as detoxification and lipid metabolism. ( ) Sarcoplasmic reticulum in muscle cells is one of its specialized forms. 29. Compared to the cytosol, which of the following is generally true about the lumen of the endoplasmic reticulum in our cells? A. It has a higher pH. B. It has a more reducing environment. C. It has a much higher calcium ion concentration. D. It has a higher density of ribosomes. E. All of the above. 30.

The signal-recognition particle (SRP) ... A. is a heterodimeric protein. B. transiently inhibits translation and polypeptide elongation by binding to and inhibiting the elongation factors. C. accompanies the nascent polypeptide all the way into the ER lumen.

D. binds GTP. E. is permanently attached to the cytosolic face of the ER membrane, thus bringing the ribosomes into close proximity of the translocon. 31. You set up an in vitro translation system containing the entire translation machinery but devoid of any component of the endoplasmic reticulum (ER) targeting machinery. To this system, you can add mRNA encoding either a 20 kD secretory protein or a 20 kD cytosolic protein. You perform in vitro translation in the presence of radioactively labeled methionine, with or without the addition of saturating amounts of SRP or microsomes, as indicated below. After separating the protein products by SDS-PAGE, and visualizing the radioactivity by autoradiography, you obtain the following results. The presence or absence of each component in the reaction is indicated at the top of the corresponding lane(s) by + and –, respectively. The numbers on the left indicate the apparent molecular mass (×1000) of spots on the gel. Which protein (X or Y) is the secretory protein? Which of the reactions (1 or 2) contained SRP? 1 – ?

Microsomes SRP X mRNA Y mRNA

20 17 -

A. X; 1 B. X; 2 C. Y; 1 D. Y; 2

+ –

2 – ? – +

+ –

3 + + – +

+ –

– +

32. Which of the following is NOT a likely ER signal sequence recognized by the signalrecognition particle? All sequences are written with their N-terminus on the left. A. Met-Lys-Leu-Ser-Leu-Val-Ala-Ala-Met-Leu-Leu-Leu-Leu-Ser-Ala-Ala-Arg-Ala B. Met-Glu-Met-Phe-Gln-Gly-Leu-Leu-Leu-Leu-Leu-Leu-Leu-Leu-Ser-Met-Gly-GlyThr-Trp-Ala C. Met-Lys-Ala-Lys-Leu-Leu-Val-Leu-Leu-Tyr-Ala-Phe-Val-Ala-Gly-Asn D. Met-Met-Ala-Ala-Gly-Pro-Arg-Thr-Ser-Leu-Leu-Leu-Ala-Phe-Ala-Leu-Leu-CysLeu-Pro-Trp-Thr-Gln-Val-Val E. Met-Leu-Ser-Leu-Arg-Gln-Ser-Ile-Arg-Phe-Phe-Lys-Pro-Ala-Thr-Arg-Thr-Leu-SerSer-Arg-Tyr-Leu 33. Rough microsomes can be subjected to a “salt extraction” procedure in which a high salt concentration is used to remove membrane-associated ribosomes and peripheral proteins. Such salt-extracted microsomes are known to be translocation-incompetent, meaning that when present co-translationally in vitro, they fail to protect translated proteins from protease digestion. However, adding back an 11S particle (S is the sedimentation coefficient) purified from the saltwash fraction is sufficient to restore the protein-translocation activity of the salt-extracted microsomes. Which of the following do you think is true regarding the 11S particle? A. It is composed of 21 proteins. B. It is a digestion product of ER-associated ribosomes. C. It is normally assembled in the nucleus and exported to the cytoplasm by exportins. D. It is an ER integral membrane protein that can interact with the translocon. E. It requires high salt concentration for its function in vivo. 34. Mitochondrial hsp70 is to matrix protein import what ... is to post-translational ER protein import. A. Ribosome B. Sec61 C. Sec63 D. BiP E. PDI 35. The signal-recognition particle is not the only factor that evaluates the authenticity of endoplasmic reticulum (ER) signal sequences in proteins. The Sec61 complex is also able to recognize the signal sequences and “opens” after binding to them. Even single point mutations within the signal sequence of a protein can render the protein unable to enter the ER efficiently,

as the Sec61 complex does not readily open in response to the mutant sequence. However, “suppressor” mutations in genes encoding components of the translocation pathway, including the Sec61 subunits, can partially restore the wild-type localization of proteins with mutant signal sequences. Many such suppressor mutations (also called prl mutations) map to or near the “plug” domain in the Sec61 translocon. These mutations, including the deletion of the entire plug, generally result in destabilization of the closed conformation of the translocon and favor its open conformation. Your friend has mutated a certain residue in the plug domain of the yeast Sec61. She has just finished measuring the translocation efficiency of an ER protein with either a wildtype or a mutant (defective) signal sequence, either in wild-type or in her Sec61-mutant cells, and has obtained the following results. Based on these early results, does her Sec61 mutation show a prl phenotype? Write down Yes or No as your answer. Wild-type Sec61

100% -

Mutant Sec61

Relative ER translocation efficiency

0% -

Wild-type signal sequence

Mutant signal sequence

36. Indicate whether the C-terminus (C) or the N-terminus (N) of each of the following proteins is expected to be located in the cytosol upon membrane integration of the protein. Your answer would be a three-letter string composed of letters C and N only, e.g. CCC. ( ) A single-pass transmembrane protein that has one N-terminal signal sequence and one internal stop-transfer signal ( ) A single-pass transmembrane protein that has one internal signal sequence that is preceded by a patch of positively charged residues ( ) An ER tail-anchored protein 37. Consider a transmembrane protein with the following topology that has an internal signal sequence (helix 1). If you fuse a canonical ER signal sequence at the N-terminus of this protein, how would you expect the topology to change? The ER lumen is at the bottom in all drawings.

For simplicity, assume that the effect of charged residues flanking the transmembrane helices is negligible in this case. N C Cytosol 1 2

ER lumen

N C 1 2

C 1 2

N B.

N 1 2

1 2

N D.

N 1 2

C E.

38. The following diagram depicts the topology of a multipass transmembrane protein in the endoplasmic reticulum (ER) membrane. Which set of helices act as stop-transfer signals in this protein? Cytosol C

N 1

ER lumen A. 1, 2, 3 B. 4, 5, 6 C. 1, 3, 5 D. 2, 4, 6 E. 1, 6

39. Considering the following hydropathy plot for a multipass transmembrane protein in the plasma membrane, are the protein termini expected to be located inside (I) or outside (O) of the cell? Which of the regions indicated as 2 and 3 in the plot is expected to have more positively charged residues? The position of positively charged residues near the first transmembrane helix is indicated by an asterisk on the plot.

+25 Hydropathy index

–25 1

100 Residue number

3 200

A. I; 2 B. I; 3 C. O; 2 D. O; 3 40. Indicate true (T) and false (F) statements below regarding N-linked glycosylation of proteins. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) N-linked glycosylation can be carried out co-translationally, possibly at multiple asparagine residues on the same protein molecule. ( ) N-linked glycosylation is a gradual process, with step-by-step addition and trimming events that commence with the addition of N-acetylglucosamine to an asparagine side chain. ( ) Most proteins synthesized in the rough ER are N-glycosylated, and some of them require this modification for their correct folding. ( ) Once a protein is properly folded in the ER, its attached oligosaccharides are quickly removed by an N-glycanase, although it may be glycosylated again later. 41. Fill in the blank in the following paragraph regarding the recognition of tail-anchored proteins by their targeting machinery. DO NOT use abbreviations.

“When the hydrophobic transmembrane α helix at the C-terminus of an ER tail-anchored protein emerges from the ribosome, the tailanchored-binding domain (TABD) of the Get3 ATPase binds to it. The job of TABD in Get3 is similar to that of the flexible hydrophobic domain in the SRP54 protein of the signal-recognition particle, in that it also has to recognize a degenerate set of hydrophobic α- helices. Therefore, it is not surprising that, just like the corresponding domain in SRP54, the binding site in Get3 is rich in ... residues.” 42.

Misfolded proteins in the ER may actively undergo any of the following EXCEPT ... A. binding to chaperones such as BiP to allow unfolding and refolding. B. binding to chaperones such as calnexin to prevent aggregation. C. being transported back to the cytosol for degradation. D. being expelled to the Golgi apparatus for disposal. E. being glycosylated and binding to lectins.

43.

Proteins of the protein disulfide isomerase (PDI) family share common domains that

harbor an active-site CXXC motif (C = cysteine; X = one of several residues). During the course of a redox reaction, one of the active-site cysteines in its reduced form can attack a disulfide bond in a substrate protein (that could itself be another PDI family member) to form a mixed disulfide between the enzyme and its substrate. This is then attacked by the other active-site cysteine, releasing the substrate in reduced form. The reverse of these reactions can occur instead in order to make disulfide bonds in target proteins. CXXA mutant PDI family proteins—in which one of the active-site cysteines is mutated to an alanine—can be trapped in the intermediate mixed disulfide state for an elongated time, facilitating the identification of their substrate proteins. Using these mutations in each of three interacting PDI family proteins—A, B, and C (one mutation at a time)—you have discovered that the mutant B interacts with A and C, while mutant A interacts with C but not with B. Finally, mutant C interacts with neither of the other two. The following diagram shows the thermodynamically favorable flow of electrons (plus protons) in the cascade involving these three proteins. Based on your results above, what proteins correspond to 1, 2, and 3 in the diagram, respectively? Your answer would be a threeletter string composed of letters A to C only, e.g. CBA.

1 Electron donor

2 3 Electron acceptor

44. Indicate whether each of the following descriptions better matches the ATF6 (A), the IRE1 (I), or the PERK (P) branch of the ER unfolded protein response. Your answer would be a four-letter string composed of letters A, I, and P only, e.g. APII. ( ) It involves a noncanonical cytoplasmic splicing process. ( ) Its sensor is a latent transcription regulator. ( ) It involves regulated proteolysis of the sensor protein in the Golgi apparatus. ( ) Its sensor bears both kinase and endoribonuclease activities. 45. A protein is covalently attached to glycosylphosphatidylinositol. Which of the following is typically NOT true regarding this protein? A. The linkage of the anchor to the C-terminus of the protein occurs in the ER. B. The attachment of the anchor coincides with cleavage of a C-terminal transmembrane segment of the protein precursor. C. The protein is likely to be an ER resident, helping with the folding of nascent imported proteins. D. A phospholipase can cleave the protein from the membrane. E. The anchor affects the localization of the protein in the membrane. 46. Indicate whether each of the following occurs on the cytosolic side of the ER membrane (C), on the lumenal side of the ER membrane (L), or in the Golgi apparatus (G). Your answer would be a five-letter string composed of letters C, L, and G only, e.g. GGGLC. ( ) Assembly of GlcNAc oligosaccharides on dolichol phosphate ( ) Synthesis of phosphatidylcholine ( ) Synthesis of sphingomyelin ( ) Ubiquitylation of misfolded ER protein ( ) Synthesis of phosphatidic acid

Answers 1. Answer: E Difficulty: 2 Section: The Compartmentalization of Cells Feedback: The N-terminal ER signal sequence is recognized first and used to target the protein to the endoplasmic reticulum co-translationally. 2. Answer: A Difficulty: 1 Section: The Compartmentalization of Cells Feedback: The cytosol comprises about half the total volume of the cell and is the site of many important cellular functions. Even though a hepatocyte has an expanded endoplasmic reticulum and a large number of peroxisomes and mitochondria, these compartments collectively account for less than half the total cell volume. Please refer to Table 12–1. 3. Answer: D Difficulty: 1 Section: The Compartmentalization of Cells Feedback: The plasma membrane accounts for only a small percentage of the total cell membrane area, while the endoplasmic reticulum (ER) membranes (including those of the smooth and rough ER) comprise about half the total membrane area. In cells such as hepatocytes, which are rich with mitochondria, the inner mitochondrial membrane also has a notable surface area, due to its extensive invaginations called cristae. Please refer to Table 12–2. 4. Answer: B Difficulty: 3 Section: The Compartmentalization of Cells Feedback: Since cell surface area and volume are proportional to the square and the cube of the cell radius, respectively, the surface-to-volume ratio for cell A is 10 times higher than that of the eukaryotic cell B. However, the total cell membrane area for cell B is 50 times its plasma membrane surface area, making the ratio of total cell membrane area to volume higher (five times higher) for the larger eukaryotic cell. 5. Answer: CCEEE Difficulty: 2 Section: The Compartmentalization of Cells Feedback: The interior of the nucleus (where the chromatin resides) is topologically equivalent to the cytosol, while the lumen of organelles such as the endoplasmic reticulum, lysosomes, and peroxisomes is equivalent to the cell exterior. 6. Answer: GGTVT Difficulty: 2 Section: The Compartmentalization of Cells

Feedback: While nuclear traffic uses gated transport, transport into the endoplasmic reticulum (and back) and into mitochondria, plastids, and peroxisomes is carried out using transmembrane translocators. Finally, proteins can move between the various membrane-enclosed organelles of the intracellular membrane system, as well as to and from the plasma membrane, by vesicular transport. 7. Answer: FTFF Difficulty: 2 Section: The Compartmentalization of Cells Feedback: Plastids are present in plants, algae, and some protozoa; both plants and algae are capable of photosynthesis using the special version of a plastid called a chloroplast. Mitochondria and chloroplasts probably evolved from bacteria that were engulfed by ancient eukaryotic cells and established a symbiotic relationship. If removed, these and many other membrane-enclosed organelles in a eukaryotic cell cannot be constructed by the cell de novo. The perinuclear space is continuous with the endoplasmic reticulum lumen and is topologically equivalent to the cell exterior. 8. Answer: endoplasmic reticulum Difficulty: 1 Section: The Compartmentalization of Cells Feedback: The endoplasmic reticulum (ER) membrane comprises about half the total area of membrane in a eukaryotic cell. The rough ER has bound ribosomes engaged in synthesizing proteins that will be associated with the secretory or endocytic pathways. 9. Answer: B Difficulty: 3 Section: The Compartmentalization of Cells Feedback: This genetic screen identifies mutations that interfere with the ER targeting of the vital protein (which has been fused to the first signal sequence) while keeping the ER targeting of the toxic protein (with the second signal sequence) almost intact, resulting in cell viability. Therefore, the mutations are expected to affect genes that encode proteins involved in the first pathway but not the second. 10. Answer: TTFT Difficulty: 2 Section: The Transport of Molecules between the Nucleus and the Cytosol Feedback: The outer nuclear membrane is continuous with the inner nuclear membrane, but has different proteins and is associated with ribosomes. The endoplasmic reticulum lumen is continuous with the perinuclear space between the nuclear membranes. Since assembly of ribosomal protein and RNA takes place inside the nucleus, the protein components have to be imported before the assembled subunits can be exported back to the cytoplasm. 11. Answer: D Difficulty: 2

Section: The Transport of Molecules between the Nucleus and the Cytosol Feedback: The structure of the nuclear pore complex usually shows an eightfold symmetry and contains cytoplasmic fibrils and a nuclear “basket.” Its diameter is approximately 100 nm and differs in different organisms. 12. Answer: BDCA Difficulty: 3 Section: The Transport of Molecules between the Nucleus and the Cytosol Feedback: While small molecules and proteins of up to about 5 kD in mass can freely diffuse through the nuclear pores, diffusion of larger proteins becomes slower with increasing size, with the NPC acting as a diffusion barrier for large proteins that are not actively transported. Unlike passive diffusion, active transport across NPCs can increase the concentration gradient. 13. Answer: E Difficulty: 3 Section: The Transport of Molecules between the Nucleus and the Cytosol Feedback: Certain proteins including nuclear import receptors can enter the nucleus unaccompanied. Importins should also be constantly exported to help with additional rounds of cargo import. If at least one component of a complex can interact with a nuclear receptor, the whole complex may be transported. Nuclear pore complexes can support two-way traffic. 14. Answer: B Difficulty: 2 Section: The Transport of Molecules between the Nucleus and the Cytosol Feedback: A high-affinity binding partner in the nucleus inevitably shifts the steady-state distribution of the protein toward this compartment. Once the partner is eliminated, the now-dominating export route leads to accumulation of the protein in the cytosol. An effect on gene expression does not depend on localization alone. 15. Answer: A Difficulty: 3 Section: The Transport of Molecules between the Nucleus and the Cytosol Feedback: The directionality of nuclear protein transport pathways relies on the asymmetrical distribution of the two conformations of Ran GTPase, which in turn depends on the activities of nuclear Ran-GEFs and cytosolic Ran-GAPs. Reversing the locations of these activities would reverse the direction of the transport pathways. 16. Answer: TFTT Difficulty: 2 Section: The Transport of Molecules between the Nucleus and the Cytosol Feedback: The karyopherin family includes many nuclear transport receptors and adaptors. FG-repeats are found in several proteins of the nuclear pore complex, and can

interact with karyopherins. Due to the presence of Ran-GEF anchored to the nuclear chromatin, GTP-bound Ran is more abundant in the nucleus. 17. Answer: C Difficulty: 3 Section: The Transport of Molecules between the Nucleus and the Cytosol Feedback: In the nucleus, Ran-GTP binds to the importin in an importin–cargo complex, causing cargo release. This is in contrast to the binding of Ran-GTP to exportin, which promotes cargo binding and the formation of a ternary export complex. In the cytoplasm, Ran-GDP dissociates from the receptors, allowing importin to bind to its cargo and exportin to release its cargo. 18. Answer: C Difficulty: 2 Section: The Transport of Molecules between the Nucleus and the Cytosol Feedback: For a protein that normally shuttles back and forth between the nucleus and the cytosol, nuclear accumulation can be achieved by biasing the import/export balance toward more net import. 19. Answer: FFFF Difficulty: 2 Section: The Transport of Proteins into Mitochondria and Chloroplasts Feedback: In contrast to other outer membrane proteins, the abundant β-barrel proteins of the outer mitochondrial membrane require the SAM complex for their correct integration, following TOM-mediated import into the intermembrane space. Just one signal sequence is enough for targeting to the matrix, and it forms an amphiphilic helix. Prior to import, mitochondrial proteins are kept unfolded in the cytosol by chaperones. 20. Answer: ABF Difficulty: 1 Section: The Transport of Proteins into Mitochondria and Chloroplasts Feedback: In the process of protein import into the mitochondrial matrix, ATP is hydrolyzed by hsp70-family chaperone proteins in the cytosol and inside the matrix. The energy in the electrochemical H+ gradient across the inner mitochondrial membrane is another source of energy to drive the import. 21. Answer: TTFF Difficulty: 2 Section: The Transport of Proteins into Mitochondria and Chloroplasts Feedback: Being a β-barrel protein, the import and integration of Tom40 (and therefore the TOM complex) depend on functional TOM and SAM complexes. 22. Answer: B Difficulty: 3 Section: The Transport of Proteins into Mitochondria and Chloroplasts

Feedback: Oxidation of the imported proteins creates reduced Mia40, which ultimately donates its electrons to the electron-transport chain to reset to its oxidized form for another round of import. 23. Answer: A Difficulty: 1 Section: The Transport of Proteins into Mitochondria and Chloroplasts Feedback: Metabolite transporters such as ATP-ADP translocase use the TIM22 pathway for their targeting to the inner mitochondrial membrane. 24. Answer: CCMBB Difficulty: 2 Section: The Transport of Proteins into Mitochondria and Chloroplasts Feedback: Chloroplasts set up their H+ gradients across the membranes of their thylakoids, unique compartments not found in mitochondria. ATP hydrolysis by chaperones in both organelles (as well as GTP hydrolysis by the TOC complex of chloroplasts) drives protein import in these organelles. 25. Answer: TTTF Difficulty: 2 Section: Peroxisomes Feedback: Peroxisomal proteins are either delivered via ER-derived vesicles or translocated post-translationally straight from the cytosol. In the latter case, the protein does not have to be unfolded prior to import. 26. Answer: D Difficulty: 2 Section: Peroxisomes Feedback: Decomposition of hydrogen peroxide (produced as a result of cellular reactions such as those presented in A and B) into water and molecular oxygen is an important role fulfilled by catalase. The enzyme can also engage in peroxidation reactions such as the conversion of ethanol to acetaldehyde, but it uses H2O2, not NAD+, as the oxidant. 27. Answer: BDAEC Difficulty: 2 Section: Peroxisomes Feedback: The monoubiquitylation of Pex5 is transient and is required for its release from the organelle, which is catalyzed by the ATPases. 28. Answer: SSRSS Difficulty: 1 Section: The Endoplasmic Reticulum Feedback: While the rough ER is mostly made up of stacks coated with ribosomes that are engaged in protein synthesis and import, the tubular smooth ER has specialized functions such as regulation of calcium storage and lipid metabolism.

29. Answer: C Difficulty: 2 Section: The Endoplasmic Reticulum Feedback: While it is usually slightly more acidic and more oxidizing than the cytosol, the lumen of the ER is also distinct by containing a concentration of calcium orders of magnitude higher than that of the cytosol. There are no ribosomes in the ER lumen. 30. Answer: D Difficulty: 2 Section: The Endoplasmic Reticulum Feedback: SRP is a ribonucleoprotein particle that recognizes endoplasmic reticulum (ER) signal sequences in nascent polypeptides, stalls the ribosome by inhibiting elongation factor binding, and docks the ribosome onto the translocon channel for protein import into the ER. Both SRP and its receptor on the ER membrane can bind and hydrolyze GTP. 31. Answer: A Difficulty: 3 Section: The Endoplasmic Reticulum Feedback: The cytosolic protein is unaffected by the presence or absence of microsomes or SRP. Addition of both SRP and microsomes to the reaction (lane 3) results in the microsomal import of the secretory protein and the cleavage of its signal peptide, while addition of SRP alone (lane 1) inhibits peptide elongation beyond a short polypeptide length, corresponding to the length of the signal sequence plus the fragment that is still in the ribosomal exit tunnel. 32. Answer: E Difficulty: 3 Section: The Endoplasmic Reticulum Feedback: Note the presence of a positively charged residue (arginine or lysine) at regular intervals in sequence E, which places these residues on one side of an amphiphilic helix, which is part of a protein destined for mitochondria. In the endoplasmic reticulum signal sequences, the uninterrupted hydrophobic core is usually preceded immediately by positively charged residues. 33. Answer: C Difficulty: 3 Section: The Endoplasmic Reticulum Feedback: It is the signal-recognition particle. 34. Answer: D Difficulty: 2 Section: The Endoplasmic Reticulum Feedback: The hsp70-like endoplasmic reticulum (ER) chaperone, BiP, helps pull the proteins into the ER lumen in the post-translational protein import pathway.

35. Answer: No Difficulty: 3 Section: The Endoplasmic Reticulum Feedback: The examined Sec61 mutation seems to be an enhancer, rather than a suppressor, of the signal-sequence mutation. 36. Answer: CNN Difficulty: 3 Section: The Endoplasmic Reticulum Feedback: In the first case, the N-terminal signal sequence would be cleaved, placing the N-terminus in the endoplasmic reticulum lumen and the C-terminus in the cytosol. The “positive inside” rule applies to the second case, placing its N-terminus in the cytosol. Tail-anchored proteins always have their N-terminal bulk in the cytosol. 37. Answer: D Difficulty: 3 Section: The Endoplasmic Reticulum Feedback: Introducing the N-terminal signal sequence (which would be cleaved) converts helix 1 into a stop-transfer signal, orienting the entire protein inside-out. This is a simplification, however; for example, the effect of charged residues in the loop regions also requires consideration. 38. Answer: D Difficulty: 3 Section: The Endoplasmic Reticulum Feedback: A stop-transfer signal directs its following segment in the protein to the cytosolic side of the membrane. 39. Answer: A Difficulty: 3 Section: The Endoplasmic Reticulum Feedback: The N-terminal position of positively charged residues around the first helix suggests that it will be oriented with this terminus facing inside. This therefore places loop 2 in the cell interior as well, which would be consistent with this loop bearing a larger positive charge than the extracellular loops 3 and 1 (not indicated). 40. Answer: TFTF Difficulty: 2 Section: The Endoplasmic Reticulum Feedback: In N-linked glycosylation, the precursor oligosaccharide assembled on a dolichol anchor is transferred en bloc to certain asparagine residues in most ER proteins. Trimming and addition of further sugars then follow, and some are important for ER quality control. Correctly folded proteins usually carry oligosaccharides that are distinct from those of unfolded or misfolded proteins. 41. Answer: methionine

Difficulty: 2 Section: The Endoplasmic Reticulum Feedback: Thanks to their unbranched and flexible side chains, methionine residues lining the binding groove for the hydrophobic helix in both of these proteins provide a sufficiently plastic binding site that can accommodate helices of different sequence, size, and shape. 42. Answer: D Difficulty: 2 Section: The Endoplasmic Reticulum Feedback: Misfolded proteins are carefully monitored by the ER quality control system and are prevented from advancing in the secretory pathways for as long as they are not correctly folded. 43. Answer: BAC Difficulty: 3 Section: The Endoplasmic Reticulum Feedback: The presence of mixed disulfide suggests that the free cysteine in the active site of the mutant (electron donor) has attacked a substrate protein (acceptor) to reduce it. The electron donor (B) in the cascade is the most reducing member, and prefers to stay in its oxidized form. 44. Answer: IAAI Difficulty: 1 Section: The Endoplasmic Reticulum Feedback: Upon regulated proteolytic cleavage, the released ATF6 fragment enters the nucleus and alters the expression of its target genes. The IRE1 protein is a kinase/endoribonuclease that regulates the splicing of a transcription regulatory protein. 45. Answer: C Difficulty: 2 Section: The Endoplasmic Reticulum Feedback: GPI-anchored proteins are made in the endoplasmic reticulum and almost exclusively found on the extracellular leaflet of the plasma membrane. 46. Answer: CCGCC Difficulty: 2 Section: The Endoplasmic Reticulum Feedback: Synthesis of many phosphoglycerides occurs on the cytoplasmic leaflet of the ER membrane. Sphingomyelin, on the other hand, is made in the Golgi apparatus from ER-made precursors.

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION Chapter 13 INTRACELLULAR MEMBRANE TRAFFIC Copyright © 2015 by W.W. Norton & Company, Inc. 1. Reconstitution of vesicle transport in cell-free systems was historically carried out in the presence of isolated “donor” and “acceptor” Golgi stacks. The donor stacks are isolated from cells expressing a viral protein and lacking a processing enzyme. The protein can be transported to the acceptor stacks (that do contain the processing enzyme) and be processed only in the presence of an energy source (such as ATP) and a cytosolic fraction. The transport vesicles that mediate this process are visualized by microscopy. Various compounds can be added to this system to help understand the mechanism of transport. Two such compounds were added in early experiments in the 1980s, both of which blocked transport of the viral protein and resulted in the accumulation of transport vesicles. However, the accumulated vesicles following treatment with one of these compounds (A) appeared to be coated, while those resulting from treatment with the other compound (B) did not. If one of these compounds is a nonhydrolyzable GTP analog and the other one is N-ethylmaleimide, which compound (A or B) do you think represents the GTP analog? Write down A or B as your answer. 2.

Indicate whether each of the following descriptions better applies to COPI- (1), COPII-

(2), or clathrin- (3) coated vesicles. Your answer would be a four-digit number composed of digits 1 to 3 only, e.g. 1322. ( ) They mediate transport from the ER to the cis Golgi network. ( ) Their coat protein forms a three-legged structure called a triskelion. ( ) They are pinched off from their donor compartment by a dynamin collar. ( ) They are involved in retrograde transport in the Golgi apparatus. 3. Indicate which one of the indicated types of coat proteins (A, B, and C) in the schematic diagram below corresponds to COPI, COPII, and clathrin, respectively. Your answer would be a three-letter string composed of letters A to C only, e.g. CBA.

A B C

4. Consider a perfectly assembled clathrin cage that is composed of 12 pentagons and 20 hexagons and therefore resembles a soccer ball. How many clathrin heavy chains are there in this cage? Write down the number as your answer, e.g. 24. 5. Adaptor proteins select cargo proteins that will be incorporated into clathrin-coated vesicles. An adaptor protein such as AP2 … A. can induce membrane curvature even before clathrin molecules bind. B. acts as a coincidence detector, assembling only when a number of requirements are met. C. binds to phosphoinositides in the cytosolic leaflet of the membrane. D. alternates between a locked cytosolic form and an unlocked membrane-bound form. E. All of the above. 6. Phosphoinositides mark different cellular membranes and play key roles in protein trafficking inside the cell. Among them, PI(4,5)P2 is involved in receptor-mediated endocytosis as well as phagocytosis at the plasma membrane. This phosphoinositide … A. is bound by the adaptor protein AP2. B. is bound by the GTPase dynamin. C. is depleted from clathrin-coated vesicles to promote their uncoating. D. All of the above.

7. Indicate whether each of the following descriptions better applies to COPI- (1), COPII(2), or clathrin- (3) coated vesicles. Your answer would be a four-digit number composed of digits 1 to 3 only, e.g. 1322. ( ) They are uncoated by an hsp70 family protein, which is stimulated by the binding of auxilin. ( ) They keep their coat proteins for a relatively long time, until they dock onto their target membrane. ( ) Their uncoating depends on activation of an ARF-GAP. ( ) Their uncoating depends in part on activation of a Sar1-GAP. 8.

Sort the following events to reflect the order in which they occur during the formation of

vesicles from the ER destined for the Golgi apparatus. Your answer would be a four-letter string composed of letters A to D only, e.g. DACB. (A) Sar1 GTP hydrolysis (B) Sar1 GTP binding (C) Sar1–Sec23 binding (D) Sar1 membrane association 9.

Indicate true (T) and false (F) statements below regarding intracellular vesicle transport.

Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF ( ) The plasma membrane is stiffer and flatter than most intracellular membranes, and its deformation generally requires greater force. ( ) Large molecules such as collagen fibers of about 300 µm in length are highly packed and coiled to fit into normal COPII vesicles of about 80 nm in diameter. ( ) Rab proteins are recruited to membranes following activation by Rab-GEFs. ( ) A Rab and its effector are always on two different membranes that will then fuse. 10.

If this protein is unable to hydrolyze its bound GTP, invaginated clathrin-coated pits

accumulate but fail to pinch off from the plasma membrane. In neurons, long vesicle necks collared by the protein are observed and presynaptic endocytosis is blocked. This protein … A. contains a PI(4,5)P2 binding domain and a GTPase domain. B. recruits other proteins to the neck of the vesicle. C. may change the membrane lipid composition by recruiting lipid-modifying enzymes. D. may directly distort the membrane using the energy from GTP hydrolysis. E. All of the above.

11. Sort the following events to reflect the order in which they occur during vesicle docking onto a target membrane, starting with an inactive Rab in the cytosol. Your answer would be a four-letter string composed of letters A to D only, e.g. DACB. (A) Rab is bound to its effector (tethering protein) on the target membrane. (B) Rab is bound to its Rab-GDI. (C) Rab is bound to the membrane in its GTP-bound form. (D) Rab dissociates from the membrane. 12. Rab5 and Rab7 constitute a Rab cascade in the process of endosome maturation. One of the Rab5 effectors is a Rab7-GEF, while one of the Rab7 effectors is a Rab5-GAP. Which of these proteins would you expect to find in early endosomes? Write down Rab5 or Rab7 as your answer. 13. Indicate whether each of the following descriptions better applies to t-SNAREs (T) or vSNAREs (V). Your answer would be a three-letter string composed of letters T and V only, e.g. TVV. ( ) They are usually located on the target membrane. ( ) They are composed of a single polypeptide chain. ( ) They are usually associated with inhibitory proteins that can be released by Rab proteins. 14. The transmembrane protein Tango1 is a packaging protein that helps some secretory proteins leave the endoplasmic reticulum (ER) after synthesis. Knocking down Tango1 by RNA interference impairs the incorporation of collagen VII, but not collagen I, into transport vesicles destined for the Golgi apparatus. The protein contains a lumenal N-terminal SH3 domain and a cytosolic C-terminal proline-rich domain. Which of the following proteins would you expect to interact with the N-terminal and C-terminal domains of Tango1, respectively? A. Procollagen VII; COPI components B. Procollagen VII; COPII components C. Procollagen I; COPI components D. Procollagen I; COPII components E. COPI components; procollagen I 15. In the following schematic diagram depicting the formation of vesicular tubular clusters between the ER and the CGN, what major coat proteins are indicated by 1 and 2, respectively?

A. COPI; COPII B. COPI; clathrin C. COPII; COPI D. COPII; clathrin E. Clathrin; COPII 16.

The cytoplasmic C-terminal KKXX sequence of transmembrane proteins interacts with

… A. t-SNAREs B. v-SNAREs C. AP2 D. COPI coatomers E. COPII coatomers 17. A schematic drawing of the secretory and endocytic pathways is presented below. Indicate which component in the drawing (A to J) corresponds to each of the following. Your answer would be a 10-letter string composed of letters A to J only, e.g. HICDJABFGE.

J I A H

B C D

( ) Early endosome ( ) Late endosome ( ) ER ( ) Lysosome ( ) cis Golgi cisterna ( ) medial Golgi cisterna ( ) trans Golgi cisterna ( ) cis Golgi network (CGN) ( ) trans Golgi network (TGN) ( ) Secretory vesicle 18. The Golgi apparatus is made up of an ordered series of compartments. To process Nlinked oligosaccharides, different Golgi compartments carry different enzymes that alter the sugar chains sequentially. Which Golgi cisternae are normally responsible for the addition of galactose and sialic acid, respectively, in complex oligosaccharides? A. cis cisterna; medial cisterna B. cis cisterna; trans cisterna C. medial cisterna; trans cisterna

D. medial cisterna; medial cisterna E. trans cisterna; trans cisterna 19. Consider two N-linked oligosaccharide chains on the same protein. The first chain contains three mannose residues and three negatively charged sialic acid residues, as well as other residues. The second chain contains only two N-acetylglucosamine and eight mannose residues. Which chain seems to have been added to the protein such that it is NOT fully accessible to the processing enzymes in the Golgi apparatus? Which chain is Endo H-sensitive? A. Chain one; chain one B. Chain one; chain two C. Chain two; chain one D. Chain two; chain two 20. Indicate whether each of the following descriptions better applies to N-linked (N) or Olinked (O) glycosylation. Your answer would be a four-letter string composed of letters N and O only, e.g. OONO. ( ) It is abundant in proteoglycans. ( ) It involves the attachment of a preassembled block of oligosaccharide onto a protein. ( ) It is attached to a serine or threonine residue in the protein. ( ) It involves heavily sulfated sugars. 21. Indicate true (T) and false (F) statements below regarding glycosylation of proteins in the endoplasmic reticulum and the Golgi apparatus. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF ( ) Glycosylation can promote protein folding. ( ) The glycosylation state of a protein can determine its fate along the secretory pathway. ( ) Glycosylation makes a protein more accessible to proteases and other proteins. ( ) Glycosylated proteins are generally more flexible. 22. Indicate whether each of the following descriptions better applies to the cisternal maturation model (C) or the vesicle transport model (V) for the organization of the Golgi apparatus. Your answer would be a four-letter string composed of letters C and V only, e.g. CVCV. ( ) Golgi cisternae are static organelles. ( ) Golgi cisternae exchange material exclusively by retrograde vesicular transport.

( ) A cis Golgi cisterna becomes a medial cisterna which becomes a trans cisterna. ( ) Any protein that passes through the Golgi apparatus should be incorporated into transport vesicles several times. 23. You have engineered a fusion protein composed of a cis Golgi resident protein and the green fluorescent protein. Similarly, you have made a fusion protein composed of a trans Golgi protein and the red fluorescent protein. You express these proteins in cells and follow individual Golgi cisternae using fluorescence microscopy. You observe that each individual cisterna emits a green, faint yellow, or red fluorescent signal at different times: it initially shows green fluorescence, but the green fluorescent signal fades away with time, and instead red fluorescence appears in the same cisterna. Additionally, you do not observe any red fluorescent cisterna that becomes green over time. Do these observations agree better with the cisternal maturation model (C) or the vesicle transport model (V) for the organization of Golgi stacks? Write down C or V as your answer. 24. Aspartic acid residues resemble phosphorylated serine residues in proteins and also carry a similar negative charge. As a result, a protein in which serine residues are mutated to aspartic acid residues may mimic the function of the corresponding wild-type protein when those serines are phosphorylated. Given that phosphorylation on serine residues in a number of Golgi matrix proteins occurs during mitosis, would you expect that mutation of these serines to aspartic acids would prevent Golgi stack formation in interphase (1) or prevent Golgi dispersion in mitosis (2)? Write down 1 or 2 as your answer. 25. Indicate whether each of the following descriptions better applies to a BAR domain (B), the outer shell of COPI coatomer (C), or a dynamin oligomer (D). Your answer would be a threeletter string composed of letters B, C, and D only, e.g. CBD. ( ) It resembles a crescent. ( ) It resembles a collar. ( ) It resembles a cage. 26. An animal cell has been wounded and has a small rupture in its plasma membrane. Which of the following is more likely to happen next? A. The cell rapidly cleaves by cytokinesis. B. The rate of receptor-mediated endocytosis is increased. C. The rate of exocytosis is increased. D. The rate of pinocytosis is increased.

E. All of the above. 27.

Lysosomes are the principal site of cellular digestion. They … A. normally maintain a pH of about 2.0 to 2.5. B. contain F-type ATPases that pump protons into the organelles. C. contain heavily glycosylated membrane proteins. D. are homogeneous in size and shape. E. All of the above.

28.

How many protons [or, more accurately, hydronium ions (H3O+)] are there in a spherical

lysosome that is about 0.6 µm in diameter and maintains an interior pH of about 5.0? Recall that the volume of a sphere of radius r is calculated as V = 4/3 × π × r3, and that pH = –log [H+]. Avogadro’s constant is approximately 6 × 1023 molecules/mole. A. About 700 ions B. About 7000 ions C. About 70,000 ions D. About 700,000 ions E. About 7,000,000 ions 29.

What is the cellular function of plant vacuoles? A. They store nutrients and waste products. B. They help increase cell size and maintain turgor pressure. C. They are used for degradation of cytoplasmic components. D. They help maintain cytosolic pH. E. All of the above.

30. You are studying the cellular basis of petal coloration in the flowering plant Ipomoea tricolor. These plants have colorful petals due to the presence of pH-sensitive vacuolar anthocyanins that change color from red/purple in acidic pH to blue at neutral pH. You treat petals with either vanadate (a specific inhibitor of P-ATPases) or bafilomycin (a specific inhibitor of V-ATPases) or both and compare the color with that of control petals, obtaining the results shown in the table below. Which of the following conclusions is consistent with these observations? Treatment

Color

Control Vanadate Bafilomycin Vanadate + bafilomycin

Red Blue Blue Blue

A. Both P- and V-ATPases are required to sufficiently acidify the vacuoles in petal cells. B. P-ATPases are sufficient for acidification of the vacuoles in petal cells. C. V-ATPases are sufficient for acidification of the vacuoles in petal cells. D. Neither P- nor V-ATPases are necessary for vacuole acidification in petal cells. 31.

Which of the following pathways does NOT directly deliver materials to lysosomes? A. Endocytosis B. Exocytosis C. Phagocytosis D. Autophagy E. Macropinocytosis

32.

A mitochondrion has just been engulfed by a cup-shaped isolation membrane as it

undergoes mitophagy. Before lysosomal fusion, how many membranes separate the cytosol from the matrix of the engulfed mitochondrion? A. 2 B. 3 C. 4 D. 5 33. In Drosophila melanogaster, loss-of-function mutations in either Pink1 or Parkin show similar phenotypes including impaired ability to fly, male sterility, and degeneration of dopaminergic neurons. Transgenic overexpression of Parkin in mutants lacking Pink1 significantly ameliorates the loss-of-function phenotype, but overexpression of Pink1 cannot rescue the Parkin loss-of-function phenotype. According to these findings, which protein is more likely to act upstream of the other one? Assuming that the Parkin loss-of-function phenotype is merely due to defects in autophagy, would you expect Parkin overexpression to also rescue an ATG9 loss-of-function phenotype? A. Parkin acts upstream of Pink1; yes

B. Parkin acts upstream of Pink1; no C. Pink1 acts upstream of Parkin; yes D. Pink1 acts upstream of Parkin; no 34. How does the affinity of M6P receptor proteins for the mannose 6-phosphate marker change between the TGN and early endosome? Which coat protein is mainly responsible for their transport from the TGN to the endosome? A. The affinity is higher in the TGN; clathrin B. The affinity is higher in the TGN; retromer C. The affinity is higher in endosomes; clathrin D. The affinity is higher in endosomes; retromer 35.

Which of the following is NOT correct regarding M6P receptors and KDEL receptors? A. They both shuttle back and forth between different membrane-enclosed compartments. B. They are both transmembrane proteins. C. They both release their soluble binding targets at lower pH. D. They both prevent the escape of proteins to the cell exterior by the “default” pathway.

36. You know that a particular sequence at the C-terminus of the lectin ERGIC53 enhances its exit from the endoplasmic reticulum (ER). You create two mutant versions of the protein, one without the suspected sequence (1) and one in which the sequence is replaced with an ER retention signal (2). You transfect a fibroblast cell line with a plasmid that encodes either wildtype ERGIC53 (0) or one of the engineered versions of it (1 or 2). After inducing the expression of the proteins, you lyse the cells and either treat the lysate with endoglycosidase H (Endo H) or leave the lysate untreated, as indicated below. You then separate the proteins by SDS-PAGE and perform a Western blot to detect the bands corresponding to ERGIC53. Your results are represented in the following drawing. Which lanes (A to C) do you expect to correspond to each of the proteins 0, 1, and 2, respectively? Your answer would be a three-letter string composed of letters A to C only, e.g. CAB.

Expressed protein: Treated with Endo H?

37.

A No

B Yes

C Yes

Yes

What is the effect of defective or missing N-acetylglucosamine phosphotransferase on

lysosomal protein sorting? A. Lysosomal proteins are secreted from the cell. B. Lysosomal proteins are retained in the Golgi network. C. Lysosomal proteins are retained in the ER. D. Lysosomal proteins remain tightly bound to M6P receptors. E. Nonfunctional lysosomal proteins accumulate in the lysosome. 38. Indicate whether each of the following descriptions better applies to receptor-mediated endocytosis (R), phagocytosis (F), pinocytosis (P), or macropinocytosis (M). Your answer would be a four-letter string composed of letters R, F, P, and M only, e.g. MMPF. ( ) This pathway depends on caveolin and cavin proteins. ( ) Pathogenic particles such as SV40, papillomavirus, and cholera toxin enter the cell via this pathway. ( ) In animals, this pathway is normally limited to professional cells such as neutrophils. ( ) This pathway proceeds by the formation of highly ruffled regions in the plasma membrane which then collapse, resulting in fluid uptake. 39. By forming over 100 pinocytic vesicles per minute, a macrophage ingests over 10% of its own volume of fluid every half an hour. This is equivalent to ingesting about 100% of the macrophage’s plasma membrane. If the average volume of a macrophage is about 450 µm3 and its average surface area is about 750 µm2, what is the average diameter of the early pinocytic vesicles in the cell? Assume spherical vesicles. The formulas for the volume (V) and surface area (S) of a sphere of radius r are V = 4/3 × π × r3, and S = 4 × π × r2, respectively. Write down your answer in nanometers, with no decimals, e.g. 830 nm.

40.

Which endocytic process is best depicted in the following schematic diagram?

A. Receptor-mediated endocytosis B. Pinocytosis C. Macropinocytosis D. Phagocytosis E. Entosis 41. How can the formation of tubular structures, as opposed to spherical vesicles, in the early endosomes facilitate recycling of membrane proteins such as cell-surface receptors? A. By increasing surface area-to-volume ratio B. By decreasing surface area-to-volume ratio C. By enhancing the formation of intralumenal vesicles D. By preventing the formation of intralumenal vesicles 42.

During maturation of early endosomes to late endosomes, … A. the vacuolar domain of the endosome is shed, whereas the tubular domain is retained. B. the endosome migrates along actin filaments away from the cell interior. C. the lumen of the endosome becomes more acidic. D. intralumenal vesicles disappear. E. All of the above.

43. Indicate true (T) and false (F) statements below regarding receptor-mediated endocytosis of LDL particles. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) LDL receptors are normally degraded in the lysosome along with their LDL ligands. ( ) LDL receptors that do not bind to extracellular LDL cannot be internalized in clathrin-coated vesicles. ( ) A mutation that impairs the attachment of an LDL receptor to a clathrin-coated pit would cause depletion of blood LDL levels. ( ) LDL receptors at the plasma membrane are usually concentrated in clathrin-coated pits. 44. Indicate whether each of the following membrane proteins releases its protein ligand (R), remains bound to its protein ligand (B), or is degraded in the lysosome along with its protein ligand (D) following receptor-mediated endocytosis. Your answer would be a three-letter string composed of letters R, B, and D only, e.g. RBD. ( ) LDL receptor ( ) Transferrin receptor ( ) EGF receptor 45. Indicate whether each of the following processes is topologically similar (Y) or not similar (N) to formation of intralumenal vesicles in multivesicular bodies. Your answer would be a four-letter string composed of letters Y and N only, e.g. YYYY. ( ) Fission of peroxisomes ( ) Formation of clathrin-coated vesicles ( ) Budding of HIV from infected cells ( ) Cytokinesis 46.

In mammals, colostrum (or “first milk”) is produced by the mother in late pregnancy and

shortly after giving birth to feed the newborn. In addition to nutrients, it is particularly rich in antibodies that are absorbed through the intestinal epithelium and support the weak immune system of the infant. Occasionally, some infants suffer from either respiratory acidosis or alkalosis. In acidosis, the blood pH becomes acidic due to lack of sufficient ventilation, while the opposite happens in alkalosis due to hyperventilation. Considering the molecular mechanism of transcytosis in intestinal epithelia, which of these conditions—acidosis (C) or alkalosis (L)— would you expect to interfere more with antibody absorption by transcytosis in these infants? Write down C or L as your answer.

47. You are viewing a sample of pond water under the microscope. The sample contains a variety of microorganisms, some moving faster than others. You spot a Paramecium that is being followed by a Didinium almost as large as itself; the Paramecium initially swims away and tries to escape, but Didinium, which moves in faster bursts, finally stops it and attaches firmly onto the side of the Paramecium and proceeds to eat it. What type of endocytosis will ensue? Write down the name of the process as your answer. Do not use abbreviations. 48. Sort the following events to reflect the order in which they occur during engulfment of an antibody-coated microorganism by a neutrophil in the blood. Your answer would be a four-letter string composed of letters A to D only, e.g. DACB. (A) Rho-GEF activation (B) Activation of Fc receptors (C) Local PI(4,5)P2 production (D) Local PI(3,4,5)P3 production 49. Indicate true (T) and false (F) statements below regarding secretory vesicles. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) Proteins destined for secretory vesicles (for regulated secretion) often aggregate in the

50.

lumen of the TGN. ( ) The lumen of secretory vesicles is generally less acidic than the lumen of the TGN from which they originate. ( ) A secretory vesicle starts recycling Golgi components only after it has budded from the TGN and its coat proteins have disassembled. ( ) Secretory vesicles recycle Golgi components to the TGN in small clathrin-coated vesicles. Insulin is a secretory protein made by the β cells in the pancreas. This protein … A. is released by the β cells in response to increased intracellular Ca2+ concentration. B. is first synthesized as preproinsulin. C. is stored as aggregates inside secretory vesicles in the β cells. D. undergoes proteolytic cleavage before secretion. E. All of the above.

51. In the simplified diagram below depicting exocytosis of synaptic vesicles, which components (A to D) correspond to each of the following? Your answer would be a four-letter string composed of letters A to D only, e.g. DBCA.

A B D C

Membrane fusion

C ( ) Ca2+ ( ) Complexin ( ) Synaptotagmin ( ) SNARE complex 52. Neurotransmitters are retrieved from the synaptic cleft soon after their release from the presynaptic axon terminal. Do you think treatment of neurons with the drug dynasore, a dynamin inhibitor, would affect this retrieval positively (P) or negatively (N)? Write down P or N as your answer. 53. A protein that is normally not expressed in epithelial cells has been induced in these cells and has been engineered such that a GPI anchor is attached to it as it is processed in the ER. Would you expect to find this protein in the apical (A) or basolateral (B) domain of the plasma membrane in these cells? Write down A or B as your answer. 54. In the highly simplified diagram below, the energy landscape of secretory vesicle fusion to the plasma membrane is shown in the presence or absence of appropriate SNARE complexes. Which curve (A or B) do you think corresponds to the presence of SNARE complexes? Write down A or B as your answer.

Free energy

Progress of fusion

Answers 1. Answer: A Difficulty: 3 Section: The Molecular Mechanisms of Membrane Transport and the Maintenance of Compartmental Diversity Feedback: GTP hydrolysis by ARF GTPases is required for the uncoating of COPI vesicles. On the other hand, the NEM-sensitive components are involved in the later steps of fusion and recycling. 2. Answer: 2331 Difficulty: 2 Section: The Molecular Mechanisms of Membrane Transport and the Maintenance of Compartmental Diversity Feedback: Typically, transport vesicles from ER to CGN are COPII-coated, whereas retrograde transport in the Golgi apparatus is mediated by COPI-coated vesicles. Clathrin-coated vesicles are involved in traffic from the plasma membrane as well between the endosomal and Golgi compartments, and require dynamin for pinching off efficiently. Clathrin triskelions assemble into a basket-like cage that envelopes these budding vesicles. 3. Answer: ACB Difficulty: 2 Section: The Molecular Mechanisms of Membrane Transport and the Maintenance of Compartmental Diversity Feedback: Please refer to Figure 13-5. 4. Answer: 180 Difficulty: 3 Section: The Molecular Mechanisms of Membrane Transport and the Maintenance of Compartmental Diversity Feedback: There is one triskelion at each vertex. The number of vertices is calculated as follows: Number of vertices = [(12 × 5) + (20 × 6)] / 3 = 60 (There are five vertices around each pentagon and six vertices around each hexagon. Each vertex is counted three times this way, the reason why we divide the sum by 3.) Since each triskelion contains three heavy chains, the number of heavy chains equals 180. 5. Answer: E Difficulty: 1 Section: The Molecular Mechanisms of Membrane Transport and the Maintenance of Compartmental Diversity Feedback: When bound to the phosphoinositide PI(4,5)P2, the adaptor protein AP2 changes conformation (or “unlocks”) and binds to cargo receptors in the membrane,

inducing membrane curvature. Since several requirements must be met for stable AP2 binding to a membrane, the protein only assembles at the right place and at the right time. 6. Answer: D Difficulty: 1 Section: The Molecular Mechanisms of Membrane Transport and the Maintenance of Compartmental Diversity Feedback: AP2 adopts an open conformation once bound to PI(4,5)P2 in the plasma membrane. Dynamin also has a PI(4,5)P2-binding domain which tethers the protein to the membrane as a clathrin-coated bud grows. Once clathrin-coated vesicles pinch off from the membrane, a PIP phosphatase that is co-packaged into the vesicles depletes PI(4,5)P2, which weakens the binding of the adaptor proteins. 7. Answer: 3212 Difficulty: 2 Section: The Molecular Mechanisms of Membrane Transport and the Maintenance of Compartmental Diversity Feedback: Uncoating of COPII-coated vesicles may occur upon docking at a target membrane well after Sar1 hydrolyzes its bound GTP and is released. Clathrin- and COPIcoated vesicles, in contrast, shed their coat soon after they pinch off, through the activation of an uncoating ATPases or an ARF-GAP, respectively. 8. Answer: BDCA Difficulty: 2 Section: The Molecular Mechanisms of Membrane Transport and the Maintenance of Compartmental Diversity Feedback: A Sar1-GEF in the endoplasmic reticulum membrane activates Sar1 by guanine nucleotide exchange; Sar1 then inserts an amphiphilic helix into the membrane, one leaflet deep. Sar1 then recruits COPII coat proteins (and also a Sar1-GAP) and initiates budding. Sar1 then hydrolyzes its GTP and dissociates following a delay; however, if a sealed coat is already formed, Sar1 dissociation may not cause uncoating. 9. Answer: TFTF Difficulty: 2 Section: The Molecular Mechanisms of Membrane Transport and the Maintenance of Compartmental Diversity Feedback: Specific packaging proteins induce the formation of especially large COPIIcoated transport vesicles that can accommodate procollagen fibers. A Rab and its effector can be on the same membrane or on different membranes. 10. Answer: E Difficulty: 2 Section: The Molecular Mechanisms of Membrane Transport and the Maintenance of Compartmental Diversity Feedback: It is dynamin.

11. Answer: BCAD Difficulty: 2 Section: The Molecular Mechanisms of Membrane Transport and the Maintenance of Compartmental Diversity Feedback: In its GDP-bound form, a Rab protein is bound to a Rab-GDP dissociation inhibitor and kept in the cytosol. Once activated by a membrane-bound Rab-GEF and bound to GTP, Rab molecules associate with the membrane through a lipid anchor and bind to their effector proteins. Finally, GTP hydrolysis results in Rab inactivation and dissociation from the membrane. 12. Answer: Rab5 Difficulty: 2 Section: The Molecular Mechanisms of Membrane Transport and the Maintenance of Compartmental Diversity Feedback: In this cascade, Rab5 is the upstream Rab that is replaced by Rab7 in the late endosomes. 13. Answer: TVT Difficulty: 2 Section: The Molecular Mechanisms of Membrane Transport and the Maintenance of Compartmental Diversity Feedback: A v-SNARE is typically found on vesicle membranes and is a single polypeptide chain, whereas a t-SNARE is typically found on target membranes and is usually composed of three proteins. The t-SNAREs are often associated with inhibitory proteins that are released only when membrane fusion is necessary. 14. Answer: B Difficulty: 2 Section: Transport from the ER through the Golgi Apparatus Feedback: ER export is mediated by COPII-coated vesicles. Tango1 is likely to interact with the type VII procollagen molecules. 15. Answer: C Difficulty: 2 Section: Transport from the ER through the Golgi Apparatus Feedback: ER-derived COPII-coated vesicles (1) fuse with one another to form the vesicular tubular clusters, which in turn bud off transport vesicles coated with COPI (2) for the retrieval of ER resident proteins. 16. Answer: D Difficulty: 1 Section: Transport from the ER through the Golgi Apparatus Feedback: By binding to the COPI coats, the retrieval signal mediates the retrograde transport of the protein back to the ER. 17. Answer: HIAJCDEBFG

Difficulty: 2 Section: Transport from the ER through the Golgi Apparatus Feedback: Please refer to Figure 13-3. 18. Answer: E Difficulty: 2 Section: Transport from the ER through the Golgi Apparatus Feedback: Addition of Gal and NANA has been shown to occur primarily in the trans cisterna. 19. Answer: D Difficulty: 2 Section: Transport from the ER through the Golgi Apparatus Feedback: Chains one and two represent complex and high-mannose oligosaccharides, respectively. 20. Answer: ONOO Difficulty: 2 Section: Transport from the ER through the Golgi Apparatus Feedback: In N-linked glycosylation, a preassembled block of oligosaccharide is attached onto an asparagine side chain in the protein. O-linked glycosylation, in which sugars are added to the hydroxyl groups in the side chains of serine, threonine, or other residues, often involves sulfated sugars and is responsible for the production of heavilyglycosylation in proteoglycans. 21. Answer: TTFF Difficulty: 2 Section: Transport from the ER through the Golgi Apparatus Feedback: Chains of sugar have limited flexibility and therefore protrude from the protein surface. As a result, the approach of other macromolecules to the protein is limited. At the same time, lectins can recognize the sugar chain and control the proper folding of the protein or send it down a particular sorting pathway. 22. Answer: VCCV Difficulty: 2 Section: Transport from the ER through the Golgi Apparatus Feedback: The cisternal maturation model views the Golgi cisterna as dynamic structures that move through the Golgi stack with cargo in their lumen. In this model, retrograde transport is responsible for carrying Golgi resident proteins back to an earlier compartment as each cisterna matures. The vesicle transport model, in contrast, describes Golgi cisternae as static structures. In this model, forward-moving vesicles carry forwardmoving cargo, and backward-moving vesicles retrieve escaped resident proteins. 23. Answer: C Difficulty: 3 Section: Transport from the ER through the Golgi Apparatus

Feedback: The cisternal maturation model predicts that each cisterna matures over time, acquiring new components (as well as red fluorescence) as it matures, and recycling old components (green fluorescence) to the less mature cisternae by retrograde vesicle transport. The vesicle transport model predicts that each cisterna remains stationary and fluoresces either red (or yellow) or green during the entire period. 24. Answer: 1 Difficulty: 3 Section: Transport from the ER through the Golgi Apparatus Feedback: The mutant proteins mimic constitutively phosphorylated proteins, and therefore recapitulate the mitotic Golgi fragmentation even in interphase. 25. Answer: BDC Difficulty: 2 Section: Transport from the trans Golgi Network to the Cell Exterior: Exocytosis Feedback: Coat proteins assemble into distinctive cages that coat many transport vesicles in the cell. Dynamin helps pinch off budding clathrin-coated vesicles by assembling like a collar around their neck. BAR-domain proteins bend membranes by binding and imposing their curved shape on the underlying membrane. 26. Answer: C Difficulty: 1 Section: Transport from the trans Golgi Network to the Cell Exterior: Exocytosis Feedback: After a rupture in the plasma membrane of a cell, a temporary cell-surface patch is made from locally-available membranes such as those of lysosomes in a process that probably involves exocytosis and homotypic vesicle-vesicle fusion. 27. Answer: C Difficulty: 1 Section: Transport from the trans Golgi Network to Lysosomes Feedback: Lysosomes vary greatly in size and shape, reflecting their various functions. They maintain a low interior pH of about 4.5–5.0 using lysosomal V-type ATPases. Their own membrane proteins are highly glycosylated to protect them from digestion by the proteases in the lumen. 28. Answer: A Difficulty: 3 Section: Transport from the trans Golgi Network to Lysosomes Feedback: The volume of the lysosome would be calculated as: V = 4/3 × π × (0.3 × 10–6 m)3 = 1.1 × 10–19 m3 = 1.1 × 10–16 L At approximately pH 5, the concentration of hydronium ions is 10–5 mol/L. Therefore, there is about 1.1 × 10–21 mole of these ions in this lysosome. This is equivalent to only approximately 660 ions. 29. Answer: E Difficulty: 1

Section: Transport from the trans Golgi Network to Lysosomes Feedback: Vacuoles in plants are equivalent to lysosomes in animal cells, and are remarkably versatile in function. 30. Answer: A Difficulty: 2 Section: Transport from the trans Golgi Network to Lysosomes Feedback: Anthocyanins can be considered as pH indicators in this experiment, which shows that blocking either type of ATPase would prevent the sufficient acidification required for red coloration. 31. Answer: B Difficulty: 1 Section: Transport from the trans Golgi Network to Lysosomes Feedback: Multiple pathways deliver materials to lysosomes for degradation. 32. Answer: C Difficulty: 2 Section: Transport from the trans Golgi Network to Lysosomes Feedback: Two membranes from the cup-shaped structure and two membranes from the original mitochondrion surround the mitochondrial matrix. After fusion of the outermost membrane with the lysosomal membrane, the three inner membranes would be digested away in the process of autophagy. 33. Answer: D Difficulty: 3 Section: Transport from the trans Golgi Network to Lysosomes Feedback: Overexpression of a protein in a pathway can often rescue the loss of function of upstream (but not downstream) proteins. Since ATG9 is involved in downstream steps of mitophagy, Parkin overexpression is not expected to rescue an ATG9 loss-of-function phenotype. 34. Answer: A Difficulty: 2 Section: Transport from the trans Golgi Network to Lysosomes Feedback: In the trans Golgi network (TGN), mannose 6-phosphate (M6P) receptor proteins bind to hydrolases that carry the M6P group and mediate their incorporation into clathrin-coated vesicles destined for endosomes. In the lower pH of the endosome lumen, the hydrolases are released and the receptor proteins are retrieved by incorporation in retromer-coated vesicles. 35. Answer: C Difficulty: 2 Section: Transport from the trans Golgi Network to Lysosomes Feedback: Whereas mannose 6-phosphate (M6P) receptors bind more tightly to their targets at higher pH (tighter in the trans Golgi network compared to endosomes), KDEL

receptors bind more tightly to their targets at lower pH (tighter in the Golgi complex and vesicular tubular clusters compared to the endoplasmic reticulum). The receptors also use different coat proteins for transport. 36. Answer: CAB Difficulty: 3 Section: Transport from the trans Golgi Network to Lysosomes Feedback: Endo H-resistance is a sign that the protein has been transferred to the Golgi apparatus and has undergone processing. With wild-type protein (0, C), the protein shows partial Endo H-sensitivity. In the absence of the exit sign (1, A), transport to the Golgi is impaired, and most of the protein remains Endo H-sensitive. When an ER retention signal is also added (2, B), the protein seems to be completely Endo H-sensitive. 37. Answer: A Difficulty: 1 Section: Transport from the trans Golgi Network to Lysosomes Feedback: Because the lysosomal enzymes cannot be phosphorylated by the phosphotransferase to create the mannose 6-phosphate (M6P) marker, they are not segregated to the appropriate transport vesicles and are lost from the cell by secretion. 38. Answer: PPFM Difficulty: 1 Section: Transport into the Cell from the Plasma Membrane: Endocytosis Feedback: Caveolae, a type of pinocytic vesicles, contain caveolins and cavins as their structural proteins, and constitute the main route of entry for several pathogens. Other pathogens may enter the cell by macropinocytosis which involves the formation of ruffles. Professional phagocytes in our body engulf foreign cells and particles by phagocytosis. 39. Answer: 360 nm Difficulty: 3 Section: Transport into the Cell from the Plasma Membrane: Endocytosis Feedback: The cell internalizes 3000 vesicles every half an hour (i.e. 100 vesicles/min × 30 min), which is equivalent to 45 µm3 in volume (i.e. 10% × 450 µm3) and 750 µm2 in surface area (i.e. 100% × 750 µm2). Thus, the volume and surface area of each vesicle is, on average, about 0.015 µm3 and 0.250 µm2, respectively. The ratio of volume to surface area for a sphere is equal to r/3 where r is the radius. Hence, the diameter (2r) is equal to (0.015 µm3/0.250 µm2) × 6 = 0.36 µm = 360 nm. 40. Answer: C Difficulty: 1 Section: Transport into the Cell from the Plasma Membrane: Endocytosis Feedback: Macropinocytosis does not operate continually and is typically induced only in response to activation of cell-surface receptors by various stimuli. The ensuing activation of intracellular signaling pathways results in changes in actin dynamics and the formation

of cell-surface protrusions known as ruffles, which then collapse to form macropinosomes. 41. Answer: A Difficulty: 2 Section: Transport into the Cell from the Plasma Membrane: Endocytosis Feedback: The geometry of such tubular structures can help enrich membrane proteins over soluble proteins, due to the large membrane area enclosing a small volume. 42. Answer: C Difficulty: 1 Section: Transport into the Cell from the Plasma Membrane: Endocytosis Feedback: During maturation, the endosome sheds its tubular domains and retains its vacuolar domains, the lumen of which becomes more acidic. The endosome moves along microtubule tracks toward the cell interior, and acquires intralumenal vesicles for membrane protein degradation. It also receives new components from the trans Golgi network. 43. Answer: FFFT Difficulty: 1 Section: Transport into the Cell from the Plasma Membrane: Endocytosis Feedback: Low-density lipoprotein (LDL) receptors associate with clathrin-coated pits and are internalized irrespective of ligand binding. They normally release their bound LDL (if any) in the early endosomes and are recycled back to the plasma membrane. Mutations that interfere with receptor-mediated endocytosis of LDL particles can cause high blood cholesterol and lead to diseases such as atherosclerosis. 44. Answer: RBD Difficulty: 2 Section: Transport into the Cell from the Plasma Membrane: Endocytosis Feedback: LDL receptors release LDL in the low pH in early endosomes and are recycled back to the cell surface. Transferrin receptors do not release transferrin in early endosomes: the low pH only results in the release of the bound iron, and the transferrinreceptor complex is recycled. In contrast, most EGF receptors are not recycled and are degraded in the lysosomes along with EGF. 45. Answer: NNYY Difficulty: 2 Section: Transport into the Cell from the Plasma Membrane: Endocytosis Feedback: Budding of HIV particles and cytokinesis are both equivalents to the formation of intralumenal vesicles, as both involve budding away from the cytosolic surface of the membrane. 46. Answer: C Difficulty: 3 Section: Transport into the Cell from the Plasma Membrane: Endocytosis

Feedback: Acidification of the blood results in reduced antibody absorption because the antibody receptors do not efficiently release their bound antibodies on the basal side of the epithelial cells. 47. Answer: Phagocytosis Difficulty: 1 Section: Transport into the Cell from the Plasma Membrane: Endocytosis Feedback: Protozoa such as Didinium carry out phagocytosis as a form of feeding. This is not the case in multicellular organisms, in which few cells are able to ingest large food particles efficiently, and breakdown of food particles occurs mostly extracellularly in the digestive tract. 48. Answer: BACD Difficulty: 2 Section: Transport into the Cell from the Plasma Membrane: Endocytosis Feedback: The ordered generation and conversion of phosphoinositides guides sequential steps of phagocytosis. First, PI(4,5)P2 is produced by a kinase to induce actin polymerization and engulfment of the target particle. Then, another kinase converts PI(4,5)P2 to PI(3,4,5)P3, which is required for the closure of the phagosome. 49. Answer: TFFT Difficulty: 1 Section: Transport from the trans Golgi Network to the Cell Exterior: Exocytosis Feedback: Secretory vesicles bud from the trans Golgi network (TGN) in a process that involves highly concentrating the enclosed secretory proteins into aggregates. The immature vesicles become more and more acidic, fuse with each other, and send clathrincoated vesicles back to the TGN, sometimes even before they have completed budding from the TGN themselves. 50. Answer: E Difficulty: 2 Section: Transport from the trans Golgi Network to the Cell Exterior: Exocytosis Feedback: Starting from a pre-pro-protein (e.g. preproinsulin), the pre-peptide is cleaved off in the ER, and the pro-peptide is cleaved later normally during maturation of secretory vesicles. The formation and maturation of secretory vesicles at the TGN involves selective segregation and concentration of secretory proteins such as insulin. 51. Answer: ACBD Difficulty: 2 Section: Transport from the trans Golgi Network to the Cell Exterior: Exocytosis Feedback: Calcium binds to synaptotagmin and activates the protein, which then displaces complexin to relieve the block that complexin places on the metastable SNARE complexes; this release leads to membrane fusion. 52. Answer: N Difficulty: 2

Section: Transport from the trans Golgi Network to the Cell Exterior: Exocytosis Feedback: Dynamin is required to complete the endocytosis of the clathrin-coated, neurotransmitter-containing vesicles. 53. Answer: A Difficulty: 2 Section: Transport from the trans Golgi Network to the Cell Exterior: Exocytosis Feedback: By being preferentially concentrated in lipid rafts, the glycosylphosphatidylinositol (GPI)-anchored protein is incorporated into transport vesicles destined for the apical membrane. 54. Answer: B Difficulty: 1 Section: Transport from the trans Golgi Network to the Cell Exterior: Exocytosis Feedback: Similar to enzymes and other catalysts, SNARE complexes facilitate membrane fusion by lowering the free-energy barrier. They also transition into a stable conformation at the end of the process, requiring the energy of ATP hydrolysis to reset. Thus, they also push the process forward by stabilizing the products.

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION CHAPTER 14: ENERGY CONVERSION: MITOCHONDRIA AND CHLOROPLASTS Copyright © 2015 by W.W. Norton & Company, Inc. 1. Indicate true (T) and false (F) statements below regarding mitochondria. Your answer would be a four-letter string composed of letters T and F only, e.g. TFTT. ( ) Mitochondria are small round organelles that are often associated with actin filaments of the cytoskeleton. ( ) In sperm cells, mitochondria are seen wrapping tightly around the nucleus. ( ) Mitochondria are large enough to be seen with modern light microscopy, and can occupy as much as 20% of cytoplasmic volume. ( ) The outer mitochondrial membrane is freely permeable to ions and small molecules. 2. You have isolated mitochondria from a liver tissue sample, suspended them in a hypotonic buffer that causes them to swell and burst their outer membrane, and then added sucrose to a final concentration of about 25%. You then layer the suspension on a density gradient, ultracentrifuge, collect the different fractions, and analyze their protein content. Which of the following proteins would you expect to be highly enriched in the lower density and higher density fractions, respectively? A. Cytochrome oxidase; porin B. Cytochrome oxidase; ATP synthase C. ATP synthase; cytochrome oxidase D. Porin; ATP synthase 3. Indicate whether each of the following descriptions better matches the outer mitochondrial membrane (O), the intermembrane space (S), the crista membrane portion of the inner mitochondrial membrane (I), or the matrix (M). Your answer would be a four-letter string composed of letters O, S, I, and M only, e.g. OMIM. ( ) It is where NADH is produced. ( ) It is composed of 75% protein by weight. ( ) It is where the respiratory chain is located. ( ) It has the same electrochemical potential for H+ as the cytoplasm.

4. For mitochondria that are active in respiration, indicate whether the movement of each of the following molecules into (I) or out of (O) the matrix is thermodynamically favorable. Your answer would be a four-letter string composed of letters I and O only, e.g. IOOI. ( ) O2 ( ) H+ ( ) ADP ( ) CO2 5. In addition to their respiratory function, mitochondria have other important roles in cellular metabolism. Which of the following processes is NOT carried out mainly by mitochondria? A. Biosynthesis of cardiolipin B. Biosynthesis of fatty acids C. Catabolism of amino acids D. Biosynthesis of heme E. Biosynthesis of iron–sulfur clusters 6. A hypothetical electron-transport chain contains an electron donor (1), an electron carrier (2), and an electron acceptor (3). Which of these molecules has the highest redox potential? Write down 1, 2, or 3 as your answer. 7. Sort the following molecules to reflect the order in which they carry electrons in the respiratory chain. Your answer would be a seven-letter string composed of letters A to G only, e.g. GFADCBE. (A) NADH dehydrogenase complex (B) Cytochrome c oxidase complex (C) Cytochrome c reductase complex (D) Cytochrome c (E) Ubiquinone (F) NADH (G) H2O 8. Indicate whether each of the following descriptions better applies to cytochrome c oxidase complex (O), cytochrome c reductase complex (R), or NADH dehydrogenase complex (N). Your answer would be a four-letter string composed of letters O, R, and N only, e.g. RONN. ( ) It is the largest of the three respiratory complexes.

( ) It accounts for the majority of O2 uptake in most cells. ( ) It employs the Q cycle to increase proton pumping. ( ) It contains separate modules for electron transport and proton pumping. 9. The free-energy change of ATP hydrolysis to ADP inside the cell is about –50 kJ/mole. The standard free-energy change for the same reaction is about –30 kJ/mole. This necessitates that [ATP] inside the cell be … A. higher than that under standard conditions. B. greater than [ADP]. C. greater than [ADP] × [Pi] D. greater than [ADP] × [Pi] × Keq E. greater than [ADP] × [Pi] / Keq 10. The free-energy change for ATP hydrolysis to ADP inside the cell is about –50 kJ/mole. The complete cellular oxidation of glucose to carbon dioxide and water has a free-energy change of about –3000 kJ/mole. What is the efficiency of the cellular metabolism in harvesting this amount of free energy in the form of ATP synthesis? A. About 10% B. About 20% C. About 30% D. About 50% E. About 70%

Reference: ATP Synthase Questions 11-14 Consider the remarkable structure of the F1Fo ATP synthase shown in the following schematic drawing. Answer the following question(s) based on the structure.

2 3

5 4

11. In the schematic drawing of the structure of ATP synthase shown above, indicate whether each of the numbered components of the machine (from 1 to 5, respectively) is rotary (R) or stationary (S). Your answer would be a five-letter string composed of letters R and S only, e.g. RRSRS. 12. In Escherichia coli, the c ring of ATP synthase is composed of 12 c subunits. For one fully productive revolution of this ring, how many protons are transported across the membrane for each molecule of ATP produced? Write down your answer as a number, rounded to the nearest integer, e.g. 17. 13. With about 7000 c-ring revolutions per minute, on average how many moles of these pumps should be working at each moment in an entire human body to make about 50 kilograms of ATP per day? The molecular weight of ATP is about 500. A. About 300 millimoles B. About 3 millimoles C. About 30 micromoles D. About 0.3 micromoles 14.

As seen from the F1 head in the matrix, the rotation of the c ring is counterclockwise in

the ATP-synthesis mode of the motor and clockwise in the ATP-hydrolysis mode of the motor. As shown in the above drawing, the “a subunit” (labeled 2) is part of the stator embedded in the membrane, and it forms the entry and exit sites for the protons that are bound by the rotating c ring. How do you think the two half-channels of the a subunit are arranged as seen from the c ring? Write down A or B from the following schematic diagram as your answer.

Cristae space Inner membrane Matrix

15.

ATP synthase molecules in mitochondria form dimers that are localized mostly to sharp

cristae ridges. What should happen if subunits of the synthase that are required for dimerization are mutated in yeast? A. Oxygen consumption would drop B. Cristae ridges would disappear C. Cell growth would slow down D. Monomeric ATP synthases would distribute randomly over the inner membrane E. All of the above 16. In the following schematic drawing of the inner and outer mitochondrial membranes, in what region (A to D) would you expect to find more ATP synthase dimers?

Cytosol

B C

Matrix

17. In actively respiring mitochondria, where in the following schematic drawing of the inner and outer mitochondrial membranes would you expect to find the lowest local pH?

Cytosol

B C

Matrix

18. The synthetic toxin 2,4-dinitrophenol can uncouple ATP synthesis from mitochondrial respiration by decreasing the permeability of the inner membrane to protons. What would be the effect of dinitrophenol treatment on the amount of ATP produced by mitochondria and on the rate of ATP transport across the inner membrane, respectively? A. Positive; positive B. Positive; negative C. Negative; positive D. Negative; negative 19. Which of the following molecules can serve as the terminal electron acceptor in bacterial electron-transport chains? A. Oxygen B. Sulfate C. Fumarate D. Nitrite E. All of the above 20. A facultative anaerobic bacterium can survive both aerobic and anaerobic conditions. Under both conditions, the cell maintains a proton-motive force across the plasma membrane that

drives many cellular processes including nutrient import. Under aerobic conditions, the protonmotive force is produced by a respiratory chain. What is the mode of action of the plasma membrane ATP synthase in this bacterium under aerobic conditions: ATP-hydrolysis mode (H) or ATP-synthesis mode (S)? Write down H or S as your answer. 21. Indicate which letter (A to D) in the schematic drawings below of a mitochondrion and a chloroplast corresponds to each of the following features. Your answer would be a four-letter string composed of letters A to D only, e.g. BDCA. Each letter should be used once. C

A B

( ) Crista ( ) Stroma ( ) Matrix ( ) Thylakoid 22.

Indicate whether each of the following descriptions better applies to the light (L) or dark

(D) reactions in plant chloroplasts. Your answer would be a five-letter string composed of letters L and D only, e.g. DDDLD. ( ) It produces the sugar glyceraldehyde 3-phosphate. ( ) It involves the electron-transfer chain embedded in the thylakoid membrane. ( ) It involves O2 production. ( ) It involves fixation of CO2. ( ) It generates ATP.

23.

The electrons used in carbon fixation by chloroplasts ultimately come from … A. Atmospheric oxygen B. Atmospheric carbon dioxide C. Water D. Glyceraldehyde 3-phosphate

24. In the following simplified diagram of the Calvin cycle, which step (A to E) is catalyzed by the abundant enzyme Rubisco? ATP 3C A CO2

3C NADPH

E B 5C 3C

C 3C

ATP Glyceraldehyde 3-phosphate

25. Indicate true (T) and false (F) statements below regarding carbon fixation in plant cells. Your answer would be a five-letter string composed of letters T and F only, e.g. TTFTF. ( ) Carbon fixation in photosynthetic cells eventually generates glucose, the major form of sugar that is transported to other plant tissues. ( ) Plant cells can generate fat droplets in their chloroplasts using glyceraldehyde 3phosphate. ( ) ATP generated in the light cycle is the major source of ATP used by the plant cell to power all of its biochemical reactions. ( ) Plant cells can generate starch granules in their chloroplasts using glyceraldehyde 3phosphate. ( ) Pyruvate is generated in the chloroplast by the glycolytic pathway.

26. The cytochrome b6-f complex that transfers electrons between the two photosystems in photosynthesis is structurally and functionally similar to which of the complexes in the mitochondrial electron-transport chain? A. NADH dehydrogenase B. Succinate dehydrogenase C. Cytochrome c reductase D. Cytochrome c oxidase E. ATP synthase 27.

Some photosynthetic bacteria have only one type of photosystem in their plasma

membrane. Which of the following is likely to be true regarding such bacteria? A. They absorb purple light only. B. Their photosystem is unrelated to those in cyanobacteria and plants. C. They do not use water as the electron donor. D. They lack ATP synthase. E. All of the above. 28.

When the special pair in a photosystem is excited by a quantum of light, charge

separation can occur. Where does this take place? What is the charge of the ionized chlorophyll? A. In the antenna complex; positive B. In the antenna complex; negative C. In the reaction center; positive D. In the reaction center; negative 29. Which of the following is true regarding light-harvesting complexes in plant chloroplasts? A. They contain chlorophyll and other pigments. B. They are found in both photosystem I and photosystem II. C. They can protect the cell by preventing the generation of reactive oxygen species. D. They cannot carry out charge separation. E. All of the above. 30. Sort the following molecules to reflect the order in which they transfer electrons in noncyclic photophosphorylation in plants. Your answer would be an eight-letter string composed of letters A to H only, e.g. FCDEHGBA.

(A) Ferredoxin-NADP+ reductase (B) Plastoquinone (C) Ferredoxin (D) Cytochrome b6-f complex (E) H2O (F) Plastocyanin (G) Photosystem I (H) Photosystem II 31.

Indicate whether each of the following descriptions better applies to photosystem I (1) or

photosystem II (2). Your answer would be a four-digit number composed of digits 1 and 2 only, e.g. 2212. ( ) It reduces plastoquinone. ( ) It uses a manganese cluster to oxidize water. ( ) It is confined to unstacked thylakoids. ( ) It contains the special pair P680. 32.

The water-splitting step in photosynthesis … A. occurs on the stromal side of the thylakoid membrane. B. is catalyzed by an iron–sulfur cluster. C. consumes H+ and therefore contributes to the proton gradient across the thylakoid membrane. D. generates all the O2 in the Earth’s atmosphere. E. All of the above.

33. Photosynthesis provides the strongest known electron donor as well as the strongest known electron acceptor in all biology. What do you think the donor and the acceptor are, respectively? A. Excited P700; excited P680 B. Excited P700; ionized P680 C. Ionized P700; excited P680 D. Ionized P700; ionized P680 34. Which of the following does NOT occur in the normal process of light-driven production of ATP and NADPH in plant chloroplasts?

A. The cytochrome b6-f complex pumps protons from the stroma into the thylakoid lumen. B. The oxygen-evolving enzyme generates protons in the thylakoid lumen. C. The ferredoxin-NADP+ reductase generates protons in the thylakoid lumen. D. The ATP synthase transports protons from the thylakoid lumen to the stroma. 35. Indicate whether each of the following statements better describes ATP production in chloroplasts (C) or in mitochondria (M). Your answer would be a four-letter string composed of letters C and M only, e.g. MCMM. ( ) The electrochemical gradient that drives ATP production is dominated by the electrical component. ( ) It involves a significant pH difference across the membrane. ( ) The ATP synthase molecules form dimers. ( ) The ATP synthase molecules are distributed randomly in flat membrane regions. 36. Order the following metabolic innovations to reflect the most likely order in which they evolved during biological evolution. Your answer would be a four-letter string composed of letters A to D only, e.g. BDCA. (A) Using light energy to generate reducing power (B) Splitting of water molecules (an electron source) (C) Fermentation of organic material (D) Reduction of molecular oxygen (a terminal electron acceptor) 37. What is the main reason why atmospheric oxygen levels did not rise immediately after the mass production of the gas by photosynthetic bacteria? A. Because oxygen was rapidly consumed by aerobic microorganisms. B. Because oxygen was dissolved in water. C. Because large amounts of ferrous ions reduced the oxygen molecules. D. Because oxygen was rapidly recycled by early photosynthetic bacteria. 38.

Which of the following probably occurred first on Earth? A. H2O photosynthesis B. H2S photosynthesis C. O2 respiration D. Eukaryotic photosynthesis

39. Indicate true (T) and false (F) statements below regarding the evolution of photosynthesis and oxidative phosphorylation. Your answer would be a five-letter string composed of letters T and F only, e.g. TTFTF. ( ) Chloroplasts are thought to descend from ancient purple sulfur bacteria. ( ) Mitochondria are thought to descend from ancient α-proteobacteria. ( ) The first photosynthetic cyanobacteria evolved before the first aerobic proteobacteria. ( ) All photosynthetic and aerobic bacteria are thought to have evolved from a common ancestor capable of fermentation and membrane electron transport. 40.

Two haploid budding yeast cells are allowed to mate. One of them carries a mutation in

its mitochondrial DNA that makes the yeast cell resistant to an antifungal drug. If the resulting diploid zygote is allowed to propagate (in the absence of the drug), how do you predict that the fraction of drug-resistant cells will change over time in the population? Choose the best curve (A to F) from the following graph. Note that the “population” is only composed of one cell at the beginning of the experiment. The curves are smoothened to cancel stochastic fluctuations that happen at such low population sizes.

100 -

Fraction of antibioticresistant cells (%)

B C D E F 00

Time (generations)

41. Indicate whether each of the following descriptions better applies to chloroplasts (C) or mitochondria (M). Your answer would be a five-letter string composed of letters C and M only, e.g. MMMMC. ( ) Extensive editing occurs on their RNAs. ( ) They utilize dynamin-like GTPases to divide from the outside.

( ) The organization of gene clusters in their genome is strikingly similar to cyanobacteria. ( ) They typically have larger genomes. ( ) They have a more ancient endosymbiotic relationship with their host. 42. Indicate whether each of the following descriptions better applies to the genome of mitochondria (M) or nuclei (N) in vertebrates. Your answer would be a four-letter string composed of letters M and N only, e.g. MNMN. ( ) It has a higher percentage of noncoding DNA. ( ) It has a stricter codon usage. ( ) Its genes are present at higher copy numbers per cell. ( ) Its evolutionary clock ticks much faster. 43. Indicate true (T) and false (F) statements below regarding the genetic systems of mitochondria and chloroplasts. Your answer would be a four-letter string composed of letters T and F only, e.g. TFTT. ( ) Most proteins in these organelles are encoded by the organelle’s genome. ( ) Mammalian mtDNA can make up less than 1% of the total cellular DNA. ( ) In some highly specialized animal cells, mtDNA can comprise as much as 99% of the cellular DNA. ( ) The genetic systems of these organelles are most similar to extremophilic archaea. 44.

Which of the following is NOT universally encoded by the mitochondrial DNA? A. Small ribosomal RNA B. Large ribosomal RNA C. A cytochrome oxidase subunit D. Transfer RNA

45.

Which of the following is NOT correct regarding the fusion and fission of mitochondria? A. They both require dynamin-related GTPases. B. Fission normally proceeds in a single step in which both outer and inner membranes are severed. C. They both proceed in a single step in which both outer and inner membranes are fused or severed. D. They both require GTP hydrolysis for force generation.

46.

In mammals, mitochondria are inherited … A. in a non-Mendelian fashion. B. maternally. C. cytoplasmically. D. uniparentally. E. All of the above

47. Consider two organisms, one with a much larger mitochondrial genome than the other. Which one is expected to show more deviations from the universal genetic code: the one with the larger mitochondrial genome (L) or the one with a smaller genome (S)? Write down L or S as your answer. 48.

Who has a different mitochondrial DNA in a family? A. The son B. The father C. The mother D. The daughter E. The maternal grandmother

49. Mitotic segregation of organelles can lead to so-called homoplasmy, when all organelles in each cell carry the same genome. In contrast, heteroplasmy describes the presence of organelles within the same cell that are different with respect to their genomic sequences. Would you expect a defect in the mitochondrial fusion machinery to favor homoplasmy (M) or heteroplasmy (T)? Write down M or T as your answer.

Answers 1. Answer: FFTT Difficulty: 1 Section: The Mitochondrion Feedback: Mitochondria form dynamic networks that are often associated with microtubules. They are found in specialized cells to localize to sites of greatest ATP consumption, such as the tail in sperm cells. 2. Answer: D Difficulty: 3 Section: The Mitochondrion Feedback: The low-density fractions would mostly contain outer-membrane components such as porins, while the high-density fractions correspond to the matrix and inner membrane. 3. Answer: MIIS Difficulty: 2 Section: The Mitochondrion Feedback: The electrochemical gradient of protons is established across the inner membrane, with higher potential on the outer side. However, there is no such gradient across the outer membrane, which is freely permeable to protons. The crista membrane houses the respiratory-chain proteins and other proteins at a very high density. The citric acid cycle proteins that generate NADH, on the other hand, are located in the matrix, which is approximately 50% protein by weight. 4. Answer: IIIO Difficulty: 2 Section: The Mitochondrion Feedback: In mitochondria that are active in respiration, O2 and ADP are constantly consumed to produce H2O and ATP, respectively. This creates concentration gradients that favor the movement of these molecules into the organelle. The opposite is true for CO2 which is generated inside the matrix in decarboxylation reactions of the citric acid cycle. Protons are actively pumped out of the matrix by the proteins of the electrontransport chain, and power ATP production when they move into the matrix down their electrochemical gradient. 5. Answer: B Difficulty: 2 Section: The Mitochondrion Feedback: Fatty acid biosynthesis is confined to the cytosol, where the citrate exported from mitochondria is used to produce acetyl CoA in the initial step. 6. Answer: 3 Difficulty: 2 Section: The Proton Pumps of the Electron-transport Chain

Feedback: Higher redox potential corresponds to an increased affinity for electrons. Consecutive electron carriers along the electron-transport chain have increasingly higher redox potentials. 7. Answer: FAECDBG Difficulty: 2 Section: The Proton Pumps of the Electron-transport Chain Feedback: This chain transports electrons from NADH to oxygen. 8. Answer: NORN Difficulty: 2 Section: The Proton Pumps of the Electron-transport Chain Feedback: NADH dehydrogenase complex is the largest respiratory enzyme complex embedded in the inner mitochondrial membrane and contains separate modules for electron transport (from NADH to ubiquinone) and proton pumping (across the membrane out of the matrix). The cytochrome c reductase complex reduces cytochrome c using electrons delivered by ubiquinol by the Q cycle mechanism. Finally, the cytochrome c oxidase complex accepts electrons from cytochrome c and reduces O2. 9. Answer: E Difficulty: 2 Section: ATP Production in Mitochondria Feedback: The equilibrium constant for the hydrolysis reaction (not considering water) is: Keq = [ADP]eq × [Pi]eq / [ATP]eq ΔG will have a greater negative value than ΔG° if and only if concentrations deviate from Keq such that [ADP] × [Pi] / [ATP] < Keq. 10. Answer: D Difficulty: 2 Section: ATP Production in Mitochondria Feedback: With 100% efficiency, about 60 moles of ATP would be produced by oxidizing one mole of glucose. Considering that only 30 moles of ATP are produced in the cell, the yield is about 50%. 11. Answer: RSSSR Difficulty: 2 Section: ATP Production in Mitochondria Feedback: The rotary part of the motor is composed of the membrane-embedded ring and its attached stalk. 12. Answer: 4 Difficulty: 2 Section: ATP Production in Mitochondria Feedback: A full revolution of the ring would transport 12 protons and would generate 3 molecules of ATP from ADP and Pi. This is equivalent to 4 protons per ATP molecule.

13. Answer: C Difficulty: 3 Section: ATP Production in Mitochondria Feedback: The number of pumps is calculated as: (50,000 g/day ÷ 500 g/mole ATP) ÷ (7000 revolution/min × 60 min/h × 24 h/day ÷ 3 ATP/revolution/pump) = (100 mole.ATP/day) ÷ (3,360,000 ATP/pump/day) = ~3 × 10–5 mole.pump 14. Answer: B Difficulty: 3 Section: ATP Production in Mitochondria Feedback: In the ATP-synthesis mode, protons move from the cristae space into the matrix (downward in the diagram), which rotates the c ring counterclockwise as seen from the F1 head on the matrix side of the crista membrane. Thus, in part B of the diagram shown, protons will enter the right half-channel, bind to a c subunit, rotate along with the c ring, and only be able to exit from the left half-channel after a complete turn of the ring. If the channels were arranged as in part A, the protons could pass through the membrane after only a tiny rotation of the ring, so that many more protons would need to pass through the ATP synthase for every ATP molecule that is synthesized by the F1 head. 15. Answer: E Difficulty: 3 Section: ATP Production in Mitochondria Feedback: Dimerization of ATP synthase has a key role in controlling the curvature of the inner membrane. 16. Answer: D Difficulty: 1 Section: ATP Production in Mitochondria Feedback: The ATP synthase dimers tend to localize at the tip of the cristae where the membrane curvature is greatest. 17. Answer: E Difficulty: 2 Section: ATP Production in Mitochondria Feedback: The ATP synthase dimers create proton “sinks” or “traps” at the cristae tips, which helps them make efficient use of the protons that are pumped out of the matrix. 18. Answer: D Difficulty: 2 Section: ATP Production in Mitochondria

Feedback: By diminishing the electrochemical gradient across the inner membrane, the toxin interferes with ATP synthesis and ATP/ADP exchange, both of which are driven by the gradient. 19. Answer: E Difficulty: 2 Section: ATP Production in Mitochondria Feedback: Various compounds are used in different bacteria as terminal electron acceptors. 20. Answer: S Difficulty: 2 Section: ATP Production in Mitochondria Feedback: Under aerobic conditions, the proton gradient established by the respiratory chain can drive ATP synthesis, whereas under anaerobic conditions, the ATP synthase pumps protons out (by ATP hydrolysis) to maintain the gradient. 21. Answer: ACBD Difficulty: 1 Section: Chloroplasts and Photosynthesis Feedback: The stroma and matrix are analogous, whereas the thylakoid space is an extra compartment not found in mitochondria. 22. Answer: DLLDL Difficulty: 2 Section: Chloroplasts and Photosynthesis Feedback: The light reactions (or photosynthetic electron-transfer reactions) occur in complexes embedded in the thylakoid membrane. The dark reactions (carbon-fixation reactions) begin in the stroma. 23. Answer: C Difficulty: 1 Section: Chloroplasts and Photosynthesis Feedback: Two water molecules provide four electrons and four protons and produce one molecule of oxygen in the electron-transfer reactions. 24. Answer: E Difficulty: 2 Section: Chloroplasts and Photosynthesis Feedback: Rubisco catalyzes the carboxylation of ribulose 1,5-bisphosphate (a fivecarbon sugar) as the first reaction in carbon fixation. 25. Answer: FTFTF Difficulty: 2 Section: Chloroplasts and Photosynthesis Feedback: Fat droplets and starch granules can be generated in chloroplasts from accumulated glyceraldehyde 3-phosphate. Sucrose is made mostly in the cytosol and is

transported as the main form of sugar to other plant tissues. Also in the cytosol, the glycolytic pathway produces metabolites that can be used to generate ATP in mitochondria, as similarly occurs in animal cells. 26. Answer: C Difficulty: 2 Section: Chloroplasts and Photosynthesis Feedback: These two complexes contain subunits that are homologous, and both pump protons using the Q cycle. 27. Answer: C Difficulty: 2 Section: Chloroplasts and Photosynthesis Feedback: The evolution of two photosystems in cyanobacteria allowed them to use water as an electron donor to reduce NADP+. 28. Answer: C Difficulty: 2 Section: Chloroplasts and Photosynthesis Feedback: Ionization of chlorophyll occurs primarily in the reaction center and results in the transfer of an electron from the pigment to an electron carrier. In the antenna complex, excitation normally leads to resonance energy transfer. 29. Answer: E Difficulty: 2 Section: Chloroplasts and Photosynthesis Feedback: Light-harvesting antenna complexes, found in both photosystems, contain chlorophyll molecules as well as orange carotenoid pigments. They help prevent the formation of harmful oxygen radicals. They collect the energy of a sufficient number of photons for photosynthesis, and pass the energy to the special pair in the reaction center, where the charge-separation step takes place. 30. Answer: EHBDFGCA Difficulty: 2 Section: Chloroplasts and Photosynthesis Feedback: Please refer to Figure 14–52. 31. Answer: 2212 Difficulty: 2 Section: Chloroplasts and Photosynthesis Feedback: Whereas photosystem I (with the special pair P700) is confined to unstacked stroma thylakoids, the stacked grana thylakoids contain photosystem II (with special pair P680). The excited P680 chlorophyll in photosystem II withdraws electrons from water held by a manganese cluster linked to the protein. The electrons are eventually passed on to plastoquinone. 32. Answer: D

Difficulty: 2 Section: Chloroplasts and Photosynthesis Feedback: This step is catalyzed by an enzyme that contains a manganese cluster and is located on the lumenal side of the membrane. It generates O2 as well as H+ in the lumen. 33. Answer: B Difficulty: 3 Section: Chloroplasts and Photosynthesis Feedback: P680 in photosystem II, when ionized (positive charge), can accept electrons from water, a molecule with a highly positive redox potential. In photosystem I, excited P700 can drive the reduction of NADP+, a molecule with a highly negative redox potential. 34. Answer: C Difficulty: 2 Section: Chloroplasts and Photosynthesis Feedback: The production of NADPH occurs on the stromal side of the thylakoid membrane and consumes two protons per molecule of NADPH generated. 35. Answer: MCMC Difficulty: 1 Section: Chloroplasts and Photosynthesis Feedback: Whereas a pH difference predominates in chloroplasts, the contribution of membrane potential to the proton-motive force is greater in mitochondria. Chloroplast ATP synthases do not form dimers. The mitochondrial ATP synthase complexes, in contrast, form long rows of dimers along the cristae ridges and are not distributed randomly in the membrane. 36. Answer: CABD Difficulty: 2 Section: Chloroplasts and Photosynthesis Feedback: Aerobic metabolism evolved after molecular oxygen became abundant due to the activity of ancient cyanobacteria. Fermentation is thought to have arisen before the harnessing of light energy to produce reducing power. 37. Answer: C Difficulty: 2 Section: Chloroplasts and Photosynthesis Feedback: Only after the supply of ferrous iron (Fe2+) was exhausted did the level of O2 in the atmosphere begin to rise steeply. 38. Answer: B Difficulty: 2 Section: Chloroplasts and Photosynthesis Feedback: The earliest forms of photosynthesis used molecules such as H2S. 39. Answer: FTTT Difficulty: 2

Section: Chloroplasts and Photosynthesis Feedback: Aerobic proteobacteria evolved from ancestral purple nonsulfur bacteria only after the atmosphere of the Earth became oxidizing. Chloroplasts are thought to have arisen from cyanobacteria. 40. Answer: B Difficulty: 2 Section: The Genetic Systems of Mitochondria and Chloroplasts Feedback: The zygote is resistant to the drug. However, over generations, mitotic segregation of mitochondrial DNA will randomly give rise to cells that only contain the wild-type, non-resistant organelles, and this should occur in about half of the population. 41. Answer: CMCCM Difficulty: 2 Section: The Genetic Systems of Mitochondria and Chloroplasts Feedback: The chloroplast was derived later than the mitochondrion in the evolutionary history of the eukaryotic cell. Chloroplasts are generally larger than mitochondria and contain larger genomes. They share many striking similarities with bacteria, including similar genome organization and similar division mechanisms. Despite this, RNA editing and RNA processing are prevalent in chloroplasts and are catalyzed by proteins not found in bacteria. 42. Answer: NNMM Difficulty: 2 Section: The Genetic Systems of Mitochondria and Chloroplasts Feedback: Mitochondria have a dense packing of genes in their genome, a relaxed codon usage, and occasional variant genetic codes. Mutation rate is much higher for mitochondrial DNA, making it useful for phylogenetic studies on relatively recent evolutionary changes. 43. Answer: FTTF Difficulty: 2 Section: The Genetic Systems of Mitochondria and Chloroplasts Feedback: The genetic systems of mitochondria and chloroplasts resemble those of prokaryotes. However, intracellular gene transfer to the nucleus over an evolutionary time scale has shrunk their genomes to a level such that most of their proteins are encoded by the nucleus in modern organisms. Depending on the number of mitochondria and their chromosome copy number, mitochondrial DNA (mtDNA) can account for less than 1% to about 99% of total DNA weight in a cell. 44. Answer: D Difficulty: 2 Section: The Genetic Systems of Mitochondria and Chloroplasts Feedback: Common to all mitochondrial genomes that have been sequenced so far are genes encoding ribosomal RNAs, the largest subunits of the cytochrome oxidase

complex, and cytochrome b. tRNA genes are common to most, but are not in the most minimal organelle genomes. 45. Answer: C Difficulty: 1 Section: The Genetic Systems of Mitochondria and Chloroplasts Feedback: Whereas fission proceeds by severing the two organelle membranes in one step, fusion is carried out by two separate machineries, one on each membrane, that function sequentially. 46. Answer: E Difficulty: 1 Section: The Genetic Systems of Mitochondria and Chloroplasts Feedback: A mammalian zygote’s mitochondria almost exclusively come from the egg. 47. Answer: S Difficulty: 2 Section: The Genetic Systems of Mitochondria and Chloroplasts Feedback: Apparently, the unusually small number of genes in mitochondrial genomes allows random drift to result in deviations of the genetic code. 48. Answer: B Difficulty: 1 Section: The Genetic Systems of Mitochondria and Chloroplasts Feedback: Mitochondria are inherited maternally, and a genetic disease caused by mutations in the mitochondrial genome in a founder female would be inherited by all descendants. 49. Answer: M Difficulty: 2 Section: The Genetic Systems of Mitochondria and Chloroplasts Feedback: Mitochondrial fusion works to ensure an even distribution of mitochondrial DNA throughout the cell. Fusion can reduce the effect of mitotic segregation and favor heteroplasmy. Thus, a defect in fusion should favor homoplasmy

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION CHAPTER 15: CELL SIGNALING Copyright © 2015 by W.W. Norton & Company, Inc. 1. A cell expresses a transmembrane protein that is cleaved at the plasma membrane to release an extracellular fragment. The fragment binds to receptor proteins on nearby cells and activates signaling pathways resulting in altered gene expression patterns in the cells. What form of intercellular signaling does this represent? A. Contact-dependent signaling B. Paracrine signaling C. Synaptic signaling D. Endocrine signaling E. Autocrine signaling 2.

Which of the following is NOT a common second messenger in cell signaling? A. Ca2+ B. Cyclic adenosine monophosphate C. Diacylglycerol D. Tyrosine E. Inositol trisphosphate

Which of the following events normally activates a GTP-binding protein? A. GTP hydrolysis by the protein B. Activation of an upstream GTPase-activating protein C. Activation of an upstream guanine nucleotide exchange factor D. Phosphorylation of a bound GDP molecule by an upstream phosphorylase E. Pi release after GTP hydrolysis

4. In which of the following schematic drawings of signaling pathways does the activation of the receptor lead to gene expression? Activating and inhibitory steps are indicated with (+) and (–), respectively. (+)

Activated receptor

(–)

Signaling proteins

(+)

Transcription activator

Gene Expression

(–)

(+)

(–)

(+)

(–)

(+)

Gene Expression

(+)

(–)

(+)

(–)

(+)

(–)

(+)

Gene Expression

(–)

Reference: Examples of Cellular Response Curves Questions 5-8 In the following graphs, the cellular response, as measured in real time by the concentration of a certain active effector protein, is plotted over time for five cell types (A to E) that are treated with three different concentrations of a signal molecule. During the time period indicated by the horizontal gray bar, the signal molecule is present in the culture media at a concentration of 1 nM (dotted curve), 5 nM (gray curve), or 25 nM (solid curve). Answer the following question(s) based on these graphs.

100 -

Cellular response (% activation of effector protein)

100 -

7 Time (hour)

Which of the cell types A to E shows the lowest sensitivity to the signal?

Which of the cell types A to E shows a response with the highest persistence?

Which of the cell types A to E shows the fastest signal adaptation?

Which of the cell types A to E shows the widest dynamic range of signal concentration?

phosphorylated

Percentage of molecules

9. Consider a signaling protein that can be phosphorylated by a dual-specificity protein kinase at two independent sites (one tyrosine and one serine residue), such that each site is phosphorylated in about 1% of the protein molecules under normal conditions. Phosphorylation at the first site has a short half-life, since a fast tyrosine phosphatase removes the phosphate soon after the residue is phosphorylated, whereas dephosphorylation at the second site—carried out by a serine/threonine phosphatase—is relatively slow. Upon activation of a signaling pathway, the concentration of the dual-specificity kinase increases several-fold rapidly. The phosphorylation state of the tyrosine and serine residues following stimulation is compared in the following graph. Which curve (1 or 2) do you think corresponds to the serine residue? Write down 1 or 2 as your answer. 100

0 0

Time after stimulation (min)

10. The Src kinase can be regulated by at least two mechanisms. Kinases of the CSK family inactivate Src by phosphorylating a C-terminal tyrosine residue. On the other hand, binding of activating ligands (including some activated receptor tyrosine kinases, or RTKs) results in autophosphorylation at a tyrosine residue near the active site, stimulating Src activity. Full activation of Src therefore requires both dephosphorylation at the inhibitory tyrosine and binding to an RTK. Which of the following logic gates (A to E) correctly models the activation of Src kinase (output) as a function of the activity of the upstream regulators (inputs)? Inputs CSK activity OFF OFF

Output of logic gates RTK activity OFF ON

→ →

ON OFF

OFF ON

OFF OFF

ON ON

OFF ON

→ →

OFF OFF

OFF ON

OFF OFF

ON OFF

OFF ON

11. Consider the signaling pathway depicted below, which involves two different receptors. Assume that a protein is active only when it is activated but not inhibited by its upstream signaling molecules. Activation and inhibition are indicated by normal and blunt arrows, respectively. Under which of the following conditions is target-gene expression induced?

Signal 1

Signal 2

GPCR

RTK

G protein

Enzyme

Kinase

Adaptor

Second messenger Kinase

Binding protein Transcription regulator Gene expression and cellular response

A. Only in the presence of both signals (1 and 2) B. In the presence of either or both signals (1 or 2) C. In the presence of either signal but not both of them D. Only in the absence of both signals E. Only in the presence of signal 2 but not signal 1 12. Some signaling pathways in the cell are chiefly based on inhibition rather than activation. For example, proteins that initiate programmed cell death by apoptosis are inhibited by antiapoptotic proteins, which can in turn be inhibited by a group of pro-apoptotic proteins, and so on.

Consider the following networks of inhibitory interactions, in which each protein is inactive if at least one of its inhibitors is active. In which pathway is the activation of either upstream protein (1 or 2) sufficient to inhibit apoptosis?

Apoptosis

1 Apoptosis

B. 2

1 Apoptosis

1 D.

Apoptosis

13. In the following simplified diagram, the activation of a hypothetical allosteric protein is plotted as a function of effector molecule concentration. Which curve (A to E) do you think corresponds to a condition in which activation requires the highest number of effector molecules bound to each protein?

% of maximum activation of target allosteric protein

A B

D E

Relative concentration of effector molecules

14. Which of the following molecular mechanisms is NOT used in the cell in order to produce abrupt responses to a signal? A. Activation of a protein only when bound to multiple molecules of an activating ligand B. Activation of a protein only when it is phosphorylated at multiple sites C. Activation of a protein by simultaneously activating its activator(s) and inactivating its inhibitors D. A positive feedback loop in which an activated protein positively regulates its activator E. A negative feedback loop in which an activated protein inhibits its activator without delay 15. A certain effector protein can be activated by phosphorylation at a key tyrosine residue. An upstream kinase rapidly phosphorylates this tyrosine in the presence of a signal. However, the kinase also phosphorylates and activates a slow-acting phosphatase that can dephosphorylate the phosphotyrosine. Which curve in the following graph would you expect to represent the activity of the effector molecule over time? The input signal is present during the period indicated in gray. The dashed line represents the response in the absence of the phosphatase.

Relative response (log scale)

100 A B C 1

D E

0.01 Time

16. You grow a tumor-derived cell line in a suspension culture, in which the cells are seen either as single cells or in cell clusters. The cells have been engineered to express green fluorescent protein (GFP) under the control of the mitogen-activated protein kinase (MAP kinase) pathway. To your culture media, you add different concentrations of fibroblast growth factor (FGF)—which is known to activate the MAP kinase pathway in these cells—and briefly incubate the cells before harvesting them. You also collect, under a microscope, hundreds of individual cell clusters from each suspension. You then disaggregate the cells in all of your samples and use a FACS (fluorescence-activated cell sorting) machine to measure the GFP signal intensity in individual cells. Finally, you plot the results as histograms, as shown below. No FGF added FGF added at intermediate concentration

Percentage of cells

Percentage of clusters

FGF added at high concentration

Average GFP intensity per cluster

GFP signal intensity per cell

According to these results, which of the following schematic drawings better represents the expression of GFP (a proxy for MAP kinase activation) in the culture after intermediate FGF stimulation? In these drawings, higher GFP expression is represented by a darker shade.

17. Cells can adapt to extracellular signal molecules in a variety of ways. Interestingly, exposure to a signal molecule can sometimes not only desensitize the cell to this molecule, but also to one or more other signal molecules that are recognized by different receptors, even if those receptors can still bind to their ligand at the cell surface and initiate signaling. Which subset of the following adaptation mechanisms (for the original signal) can in principle account for this type of desensitization? Choose all that apply. Your answer would be a string composed of letters A to E only, in alphabetical order; e.g. BDE. (A) Receptor sequestration (B) Receptor down-regulation (C) Receptor inactivation (D) Inactivity of signaling protein (E) Production of inhibitory protein 18.

For the α subunit of a trimeric G protein, … A. a G-protein-coupled receptor (GPCR) acts as a guanine nucleotide exchange factor (GEF), whereas a regulator of G protein signaling (RGS) can act as a GTPaseactivating protein (GAP).

B. a GPCR acts as a GAP, whereas an RGS can act as a GEF. C. both a GPCR and an RGS can act as a GEF. D. both a GPCR and an RGS can act as a GAP. 19. Indicate true (T) and false (F) statements below regarding G-protein-coupled receptors (GPCRs). Your answer would be a five-letter string composed of letters T and F only, e.g. TTFFF. ( ) All GPCRs share a similar structure composed of seven transmembrane helices. ( ) All GPCR ligands (signal molecules) have a similar structure. ( ) GPCRs have only been found in multicellular organisms, consistent with their role in intercellular signaling. ( ) The hormone insulin is recognized by a GPCR on the surface of its target cells. ( ) Once activated, a GPCR molecule can activate multiple molecules of G protein to amplify the incoming signal. 20. Whereas the cholera toxin ADP-ribosylates the α subunit of stimulatory G protein (Gs), thereby blocking GTP hydrolysis, pertussis toxin ADP-ribosylates the α subunit of inhibitory G protein (Gi) and prevents interaction with the receptor. What is the effect of these toxins on the concentration of intracellular cAMP? A. Cholera toxin tends to increase cAMP concentration, whereas pertussis toxin tends to decrease cAMP concentration. B. Cholera toxin tends to decrease cAMP concentration, whereas pertussis toxin tends to increase cAMP concentration. C. They both tend to increase cAMP concentration. D. They both tend to decrease cAMP concentration. 21. Sort the following events to reflect the normal order in which they occur in G-proteincoupled receptor signaling leading to transcription of genes with cAMP response elements. Your answer would be a four-letter string composed of letters A to E only, e.g. ACBD. (A) Binding of CREB to PKA (B) Binding of cAMP to PKA (C) Dissociation of PKA into catalytic and regulatory subunits (D) Activation of adenylyl cyclase 22. Which of the following is a major consequence of activation of phospholipase C-β (PLCβ) by the Gq trimeric GTPase?

A. Elevation of intracellular cAMP levels, leading to the activation of protein kinase A B. Elevation of PIP3 levels in the plasma membrane, leading to the activation of protein kinase B C. Elevation of intracellular Ca2+ levels, leading to the activation of protein kinase C D. Elevation of IP3 in the plasma membrane, leading to the activation of protein kinase D E. Elevation of intracellular cGMP levels, leading to the activation of protein kinase G 23. Intracellular Ca2+ oscillations, initiated by the activation of IP3 receptors and orchestrated via intricate positive and negative feedback loops, can be subject to frequency modulation, an example of which is the increase in oscillation frequency with increasing stimulus strength. Caffeine is known to bind to and sensitize the ryanodine receptors, resulting in their opening in response to lower calcium ion concentrations. Do you think caffeine would tend to increase (I) or decrease (D) the frequency of calcium ion oscillations in stimulated cells? Write down I or D as your answer. 24.

Fill in the blank in the following paragraph. “Cytosolic Ca2+ can be sensed by ..., a small, conserved, dumbbellshaped protein with four Ca2+-binding sites. Upon Ca2+ binding, it binds and activates dozens of proteins by changing its conformation in different ways.”

25. Indicate whether each of the following descriptions applies to visual signal transduction (V), olfactory signal transduction (O), or both (B). Your answer would be a four-letter string composed of letters V, O, and B only, e.g. VOVV. ( ) It involves cGMP as a second messenger . ( ) A rise in cyclic mononucleotide concentration follows the arrival of the signal. ( ) The signal is received by a G-protein-coupled receptor. ( ) An incoming signal leads to membrane hyperpolarization. 26. Sort the following events into the order that they take place in phototransduction in response to a flash of light. Your answer would be a four-letter string composed of letters A to E, e.g. BCDEA. (A) The cGMP level drops. (B) The cGMP level rises.

(C) The Ca2+ level drops. (D) Retinal isomerizes to an all-trans configuration. (E) Gt exchanges its bound nucleotide. 27. Which of the following is NOT part of a negative feedback mechanism in adaptation to light in retinal rod cells? A. Phosphorylation of G-protein-coupled receptor B. Binding of arrestin to G protein C. Stimulation of guanylyl cyclase by decreased Ca2+ levels D. Binding of regulator of G protein signaling to transducin 28. Indicate true (T) and false (F) statements below regarding cellular signaling mediated by nitric oxide (NO). Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) Once produced, NO can diffuse to neighboring cells. ( ) NO has known roles in cGMP-independent as well as cGMP-dependent signaling. ( ) NO normally decreases cGMP concentration by activating cGMP phosphodiesterase. ( ) The drug Viagra® counteracts the effects of NO on penile blood vessels. 29. Consider visual transduction in rod photoreceptors in the vertebrate retina. Which of the following steps does NOT normally amplify the signal in this pathway? A. Activation of transducin by active rhodopsin B. Blockage of Na+ influx by cation-channel closure C. Cation-channel closure due to cGMP depletion D. Depletion of cGMP by active cGMP phosphodiesterase E. All of the above DO involve amplification 30.

A genetic form of “night blindness” (i.e. poor vision in dim light) is caused by mutations

in genes encoding rhodopsin kinase (RK) and arrestin. The mutations reduce the dynamic range of light perception, and the affected individuals have a prolonged light response and adapt very slowly to low light (when entering a dark theater room, for example). Which of the following mutations do you expect to be the cause? A. Loss-of-function RK mutations and loss-of-function arrestin mutations B. Loss-of-function RK mutations and gain-of-function arrestin mutations C. Gain-of-function RK mutations and loss-of-function arrestin mutations D. Gain-of-function RK mutations and gain-of-function arrestin mutations

31. How is the activation of epidermal growth factor receptor (EGFR) different from that of insulin receptor (IR)? A. EGFR is a dimer in its inactive and active forms, whereas IR dimerizes only when active. B. EGFR activation requires dimerization, whereas IR can become activated as a monomer. C. EGFR is activated by transautophosphorylation, whereas IR activation is phosphorylation-independent and is a simple consequence of receptor dimerization. D. Unlike IR activation, EGFR activation involves an asymmetric arrangement of the intracellular kinase domains. E. Both C and D above. 32. Indicate true (T) and false (F) statements below regarding the ubiquitin ligase c-Cbl. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) It contains an SH2 domain, and can therefore activate RTK signaling. ( ) It ubiquitylates activated RTKs. ( ) Its overactivation can lead to prolonged RTK signaling and promote the development of cancer. ( ) It is part of a negative feedback loop to down-regulate RTKs. 33. Consider a signaling protein that is only made up of one SH2 domain and two SH3 domains. This protein is most likely ... A. a monomeric G protein. B. a guanine nucleotide exchange factor. C. a kinase associated with receptor tyrosine kinase signaling. D. an adaptor protein. E. a negative regulator of receptor tyrosine kinase signaling. 34. What is the major way by which the monomeric G protein Ras is activated in receptor tyrosine kinase signaling? A. Activation of Ras-GAP B. Activation of Ras-GEF C. Inactivation of Ras-GAP D. Inactivation of Ras-GEF E. Inactivation of Ras-GDI

35. Indicate true (T) and false (F) statements below regarding the Ras–MAP-kinase signaling pathway. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) The immediate early genes turn on a few hours after activation of the RTK. ( ) Ras-GTP binds directly to, and activates, the upstream kinase in the MAP kinase module. ( ) Signal transduction through the Ras–MAP-kinase pathway only leads to a transient response. ( ) In the MAP kinase module, Raf can be phosphorylated by Erk to create a negative feedback loop. 36. Genetic screens for enhancers and suppressors of temperature-sensitive loss-of-function mutations in sevenless (sev) in Drosophila melanogaster led to the discovery of some other genes involved in the Ras–MAP-kinase pathway. Would you expect loss-of-function mutations in genes encoding each of the following proteins to be an enhancer (E) or suppressor (S) of sev? An enhancer mutation exacerbates the effect of the original mutation, whereas a suppressor mutation alleviates it. Note that loss-of-function mutations can be just partial defects and not necessarily null mutations. Your answer would be a four-letter string composed of letters E and S only, e.g. EEEE. ( ) Sos ( ) Ras ( ) Ras-GAP ( ) Raf 37. What is the effect of using scaffold proteins on precision and amplification capacity in cell signaling? A. Both precision and amplification are improved. B. Precision is improved, but amplification is limited. C. Precision is compromised, but amplification is improved. D. Both precision and amplification are limited. 38. In the following schematic diagram showing the initial signaling steps in growth cone collapse in motor neurons, indicate which protein (A to E) corresponds to the following. Your answer would be a four-letter string composed of letters A to E only, e.g. CBAE.

( ) Ephrin ( ) Ephexin ( ) RhoA GTPase ( ) Cytoplasmic tyrosine kinase 39. Which of the following phosphoinositides is the preferred binding target for a PHdomain-containing protein?

40.

PI 3-kinase ... A. is a membrane-associated tyrosine kinase. B. activates PDK1 by phosphorylating a serine residue on the protein. C. is counteracted by PTEN phosphatase. D. is only known to be activated by receptor tyrosine kinases. E. inhibits Akt.

41.

Which of the following proteins lacks a PH domain? A. Sos (a Ras-GEF) B. Grb2 (an RTK signaling adaptor) C. Pleckstrin (a protein kinase C substrate) D. Akt (protein kinase B) E. PDK1 (phosphoinositide-dependent kinase)

42. Indicate true (T) and false (F) statements below regarding the mTOR complexes. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) mTOR in complex 1 contains the protein raptor, is sensitive to rapamycin, and stimulates cell growth. ( ) Akt activation is stimulated by mTOR in complex 2, which contains the protein rictor. ( ) Akt activates mTOR in complex 2 by activating a Rheb-GAP called Tsc2. ( ) mTOR in complex 1 is activated in the presence of growth factors. 43. In the following diagram showing five parallel signaling pathways downstream of a receptor tyrosine kinase (RTK) and a G-protein-coupled receptor (GPCR), indicate which boxes (A to E) correspond to the following proteins. Your answer would be a five-letter string composed of letters A to E only, e.g. CDBAE.

D E B A

( ) Cyclic AMP ( ) Protein kinase C ( ) Sos ( ) PI(3,4,5)P3 ( ) Ca2+ 44.

Proteins of the Src family (e.g. Src, Fyn, Lck, and Hck) have important roles in several

signaling pathways in mammals. A typical Src family protein ... A. is a cytosolic tyrosine kinase. B. contains SH2 and SH3 domains in addition to a kinase domain. C. has covalently attached lipid chains. D. is located on the cytoplasmic side of the plasma membrane. E. All of the above.

45. The SH2 domain of STAT proteins plays a key role in cytokine receptor signaling. This domain can bind to phosphotyrosine ... A. on an activated receptor molecule, or (intramolecularly) on the same STAT molecule. B. on a Janus kinase molecule, or (intramolecularly) on the same STAT molecule. C. on an activated receptor molecule, or on another STAT molecule in a dimer. D. on a Janus kinase molecule, or on another STAT molecule in a dimer. E. on the same STAT molecule or on another STAT molecule in a dimer. 46. Which of the following mechanisms is NOT employed by I-Smads to negatively regulate Smad signaling? A. Competing with R-Smads for binding sites on the receptor, decreasing R-Smad phosphorylation. B. Recruiting the ubiquitin ligase Smurf, leading to receptor degradation. C. Binding to caveolin, inhibiting the formation of caveolae. D. Binding to the co-Smad, Smad4, inhibiting it. E. Recruiting protein phosphatases, leading to receptor inactivation. 47. What two cell-surface receptors are represented in the two simplified diagrams below (from left to right)?

Cytosolic tyrosine kinase

Cytosolic serine/threonine kinase domain

Phosphorylated transcription regulators

A. TGFβ receptor and TNF receptor.

B. Cytokine receptor and TNF receptor. C. TNF receptor and TGFβ receptor. D. TGFβ receptor and cytokine receptor. E. Cytokine receptor and TGFβ receptor. 48. How does the expression of Delta on the surface of a cell activate the expression of certain genes in the nucleus of its neighboring cell? A. Delta binding activates Notch, which activates a transcriptional activator through the JAK–STAT pathway. B. Delta binding leads to the stabilization of a cytoskeleton-associated transcriptional activator. C. Delta binding releases the intracellular tail of Notch, which enters the nucleus and converts a transcriptional repressor into a transcriptional activator. D. Delta binding leads to Notch-mediated recruitment of protein complexes to the plasma membrane, resulting in the degradation of a transcriptional repressor. E. Delta binding leads to the proteolytic cleavage of Notch and inhibition of its activity as a transcriptional repressor, leading to the activation of target genes. 49.

Which of the following represents the active form of β-catenin in cells stimulated with

Wnt? A. Phosphorylated by GSK3 and CK1 B. Bound to APC C. Bound to Groucho D. Bound to LRP E. None of the above 50. Cancer-related genes can be classified into two major groups. Proto-oncogenes encode proteins that normally act in promoting cell growth and proliferation. Gain-of-function mutations in proto-oncogenes can therefore transform the cells into a cancerous state. Tumor suppressor genes, on the other hand, encode proteins that normally function to keep cell proliferation in check. Thus, loss-of-function mutations in tumor suppressors can lead to cancer by eliminating the inhibitory effect of these genes. Mutations in genes encoding the components of the canonical Wnt signaling pathway are commonly found in colorectal tumors. Indicate whether the gene for each of the following is a proto-oncogene (P) or tumor suppressor (T). Your answer would be a four-letter string composed of letters P and T only, e.g. PPPP. ( ) Wnt

( ) APC ( ) Axin ( ) β-Catenin 51. Regulation of the Ci (Cubitus interruptus) protein in the Hedgehog signaling pathway is reminiscent of that of β-catenin in the Wnt pathway. Which of the following features is shared among these two proteins? A. Both proteins are ubiquitylated when the signal (Hedgehog or Wnt) is present. B. Both proteins are able to enter the nucleus only in the absence of the signal. C. Both proteins can be phosphorylated by GSK3 and CK1 in the absence of the signal. D. Both proteins are fully degraded by the proteasome in the absence of the signal. E. None of the above. 52. Indicate whether each of the following occurs inside the cell in the presence (P) or absence (A) of stimulation with a Hedgehog signal. Your answer would be a four-letter string composed of letters P and A only, e.g. PPPP. ( ) Sequestration of Smoothened in intracellular vesicles ( ) Accumulation of unprocessed Ci in the nucleus ( ) Phosphorylation of Smoothened ( ) Recruitment of Fused and Costal2 to the plasma membrane 53. In the following simplified diagram of the NFκB-dependent signaling pathway, indicate which boxes (A to C) correspond to the NF-κB, IκB, and IKK proteins, respectively. Your answer would be a three-letter string composed of letters A to C only, e.g. BCA. IL1 receptor activation

A B C Expression of inflammatory response genes

54. Activated STATs can induce the expression of SOCS proteins, which in turn inhibit the JAK-dependent phosphorylation of the STATs. Under sustained serum stimulation of mouse fibroblast cells, a number of oscillations can be observed as a consequence of this delayed negative feedback loop, as shown in the following simplified graph. According to this graph, approximately how long is the delay in the feedback response, and what process is mainly responsible for it? Phosphorylated STAT protein SOCS nuclear mRNA SOCS cytosolic mRNA SOCS protein

Relative amount

100

0 0

4 Serum added

Time (h)

A. 1 hour; transcription B. 2 hours; transcription C. 1 hour; mRNA export D. 1 hour; mRNA translation E. 2 hours; mRNA translation 55.

All nuclear receptors ... A. are cytosolic proteins that enter the nucleus upon ligand binding. B. have ligand-binding and DNA-binding domains, and can directly bind to DNA. C. are transcriptional activators when bound to their ligand. D. bind to steroid hormones. E. are transcriptional repressors in the absence of their ligand.

Phosphorylated (%)

100

Phosphorylated (%)

56. The phosphorylation state of a human protein follows the oscillatory pattern shown in the first graph below. The protein is phosphorylated by two kinases (1 and 2), both of which are activated in an oscillatory pattern. Kinase 1 oscillates in congruence with the circadian rhythm, while kinase 2 oscillates at a higher frequency and is independent of the circadian clock. The second graph shows the phosphorylation state of the same protein in mutant cells lacking one of the kinases mentioned above. Which kinase (1 or 2) do you think is missing in these mutant cells? Write down 1 or 2 as your answer.

100

Wild type

0 0

48 Time (hours)

Mutant

0 0

57. You have studied the circadian clock in a certain mouse strain by monitoring the physical activity (e.g. wheel-running) of a few mice over an extended period of time. For the first week, you kept the mice under normal 12 hr light /12 hr dark cycles; they were then transferred to constant darkness. You have plotted the average activity of the mice in the following graph. The light and dark conditions are represented with white and gray backgrounds, respectively. The activity in each day is plotted in black in one row of the graph, and consecutive days are arranged from the top to the bottom. The horizontal axis represents the time of the day. Based on these results, what is the most accurate estimate for the free-running period of the circadian clock in this mouse strain?

Day 1

Activity

Day 7

Day 20 0

Time (hours) A. About 25 hr B. About 24.5 hr C. About 24 hr D. About 23.5 hr E. About 23 hr 58. Indicate true (T) and false (F) statements below regarding signaling in plants. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) Since multicellularity evolved before the divergence of the plant and animal lineages, plants have homologs of most animal signaling proteins, such as nuclear receptors, Ras, JAK, and Notch. ( ) Receptor tyrosine kinases are the largest class of cell-surface receptors in plant cells. ( ) Both plants and animals use cyclic GMP, NO, and Ca2+ for cell signaling. ( ) Leucine-rich repeat (LLR) receptor kinases are the most abundant type of receptor serine/threonine kinases in plants.

59. Fill in the blank in the following paragraph regarding plant hormones. DO NOT use abbreviations. “As simple a signaling molecule as it is, ... is also a key regulator of plant physiology. It can promote processes such as fruit ripening, flower opening, and leaf abscission; it can also be produced in response to various types of stress including drought, flooding, and bacterial and viral infections.” 60.

Ethylene receptors in plant cells ... A. are located primarily at the plasma membrane. B. are single-pass transmembrane proteins. C. interact with CTR1, which is closely related to the monomeric G protein Ras. D. have a copper-containing ethylene-binding domain. E. All of the above.

61. In the following schematic diagrams of ethylene signaling in plants, 1 denotes ethylene, 2 denotes ethylene receptor, 3 denotes CTR1, 4 denotes EIN3, and 5 represents the expression of ethylene-responsive genes. Which diagram better depicts the signaling pathway? Activation and inhibition are indicated as normal and blunt arrows, respectively.

62. Indicate whether each of the following descriptions better applies to phytochromes (P), cryptochromes (C), or phototropins (T). Your answer would be a four-letter string composed of letters P, C, and T only, e.g. TPCC. ( ) They respond to red light. ( ) They are found in animals as well as in plants. ( ) They are flavoproteins. ( ) They are dimeric, cytosolic serine/threonine protein kinases that are normally activated by autophosphorylation. 63.

Arabidopsis thaliana seedlings show positive phototropism: they bend and grow toward

the source of (blue) light. However, phototropism is not observed if the seedlings are treated with an auxin efflux inhibitor called NPA. Knowing that auxin stimulates elongation of the cells in the growing shoot, these observations are consistent with a model in which, in the presence of a lateral light, ... A. auxin efflux transporter proteins become localized to the side of the cell that is facing the light source. B. auxin efflux transporter proteins become localized to the side of the cell that is away from the light source. C. auxin influx transporter proteins become localized to the side of the cell that is facing the light source. D. auxin influx transporter proteins become localized to the side of the cell that is away from the light source. E. auxin efflux transporter proteins become uniformly distributed around the cell.

Answers 1. Answer: B Difficulty: 1 Section: Principles of Cell Signaling Feedback: Paracrine signaling acts on neighboring cells via local mediators. 2. Answer: D Difficulty: 1 Section: Principles of Cell Signaling Feedback: Second messengers are small chemicals generated in large amounts in response to receptor activation in signaling pathways and diffuse away from their source to spread the signal. Some are water-soluble [e.g. calcium ions and cyclic adenosine monophosphate (cAMP)] and some are lipid-soluble [e.g. diacylglycerol (DAG) and inositol trisphosphate (IP3)]. 3. Answer: C Difficulty: 1 Section: Principles of Cell Signaling Feedback: Activation of an upstream guanine nucleotide exchange factor (GEF) activates a GTPase by facilitating its conversion into a GTP-bound form through guanine nucleotide exchange. 4. Answer: A Difficulty: 2 Section: Principles of Cell Signaling Feedback: In principle, an even number of inhibitory steps in a sequence is equivalent to activation. 5. Answer: E Difficulty: 3 Section: Principles of Cell Signaling Feedback: This cell type responds only to the highest signal concentration tested. 6. Answer: A Difficulty: 3 Section: Principles of Cell Signaling Feedback: This cell type seems to show a permanent response to the signal. 7. Answer: C Difficulty: 3 Section: Principles of Cell Signaling

Feedback: This cell type reduces its response significantly even in the presence of the signal. 8. Answer: C Difficulty: 3 Section: Principles of Cell Signaling Feedback: The wide dynamic range allows this cell type to differentially respond to the signal over a broad concentration range. 9. Answer: 2 Difficulty: 2 Section: Principles of Cell Signaling Feedback: In this example, the slower dephosphorylation (longer half-life) results in a slower change in serine phosphorylation, compared to tyrosine phosphorylation, upon stimulation of the signaling pathway. 10. Answer: C Difficulty: 3 Section: Principles of Cell Signaling Feedback: The activation of Src can be described as “(RTK) AND NOT (CSK).” 11. Answer: C Difficulty: 3 Section: Principles of Cell Signaling Feedback: In this imaginary example, each of the two parallel pathways depicted can stimulate gene expression through the same transcription regulator; however, when both signals are present, each pathway inhibits the other one, leading to no response. 12. Answer: B Difficulty: 3 Section: Principles of Cell Signaling Feedback: Pathway A is equivalent to an OR gate; either protein (1 or 2) can trigger apoptosis. Pathway B is equivalent to a NOR gate; either 1 or 2 can inhibit apoptosis. Pathway C is equivalent to a gate that ignores its second input (protein 2). Pathway D is equivalent to an AND gate; both proteins 1 and 2 in their active form are required to trigger apoptosis. 13. Answer: C Difficulty: 2 Section: Principles of Cell Signaling

Feedback: The sharpness of the sigmoidal activation response increases with an increase in the number of effector molecules that must be bound simultaneously for complete activation of the target protein. 14. Answer: E Difficulty: 3 Section: Principles of Cell Signaling Feedback: Negative feedback generally dampens the response, while positive feedback loops and cooperative activation sharpen it. 15. Answer: B Difficulty: 3 Section: Principles of Cell Signaling Feedback: The upstream kinase both activates its target (directly) and slowly inactivates it (indirectly). This results in an attenuated response under sustained stimulation. 16. Answer: C Difficulty: 3 Section: Principles of Cell Signaling Feedback: The histograms from the FACS data show a gradual increase in the average GFP signal per cluster (left plot) and an all-or-none activation of the pathway in individual cells (right plot) with increasing concentrations of the signal, consistent with the model shown in C. 17. Answer: DE Difficulty: 2 Section: Principles of Cell Signaling Feedback: If the signaling pathways for the signals converge (or overlap), adaptation to one signal can also desensitize the cell to the others by affecting the common signaling molecules. In this example, convergence occurs downstream of the receptors; therefore, the first three mechanisms do not apply. Note that the first three adaptation mechanisms act on the receptor for the original signal, not on other receptors. 18. Answer: A Difficulty: 2 Section: Signaling through G-Protein-Coupled Receptors Feedback: The GPCR and the RGS act in opposite directions with regards to the activation of trimeric G proteins, by acting as a GEF and a GAP, respectively. 19. Answer: TFFFT Difficulty: 1 Section: Signaling through G-Protein-Coupled Receptors

Feedback: All GPCRs (found in multicellular as well as unicellular organisms) belong to a large family of proteins that share a similar structure with seven transmembrane helices. Despite this similarity, they recognize a wide variety of ligands. The GPCR activates the α subunit of a trimeric G protein, which is then released; this is usually followed by the binding of another α subunit to repeat the cycle for as long as the receptor is activated. Insulin is recognized by a receptor tyrosine kinase. 20. Answer: C Difficulty: 2 Section: Signaling through G-Protein-Coupled Receptors Feedback: While cholera toxin causes constitutive Gs activation, pertussis toxin prevents Gi activation. Both of these result in a net activation of adenylyl cyclase, leading to higher cAMP levels. Note that adenylyl cyclase is not the only target of Gi. 21. Answer: DBCA Difficulty: 1 Section: Signaling through G-Protein-Coupled Receptors Feedback: Upon cAMP production by activated adenylyl cyclase, cAMP-dependent protein kinase (PKA) is activated through the release of its catalytic subunits as a result of cAMP binding to the regulatory subunits. The catalytic subunits then enter the nucleus where they phosphorylate CREB (cAMP response element-binding protein), which recruits other proteins to activate transcription of target genes. 22. Answer: C Difficulty: 1 Section: Signaling through G-Protein-Coupled Receptors Feedback: Stimulation of protein kinase C (PKC) activity near the plasma membrane results from elevated cytosolic Ca2+ [induced by inositol 1,4,5-trisphosphate (IP3) production] and from binding to diacylglycerol and negatively charged phospholipids. IP3 and diacylglycerol are second messengers in the cytosol and plasma membrane, respectively, and are produced from phosphatidylinositol 4,5-bisphosphate by the action of PLCβ. 23. Answer: I Difficulty: 3 Section: Signaling through G-Protein-Coupled Receptors Feedback: The effect of channel sensitization to Ca2+ is comparable to that of increased Ca2+ concentration, i.e. higher-frequency oscillations. The frequency as well as the amplitude of the Ca2+ spikes can also be modulated by other mechanisms. 24. Answer: calmodulin

Difficulty: 1 Section: Signaling through G-Protein-Coupled Receptors Feedback: Calmodulin is an important Ca2+-binding protein that helps mediate many cellular responses to Ca2+ signals. 25. Answer: VOBV Difficulty: 2 Section: Signaling through G-Protein-Coupled Receptors Feedback: Both smell and vision depend on G-protein-coupled receptors (GPCR) that egulate ion channels. Stimulation of olfactory receptors results in the opening of cyclicAMP-gated cation channels (through the activation of Golf) and membrane depolarization in the neuron. In vertebrate photoreceptors, in contrast, GPCR activation hyperpolarizes the cell: it closes cyclic-GMP-sensitive cation channels through the activation cyclic GMP phosphodiesterase by Gt. 26. Answer: DEACB Difficulty: 2 Section: Signaling through G-Protein-Coupled Receptors Feedback: Light stimulation of rod photoreceptors isomerizes 11-cis retinal in rhodopsin to an all-trans state. This changes the conformation of opsin and activates transducin (Gt) by nucleotide exchange. Gt then activates cyclic GMP phosphodiesterase to lower cyclic GMP levels and hence close cyclic-GMP-sensitive cation channels. Through a negative feedback loop, the resulting decrease in intracellular Ca2+ concentration stimulates rapid replenishment of cyclic GMP. 27. Answer: B Difficulty: 2 Section: Signaling through G-Protein-Coupled Receptors Feedback: Following the phosphorylation of the receptor by rhodopsin kinase, arrestin binds to it to prevent transducin activation. Other mechanisms, such as binding of RGS (regulator of G protein signaling) to the G protein α subunit or activation of guanylyl cyclase, are also involved in visual adaptation in rod cells. 28. Answer: TTFF Difficulty: 2 Section: Signaling through G-Protein-Coupled Receptors Feedback: Dissolved NO passes readily across cell membranes and can diffuse from its site of production to neighboring smooth muscle cells, where it increases cyclic GMP concentration by activating guanylyl cyclase. The drug Viagra® and its relatives inhibit

cyclic GMP phosphodiesterase in the penis, allowing the effects of NO to last longer. NO can also signal cells independently of cyclic GMP. 29. Answer: C Difficulty: 2 Section: Signaling through G-Protein-Coupled Receptors Feedback: A single quantum of light can result in the hydrolysis of more than 105 cyclic GMP molecules in the visual transduction cascade. This closes only a few hundred cyclic-GMP-sensitive channels. 30. Answer: A Difficulty: 2 Section: Signaling through G-Protein-Coupled Receptors Feedback: Mice or humans with defects in RK, or its downstream regulator arrestin, have a prolonged light response, incapable of adapting quickly to sudden drops in ambient light intensity. 31. Answer: D Difficulty: 2 Section: Signaling through Enzyme-Coupled Receptors Feedback: Dimerization of the insulin receptor brings the two kinase domains close to each other such that they can phosphorylate each other in the active site. In the case of the EGF receptor, the kinase is activated by the conformational change induced by asymmetric interactions between the two kinase domains outside their active sites. 32. Answer: FTFT Difficulty: 2 Section: Signaling through Enzyme-Coupled Receptors Feedback: c-Cbl is a ubiquitin ligase that is recruited to activated receptor tyrosine kinases (RTKs) through its SH2 domain and inactivates the receptors in a negative feedback loop. Mutations that inactivate this feedback mechanism cause prolonged RTK signaling and can promote the development of cancer. 33. Answer: D Difficulty: 1 Section: Signaling through Enzyme-Coupled Receptors Feedback: An example of such a protein is Grb2, an adaptor protein involved in signaling by receptor tyrosine kinases. 34. Answer: B Difficulty: 1 Section: Signaling through Enzyme-Coupled Receptors

Feedback: Ras is activated by guanine nucleotide exchange (GDP release and GTP binding), which is catalyzed by upstream Ras guanine nucleotide exchange factors (RasGEFs). In contrast, Ras GTPase-activating proteins (Ras-GAPs) increase the rate of GTP hydrolysis by Ras and therefore inactivate it. Guanine nucleotide dissociation inhibitors (GDIs) prevent the association of some GTPases with their GEFs, thereby keeping the GTPases inactive. 35. Answer: FTFT Difficulty: 2 Section: Signaling through Enzyme-Coupled Receptors Feedback: The mitogen-activated protein kinase (MAP kinase) module in the mammalian Ras-MAP-kinase signaling pathway is composed of Raf, Mek, and Erk. Stimulation of receptor tyrosine kinases (RTKs) at the cell surface can activate Ras, which then triggers the MAP kinase cascade. Once activated, Erk relays the signal downstream, leading to transient or long-lasting responses. Erk can also regulate Raf activity through negative feedback loops. The immediate early genes are typically activated minutes after RTK stimulation. 36. Answer: EESE Difficulty: 2 Section: Signaling through Enzyme-Coupled Receptors Feedback: Activation of Sevenless signaling leads to the activation of Sos, Ras, and Raf proteins. Consistently, loss-of-function mutations in genes encoding these proteins would exacerbate the effect of sev loss-of-function mutations. Defects in Ras-GAP, however, would suppress the effect. 37. Answer: B Difficulty: 1 Section: Signaling through Enzyme-Coupled Receptors Feedback: Scaffold proteins help prevent cross-talk and improve precision in signaling pathways, but they also limit diffusion of pathway components and thereby reduce the opportunities for amplification and spreading of the signal to different parts of the cell. 38. Answer: ACED Difficulty: 1 Section: Signaling through Enzyme-Coupled Receptors Feedback: Please refer to Figure 15–51. 39. Answer: E Difficulty: 2 Section: Signaling through Enzyme-Coupled Receptors

Feedback: PH domain preferentially bind to PI(3,4,5)P3 (E) produced by PI 3-kinase. 40. Answer: C Difficulty: 1 Section: Signaling through Enzyme-Coupled Receptors Feedback: The plasma-membrane-bound PI 3-kinase principally phosphorylates inositol phospholipids rather than proteins. It can be activated downstream of receptor tyrosine kinases or G-protein-coupled receptors, and generates docking sites for various signaling proteins. PTEN phosphatase counteracts PI 3-kinase by dephosphorylating the 3 position of the inositol ring. 41. Answer: B Difficulty: 2 Section: Signaling through Enzyme-Coupled Receptors Feedback: Mainly functioning in protein-protein interaction, PH domains are found in many signaling proteins, including pleckstrin, Akt, PDK1, and Sos. 42. Answer: TTFT Difficulty: 2 Section: Signaling through Enzyme-Coupled Receptors Feedback: mTOR complex 1 is rapamycin-sensitive, and is activated in the presence of growth factors, mainly via the PI-3-kinase-Akt pathway, to stimulate cell growth. Akt phosphorylates and inhibits Tsc2, a Rheb-GAP. This activates Rheb, which then activates mTOR in complex 1. 43. Answer: ACDEB Difficulty: 2 Section: Signaling through Enzyme-Coupled Receptors Feedback: Please refer to Figure 15–55. 44. Answer: E Difficulty: 1 Section: Signaling through Enzyme-Coupled Receptors Feedback: Src family proteins are cytosolic tyrosine kinases exemplified by Src. They contain SH2 and SH3 domains and can associate with the cytoplasmic side of the plasma membrane partly by covalently attached lipid chains. 45. Answer: C Difficulty: 1 Section: Signaling through Enzyme-Coupled Receptors

Feedback: Each STAT has an SH2 domain with two functions: it mediates STAT binding to an activated cytokine receptor; after JAK phosphorylation of STAT, it mediates binding between two STAT molecules in a dimer. 46. Answer: C Difficulty: 2 Section: Signaling through Enzyme-Coupled Receptors Feedback: Activated TGFβ receptors can be endocytosed by two distinct routes: one is dependent on clathrin-coated vesicles and leads to further activation, whereas the other one depends on caveolae and leads to inactivation. Inhibiting the formation of caveolae can therefore activate, rather than, inhibit TGFβ signaling. 47. Answer: E Difficulty: 2 Section: Signaling through Enzyme-Coupled Receptors Feedback: Both the cytokine receptor (left) and the TGFβ receptor (right) are enzymecoupled receptors and are noncovalently associated with cytosolic tyrosine kinases and serine/threonine kinases, respectively. Activated STATs form dimers, whereas activated R-Smads can form dimers or trimers with a co-Smad. Both of these phosphorylated complexes enter the nucleus and regulate the expression of their target genes. 48. Answer: C Difficulty: 1 Section: Alternative Signaling Routes in Gene Regulation Feedback: Upon Notch binding to Delta, a plasma-membrane-bound protease cleaves off a cytoplasmic segment of Notch, which enters the nucleus to activate the expression of target genes. The fragment acts by binding to a DNA-binding protein and converting it from a transcriptional repressor into a transcriptional activator. 49. Answer: E Difficulty: 2 Section: Alternative Signaling Routes in Gene Regulation Feedback: Wnt binding to Frizzled and LRP results in the disassembly of the β-catenin degradation complex (containing APC, axin, CK1, and GSK3), which allows β-catenin to accumulate and enter the nucleus, where it binds to LEF1/TCF, displacing the corepressor Groucho. 50. Answer: PTTP Difficulty: 2 Section: Alternative Signaling Routes in Gene Regulation

Feedback: Uncontrolled cell growth and proliferation can result if β-catenin activity is not properly limited. Thus, genes encoding Wnt and β-catenin are proto-oncogenes, whereas genes encoding APC and axin are tumor suppressors. Loss-of-function mutations in the Apc gene are found in 80% of human colon cancers. 51. Answer: C Difficulty: 2 Section: Alternative Signaling Routes in Gene Regulation Feedback: Both Ci and β-catenin are phosphorylated by GSK3 and CK1 in the absence of stimulation in Hedgehog and Wnt signaling pathways, respectively. They are both ubiquitylated and undergo proteolysis in the absence of signal. β-catenin is stabilized and can enter the nucleus only in the presence of Wnt. Ci enters the nucleus in the presence of Hedgehog and acts as a transcriptional activator, but its fragment (produced from partial proteolysis) can also enter the nucleus and act as a transcriptional repressor in the absence of signal. 52. Answer: APPP Difficulty: 2 Section: Alternative Signaling Routes in Gene Regulation Feedback: In the presence of the Hedgehog signal, Smoothened is phosphorylated and recruited to the plasma membrane, where it recruits the complex containing Ci, Fused, and Costal and inhibits Ci proteolysis. This results in the accumulation of unprocessed Ci in the nucleus and transcriptional activation of target genes. 53. Answer: CBA Difficulty: 2 Section: Alternative Signaling Routes in Gene Regulation Feedback: Various cell-surface receptors can activate NFκB in similar ways. IκB proteins bind tightly to NFκB and prevent their nuclear entry in the absence of the signal. In the presence of signal, IKK phosphorylates IκB to mark it for degradation. This frees NFκB to enter the nucleus and activate transcription of its target genes. 54. Answer: C Difficulty: 3 Section: Alternative Signaling Routes in Gene Regulation Feedback: The major contribution to the delay appears to be that of mRNA export from the nucleus to the cytosol, corresponding to the lag in the solid gray curve compared to the dashed black curve. The former curve peaks about an hour after the latter. 55. Answer: B Difficulty: 1

Section: Alternative Signaling Routes in Gene Regulation Feedback: Nuclear receptors have at least a ligand-binding domain and a DNA-binding domain. Some of them only enter the nucleus after ligand binding, whereas others are present inside the nucleus even in the absence of their ligand. Their ligands are typically small hydrophobic molecules such as steroid hormones and thyroid hormones. Some nuclear receptors are transcriptional activators, whereas others are transcriptional repressors. 56. Answer: 2 Difficulty: 3 Section: Alternative Signaling Routes in Gene Regulation Feedback: The period of the circadian rhythm is about 24 hours and would be conserved in mutants that lack kinase 2 but retain kinase 1. These mutants have nevertheless lost the higher-frequency oscillations caused by kinase 2. 57. Answer: D Difficulty: 3 Section: Alternative Signaling Routes in Gene Regulation Feedback: The clock shifts by about 6 hours between days 8 and 20 (in 12 days), i.e. by about 0.5 hr/day. The free-running period, therefore, appears to be about 23.5 hours. 58. Answer: FFTT Difficulty: 2 Section: Signaling in Plants Feedback: Multicellularity and cell communication evolved independently in plants and animals. However, plants and animal share common signaling molecules such as cyclic GMP and nitric oxide. Receptor serine/threonine kinases are the largest class of cellsurface receptors in plant cells. Among them, LRR receptor kinases constitute a major group. 59. Answer: ethylene Difficulty: 1 Section: Signaling in Plants Feedback: Ethylene is a small gas molecule that can influence plant development in various ways. 60. Answer: D Difficulty: 1 Section: Signaling in Plants Feedback: Ethylene receptors in plant cells are dimeric multipass transmembrane proteins primarily located in the endoplasmic reticulum. In addiction to a copper-containing

ethylene-binding domain, they contain a cytosolic domain that interacts with CTR1, a serine/threonine kinase related to Raf, an upstream kinase in the MAP kinase module in animal cells. 61. Answer: E Difficulty: 2 Section: Signaling in Plants Feedback: In the ethylene signaling pathway in plant cells, ethylene receptor and CTR1 are active in the absence of ethylene. Active CTR1 stimulates the degradation of of EIN3, preventing it from activating the transcription of ethylene-responsive genes. Ethylene binding inactivates the receptors, resulting in EIN3 accumulation and gene expression. 62. Answer: PCCP Difficulty: 1 Section: Signaling in Plants Feedback: Phytochromes are cytosolic serine/threonine protein kinases that are activated by red light. Cryptochromes are flavoproteins that are sensitive to blue light. They are found in animal as well as in plant cells. 63. Answer: B Difficulty: 2 Section: Signaling in Plants Feedback: In the presence of a lateral light, auxin efflux transporter proteins become primarily localized to the side of the cell that is away from the light source. This redirects the auxin flux mainly to the darker part of the growing shoot, where it stimulates the elongation of the cells (in contrast to its inhibitory effect on the elongation of root epidermal cells seen in gravitotropism), bending the shoot towards the light source.

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION CHAPTER 16: THE CYTOSKELETON Copyright © 2015 by W.W. Norton & Company, Inc. 1. Indicate if each of the following structures is based on actin filaments (A), microtubules (M), or intermediate filaments (I). Your answer would be a five-letter string composed of letters A, M, and I only; e.g. AAAMM. ( ) The cell cortex ( ) The mitotic spindle ( ) The nuclear lamina ( ) Cilia ( ) Filopodia 2. Indicate if each of the following changes occurring during mitosis in a fibroblast is the result of the reorganization of actin filaments (A), microtubules (M), or intermediate filaments (I). Your answer would be a five-letter string composed of letters A, M, and I only; e.g. AIAAM. ( ) The cell rounds up. ( ) The endoplasmic reticulum collapses. ( ) The Golgi apparatus fragments. ( ) The primary cilium is resorbed (disappears). ( ) The contractile ring forms and constricts. 3.

Which of the following cytoskeletal filaments are abundant in an animal cell nucleus? A. Microfilaments B. Microtubules C. Septins D. Intermediate filaments E. Spectrin filaments

4. From left to right, indicate if each of the following schematic drawings represents the typical overall organization of actin filaments (A), microtubules (M), or intermediate filaments (I) in an animal cell. Your answer would be a three-letter string composed of letters A, M, and I only; e.g. AMI.

5. Indicate true (T) and false (F) statements below regarding the cell cytoskeleton. Your answer would be a four-letter string composed of letters T and F only; e.g. TTFF. ( ) The three major building blocks of cytoskeletal filaments can bind to and hydrolyze nucleotides. ( ) The building blocks of microfilaments and microtubules are globular proteins, whereas those of intermediate filaments are themselves filamentous proteins. ( ) Intermediate filaments are typically thicker than actin filaments but thinner than microtubules. ( ) Plant cells lack microtubules. 6. Fill in the blank: Each microtubule is typically made of thirteen parallel …. that associate laterally to form a hollow tube. 7. Indicate if each of the following descriptions matches actin filaments (A), microtubules (M), or intermediate filaments (I). Your answer would be a four-letter string composed of letters A, M, and I only; e.g. AAMM. ( ) They form hollow structures with multiple lateral interactions. ( ) They form strong structures that are more resilient than the other two cytoskeletal filaments. ( ) Their subunits bind GTP and hydrolyze it. ( ) They form coiled-coil interactions between the subunits. 8.

Bacteria contain homologs of cytoskeletal filament subunits …

A. except those of the intermediate filaments. B. but these homologs are incapable of nucleotide binding and hydrolysis. C. that are less diverse in their function relative to their eukaryotic counterparts. D. that can have different functions to those of their eukaryotic counterparts. E. that are dispensable for cell growth and proliferation. 9. Persistence length for a cytoskeletal filament is the minimum filament length at which random thermal fluctuations are likely to cause it to bend. Which of the following comparisons are true, considering the persistence lengths of (a) an actin filament, (b) a bundle of cross-linked actin filaments, and (c) a microtubule? A. (a > b) and (a > c) B. (a > b) and (a < c) C. (a < b) and (a > c) D. (a < b) and (a < c) E. (a > b) and (b > c) 10. You have prepared actin filament seeds in a microfluidic chamber through which you then pass either unlabeled or fluorescently labeled actin subunits in alternation several times, keeping the total monomer concentration constant. You then observe the resulting filaments under a fluorescence microscope. If the filaments appear as drawn schematically in the example below (in which fluorescence is indicated by black color), which end do you think is the plus end—(a) or (b)? Was the concentration of free actin subunits below or above the critical concentration (Cc) for polymerization at the barbed end?

(b) (a)

A. (a); below B. (b); below C. (a); above D. (b); above

11. F-actin is not a straight polymer but can be considered a double-helical assembly. The two protofilaments twist around each other and cross over approximately every 13 actin subunits, and the subunits in each protofilament are repeated every 2.8 nm along the helix. F-actin is almost 8 nm wide. Different myosin motors have different “step lengths” on actin depending on their function. Which of the following step lengths would you expect to belong to a processive myosin motor that carries large cellular cargoes such as endocytic vesicles? A. About 2.8 nm B. About 8 nm C. About 13 nm D. About 27 nm E. About 36 nm 12. In the polymerization in vitro of actin filaments and microtubules from their subunits, what does the “lag phase” correspond to? A. Nucleation B. Reaching steady state C. Nucleotide exchange D. ATP or GTP hydrolysis E. Treadmilling 13. The time courses of seeded actin polymerization under two different conditions are compared in the following graph. If the overall kon for polymerization is known to be the same under both of these conditions, which curve—(1) or (2)—corresponds to the condition with a higher koff rate constant? Which one corresponds to a higher Cc for polymerization?

Percentage of actin subunits in filaments

(1) (2)

Time after salt addition

A. Curve (1); curve (1) B. Curve (1); curve (2) C. Curve (2); curve (1) D. Curve (2); curve (2) E. Curve (2); both curves have the same Cc. 14. If the concentration of free subunits is C, under which condition does the growth of a cytoskeletal filament proceed spontaneously? A. C > 1/kon B. C < 1/koff C. C > kon/koff D. C > koff/kon E. C > Cc×kon/koff 15. Consider the ATP-bound and ADP-bound forms of actin and the polarized nature of the actin filaments. In the following diagram that shows the various actin polymerization rate constants (kon and koff values), which monomer corresponds to an ADP-bound actin incorporated at the plus end? (A)

12 µM–1 sec–1

0.3 sec–1

1.5 sec–1

1.5 µM–1.sec–1

4 µM–1 sec–1 0.8 sec–1

8 sec–1

(B)

16.

(D)

In the following graph of actin elongation rates under different subunit concentrations,

which of the following corresponds to the slope of the line?

Elongation rate

grow

0 shrink Subunit concentration

A. kon B. koff C. kon/koff D. koff/kon E. kon – koff 17.

According to the following graph, which shows the elongation rate at the plus and minus

ends of actin filaments as a function of actin subunit concentration, at what concentration (A to E) does the total length of the filament remain more or less constant with time (i.e. steady-state treadmilling occurs)?

Elongation rate

grow

0 shrink A BC DE Subunit concentration

18.

Which of the following drugs is toxic for our cells? A. Cytochalasin B, which caps the plus end of actin filaments and prevents actin polymerization. B. Phalloidin, which binds along actin filaments and stabilizes them. C. Nocodazole, which binds to tubulin subunits and prevents microtubule polymerization. D. Colchicine, which caps microtubule ends and leads to their depolymerization. E. All of the above.

19.

The actin-nucleating protein formin has flexible “whiskers” containing binding sites that

help recruit actin subunits in order to enhance polymerization by this protein. What protein would you expect to bind to these sites? A. Thymosin B. Profilin C. Cofilin D. Gelsolin E. Tropomodulin 20.

Which of the following is an actin homolog? A. Arp2 B. Arp3 C. MreB D. ParM E. All of the above

21. Which of the following actin-binding proteins cannot bind to the same actin filament simultaneously? A. Gelsolin and tropomodulin B. Tropomyosin and tropomodulin C. Profilin and tropomodulin D. Cofilin and CapZ E. Formin and CapZ 22. In the following graph, the elongation rate of pure actin filaments as a function of actin subunit concentration is shown as a dashed line. Which one of the other five lines (A to E) would you think better shows what happens when a plus-end capping protein such as CapZ is present?

grow

Elongation rate

A B C D 0 E

shrink

Subunit concentration 23. Which of the actin-binding proteins (1 to 5) in the following schematic drawing represents tropomyosin, α-actinin, Arp 2/3 complex, CapZ, and myosin, respectively? Your answer would be a five-digit number composed of digits 1 to 5 only; e.g. 15324. (1) (+) end (3)

(–) end (4)

(–) end

(2) (5) (+) end

24. Cofilin binds preferentially to ADP-containing actin filaments rather than to ATPcontaining filaments. Consequently, this protein … A. competes with profilin for actin binding. B. binds to older actin filaments. C. binds to the plus end of actin filaments. D. stabilizes actin filaments. E. All of the above.

25. Actin filaments that are held together by the cross-linking protein fimbrin are not contractile. This is probably because … A. the very weak cross-linking by this protein cannot convert myosin II activity into a coherent contraction. B. fimbrin arranges the actin filaments in parallel bundles in which all the plus ends point to the same direction. C. the very tight packing of actin filaments by this small protein excludes myosin II filaments and other large proteins. D. fimbrin arranges the actin filaments into gel-like networks in which myosin II activity does not produce contraction. E. fimbrin is a large protein that binds to several actin filaments and resists contraction. 26. In the dendritic networks of actin filaments in lamellipodia, nucleation of actin polymerization is mostly performed by … A. ERM proteins B. WASp proteins C. Formin D. Arp 2/3 complex E. γ-TuRC 27. In the structure of which of the following proteins are the actin-filament binding sites furthest apart in space? A. Spectrin B. Filamin C. α-Actinin D. Fimbrin E. Myosin II 28.

The proteins of the ERM family such as moesin … A. bind to and organize the cortical actin cytoskeleton. B. interact with transmembrane proteins. C. affect cortical stiffness and cell shape. D. affect the localization and activity of cell signaling molecules. E. All of the above.

29. The pathogen Listeria monocytogenes can hijack the actin cytoskeleton in human cells and spread inside the host. Indicate true (T) and false (F) statements below about this phenomenon. Your answer would be a four-letter string composed of letters T and F only; e.g. TTFF. ( ) The movement can be reconstituted in vitro by placing the bacteria in a mixture of actin, formin, gelsolin, and capping protein. ( ) Cofilin counteracts the movement by depolymerizing actin filaments. ( ) The actin filaments grow with their minus ends pointed toward the bacterium. ( ) The movement depends on myosin activity to transport the bacteria on the actin filaments. 30. Indicate true (T) and false (F) statements below regarding cytoskeletal motor proteins. Your answer would be a five-letter string composed of letters T and F only, e.g. TTTFF. ( ) All myosin motors move toward the plus end of actin filaments. ( ) All myosin motors move toward the minus end of actin filaments. ( ) All kinesin motors move toward the plus end of microtubules. ( ) All kinesin motors move toward the minus end of microtubules. ( ) All dynein motors move toward the minus end of microtubules. 31. If myosin II heads are attached to a glass slide and actin filaments are allowed to bind to them, the filaments will glide on the surface … A. toward their plus end in the presence of ATP. B. toward their plus end in the presence of GTP. C. toward their minus end in the presence of ATP. D. toward their minus end in the presence of GTP. E. toward their plus end in the presence of ADP. 32.

Sort the following events to show the sequence in the mechanochemical cycle of myosin

II, following ATP binding by the head domain. Your answer would be a four-digit number composed of digits 1 to 4 only, e.g. 1342. (A) Myosin head binds tightly to actin. (B) Power stroke is triggered. (C) Cocking occurs and the head is displaced along the actin filament. (D) The binding affinity of myosin for actin is reduced.

33. In the presence of an ATP analog that can bind myosin normally but cannot be hydrolyzed, … A. a myosin cannot bind tightly to an actin filament and cannot move on it. B. a myosin cannot be released from an actin filament and cannot move on it. C. a myosin performs the power stroke but cannot be released from an actin filament. D. a myosin performs the power stroke but cannot bind tightly to an actin filament. E. a myosin is able to perform normally. 34. After an animal dies, its muscles start to stiffen before the decomposition of tissues relaxes the muscles again. Which of the following would you expect to explain this muscle stiffening (i.e. rigor mortis)? A. The myosin II heads in muscle fibers remain attached to actin filaments due to the absence of Ca2+ in these cells. B. The ATPase activity of muscle myosin II is inhibited by the elevated Ca2+. C. The myosin II heads in muscle cells remain attached to actin filaments due to the elevated Ca2+ in the muscle fibers. D. Titin molecules unfold, preventing muscle relaxation. E. The myosin II heads in muscle cells remain attached to actin filaments due to covalent cross-linking. 35. You have used “optical tweezers” to study the mechanics of myosin movement in a newly discovered member of the myosin II subfamily. Your results indicate that each myosin exerts a force of about 3 pN along the length of the actin filament, and displaces the filament by about 10 nm in each cycle of ATP hydrolysis. Assuming that the free-energy change for ATP hydrolysis is –50 kJ/mol under your experimental conditions, what is the efficiency of the myosin motor in converting the free energy to mechanical work? Remember that under a constant force F (in newtons), work (W; in joules) is calculated as W = F × d, where d is displacement (in meters) in the direction of the force. Avogadro’s number is approximately 6 × 1023 molecules/mole. Write down your answer as a percentage with no decimals, e.g. 99%. 36. Consider the structure of a sarcomere. Which of its features does NOT shorten during skeletal muscle contraction? A. The dark band B. The light band C. The distance from the M line to the Z disc D. The distance between two consecutive Z discs

E. The extension of the titin molecules 37. Skeletal muscle cells consume enormous amounts of ATP. From the following list, choose the two proteins that directly hydrolyze most of this ATP. Your answer would be a twoletter string composed of letters A to F only, in alphabetical order, e.g. AF. A. Troponin complex B. Ca2+-release channel C. Myosin light chains D. Myosin heavy chain E. Myosin light-chain kinase F. Sarcoplasmic reticulum Ca2+-pump 38. Heart conditions known as cardiomyopathies include two main subtypes, namely the hypertrophic and dilated cardiomyopathies, in which some portions of the cardiac muscle become stiff or dilated, respectively. Drugs such as the calcium channel blocker verapamil are used to treat the former, while the Na+-K+ ATPase inhibitor digoxin (which elevates intracellular calcium concentration in cardiac muscle cells) is sometimes used for the latter. Mutations in the subunits of the troponin complex are among the genetic causes of these diseases, and can be grouped into mutations that result in either decreased (D) or increased (I) calcium-ion sensitivity of actin–myosin contraction. Which of these mutations would you expect to be associated with hypertrophic cardiomyopathy? Write down D or I as your answer. 39. Sort the following events to reflect the sequence during smooth muscle contraction upon delivery of muscle stimulation. Your answer would be a five-digit number composed of digits 1 to 5 only, e.g. 13452. (A) Activation of the myosin light-chain kinase (B) Interaction of myosin head with actin (C) Phosphorylation of myosin (D) Calcium release into the cytoplasm (E) Activation of calmodulin 40. In which of the following structures are actin–myosin II bundles regulated by the troponin complex? A. Stress fiber B. Circumferential belt

C. Contractile ring D. Adherens junction E. None of the above

41. A small unique insert found near the end of the motor domain in myosin VI has been linked to the singular ability of this myosin to move toward the pointed end of actin filaments. If the unique insert is removed by genetic engineering, the resulting mutant myosin VI is a plus-end directed motor. This observation indicates that … A. the unique insert is necessary and sufficient for minus-end directionality. B. the unique insert is necessary for motor activity. C. the unique insert is necessary for plus-end directed directionality. D. the unique insert is not sufficient for minus-end directionality. E. None of the above. 42. The duty ratio for a motor protein is defined as the fraction of time in each cycle of activity of the motor in which the head is bound with high affinity to its cytoskeletal track. Which of the following is reasonable given this definition? A. Myosin V has a higher duty ratio than myosin II, because Pi release is the ratelimiting step in its cycle. B. Myosin II has a higher duty ratio than myosin V, because Pi release is the ratelimiting step in its cycle. C. Myosin V has a higher duty ratio than myosin II, because ADP release is the ratelimiting step in its cycle. D. Myosin II has a higher duty ratio than myosin V, because ADP release is the ratelimiting step in its cycle. 43.

A microtubule appears as a left-handed helix due to an approximately 0.9-nm stagger in

the lateral contacts between adjacent protofilaments. In the lateral contacts, α- and β-tubulins in one protofilament interact with α- and β-tubulins, respectively, in the neighboring protofilament, except for a longitudinal discontinuity along the microtubule called the seam. Along the seam, lateral contacts have to be made between different tubulins (i.e. α-tubulin with β-tubulin). Which of the following do you think is acceptable as the repeat distance of tubulin monomers along a protofilament? A. 4 nm B. 5 nm

C. 6 nm D. 7 nm E. 10 nm 44. In the following graph that shows changes in the lengths of two microtubules over time, which time point corresponds to a catastrophe for both microtubules? Which trace corresponds to a microtubule with greater dynamic instability?

Microtubule length

long

(a)

(b)

short 0

t2 Time

A. t1; trace (a) B. t1; trace (b) C. t2; trace (a) D. t2; trace (b) 45.

In contrast to growing microtubules, shrinking microtubules … A. have a GTP cap at their plus end. B. have strong lateral interactions at their plus ends. C. have curved protofilaments at their plus ends. D. cannot be rescued unless microtubule-stabilizing proteins bind and inhibit depolymerization. E. All of the above.

46.

The γ-tubulin ring complex is to microtubules what … is to actin filaments.

A. the Arp 2/3 complex B. the dynactin complex C. the troponin complex D. formin E. contractile ring 47.

What is the major microtubule-organizing center in animal cells? A. The γ-tubulin ring complex B. The centrosome C. The cell cortex D. The primary cilium E. The spindle pole body

48. Which of the following proteins do you expect to be enriched near the plus end of microtubules? A. Dynein B. XMAP215 C. γ-Tubulin D. Katanin E. All of the above 49. Indicate if each of the following descriptions applies to (1) EB1, (2) kinesin-1, (3) kinesin-13, or (4) katanin. Your answer would be a four-digit number composed of digits 1 to 4 only, e.g. 1432. ( ) This is a microtubule-severing protein that can release microtubules from the microtubule-organizing centers. ( ) It increases the frequency of catastrophe by deforming microtubule protofilaments. ( ) This is a conventional motor protein that moves toward the plus end of a microtubule. ( ) It recognizes the structure of a growing microtubule end and binds to it, helping other proteins to also bind to the plus end. 50. Indicate whether each of the following descriptions applies to myosins (M), kinesins (K), or dyneins (D). Your answer would be a five-letter string composed of letters M, K, and D only, e.g. MMMDD. ( ) They have larger structures than the other two. ( ) They are generally faster than the other two.

( ) They are structurally unrelated to the other two. ( ) They walk on a different cytoskeletal filament than the other two. ( ) They are all minus-end directed. 51. Unlike a myosin head, a kinesin head is tightly bound to its cytoskeletal track when bound to ... A. ATP. B. ADP. C. no nucleotide. D. GTP. E. GDP. 52. A dimeric kinesin-1 molecule has 8-nm steps and can move at rates of about 1 µm/sec. Olympic 100-meter sprinters typically run at about 180 steps per minute and can reach speeds of about 42 km/h. With the same step size, if the Olympic runner had a step frequency of a kinesin1 molecule, how fast could she run? Write down your answer in km/h, e.g. 52 km/h. 53. Dynamitin is a subunit of the dynactin complex. Its overexpression leads to the disassembly of the complex, and it is therefore considered a dynactin inhibitor. Which of the following processes would happen if dynamitin is overexpressed in a cell? A. The motile cilia (if any) would stop beating because the axonemal dyneins could not attach to their neighboring microtubules. B. The Golgi apparatus would fragment and become dispersed because the cytoplasmic dyneins could not associate with the Golgi membranes. C. The endoplasmic reticulum would collapse because it could not recruit enough kinesins. D. The nuclear envelope would disintegrate because the nuclear lamins could not interact with the cytoplasmic cytoskeleton. E. The cortex would lose its integrity because the Arp 2/3 complex would also be inhibited. 54. Using time-lapse fluorescence microscopy, you have recorded the transport of fluorescently labeled mitochondria in a region of an axon in a fruit-fly larva; you have plotted the results in the following graph, in which the vertical axis is time (from the top to the bottom), the horizontal axis is the position along the axon, and each trace represents one mitochondrion. Is the retrograde transport or the anterograde transport of mitochondria faster in these axons? Does the

trace indicated by an asterisk correspond to a mitochondrion that was transported by a dynein or a kinesin? 0

Time

toward the cell body

Position

toward the tip of the axon

A. Anterograde; kinesin B. Anterograde; dynein C. Retrograde; kinesin D. Retrograde; dynein 55. “Headless” kinesin mutants only contain the stalk (middle) and tail domains and can therefore dimerize with their wild-type kinesin partners. However, since they lack the motor (head) domain, the resulting dimers are unable to carry out processive transport of their cargoes and the mutation thus behaves as “dominant negative,” meaning that the mutant not only is nonfunctional, but can also interfere with the function of its wild-type counterparts. If a headless mutant of a kinesin heavy chain involved in melanosome movement is overexpressed in fish melanocytes, what would you predict happens in these cells? A. Pigment dispersion would be inhibited and there would be more tug-of-war between the motors. B. Pigment aggregation would be inhibited and there would be more tug-of-war between the motors. C. Pigment dispersion would be inhibited and there would be less tug-of-war between the motors. D. Pigment aggregation would be inhibited and there would be less tug-of-war between the motors.

56. Indicate true (T) and false (F) statements below regarding the microtubule cytoskeleton in neurons. Your answer would be a five-letter string composed of letters T and F only, e.g. TTFFF. ( ) All microtubules in an axon are normally oriented in the same direction. ( ) All microtubules in a dendrite are normally oriented in the same direction. ( ) Each microtubule in an axon has one end at the cell body and its other end at the axon terminal. ( ) Anterograde transport in axons is exclusively carried out by kinesins. ( ) Anterograde transport (toward the tip) in dendrites is exclusively carried out by kinesins. 57.

Indicate whether each of the following descriptions applies to cilia (C), flagella (F), or

both (B). Your answer would be a four-letter string composed of letters C, F, and B only, e.g. CCFB. ( ) They are short and present at high numbers per cell. ( ) They have a whiplike motion that resembles breaststroke in swimming. ( ) They are based on the axoneme structure. ( ) They are found in the epithelial cells of the human respiratory tract. 58.

In the presence of ATP in a flagellum, an axonemal dynein that is interacting through its

tail with the A microtubule of a peripheral doublet can push this doublet toward the ...(1) of the flagellum, but due to the presence of linking proteins such as ...(2), this force is converted into a bending motion. A. (1) tip; (2) nexin B. (1) base; (2) nexin C. (1) tip; (2) nebulin D. (1) base; (2) nebulin E. (1) tip; (2) tau 59. Indicate true (T) and false (F) statements below regarding the primary cilia. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) Primary cilia are found on the surface of almost all cell types in our bodies. ( ) Primary cilia are motile. ( ) Primary cilia are made in interphase at basal bodies. ( ) Primary cilia are thought to function in sensing and responding to external signals.

60. Indicate whether each of the following descriptions matches actin filaments (A), microtubules (M), or intermediate filaments (I). Your answer would be a three-letter string composed of letters A, M, and I only, e.g. MMA. ( ) They have the smallest persistence length. ( ) They lack polarity. ( ) They have the highest tensile strength. 61.

Keratins are intermediate filaments that ... A. are localized inside the nucleus. B. are composed of 50% type I and 50% type II keratin proteins. C. are bundled in the epidermis to make cells more flexible. D. anchor the intermediate filament network at the adherens junctions. E. All of the above.

62.

What are the typical consequences of mutations in keratins and their associated proteins? A. Neurodegeneration as a result of interference with normal axonal transport. B. Muscle development defects as a result of sarcomere disorganization. C. Skeletal and cardiac abnormalities as a result of a weakened nuclear envelope. D. Cornea disorders as a result of cell rupture from mechanical trauma. E. Developmental defects as a result of abnormal signaling related to the primary cilia.

63. In a cross section of a vertebrate axon, longitudinally organized cytoskeletal proteins can be seen as dots, as shown in the schematic drawing below. What type of filaments do these dots represent?

A. Microtubules and microfilaments B. Microtubules and vimentins C. Microtubules and neurofilaments D. Microtubules and keratins

E. Microtubules and septins 64. These proteins are found in the budding yeast Saccharomyces cerevisiae; they form “neck filaments” between a mother cell and its growing bud, and help polarize protein distribution between the two. These proteins ... A. can polymerize to form filaments and sheets. B. bind GTP. C. are also involved in contractile ring formation during cytokinesis in animal cells. D. form filaments that are thought to be nonpolar. E. All of the above. 65. SUN and KASH proteins embedded in the nuclear envelope provide connections between the organization of the nucleus and the cytoplasm. Which of the following pairs of proteins DO NOT bind directly to each other in these connections? A. Nuclear lamina and SUN proteins B. Plectins and KASH proteins C. KASH proteins and microfilaments D. KASH proteins and SUN proteins E. SUN proteins and cytoplasmic intermediate filaments. 66.

Which of the following is NOT an example of a crawling cell? A. Macrophage B. Osteoclast C. Keratocyte D. Neural crest cell E. Sperm

67.

Indicate whether each of the following structures has actin organized mostly in a one- (1),

two- (2), or three- (3) dimensional arrangement. Your answer would be a four-digit number composed of digits 1 to 3 only, e.g. 1333. ( ) Stress fibers ( ) Invadopodia ( ) Lamellipodia ( ) Filopodia 68.

How is membrane protrusion by blebbing different from that by lamellipodia?

A. Blebbing usually occurs on a rigid substratum such as glass, whereas lamellipodia form on pliable substrata. B. Blebbing is mostly observed in vitro, whereas lamellipodia are observed both in vitro and in vivo. C. Blebbing does not involve myosin II activity, but formation of lamellipodia does. D. Blebbing requires loss of membrane interaction with actin filaments, whereas lamellipodia require a close interaction between the two. E. All of the above. 69.

Consider an actin subunit that has just been incorporated into an actin filament at the

leading edge of a lamellipodium. Before its ATP is hydrolyzed, how does its distance from the leading front edge of the plasma membrane change over time? How does its distance from the Factin minus end change over time? A. Decreases; decreases B. Decreases; remains constant C. Decreases; increases D. Increases; decreases E. Increases; remains constant 70. In the following schematic drawing, in which direction (1 or 2) is the keratocyte migrating? What is the approximate length of the cell indicated by the bar on the left?

A. Direction 1; about 20 µm B. Direction 1; about 2 µm C. Direction 2; about 20 µm D. Direction 2; about 2 µm E. Either direction; about 2 µm 71. If a certain isoform of myosin II is depleted from a cell, stress fibers are lost and focal adhesions disappear. If these cells are placed on a surface coated with an array of flexible pillars to measure traction forces, would you expect the traction to increase (I) or decrease (D) in these cells compared to wild-type cells? Write down I or D as your answer. 72. In lamellipodia, actin polymerization is nucleated by ...(1), while depolymerization is catalyzed by ...(2). A. (1) formin; (2) gelsolin B. (1) Arp 2/3 complex; (2) gelsolin C. (1) formin; (2) cofilin D. (1) Arp 2/3 complex; (2) cofilin E. (1) formin; (2) capping protein 73. Indicate whether each of the following descriptions matches Cdc42 (C), Rac (R), or Rho (H) from the Rho family of monomeric GTPases. Your answer would be a three-letter string composed of letters C, R, and H only, e.g. HRR. ( ) It activates formin. ( ) It inhibits myosin II activity. ( ) When constitutively active, it induces the formation of many prominent stress fibers. 74.

In the following schematic drawing of a polarized neutrophil engaged in chemotaxis, in

what region (a or b) does Rac activity dominate? What other member of the Rho family dominates at the other region?

Bacterial peptides

a b A. Region (a); Rho B. Region (a); Cdc42 C. Region (b); Rho D. Region (b); Cdc42 75. Considering the diagram below, which summarizes two signaling pathways initiated by activated GTPases of the Rho family, what letter (A to D) corresponds to each of the following? Your answer would be a four-letter string composed of letters A to D only, e.g. ABCD.

C ( ) Rac-GTP ( ) Rho-GTP ( ) Lamellipodia formation ( ) Stress-fiber formation

Answers 1. Answer: AMIMA Difficulty: 1 Section: Function and Origin of the Cytoskeleton Feedback: Microtubules form cilia and flagella and are also responsible for the formation of the bipolar mitotic spindle. Actin filaments underlie the plasma membrane of animal cells and form dynamic structures such as filopodia. The nuclear lamina is an example of structures formed by intermediate filaments to provide the cell interior with mechanical strength. 2. Answer: AMMMA Difficulty: 2 Section: Function and Origin of the Cytoskeleton Feedback: The change in the organization of the actin-based cell cortex leads to round cells that are later cleaved into two daughter cells by the actin-based contractile ring. The network of endoplasmic reticulum membranes and Golgi stacks lose their interphase distribution during mitosis as a result of a profound reorganization of the microtubules, which also affects the microtubule-based primary cilia. 3. Answer: D Difficulty: 1 Section: Function and Origin of the Cytoskeleton Feedback: The lamins (one type of intermediate filament) form a meshwork beneath the nuclear envelope called the nuclear lamina. 4. Answer: MIA Difficulty: 2 Section: Function and Origin of the Cytoskeleton Feedback: While the microtubule network usually has an overall radial distribution during interphase, with the majority of the microtubules originating from the cell center, the actin cytoskeleton forms the cell cortex near the cell periphery and is also found in other places in the cell, forming structures such as stress fibers. The intermediate filaments extend across the cytoplasm and give mechanical strength to the cell. 5. Answer: FTTF Difficulty: 2 Section: Function and Origin of the Cytoskeleton Feedback: Actin and tubulin are globular proteins capable of nucleotide triphosphate binding and hydrolysis and are found in all eukaryotes. The intermediate filaments, on

the other hand, are made of filamentous subunits and are not as well conserved. Their thickness is intermediate between that of the other two cytoskeletal filaments. 6. Answer: protofilaments Difficulty: 1 Section: Function and Origin of the Cytoskeleton Feedback: Single microtubules are normally composed of 13 protofilaments—linear strings of tubulin subunits joined end-to-end. 7. Answer: MIMI Difficulty: 2 Section: Function and Origin of the Cytoskeleton Feedback: Microtubules form hollow tubes whose walls are composed of tubulin subunits. Tubulins can bind GTP. Intermediate filament subunits form α-helical coiledcoils. These filaments form strong ropelike structures that tolerate mechanical stress, such as stretching and bending, to a greater extent compared to actin filaments and microtubules. 8. Answer: D Difficulty: 2 Section: Function and Origin of the Cytoskeleton Feedback: Bacteria contain homologs of all three types of cytoskeletal filaments. These homologs are more diverse than their eukaryotic versions in terms of functional and structural features. 9. Answer: D Difficulty: 3 Section: Actin and Actin-binding Proteins Feedback: A bundle formed from several cross-linked actin filaments has a higher persistence length than each individual filament. Microtubules have a higher persistence length compared to actin filaments. 10. Answer: D Difficulty: 3 Section: Actin and Actin-binding Proteins Feedback: Since the polymerization rate is higher at the plus (barbed) end, the fluorescent segments at this end are longer than those at the minus (pointed) end. Both ends have grown (at different rates), which means the subunit concentration has been above the critical concentration. 11. Answer: E Difficulty: 3

Section: Actin and Actin-binding Proteins Feedback: The protofilaments cross over approximately every 36 nm (i.e. 13 subunits × 2.8 nm/subunit), which means a processive myosin of this step size does not have to rotate around the filament as it moves along it. If the step size was not a multiple of the helical repeat, carrying large cargoes would involve a screw motion and be less efficient. 12. Answer: A Difficulty: 2 Section: Actin and Actin-binding Proteins Feedback: The lag phase corresponds to the rate-limiting nucleation step. If this step is bypassed by introduction of preformed filament seeds, the lag phase disappears. 13. Answer: D Difficulty: 3 Section: Actin and Actin-binding Proteins Feedback: The condition corresponding to curve (2) has a higher koff and hence a higher Cc, manifested as a lower steady-state percentage of incorporated subunits, as well as an overall lower net polymerization rate before reaching the steady state. 14. Answer: D Difficulty: 3 Section: Actin and Actin-binding Proteins Feedback: Cc = koff/kon. At concentrations of monomers C > Cc, the free-energy change for polymerization is negative and growth can proceed spontaneously. 15. Answer: D Difficulty: 3 Section: Actin and Actin-binding Proteins Feedback: The plus end (right) has higher polymerization rate constants. It also has a lower Cc value (i.e. a lower koff/kon ratio). ATP-bound actin (B and C) has the higher tendency for polymerization at both ends. 16. Answer: A Difficulty: 3 Section: Actin and Actin-binding Proteins Feedback: The elongation rate (R) at each subunit concentration (C) follows the linear equation R = kon × C – koff , in which kon represents the slope. 17. Answer: B Difficulty: 3 Section: Actin and Actin-binding Proteins

Feedback: The line with the greater slope corresponds to polymerization or depolymerization at the plus end. Within the treadmilling range (concentrations between A and E), when the positive elongation rate at the plus end exactly cancels out the negative rate at the minus end, the length of the filament remains constant. This does not happen when the monomer concentration (C) is exactly in the middle of the range (since the two lines have different slopes), but occurs around B for which the sum of the elongation rates for plus and minus ends equals zero. 18. Answer: E Difficulty: 2 Section: Actin and Actin-binding Proteins Feedback: These drugs, whether they stabilize or destabilize the cytoskeletal filaments, interfere with the dynamics of the cytoskeleton and are therefore cytotoxic. 19. Answer: B Difficulty: 3 Section: Actin and Actin-binding Proteins Feedback: Profilin competes with thymosin for the binding of actin subunits, but unlike thymosin, it does not prevent polymerization. Recruitment of profilin–actin therefore facilitates the rapid formin-mediated actin polymerization. 20. Answer: E Difficulty: 1 Section: Actin and Actin-binding Proteins Feedback: Arp2 and Arp3 are actin homologs in the Arp 2/3 complex, which nucleates actin filaments by taking advantage of the structural similarities between these molecules and the actin subunits. MreB and ParM are bacterial actin homologs. 21. Answer: E Difficulty: 3 Section: Actin and Actin-binding Proteins Feedback: Both formin and CapZ bind to the plus end and have incompatible functions. 22. Answer: D Difficulty: 3 Section: Actin and Actin-binding Proteins Feedback: With pure subunits, the (dashed) line represents the overall growth rate (i.e. at both ends) of the actin filaments. Addition of a plus-end capping protein suppresses the contribution of the plus end to the overall growth, leading to a slower growth and a higher overall Cc that is close to that of the minus end. 23. Answer: 42351

Difficulty: 3 Section: Actin and Actin-binding Proteins Feedback: Tropomyosin binds to several adjacent actin subunits along the actin filament and stabilizes the filament. α-Actinin cross-links actin filaments, commonly in an antiparallel fashion. The Arp 2/3 complex nucleates actin filaments and can remain bound at the minus end. CapZ binds to the plus end and prevents polymerization. The myosin motor moves on the actin filament. 24. Answer: B Difficulty: 2 Section: Actin and Actin-binding Proteins Feedback: Cofilin induces twisting of older actin filaments and destabilizes them. ATP hydrolysis by incorporated actin is usually slower than filament assembly; therefore the new actin filaments in the cell are more resistant to depolymerization brought about by cofilin. 25. Answer: C Difficulty: 2 Section: Actin and Actin-binding Proteins Feedback: The tight packing caused by the small bundling protein fimbrin apparently excludes myosin II from the bundle. 26. Answer: D Difficulty: 1 Section: Actin and Actin-binding Proteins Feedback: The Arp 2/3 complex creates a branched network of actin filaments and helps establish the structure of lamellipodia. 27. Answer: A Difficulty: 1 Section: Actin and Actin-binding Proteins Feedback: Spectrin is a long and flexible protein complex in which the two binding sites for actin are about 200 nm apart. This distance is about an order of magnitude greater than that of the other proteins listed. 28. Answer: E Difficulty: 1 Section: Actin and Actin-binding Proteins Feedback: Proteins of the ERM family play crucial roles including connecting the cortical actin cytoskeleton to the plasma membrane and to signaling proteins. 29. Answer: FFFF

Difficulty: 1 Section: Actin and Actin-binding Proteins Feedback: The bacterium recruits and activates the Arp 2/3 complex at its surface. The force generated by actin polymerization at the plus end pushes the bacterium, while cofilin and capping protein ensure continuous motion and prevent the depletion of free actin subunits by promoting filament turnover. 30. Answer: FFFFT Difficulty: 1 Section: Microtubules Feedback: Both plus-end directed and minus-end directed myosins and kinesins have been identified; however, all known dyneins move toward the minus end of microtubules. 31. Answer: C Difficulty: 3 Section: Myosin and Actin Feedback: Gliding occurs due to many individual steps taken by bound myosin heads in the presence of ATP. Since myosin II moves toward the plus end of actin filaments and is immobilized on the glass slide, the actin filaments will glide toward their minus end. 32. Answer: DCAB Difficulty: 2 Section: Myosin and Actin Feedback: This sequence of events moves the head on the actin filament. 33. Answer: A Difficulty: 3 Section: Myosin and Actin Feedback: The power stroke and tight binding to the actin filament require ATP hydrolysis and inorganic phosphate (Pi) release. However, the release of myosin from the actin filament only requires ATP binding. 34. Answer: C Difficulty: 3 Section: Myosin and Actin Feedback: Rigor mortis occurs after death due to the availability of myosin binding sites on actin as a result of elevated Ca2+. Myosins remain attached to actin once they bind and contract the sarcomeres. 35. Answer: 36% Difficulty: 3 Section: Myosin and Actin

Feedback: The work performed by one myosin head in each cycle is: W = F × d = 3 pN × 10 nm = 3 × 10–20 N.m = 3 × 10–20 J Multiplying this by Avogadro’s number gives 1.8 × 104 J/mol, which is equivalent to 36% of the free-energy change of ATP hydrolysis. 36. Answer: A Difficulty: 2 Section: Myosin and Actin Feedback: The dark band represents the thick myosin filaments and its length remains fairly constant, whereas the light band and the total sarcomere length shorten by several percent. 37. Answer: DF Difficulty: 2 Section: Myosin and Actin Feedback: The filament sliding and calcium-ion pumping carried out by these two ATPases require a large supply of ATP in these cells. 38. Answer: I Difficulty: 3 Section: Myosin and Actin Feedback: Increased sensitivity to Ca2+ may result in hypertrophy by increasing cardiac muscle contraction, which can be counteracted by drugs that block calcium channels thus reducing the concentration of calcium available to the troponin complex. 39. Answer: DEACB Difficulty: 2 Section: Myosin and Actin Feedback: This sequence of events results in a slow but sustained contraction in smooth muscle cells. 40. Answer: E Difficulty: 1 Section: Myosin and Actin Feedback: The troponin complex regulates the contraction of sarcomeres in skeletal and heart muscle cells. Non-muscle contractile bundles are regulated by myosin phosphorylation. 41. Answer: E Difficulty: 2 Section: Myosin and Actin

Feedback: The result shows that the unique insert is necessary for the minus-end directionality of myosin VI, but not necessary for its motor activity. The experiment does not address the sufficiency of the insert. 42. Answer: C Difficulty: 3 Section: Myosin and Actin Feedback: Myosin V is a processive motor and, not surprisingly, its duty ratio is significantly higher than that of myosin II. This is consistent with ADP release or ATP binding being slow (and rate-limiting) in the myosin V cycle, making the motor stay longer in the attached post-stroke state. In myosin II, inorganic phosphate (Pi) release is rate-limiting instead, keeping the motor in the released or cocked state for a longer fraction of the cycle. 43. Answer: A Difficulty: 3 Section: Microtubules Feedback: Thirteen protofilaments have a total of thirteen staggers between them. Thus, a full turn of the helix has a rise of approximately 12 nm (= 13 staggers × 0.9 nm/stagger). Since there is a seam, this rise should correspond to an odd number of monomer distances, i.e. 1, 3, 5, etc. In reality, the number is 3, equivalent to an axial tubulin displacement of 4 nm (= 12 nm/3). 44. Answer: D Difficulty: 2 Section: Microtubules Feedback: Catastrophe is the transition from growth to shrinkage, while rescue is the transition from shrinkage to growth. The transition frequencies are generally higher in more dynamic microtubules, which also tend to be shorter. 45. Answer: C Difficulty: 2 Section: Microtubules Feedback: Without a GTP cap, shrinking microtubules have curved protofilaments at their plus ends with weak lateral interactions. Rescue can also happen randomly without the help of any associated protein. 46. Answer: A Difficulty: 2 Section: Microtubules

Feedback: They both nucleate the filaments and remain bound to the minus end. They also both employ homologs of the filament subunits to catalyze nucleation. 47. Answer: B Difficulty: 1 Section: Microtubules Feedback: The centrosome is the major microtubule-organizing center (MTOC) in most animal cells. The γ-tubulin ring complexes associated with the centrosome nucleate microtubules that emanate from the center of the cell toward its periphery. 48. Answer: B Difficulty: 1 Section: Microtubules Feedback: XMAP215 behaves as a plus-end tracking protein (+TIP), and stabilizes the microtubules by binding near the plus end and lowering catastrophe frequency. 49. Answer: 4321 Difficulty: 1 Section: Microtubules Feedback: Kinesin-1, also called the “conventional kinesin”, is a plus-end directed microtubule motor. Katanin, named after the Japanese word for “sword” is an ATPdependent microtubule-severing protein. EB1 and kinesin-13 are +TIPs. The latter also acts as a catastrophe factor. 50. Answer: DDDMD Difficulty: 2 Section: Microtubules Feedback: Dyneins are fast, minus-end directed, microtubule-based motors that have large structures unrelated to those of kinesins (also microtubule-based) and myosins (actin-based) and are also faster than these other two motor proteins. 51. Answer: A Difficulty: 2 Section: Microtubules Feedback: In the mechanochemical cycle of kinesins, an ATP-bound head binds tightly to its microtubule track. Myosins release their actin filament once bound to ATP. 52. Answer: 1750 km/h Difficulty: 3 Section: Microtubules Feedback: The step frequency of kinesin-1 is about 125 steps per second [= (1000 nm/sec)/(8 nm/step)], while that of a runner is only about 3 steps per second [= (180

steps/min)/(60 sec/min)]. If the runner had the step frequency of the kinesin motor, they could run at 1750 km/h (= 42 km/h × 125/3). This is a supersonic speed, faster than bullets and regular passenger planes. 53. Answer: B Difficulty: 2 Section: Microtubules Feedback: Cytoplasmic dyneins rely on the dynactin complex to associate with their cargoes, including the Golgi components, which are held near the center of the cell by the action of these motors. 54. Answer: C Difficulty: 2 Section: Microtubules Feedback: In this graph, the traces with a negative slope and indicated by thicker lines represent anterograde movement toward the axon terminal mediated by kinesins; those traces with a positive slope represent retrograde motion mediated by dynein. The speed is inversely related to the slope in this graph, and is higher for the retrograde movement. 55. Answer: C Difficulty: 2 Section: Microtubules Feedback: Kinesins normally carry out the outward transport of melanosomes toward the plus end of microtubules at the cell periphery, by winning a tug-of-war against dyneins that would otherwise transport the pigments back to the center of the cell. The headless mutant inhibits kinesin, resulting in inhibition of the outward movement. 56. Answer: TFFTF Difficulty: 2 Section: Microtubules Feedback: Essentially all microtubules in axons are oriented with their plus end pointed at the axon terminal. Anterograde movements on these microtubules are exclusive to kinesins. In dendrites, however, the microtubules are of mixed polarity. The microtubules are too short to span the entire length of axons or even dendrites. 57. Answer: CCBC Difficulty: 2 Section: Microtubules Feedback: Cilia are found in large numbers on ciliated cells such as those lining our respiratory tract. They are relatively short and beat with a whiplike motion. Axoneme constitutes the structural core of both cilia and flagella.

58. Answer: B Difficulty: 2 Section: Microtubules Feedback: The dynein tail domain is bound to the A microtubule of one doublet, while the motor domain interacts with the B microtubule of the neighboring doublet. The former doublet is pushed toward its minus end (and the base of the flagellum) by the minus-end directed movement of the motor domain on the latter doublet. Proteins like nexin prevent such sliding, directing the force into a bending motion. 59. Answer: TFTT Difficulty: 2 Section: Microtubules Feedback: Primary cilia are nonmotile interphase structures with a very widespread distribution among various cells. They are thought to have sensory and signaling functions. 60. Answer: III Difficulty: 2 Section: Intermediate Filaments and Septins Feedback: The highly resilient intermediate filaments have smaller persistence length compared to other cytoskeletal filaments, and are not polar. 61. Answer: B Difficulty: 1 Section: Intermediate Filaments and Septins Feedback: Keratins give cells mechanical strength, and provide tough coverings for animals. Every keratin filament is composed of an equal mix of type I (acidic) and type II (neutral/basic) keratins. These filaments are also involved in anchoring the cell cytoskeleton at desmosomes and hemidesmosomes. 62. Answer: D Difficulty: 1 Section: Intermediate Filaments and Septins Feedback: Cell rupture caused by mechanical trauma, and disorganization or clumping of the keratin filaments, is the hallmark of diseases (such as cornea diseases and skin blistering) caused by mutations in keratins and their associated proteins. 63. Answer: C Difficulty: 1 Section: Intermediate Filaments and Septins

Feedback: Both of these filament types are found in an axon and are critical for its structure and function. 64. Answer: E Difficulty: 1 Section: Intermediate Filaments and Septins Feedback: They are septins. 65. Answer: E Difficulty: 2 Section: Intermediate Filaments and Septins Feedback: SUN proteins in the inner nuclear membrane bind to KASH proteins in the outer nuclear membranes. Through this interaction, nuclear lamina and chromosomes (which bind to SUN proteins) are connected to actin filaments, microtubule motors, or plectins (which bind to KASH proteins) in the cytoplasm. 66. Answer: E Difficulty: 1 Section: Cell Polarization and Migration Feedback: Sperm cell motility is achieved through flagellum-mediated swimming. 67. Answer: 1321 Difficulty: 2 Section: Cell Polarization and Migration Feedback: Filopodia and stress fibers are linear structures, lamellipodia form flat, crosslinked networks, while invadopodia protrude across tissue barriers as footlike protrusions. 68. Answer: D Difficulty: 1 Section: Cell Polarization and Migration Feedback: Blebbing is mostly observed in vivo where the substratum is pliable. Both lamellipodia formation and blebbing require the activity of myosin II at some point. One of the key differences between the two is in their requirements for membrane interaction with cortical actin filaments. 69. Answer: D Difficulty: 3 Section: Cell Polarization and Migration Feedback: The actin filaments at the leading edge form a web, the whole of which undergoes treadmilling. An incorporated actin in this web remains more or less immobile relative to the substratum, but is constantly moving away from the polymerizing plus end (and the membrane) and getting closer to the depolymerizing minus end.

70. Answer: C Difficulty: 2 Section: Cell Polarization and Migration Feedback: The lamellipodium forms at the front of the cell toward where it is migrating. The cell is about 20 µm across and can move as fast as 30 µm/sec. 71. Answer: D Difficulty: 2 Section: Cell Polarization and Migration Feedback: With the loss of the stress fibers and focal adhesions as a result of the depletion of the myosin isoform, the traction forces on the substratum are expected to decrease compared to normal cells. 72. Answer: D Difficulty: 2 Section: Cell Polarization and Migration Feedback: The Arp 2/3 complex nucleates a two-dimensional web of actin filaments in a lamellipodium. Behind the leading edge, cofilin disassembles older filaments which contain ADP-actin. 73. Answer: HCH Difficulty: 2 Section: Cell Polarization and Migration Feedback: The Rho family proteins regulate the actin cytoskeleton and cell polarization in different ways by acting on different but overlapping sets of target proteins. 74. Answer: C Difficulty: 2 Section: Cell Polarization and Migration Feedback: At the front of the cell (toward the source of the bacterial peptides), the Rac pathway dominates and limits the activity of the Rho GTPase to the other end of the cell. 75. Answer: BADC Difficulty: 2 Section: Cell Polarization and Migration Feedback: Rac-GTP (B) promotes lamellipodium formation (D) by activating WASp family members, which leads to the nucleation of branched actin networks by the Arp 2/3 complex, as well as by activating filamin, a gel-forming actin cross-linker. Rho-GTP (A), on the other hand, activates formin to create parallel actin bundles, inhibits cofilin to stabilize actin filaments, and increases myosin motor activity, which all lead to formation of more stress fibers (C).

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION Chapter 17: THE CELL CYCLE Copyright © 2015 by W.W. Norton & Company, Inc. 1. Which of the following simplified diagrams better shows the changes in the concentrations of three major cyclin–Cdk complexes (G1/S-Cdk, S-Cdk, and M-Cdk) in the cell in different stages of the cell cycle? G1

2. Resveratrol is a natural compound found in red grapes (and red wine) and is thought to have beneficial effects in mammals, such as preventing tumor growth and delaying age-related diseases. In vitro, resveratrol and its derivatives have been shown to cause cell-cycle arrest in S phase and induce apoptosis. You have analyzed the DNA content of cultured cells in the presence and absence of these drugs using fluorescence-activated cell sorting. Compared with control cells (dashed line), which of the following curves do you think better represents the DNA content of cells treated with these compounds?

Number of cells

Relative cellular DNA content

Number of cells

Relative cellular DNA content

Reference: FACS Data Analysis (Questions 3-4) You have been studying the effect of loss-of-function mutations in the Cdk inhibitor protein (CKI) p21. You add the drug fucoxanthin to cell cultures harboring either wild-type or mutant

versions of the p21 gene. Fucoxanthin is known to induce cell-cycle arrest in G1. After a day, you add the thymidine analog BrdU to the culture media, collect the cells after an hour, treat them with anti-BrdU antibody and the fluorescent DNA stain DAPI, and finally subject them to fluorescence-activated cell sorting (FACS). The FACS data can be viewed as a two-dimensional dot plot composed of thousands of dots, in which each cell is represented by one dot at coordinates that correspond to the intensities of the DAPI fluorescence signal (X axis) and BrdU fluorescence signal (Y axis) for that cell. Answer the following question(s) according to the simplified dot plot below, generated from your experiment. b

2 DAPI signal intensity

BrdU signal intensity

2 DAPI signal intensity

3. Indicate which boxed region (1, 2, or 3) in the FACS plots corresponds better to each of the following phases of the cell cycle. Your answer would be a three-digit number composed of numbers 1 to 3, with each number used once, e.g. 312. ( ) G1 phase ( ) S phase ( ) G2 and M phases

4. Which one of the FACS plots (a or b) would you expect to correspond to the loss-offunction p21 mutants? Write down a or b as your answer.

5. Sort the following schematic diagrams (A to F) to reflect the order of events in a typical eukaryotic M phase. An interphase cell in G2 phase is drawn on the left for comparison. Your answer would be a six-letter string composed of letters A to F only, e.g. BCEADF.

G2 phase

Consider two mammalian cells, one in G1 and the other in G0 (stationary) phase. If they

are stimulated to pass the restriction point by the addition of an extracellular proliferation signal, but the signal is then immediately removed, what would you expect to happen? A. Both cells will replicate their DNA. B. Only the G1 cell will replicate its DNA. C. Only the G0 cell will replicate its DNA. D. Only the G1 cell will start to replicate its DNA, but will stop halfway through the replication and will not reach G2. E. Neither of the cells will replicate their DNA. 7. Which of the time points (A to E) in the following schematic drawing of the mammalian cell cycle represents the restriction point? E C

A M

G2 G1

Answer: B Difficulty: 1 Section: The Cell-Cycle Control System Feedback: Under favorable conditions and in the presence of signals to grow and divide, cells in early G1 (or G0) progress through a commitment point near the end of G1 known as Start (in yeasts) or the restriction point (in mammalian cells). 8. Indicate whether each of the following descriptions better applies to a Wee1 protein (W) or a Cdc25 protein (C). Your answer would be a four-letter string composed of letters W and C only, e.g. WWCW. ( ) It is a protein kinase. ( ) It activates M-Cdk complexes. ( ) It is activated by M-Cdk complexes. ( ) Its loss in fission yeast results in small cell size. 9. Mammalian Cdk inhibitor proteins (CKIs) can be grouped into two families based on their structural and functional differences. The Cip/Kip family proteins (e.g. p21) have a broad binding specificity. These proteins bind preferentially to already formed cyclin–Cdk complexes and thus enhance complex formation. However, they inhibit the kinase activity of most complexes (e.g. S-Cdks), except in the case of G1-Cdk complexes where no inhibition occurs. Consequently, Cip/Kip family proteins have an overall positive effect on Cdk4/6 activity due to their help in bringing the subunits together. In contrast, the inhibitors of the INK4 family (e.g. p16) bind only to the Cdk subunit of G1-Cdks and prevent binding of both the G1 cyclins and the Cip/Kip family CKIs. Based solely on these findings, would you expect p16 to activate (A) or inactivate (I) the S-Cdks in the presence of limited amounts of p21? 10. In the fission yeast Schizosaccharomyces pombe, the proteins Cut1 and Cut2 form a complex that is catalytically inactive. At the onset of anaphase, Cut2 is polyubiquitylated by a large E3 complex containing Cut4, Cut9, Cut23, and other proteins, and is subsequently destroyed. Cut1 then cleaves Rad21, a non-SMC component of a complex that also contains two SMC proteins, thus allowing sister-chromatid separation. Mutations in the cut genes lead to the cut phenotype, in which the cell attempts cytokinesis without chromosome segregation. Indicate which of the following proteins or protein complexes corresponds to or contains the product of the genes cut1 (A), cut2 (B), cut4 (C), and rad21 (D). Your answer would be a four-letter string composed of letters A to D only, e.g. DCAB. ( ) Securin ( ) Cohesin ( ) Anaphase-promoting complex

( ) Separase 11. The following schematic diagram shows the activation of M-Cdk in mitosis. Indicate which proteins below correspond to those indicated as A to D in the diagram. Your answer would be a four-letter string composed of letters A to D only, e.g. BACD.

D A B

( ) Wee1 ( ) CAK ( ) Cdc25 ( ) M-cyclin 12. Indicate whether each of the following occurs mainly in G1 phase (G), S phase (S), or G2 phase (H) of the cell cycle. Your answer would be a four-letter string composed of letters G, S, and H only, e.g. HGSS. ( ) DNA helicase activation ( ) DNA helicase deposition on DNA at the replication origins ( ) ORC phosphorylation ( ) Licensing of replication origins 13.

Which of the following events occurs in mitotic metaphase? A. Nuclear envelope breakdown B. Nuclear envelope reassembly C. Chromosome attachment to spindle microtubules for the first time D. Chromosome alignment at the spindle equator

E. Mitotic spindle assembly 14. Consider two kinesin motor proteins at the mitotic spindle midzone: kinesin-5 is a tetrameric motor that walks toward the plus end of both microtubules to which it is attached via its motor domains; kinesin-14, on the other hand, walks toward the minus end of one microtubule while it is attached to another microtubule via its tail domain. How do these motors affect the length of the spindle? A. They both work to shorten the spindle. B. Kinesin-5 works to shorten the spindle whereas kinesin-14 works to lengthen it. C. Kinesin-5 works to lengthen the spindle whereas kinesin-14 works to shorten it. D. They both work to lengthen the spindle. 15.

How is centrosome duplication similar to DNA replication? A. They both use a semiconservative mechanism. B. They are initiated at around the same time in the cell cycle, near the G1/S transition. C. They are both controlled in such a way that they replicate once and only once per cell cycle. D. They are both separated from their sister copies in mitosis. E. All of the above.

16.

Which of the following is more directly driven by active M-Cdk? A. Centrosome maturation B. Centrosome duplication C. Nuclear envelope reassembly D. Inactivation of APC/C E. Cell cleavage

17.

Indicate true (T) and false (F) statements below regarding mitosis and the changes that it

brings about compared to interphase. Your answer would be a four-letter string composed of letters T and F only, e.g. TFTF. ( ) Microtubules become greatly stabilized in mitosis compared to interphase. ( ) The number of γ-tubulin ring complexes in the centrosomes increases greatly during mitosis compared to interphase. ( ) Chromosomes stabilize mitotic microtubules through activation of Ran monomeric G protein. ( ) A bipolar spindle can be formed even in the absence of centrosomes.

18. Fill in the blank: “The … is a large structure formed at the centromeric region of each eukaryotic chromosome. It captures spindle microtubules in mitosis, and therefore serves to attach the chromosomes to the spindle poles.” 19. Which one of the following chromosomes has formed stable attachment(s) to the spindle microtubules?

20. The cell cycle can be arrested in mitosis when a single sister-chromatid pair is monooriented on the mitotic spindle. If at this point a glass microneedle is used to pull the monooriented chromosome toward the pole to which it is not attached, the cell proceeds to anaphase. This observation confirms that … A. mono-oriented chromosomes are stable. B. mechanical tension drives the formation of bi-oriented chromosomes. C. bi-oriented chromosomes activate the spindle assembly checkpoint. D. lack of mechanical tension at the kinetochore in at least one chromosome prevents entry into anaphase. E. mechanical tension at the kinetochore in at least one chromosome is required for entry into anaphase.

21. Indicate whether each of the following phosphorylation events typically activates (A) or inactivates (I) the protein that is being phosphorylated. Your answer would be a four-letter string composed of letters A and I only, e.g. IIIA. ( ) Phosphorylation of Cdc25 by M-Cdk ( ) Phosphorylation of condensin subunits by M-Cdk ( ) Phosphorylation of kinesin-5 by Aurora-A ( ) Phosphorylation of Ndc80 subunits by Aurora-B 22. In the following schematic drawing of a vertebrate prometaphase chromosome as seen under a microscope, indicate whether the chromosome is more likely to be closer to pole A or pole B. Write down A or B as your answer.

Toward pole A

Toward pole B

Reference: Chromosome Movement in Anaphase (Questions 23-24) To study chromosome movement during anaphase in mammals, you have injected fluorescently labeled tubulin subunits into cells such that microtubules can be seen under a fluorescence microscope. You then use a relatively strong laser beam to bleach the fluorescent dyes in a limited area of a metaphase cell, as indicated in the schematic diagram below, and follow the progression of mitosis under the microscope by time-lapse imaging. You then use the images to measure the change in distance between various components, and plot the results in the graph below. Answer the following question(s) based on these results.

Pole A

Chromosomes

Pole B

Bleached box

Distance (µm)

Distances: Pole A – pole B Chromosome – pole B Bleached box – pole B

0 0

Time after anaphase entry (min)

23. Based on the results from the experiment, which is dominant in this cell: anaphase A or anaphase B? Write down A or B as your answer. 24. Based on the results from your experiment, which force is dominant in chromosome movement in this cell: polar ejection force (E), microtubule flux (F), or kinetochore microtubule plus-end depolymerization (P)? Write down E, F, or P as your answer. 25. Imagine a prometaphase chromosome pair that is attached to one spindle pole. Which of the following would happen if both arms of the chromosome are severed with a strong laser beam? A. All chromosome fragments (the centromere-containing fragment and the arm fragments) would be pushed away from the pole.

B. Only the two arm fragments would be pulled toward the pole. The kinetochorecontaining fragment would remain stationary. C. The kinetochore-containing fragment would be pulled toward the pole. The two arm fragments would move away from the pole. D. The kinetochore-containing fragment would move away from the pole, but the two arm fragments would be pulled toward the pole. E. All chromosome fragments would be pulled toward the pole. 26. If cells that have started mitosis are treated with nocodazole, a drug that depolymerizes microtubules, what would you expect to happen? Where would you expect Mad2 protein to be localized? A. The cells would arrest in prometaphase; Mad2 would be localized to the spindle poles. B. The cells would arrest in telophase; Mad2 would be localized to the spindle poles. C. The cells would immediately enter anaphase, finish mitosis, and enter G1; Mad2 would be localized to the spindle poles. D. The cells would arrest in prometaphase; Mad2 would be localized to almost all kinetochores. E. The cells would arrest in telophase; Mad2 would be localized to almost all kinetochores. 27. Treatment of dividing cells with a low dose of the antifungal drug benomyl, which destabilizes microtubules, slows down correct spindle assembly. But at such doses, the spindle is eventually formed and the cells survive. However, mutations in some genes confer benomyl sensitivity: the mutant cells die because they fail to arrest the cell cycle in the presence of unattached kinetochores and progress through anaphase, with disastrous consequences. Which of the following would you expect to be a benomyl-sensitive mutant? A. A loss-of-function mutant in the gene encoding Mad2. B. A mutation causing the overexpression of Cdc20. C. A loss-of-function mutation in the gene encoding a kinase that inhibits Cdc20– APC/C. D. A loss-of-function mutation in the gene encoding a tubulin subunit. E. All of the above. 28.

Which of the following motor proteins are more directly involved in anaphase B? A. Kinesin-5 on interpolar microtubules and dynein on kinetochore microtubules

B. Kinesin-13 on kinetochore microtubules and dynein on astral microtubules C. Kinesins 4 and 10 on interpolar and astral microtubules and dynein on kinetochore microtubules D. Kinesin-5 on interpolar microtubules and kinesins 4 and 10 on interpolar and astral microtubules E. Kinesin-5 on interpolar microtubules and dynein on astral microtubules 29. The microtubule-binding protein Patronin binds to the minus ends of microtubules at the spindle pole and protects them from the effect of catastrophe factors. If Patronin activity is inhibited by injecting an anti-Patronin antibody into Drosophila melanogaster embryos prior to cellularization, anaphase B is suppressed and the spindles are significantly shorter. Which of the forces (1 or 2) in the following schematic diagram do you think is mostly responsible for anaphase B in this organism at this stage? Does the effect of Patronin inhibition resemble that of kinesin-5 inhibition (I) or overactivation (O)? 1 2

1 2

A. 1; I B. 1; O C. 2; I D. 2; O 30.

Mutant Saccharomyces cerevisiae cells that lack the gene encoding securin can divide

more or less normally by mitosis, without significant chromosome segregation defects. Cells harboring a nondegradable version of securin, on the other hand, arrest in metaphase as expected, since they cannot activate separase to enter anaphase. Similarly, cells lacking Cdc20 arrest in metaphase, since they cannot activate APC/C. Finally, cells lacking both securin and Cdc20 arrest in anaphase: they manage to separate sister chromatids, but do not progress much further. These results suggest that in wild-type cells, … A. degradation of securin is necessary to trigger sister-chromatid separation. B. degradation of securin is sufficient to trigger sister-chromatid separation.

C. Cdc20–APC/C is NOT necessary for sister-chromatid separation. D. Cdc20–APC/C is NOT necessary for later events in anaphase. E. All of the above. 31. Indicate true (T) and false (F) statements below regarding cytokinesis in animal cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) The force for cytokinesis is generated by kinesin motors on microtubule bundles that form the contractile ring. ( ) As the contractile ring constricts, its thickness increases to keep a constant volume. ( ) The midbody forms from bundles of actin and myosin II. ( ) Local activation of Ran GTPase triggers the assembly and contraction of the contractile ring. 32. Formin nucleates the growth of parallel actin bundles in the cell. Additionally, myosin motor activity is positively regulated by phosphorylation. The monomeric G protein RhoA is important in cytokinesis, because it directly or indirectly … A. activates formins and inactivates myosin light-chain phosphatase. B. inactivates formins and activates myosin light-chain kinases. C. activates formins as well as myosin light-chain phosphatases. D. inactivates formins as well as myosin light-chain kinases. E. None of the above. Reference: Contractile-Ring Positioning (Questions 33-35) Three models for contractile-ring positioning in animal cells are presented in the schematic diagrams below. Answer the following question(s) according to these models.

Model 1

Model 2

Model 3

33. In classic experiments carried out in the 1960s and schematized below, an egg of the sand dollar Echinarachnius parma was made into a torus shape using a glass bead on the end of a needle, to study the influence of microtubules on cleavage-furrow positioning. The cells were allowed to undergo two rounds of mitosis. After the second mitosis, an extra furrow (indicated by a red arrowhead) was observed between the two asters that did not have a spindle between them. Which of the models (1 to 3) is more consistent with this observation? Write down 1, 2, or 3 as your answer.

34.

In experiments on the flattened eggs of the sea urchin Tripneustes gratilla, cytokinesis

was analyzed after placing a small physical barrier (e.g. a glass needle or an oil droplet) between the spindle and various areas of the cell cortex. Blocks that were located over the equatorial plane prevented furrow formation in the proximal membrane, whereas blocks located in other areas did not affect furrowing. Which model (1 to 3) is NOT supported by these results? Write down 1, 2, or 3 as your answer.

Flattened egg before cytokinesis

Normal furrow

Furrow after insertion of a needle perpendicular to the equatorial plane

Furrow after insertion of a needle parallel to the equatorial plane

35. In the early embryos of Caenorhabditis elegans, defects in the formation of astral microtubules increase myosin activity throughout the cell cortex. If the spindle is forced to one side of the cell, the cortical myosin activity is observed mostly at the opposite side of the cell. Which model (1, 2, or 3) better predicts these observations? Write down 1, 2, or 3 as your answer. 36. Indicate true (T) and false (F) statements below regarding cytokinesis in eukaryotic cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) The preprophase band, composed of microtubules and actin filaments, marks the site of cytokinesis in plant cells. ( ) In some cell types, the site of contractile-ring formation is determined before mitosis. ( ) The early cell plate in dividing plant cells appears after phragmoplast formation. ( ) The membrane required for the newly formed cell plate in plant cells is provided by endocytosis from the equatorial plasma membrane.

37. In most mammalian cells, low M-cyclin protein levels are maintained during G1 phase. What is mainly responsible for maintaining these low levels? A. Cdc20–APC/C B. Cdh1–APC/C C. Skp2–SCF D. β-trCP–SCF E. p27 38. In the following schematic diagram of an early Drosophila embryo, which step (A to C) corresponds to cellularization? Write down A, B, or C as your answer.

Fertilized egg

39. Indicate whether each of the following, when active, tends to activate (A) or inactivate (I) M-Cdks. Your answer would be a four-letter string composed of letters A and I only, e.g. IIAI. ( ) M-Cdk ( ) Cdc20–APC/C ( ) Cdh1–APC/C ( ) Sic1 40.

How is Cdc20–APC/C similar to Cdh1–APC/C? A. They are both active throughout interphase. B. They are both inhibited by M-Cdk. C. They both inhibit M-Cdk activity. D. They are both activated suddenly at the onset of mitosis. E. They are both inactivated soon after anaphase.

41. Which division in meiosis is more similar to mitosis? In which division do sister chromatids separate from each other? A. Meiosis I; meiosis I B. Meiosis I; meiosis II C. Meiosis II; meiosis I D. Meiosis II; meiosis II 42.

Which stage is usually the longest in meiosis? A. Prophase I B. Prometaphase I C. Telophase I D. Prophase II E. Prometaphase II

43. Indicate true (T) and false (F) statements below regarding meiosis in eukaryotic cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) Bivalents form before prophase I. ( ) Crossing-over begins before the synaptonemal complex assembly. ( ) Chiasmata can first be seen under the microscope after the disassembly of the synaptonemal complexes. ( ) All recombination events lead to crossovers. 44. In the following schematic diagram of a meiotic bivalent in diplotene, indicate whether each of the following chromatid pairs have undergone DNA exchange (E) or not (N). The different color of the final form of chromatid 3 is for clarity only. Your answer would be a fourletter string composed of letters E and N only, e.g. NENE.

1 2 3 4 ( ) Chromatids 1 and 3 ( ) Chromatids 1 and 4 ( ) Chromatids 2 and 3 ( ) Chromatids 2 and 4 45. Loss of the gene encoding shugoshin in many multicellular organisms leads to sterility, suggesting defects in meiosis. What would you expect to occur in meiotic cells lacking shugoshin? A. The homologs fail to separate in anaphase I. B. The sister chromatids fail to separate in anaphase II. C. All chromatids separate prematurely in anaphase I. D. Removal of cohesion between homolog arms fails in prophase I. E. Removal of cohesion between sister chromatids fails in prophase II. 46. Indicate whether each of the following descriptions better applies to mitogens (M), growth factors (G), or survival factors (S). Your answer would be a four-letter string composed of letters M, G, or S only, e.g. GMSG. ( ) They unblock cell-cycle progression ( ) They suppress apoptosis ( ) They trigger a wave of G1/S-Cdk activity ( ) They inhibit the degradation of cellular macromolecules

47. Which one better supports cell proliferation when added to fibroblast cultures: serum or plasma? This activity is due to the presence of mitogens in this fluid. What is responsible for making these mitogens? A. Serum; red blood cells B. Serum; platelets C. Plasma; red blood cells D. Plasma; white blood cells E. Plasma; platelets 48.

Which of the following is an inhibitory extracellular signal for cell proliferation? A. EGF B. TGFβ C. Erythropoietin D. IGF E. PDGF

49.

Which of the following cell populations in our body has the highest mitotic index? A. Neurons B. Hepatocytes C. Red blood cells D. Fibroblasts E. Skeletal myocytes

50. Which of the following proteins is the product of an immediate early gene expressed following mitogenic stimulation of cell-cycle entry? A. E2F B. Rb C. Myc D. G1-cyclins E. All of the above 51. Which of the following events occurs at the onset of S phase in a mitogen-stimulated vertebrate cell? A. Activation of Cdh1–APC/C B. Activation of Rb C. Activation of geminin

D. Activation of Cdc20 E. All of the above 52. Which of the following events contributes to driving the mammalian cell past the restriction point of the cell cycle? A. Phosphorylation of Cdh1–APC/C by G1/S-Cdk B. Destruction of CKIs that target S-Cdks C. Phosphorylation of Rb by G1-Cdk, G1/S-Cdk, and S-Cdk D. Activation of E2F gene expression by active E2F protein E. All of the above 53. In the following schematic diagram of a typical eukaryotic cell cycle, choose two major time points (among A to E) at which the cell-cycle control system normally arrests the cycle if DNA damage is detected. Your answer would be a two-letter string composed of letters A to E only in alphabetical order, e.g. CE. E C

A M

G2 G1

54. A cell has been subjected to ultraviolet irradiation, causing a significant number of mutations in the genome. Which of the following would you NOT expect to occur as a result? A. Activation of the protein kinase ATR B. Activation of the protein kinase Chk1 C. Inactivation of the protein phosphatase Cdc25 D. Binding of p53 to Mdm2

E. Stabilization of p53 55. In a multicellular organism such as a mammal, loss-of-function mutations in many genes can contribute to the development of cancer. These genes are therefore called tumor suppressors. In their absence, the cell fails to stop progression through the cell cycle under conditions in which normal cells would arrest, paving the way for tumorigenesis. Indicate whether each of the following proteins is (Y) or is not (N) expected to be the product of a tumor suppressor gene based on its function in the cell cycle. Your answer would be a four-letter string composed of letters Y and N only, e.g. YNNN. ( ) Rb ( ) Myc ( ) p53 ( ) p21 56. Indicate true (T) and false (F) statements below regarding cellular control of growth and division. Your answer would be a four-letter string composed of letters T and F only, e.g. TFTF. ( ) Replicative cell senescence usually arises due to the accumulation of mutations in genes encoding S- and M-cyclins. ( ) Replicative cell senescence is caused by the induction of apoptosis by p53. ( ) Excessive mitogenic stimulation can result in Mdm2 activation and the induction of apoptosis or cell-cycle arrest. ( ) The Mdm2 inhibitor Arf induces cell-cycle progression. 57. You have synchronized a culture of HeLa cells so that the cells are all at the same stage in the cell cycle. You then treat the cells with serum containing nutrients and growth factors in the presence or absence of the drug rapamycin. The distribution of cell sizes in the two cell populations is shown below. Based on these results, do you think rapamycin is an activator or inhibitor of the TOR complex? Is the S6 kinase expected to be up-regulated or down-regulated in the presence of rapamycin?

Cell number

Before serum addition 5 hours after serum addition with rapamycin 5 hours after serum addition without rapamycin

Cell size A. Activator; up-regulated B. Activator; down-regulated C. Inhibitor; up-regulated D. Inhibitor; down-regulated

Answers 1. Answer: A Difficulty: 2 Section: The Cell-Cycle Control System Feedback: The G1/S-Cdk activity (solid black curve) peaks in late G1 and falls in S phase. The S-Cdk levels (solid gray curve) remain elevated through S and G2 until early M phase. M-Cdk (dashed gray curve) accumulates before entry into mitosis, but their level falls in mid-mitosis. 2. Answer: B Difficulty: 3 Section: The Cell-Cycle Control System Feedback: S-phase arrest will increase the proportion of cells in this phase; that is, cells with a relative cellular DNA content of between 1 and 2. 3. Answer: 213 Difficulty: 3 Section: The Cell-Cycle Control System Feedback: After a short incubation with BrdU, cells that are replicating their DNA (in S phase) would incorporate the compound and would show a high BrdU signal compared to cells that are in G1, G2, or M phases. The DAPI signal intensity represents the total DNA content, which increases from G1 to S to G2 phases. 4. Answer: a Difficulty: 3 Section: The Cell-Cycle Control System Feedback: Given the significant difference between the results from wild-type and p21 mutant cells, fucoxanthin appears to arrest the cell cycle (in G1) by activating a p21dependent pathway. Loss of p21 (plot a) therefore blocks the cell-cycle arrest induced by the drug. 5. Answer: CFEDAB Difficulty: 2 Section: Overview of the Cell Cycle Feedback: The M phase includes mitotic prophase (C), prometaphase (F), metaphase (E), anaphase (D), telophase (A), and finally cytokinesis (leading to B). 6. Answer: A Difficulty: 1 Section: Overview of the Cell Cycle Feedback: Once a cell passes the restriction point, it is committed to DNA replication (and cell division) even if the stimulating signals are removed. 7. Answer: B Difficulty: 1 Section: The Cell-Cycle Control System Feedback: Under favorable conditions and in the presence of signals to grow and divide, cells in early G1 (or G0) progress through a commitment point near the end of G1 known as Start (in yeasts) or the restriction point (in mammalian cells). 8. Answer: WCCW Difficulty: 2

Section: The Cell-Cycle Control System Feedback: Wee1 kinases phosphorylate and inactivate cyclin–Cdk complexes; Cdc25 phosphatases do the opposite. In a positive feedback loop at the onset of mitosis, M-Cdk inactivates its inactivator, Wee1, and activates its activator, a Cdc25. The small cell size in wee mutants can result from Wee1 depletion. 9. Answer: I Difficulty: 3 Section: The Cell-Cycle Control System Feedback: In the absence of p16, p21 is sequestered by the G1-Cdk complex. Active p16, however, prevents p21 from binding, making it available to inhibit the S-Cdk complexes. 10. Answer: BDCA Difficulty: 2 Section: S Phase Feedback: Rad21 is a Scc1 homolog, which is cleaved by separase in the metaphase– anaphase transition. 11. Answer: CBDA Difficulty: 2 Section: Mitosis Feedback: Activation of M-Cdk is sharpened by positive feedback loops in which M-Cdk activates its activator (Cdc25) and inactivates its inactivator (Wee1) by phosphorylation. 12. Answer: SGSG Difficulty: 2 Section: S Phase Feedback: The loading of inactive DNA helicases onto the replication origins and the formation of the prereplicative complex (preRC) occur in late mitosis and early G1. In S phase, the helicases are activated to initiate DNA replication, while the origin is prevented from refiring by origin recognition complex (ORC) phosphorylation and other mechanisms. 13. Answer: D Difficulty: 1 Section: Mitosis Feedback: The alignment of chromosomes at the cell equator is the hallmark of metaphase. 14. Answer: C Difficulty: 2 Section: Mitosis Feedback: While kinesin-5 pushes the two poles of the spindle apart by walking toward the plus end of antiparallel microtubules, kinesin-14 pulls the two poles toward each other by walking toward the minus end of interpolar microtubules in the midzone. 15. Answer: E Difficulty: 1 Section: Mitosis Feedback: The centrosomes, like the chromosomes, are duplicated only once during each cell cycle. Centrosome duplication is triggered by the activation of G1/S-Cdk. Both

centrosome duplication and DNA replication employ a semiconservative mechanism. The two new centrosomes remain together on one side of the cell until entry into mitosis. 16. Answer: A Difficulty: 1 Section: Mitosis Feedback: The M-Cdk is responsible for most events in early mitosis, including spindle assembly and centrosome maturation. It is inactivated later in mitosis, resulting in the events of telophase and cytokinesis. 17. Answer: FTTT Difficulty: 2 Section: Mitosis Feedback: Mitosis is accompanied by a dramatic reorganization of the microtubule cytoskeleton, with a significant increase in microtubule instability. At the same time, centrosome maturation greatly increases microtubule nucleation. The capture of microtubules by chromosomes stabilizes microtubules in part by activating Ran, a monomeric GTPase. The ability of mitotic chromosomes to stabilize and organize microtubules makes it possible to form bipolar spindles even in the absence of centrosomes. 18. Answer: kinetochore Difficulty: 1 Section: Mitosis Feedback: Kinetochores are giant protein structures that attach sister chromatids to the mitotic spindle. 19. Answer: C Difficulty: 1 Section: Mitosis Feedback: Bi-oriented chromosomes are stabilized due to the unique tension created. 20. Answer: D Difficulty: 2 Section: Mitosis Feedback: This important experiment confirms the importance of mechanical tension within the kinetochores in the spindle assembly checkpoint mechanism. 21. Answer: AAAI Difficulty: 2 Section: Mitosis Feedback: M-Cdk activates the Cdc25 phosphatase in a positive feedback loop. It also phosphorylates condensin subunits, stimulating DNA binding and supercoiling activity. Along with Aurora-A, it also phosphorylates kinesins such as kinesin-5, stimulating microtubule binding and motor activity. Without tension in the kinetochore, Aurora-B phosphorylates and inactivates components of Ndc80, lowering its affinity for microtubule plus ends. 22. Answer: A Difficulty: 2 Section: Mitosis

Feedback: The polar wind, mediated by plus-end directed kinesin motors, gets stronger as the chromosome approaches a pole, and pushes the chromosome arms toward the opposite pole. 23. Answer: A Difficulty: 2 Feedback: In this example, the two poles are pushed away from each other only slightly during anaphase. 24. Answer: F Difficulty: 3 Feedback: In these cells, the chromosomes seem to approach the nearby pole as fast as the bleached zone approaches the pole, meaning that the microtubule subunit flux is enough to explain the poleward chromosome movement. 25. Answer: C Difficulty: 2 Section: Mitosis Feedback: The result of such an experiment has provided evidence for polar ejection forces acting on chromosome arms. 26. Answer: D Difficulty: 2 Section: Mitosis Feedback: A cell treated with nocodazole fails to assemble a functional spindle, and therefore arrests in prometaphase with the chromosomes left unattached. Mad2 is recruited to unattached kinetochores and activates the spindle assembly checkpoint, preventing the onset of anaphase. 27. Answer: E Difficulty: 3 Section: Mitosis Feedback: All of these mutations make the cells more sensitive to the benzimidazole drug. 28. Answer: E Difficulty: 1 Section: Mitosis Feedback: Two major forces act in anaphase B. Kinesin-5 pushes the two poles away from each other, and dynein pulls the poles toward the cell periphery. 29. Answer: A Difficulty: 3 Section: Mitosis Feedback: The force generated by kinesin-5 to lengthen the spindle in anaphase B is counteracted by catastrophe factors at the poles, which are inhibited by Patronin. Inhibition of Patronin would therefore result in anaphase B suppression. 30. Answer: A Difficulty: 3 Section: Mitosis Feedback: Both Cdc20 activation and securin destruction are necessary for sisterchromatid separation because cells in which either of these fail to occur are observed to

arrest in metaphase. Securin degradation does not appear to be sufficient for sisterchromatid separation, as the cells lacking securin are more or less healthy, meaning other mechanisms also control the separation. Cdc20 appears to be necessary for anaphase beyond sister-chromatid separation, since the double knockouts still arrest in anaphase despite lacking securin. 31. Answer: FFFF Difficulty: 1 Section: Cytokinesis Feedback: The contractile ring is based on actin and myosin II, constricts in a dynamic fashion, and gradually disassembles. RhoA GTPase helps assemble this structure. 32. Answer: A Difficulty: 1 Section: Cytokinesis Feedback: This G protein activates formin as well as multiple protein kinases, including the Rho-activated kinase Rock. Rock phosphorylates regulatory myosin light chains and also inhibits myosin light-chain phosphatase by phosphorylation. 33. Answer: 1 Difficulty: 2 Section: Cytokinesis Feedback: The astral stimulation model (1) predicts this outcome. The central spindle stimulation model (2) predicts the formation of two furrows when there are two spindles. However, it is also likely that a midzone-like structure formed between the two spindles in these experiments, and so model 2 could also be correct here. The astral relaxation model (3) predicts the formation of two furrows in the first division. 34. Answer: 3 Difficulty: 2 Section: Cytokinesis Feedback: The astral relaxation model (3) cannot explain these results. 35. Answer: 3 Difficulty: 2 Section: Cytokinesis Feedback: These observations are in better agreement with the astral relaxation model (3). 36. Answer: TTTF Difficulty: 2 Section: Mitosis Feedback: In plant cells, the preprophase band assembles just before mitosis and marks the site of cytokinesis. The assembly of the cell plate between the two daughter cells is then guided by phragmoplast. The membrane required for the formation of cell plate comes from the Golgi apparatus. 37. Answer: B Difficulty: 1 Section: Mitosis Feedback: Cdh1–APC/C activity increases in late mitosis and keeps the M-Cdk activity low via M-cyclin degradation during G1.

38. Answer: C Difficulty: 1 Section: Cytokinesis Feedback: Cellularization is the coordinated cytokinesis that occurs after 13 mitoses in early fruit-fly development. 39. Answer: AIII Difficulty: 2 Section: Mitosis Feedback: Both Cdc20–APC/C and Cdh1–APC/C target M-cyclins for degradation. MCdk, in turn, activates the former and inactivates the latter. Sic1 in budding yeast is an example of a CKI that also inhibits M-Cdks. M-Cdk activates the expression of M-cyclin genes in a positive feedback loop, although its activity is also regulated by negative feedback, for example through Cdc20–APC/C mentioned above. 40. Answer: C Difficulty: 2 Section: Mitosis Feedback: Although regulated differently, both APC/C complexes target M-cyclins for degradation. 41. Answer: D Difficulty: 2 Section: Meiosis Feedback: Meiosis II resembles mitosis, in which sister chromatids segregate during anaphase. 42. Answer: A Difficulty: 1 Section: Meiosis Feedback: Prophase I is by far the longest stage in meiosis. 43. Answer: FTTF Difficulty: 2 Section: Meiosis Feedback: Bivalents form and recombination begins during leptotene in prophase I. The synaptonemal complex is formed during zygotene and pachytene, and disassembled in diplotene, which is when chiasmata become visible. Only some recombination events lead to crossovers; that is, reciprocal exchange of DNA. 44. Answer: NNEN Difficulty: 2 Section: Meiosis Feedback: There are two chiasmata in this drawing, both corresponding to exchange between chromatids 2 and 3. 45. Answer: C Difficulty: 1 Section: Meiosis Feedback: In the absence of shugoshin, cohesin complexes at the centromeric regions are no longer protected from phosphorylation. The phosphorylated cohesin subunits are then

cleaved by separase in anaphase I. Consequently, the sister chromatids segregate randomly in anaphase II. 46. Answer: MSMG Difficulty: 1 Section: Control of Cell Division and Cell Growth Feedback: Mitogens stimulate cell division by promoting cell cycle progression. Survival factors promote cell survival by suppressing apoptosis. Growth factors stimulate increase in cell mass by promoting anabolic metabolism. 47. Answer: B Difficulty: 1 Section: Control of Cell Division and Cell Growth Feedback: When blood clots, the incorporated platelets release the contents of their secretory vesicles, containing platelet-derived growth factor (PDGF) among other proteins. 48. Answer: B Difficulty: 1 Section: Control of Cell Division and Cell Growth Feedback: Transforming growth factor-β (TGFβ) and its relatives are amongst the bestunderstood inhibitory signal proteins. 49. Answer: D Difficulty: 1 Section: Control of Cell Division and Cell Growth Feedback: Since they spend less time than the other cells in G1, fibroblasts have a high mitotic index. Note that not all cells with a rapid turnover have a high mitotic index. 50. Answer: C Difficulty: 1 Section: Control of Cell Division and Cell Growth Feedback: Once produced, Myc activates the transcription of delayed-response genes such as G1-cyclins. These in turn activate transcription regulators of the E2F family. 51. Answer: C Difficulty: 1 Section: Control of Cell Division and Cell Growth Feedback: Once the activity of Cdh1–APC/C falls in late G1, geminin starts to accumulate. 52. Answer: E Difficulty: 1 Section: Control of Cell Division and Cell Growth Feedback: All of these processes ensure a sharp transition from G1 to S phase, as well as sustained S-Cdk activity afterward. 53. Answer: BC Difficulty: 1 Section: Control of Cell Division and Cell Growth Feedback: These two transitions are the major time points at which DNA damage is detected and cell-cycle arrest occurs, although the cell can sense DNA damage and arrest the cell cycle at other points throughout the cycle, including in S phase.

54. Answer: D Difficulty: 2 Section: Control of Cell Division and Cell Growth Feedback: Activation of the DNA damage response involves the two upstream kinases ATM and ATR, which phosphorylate and activate the Chk1 and Chk2 kinases. Cdc25 phosphatases can be inhibited by phosphorylation in this pathway, whereas p53 phosphorylation stabilizes the protein by reducing its interaction with its inhibitor Mdm2. 55. Answer: YNYY Difficulty: 2 Section: Control of Cell Division and Cell Growth Feedback: Loss-of-function mutations in p53, Rb, and p21 are found in human cancers. Myc, on the other hand, has the opposite effect: gain-of-function mutations in Myc can drive cancer. 56. Answer: FFFF Difficulty: 2 Section: Control of Cell Division and Cell Growth Feedback: Replicative cell senescence is a result of the shortening of telomeres in somatic cells that lack telomerase activity. Eventually, the unprotected chromosome ends trigger cell-cycle arrest or apoptosis through p53 activation. Excessive stimulation by mitogens can also activate p53, since the cell-cycle inhibitor Arf is activated under excessive Myc or Ras activity and inhibits the p53 inhibitor Mdm2. 57. Answer: D Difficulty: 3 Section: Control of Cell Division and Cell Growth Feedback: Rapamycin inhibits the mammalian TOR complex that is activated in response to growth factors in a PI 3-kinase-dependent manner. As a result, cell growth is retarded in a number of ways, including by inhibition of TOR-mediated activation of S6 kinase.

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION CHAPTER 18: CELL DEATH Copyright © 2015 by W.W. Norton & Company, Inc. 1. Which of the following morphological changes is NOT typically seen in a cell that is undergoing apoptosis? A. The cell rounds up. B. The nuclear envelope disassembles. C. The cell swells. D. Large cells break up into membrane-enclosed fragments. E. The nuclear chromatin breaks into fragments. 2. Indicate true (T) and false (F) statements below regarding programmed cell death by apoptosis. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) Apoptosis is the final fate of almost all cells in an adult animal. ( ) Even perfectly healthy cells may undergo apoptosis. ( ) DNA damage that cannot be repaired inhibits apoptotic pathways. ( ) Apoptosis is the main form of programmed cell death in plant cells as well as in animal cells. 3.

Initiator and executioner caspases share all of the following features EXCEPT that ... A. they are cysteine proteases (they have a cysteine residue at their active site). B. their inactive form is a monomer. C. they undergo cleavage during activation. D. their active form is a dimer. E. they are inhibited by IAPs.

4. v-FLIPs are viral proteins that were first identified as modulators of apoptosis; they contain two death effector domains, which are also found in some initiator caspases such as procaspase-8. These v-FLIP proteins can be recruited to the DISC through the binding of the death effector domain to similar domains in the adaptor proteins, but are otherwise catalytically inactive. What do you think is the effect of v-FLIP expression in the host cell? A. It promotes apoptosis mainly via the extrinsic pathway. B. It inhibits the extrinsic pathway of apoptosis.

C. It activates only the mitochondrial pathway of apoptosis. D. It inhibits the intrinsic pathway of apoptosis. E. It enhances the caspase cascades in both the intrinsic and extrinsic pathways. 5. Fill in the blank: In the intrinsic pathway of apoptosis, a protein called ... is released from the mitochondria into the cytosol and binds to the adaptor protein Apaf1, causing it to oligomerize into a wheel-like assembly called an apoptosome, which then recruits initiator caspase-9 proteins. 6.

The Bcl2 family is comprised of anti-apoptotic (A), BH3-only (B), and effector (E)

proteins. In the following diagram representing the regulation of the intrinsic pathway of apoptosis, what class of activated Bcl2 family proteins (A, B, or E) corresponds to boxes 1 to 3, respectively? Your answer would be a three-letter string composed of letters A, B, and E only, e.g. ABE. 1

Released mitochondrial intermembrane proteins

Initiator caspases

Which of the following proteins activates the mitochondrial pathway of apoptosis? A. The tumor suppressor protein p53, when activated in response to extensive DNA damage. B. The BH3-only protein Bid, when cleaved by the initiator caspase-8 (from the extrinsic pathway). C. The anti-IAP protein Omi, when activated by dephosphorylation. D. The BH3-only protein Bad, when activated by dephosphorylation. E. All of the above.

8. Indicate whether each of the following mutations would likely promote (P) or inhibit (I) apoptosis in cells harboring the mutation(s). Your answer would be a four-letter string composed of letters P and I only, e.g. PPPI.

( ) Mutations in the pro-apoptotic effector Bcl2 family proteins Bax and Bak that prevent their association with the outer mitochondrial membrane. ( ) A mutation in the BIR domain of the IAP protein DIAP1 that prevents binding to either caspases or anti-IAP proteins. ( ) A mutation in the anti-IAP protein Reaper that prevents its binding to the IAP proteins. ( ) A mutation in the CARD domain of caspase-9 that prevents its binding to Apaf1. 9. Soon after the discovery of nerve growth factor (NGF), researchers injected newborn mice with rabbit antiserum (i.e. serum that contains antibodies) against NGF. They observed massive nerve cell death compared to appropriate control injections. Up to 99% of the neurons in some parts of the developing peripheral nervous system died after about a week of daily injections. These results suggest that ... A. NGF signaling sensitizes developing nerve cells to apoptotic signals. B. developing neurons undergo apoptosis in the presence of rabbit proteins. C. developing neurons require NGF for apoptosis. D. developing neurons undergo cell death in the absence of NGF. E. NGF signaling is sufficient for survival of the developing neurons. 10. Apoptotic cells are efficiently phagocytosed by neighboring cells or macrophages. Which of the following DOES NOT normally happen in this process? A. The apoptotic cell releases some of its cytoplasmic content to induce a local inflammatory response. B. The apoptotic cell exposes phosphatidylserine at its surface, which interacts with receptor proteins on the surface of phagocytes via “bridging” proteins. C. The apoptotic cell loses or inactivates “don’t eat me” signals. D. The apoptotic cell rounds up and detaches from its neighbors, which facilitates phagocytosis. 11. Indicate true (T) and false (F) statements below regarding apoptosis in disease. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) Either excessive or insufficient apoptosis can contribute to disease. ( ) Excessive apoptosis, in many cases, leads to autoimmune disease and cancer. ( ) In about half of human cancers, the tumor suppressor protein p53 is mutated. ( ) Drugs that interfere with the function of Bcl2 family proteins such as Bax and Bak may treat cancers by stimulating apoptosis.

12. For each of the following proteins that are proteolytically cleaved in apoptosis, indicate whether the cleavage is first carried out by an initiator (I) or executioner (E) caspase. Your answer would be a five-letter string composed of letters I and E only, e.g. EEEII. ( ) The initiator caspase-2 ( ) The executioner caspase-3 ( ) The BH3-only protein Bid ( ) The endonuclease inhibitor iCAD ( ) The nuclear protein Lamin A

Answers: 1. Answer: C Difficulty: 1 Section: Cell Death Feedback: While cells undergoing necrosis normally swell and burst, apoptotic cells shrink and condense. 2. Answer: FTFF Difficulty: 2 Section: Cell Death Feedback: Even though the number of apoptotic cell deaths in developing and adult animal tissues is astonishing, not all cells die in this way. Apoptosis is found primarily in animals. It can be triggered in perfectly healthy cells that are no longer needed. Cells with extensive DNA damage can kill themselves by undergoing apoptosis. 3. Answer: B Difficulty: 2 Section: Cell Death Feedback: Caspases are cysteine proteases that are normally activated in apoptosis through proteolytic cleavage by other caspase molecules and formation of active dimers. Both initiator and executioner caspases can be inhibited by inhibitors of apoptosis (IAPs). While inactive initiator caspases are monomers, inactive executioner caspases are dimers. 4. Answer: B Difficulty: 2 Section: Cell Death Feedback: By binding to the adaptor proteins, these viral proteins interfere with caspase activation at the death-inducing signaling complex (DISC), and therefore make the cells resistant to Fas-mediated apoptosis, which is normally triggered by cytotoxic lymphocytes and proceeds via the extrinsic pathway. 5. Answer: cytochrome c Difficulty: 1 Section: Cell Death Feedback: In the mitochondrial pathway of apoptosis, caspase activation occurs after cytochrome c is released into the cytosol and induces the formation of apoptosomes. 6. Answer: BAE Difficulty: 2 Section: Cell Death

Feedback: The pro-apoptotic effector Bcl2 family proteins such as Bax and Bak induce the release of cytochrome c and other mitochondrial intermembrane proteins into the cytosol. Anti-apoptotic Bcl2 family proteins, such as Bcl2 and BclXL, inhibit these effector proteins and are in turn inhibited by the pro-apoptotic BH3-only proteins such as Bid and Bim. Some BH3-only proteins may also directly activate Bax and Bak (not shown). 7. Answer: E Difficulty: 2 Section: Cell Death Feedback: Activated BH3-only proteins or anti-IAP proteins can result in the activation of the intrinsic pathway of apoptosis. In response to DNA damage that cannot be repaired, the tumor suppressor protein p53 activates the expression of genes encoding BH3-only proteins (such as Puma). 8. Answer: IPII Difficulty: 3 Section: Cell Death Feedback: Association of Bax and Bak with the mitochondrial membrane is necessary (although not sufficient) for the activation of the intrinsic pathway of apoptosis. The same is true for the association of procaspase-9 with Apaf1 in apoptosomes. Interfering with the binding of IAPs to the caspases promotes apoptosis, while interfering with the binding of IAPs to anti-IAPs represses apoptosis. 9. Answer: D Difficulty: 3 Section: Cell Death Feedback: The antiserum injection results in the perturbation of NGF function, which is similar to knocking-down NGF using genetic techniques. The increase in cell death after injection suggests that NGF is important for nerve cell survival. We now know that the cells undergo apoptosis in the absence of the survival factor. 10. Answer: A Difficulty: 1 Section: Cell Death Feedback: Apoptosis is carried out in a remarkably tidy process and the cell is phagocytosed without triggering an inflammatory response. 11. Answer: TFTF Difficulty: 2 Section: Cell Death

Feedback: Contribution to disease can come from either excessive apoptosis (e.g. in strokes) or insufficient apoptosis (e.g. in some autoimmune disease and cancers). The p53 protein normally arrests the cell cycle or initiates apoptosis in response to DNA damage, which explains why the loss of p53 function is of benefit to, and is widely observed in, cancer cells. 12. Answer: IIIEE Difficulty: 3 Section: Cell Death Feedback: Initiator caspases are cleaved and activated when brought into proximity. They then cleave and activate their downstream executioner caspases and initiate caspase cascades. Whereas very few other substrates, including Bid, are also cleaved by initiator caspases, the great majority of over a thousand identified caspase substrates are thought to be cleaved by executioner caspases to orchestrate apoptosis.

MOLECULAR BIOLOGY OF THE CELL: SIXTH EDITION CHAPTER 19: CELL JUNCTIONS AND THE EXTRACELLULAR MATRIX Copyright © 2015 by W.W. Norton & Company, Inc. 1. Indicate whether each of the following descriptions better applies to connective (C) or epithelial (E) tissues. Your answer would be a four-letter string composed of letters C and E only, e.g. EEEE. ( ) Cells are usually distributed sparsely in the extracellular matrix. ( ) Gap junctions are rarely found. ( ) Direct cell–cell attachments are common. ( ) Cells are tightly associated into sheets. 2. In the following schematic diagram of a simple columnar epithelium lining the digestive tract, indicate which position (A to F) along the basal–apical axis better corresponds to each of the following features. Your answer would be a six-letter string composed of letters A to F only, e.g. FEDABC. A B C D E F

Gut lumen

( ) Basal lamina ( ) Cell apex

Connective tissue

( ) Adherens junctions ( ) Gap junctions ( ) Hemidesmosomes ( ) Tight junctions 3. Which of the following cell junctions uses cadherin cell adhesion molecules to anchor the actin cytoskeleton? A. Tight junction B. Adherens junction C. Desmosome D. Hemidesmosome E. Gap junction 4.

Which of the following is NOT an anchoring junction? A. Adherens junction B. Desmosome C. Actin-linked cell–matrix junction D. Hemidesmosome E. Tight junction

5. Indicate true (T) and false (F) statements below regarding cadherins. Your answer would be a four-letter string composed of letters T and F only, e.g. TFTT. ( ) Cadherins are present in all multicellular organisms. ( ) Adhesion by cadherins is dependent on Ca2+ ions. ( ) In contrast to nonclassical cadherins, classical cadherins are more closely related in sequence. ( ) E, N, and P cadherins are mainly found in desmosomes. 6. You have isolated epithelial cells from mouse mammary glands and have grown them in a three-dimensional gelatinous matrix in the presence or absence of a specific inhibitor that targets a transmembrane metalloprotease. The protease is capable of cleaving cadherin near the base of its extracellular domain to release an N-terminal fragment containing the extracellular cadherin domains. The distribution of the cells in the matrix, as seen under a light microscope, one day after the transplantation of the isolated tissue is presented in the following schematic drawings. Which condition (1 or 2) do you think corresponds to the presence of the protease inhibitor? Write down 1 or 2 as your answer.

7. Cadherin molecules at the cell surface are often clustered side-to-side to create a molecular Velcro that attaches the cell to another cell or to the extracellular matrix. In the following simplified drawings, two cells are initially attached via such a Velcro. When one of the cells moves in either of three different ways (A to C), it is faced with a mechanical resistance from the cadherin interactions. The resistance against which type of movement do you think is the weakest? Write down A, B, or C as your answer.

Two neighboring cells

A The cell moves away from its neighbor

B The cell slides past its neighbor

C The cell rolls over its neighbor

8. Cells expressing either N-cadherin (A), high levels of E-cadherin (B), or low levels of Ecadherin (C) have sorted themselves out on a substratum in a cadherin-dependent manner as shown in the schematic drawing below. Which cells (A to C) would you expect to correspond to white, gray, and black in the drawing, respectively? Your answer would be a three-letter string composed of letters A to C only, e.g. CAB.

Overproduction of cadherins such as E-cadherin … A. is often found in cancers originating from epithelia. B. is induced by the transcription regulatory protein Twist. C. induces epithelial–mesenchymal transition. D. leads to stronger cell–cell adhesion. E. All of the above.

10. Consider two cells attached to each other via adherens junctions. The forces generated by the actin cytoskeleton in one cell … A. are dependent on myosin activity. B. are normally balanced by similar forces in the neighboring cell. C. can strengthen the adherens junction by unfolding catenin proteins. D. All of the above. 11. Consider a sheet of epithelial tissue that invaginates to form a tube along the anterior– posterior axis of a developing vertebrate embryo. The tube is then closed concomitant with a “convergence and extension” process similar to that observed in germ-band extension in Drosophila melanogaster. During this process, the tube elongates and narrows before pinching off from the sheet. Which axis (1 or 2) in the following schematic drawing of the tube cells do you think corresponds to the anterior–posterior axis? Write down 1 or 2 as your answer.

1 2

Cells of the epithelial tube 12.

Fill in the blank in the following paragraph. Do not use abbreviations. “Particularly abundant in cells of heart muscle and the epidermis, … are structurally similar to adherens junctions and contain a number of homologous components. However, their cadherin molecules are linked to intermediate filaments instead of the actin cytoskeleton.”

13. A researcher has grown monolayers of cells from either of two epithelial cell lines on a permeable supporting membrane that separates two electrolyte chambers. One of the cell lines (A) has been derived from the mammalian kidney and forms epithelial sheets resembling those lining the renal collecting duct and the urinary bladder. The other (B) is derived from the mammalian small intestine. The researcher has measured the voltage (V) and electrical current (I) across the monolayer and has plotted the results in the graph below. Which line (1 or 2) in the graph do you think corresponds to cell line A? Recall that current (I) is related to the applied voltage (V) by the equation I = V/R, in which R is the electrical resistance. Write down 1 or 2 as your answer.

100-

V (millivolts)

2 00

150 I (microamperes)

14. Indicate whether each of the following descriptions better applies to adherens junctions (A), desmosomes (D), gap junctions (G), or tight junctions (T). Your answer would be a fourletter string composed of letters A, D, G, and T only, e.g. AAAD. ( ) They play a direct role in structural polarization of epithelial cells. ( ) They couple cells both electrically and metabolically. ( ) They are formed from hexameric hemichannels that can assemble into heterotypic complexes. ( ) They create semipermeable paracellular pores. 15. In the following schematic drawing of connexons as seen from the cytosol of a cell, what is the approximate diameter of each of the pores? Which of the indicated connexons (1 or 2) is more likely to have been incorporated more recently?

1 2

Pore

A. About 20 nm; connexon 1 B. About 4 nm; connexon 1 C. About 1.5 nm; connexon 1 D. About 4 nm; connexon 2 E. About 1.5 nm; connexon 2 16.

Plasmodesmata in plant cells are functionally similar to animal cells’ … A. adherens junctions. B. desmosomes. C. gap junctions. D. hemidesmosomes. E. tight junctions.

17.

In insects such as Drosophila melanogaster, electrical synapses are abundant in the

ventral nerve cord. Unlike a chemical synapse, the cytosols of pre- and postsynaptic cells in an electrical synapse are connected, allowing action potentials to spread rapidly and without delay. What junctional proteins in the flies are chiefly responsible for this unique feature of the electrical synapses? A. Cadherins B. Claudins C. Connexins

D. Innexins E. Integrins 18.

Selectins … A. mainly mediate cell–matrix attachments. B. are members of the Ig superfamily of cell adhesion molecules. C. do NOT require Ca2+ for their adhesive function, unlike integrins. D. are carbohydrate-binding proteins. E. All of the above.

19.

Malignant cancer cells that have entered the bloodstream can imitate lymphocytes in

exiting the bloodstream and entering underlying tissues in the process of metastasis. This “lymphocyte mimicry” can be summarized as … A. an initial cadherin-dependent rolling followed by an integrin-dependent adhesion and emigration. B. an initial cadherin-dependent rolling followed by an I-CAM-dependent adhesion and emigration. C. an initial cadherin-dependent rolling followed by a selectin-dependent adhesion and emigration. D. an initial selectin-dependent rolling followed by a cadherin-dependent adhesion and emigration. E. an initial selectin-dependent rolling followed by an I-CAM-dependent adhesion and emigration. 20. In the following schematic diagram, the estimated attractive or repulsive force between two adhesive cells is plotted as a function of distance between the cells. The dashed curve represents a condition under which the cells express cadherin but not N-CAM. The other two curves represent conditions under which both cadherin and heavily sialylated N-CAM are expressed. In one of the conditions, however, the ionic strength of the medium has been artificially increased by addition of salt. Which curve (1 or 2) do you think represents the highsalt condition? Write down 1 or 2 as your answer.

Repulsion

Attraction

Force

0 Intercellular distance 21. Indicate whether each of the following descriptions better applies to glycosaminoglycans (G), fibrous proteins (F), or glycoproteins (P) of the extracellular matrix in animal cells. Your answer would be a six-letter string composed of letters G, F, and P only, e.g. GFPGFP. ( ) They include laminin and fibronectin. ( ) They primarily include collagens. ( ) They include hyaluronan. ( ) They are found in proteoglycans. ( ) They occupy large volumes and form hydrated gels. ( ) They constitute the major proteins of the matrix. 22. Indicate true (T) and false (F) statements below regarding glycosaminoglycan chains in the extracellular matrix. Your answer would be a four-letter string composed of letters T and F only, e.g. TFTT. ( ) They are highly branched polysaccharides. ( ) They are strongly hydrophilic and absorb a large amount of water. ( ) They all contain N-acetylglucosamine or N-acetylgalactosamine. ( ) They are highly positively charged. 23. Fill in the blank in the following paragraph regarding the extracellular matrix. Do not use abbreviations. “The extracellular matrix of connective tissues in animals is primarily made and secreted by the … family of cells, although they may have more specific names in certain tissues. Being

common in connective tissues, these cells can also migrate and proliferate if need be.” 24. How is hyaluronic acid different from other glycosaminoglycans of the extracellular matrix? A. It lacks sulfate groups. B. It is generally not covalently linked to proteins. C. It is not assembled in the Golgi apparatus. D. It can be several megadaltons in mass. E. All of the above. 25. In the following schematic drawings of aggrecan and decorin, which one (1 or 2) corresponds to aggrecan? Which one (1 or 2) is synthesized primarily by chondrocytes in cartilage?

A. 1; 1 B. 1; 2 C. 2; 1 D. 2; 2 26. A giant aggrecan aggregate can have a molecular mass of 108 daltons or more, and can be as large as a bacterium. Its structure resembles a brush. What is responsible for attaching the bristles of the brush to its core? A. Covalent bonds between aggrecan core protein and keratan or chondroitin sulfate. B. Noncovalent bonds between aggrecan core protein and keratan or chondroitin sulfate. C. Covalent bonds between aggrecan core protein and hyaluronan. D. Noncovalent bonds between aggrecan core protein and hyaluronan.

27. Which of the following extracellular matrix components contains a transmembrane protein? A. Type I collagen B. Perlecan C. Aggrecan D. Syndecan E. Decorin 28.

Collagens are extremely rich in … A. alanine and valine. B. glycine and proline. C. lysine and arginine. D. serine and threonine. E. tyrosine and phenylalanine.

29. Type IV collagen molecules can interact at their C-termini to form dimers and at their Ntermini to form tetramers. What category do you think this type of collagen belongs to? A. Fibrillar B. Fibril-associated C. Network-forming D. Proteoglycan core E. Anchorage fibril 30.

Which collagens are the most common in our body? A. Fibrillar B. Fibril-associated C. Network-forming D. Proteoglycan core E. Anchorage fibril

31. Which components of the extracellular matrix are mostly responsible for its ability to resist compressive and tensile forces, respectively? A. Collagens and glycoproteins B. Glycoproteins and collagens C. Glycosaminoglycans and glycoproteins D. Glycosaminoglycans and collagens

E. Collagens and glycosaminoglycans 32. Which of the following is true regarding fibril-associated collagens such as type IX collagen? A. They aggregate with one another to form long fibrils in the extracellular space. B. They lack the triple-stranded helical structure found in other collagens. C. They are more flexible than collagens of types I to IV. D. They are the most common among collagens. E. All of the above. 33.

The size distribution of exons coding for collagen helical domains in various vertebrates

reveals a unique pattern. This pattern provides clues to the evolutionary history of the genes encoding fibrillar collagen α chains. What is this pattern? A. The length of most exons is a multiple of 100 nucleotides. B. Most exons code for 54, or a multiple of 54, Gly-X-Y repeats. C. The length of most exons is a multiple of 30 nucleotides. D. Most exons code for 6, or a multiple of 6, Gly-X-Y repeats. E. None of the above. 34. The elasticity of the components of the extracellular matrix can be visualized in a stress– strain curve, in which the tensile stress (corresponding to load) applied to a tissue sample is plotted against the strain (corresponding to deformation) measured in the tissue. In the following stress–strain graph, the elasticity of an intact connective tissue is represented by a dashed curve. The other two curves correspond to the same tissue that has been treated with either formic acid (to digest collagen) or trypsin (to digest elastin). Which curve (1 or 2) do you think corresponds to trypsin treatment?

Stress (Pa)

Strain (%) 35. This large glycoprotein of the extracellular matrix is associated with elastin fibers and is essential for their integrity. Its deposition in the developing connective tissues often precedes the appearance of elastin. It is part of microfibrils that cover the elastic fibers. This protein is … A. collagen. B. fibronectin. C. fibrillin. D. filamin. E. laminin. 36. Multidomain glycoproteins in the extracellular matrix have binding sites for many other macromolecules. These glycoproteins … A. can serve as tracks along which cells can migrate. B. can bind to growth factors and influence cell signaling. C. can bind to cell-surface receptors. D. can serve as repellents to forbid cell migration. E. All of the above. 37. Indicate true (T) and false (F) statements below regarding fibronectin in the extracellular matrix. Your answer would be a four-letter string composed of letters T and F only, e.g. TFTT. ( ) Fibronectin carries asparagine-linked oligosaccharides. ( ) Fibronectin often forms homodimers by cross-linking through covalent attachment of lysine residues. ( ) Fibronectin binds to integrins, collagens, and glycosaminoglycans through its type III repeats.

( ) Fibronectin molecules can assemble into fibrils only in the vicinity of cells. 38. Neuromuscular junctions of vertebrates are special types of chemical synapses formed from close association between motor neuron axon terminals and muscle fibers. A thin layer of basal lamina separates the two cell membranes at the neuromuscular junction and is important for its organization and maintenance. Which of the following would you NOT expect to find in the basal lamina in these structures? A. Nidogen B. Laminin C. Type I collagen D. Type IV collagen E. Perlecan 39. Indicate true (T) and false (F) statements below regarding the interaction of cells with the extracellular matrix using matrix receptors. Your answer would be a four-letter string composed of letters T and F only, e.g. TFTT. ( ) Integrins are the only known matrix receptors in animal cells. ( ) Integrins can transmit signals in both directions across the membrane; that is, both outside-in and inside-out. ( ) Tension can increase the binding affinity of an integrin for its intracellular and extracellular ligands. ( ) Integrins can convert molecular signals into mechanical ones. 40. Indicate whether each of the following descriptions better applies to the intracellular (I) or extracellular (E) domain of integrins. Your answer would be a four-letter string composed of letters I and E only, e.g. EEEE. ( ) It binds to talin. ( ) It binds to the RGD tripeptide motif. ( ) It is bulkier. ( ) It contains divalent cation-binding domains. 41. A researcher has coated the wells in a multiwell plate with human plasma fibronectin. She introduces integrin-expressing cells into these wells in the presence of different concentrations of a synthetic peptide. She then washes the wells to remove unbound cells and counts the number of cells that remain attached to each well. The results of her experiment are presented in the following schematic graph. The synthetic peptide sequences used in these

Number of remaining cells

experiments were GRGDSPC (1) or GGDRSLN (2). Which one of these peptides (1 or 2) would you expect to correspond to curve a in the graph? Which curve (a or b) would you expect if a synthetic peptide with the sequence GDGRSPC was used? C is cysteine, D is aspartic acid, G is glycine, L is leucine, N is asparagine, P is proline, R is arginine, and S is serine.

Peptide concentration used A. 1; a B. 1; b C. 2; a D. 2; b 42. EDTA is a metal-ion “chelator” that is commonly used to lower the concentration of free divalent cations in solution. In the lab, your friend has cultured a human cell line on elastic membranes that have been coated with fibronectin and attached to a device that can stretch the membrane. She has quantified the binding of talin to integrin under different conditions, and has presented the results in the following schematic graph, which shows the binding in the presence or absence of EDTA in the cell-culture medium. Which experiment (A or B) do you think was performed in the presence of EDTA? Write down A or B as your answer.

100 -

Stretched

Bound integrin (%)

No stretch

43. The following schematic drawing shows two integrin molecules. Which molecule (a or b) is depicted in its active configuration? Which side of the plasma membrane (1 or 2) represents the cytosol?

b 1 Plasma membrane

A. a; 1 B. a; 2 C. b; 1 D. b; 2 44.

In the active state of the integrin dimer, … A. both intracellular and extracellular binding sites are exposed.

B. the intracellular binding sites are inaccessible, while the extracellular binding sites are exposed. C. the intracellular binding sites are exposed, while the extracellular binding sites are inaccessible. D. both intracellular and extracellular binding sites are inaccessible. 45. Platelets that express a constitutively active Rap1 are able to activate integrins even in the absence of thrombin stimulation. In contrast, cells expressing a binding-deficient version of talin are unable to perform integrin activation. In platelets expressing both of these mutant proteins, what would you expect to observe: the Rap1 gain-of-function phenotype (R) or the talin loss-offunction phenotype (T), as described above? Write down R or T as your answer. 46.

How does talin activate the inside-out signaling through integrin? A. It binds to both integrin subunits to bring them together in an interlocked conformation. B. It is a protein kinase that phosphorylates both integrin subunits and induces activation by conformational change. C. It competes with the α subunit for binding to the β subunit, therefore blocking intimate α-β linkage. D. It binds to the α subunit, and dissociates the dimer into individual subunits that are functionally active.

47. Which of the following molecules is capable of tension-sensing by exposing binding sites for other proteins when mechanically stretched? A. α-Catenin B. Talin C. Fibronectin D. All of the above 48. The chemical compound PMA is a potent stimulator of the epithelial–mesenchymal transition in epithelial cells. The activity of two of the following types of proteins is typically upregulated in epithelial cells treated with PMA, while that of the other two is mainly downregulated. Choose the two proteins whose activity you think is stimulated by PMA. Your answer would be a two-letter string composed of letters A to D only in alphabetical order, e.g. BC. (A) Cadherin (B) Connexin

The focal adhesion kinase … A. is a cytosolic serine/threonine kinase. B. binds directly to integrin dimers. C. can be phosphorylated by Src kinases. D. prevents the relay of outside-in integrin signaling. E. All of the above.

50.

Indicate true (T) and false (F) statements below regarding integrin signaling. Your

answer would be a four-letter string composed of letters T and F only, e.g. TFTT. ( ) Integrins usually bind their ligands with a higher affinity compared to other cellsurface receptors. ( ) Unlike cadherins, integrins usually function as individual dimers and do not cluster. ( ) Anchorage dependence is mainly mediated via integrin signaling. ( ) Integrins can activate the Ras–MAPK pathway. 51.

A cell’s response is not the same when placed on a rigid compared to a soft matrix.

Indicate whether each of the following is expected to occur in a cell that is placed on a rigid matrix (R) or a soft matrix (S). Your answer would be a three-letter string composed of letters R and S only, e.g. RSR. ( ) More tension is generated in cell–matrix contacts. ( ) More vinculin proteins would bind to the C-terminal tail of talin. ( ) More actin filaments are recruited to the site of cell–matrix adhesion. 52. Indicate whether each of the following descriptions better applies to the plant cell wall (W) or the extracellular matrix of animal cells (M). Your answer would be a three-letter string composed of letters W and M only, e.g. WMM. ( ) It is generally thicker and stronger. ( ) It is more rigid. ( ) It has higher nitrogen content. 53. What is the most common additional polymer present in the plant secondary, but not primary, cell wall? A. Cellulose

B. Pectin C. Lignin D. Starch 54. Indicate whether each of the following descriptions better applies to cellulose microfibrils (C) or pectins (P) in plant cell walls. Your answer would be a five-letter string composed of letters C and P only, e.g. CPPPP. ( ) It is made into a network by cross-linking glycans. ( ) It is particularly abundant in the middle lamella. ( ) Divalent ions such as calcium can cross-link it to form gels. ( ) It forms more homogeneous polymers. ( ) It contains many negatively charged galacturonic acid units. 55. In the following schematic diagram of a cubic plant cell that has just left the meristem zone, the cellulose microfibrils are shown as black lines. In which direction (1 or 2) do you think the cell will elongate when the cell grows? If the cell wall is momentarily dissolved, in what direction (3 or 4) would water tend to flow?

1 2

3 4

A. 1; 3 B. 1; 4 C. 2; 3 D. 2; 4

56. Cellulose is deposited onto the plant cell wall in highly ordered crystalline aggregates by the cellulose synthase complex embedded in the plasma membrane. Would you expect cell wall deposition to continue following treatment of plant cells with a drug that depolymerizes microtubules? Would you expect these cells to be able to switch the orientation of the microfibril pattern between successive lamellae? A. Yes; yes B. Yes; no C. No; yes D. No; no

Answers: 1. Answer: CCEE Difficulty: 1 Section: Cell–Cell Junctions Feedback: Direct cell–cell attachments are rare in connective tissue compared to epithelial tissue. Cell–matrix attachments are instead dominant in connective tissues. 2. Answer: FACDEB Difficulty: 2 Section: Cell–Cell Junctions Feedback: The adhesion belt (C) is usually formed beneath the strands of tight junctions (B) near the apex (A). Gap junctions (D) are usually more basally located. Finally, hemidesmosomes (E) link the cytoskeleton to the underlying basal lamina (F) at the basal surface of the cells. 3. Answer: B Difficulty: 1 Section: Cell–Cell Junctions Feedback: In adherens junctions, classical cadherins from two neighboring cells interact to link the actin cytoskeletons of the cells together. In these structures, actin filaments bind cadherin indirectly via intracellular adaptor proteins. 4. Answer: E Difficulty: 1 Section: Cell–Cell Junctions Feedback: An anchoring junction links the cytoskeleton of a cell to that of its neighbor (cell–cell junction) or to the extracellular matrix (cell–matrix junction). 5. Answer: FTTF Difficulty: 1 Section: Cell–Cell Junctions Feedback: Cadherins form a diverse family of Ca2+-dependent cell adhesion molecules found in all animals but not in plants and fungi. The more closely related classical cadherins include the E, N, and P cadherins and are found in adherens junctions. 6. Answer: 1 Difficulty: 2 Section: Cell–Cell Junctions Feedback: The inhibitor prevents the shedding of cadherin induced by the protease, and therefore facilitates cell–cell adhesion. This helps the cells clump together. 7. Answer: C Difficulty: 2 Section: Cell–Cell Junctions

Feedback: According to the Velcro model, clustered cadherin molecules would provide cell adhesions with great shear strength (B) as well as tensile strength (A), while simultaneously allowing “peeling” (C) to occur easily. 8. Answer: CAB Difficulty: 2 Section: Cell–Cell Junctions Feedback: Cells expressing N-cadherin (gray) sort out from cells expressing E-cadherin. Cells expressing high levels of E-cadherin (black) sort out from cells expressing low levels (white), which end up mostly at the periphery because of their weaker adhesion. 9. Answer: D Difficulty: 1 Section: Cell–Cell Junctions Feedback: Loss of cadherin function is often found in malignant cancer cells. Twist negatively regulates cadherin and induces epithelial–mesenchymal transition. A higher level of cadherin typically leads to stronger cadherin-dependent adhesions. 10. Answer: D Difficulty: 1 Section: Cell–Cell Junctions Feedback: Adherens junctions are subjected to pulling forces generated by the attached contractile bundles of actin filaments and non-muscle myosin II. The protein α-catenin can be stretched to an extended conformation by this force, recruiting more actin to the junction to strengthen it. Through such mechanotransduction mechanisms, pulling on a junction in one cell increases the contractile force generated in the attached cell. 11. Answer: 2 Difficulty: 2 Section: Cell–Cell Junctions Feedback: The remodeling of cell–cell adhesions proceeds by contraction of actin– myosin bundles and formation and expansion of new cadherin-based adhesions along the tube. This process extends the tube and accompanies its closure. 12. Answer: desmosomes Difficulty: 1 Section: Cell–Cell Junctions Feedback: Desmosomes provide epithelial cells with mechanical strength and are plentiful in tissues that are subject to high levels of mechanical stress. 13. Answer: 1 Difficulty: 3 Section: Cell–Cell Junctions Feedback: The transepithelial electrical resistance is represented by the slopes of the two lines and is expected to be higher for cell line A and the epithelial cells lining the urinary

bladder. These cells are orders of magnitude less permeable to small ions compared to intestinal epithelia. 14. Answer: TGGT Difficulty: 1 Section: Cell–Cell Junctions Feedback: Tight junctions form a fence between plasma membrane domains in a polarized cell, contributing to the difference in apical and basolateral surface proteins. They also form seals between cells in an epithelial sheet to confer particular permeability properties on the sheet. Gap Junctions, formed from aligned hexameric hemichannels on the plasma membranes of two neighboring cells, form molecular sieves that that allow the cells to share small molecules and ions. 15. Answer: C Difficulty: 1 Section: Cell–Cell Junctions Feedback: The pore formed by the intercellular channels in gap junctions is about 1.5 nm in diameter, allowing the diffusion of only small molecules. Gap-junction plaques are dynamic structures; new channels are added at the periphery and older channels are removed from near the center of the plaque. 16. Answer: C Difficulty: 1 Section: Cell–Cell Junctions Feedback: Like gap junctions, plasmodesmata connect the cytosols of neighboring cells in tissues, allowing small metabolites and signaling molecules to diffuse between the cells. 17. Answer: D Difficulty: 1 Section: Cell–Cell Junctions Feedback: Gap junctions at electrical synapses allow action potentials to spread without delay. In invertebrates, they are made of innexin subunits. 18. Answer: D Difficulty: 1 Section: Cell–Cell Junctions Feedback: Selectins are lectins that mediate transient, calcium-dependent cell–cell adhesions. 19. Answer: E Difficulty: 2 Section: Cell–Cell Junctions Feedback: The “rolling” step is accompanied by relatively weak selectin-mediated adhesions and is followed by stronger adhesions as a result of integrin binding to immunoglobulin (Ig) superfamily proteins such as I-CAMs.

20. Answer: 2 Difficulty: 3 Section: Cell–Cell Junctions Feedback: The overall force is the sum of cadherin-dependent attractive forces, N-CAMdependent electrostatic repulsion, and steric exclusions. In the presence of N-CAM with long polysialic acid chains, the repulsive forces are dominant: the chains are negatively charged and occupy a large volume. Increasing the ionic strength weakens the repulsive forces. 21. Answer: PFGGGF Difficulty: 1 Section: The Extracellular Matrix of Animals Feedback: Glycosaminoglycan chains, such as hyaluronan or those covalently linked to core proteins in proteoglycans, occupy large amounts of space and form hydrated gels that resist compression. Fibrous proteins, primarily collagens, constitute a significant fraction of the total protein mass in the extracellular matrix. Noncollagen glycoproteins, such as fibronectin and laminin, carry conventional N-linked oligosaccharides. 22. Answer: FTTF Difficulty: 1 Section: The Extracellular Matrix of Animals Feedback: Glycosaminoglycans (GAGs) are unbranched and highly anionic. 23. Answer: fibroblast Difficulty: 1 Section: The Extracellular Matrix of Animals Feedback: In most connective tissues, the extracellular matrix components are secreted by fibroblasts. 24. Answer: E Difficulty: 1 Section: The Extracellular Matrix of Animals Feedback: Being the simplest of the glycosaminoglycans (GAGs), hyaluronan chains can be remarkably long. Not typical of GAGs, it contains no sulfated sugars, is not generally linked to proteins covalently, and is synthesized and spun out directly from the cell surface by a transmembrane enzyme. 25. Answer: D Difficulty: 1 Section: The Extracellular Matrix of Animals Feedback: Aggrecan (2) is a major component of cartilage. 26. Answer: D Difficulty: 1 Section: The Extracellular Matrix of Animals

Feedback: This brushlike structure is an order of magnitude larger than the bottlebrush structure of the aggrecan proteoglycan itself. Each bristle represents an aggrecan that is noncovalently attached to a central hyaluronan molecule. 27. Answer: D Difficulty: 1 Section: The Extracellular Matrix of Animals Feedback: Syndecans are found at the plasma membrane in many cell types, where they regulate cell adhesions and signaling. 28. Answer: B Difficulty: 1 Section: The Extracellular Matrix of Animals Feedback: Both glycine (small size) and proline (unique backbone angles) are important in the formation of the triple-stranded helix in collagen. 29. Answer: C Difficulty: 2 Section: The Extracellular Matrix of Animals Feedback: This type of interaction creates sheetlike networks of type IV collagens in basal lamina. 30. Answer: A Difficulty: 1 Section: The Extracellular Matrix of Animals Feedback: Type I collagen, which is a fibrillar collagen, is by far the most common collagen in the body. 31. Answer: D Difficulty: 1 Section: The Extracellular Matrix of Animals Feedback: While glycosaminoglycans of the extracellular matrix provide a tissue with resistance to compression, collagen fibrils form structures that can confer great tensile strength to the tissue. 32. Answer: C Difficulty: 1 Section: The Extracellular Matrix of Animals Feedback: Fibril-associated collagens bind in a periodic manner to fibrils made of fibrillar collagens (e.g. type I). The presence of nonhelical interruptions within their triple-stranded helical structure makes these fibril-associated collagens more flexible than fibrillar collagen molecules. 33. Answer: D Difficulty: 2 Section: The Extracellular Matrix of Animals

Feedback: The size of the coding exons for collagen helical domains is often 54 or a multiple of 54 nucleotides. This suggests that duplications of an ancestral gene (that was 54 nucleotides long and coded for six Gly-X-Y repeats) gave rise to the helical domains in modern fibrillar collagens. 34. Answer: 1 Difficulty: 3 Section: The Extracellular Matrix of Animals Feedback: The elastic fibers (composed of elastin) give tissues their elasticity. Under increasing load, these fibers stretch to a greater extent compared to collagen fibers. Therefore, the stress-strain curve for elastin fibers (2) has a lower slope compared to that for collagen fibers (1). When elastin fibers are digested by trypsin, the curve for the tissue approaches that of the trypsin-resistant collagen. 35. Answer: C Difficulty: 1 Section: The Extracellular Matrix of Animals Feedback: The elastin core in elastin fibers is covered with a sheath of microfibrils which are elastic in their own right and seem to provide scaffolding to guide elastin deposition in developing tissues. Fibrillin is a glycoprotein in microfibrils that is essential for the integrity of elastic fibers. 36. Answer: E Difficulty: 1 Section: The Extracellular Matrix of Animals Feedback: The extracellular matrix contains a large and varied set of multidomain glycoproteins such as fibronectin. They help organize the matrix and cell–matrix attachments. They also guide and control cell migration during development. Their binding to peptide growth factors and small molecules influences cell signaling within a tissue. 37. Answer: TFFT Difficulty: 2 Section: The Extracellular Matrix of Animals Feedback: Fibronectin is a matrix multidomain glycoprotein and can form fibronectin fibrils on the surface of cells. This assembly is regulated by the tension exerted by the cells via surface proteins such as integrins. Type III fibronectin repeats are responsible for integrin binding, whereas binding to collagen and heparin is mediated by type I and type II repeats. Fibronectin heterodimers are held together by C-terminal disulfide bonds. 38. Answer: C Difficulty: 2 Section: The Extracellular Matrix of Animals

Feedback: Laminin and type IV collagen are major and ubiquitous components of the basal lamina. Glycoproteins such as nidogen and proteoglycans such as perlecan are also commonly found in basal laminae. 39. Answer: FTTT Difficulty: 1 Section: Cell–Matrix Junctions Feedback: Integrins are the principal matrix receptors in animal cells. They can transmit mechanical and molecular signals in both directions across the plasma membrane. 40. Answer: IEEE Difficulty: 1 Section: Cell–Matrix Junctions Feedback: The large N-terminal extracellular domain of integrins can bind to proteins containing the RGD or other sequences. The binding is modulated by the concentration of divalent ions such as Ca2+. The intracellular domain can engage the actin cytoskeleton or the intermediate filaments by using adaptor proteins such as talin and plectin, respectively. 41. Answer: C Difficulty: 3 Section: Cell–Matrix Junctions Feedback: The synthetic peptide containing the RGD sequence (1) competes with fibronectin for integrin binding. At higher concentrations, its presence results in fewer cells attaching to the fibronectin on the well surface, as shown in curve B. A sequence that lacks the RGD motif would be expected to result in curve A. 42. Answer: B Difficulty: 2 Section: Cell–Matrix Junctions Feedback: Dependence of the binding of integrin on the presence of divalent cations is reflected in the inhibition of integrin binding in the presence of the chelating agent. 43. Answer: A Difficulty: 2 Section: Cell–Matrix Junctions Feedback: Integrins have two major activity states; in the active form, the large extracellular domains are more extended and further apart from each other. 44. Answer: A Difficulty: 1 Section: Cell–Matrix Junctions Feedback: In the active state of integrins, the two subunits in each dimer are unhooked at the membrane to expose the binding sites at both sides of the membrane. 45. Answer: T Difficulty: 3

Section: Cell–Matrix Junctions Feedback: Since talin functions downstream of Rap1, the talin loss-of-function phenotype is also observed in the double mutants. 46. Answer: C Difficulty: 2 Section: Cell–Matrix Junctions Feedback: Talin activates integrin by tilting the transmembrane domains of the two integrin subunits away from each other, favoring the active state. 47. Answer: D Difficulty: 1 Section: Cell–Matrix Junctions Feedback: Fibronectin is a mechanical sensor in the extracellular matrix, while α-catenin and talin are intracellular sensors mainly involved in adherens junctions and focal adhesions, respectively. 48. Answer: CD Difficulty: 2 Section: Cell–Matrix Junctions Feedback: Activation of some integrins and matrix metalloproteases occurs in the epithelial–mesenchymal transition, whereas cadherin-based adherens junctions and connexin-based gap junctions are down-regulated in this process. This allows the cells to lose their epithelial character. 49. Answer: C Difficulty: 2 Section: Cell–Matrix Junctions Feedback: Focal adhesion kinase (FAK) is a tyrosine kinase recruited to active integrins via adaptor proteins such as talin and paxillin; it is autophosphorylated and also phosphorylated by recruited Src kinases, helping to relay signals into the cell. 50. Answer: FFTT Difficulty: 2 Section: Cell–Matrix Junctions Feedback: Integrins usually bind to their ligands weakly; their high concentration in the membrane and their clustering affect their collective affinity. 51. Answer: RRR Difficulty: 2 Section: Cell–Matrix Junctions Feedback: When the cell is placed on a rigid substratum that strongly resists pulling forces applied at the cell-matrix junctions, the increased tension is sensed by the junction and more proteins are recruited to the junction as a result. 52. Answer: WWM Difficulty: 2

Section: The Plant Cell Wall Feedback: Plant cell wall is generally thicker, stronger, and more rigid compared to extracellular matrix surrounding animal cells. It has a very low protein content and is instead rich in polymers such as cellulose and lignin which can be made from “cheap” carbon-based structural building blocks. 53. Answer: C Difficulty: 2 Section: The Plant Cell Wall Feedback: Lignin is a complex network of aromatic polymers found in woody tissues. 54. Answer: CPPCP Difficulty: 2 Section: The Plant Cell Wall Feedback: In plant cells walls, cellulose microfibrils are homogeneous polymers that are cross-linked into a complex network by a heterogeneous set of cross-linking glycans. Pectins constitute a heterogeneous group of branched polysaccharides which are rich in galacturonic acid residues and are abundant in the middle lamella. They can turn into a semirigid gel by the addition of Ca2+. 55. Answer: C Difficulty: 3 Section: The Plant Cell Wall Feedback: The orientation of cellulose microfibrils in the diagram favors growth, powered by turgor pressure, in the horizontal direction. Without the cell wall, water would flow into the cell, potentially bursting it open. 56. Answer: B Difficulty: 2 Section: The Plant Cell Wall Feedback: Synthesis of cellulose and its deposition as microfibrils would continue after such treatment; however, it is thought that in the absence of cortical microtubules, the process is not spatially restrained, and therefore the microfibril pattern is locked: new microfibrils are deposited in the same orientation as the layer put down previously.

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION CHAPTER 20: CANCER Copyright © 2015 by W.W. Norton & Company, Inc. 1. In the following simplified diagram of cell divisions in a multicellular species, the germcell and somatic-cell lineages are depicted. Which of the indicated cells (1 or 2) represents the germ line? Write down 1 or 2 as your answer.

…

2. Consider a healthy adult animal in which 1015 cell divisions have taken place since birth. Spontaneous mutations can occur at a rate of approximately one nucleotide out of about ten billion nucleotides every time DNA is replicated. The animal has a diploid genome size of about 2 billion nucleotide pairs. Assuming that only about 5% of mutations occur within genes or gene regulatory sequences, and further assuming that about 0.1% of those may cause cancer, how many potential cancer-causing mutations has the animal been able to successfully suppress (i.e. has been able to survive) during its lifetime? A. Ten thousand B. One hundred thousand C. Ten million D. One billion E. Ten billion 3. Indicate true (T) and false (F) statements below regarding the properties of cancer cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) Cancer cells invade and colonize territories that normally belong to other cells. ( ) Unlike in normal tissues, cell death is extremely rare in tumors. ( ) Cancer cells grow and proliferate in defiance of normal restraints. ( ) Malignant tumors are composed of cells that grow and proliferate, but still have not acquired invasiveness.

4. Indicate whether each of the following cancers can be best classified as a carcinoma (C), sarcoma (S), or neither of the two (N). Your answer would be a four-letter string composed of letters C, S, and N only, e.g. SSNC. ( ) Breast cancer ( ) Lung cancer ( ) Colorectal cancer ( ) Myeloma 5.

A benign neoplasia of cartilage is called a … A. Chondrocarcinoma B. Chondroma C. Chondrosarcoma D. Adenochondroma E. Chondromelanoma

6. Among the following cancers, one is currently leading to the most number of deaths in the United States and in the rest of the world. In the United States, it contributes to cancer mortality more than the next three killing cancers combined. Worldwide, it claims more than 1.5 million lives every year. Which cancer is this? A. Lung cancer B. Breast cancer C. Colon cancer D. Pancreatic cancer E. Stomach cancer 7. X-chromosome inactivation in female mammals occurs mostly randomly early in development, resulting in a heterogeneous cell population, with each cell having inactivated one or the other of its X chromosomes and passing on the same X-inactivation choice to its offspring. The inactivated X chromosome is generally hypermethylated and transcriptionally inactive. You are studying a newly discovered type of colon tumor in women that has a morphology distinct from that of other colon adenomas. You extract chromosomal DNA from the tumor cells. You then either keep the DNA untreated, or digest the DNA with a methylation-sensitive restriction enzyme that only cleaves its recognition DNA sequence if the sequence is not methylated. Finally, you amplify by polymerase chain reaction (PCR) a locus on the X chromosome known to be polymorphic in length (i.e. it is expected to be of different sizes in different X

chromosomes). The locus has a restriction site for the mentioned enzyme, such that cleavage would prevent PCR amplification. You quantify the amount of PCR products corresponding to shorter and longer versions of the locus, and obtain the results shown in the following table. Do these data appear to be in better agreement with a monoclonal (M) or a polyclonal (P) origin of cancer? A monoclonal origin would mean that all cells in the tumor are the clonal descendants of a single abnormal cell, while a polyclonal tumor is composed of cells from different lineages. Write M or P as your answer.

Healthy tissue sample

Tumor sample

Undigested sample

Longer PCR product Shorter PCR product

49% 51%

Digested sample

Longer PCR product Shorter PCR product

47% 53%

99% 1%

8. Indicate true (T) and false (F) statements below regarding cancer. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) Cancer can be induced by infectious agents such as viruses. ( ) The earlier a cancer is diagnosed, the better the chances are for a cure. ( ) Most cancers originate from a single aberrant cell. ( ) A single mutation is NOT enough to turn a normal cell into a cancer cell. 9. The requirement for accumulation of multiple mutations in cancer progression is manifested in the normalized percentage of new cases of cancer diagnosed in different age groups. Which of the curves A to E in the following graph better represents the incidence of human cancers as a function of age?

Rate of new cancer incidence per 100 individuals

A B C D E

0 Birth

80 years Age

10. Indicate whether each of the following descriptions better applies to a cancer cell (C) or a normal adult cell (N). Your answer would be a four-letter string composed of letters C and N only, e.g. CCNC. ( ) Higher lactate production ( ) Higher oxidative phosphorylation ( ) Contact inhibition ( ) Anchorage independence 11. In medical oncology, PET (positron emission tomography) is used to selectively image tumors in the body and to monitor cancer progression and response to treatment. Before performing a PET scan, the patient should fast for at least several hours for blood glucose to be sufficiently low. At the time of the scan, the positron-emitting glucose analog fluorodeoxyglucose (FDG) is injected into the bloodstream and the patient is asked to wait for up to an hour while avoiding physical activity. Finally, the scanner moves slowly over the body to reveal the location of possible tumors. Why do you think the patient should avoid physical activity before the scan? A. To prevent the Warburg effect B. To accelerate glucose uptake by the tumor cells C. To prevent the absorption of the radioactive tracer by healthy tissues D. To promote fermentation in healthy tissues E. All of the above 12. Compared to cells of a normal tissue, which of the following occurs less frequently in cells within a solid tumor?

A. Apoptosis B. Necrosis C. Cell division D. Mitotic recombination E. Stress 13. Indicate true (T) and false (F) statements below regarding cell proliferation in human somatic cancer cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) Cancer cells show replicative cell senescence. ( ) Cancer cells maintain their telomeres by inhibiting the enzyme telomerase. ( ) Some cancer cells do not rely on telomerase for telomere lengthening. ( ) Most cancer cells lack telomeres. 14.

Which of the following proteins is NOT encoded by a proto-oncogene? A. Src B. Ras C. EGF receptor D. Myc E. E-cadherin

15. Mutations in two important cancer-critical genes, encoding p53 and Rb, respectively, are commonly found in cancers. What type of mutations are these expected to be? A. Loss-of-function mutations in both genes B. Loss-of-function mutation in p53 and gain-of-function mutation in Rb C. Gain-of-function mutation in p53 and loss-of-function mutation in Rb D. Gain-of-function mutations in both genes 16. Indicate whether each of the following descriptions better applies to proto-oncogenes (P) or tumor suppressor genes (T). Your answer would be a four-letter string composed of letters P and T only, e.g. PPPT. ( ) Cancer mutations in these genes are usually recessive. ( ) Cancer mutations in these genes include gene duplications. ( ) Cancer mutations in these genes are responsible for most hereditary cancers. ( ) Cancer mutations in these genes are commonly in the form of nonsense (truncating) mutations that abort protein synthesis.

17. The homologous chromosome pairs in our cells do not carry identical sequences in all loci. This heterozygosity (difference between the two copies) can be altered in cancer: in fact, loss of heterozygosity at many loci is observed in cancer cells, through an increase in either homozygosity (two identical copies) or hemizygosity (i.e. loss of one copy). Researchers can take advantage of this loss of heterozygosity in cancer cells to identify genomic loci that contain cancer-critical genes. What type of gene would you expect to find in chromosomal regions with a loss of heterozygosity? Proto-oncogenes (P) or tumor suppressor genes (T)? Write down P or T as your answer. 18.

The immortalized non-malignant mouse cell line NIH-3T3 was derived from normal

mouse fibroblasts in the early 1960s. These cells are able to readily take up exogenous DNA and are prone to transformation by cancer-causing agents, including some retroviruses. DNA extracted from a human bladder carcinoma line is able to transform these cells, as judged by a significant increase in the number of foci (cell clumps) in the cell-culture plates when the DNA is added. The malignant cells contain human DNA, and the DNA can be shown by sequence analysis to contain a single mutant gene that is present in the original bladder carcinoma cell line. The gene codes for a monomeric G protein and was one of the first cancer-critical genes to be identified in this way. The protein encoded by this gene is … A. Src B. Ras C. Myc D. p53 E. Apc 19. Retinoblastoma is an early-onset cancer of the retina with a rapid progression, and is mostly diagnosed in children. In its hereditary form, multiple eye tumors usually arise in both eyes, while the nonhereditary form usually causes fewer tumors in only one eye. Treatment may involve a combination of chemotherapy, radiotherapy, and other therapies and the majority of patients can be cured if given the right treatment. However, survivors of one form of retinoblastoma (and not the other form) have a markedly increased frequency of subsequent neoplasms that can lead to other cancers later in life, especially soft-tissue sarcomas. These patients should therefore be closely monitored throughout their lives. Which gene is affected by the primary driver mutation in this cancer as well as the later sarcomas? Which form of retinoblastoma do you think is associated with a higher risk of subsequent neoplasms? A. p53; hereditary

B. Rb; hereditary C. Ras; nonhereditary D. p53; nonhereditary E. Rb; nonhereditary 20. The effect of the deletion of one copy of the gene encoding p53 is different from the effects caused by other p53 mutations. For example, some loss-of-function mutations in the DNA-binding domain of p53 cripple its function as a transcription regulator. Such a mutation in only one copy of the p53 gene can be enough to confer a p53 loss-of-function phenotype, even when the other copy of the gene on the homologous chromosome is wild type. This is because … A. p53 is a proto-oncogene. B. these mutations are recessive. C. p53 is a tumor suppressor. D. p53 forms a tetramer. E. p53 can induce apoptosis. 21. You have analyzed a large set of human cancer-critical genes for a selected group of carcinomas, classifying each of the genes based on whether they are known to undergo somatic or germ-line mutations, as well as based on whether they are dominant or recessive. You then group them and plot the statistics in the following histograms. Which group (a or b) do you think represents somatic, as compared to germ-line, mutations? Which group (1 or 2) do you think represents dominant, as compared to recessive, mutations? (Note that the sum of percentages of somatic and germ-line mutations is more than 100%, since some genes are mutated in both somatic and germ cells).

Percentage of total genes studied

100

0 1

2 b

Mutation groups A. a; 1 B. a; 2 C. b; 1 D. b; 2

22. Indicate true (T) and false (F) statements below regarding the mutational landscape of cancer cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) In each cancer, usually there is one driver mutation and a large number of passenger mutations. ( ) It is estimated that about 20% of our genes are cancer-critical. ( ) Cancer-critical genes can encode metabolic enzymes or components of the RNA splicing machinery. ( ) The karyotype is often severely disordered in cancer cells. 23. Myelomas are cancers of blood plasma cells—white blood cells that are normally responsible for producing large quantities of antibodies. In the following “Circos plot” for myelomas in a hypothetical mammalian genome, the interchromosomal rearrangements are indicated by red lines and variations in copy numbers are indicated in blue. The positions of named genes are indicated with arrows. On which chromosome do you expect to find the antibody genes? Write down the chromosome number (1 to 5) as your answer.

Wnt

Ras

1 Myc

FGF receptor

Rb 2 p53

24.

Three fundamental controls seem to have been subverted in essentially every type of

cancer. Choose these three among the following regulatory axes. Your answer would be a threeletter string composed of letters A to F only, in alphabetical order, e.g. BDF. (A) Wnt pathway (B) Rb pathway (C) RTK/Ras/PI3K pathway (D) p53 pathway (E) Hippo pathway (F) GPCR/PKA pathway 25. For each of the following genes involved in regulation of cell growth through the mTOR pathway, indicate whether the gene is activated (A) or inactivated (I) in cancer cells compared to normal healthy cells. Your answer would be a four-letter string composed of letters A and I only, e.g. AAAA. ( ) mTOR ( ) Akt ( ) PTEN

( ) PI3K 26.

Genetically knocking out both copies of the p53 gene in rats … A. is embryonic lethal. B. results in a lower malignancy rate, but the rats are otherwise seemingly normal. C. results in a higher rate of cancer onset, but the rats are otherwise seemingly normal. D. increases cell death by apoptosis, leading to developmental defects. E. does not have any effect unless the rats live outside of the laboratory and are exposed to various types of stress.

27.

Which of the following can lead to p53 stabilization and activation? A. Hypoxia B. Overexpression of Myc C. DNA damage D. Telomere loss E. All of the above

28. From an immortalized human HeLa cell line with wild-type p53 genes, you have derived a line that lacks both copies of the gene. You then treat the original and derived cells with the anticancer drug doxorubicin, which can activate the p53 pathway in the cell by stalling DNA replication forks and inducing double-strand breaks in DNA. You measure cell proliferation in the presence of different doses of the drug in each of the two cell lines, and plot the results as shown in the graph below. Which cell line (A or B) do you expect to be the original HeLa line? Write down A or B as your answer.

Relative cell proliferation

1 A

0 0.1

100

Concentration of doxorubicin (nM, log scale) 29. Indicate true (T) and false (F) statements below regarding cancer. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) Cancers become less and less heterogeneous as they progress. ( ) Knocking out Ras or Myc genes individually leads to a higher incidence of cancers in mice, and knocking out both genes simultaneously has an even stronger phenotype. ( ) Wnt signaling is important in colon epithelial cells and, correspondingly, mutations in genes in the Wnt pathway are present in most colorectal cancers. ( ) Genome destabilization in a subset of colorectal cancers that have defects in DNA mismatch repair takes the form of chromosome breaks, translocations, and deletions. 30. Which of the following sequential barriers to metastasis is the easiest to overcome for cancer cells in general? A. Vessel entry through acquisition of local invasiveness B. Exit from the blood into a remote tissue or organ C. Survival of cells in the foreign tissue D. Initial growth of cells in the foreign tissue E. Persistence of growth in the remote site 31. Carcinoma cells that have acquired malignancy and started local invasiveness to begin metastasis … A. decrease the expression of E-cadherin and undergo mesenchymal–epithelial transition. B. increase the expression of E-cadherin and undergo mesenchymal–epithelial transition.

C. decrease the expression of E-cadherin and undergo epithelial–mesenchymal transition. D. increase the expression of E-cadherin and undergo epithelial–mesenchymal transition. 32.

According to the cancer stem-cell model for tumor growth and propagation, … A. transit amplifying cells are incapable of cell division, which is reserved for the cancer stem cells. B. transit amplifying cells constitute the great majority of the cells in the tumor. C. if each stem cell divides to create one stem cell and one transit amplifying cell, the abundance of stem cells in the tumor will increase exponentially over time. D. tumors are genetically heterogeneous, even though they show high phenotypic homogeneity. E. the transit amplifying cells, even though each acts transiently, carry the whole responsibility for maintenance of the tumor in the long term.

33. Suppose you are studying tumor heterogeneity in a certain type of melanoma. You have used fluorescence-activated cell sorting (FACS) to specifically isolate those melanoma tumor cells that either do (first category) or do not (second category) express a specific marker present in normal stem cells in the tissue of origin (i.e. the melanocyte stem cells). You implant the same number of cells from each of these categories into severely immunodeficient mice and compare the tumor-formation efficiencies after several weeks, which turn out to be significantly higher for the first category. You then analyze the new tumors using FACS, and find out that the majority of the cells in the tumors that originated from the first category of cells harbor the stem-cell marker, whereas the majority of the cells in the tumors that originated from the second cell category lack the marker, just like their respective founder cells. Do these observations support the existence of cancer stem cells? Write down Yes or No as your answer. 34.

Indicate true (T) and false (F) statements below regarding colorectal cancers. Your

answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) Observation of polyps in the colon epithelium of a patient is an indication of a malignant carcinoma. ( ) Progression of colorectal cancer is very slow and normally takes over 10 years to turn into malignancy. ( ) Colorectal cancers are usually diagnosed later in life. ( ) Invasive colorectal cancer cells usually metastasize to lymph nodes via lymphatic vessels and then into the bloodstream.

35. Mutation in which of the following genes is most prevalent in human colorectal cancer cells? A. K-Ras B. β-Catenin C. Apc D. p53 E. MLH 36.

The Rb gene in retinoblastomas is similar to the Apc gene in polyposis colon carcinomas

in that both genes … A. are tumor suppressors. B. are mutated in one copy in all cells of patients with a hereditary form of the cancer. C. are in a locus that shows loss of heterozygosity in the hereditary form of the cancer. D. should be inactivated in both copies to cause the nonhereditary form of the cancer. E. All of the above. 37. The genotypes of 400 colorectal cancer tumors are tabulated below, where the number of tumors with or without mutations in each of the two cancer-critical genes, β-catenin and Apc, are indicated. Which row (a or b) corresponds to those with a mutant Apc gene? Which column (1 or 2) corresponds to those with a mutant β-catenin gene? β-catenin 1 2 Apc

A. a; 1 B. a; 2 C. b; 1 D. b; 2

a b

351 13

2 34

38. The following simplified diagram shows the typical sequence of genetic changes in a developing colorectal carcinoma. Indicate which event (A to C) corresponds to the following changes. Your answer would be a three-letter string composed of letters A to C only, e.g. CAB. Normal epithelium A Early adenoma B Intermediate and late adenomas C Adenocarcinoma and metastases ( ) Activation of K-Ras ( ) Loss of p53 ( ) Loss of Apc 39. You have karyotyped cells from two colorectal tumor samples, one from a hereditary nonpolyposis colorectal cancer (HNPCC) patient, and the other from a familial adenomatous polyposis coli (FAP) patient. One group of the karyotypes shows gross chromosomal abnormalities with extra or deleted chromosomes and several translocations and deletions. The other group, however, is almost normal, and comparable to noncancerous samples. Which group would you expect to have loss-of-function mutations in the DNA mismatch repair system genes MSH2 and MLH1 as their primary driver mutations? A. HNPCC, which has an almost normal karyotype B. HNPCC, which has a grossly abnormal karyotype C. FAP, which has an almost normal karyotype D. FAP, which has a grossly abnormal karyotype 40. Which of the following is estimated to be the leading cause of death from cancer in the United States?

A. UV light B. Obesity C. Smoking D. Alcohol consumption E. Sedentary lifestyle 41. You are studying the rising incidence of a certain subtype of cervical cancer in Oceania, and are curious to know whether environmental factors are the dominant cause of the disease. You collect the incidence statistics from indigenous populations as well as from two different immigrant populations in three different countries, as shown in the following table. Do these data appear to be consistent with a dominant role of environmental risk factors (E) or a genetic background (G) for this type of cancer? Write down E or G as your answer.

Populations

Annual age-adjusted incidence rate in females (per 100,000 women)

Australia (entire population) Indigenous inhabitants Japanese immigrants (first generation) Melanesian immigrants (first generation)

2.2 2.1 2.0 2.1

Japan (entire population) Indigenous inhabitants Australian immigrants (first generation) Melanesian immigrants (first generation) Melanesia (entire population) Indigenous inhabitants Australian immigrants (first generation) Japanese immigrants (first generation)

3.5 3.6 3.7 3.6 12.9 12.6 12.9 12.9

42. The Ames test is used to test the mutagenicity of a compound suspected to be a carcinogen. In a simple form of the test, the carcinogen is first mixed with a rat liver extract. A disc of filter paper is soaked with this mixture and placed on a culture of a strain of Salmonella typhimurium that is defective in a gene involved in the synthesis of histidine, an amino acid that is essential for cell growth and proliferation. The strain is thus normally unable to grow into visible colonies when the histidine in the culture medium is depleted. In the presence of a mutagen, however, mutations (often “reverse mutations” in the same gene) can enable the

bacteria to produce histidine on their own, and therefore grow into colonies. The results of the Ames test for three compounds A, B, and C—each used at the same concentration—are shown in the schematic diagram below. Colonies are indicated with black dots, and the disc is indicated with a white circle at the center of each plate. Which compound (A to C) appears to be a stronger mutagen in this assay? Write down A, B, or C as your answer.

43. Which of the molecules (A or B) in the following drawing is a more potent mutagen? Write down A or B as your answer.

(A)

(B)

44. Indicate true (T) and false (F) statements below regarding cancer incidence and cancer prevention. Your answer would be a four-letter string composed of letters T and F only, e.g. TFFF. ( ) Over half of all cancers are preventable by lifestyle changes.

( ) The age-adjusted cancer death rates have increased steadily since 1900, mostly due to the industrial way of life. ( ) Currently, more than half of all cancer patients survive at least five years postdiagnosis. ( ) About half of all cancers are thought to arise by infection with viruses, bacteria, or parasites. 45. Indicate whether each of the following viruses is mostly associated with cervical cancer (C), Kaposi’s sarcoma (K), liver cancer (L), or stomach cancer (S). Your answer would be a four-letter string composed of letters C, K, L, and S, e.g. CKLS. ( ) Hepatitis-B virus (HBV) ( ) Human immunodeficiency virus (HIV) ( ) Human papillomavirus (HPV) ( ) Helicobacter pylori 46.

Carcinoma of the uterine cervix in humans … A. can be largely prevented by vaccination. B. is caused by a retrovirus. C. is caused by activation of a viral Src kinase. D. rapidly progresses (from uterine warts) to malignancy. E. All of the above.

47.

Most DNA tumor viruses inhibit the products of … A. Apc and Rb B. Brca1 and Rb C. p53 and Rb D. Brca1 and Apc E. Brca1 and p53

48. PARP inhibitors can efficiently kill many breast cancer cells that lack functional Brca1 or Brca2 genes. How do these drugs accomplish this? A. By increasing the ability of p53 in cancer cells to limit cell proliferation B. By inhibiting proteins that are normally inhibited by the Brca1 or Brca2 gene products C. By increasing the occurrences of homologous recombination D. By inhibiting a DNA repair pathway

E. By increasing the ability of the cancer cells to repair the mutations in their cancercritical genes 49. Once the molecular aberrations in a cancer are understood, drugs can be designed with a rational approach to treat the cancer. Which of the following is NOT true regarding such drugs? A. The drug imatinib (Gleevec®) can inhibit the chimeric tyrosine kinase Bcr-Abl in chronic myelogenous leukemia (CML). B. Even if a protein is not the product of a cancer-critical gene, drugs that specifically block its activity can be effective in curing cancer. C. Imatinib is most effective in treating chronic myelogenous leukemia (CML) in its acute blast-crisis phase. D. Protein kinases have turned out to be relatively easy to inhibit with small molecules. E. The success of imatinib relies on the phenomenon of oncogene dependence in cancer cells. 50. Indicate whether each of the following descriptions better applies to trastuzumab (T) which targets Her2, imatinib (I) which targets Bcr-Abl, or ipilimumab (P) which targets the CTLA4 protein. Your answer would be a three-letter string composed of letters T, I, and P only, e.g. TTP. ( ) It is NOT an antibody. ( ) It counters the immunosuppressive microenvironment of tumors. ( ) It does not bind to a cancer cell component.

Answers: 1. Answer: 1 Difficulty: 1 Section: Cancer as a Microevolutionary Process Feedback: Only the germ line contributes to the organism in the next generation; however, by sacrificing their survival to support the germ cells, the somatic cells help to propagate copies of their own genome. 2. Answer: E Difficulty: 3 Section: Cancer as a Microevolutionary Process Feedback: The number of cancer-related mutations is calculated from these assumptions as: (1015 cell divisions) × (10–10 mutations/nucleotide/division) × (2 × 109 nucleotides/genome) × (5 × 10–2 genic mutation/mutation) × (10–3 cancerous mutations/genic mutations) = 1010 cancerous mutations/genome. 3. Answer: TFTF Difficulty: 1 Section: Cancer as a Microevolutionary Process Feedback: Unrestrained growth and division plus invasiveness are two key features of true malignant cancer cells. Cells within a tumor still die in large numbers. 4. Answer: CCCN Difficulty: 2 Section: Cancer as a Microevolutionary Process Feedback: It is not surprising that many cancers are derived from epithelial cells (i.e. are carcinomas) as these cells are more proliferative and also more exposed to the environment. 5. Answer: B Difficulty: 1 Section: Cancer as a Microevolutionary Process Feedback: Benign and malignant tumors of cartilage are termed chondroma and chondrosarcoma, respectively. 6. Answer: A Difficulty: 2 Section: Cancer as a Microevolutionary Process Feedback: After the cancers of the respiratory system (lung), the next leading cancers by mortality in the US are those of the digestive organs, reproductive tract, and breast. 7. Answer: M Difficulty: 3 Section: Cancer as a Microevolutionary Process

Feedback: The individual is heterozygous in the locus, represented in a ratio close to 1:1 for the PCR products of the longer to shorter alleles when an undigested DNA is used. When fully digested, however, only the methylated DNA is expected to be represented in the PCR products. Whereas the polyclonal healthy tissue sample still has a mostly unbiased 1:1 ratio, the monoclonal tumor cells (all with the same inactive X chromosome) have a skewed ratio. 8. Answer: TTTT Difficulty: 1 Section: Cancer as a Microevolutionary Process Feedback: A single mutation is not enough to change a normal cell into a cancer cell: most cancers develop gradually from a single aberrant cell by the accumulation of a number of genetic and epigenetic changes over time. Treatment is generally easier if the cancer is diagnosed at earlier stages. Some cancers can be induced by infectious agents. 9. Answer: E Difficulty: 2 Section: Cancer as a Microevolutionary Process Feedback: The incidence of cancers grows exponentially as a function of age in adulthood, although it is thought to reach a plateau or even decline after the age of 80. 10. Answer: CNNC Difficulty: 2 Section: Cancer as a Microevolutionary Process Feedback: Transformed cells display an altered growth control (e.g. lack of contact inhibition or anchorage dependence) and an altered sugar metabolism (e.g. deemphasized oxidative phosphorylation, increased glucose uptake, and increased lactic acid fermentation). 11. Answer: C Difficulty: 2 Section: Cancer as a Microevolutionary Process Feedback: When the blood glucose level is low, the added radioactive glucose analog is preferentially and rapidly taken up by tumor cells as a result of the Warburg effect. Physical activity stimulates uptake by muscle cells also (including heart muscle), which interferes with the imaging. 12. Answer: A Difficulty: 1 Section: Cancer as a Microevolutionary Process Feedback: Although cancer cells tend to avoid apoptosis, they still die on a massive scale by necrosis in solid tumors. 13. Answer: FFTF Difficulty: 2 Section: Cancer as a Microevolutionary Process

Feedback: Mammalian cancer cells avoid replicative cell senescence (which generally depends on telomere shortening) in two ways: they can either maintain telomerase activity to prevent telomere shortening, or evolve an alternative mechanism based on homologous recombination to lengthen their telomeres. 14. Answer: E Difficulty: 1 Section: Cancer-critical Genes Feedback: Loss-of-function mutations in the tumor suppressor E-cadherin promote the epithelial–mesenchymal transition and local invasiveness. 15. Answer: A Difficulty: 1 Section: Cancer-critical Genes Feedback: Both p53 and Rb are coded by tumor suppressor genes, and their inactivation promotes cancer. 16. Answer: TPTT Difficulty: 2 Section: Cancer as a Microevolutionary Process Feedback: Truncating mutations in tumor suppressor genes can create recessive, loss-offunction mutants associated with most hereditary cancers. A proto-oncogene, in contrast, can be activated by a limited set of dominant, gain-of-function mutations (e.g. missense point mutations or other accidents such as gene duplication, which can produce an abnormally large amount of gene product). 17. Answer: T Difficulty: 3 Section: Cancer as a Microevolutionary Process Feedback: Once the first copy of a tumor suppressor gene is lost or inactivated, the remaining copy is commonly lost by a less specific mechanism, leading to loss of heterozygosity in the gene as well as in its neighboring loci. This represents a way of finding loci that contain tumor suppressor genes. 18. Answer: B Difficulty: 1 Section: Cancer-critical Genes Feedback: Ras was also found earlier to be the oncogene carried by some sarcoma viruses, and the identification of the same gene (with similar mutations) in both cases was an important discovery that advanced our understanding of the molecular biology of cancer. 19. Answer: B Difficulty: 2 Section: Cancer-critical Genes

Feedback: Due to a germ-line mutation in the Rb gene, people with hereditary retinoblastoma are at a higher risk of cancers in older age. Note that this is different from “relapse” of partially treated retinoblastomas, which can be accounted for by the survival of a few original cancer cells after the initial treatment. 20. Answer: D Difficulty: 2 Section: Cancer-critical Genes Feedback: The dominant negative effect of p53 mutations is due to the fact that defective p53 monomers can block the function of the tetramers in which they participate. 21. Answer: A Difficulty: 3 Section: Cancer-critical Genes Feedback: Most cancers arise from dominant somatic mutations in cancer genes; however, most germ-line mutations are recessive, which enables them to be carried through generations. 22. Answer: FFTT Difficulty: 2 Section: Cancer-critical Genes Feedback: The number of driver mutations in most cancers is estimated to be on the order of 10. About 300 genes (less than 2% of our genes) are strongly suspected to be cancercritical. They encode proteins of various functions. In addition to DNA sequence changes, chromosomal breaks and translocations are common in cancer cells. 23. Answer: 5 Difficulty: 3 Section: Cancer-critical Genes Feedback: Translocation between immunoglobulin loci and proto-oncogene loci activates the proto-oncogenes and can transform the cell. These translocations are common in myelomas, where each one takes advantage of the powerful transcription potential of the immunoglobulin genes. 24. Answer: BCD Difficulty: 1 Section: Cancer-critical Genes Feedback: Genes in each of these pathways are mutated in most cancer cells that have been analyzed. 25. Answer: AAIA Difficulty: 2 Section: Cancer-critical Genes Feedback: Activation of the kinase subunit of mTOR (stimulated by PI3K and Akt kinases, and inhibited by PTEN phosphatases) can lead to an increase in nutrient uptake and utilization in cancer cells.

26. Answer: C Difficulty: 2 Section: Cancer-critical Genes Feedback: Most rats (or mice) with a homozygous knockout of the gene encoding p53 develop cancer within a few months after birth but appear normal otherwise, suggesting that p53 function is required only in special circumstances. 27. Answer: E Difficulty: 1 Section: Cancer-critical Genes Feedback: Cells raise their concentration of p53 in response to a whole range of conditions. 28. Answer: B Difficulty: 3 Section: Cancer-critical Genes Feedback: Cells in the original HeLa line better respond to the drug since they can trigger apoptosis through the p53 pathway. Knocking out p53 makes the cells more resistant to this effect of the drug. 29. Answer: FFTF Difficulty: 2 Section: Cancer-critical Genes Feedback: Cancer risk is increased by gain-of-function mutations in Ras and Myc, not by their knockout. It is not surprising that in cancer cells with defects in DNA mismatch repair, many point mutations are found throughout the genome. Cancer progression coincides with increasing heterogeneity of the tumor cell population. 30. Answer: B Difficulty: 1 Section: Cancer-critical Genes Feedback: Traveling through the body’s circulation, which includes survival in the circulation, arrest in capillaries or small vessels, and extravasation into remote tissues, is generally more efficient than the prior phase of escape from the parent tissue and the later phase of colonization of a remote site. 31. Answer: C Difficulty: 1 Section: Cancer-critical Genes Feedback: The transition in cancer cells to become invasive involves shifting to a less adhesive and more motile character, resembling the epithelial–mesenchymal transition in normal development. A key part in this process involves switching off the expression of E-cadherin, which is a cell adhesion molecule. 32. Answer: B Difficulty: 1

Section: Cancer-critical Genes Feedback: Cancer stem cells can be responsible for tumor growth and maintenance, yet remain only a small part of the tumor cell population. 33. Answer: No Difficulty: 3 Section: Cancer-critical Genes Feedback: Cancer stem cells are expected to constitute a minority of the tumor cell population. Even though the transplantation is more efficient with the subpopulation of cells that expresses the stem-cell marker, this might simply reflect the heterogeneity in the original tumor, with these cells harboring a higher ability to found new tumors. 34. Answer: FTTT Difficulty: 2 Section: Cancer-critical Genes Feedback: The adenomatous polyps are believed to be the precursors of most colorectal cancers, but progression into malignancy is slow and the average time from detection of the polyp to cancer diagnosis is about 12 years. This usually happens later in life. Colorectal cancers tend to metastasize through lymphatic nodes after spreading through the layers of tissues lining the gut. 35. Answer: C Difficulty: 1 Section: Cancer-critical Genes Feedback: The Apc tumor suppressor gene is mutated (and is defective) in over 80% of colorectal cancers. 36. Answer: E Difficulty: 2 Section: Cancer-critical Genes Feedback: Apc and Rb are tumor suppressors, and inactivation of both copies of each gene occurs in colorectal cancers and retinoblastomas, respectively. The hereditary forms, however, are easier to develop since the person receives only one functional copy to begin with, which can become lost or defective. 37. Answer: B Difficulty: 2 Section: Cancer-critical Genes Feedback: The majority of colorectal carcinoma cells have a loss-of-function mutation in Apc. Among those that do not, the majority have a gain-of-function mutation in βcatenin. Having both mutations is not common, implying that the activation of the Wnt pathway, and not the individual mutations, confers a selective advantage to the tumor cells. 38. Answer: BCA Difficulty: 2

Section: Cancer-critical Genes Feedback: This oversimplified diagram provides a general correspondence between mutations and the stages of cancer progression. In most cases, inactivating Apc mutations appear to occur early, followed by later activation of K-Ras and inactivation of p53. Many other mutations are generally involved. Note that genetic and epigenetic instability rises during the process. 39. Answer: A Difficulty: 2 Section: Cancer-critical Genes Feedback: HNPCC tumor cells show an almost normal karyotype, in sharp contrast to FAP tumor cells. 40. Answer: C Difficulty: 1 Section: Cancer Prevention and Treatment Feedback: Smoking is estimated to account for about 33% of cancers and to kill near 190,000 people in the US every year as a result. 41. Answer: E Difficulty: 3 Section: Cancer Prevention and Treatment Feedback: The incidence rates for this type of cancer in each immigrant population mirrors that of the host country, suggesting an overall dominance of environmental factors. 42. Answer: B Difficulty: 3 Section: Cancer Prevention and Treatment Feedback: More colonies (revertants) appear in the presence of a stronger mutagen, with the distribution of colonies reflecting the dose–response relationship for the mutagen (stronger mutagens will affect bacteria further away from the disc). 43. Answer: B Difficulty: 1 Section: Cancer Prevention and Treatment Feedback: Unfortunately, carcinogens such as aflatoxin B1 are rendered more potent by the P-450 detoxification system in the liver. 44. Answer: TFTF Difficulty: 2 Section: Cancer Prevention and Treatment Feedback: Except for the increase in cancers caused by smoking, age-adjusted death rates for most common cancers have either stayed the same since half a century ago, or have even declined significantly. Survival rates have also improved: currently, more than twothirds of cancer patients live more than five years from the time of diagnosis. Fifty

percent of cancers could be prevented by changes in lifestyle such as altered smoking, eating, and exercise habits. Only a small proportion of human cancers (perhaps ~15%) are thought to arise from infection. 45. Answer: LKCS Difficulty: 1 Section: Cancer Prevention and Treatment Feedback: A small but significant proportion of human cancers is thought to arise from infections. 46. Answer: A Difficulty: 1 Section: Cancer Prevention and Treatment Feedback: Cervical cancer is mostly a result of papillomavirus infection transmitted sexually. Accidentally, genes in this DNA virus that interfere with p53 function are expressed in basal epithelial cells of the uterine cervix, leading to lesions that can very slowly develop into cancer. 47. Answer: C Difficulty: 1 Section: Cancer Prevention and Treatment Feedback: Small DNA tumor viruses encode proteins that interfere with the Rb and p53 pathways, allowing the replication of the viral genome. 48. Answer: D Difficulty: 2 Section: Cancer Prevention and Treatment Feedback: PARP inhibitors interfere with a DNA repair pathway that becomes essential for cancers that lack the alternative Brca-dependent pathway. 49. Answer: C Difficulty: 1 Section: Cancer Prevention and Treatment Feedback: Imatinib is not very successful in treating CML once the cancer has advanced to a terminal phase, blast crisis, and no longer behaves like a “chronic” leukemia. 50. Answer: IPP Difficulty: 2 Section: Cancer Prevention and Treatment Feedback: By binding to an inhibitory receptor on T cells, the monoclonal antibody ipilimumab helps block immunosuppression in tumors. Trastuzumab is also a monoclonal antibody, but it directly binds to EGF receptors on some cancer cells to block their function.

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION CHAPTER 21: DEVELOPMENT OF MULTICELLULAR ORGANISMS Copyright © 2015 by W.W. Norton & Company, Inc. 1. Plants and animals use different developmental strategies and have very different ways of life. In which of the following fundamental cellular processes during development are plants most different to animals? A. Cell proliferation B. Cell–cell interactions C. Cell specialization D. Cell movement 2. The simplified drawing below depicts early stages of animal development. Indicate which letter (A to E) in the drawing corresponds to each of the following terms. Your answer would be a five-letter string composed of letters A to E only, e.g. ECDBA. D C

( ) Blastula ( ) Ectoderm ( ) Endoderm ( ) Mesoderm ( ) Gastrula

3. Indicate whether each of the following organs or tissues arises from ectoderm (C), mesoderm (M), or endoderm (N). Your answer would be a four-letter string composed of letters C, M, and N only, e.g. MMCC. ( ) Blood ( ) Liver and pancreas ( ) Brain ( ) Bone and cartilage 4. Indicate true (T) and false (F) statements below regarding animal development. Your answer would be a four-letter string composed of letters T and F only, e.g. FFFF. ( ) A fertilized egg is totipotent. ( ) Differences in their regulatory DNA can largely explain the differences between animal species. ( ) Inductive signaling is mostly mediated through G-protein-coupled receptors. ( ) A cell’s response to a signal depends on its exposure to other signals at that present time as well as in the past. 5. Imagine a morphogen gradient established from left to right in a field of cells in a developing tissue, as shown in the following schematic diagram. Below a first threshold of morphogen concentration, cells do not respond to the morphogen and express gene “red” by default. Cells exposed to morphogen concentrations above this threshold respond by expressing gene “white” instead, while those exposed to even higher concentrations, above a second threshold, express gene “blue.” As shown, the initial pattern resembles a French flag with equally wide blue, white, and red expression domains. With no other change, if the diffusion rate of the morphogen is increased (by a modification that decreases its affinity for heparan sulfate proteoglycans, for example), the gradient profile changes from the gray curve to the black curve, as indicated. Under this new condition, indicate whether each of the following would be expected to increase (I), decrease (D), or remain unchanged (U) in its range. Your answer would be a three-letter string composed of letters I, D, and U only, e.g. UUI.

Morphogen concentration

Position

Initial pattern ( ) Blue expression domain ( ) White expression domain ( ) Red expression domain 6. Indicate whether each of the following descriptions better applies to patterning by lateral inhibition (L), reaction-diffusion systems (R), or sequential induction (S). Your answer would be a four-letter string composed of letters L, R, and S only, e.g. SSRR. ( ) It does NOT generate asymmetrical patterns from an initial noisy field. ( ) It is based on short-range activation and long-range inhibition. ( ) It can readily generate complex patterns resembling the spots of a leopard or stripes of a zebra. ( ) It is commonly mediated by Notch signaling. 7. Indicate whether each of the following descriptions better applies to these poles established in an early embryo: anterior (A), posterior (P), dorsal (D), ventral (V), animal (N), or vegetal (G). Your answer would be a six-letter string composed of letters A, D, G, N, P, and V only, e.g. ADGPVN. ( ) It defines the part to become internalized in gastrulation. ( ) It defines the parts to remain external. ( ) It defines the location of the future head. ( ) It defines the location of the future tail. ( ) It defines the location of the future belly. ( ) It defines the location of the future back.

8. Sort the following organisms from the least to the most degree of asymmetry in the unfertilized egg (i.e. pre-defined axes of polarization). Your answer would be a three-letter string composed of letters A to C only, e.g. BCA. (A) M. musculus (B) D. melanogaster (C) X. laevis 9. Sort the following primary axes in the order that they are established during the development of Xenopus laevis. Your answer would be a three-letter string composed of letters A to C only, e.g. ACB. (A) A-P (B) D-V (C) A-V 10. Cortical rotation following fertilization in X. laevis places the …(1) pole at the point of sperm entry, while Wnt11 mRNA is transported to the …(2) pole. A. 1: anterior; 2: posterior B. 1: posterior; 2: anterior C. 1: animal; 2: vegetal D. 1: dorsal; 2: ventral E. 1: ventral; 2: dorsal 11.

Which of the following is true regarding maternal-effect genes? A. Bicoid and Nanos are maternal-effect genes. B. A female homozygous for a loss-of-function maternal-effect gene mutation can be fully normal, but her offspring will show the phenotype. C. The offspring of a female homozygous for a loss-of-function maternal-effect gene mutation will show the phenotype regardless of the paternal genotype. D. The second-generation offspring of a male homozygous for a loss-of-function maternal-effect gene mutation can show the phenotype. E. All of the above.

12. Indicate whether each of the following descriptions better applies to egg-polarity genes (E), gap genes (G), pair-rule genes (P), or segment-polarity genes (S). Your answer would be a four-letter string composed of letters E, G, P, and S only, e.g. SSGG. ( ) Mutations in these genes show a maternal effect.

( ) Mutations in these genes leave the embryo with only half the number of normal segments. ( ) Mutations in these genes produce a normal number of segments but with part of each segment replaced by a mirror-image duplicate of other parts of the segment. ( ) Mutations in these genes can eliminate one or two groups of adjacent segments altogether. 13. In a developing Drosophila melanogaster embryo, a hierarchy of gene regulatory interactions subdivides the embryo to regulate progressively finer details of patterning. For each of the following proteins, indicate which expression pattern (1 to 4) in the schematic drawing below is more appropriate. Your answer would be a four-digit number composed of digits 1 to 4 only, with each digit used once, e.g. 3412.

4 anterior

( ) Hunchback (product of a gap gene) ( ) Engrailed (product of a segment-polarity gene) ( ) Bicoid (product of an egg-polarity gene) ( ) Even-skipped (product of a pair-rule gene)

posterior

14. Indicate whether each of the following groups of genes typically creates a transient pattern in the developing embryo (T) or a long-lived pattern that is preserved (L). Your answer would be a five-letter string composed of letters T and L only, e.g. TTTLT. ( ) Egg-polarity genes ( ) Pair-rule genes ( ) Hox genes ( ) Gap genes ( ) Segment-polarity genes 15.

Which of the following schematic drawings better depicts the regulatory network that

maintains Engrailed (en), Hedgehog (hh), and Wingless (wg) expression following cellularization in a developing Drosophila embryo?

A anterior

posterior

16. In Drosophila melanogaster, the expression of genes Ultrabithorax and Antennapedia can normally be observed in the third thoracic segment which bears a pair of legs as well as a pair of halteres. Homeotic mutations associated with these two genes can give rise to remarkable disturbances in the organization of the adult fly: two pairs of wings in the case of Ultrabithorax, and legs in the place of antennae in the case of Antennapedia. Would you expect these to be gain-of-function or loss-of-function mutations? A. Loss-of-function for both genes B. Loss-of-function for Ultrabithorax and gain-of-function for Antennapedia C. Gain-of-function for Ultrabithorax and loss-of-function for Antennapedia D. Gain-of-function for both genes

17. The gene clusters known as the Bithorax complex and the Antennapedia complex contain… A. a subset of segment-polarity genes. B. segmentation genes. C. homeobox-containing genes. D. genes encoding chromatin repressors. E. genes encoding chromatin remodelers. 18. Indicate true (T) and false (F) statements below regarding Hox genes. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) The order of expression of Hox genes along the body corresponds to their order in the Hox complex. ( ) Generally, the more anterior of the Hox genes dominate (or suppress) posterior Hox genes. ( ) When a posterior Hox gene is artificially expressed in an anterior region of the embryo, the tissue maintains its anterior character. ( ) Hox genes control the A-P axis in both vertebrates and invertebrates. 19.

Which of the following is correct regarding the regulation and maintenance of Hox gene

expression in Drosophila? A. Proteins of the Polycomb and Trithorax groups maintain inactive and active states of Hox gene expression, respectively. B. If Polycomb or Trithorax group genes are defective, Hox gene expression patterns are still initiated, but cannot be correctly maintained. C. The remodeled chromatin at the Hox complex is heritable through cell generations. D. If all the Hox genes in an embryo are deleted, segmentation still occurs but distinct segment identities are lost. E. All of the above. 20. Indicate whether each of the following proteins is most concentrated near the dorsal (D) or ventral (V) side of the Drosophila embryo. Your answer would be a four-letter string composed of letters D and V only, e.g. DDDV. ( ) Decapentaplegic ( ) Dorsal (nuclear fraction) ( ) Toll (active form) ( ) Twist

21. Cactus is a maternal-effect gene coding for an inhibitory protein in the Drosophila Toll signaling pathway that binds to Dorsal in the cytosol and keeps it from nuclear entry. Would you expect maternal loss-of-function mutations in Cactus to give rise to dorsalized or ventralized embryos? What about loss-of-function mutations in Decapentaplegic (Dpp)? A. Dorsalized; dorsalized B. Dorsalized; ventralized C. Ventralized; dorsalized D. Ventralized; ventralized 22.

Assuming that the only function of Short gastrulation (Sog) in fruit flies is to regulate

Decapentaplegic (Dpp), which of the following genetic interactions would you NOT expect to observe? A. Mutant embryos lacking both Sog and Dpp phenotypically resemble those lacking only Dpp. B. Partial loss of Sog suppresses the phenotype of partial loss of Dpp. C. Gain of Dpp suppresses the phenotype of partial loss of Sog. D. Gain of Sog suppresses the phenotype of gain of Dpp.

Concentration

23. The following graph qualitatively represents the gradients of Nodal and Lefty gene products in an early frog embryo. The position along which primary axis is defined by these gradients? Which curves correspond to these two proteins?

1 2 3 Position

A. D-V axis; curves 1 and 2 B. D-V axis; curves 1 and 3 C. A-V axis; curves 1 and 2 D. A-V axis; curves 1 and 3

24. In the following fate map for the Xenopus blastula, indicate which zone in the map (A to D) better corresponds to each of the following fates. Your answer would be a four-letter string composed of letters A to D only, e.g. DCAB. Animal

A Ventral

B C

Dorsal

D Vegetal

( ) Lung ( ) Blood ( ) Brain ( ) Skin 25.

The role of Chordin and Noggin in patterning in developing vertebrates is equivalent to

that of …(1) in Drosophila. They are secreted from the …(2) pole and antagonize the activity of bone morphogenetic factors that are secreted throughout the embryo. A. 1: Decapentaplegic; 2: dorsal B. 1: Decapentaplegic; 2: ventral C. 1: Short gastrulation; 2: dorsal D. 1: Short gastrulation; 2: ventral 26. Which of the following is NOT true regarding cell-type specification in developing animals? A. Members of the MyoD/myogenin family of transcription regulators drive cells to differentiate into muscle cells. B. Members of the Achaete/Scute family of transcription regulators drive the cells to form endothelial layers. C. Many proteins that induce the differentiation of particular cell types belong to the bHLH (basic helix–loop–helix) class of transcription regulators.

D. Eyeless is both necessary and sufficient to generate eye structures composed of various specialized cell types in Drosophila. E. In vertebrates, loss of the Eyeless homolog Pax6 leads to an animal without eyes. 27. The result of Notch-mediated competitive lateral inhibition in a patch of wild-type cells is depicted on the left in the following schematic drawing, in which the black cells have become specialized. Consider a genetic mutation that interferes with intracellular Notch signaling in such a way that Notch can no longer regulate Delta effectively. Which drawing (1 or 2) better represents the outcome with cells bearing this mutation? Do the specialized cells (black hexagons) have active or inactive Notch at their surface?

Wild type

A. Drawing 1; active Notch B. Drawing 1; inactive Notch C. Drawing 2; active Notch D. Drawing 2; inactive Notch 28. Which of the following evolutionary changes better explains morphological differences between different animals despite many common molecular mechanisms governing their development? A. Changes in the types of genes responsible for key developmental processes. B. Changes in regulatory DNA controlling expression of key developmental genes. C. Changes in DNA-binding domains of key transcription regulators involved in development. D. Changes in coding sequences of key developmental genes. E. Changes in copy number of key developmental genes. 29.

Which of the following is NOT true regarding vertebrate segmentation? A. The presomitic mesoderm retreats tailward as new somites are generated.

B. The period of gene-expression oscillations in the presomitic mesoderm determines the size of each somite. C. If feedback delays in the gene-expression oscillations are increased, somite size will consequently increase as well. D. Notch-mediated lateral inhibition ensures that the segmentation clock is different in neighboring cells in the presomitic mesoderm. 30. The gene encoding Hes7 in mouse contains three introns. Any or all of these introns can be deleted to alter the delay associated with transcription and splicing of Hes7 mRNA. As a result of such deletions, either the oscillation frequency of this gene’s expression can change (result 1) or the oscillatory behavior can disappear altogether (result 2). One of these two results is obtained when only one intron is deleted from the gene, while the other result is obtained when all three introns are removed. Would you expect the oscillation frequency to increase or decrease if result 1 is obtained? Which result (1 or 2) would you expect to observe when all three introns are deleted? A. Increase; result 1 B. Increase; result 2 C. Decrease; result 1 D. Decrease; result 2 31. Indicate true (T) and false (F) statements below regarding developmental timing in animals. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTT. ( ) Cell-division cycles generally serve as an intracellular timer that controls the timing of cellular differentiation. ( ) Developmental transitions are often regulated by microRNAs. ( ) Intracellular developmental programs are often followed even if the cells are taken from the developing embryo and maintained in culture. ( ) Timing of developmental transitions can be coordinated by cell–cell interactions as well as globally by hormones. 32. Does the nuclear-to-cytoplasmic ratio increase (I) or decrease (D) during cleavage in early vertebrate development? Would you expect tetraploid embryos to undergo the maternalzygotic transition earlier (E) or later (L) compared to diploids? If the availability of a transcription regulatory protein determines the timing of this transition, should the protein be a transcriptional activator (A) or repressor (R)? Write down your answer as a three-letter string using the letters in the parentheses above, e.g. IEA.

33. TSH is a pituitary hormone that stimulates the production of thyroid hormone by the thyroid gland. This production can be blocked by sodium perchlorate (SP), which inhibits the cellular import of iodine necessary for thyroid hormone synthesis. What would be the effect of TSH or SP exposure, respectively, on the timing of metamorphosis in frog larvae? A. Premature metamorphosis; premature metamorphosis B. Premature metamorphosis; delayed metamorphosis C. Delayed metamorphosis; premature metamorphosis D. Delayed metamorphosis; delayed metamorphosis 34.

The qualitative graph below shows the molecular changes accompanying vernalization in

the flowering plant Arabidopsis thaliana. Indicate which curve (A to D) in the graph better represents the temporal changes in each of the following variables. Your answer would be a fourletter string composed of letters A to D only, with each letter used once, e.g. BACD. B

A D

Autumn

Winter

Spring

Time ( ) Level of Coolair noncoding RNA ( ) Level of Flowering locus C (FlC) gene product (transcriptional repressor) ( ) Level of Flowering locus T (Ft) gene product (transcriptional activator) ( ) Level of repressive chromatin marks at the FlC locus 35. Stromal cell-derived factor 1 (SDF1) is a secreted protein that plays a major role in guiding the migration of various cells during development. It binds to a G-protein-coupled receptor on the surface of the migrating cell, which in turn activates a trimeric G protein containing a Gi subunit. Pertussis toxin (PTX) modifies and inactivates the Gi subunit. What would you expect to observe if migrating cells are treated with PTX? A. They migrate toward the source of SDF1. B. They migrate away from the source of SDF1. C. They migrate in a certain direction chosen randomly.

D. They move in small steps in random directions. E. They lose their motility. 36. The Steel factor/Kit signaling pathway plays an important role in the migration of many types of cells during development as well as in the adult animal. Not surprisingly, loss-offunction mutations in Steel factor result in cell migration defects. Selective inactivation in these cells of which of the following proteins might be expected to rescue (i.e. partially restore to normal) the defective phenotype of Steel loss-of-function mutants? A. Kit, the receptor for Steel B. Bax, an essential apoptotic protein C. Endothelin-3, a survival factor for neural crest cells D. Fibronectin, a multi-adhesive protein of the extracellular matrix E. FGF4, a growth factor 37. In classical experiments done half a century ago, the cells of early frog embryos were disaggregated and later reaggregated in desired combinations. The cells managed to rearrange and sort themselves out into an overall arrangement similar to that of a normal embryo. This effect is mainly due to … A. chemotactic cell motility. B. asymmetric cell division. C. cell adhesion. D. convergent extension. E. lateral inhibition. 38. Fill in the blank in the following paragraph describing collective cell rearrangements. Do not use abbreviations. “Cells form lamellipodia and attempt to crawl over one another, essentially pulling their neighbors inward into a narrow zone. This is accompanied by elongation along the long axis of the narrow zone. This process of … depends on Wnt signaling and is observed multiple times during the development of a vertebrate.” 39. The following schematic drawings show an epithelial sheet. Which example shows a mutant with planar-cell-polarity defects but no other defects?

40.

During branching morphogenesis in lung development, … A. FGF10 is secreted by epithelial cells at the tip of the growing epithelial tubes. B. FGF10 inhibits Shh production by the epithelial cells at the tip of the growing epithelial tubes. C. Shh inhibits FGF10 production by the epithelial cells at the tip of the growing epithelial tubes. D. Shh is produced by epithelial cells at the tip of the growing epithelial tubes. E. None of the above.

41. Which cell behavior depicted below is involved in the tube formation that underlies the development of lungs and trachea?

42. CFSE (carboxyfluorescein succinimidyl ester) is a cell-tracing molecule that, once inside a cell, is modified to yield a highly fluorescent molecule that is retained within the cell. Due to its stability, it can be used to trace cells and follow their divisions: each division dilutes the fluorescent dye twofold. You label T cells from a healthy mouse with CFSE and inject them into either a wild-type strain of mouse or a strain with T cell deficiency, both of which are immunologically compatible with the donor mouse. You later collect and analyze peripheral T cells from the recipient mice and draw the following histograms according to the fluorescent intensity per cell. Based on these results, do you think the total number of T cells in the mouse body is (1) controlled by intracellular programs (as in the thymus, for example) or is (2) regulated as a whole (as in the spleen)? Write down 1 or 2 as your answer.

Number of cells

Wild type

T cell deficient

CFSE fluorescence per cell (log scale)

43. You have obtained leaf samples from three strawberry varieties A, B, and C. You isolate the cells, extract their nuclei, and stain them with propidium iodide, a fluorescent dye that binds quantitatively to DNA. You then use a fluorescence-activated cell sorter (FACS) machine to sort the nuclei based on their fluorescence. Guessing from the results, presented in the following histograms, which variety would you expect to yield larger strawberries?

Number of nuclei

Fluorescence

A. Variety A B. Variety B C. Variety C D. They should all be the same 44. After reaching sexual maturity, the nematode Caenorhabditis elegans normally doubles in size within about two weeks. This doubling is mainly due to … A. cell division. B. DNA replication without cell division. C. reduced cell death.

D. deposition of extracellular matrix. 45. Mutations in certain components of the cell-cycle machinery in Drosophila melanogaster can be used to slow down the rate of progression through the cell cycle in the wing imaginal discs of the fly larvae, without a major effect on the rate of cell growth. As a result, compared to wild-type flies, the wing in the mutant flies would … A. have a larger size (area) with a higher number of relatively normal cells. B. have the same size, but with a higher number of relatively smaller cells. C. have the same size, but with a smaller number of relatively larger cells. D. have a smaller size with the same number of relatively smaller cells. E. have a smaller size with a smaller number of relatively normal cells. 46. Indicate whether each of the following conditions favors a larger (L) or smaller (S) tissue or body size. Your answer would be a four-letter string composed of letters L and S only, e.g. SSSS. ( ) Hippo overexpression ( ) Insulin-like growth factor overexpression ( ) Myostatin deletion ( ) Growth hormone deficiency 47. Sort the following phases in the order that they take place in neural development in vertebrates. Your answer would be a four-letter string composed of letters A to D only, e.g. DACB. (A) Neurons are assigned specific characters according to the place and time of their birth. (B) Neurons form synapses with other neurons or with muscle cells. (C) Synaptic connections are refined and adjusted. (D) Neurons extend axons and dendrites toward their target cells. 48. A cross section of a developing spinal cord in a vertebrate embryo is shown in the schematic drawing below. Indicate which feature in the drawing (labeled A to E) better matches each of the following descriptions. Your answer would be a four-letter string composed of letters A to E only, e.g. AAEB.

B E

( ) It contains motor neurons. ( ) It is the floor plate. ( ) It secretes BMP and Wnt signals. ( ) It contains sensory neurons. 49. Consider the embryonic cerebral cortex of a mammal. Near which side of the cortex are the cell bodies of radial glial cells located? To which side of the cortex do the first-born neurons eventually migrate? A. Inner surface; inner surface B. Inner surface; outer surface C. Outer surface; inner surface D. Outer surface; outer surface 50. In the following schematic drawing, a growing axon (in red) is extended to a target neuron to form three synapses. In its journey, it is guided by a variety of mechanisms. Sort the following mechanisms to reflect the order in which they guide the navigation of the growth cone toward the target neuron in this example. Ignore the final chemoattraction step toward the target neuron itself. Your answer would be a six-letter string composed of letters A to F only, e.g. FACBDE.

(A) Chemoattraction (B) Chemorepulsion (C) Cell surface adhesion (D) Contact inhibition (E) Extracellular matrix adhesion (F) Guidance by pioneer neuron 51. You have grown neurons in a culture dish containing laminin and have induced axon outgrowth. You now incubate the cells in the presence of various concentrations of two secreted proteins (1 and 2). You then fix the cells, stain them with rhodamine-phalloidin (a fluorescent

Collapsed cones (%)

dye that selectively labels actin filaments), and finally use a fluorescence microscope to examine the axons. Some of the axon tips have lost their spread morphology (with lamellipodia shrunk and filopodia retracted) and lack significant fluorescence. You call these “collapsed cones.” In the following graph, the percentage of cones that have collapsed is plotted as a function of concentration of proteins 1 and 2. Do you expect these proteins to be chemoattractants or chemorepellents in guiding axonal growth cones? Which one is more potent? 100 1 2

00 Protein concentration (pM)

200

A. Chemoattractants; protein 1 is more potent. B. Chemoattractants; protein 2 is more potent. C. Chemorepellents; protein 1 is more potent. D. Chemorepellents; protein 2 is more potent. E. Protein 1 is a chemoattractant, and protein 2 is a chemorepellent; protein 2 is more potent. 52. Robo3.1 is an alternative splice form of Robo3 that blocks the reception of Slit signals by the Robo1 and Robo2 receptors. Considering the guidance of commissural neurons crossing the ventral midline in the embryonic spinal cord of a vertebrate, would you expect Robo3.1 activity to increase (I) or decrease (D) as the growth cone crosses the midline? Write down I or D as your answer. 53. Indicate true (T) and false (F) statements below regarding the formation of orderly neural maps in the mammalian brain. Your answer would be a four-letter string composed of letters T and F only, e.g. FFFF. ( ) As a general rule, neurons that fire together avoid each other. ( ) In the tonotopic map of the auditory system, brain neurons are arranged in the auditory cortex according to the pitch of the sound they respond to, like the keys of a piano. ( ) In the optic tectum in the midbrain, brain neurons are arranged along the anteriorposterior axis according to the frequency of light they respond to, like the colors of a rainbow. ( ) If the optic nerve of a frog is cut and the eye is rotated 180° in its socket, the retinal ganglion cells reconnect to their original targets in the optic tectum, therefore creating an inverted map. 54.

If explants that can produce axons from either the anterior or posterior retina are placed

on a culture substratum that is coated by either anterior or posterior tectal membranes, … A. anterior axons grow only on the anterior tectal membranes, whereas posterior axons grow only on the posterior tectal membranes. B. anterior axons grow only on the posterior tectal membranes, whereas posterior axons grow only on the anterior tectal membranes. C. anterior axons grow only on the anterior tectal membranes, whereas posterior axons grow on both anterior and posterior tectal membranes.

D. anterior axons grow only on the posterior tectal membranes, whereas posterior axons grow on both anterior and posterior tectal membranes. E. anterior axons grow on both the anterior and posterior tectal membranes, whereas posterior axons grow only on the anterior tectal membranes. 55. Which of the following is correct regarding neuronal self-avoidance in both invertebrates and vertebrates? A. They both rely on alternative splicing to generate the diversity required to uniquely identify each neuron. B. They both result in tiling, in which not more than one neuron occupies a certain territory. C. They both rely on protocadherins. D. They both use a strategy in which homophilic recognition results in avoidance. E. All of the above. 56.

The axons and dendrites of neurons that lack DSCAM1 altogether … A. avoid other neurons as well as themselves. B. avoid other neurons but not themselves. C. avoid themselves but not other neurons. D. avoid neither themselves nor other neurons.

57. In the following graph, which line (1 to 3) better represents the changes in the level of EphA receptors on retinal axons along the anterior-posterior axis of the mouse retina? Which line better represents the changes in the abundance of EphrinA along the anterior-posterior axis of the mouse optical tectum? Write down your answer as a two-digit number, e.g. 22.

Abundance

posterior Position in retina or tectum

58. You grow neurons in culture in the presence or absence of nerve growth factor (NGF). You perform this experiment using either wild-type neurons or neurons lacking caspase-3, an effector protein required for programmed cell death by apoptosis in these cells. After a day, you examine cell survival in the culture using a microscope. The survival scores (percentage of cells surviving the treatment) are presented in the following table. Which column (a or b) corresponds to the experiment in the presence of NGF? Which row (1 or 2) corresponds to the experiment using wild-type cells? NGF present? Type of cells?

100

A. a; 1 B. a; 2 C. b; 1 D. b; 2 59. An axonal growth cone has reached a muscle fiber and is forming a neuromuscular junction. Which of the following events does NOT normally occur in this process? A. Agrin secreted from the growth cone serves as a signaling molecule affecting the muscle cell. B. LRP4 on the surface of the growth cone serves as a signaling molecule affecting the muscle cell. C. Synaptic vesicles containing the neurotransmitter acetylcholine appear in the growth cone. D. Acetylcholine receptors cluster on the muscle cell plasma membrane at the junction. E. A thin layer of basal lamina forms at the cleft between the two cell membranes at the junction. 60. Consider Agrin signaling in the formation of neuromuscular junctions in vertebrates. Shp2 is a protein tyrosine phosphatase that counteracts the tyrosine kinase activity of MuSK. Rapsyn interacts with the acetylcholine receptors and induces their agrin-dependent clustering.

Indicate whether each of the following is expected to enhance (E) or suppress (S) acetylcholine receptor clustering in this process. Your answer would be a four-letter string composed of letters E and S only, e.g. EEEE. ( ) Specific inhibition of Shp2 ( ) Loss of Rapsyn ( ) Loss of LRP4 ( ) Constitutive activation of MuSK 61. A cat’s left eye was covered during the critical period of development after eye opening. At the age of 6 months, a radioactive tracer molecule was injected into the right eye and allowed to reach the primary visual cortex. The cortex was then sectioned and subjected to autoradiography to reveal the ocular dominance columns. Which stripes would you expect to be wider: the dark stripes due to radioactivity (D) or light stripes (L)? Write down D or L as your answer.

Answers: 1. Answer: D Difficulty: 1 Section: Overview of Development Feedback: Plant cells do not actively migrate through the embryo. 2. Answer: ACEDB Difficulty: 1 Section: Overview of Development Feedback: Please refer to Figure 21–3. 3. Answer: MNCM Difficulty: 1 Section: Overview of Development Feedback: Ectoderm gives rise to the epidermis and the nervous system; endoderm gives rise to the gut tube and its appendages, such as lung, pancreas, and liver; mesoderm gives rise to muscles, connective tissues (e.g. bone and cartilage), blood, kidney, and various other components. 4. Answer: TTFT Difficulty: 1 Section: Overview of Development Feedback: A fertilized egg is totipotent, able to give rise to all the different cell types in an organism. Changes in regulatory DNA seem to be mainly responsible for the dramatic differences between one class of animals and another, even though the coding DNA has been, for the most part, highly conserved between the classes. Most known inductive signaling events during animal development are mediated by the transforming growth factor-β (TGFβ), Wnt, Hedgehog, and receptor tyrosine kinase (RTK) pathways. During development and in an adult organism, the cellular response to a signal depends not only on the identity of the signal, but also on the other signals that the cell is receiving, as well as on the previous experiences of the cell. 5. Answer: DID Difficulty: 3 Section: Overview of Development Feedback: Changes in parameters such as diffusion rate can affect the patterning of cells by a morphogen. In this example, the more shallow gradient widens the gap between positions corresponding to the first and second thresholds—i.e. the “white” expression domain—at the expense of the other two domains. 6. Answer: SRRL Difficulty: 2 Section: Overview of Development Feedback: Lateral inhibition, commonly mediated by Notch signaling, can generate asymmetry when combined with positive feedback. Reaction-diffusion systems, based on

short-range activation and long-range inhibition, are also capable of spontaneous pattern generation through positive feedback, and give rise to complex patterns such as spots and stripes. Sequential induction, on the other hand, is a strategy to refine initial patterns into more complicated ones, and cannot create asymmetrical patterns by itself. 7. Answer: GNAPVD Difficulty: 2 Section: Mechanisms of Pattern Formation Feedback: Three axes generally have to be established in the early embryo. In most animals, the animal-vegetal (A-V) axis defines which parts are to remain external in gastrulation and which are to become internalized. The anteroposterior (A-P) axis specifies the locations of the future head and tail. The dorsoventral (D-V) axis specifies the future back and belly. 8. Answer: ACB Difficulty: 2 Section: Mechanisms of Pattern Formation Feedback: The mouse egg is almost perfectly symmetrical. In contrast, fly eggs are highly polarized. Many other organisms, such as the frog, lie between these two extremes. 9. Answer: CBA Difficulty: 2 Section: Mechanisms of Pattern Formation Feedback: The animal-vegetal (A-V) asymmetry is established before fertilization. The point of sperm entry then helps determine the dorsoventral (D-V) axis. The anteroposterior (A-P) axis is established later. 10. Answer: E Difficulty: 2 Section: Mechanisms of Pattern Formation Feedback: Fertilization in Xenopus laevis, through a reorganization of the microtubule cytoskeleton, triggers a rotation of the egg cortex. This places the ventral pole at the point of sperm entry. The reorganization also leads to the transport of several cytoplasmic components, including the Wnt 11 mRNA, which becomes concentrated at the dorsal side. 11. Answer: E Difficulty: 2 Section: Mechanisms of Pattern Formation Feedback: For maternal-effect genes, it is the mother’s genome rather than the zygotic genome that is critical. 12. Answer: EPSG Difficulty: 2 Section: Mechanisms of Pattern Formation

Feedback: Egg-polarity genes in Drosophila are maternal-effect genes: their mRNA transcript is deposited by the mother into the egg, and hence it is the mother’s genome (rather than the zygote’s genome) that is critical. Mutations in gap genes result in the elimination of one or more adjacent segments altogether. Mutations in pair-rule genes cause a series of deletions affecting alternate segments, leaving the embryo with only half as many segments as usual. Finally, mutations in segment-polarity genes produce the right number of segments but with a part of each segment replaced by a mirror-image duplicate of another part of the segment. 13. Answer: 2413 Difficulty: 2 Section: Mechanisms of Pattern Formation Feedback: In the regulatory hierarchy of egg-polarity and segmentation genes, the gapgene products provide a further tier of positional signals by controlling the expression of other genes, including the pair-rule genes. The pair-rule genes, in turn, collaborate with one another and with the gap genes to set up a regular, periodic pattern of expression of the segment-polarity genes, which collaborate with one another to define the internal pattern of each individual segment. 14. Answer: TTLTL Difficulty: 2 Section: Mechanisms of Pattern Formation Feedback: Egg-polarity, gap, and pair-rule genes create transient patterns that are remembered by segment-polarity and Hox genes. 15. Answer: A Difficulty: 2 Section: Mechanisms of Pattern Formation Feedback: In each segment, cells that express the signal protein Wingless induce their neighboring cells to express the transcription regulator Engrailed, which activates the expression of the signal protein Hedgehog in these cells. This in turn activates the expression of Wingless in the first cell, maintaining the fine patterning within the segment. 16. Answer: B Difficulty: 3 Section: Mechanisms of Pattern Formation Feedback: Loss of Ultrabithorax transforms the segment into a more anterior identity and gives rise to a pair of wings in this segment. Gain of Antennapedia transforms a head segment such that it bears a leg in the place of an antenna. 17. Answer: C Difficulty: 1 Section: Mechanisms of Pattern Formation

Feedback: Hox genes lie in one or the other of two gene clusters, the Bithorax complex and the Antennapedia complex. 18. Answer: TFFT Difficulty: 1 Section: Mechanisms of Pattern Formation Feedback: The genomic location of Hox genes (which control the anteroposterior [A-P] axis in both vertebrates and invertebrates) corresponds well with their order of expression along the A-P axis. In other words, Hox genes are expressed according to their order in the Hox complex. Generally, the more posterior of the Hox genes dominate the more anterior ones, so that when both are expressed in a given segment, the segment will assume an identity dictated by the more posterior Hox genes. 19. Answer: E Difficulty: 2 Section: Developmental Timing Feedback: Hox proteins give each segment its individuality; without them, segments will lose their individual characters. Trithorax and Polycomb group proteins work together, in opposite ways, to enable the Hox complexes to maintain a permanent record of positional information. Without them, the Hox gene expression pattern is set up correctly, but cannot be maintained. 20. Answer: DVVV Difficulty: 2 Section: Developmental Timing Feedback: Activated Toll receptors on the ventral side of the embryo result in the nuclear localization of Dorsal at this side. The highest nuclear Dorsal concentration activates the expression of Twist, whereas the lowest concentration de-represses Decapentaplegic (Dpp). At intermediate concentrations, Short gastrulation (Sog) expression is induced. 21. Answer: D Difficulty: 2 Section: Developmental Timing Feedback: Both of these mutations would result in a ventralized phenotype (similar to when Dorsal is activated in the entire embryo). 22. Answer: C Difficulty: 2 Section: Developmental Timing Feedback: Given the antagonist role of Sog in regulating Dpp function, gain of Dpp is expected to enhance (rather than suppress) the phenotype of Sog loss-of-function mutations. 23. Answer: C Difficulty: 2 Section: Mechanisms of Pattern Formation

Feedback: The cells at the vegetal pole secrete two opposing morphogens, Nodal (1) and Lefty (2), the latter being more rapidly diffusing. The result is the domination of Nodal near the vegetal end, and domination of Lefty near the animal end. 24. Answer: DCBA Difficulty: 2 Section: Mechanisms of Pattern Formation Feedback: Please refer to Figure 21–28. 25. Answer: C Difficulty: 2 Section: Mechanisms of Pattern Formation Feedback: Like their Drosophila counterpart Sog, Noggin and Chordin antagonize the proteins of the bone morphogenetic protein (BMP) family (Decapentaplegic is a family member in the fruit fly). However, unlike Sog, Noggin and Chordin limit BMP activity to the ventral (rather than the dorsal) side of the embryo. This results in the inversion of the D-V axis in vertebrates compared to that of insects. 26. Answer: B Difficulty: 2 Section: Mechanisms of Pattern Formation Feedback: Members of the Achaete/Scute family of transcription regulators drive the cells to become neural, not endothelial, progenitors. 27. Answer: B Difficulty: 3 Section: Mechanisms of Pattern Formation Feedback: In lateral inhibition, Notch signaling typically leads to inhibition of the differentiation program as well as to reduction in the cell’s display of Delta. Interfering with Notch signaling would therefore impair lateral inhibition—generating patterns in which two neighboring cells undergo specialization (drawing 1)—since Notch signaling can be off in both cells. 28. Answer: B Difficulty: 1 Section: Mechanisms of Pattern Formation Feedback: Changes in regulatory DNA controlling expression of key developmental genes are largely responsible for morphological differences between animal species. 29. Answer: D Difficulty: 2 Section: Developmental Timing Feedback: A gene-expression oscillator acts as a clock to control segmentation in vertebrates. During this process, Notch signaling ensures that the oscillations are in synchrony in neighboring cells in the presomitic mesoderm (i.e. the segmentation clock is the same). Note that this is opposite to the role of Notch signaling in lateral inhibition.

30. Answer: B Difficulty: 3 Section: Developmental Timing Feedback: Deleting only one intron shortens the gestation delay for the mRNA only slightly, leading to an overall shorter feedback delay and therefore higher oscillation frequency. Deleting all three introns shortens the delay to such an extent that a sustained oscillation is no longer supported. 31. Answer: FTTT Difficulty: 1 Section: Developmental Timing Feedback: Cells rarely count cell divisions to time their development. In many cases, intracellular developmental programs help determine the time course of a cell’s development, allowing it to step through more or less the same program whether it is within the embryo or in culture. The timing of developmental transitions can also be coordinated by cell–cell interactions and a global coordinating signal. The transitions are often regulated and sharpened by microRNAs in the cell. 32. Answer: IER Difficulty: 2 Section: Developmental Timing Feedback: Rapid cleavage increases total DNA without a concomitant increase in the total cytoplasmic volume, thereby increasing the ratio of DNA to diffusible repressors. This ratio reaches a threshold at the maternal-zygotic transition (MZT). This occurs after fewer rounds of cell division if the cells are polyploid. 33. Answer: B Difficulty: 2 Section: Developmental Timing Feedback: Metamorphosis in amphibians is triggered by the secretion of thyroid hormone from the thyroid gland, which coordinates the various processes that are involved in this important developmental transition. 34. Answer: BADC Difficulty: 3 Section: Developmental Timing Feedback: As Coolair levels (curve B) peak during vernalization, closed chromatin structures (curve C) are induced at the FlC locus, gradually shutting off FlC expression (curve A). This persisting change in turn allows flowering genes of the Ft locus to be expressed under the right conditions in the spring (curve D). 35. Answer: D Difficulty: 2 Section: Morphogenesis

Feedback: SDF1 biases the site of membrane protrusions (blebs) toward its source, but the cells still display blebbing even when SDF1 signaling is perturbed. 36. Answer: B Difficulty: 2 Section: Morphogenesis Feedback: Steel factor and its receptor provide survival signals for many migrating cells. In the absence of Steel factor, therefore, the cells would undergo apoptosis. This would be prevented if Bax is also deleted. 37. Answer: C Difficulty: 2 Section: Morphogenesis Feedback: Cell adhesion molecules such as cadherins play a major role in forcing the cells into their natural arrangement. 38. Answer: convergent extension Difficulty: 1 Section: Morphogenesis Feedback: In convergent extension, the cells in a sheet crawl over one another in a coordinated fashion, causing the sheet to narrow along one axis (converge) and elongate along another (extend). 39. Answer: B Difficulty: 2 Section: Morphogenesis Feedback: In planar cell polarity, the cells in an epithelium are all arranged as if they had an arrow written on them, pointing in a specific direction in the plane of the epithelium. In the examples shown, A represents a normal epithelium, whereas B shows planar-cellpolarity defects. Note that C and D show defects in apical–basal polarity as well. 40. Answer: D Difficulty: 2 Section: Growth Feedback: Fibroblast growth factor FGF10 is secreted by the mesenchymal cells, inviting the nearby epithelial cells at the tip of the growing tube to invade the mesenchyme. The Sonic hedgehog (Shh) signal is sent in the opposite direction—from the epithelial cells back to the mesenchymal cells—where it likely inhibits FGF10 production, leading to branching of the growing bud. 41. Answer: B Difficulty: 2 Section: Growth Feedback: Budding (B) generates the tubes of developing lungs and trachea. 42. Answer: 2

Difficulty: 3 Section: Growth Feedback: The injected T cells have divided (the CFSE dye is diluted) in the T-celldeficient mouse but not in the wild-type mouse, supporting the idea that the total population of these cells is regulated as a whole. 43. Answer: B Difficulty: 3 Section: Growth Feedback: In plants, as in animals, cell size correlates with ploidy. Additionally, body and organ size in plants correlates well with ploidy. The fluorescent signal in this experiment corresponds to the nuclear DNA content. Varieties A, B, and C appear to be mostly diploid, octaploid, and tetraploid, respectively. 44. Answer: B Difficulty: 2 Section: Growth Feedback: After cell division in the somatic tissue is halted in the adult worm, endoreplication makes these cells polyploid, which is accompanied by a larger cell size and a larger overall body size. 45. Answer: C Difficulty: 2 Section: Growth Feedback: The size of the disc is not set to contain a certain number of cells; instead, it is regulated by the total disc size. 46. Answer: SLLS Difficulty: 2 Section: Growth Feedback: Growth hormone stimulates growth, and its deficiency leads to a smaller body size. It acts by inducing the expression of insulin-like growth factor, overexpression of which may lead to a larger body size in an animal. Myostatin specifically inhibits the growth and proliferation of myoblasts; consequently, loss of myostatin results in greatly enlarged muscle tissue. The Hippo pathway inhibits organ and body growth in general; overexpression of Hippo, an upstream component of this pathway, inhibits growth. 47. Answer: ADBC Difficulty: 1 Section: Neural Development Feedback: First, the neurons are assigned specific characters according to the place and time of their birth. Next, newborn neurons extend projections along specific routes toward their target cells. In the third phase, neurons form synapses (with other neurons or muscle cells) that are then refined and adjusted in the fourth phase. 48. Answer: CDAB

Difficulty: 2 Section: Neural Development Feedback: A represents the roof plate which secretes BMP and Wnt signals; D represents the floor plate which secretes Sonic hedgehog protein. B and C represent the dorsal and ventral regions of the tube, containing sensory and motor neurons, respectively. 49. Answer: A Difficulty: 2 Section: Neural Development Feedback: The radial glial cells form a scaffold in the developing cortex to guide the migrating neurons to their destined layer. The first-born neurons migrate and settle closest to the lumenal (inner) surface of the cortex, whereas last-born neurons crawl past them and settle in outer layers. The radial glial cells divide near the lumenal surface (where their cell bodies are located) to give rise to neurons and glia. 50. Answer: ECFADB Difficulty: 2 ection: Neural Development Feedback: Growth cones use various cues to navigate toward their targets. Please refer to Figure 21–73. 51. Answer: C Difficulty: 3 Section: Neural Development Feedback: Both of these proteins would be expected to be chemorepellents in guiding the migration of growth cones, as they both increase the rate of collapse. Protein 1 is a more potent repellent. 52. Answer: D Difficulty: 2 Section: Neural Development Feedback: Robo3.1 is lost as neurites cross the midline. As a result, the growth cones now respond to Slit and are repelled from the midline. 53. Answer: FTFT Difficulty: 2 Section: Neural Development Feedback: As a general rule, “neurons that fire together wire together.” Unlike the tonotopic map in the auditory cortex, in which the neurons are arranged according to the frequency of sound to which they respond, the neurons in the retinotopic map in the optic tectum are arranged according to the position of the cell bodies of the retinal ganglion cells to which they are connected. The retinotopic map can be inverted, as shown by the classic optic nerve regeneration experiments in frogs. 54. Answer: E Difficulty: 2

Section: Neural Development Feedback: In this experiment, posterior explants would only grow on anterior membranes; anterior explants, however, can grow on both membranes. 55. Answer: D Difficulty: 2 Section: Neural Development Feedback: Despite employing similar strategies for self-avoidance, neurons in vertebrates and invertebrates use different cell-surface molecules to distinguish self from non-self. DSCAM proteins in Drosophila can be produced from an extraordinary large number of different isoforms generated by alternative RNA splicing. The protocadherins in vertebrates are expressed in a large number of different combinations. In both examples, homophilic recognition results in self-avoidance. This is distinct from the phenomenon of tiling, where each neuron occupies a solitary territory, repelling its neighbors. 56. Answer: D Difficulty: 2 Section: Neural Development Feedback: Drosophila neurons that do not express any DSCAM1 isoform avoid neither themselves nor other neurons, including wild-type neurons. 57. Answer: 33 Difficulty: 3 Section: Neural Development Feedback: Neurons in the posterior retina express EphA. Their axons are repelled by high levels of EphrinA protein in the posterior tectum and thus project preferentially to the anterior tectum. 58. Answer: D Difficulty: 3 Section: Neural Development Feedback: NGF withdrawal (a) results in apoptosis, but not when caspase-3 is deleted (1). 59. Answer: B Difficulty: 3 Section: Neural Development Feedback: Signaling occurs in both directions for the formation of a neuromuscular junction. Agrin is released from the growth cone and binds to its receptor, LRP4, on the surface of the muscle cell. In addition to its role as a co-receptor, LRP4 also serves as a signaling molecule to stimulate the formation of presynaptic structures in the neuron. 60. Answer: ESSE Difficulty: 3 Section: Neural Development Feedback: Agrin binding to LRP4 activates MuSK, leading to acetylcholine receptor clustering mediated by Rapsyn. As expected, loss of LRP4 or Rapsyn negatively affects

receptor clustering and synapse formation. On the other hand, abnormal activation of MuSK or inhibition of its counteracting phosphatases leads to unnecessary clustering. 61. Answer: D Difficulty: 3 Section: Neural Development Feedback: If the left eye is deprived of visual experience during the critical period of development, its stripes (light) shrink and those of the active right eye (dark) expand.

Which of the following cells in a healthy adult human show the highest turnover? A. Neurons of the hippocampus B. Epithelial cells of the urinary tract C. Hepatocytes of the liver D. Epithelial cells of the small intestine E. Neutrophils of the blood

2. Indicate true (T) and false (F) statements below regarding the pattern of cell proliferation in the epithelium that forms the lining of the small intestine. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTT. ( ) New epithelial cells are constantly born throughout the epithelial sheet, mostly at the villi. ( ) Every type of intestinal epithelial cell is eventually discarded into the gut lumen where they die. ( ) There is a net movement of cells from the bottom of the crypts up toward the tip of the villi. ( ) The majority of intestinal epithelial cells have packed microvilli on their apical surfaces. 3. The schematic drawing below shows the cross section of a crypt in the gut epithelium of a mammal. Indicate whether each of the following descriptions better applies to the areas labeled A, B, or C. Your answer would be a four-letter string composed of letters A to C only, e.g. CCBB.

( ) The stem cells reside in this area. ( ) This area contains the most rapidly dividing cells. ( ) Cells in this area would show the least radioactivity following a pulse of tritiated thymidine. ( ) This area contains cells of the innate immune system. 4.

Stem cells … A. always divide asymmetrically to give rise to two different daughter cells. B. are terminally differentiated. C. can divide for the entire lifetime of the organism. D. divide at a relatively fast rate. E. All of the above.

In some adenomas of the colon, intestinal crypt cells appear to have proliferated

abnormally to form small tumors known as polyps. Studies on the familial type of such a disease led to the identification of a major signaling pathway (which was aberrant in the polyp cells) as being involved in the maintenance of the gut stem-cell compartment. What signaling pathway is this? Is it upregulated or downregulated in colon cancers? A. MAPK pathway; up-regulated B. MAPK pathway; down-regulated C. Wnt pathway; up-regulated

D. Wnt pathway; down-regulated E. Hedgehog pathway; down-regulated 6. Radial glial cells are stem cell-like neural progenitors in the adult mammalian brain and can give rise to a number of differentiated neural cell types. Knowing that quiescent radial glial cells uniquely express the cytoskeletal protein nestin, a researcher performed clonal analysis in mice to study the characteristics of radial glial cells and their descendants. She created transgenic mice expressing Cre recombinase under the control of the nestin regulatory sequences. She also created transgenic mice that carry the gene encoding the green fluorescent protein (GFP) under the control of a ubiquitous promoter. The GFP-encoding gene, however, is inactive unless recombination by Cre removes a blocking sequence between the promoter and the gene. After mating these mice, she obtains adult mice carrying both transgenes and injects them with a low dose of tamoxifen to activate Cre sporadically. After a month, she uses microscopy to study the cells of the brain hippocampus. She discovers that most neural cell types in the tissue, including neurons, develop from the radial glial cells. Oligodendrocytes, however, appear to derive from another stem-cell population. Based on her conclusions, which cells do you think expressed GFP after a month: neurons (N) or oligodendrocytes (O)? If a single isolated GFP-expressing cell is observed in the tissue, would you expect it to be a differentiated cell (D) or a radial glial cell (R)? If a larger clone is observed to be entirely made of radial glial cells, would a symmetric (S) or asymmetric (A) model be favored for the strategy of cell divisions in these cells? Write down your answer as a three-letter string, e.g. NDA. 7.

Each gut stem cell … A. is pluripotent. B. divides approximately once a week. C. gives rise to all four major epithelial cell types in the right proportions. D. expresses Lgr5, just like the epithelial cells that derive from it. E. All of the above.

8. Indicate whether each of the following descriptions better applies to an asymmetric (A) or a symmetric (S) stem-cell division strategy. Your answer would be a four-letter string composed of letters A and S only, e.g. AAAA. ( ) It predicts the number of stem cells in a population to remain strictly constant after several rounds of cell division. ( ) Cytoplasmic factors that control stemness are localized to one side of the cell before cytokinesis.

( ) It is a more flexible strategy and can better accommodate the effect of environmental factors. ( ) It is more consistent with observations in gut stem cells and other stem cells. 9. Considering the role of Notch and Wnt signaling pathways in epithelial cell differentiation in the mammalian intestine, indicate whether each of the following changes in “minigut” organoids in culture is expected to lead to a higher (H) or lower (L) number of stem cells. Your answer would be a two-letter string composed of letters H and L only, e.g. HH. ( ) RNA interference knockdown of Hes1, an essential downstream effector of the Notch signaling pathway ( ) Deletion of the gene encoding Apc, a gene that is mutated in many colorectal cancers 10. In each of the following examples of contact signaling between adjacent cells in the mammalian gut epithelia, indicate whether the signaling is mainly mediated by Delta (D) or Ephrin (E) signals. Your answer would be a three-letter string composed of letters D and E only, e.g. EDE. ( ) Separation of crypt cell types from villus cell types ( ) Regulation of stem-cell differentiation by Paneth cells ( ) Differentiation of transit amplifying cells into either secretory cells or absorptive cells 11. How is the organization of epithelial cells in epidermis different from that in the lining of the intestine? A. The epidermis is stratified and squamous (i.e. forms squames). B. The stem-cell niche in epidermis is provided by the basal lamina and the underlying connective tissue. C. The transit amplifying cells in the epidermis move mostly perpendicular to the plane of the epithelium. D. The cells of the epidermis die long before they are shed. E. All of the above. 12. Which of the following cell types in our body are typically renewed independently of stem cells? Your answer would be a string composed of a subset of the letters A to E in alphabetical order, e.g. BDE. (A) Red blood cells (B) β Cells of pancreatic islets (C) Hepatocytes of the liver

(D) Epithelial cells of the epidermis (E) Neutrophils of the blood 13. If some cells in an adult mammal are lost, they are lost forever and cannot be replaced. Which of the following cell types have such a limitation? Your answer would be a string composed of a subset of the letters A to E in alphabetical order, e.g. AE. (A) Photoreceptive epithelium of the retina (B) β Cells of pancreatic islets (C) Hepatocytes of the liver (D) Auditory epithelium of the inner ear (E) Olfactory epithelium of the nose 14. In addition to fibroblasts themselves, the fibroblast family of connective-tissue cells includes all of the following EXCEPT … A. Smooth muscle cells B. Chondrocytes C. Osteoblasts D. Adipocytes E. Macrophages 15. The differentiation of bone marrow stromal cells into bone or fat cells depends on chemical and mechanical signals. In each of the following conditions, indicate whether you would expect the stromal cells to differentiate into bone (B) or fat (F) cells. Your answer would be a three-letter string composed of letters B and F only, e.g. FBB. ( ) The cells are embedded in a stiff hydrogel. ( ) The cells are embedded in a soft hydrogel. ( ) The cells are treated with a Rho inhibitor. The monomeric GTPase Rho is required for the assembly of actin stress fibers and actin–myosin bundles. 16. Indicate whether each of the following descriptions better matches bone (B) or cartilage (C). Your answer would be a four-letter string composed of letters B and C only, e.g. CCCC. ( ) Its extracellular matrix is mainly composed of type I collagen and calcium crystals. ( ) Its extracellular matrix is made by osteoblasts. ( ) It grows by expansion (as opposed to surface apposition). ( ) Its cells are connected via canaliculi and form gap junctions.

17. Indicate true (T) and false (F) statements below regarding bone and bone remodeling. Your answer would be a four-letter string composed of letters T and F only, e.g. FFFF. ( ) Cells that demolish old bone matrix are related to macrophages of the immune system. ( ) Bone is over 99% nonliving material. ( ) A blood capillary can form as bone is excavated during remodeling. ( ) Tunnels made by osteoclasts are immediately and fully refilled by osteoblasts. 18. Indicate whether each of the following descriptions better applies to cardiac (C), skeletal (K), myoepithelial (M), or smooth (O) muscle cells. Your answer would be a four-letter string composed of letters C, K, M, and O only, e.g. MMKC. ( ) They are derived from the ectoderm layer, similar to epithelial cells. ( ) They form the lining of cavities such as the urinary bladder. ( ) They are responsible for most voluntary muscle movements. ( ) They form multinucleated and exceptionally long muscle fibers. 19. Which of the following statements is true regarding skeletal muscle development and regeneration in humans? A. Muscle is among those tissues that cannot be renewed once damaged. B. Myoblasts are multipotent stem cells. C. Muscle regeneration occurs as a result of stimulation of quiescent stem cells in the periphery of muscle fibers. D. A muscle that has been repaired once cannot be repaired again. 20.

Lymphatic capillaries … A. are lined with endothelial cells. B. have leakier walls compared to blood capillaries. C. can provide a path for cancer metastasis. D. All of the above.

21.

Fill in the blank in the following paragraph. Do not use abbreviations. “Growth and branching of blood vessels by the process of … creates a highly branched network of blood vessels and capillaries to ensure that almost every cell within our body has access to nutrients and oxygen from a nearby capillary. This process is

distinct from development.”

novo

vasculogenesis

during

embryonic

22. Sort the following events to reflect the order in which they occur for formation of new blood capillaries in conditions of insufficient oxygen supply. Your answer would be a four-letter string composed of letters A to D only, e.g. DACB. (A) New tip cells are formed on the sides of nearby capillaries. (B) Stalk cells divide and hollow out to form tubes. (C) VEGF is secreted by the cell. (D) The level of intracellular HIF1α increases. 23.

All blood cells in our body … A. have exceptionally short life-spans—at most a month or so. B. are produced only before adulthood. C. are ultimately generated from multipotent hematopoietic stem cells. D. are made from progenitors that themselves circulate in the blood. E. All of the above.

24.

Indicate true (T) and false (F) statements below regarding blood leukocytes. Your answer

would be a four-letter string composed of letters T and F only, e.g. TTTT. ( ) Most white blood cells use blood merely as a transportation medium, and function in other tissues. ( ) Some granulocytes are professional phagocytes. ( ) Monocytes typically give rise to macrophages and polymorphonuclear leukocytes. ( ) Natural killer cells have a lymphoid origin. 25. Sort the following cell types from high to low abundance in the blood (number of cells per unit volume) in a healthy individual. Your answer would be a four-letter string composed of letters A to D only, e.g. BADC. (A) Erythrocytes (B) Neutrophils (C) Basophils (D) T lymphocytes

26. A physician had prescribed a differential white blood cell count to help diagnose a sick child with sustained fever and weight loss. The results show a significant increase in the number of eosinophils beyond the normal range for healthy individuals. This is most likely due to … A. bacterial infection. B. parasitic infection. C. viral infection. D. kidney failure. E. internal bleeding. 27.

Blood platelets in an adult human … A. are made in lymphoid organs. B. are NOT derived from hematopoietic stem cells. C. lack nuclei. D. All of the above.

28. Indicate true (T) and false (F) statements below regarding differentiation of blood cells. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTT. ( ) Most blood cells are derived from the pluripotent stem cells by a few cell divisions. ( ) Commitment to a specific differentiation path is a stepwise process. ( ) Upon contacting connective-tissue cells of the bone marrow, hematopoietic stem cells lose their stem-cell character and differentiate. ( ) Committed hematopoietic progenitor cells typically divide rapidly. 29. Using Cre recombination, the gene encoding stem cell factor (SCF) can be deleted in specific cell types in adult mice. Deletion of this gene in which of the following cells would you expect to lead to the most significant depletion of hematopoietic stem cells in the bone marrow? A. Osteoblasts B. Osteoclasts C. Stromal cells D. Hematopoietic stem cells E. Common myeloid precursor cells 30. Which mammalian blood cell type undergoes the developmental event shown schematically in the following drawing as part of its maturation?

A. Monocyte B. Megakaryocyte C. Erythrocyte D. Lymphocyte E. Granulocyte 31. The GM progenitor cell gives rise to two main classes of white blood cells in the presence of appropriate colony-stimulating factors. Both of these cell types … A. have relatively long half-lives, being capable of living for months if not years. B. circulate in the bloodstream for most of their life-span. C. are phagocytes. D. are unaffected by colony-stimulating factors once terminally differentiated. E. All of the above. 32. Indicate true (T) and false (F) statements below regarding erythropoiesis. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) Oxygen stimulates the secretion of erythropoietin.

( ) Each human erythrocyte lives for a few days on average. ( ) All erythrocytes express “DON’T EAT ME” signals on their surface to prevent phagocytosis. ( ) It only takes 1 to 2 days for erythrocyte numbers to rise following stimulation with erythropoietin. 33.

If a hematopoietic cell is deprived of colony-stimulating factors, it typically … A. dies by apoptosis. B. differentiates randomly into several cell types. C. differentiates randomly to one cell type. D. enters a quiescent state. E. divides without differentiating.

34. Indicate true (T) and false (F) statements below regarding the regenerative capacity of the planarian Schmidtea mediterranea. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTF. ( ) When the animal is starved, some of its cells are cannibalized by other cells. ( ) The differentiated cells in the animal’s body are continually replaced. ( ) Even a single healthy cell can rescue a lethally irradiated animal from death. ( ) Only neoblasts are capable of rescuing a lethally irradiated animal from death. 35. BrdU (bromodeoxyuridine) is a thymidine analog that can be naturally incorporated into DNA (in place of thymidine) and detected by specific anti-BrdU antibodies. In an experiment to study the regeneration capabilities of Schmidtea mediterranea, you first fed the worms with BrdU-containing food for 30 minutes. After two hours, you fixed and incubated the worms with a fluorescently labeled anti-BrdU antibody, and observed its tissues with a fluorescence microscope. The result was that some nuclei are labeled with fluorescence while others are unlabeled. Would you expect to observe neoblasts mostly among cells that have (Y) or have not (N) been labeled with fluorescence? Are the neoblasts expected to be scattered throughout the body (S) or localized to a certain region (L)? If you fixed the worms after a week, instead of two hours, would you expect the number of nuclei with detectable fluorescence to increase (I), remain unchanged (U), or decrease (D) compared to these results? Write down your answer as a three-letter string, e.g. YSD. 36. A researcher is studying limb regeneration in the newt Notophthalmus viridescens. She transplants skeletal muscle cells from a transgenic newt that expresses green fluorescent protein

(GFP) in all of its cells into the limb of a wild-type recipient newt. She then amputates the recipient limb right at the site of the graft, such that the stump contains GFP-expressing cells. She repeats this experiment several times. Where would you expect to detect GFP in the resulting regenerated limbs? A. Only in muscle cells. B. Only in skin cells. C. Only in nerve cells. D. Only in muscle and skin cells. E. In all of the cells. 37.

An increasing number of lymphomas are being treated with “stem-cell transplantation”

therapy that is composed of the following overall steps. Sort these steps into the correct order. Your answer would be a three-letter string composed of letters A to C only, e.g. BAC. (A) Injection of cells into the blood (B) Harvesting and isolating selected cells from the patient’s bone marrow (C) Heavy x-ray irradiation of the patient 38. Stem cells can be obtained from self-renewing regions of an adult vertebrate brain, cultured in suspension or in monolayer, and later grafted back into a live animal. When implanted, stem cells isolated from mouse hippocampus … A. can only give rise to hippocampus neurons, even in a different location in the brain. B. can give rise to neurons in another part of the brain, e.g. in the olfactory bulb. C. are unable to proliferate unless implanted in the hippocampus, where they contribute to its neurons. D. can give rise to any cell (e.g. muscle cells) when implanted into muscle tissue. 39. Although quite inefficient, cloning by nuclear transplantation can be successful, as exemplified by cloning of the famous Dolly. Such a success implies that … A. nuclei can be reprogrammed by cytoplasmic factors in a foreign cytoplasm. B. epigenetic changes in somatic cells are not functionally irreversible. C. even a differentiated nucleus contains a complete genome, capable of supporting the development of an entire organism. D. All of the above. 40. Cells of the inner cell mass in an early mammalian embryo can be isolated and grown in culture. Which of the following is NOT true regarding these cells?

A. They are totipotent. B. They can be introduced into another developing embryo where their progeny become incorporated into the resulting adult animal, even into its germ line. C. They are stem cells. D. They can be manipulated in culture to give rise to almost any type of differentiated cell. 41. Which of the following transcription regulators constitute the master set of transcription regulatory proteins whose expression is most required for generating induced pluripotent stem cells and for maintaining pluripotency? Choose three proteins. Your answer would be a threeletter string composed of letters A to F in alphabetical order, e.g. CEF. (A) Myc (B) Klf4 (C) MyoD (D) Nanog (E) Oct4 (F) Sox2 42.

Only cells that express a puromycin-resistance (PR) gene can grow in the presence of the

drug puromycin. Ganciclovir is a nontoxic drug that is converted to a lethal toxin in a cell that expresses thymidine kinase (TK). Thy-1 is a fibroblast-specific marker protein. Which of the following selection strategies is the most reasonable one to use in experiments that attempt to generate induced pluripotent stem cells (iPS cells) from fibroblasts? A. Engineering fibroblasts to express PR under the control of the Nanog promoter, and selecting iPS cells in puromycin-containing media B. Engineering fibroblasts to express PR under the control of the Thy-1 promoter, and selecting iPS cells in puromycin-containing media C. Engineering fibroblasts to express TK under the control of the Nanog promoter, and selecting iPS cells in ganciclovir-containing media D. Engineering fibroblasts to express TK under the control of the Thy-1 promoter, and selecting iPS cells in ganciclovir-containing media E. Engineering fibroblasts to express TK under the control of the Thy-1 promoter, and selecting iPS cells in puromycin-containing media

43. Sort the following events to reflect the order in which they typically take place during the reprogramming of fibroblasts to induced pluripotent stem cells. Your answer would be a fourletter string composed of letters A to D only, e.g. ADBC. (A) Sustained endogenous expression of Oct4 (B) Gain of embryonic marker protein (C) Loss of fibroblast-specific marker protein (D) OSKM overexpression 44. Which group of factors is thought to impede iPS cell generation from a differentiated cell, inasmuch as the reprogramming efficiency is increased when the activity of the factors is decreased? A. Histone acetyl transferases B. Histone deacetylases C. Chromatin remodelers D. DNA demethylases E. All of the above 45. Which of the following tasks is possible with current stem-cell and reprogramming technology? A. Guiding ES cells to generate any desired adult cell type. B. Generating functional whole organs from iPS cells for transplantation. C. Forcing transdifferentiation of any terminally differentiated cell into another cell. D. In vitro screening for drug discovery using patient-specific iPS-derived cells as disease models. E. All of the above.

Answers: 1. Answer: D Difficulty: 1 Section: Stem Cells and Renewal in Epithelial Tissues Feedback: Epithelial cells forming the lining of the small intestine are renewed at a greater rate than any other tissue in the body of a mammal. Skin and blood self-renew at high rates as well. 2. Answer: FFTT Difficulty: 1 Section: Stem Cells and Renewal in Epithelial Tissues Feedback: The microvillus-containing absorptive epithelial cells of the small intestine (as well as goblet cells and enteroendocrine cells) travel mainly upward from their site of birth in the crypt, through the plane of the epithelial sheet, toward the tip of the villi where they die by apoptosis and are discarded into the gut lumen. However, some of the epithelial cells in the crypt remain there as stem cells. 3. Answer: CBAC Difficulty: 2 Section: Stem Cells and Renewal in Epithelial Tissues Feedback: The stem cells are located near the base of the crypt, interspersed between Paneth cells (in area C), and divide at a relatively slow pace compared to the committed precursors (in area B) that divide rapidly. The cells in area A are differentiated and no longer divide. 4. Answer: C Difficulty: 1 Section: Stem Cells and Renewal in Epithelial Tissues Feedback: Stem cells are not terminally differentiated, and can self-renew by unlimited divisions (at least for the lifetime of the animal). When they divide, which they usually do at a slow pace, each daughter cell has a choice of whether or not to remain as a stem cell. 5. Answer: C Difficulty: 1 Section: Stem Cells and Renewal in Epithelial Tissues Feedback: Inactivation of Apc, the product of which negatively regulates the Wnt signaling pathway, is found in many colon cancers. Wnt signaling is important in keeping the crypt cells in a proliferative state. 6. Answer: NRS Difficulty: 2 Section: Stem Cells and Renewal in Epithelial Tissues

Feedback: In clonal analysis, if Cre recombinase is activated in a progenitor cell and performs recombination, the cell and all of its descendants are expected to express the reporter (GFP in this case). If such a cell remains quiescent, the result would be a single isolated GFP-positive cell. The presence of multiple radial glial cells in one colony implies symmetric growth. 7. Answer: C Difficulty: 2 Section: Stem Cells and Renewal in Epithelial Tissues Feedback: The Lgr5-positive gut stem cells are multipotent and divide approximately once a day. D is incorrect because the epithelial cells that are produced from these stem cells no longer express Lgr5. 8. Answer: AASS Difficulty: 1 Section: Stem Cells and Renewal in Epithelial Tissues Feedback: The symmetric strategy is more probabilistic and flexible. It is also more consistent with the observed division patterns of stem cells in the gut and elsewhere. 9. Answer: LH Difficulty: 3 Section: Stem Cells and Renewal in Epithelial Tissues Feedback: Notch signaling originating from Paneth cells inhibits differentiation of the gut epithelial stem cells. Inhibition of Notch signaling will therefore lead to differentiation. By allowing activation of the Wnt signaling pathway, Apc deletion interferes with stemcell differentiation, producing more stem cells. 10. Answer: EDD Difficulty: 1 Section: Stem Cells and Renewal in Epithelial Tissues Feedback: Ephrin–Eph signaling drives segregation of different gut cell types, for example in the separation of crypt cells from absorptive, goblet, and enteroendocrine cells. Delta–Notch signaling (in combination with Wnt) is involved in controlling gut cell diversification and maintaining the stem-cell state. 11. Answer: E Difficulty: 1 Section: Stem Cells and Renewal in Epithelial Tissues Feedback: The architecture of epidermis is very different from that of the intestine. 12. Answer: BC Difficulty: 1 Section: Stem Cells and Renewal in Epithelial Tissues Feedback: Although pancreas and liver contain stem cells as backups, normal tissue renewal of β cells and hepatocytes is usually carried out by simply duplicating differentiated cells.

13. Answer: AD Difficulty: 1 Section: Stem Cells and Renewal in Epithelial Tissues Feedback: The loss of photoreceptive cells in the retina or the sensory hair cells of the inner ear is permanent. 14. Answer: E Difficulty: 1 Section: Fibroblasts and Their Transformations Feedback: Cells of the connective-tissue family are related by origin and are often remarkably interconvertible. The family includes fibroblasts, cartilage cells (chondrocytes), bone cells (osteoblasts), fat cells (adipocytes), and smooth muscle cells. 15. Answer: BFF Difficulty: 1 Section: Fibroblasts and Their Transformations Feedback: Differentiation into fat cells (in a soft matrix) or bone cells (in a stiff matrix) depends on a tension-sensing mechanism that is mediated by actin–myosin bundles. 16. Answer: BBCB Difficulty: 1 Section: Fibroblasts and Their Transformations Feedback: Produced by osteoblasts and growing by surface apposition, the dense and rigid bone matrix is composed of tough type I collagen fibrils and solid hydroxylapatite crystals. Osteocytes that have become trapped in this matrix are connected to each other through tiny channels called canaliculi. Cartilage tissue is composed of chondrocytes embedded in a uniform and highly hydrated matrix and grows by expanding. 17. Answer: TFTF Difficulty: 1 Section: Fibroblasts and Their Transformations Feedback: Living cells such as osteoblasts and osteoclasts account for about 15% of the weight of compact bones. Blood vessels can form and remain in tunnels carved by osteoclasts. Osteoblasts deposit new bone around them in concentric layers. 18. Answer: MOKK Difficulty: 1 Section: Genesis and Regeneration of Skeletal Muscle Feedback: Myoepithelial cells lie in epithelia and, like epithelial cells, are derived from the ectoderm layer. Smooth muscle cells line cavities such as the urinary bladder and the gut. Skeletal muscle cells have a striated appearance (similar to cardiac muscle cells), are multinucleated and very long, and power most voluntary movements of an animal. 19. Answer: C Difficulty: 1 Section: Genesis and Regeneration of Skeletal Muscle

Feedback: The satellite cells can serve as myoblasts in adult muscle and become activated to repair damage to some extent. 20. Answer: D Difficulty: 1 Section: Blood Vessels, Lymphatics, and Endothelial Cells Feedback: Lymphatic vessels are lined by endothelial cells and carry no blood. Their walls are thinner and more permeable compared to blood vessels. They can therefore provide a migration route for white blood cells, as well as cancer cells, to exit a tissue and spread to other tissues. 21. Answer: angiogenesis Difficulty: 1 Section: Blood Vessels, Lymphatics, and Endothelial Cells Feedback: Angiogenesis is the process by which blood vessels grow and branch throughout the body. 22. Answer: DCAB Difficulty: 2 Section: Blood Vessels, Lymphatics, and Endothelial Cells Feedback: Tissues requiring a blood supply can trigger branching of nearby blood vessels toward themselves by releasing vascular endothelial growth factor (VEGF). Under hypoxic conditions, hypoxia-inducible factor 1α (HIF1α) levels increase inside the cell. This transcription regulatory protein then induces the expression of VEGF, which is secreted. Endothelial cells respond to VEGF by invading in the direction set by the VEGF gradient. 23. Answer: C Difficulty: 2 Section: A Hierarchical Stem-Cell System Feedback: All blood cells are ultimately generated from multipotent hematopoietic stem cells located in the bone marrow. They are produced throughout the life of the animal and have limited life-spans, although some can be quite long-lived. 24. Answer: TTFT Difficulty: 2 Section: A Hierarchical Stem-Cell System Feedback: Monocytes give rise to macrophages and most dendritic cells. Polymorphonuclear leukocytes (neutrophils) are granulocytes. 25. Answer: ABDC Difficulty: 2 Section: A Hierarchical Stem-Cell System Feedback: Erythrocytes are by far the most abundant blood cell type. Neutrophils are the most common among white blood cells, whereas basophils are the least common. Please refer to Table 22–1.

26. Answer: B Difficulty: 2 Section: A Hierarchical Stem-Cell System Feedback: Infection with protozoa and other parasites usually causes a selective increase in the number of eosinophils, which can be detected in a differential white blood cell count. 27. Answer: C Difficulty: 2 Section: A Hierarchical Stem-Cell System Feedback: All blood cells, as well as platelets, can be made from a single pluripotent hematopoietic stem cell residing in the bone marrow. Platelets are cellular fragments pinched off from megakaryocytes. 28. Answer: FTFT Difficulty: 2 Section: A Hierarchical Stem-Cell System Feedback: Hematopoietic stem cells interact with the stromal cells in the bone marrow to maintain their stemness, while the committed progeny go on to divide many times rapidly to create specialized blood cells. Commitment of progenitor cells to the production of each of the various blood cells occurs in a stepwise manner. 29. Answer: C Difficulty: 3 Section: A Hierarchical Stem-Cell System Feedback: Stem cell factor helps define the stem-cell niche in the bone marrow stroma. It binds to Kit, a receptor tyrosine kinase, on the surface of hematopoietic stem cells to maintain their stem-cell state. Note that in mice, spleen and liver can also be home to hematopoietic stem cells. 30. Answer: C Difficulty: 2 Section: A Hierarchical Stem-Cell System Feedback: Erythroblasts extrude their nuclei (as shown in the diagram) to become immature erythrocytes (reticulocytes). In contrast to erythroblasts, the mitochondriacontaining reticulocytes can enter the bloodstream. Reticulocytes lose their mitochondria within a day or two to become mature erythrocytes. 31. Answer: C Difficulty: 2 Section: A Hierarchical Stem-Cell System Feedback: Both macrophages and neutrophils are phagocytes derived from granulocyte/macrophage (GM) progenitors. Unlike macrophages, neutrophils have a short life-span. Both of these cells primarily reside outside of the bloodstream. Their activity (phagocytosis and killing) can be regulated by colony-stimulating factors.

32. Answer: FFFT Difficulty: 2 Section: A Hierarchical Stem-Cell System Feedback: A lack of oxygen can stimulate specialized cells in the kidneys to secrete erythropoietin. Shortly afterward, the rate of release of new erythrocytes is increased, suggesting that late precursors are affected by the hormone. Each human erythrocyte lives for about 120 days. Only younger ones express signals to avoid being eaten by macrophages. 33. Answer: A Difficulty: 2 Section: A Hierarchical Stem-Cell System Feedback: Regulation of cell survival, as well as regulation of cell proliferation, is important in regulating the generation of blood cells. Hematopoietic cells fail to receive survival signals and die by apoptosis in massive numbers every day in our body. 34. Answer: TTTT Difficulty: 2 Section: Regeneration and Repair Feedback: When starved, Schmidtea undergoes degrowth, during which some of its cells are cannibalized by others. A continual cell turnover underlies this phenomenon: the differentiated cells in the animal’s body are continually replaced by division and differentiation of stem cells called neoblasts. Remarkably, a single healthy neoblast is sufficient to rescue a lethally irradiated animal from death. 35. Answer: YSI Difficulty: 3 Section: Regeneration and Repair Feedback: The pulse of BrdU leads to incorporation of the thymidine analog into the DNA of only those cells that were replicating their DNA (in S phase, prior to mitosis) around the time of the pulse. These are neoblasts and they should be scattered throughout the body to be able to regenerate all the different tissues. With time, the cells divide and differentiate and the BrdU is “diluted” and distributed into more cells, most of which are not neoblasts. 36. Answer: A Difficulty: 3 Section: Regeneration and Repair Feedback: Experiments such as this one can demonstrate the limited regeneration capacity in vertebrates: the cells are restricted according to their origin. 37. Answer: BCA Difficulty: 2 Section: Regeneration and Repair

Feedback: In the ideal case, the cells are taken from the patient’s own bone marrow, sorted to enrich for the stem cells, and injected back into the patient after a dose of x-ray irradiation that kills all of the patient’s hematopoietic stem cells and lymphoma cells. 38. Answer: B Difficulty: 2 Section: Regeneration and Repair Feedback: After being grown under the right culture conditions, neural stem cells derived from the mouse hippocampus can give rise to neurons that become correctly incorporated into the olfactory bulb when implanted in the mouse olfactory-bulb-precursor pathway. The capacity of these stem cells to adapt to new environments suggests applications in the treatment of neurodegenerative diseases. 39. Answer: D Difficulty: 1 Section: Cell Reprogramming and Pluripotent Stem Cells Feedback: The successful cloning of animals by nuclear transplantation shows that even a differentiated cell contains a complete genome in its nucleus and that the epigenetic changes in its DNA can be functionally reversed. It further shows that cytoplasmic factors in the oocyte can drive this reversal, reprogramming the transplanted nucleus to an early embryonic state. 40. Answer: A Difficulty: 1 Section: Cell Reprogramming and Pluripotent Stem Cells Feedback: Embryonic stem cells (ES cells) are not totipotent; they are pluripotent. This limitation is only minor, and requires that they are injected into preformed early embryos, as opposed to being directly implanted. 41. Answer: BEF Difficulty: 1 Section: Cell Reprogramming and Pluripotent Stem Cells Feedback: The OSK factors seem to be required for induced pluripotent stem cell induction, while Myc overexpression enhances the process. 42. Answer: A Difficulty: 3 Section: Cell Reprogramming and Pluripotent Stem Cells Feedback: This strategy is similar to the one employing the Fbx15 promoter. A disadvantage of using negative selection with TK under the control of the Thy-1 promoter is that many cells can lose fibroblast-specific gene expression but fail to express pluripotency genes like Nanog. 43. Answer: DCBA Difficulty: 2 Section: Cell Reprogramming and Pluripotent Stem Cells

Feedback: Two major waves of new gene expression are induced by OSKM (Oct4, Sox2, Klf4, and Myc) overexpression: in the first wave, genes involved in cell proliferation, metabolism, and cytoskeletal organization are induced, and fibroblast-specific markers are lost; in the second wave, which occurs only in the subset of cells that have gained an embryonic marker protein, genes required for stem-cell maintenance and embryonic development (e.g. endogenous Oct4) are switched on. Overall, the majority of cells (~99%) fail to become induced pluripotent stem cells. 44. Answer: B Difficulty: 2 Section: Cell Reprogramming and Pluripotent Stem Cells Feedback: Generally, activation of chromatin remodelers, histone acetyl transferases, and DNA demethylases leads to less repressive chromatin states that are more permissive for reprogramming. As expected from their activities, histone acetyl transferases and deacetylases have opposite effects on reprogramming efficiency. 45. Answer: D Difficulty: 2 Section: Cell Reprogramming and Pluripotent Stem Cells Feedback: Use of embryonic stem cells (ES cells) and induced pluripotent stem cells (iPS cells) for drug discovery and for analysis of genetic disease is already underway, and is a promising approach to finding treatments for such diseases.

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION CHAPTER 23: PATHOGENS AND INFECTION Copyright © 2015 by W.W. Norton & Company, Inc. 1. Indicate true (T) and false (F) statements below regarding the human microbiota. Your answer would be a four-letter string composed of letters T and F only, e.g. TFTT. ( ) The number of human cells in our body is greater than the number of bacterial, fungal, and protozoan cells of our normal flora. ( ) There are far more genes in our microbiome than in our own genome. ( ) Infectious diseases currently cause more human deaths than cardiovascular diseases and cancers combined. ( ) All of the microorganisms that constitute the normal flora are nonpathogens. 2. Indicate whether each of the following examples better applies to commensalism (C), mutualism (M), or parasitism (P) in the interaction between a host and a microbe. Your answer would be a four-letter string composed of letters C, M, and P only, e.g. CCPP. ( ) Due to poor public sanitation, a child catches cholera. ( ) A mouse is infected with a virus but shows no noticeable health defect whatsoever. ( ) Biotin and other vitamins produced by intestinal microbiota are regularly absorbed in the human intestine. ( ) Bacteria on the skin of cattle produce antifungal compounds. 3. Which of the following illnesses is NOT associated with infection with the pathogen indicated? A. Cervical cancer associated with human papillomavirus infection B. Cancer in chickens associated with Rous sarcoma virus infection C. Colon cancer associated with Vibrio cholerae infection D. Stomach cancer associated with Helicobacter pylori infection E. Atherosclerosis associated with Chlamydia pneumoniae 4. Mycobacterium tuberculosis can cause tuberculosis, a life-threatening lung infection, but can also infect an individual asymptomatically for years. It is not considered to be part of the normal flora, and can infect healthy individuals upon exposure. It can only replicate in the host and thus mainly spreads by direct human contact. Mycobacterium tuberculosis is therefore … A. a facultative primary pathogen.

B. a facultative opportunistic pathogen. C. an obligate primary pathogen. D. an obligate opportunistic pathogen. 5. In Gram staining, crystal violet (a violet dye) is used to specifically stain Gram-positive bacteria. After performing Gram staining on a bacterial sample taken from an infected animal tissue, you observe the results with a microscope. You find two major types of bacteria in the sample, as shown in the schematic drawing below. According to these results, indicate whether each of the following statements is correct (C) or incorrect (I). Your answer would be a threeletter string composed of letters C and I only, e.g. CCC.

2 µm ( ) The Gram-positive bacteria in this sample are bacilli. ( ) Crystal violet stains lipopolysaccharide (LPS). ( ) The bacteria stained violet in this sample have a thicker layer of peptidoglycan compared to the other bacteria. 6. In the following simplified diagram, three mechanisms for the horizontal transfer of virulence genes to an avirulent bacterium are depicted. Indicate which mechanism (A to C) corresponds to conjugation, transduction, and transformation, respectively. Your answer would be a three-letter string composed of letters A to C only, e.g. CAB.

How has pathogenicity in pandemic strains of Vibrio cholerae been acquired? A. By transformation B. By transduction C. By conjugation D. By vertical gene transfer

You infect human epithelial cells in culture with either nonpathogenic bacteria or Vibrio

cholerae, each without any further treatment, in the presence of MDC (an inhibitor of clathrindependent endocytosis) or in the presence of filipin (an inhibitor of a clathrin-independent endocytic pathway). You then measure the intracellular concentration of cyclic AMP (in picomoles per milligram of total cell protein) and summarize the results in the following table. Which row (1 or 2) do you think corresponds to infection with V. cholerae? From these results, does cholera toxin enter the cell in clathrin-coated vesicles? Treatment No drug MDC Filipin Infection

A. Row 1; yes B. Row 1; no C. Row 2; yes D. Row 2; no

400

390

Both Vibrio cholerae and Bacillus anthracis … A. secrete a lethal factor and an edema factor. B. secrete toxins that enter the cell in endosomes and eventually enter the cytosol after reaching the endoplasmic reticulum. C. raise cyclic AMP levels in the cytosol of their target cells. D. secrete toxins that have enzymatic adenylyl cyclase activity. E. All of the above.

10.

The type III secretion system of pathogenic bacteria … A. is a general secretion system found only in Gram-positive bacteria. B. is used to inject effector proteins directly into the cytoplasm of host cells. C. is similar to the eukaryotic secretion system. D. is similar to the bacterial conjugation apparatus. E. All of the above

11. Indicate true (T) and false (F) statements below regarding bacterial, viral, and eukaryotic pathogens. Your answer would be a four-letter string composed of letters T and F only, e.g. TTTT. ( ) Compared to bacteria and viruses, eukaryotic parasites have simpler life cycles. ( ) Most important pathogenic fungi show dimorphism, growing as either yeast or mold. ( ) Protozoan parasites often require more than one host to complete their life cycle. ( ) Plasmodium falciparum can invade human liver and red blood cells, as well as cells lining the gut in female Anopheles mosquitoes. 12. A point mutation in the gene encoding the β subunit of hemoglobin can lead to sickle-cell anemia. Due to formation of mutant hemoglobin aggregates, red blood cells in affected individuals can adopt a sickle-like shape that interferes with their normal function. Interestingly, individuals carrying this mutation are less vulnerable to malaria infection by Plasmodium falciparum, which replicates inside red blood cells. Accordingly, would you expect the frequency of the mutated hemoglobin allele to be higher in West Africa (A) or in Europe (E)? Write down A or E as your answer. 13. Indicate whether each of the following descriptions better applies to bacterial (B), fungal (F), protozoan (P), or viral (V) pathogens. Your answer would be a four-letter string composed of letters B, F, P, and V only, e.g. FFFB. ( ) They cannot be all grouped neatly into a single phylogenetic tree.

( ) They have the most complex life cycles. ( ) They are essentially fragments of nucleic acid wrapped in a protective shell of proteins and (in some cases) an outer membrane. ( ) They are responsible for diseases such as polio and smallpox. 14. Sort the following events to reflect the order in which they typically occur in viral replication after a virus enters a cell. Your answer would be a four-letter string composed of letters A to D only, e.g. BACD. (A) Replication of the viral genome and transcription of viral genes (B) Virus particle assembly (C) Progeny virion release (D) Virus particle disassembly 15. Indicate whether each of the following descriptions better applies to enveloped (E) or nonenveloped (N) viruses. Your answer would be a four-letter string composed of letters E and N only, e.g. NNNN. ( ) They include retroviruses. ( ) They include adenoviruses. ( ) To leave the cell, they normally lyse it. ( ) They are more sensitive to treatment with detergents, heat, or drying. 16.

Which of the following is a primary pathogen? A. Staphylococcus aureus B. Streptococcus pneumoniae C. Yersinia pestis D. Papillomavirus E. MRSA (methicillin-resistant S. aureus)

17. In the bacterium Yersinia pestis, which causes bubonic plague, loss of an operon that is normally responsible for iron storage diminishes the ability of the cells to form biofilms and colonize inside the iron-rich gut of infected fleas. Would you expect the efficiency of transmission of plague to increase (I), decrease (D), or remain unchanged (U) as a result of this mutation? Write down I, D, or U as your answer.

18. Indicate whether the epithelium in each of the following organs is (Y) or is not (N) kept almost sterile in a healthy human. Your answer would be a four-letter string composed of letters Y and N only, e.g. YYYY. ( ) Bladder ( ) Cervix ( ) Lower lung ( ) Colon 19.

How does Helicobacter pylori persist in the hostile environment of the stomach? A. By expressing adhesin proteins that bind to molecules on the surface of gastric epithelial cells B. By producing the enzyme urease, which locally neutralizes the gastric acid C. By chemotaxis, using flagella, toward gastric epithelial cells D. By producing toxins that target both epithelial cells and immune cells E. All of the above

20. Sort the following events to reflect the order in which they take place during the infection of intestinal epithelium by enteropathogenic E. coli. Your answer would be a five-letter string composed of letters A to E only, e.g. BCDAE. (A) Actin polymerization and pedestal formation (B) Binding of intimin to Tir (C) Folding of Tir in the plasma membrane (D) Insertion of Tir into the epithelial cell (E) Assembly of type III secretion system 21. Indicate true (T) and false (F) statements below regarding viral cell entry. Your answer would be a four-letter string composed of letters T and F only, e.g. FFTF. ( ) Endocytosis inhibitors can block HIV entry but not influenza virus entry. ( ) Adenovirus infection is blocked by membrane-fusion inhibitors. ( ) Individuals with a defective CCR5 gene are more susceptible to HIV infection. ( ) Most viruses enter the host cell by phagocytosis. 22. Which of the following conditions is NOT expected to provide protection against HIV-1 infection of helper T cells? A. A loss-of-function mutation in the CCR5 gene B. Treatment with enfuvirtide, a membrane-fusion inhibitor

C. Treatment with concanamycin A, an inhibitor of endosomal proton pumps D. Treatment with AMD-070, a CXCR4 antagonist E. Treatment with PRO-542, a soluble decoy protein containing CD4 sequences 23. To enter the host cell, intracellular bacterial pathogens can induce phagocytosis in cells that are normally nonphagocytic. This is done by two major mechanisms depicted in the following schematic diagrams (A and B). Indicate whether each of the following descriptions better applies to mechanism A or B. Your answer would be a four-letter string composed of letters A and B only, e.g. ABAA. A

( ) It is called the zipper mechanism. ( ) It depends on invasin proteins on the surface of the bacterium that bind to their receptors on the surface of the host cell. ( ) It depends on the injection of effector proteins into the host cell by a bacterial secretion system. ( ) It resembles the process of macropinocytosis. 24. Toxoplasma gondii is an intracellular eukaryotic parasite that can cause serious human infections. Which of the following is NOT true regarding this parasite?

A. It lyses its host cell to leave. B. The energy for its invasion into the host cell comes from actin polymerization in the host cell. C. Once inside the cell, it is surrounded by a host-cell membrane that is deprived of many host transmembrane proteins. D. Once inside the cell, it absorbs nutrients and small metabolites from the host cytosol. E. Its invasion mechanism is similar to that of malaria parasites. 25. Cytochalasin D (CyD) is a drug that binds to the plus end of actin filaments and prevents actin polymerization. Having identified mutations in actin that confer resistance to CyD, a researcher sets out to study the role of the actin cytoskeleton in the invasion of mammalian host cells by two intracellular parasites: the protozoan Toxoplasma gondii and the bacterium Salmonella enterica. She infects wild-type or CyD-resistant host cells with either wild-type T. gondii, CyD-resistant T. gondii, or wild-type S. enterica, each in the presence (+) or absence (–) of CyD, and measures parasite internalization and intracellular proliferation as a result. The findings are summarized in the table below, in which High or Low levels of proliferation are indicated. According to these results, which column (A to C) do you think corresponds to infection with wild-type T. gondii? Which column corresponds to CyD-resistant T. gondii? Which row (D or E) corresponds to the wild-type host? Your answer would be a three-letter string composed of letters A to E only, e.g. ABE. Infected with…

D Host: E

+ CyD

Low

High

– CyD

High

+ CyD

High

Low

High

– CyD

High

26. Trypanosoma cruzi uses two alternative strategies to invade its host cells. To examine the contribution of each of these strategies to the pathogenicity, you culture mammalian cells that can be infected with this pathogen. The cells are engineered to express green fluorescent protein (GFP) fused to a plasma membrane protein, as well as red fluorescent protein (RFP) fused to a lysosome-specific transmembrane protein. You then either add only tiny latex beads (as a negative control) or add T. cruzi protozoa (to infect the cells). After 10 minutes, you fix the cells

and examine them under a fluorescence microscope. You observe that a significant fraction (~30%) of phagosomes/vacuoles in sample 1 are RFP-positive (i.e. show RFP fluorescence at their membrane), whereas all phagosomes in sample 2 are RFP-negative. Which sample (1 or 2) is the one infected with T. cruzi? In this infected sample, would you expect the RFP-negative trypanosome-containing vacuoles to be GFP-positive or GFP-negative? A. Sample 1; GFP-positive B. Sample 1; GFP-negative C. Sample 2; GFP-positive D. Sample 2; GFP-negative 27.

After gaining entry into the host cell by the zipper mechanism, the bacterium Listeria

monocytogenes escapes phagosomes by secreting listeriolysin O. This protein … A. is active at neutral pH and cannot be degraded by the proteasome. B. is active at neutral pH and is rapidly degraded by the proteasome. C. is active at acidic pH and cannot be degraded by the proteasome. D. is active at acidic pH and is rapidly degraded by the proteasome. B. is active at basic pH and cannot be degraded by the proteasome. 28.

An intracellular pathogen uses one of three major strategies to survive and replicate once

inside the host cell, as shown in the schematic drawing below. Indicate which strategy (1, 2, or 3) each of the following pathogens employs for this purpose. Your answer would be a three-digit number composed of digits 1 to 3 only, e.g. 332.

( ) Listeria monocytogenes

( ) Legionella pneumophila ( ) Shigella flexneri 29. Modifications of membrane traffic in host cells by three bacterial pathogens are shown in the following schematic drawing. Indicate whether each of the following bacteria is represented by A, B, or C in the drawing. Your answer would be a three-letter string composed of letters A to C only, e.g. BCA.

( ) It is a common cause of food poisoning in humans. ( ) It causes the life-threatening lung infection tuberculosis. ( ) It is the cause of a type of pneumonia known as Legionnaire’s disease. 30. Consider the movement in the host cell of two intracellular pathogens, the herpes simplex virus (H) and the bacterium Listeria monocytogenes (L). Indicate whether each of the following descriptions better applies to the virus (H) or the bacterium (L). Your answer would be a fourletter string composed of letters L and H only, e.g. HLLL.

( ) It moves by a “pushing” mechanism (as opposed to a “pulling” mechanism). ( ) It can collide with the plasma membrane and create long, thin protrusions from the cell that can then be engulfed by a neighboring cell. ( ) Its movement is powered by ATP hydrolysis by kinesin and dynein motors. ( ) Its movement relies on dynamic cytoskeletal filaments. 31. In which of the following groups of viruses is an RNA-dependent RNA polymerase packaged as a structural protein in each newly made viral particle? A. DNA viruses B. Viruses with a positive [+] strand RNA genome C. Viruses with a negative [–] strand RNA genome D. Retroviruses E. Viruses with a part single- and part double-stranded DNA genome 32. Indicate true (T) and false (F) statements below regarding bacterial and viral evolution. Your answer would be a four-letter string composed of letters T and F only, e.g. FFFT. ( ) Bacteria that reside in natural soil that has not been deliberately exposed to antibiotics are normally sensitive to any antibiotic. ( ) Whereas bacterial and eukaryotic pathogens evade the immune response mainly by genetic recombination, antigenic variation in most viruses occurs by error-prone replication mechanisms. ( ) If an antibiotic targets a vital bacterial protein that cannot be altered in any way to become antibiotic-resistant and functional at the same time, the bacteria would never develop resistance to the antibiotic. ( ) It is impossible to develop effective vaccines against viral infections, since viruses evolve very rapidly.

Answers: 1. Answer: FTFF Difficulty: 1 Section: Introduction to Pathogens and the Human Microbiota Feedback: The human body with about 1013 cells hosts about 1014 microorganisms whose gene repertoires exceed that of the human genome by over 100-fold. Some of these microorganisms are potentially pathogenic. Infectious diseases caused by pathogens account for about a quarter of human deaths worldwide, second only to cardiovascular diseases. 2. Answer: PCMM Difficulty: 1 Section: Introduction to Pathogens and the Human Microbiota Feedback: In mutualism, both the host and the microbe benefit. In commensalism, the microbe benefits from the host but offers no benefit or harm in return. In parasitism, which is often the case with pathogens, the microbe benefits to the detriment of the host. 3. Answer: C Difficulty: 1 Section: Introduction to Pathogens and the Human Microbiota Feedback: Pathogens can contribute to illnesses such as cancer and cardiovascular disease. Human papillomavirus, for example, is responsible for more than 90% of cervical cancer cases. Inflammation caused by H. pylori can be a major contributor to stomach cancer. The Rous sarcoma virus causes sarcomas in chicken. Foam cells in atherosclerotic plaques often contain C. pneumoniae. 4. Answer: C Difficulty: 1 Section: Introduction to Pathogens and the Human Microbiota Feedback: Primary pathogens can cause overt disease in healthy individuals. Obligate pathogens can only replicate inside the body of their host. 5. Answer: IIC Difficulty: 2 Section: Introduction to Pathogens and the Human Microbiota Feedback: Gram-positive bacteria (the cocci in this example) have a thicker layer of peptidoglycan cell wall. LPS is found on the outside of Gram-negative bacteria. 6. Answer: CBA Difficulty: 1 Section: Introduction to Pathogens and the Human Microbiota Feedback: In natural transformation (A), naked DNA is taken in by competent bacteria. In transduction (B), bacteriophages transfer DNA from one bacterium into another. In conjugation (C), DNA is transferred from a donor to a recipient bacterium.

7. Answer: B Difficulty: 1 Section: Introduction to Pathogens and the Human Microbiota Feedback: Pathogenicity in V. cholerae is associated with infection of the bacteria with bacteriophages. Incorporation of the CTXφ bacteriophage created the Classical pathogenic strains of V. cholerae responsible for the first six worldwide cholera epidemics. The seventh was caused by a strain that picked up the CTXφ bacteriophage along with an associated bacteriophage RS1φ. 8. Answer: D Difficulty: 3 Section: Introduction to Pathogens and the Human Microbiota Feedback: Cholera toxin increases intracellular cAMP concentration through the ADPribosylation and activation of a Gs subunit that activates adenylyl cyclase, as shown in row 2 in the table. From the results obtained in the presence of the drugs, the toxin appears to show this effect in a clathrin-independent manner. 9. Answer: C Difficulty: 2 Section: Introduction to Pathogens and the Human Microbiota Feedback: Although using different mechanisms, both V. cholerae and B. anthracis raise intracellular cyclic AMP levels. B. anthracis secretes a lethal factor and an edema factor, which, unlike the cholera toxin, normally enter the host cytosol directly from the endosome. The A subunit of edema factor is an adenylyl cyclase. 10. Answer: B Difficulty: 2 Section: Introduction to Pathogens and the Human Microbiota Feedback: Some Gram-negative bacteria use type III secretion systems to deliver effector proteins into the cytosol of a host cell. There is a remarkable degree of structural similarity between the type III syringe and the base of a bacterial flagellum. 11. Answer: FTTT Difficulty: 2 Section: Introduction to Pathogens and the Human Microbiota Feedback: Fungal and protozoan parasites have complex life cycles involving multiple forms. Many protozoan parasites require more than one host to complete their life cycles. For example, fertilization between Plasmodium gametes takes place in Anopheles mosquitoes. Most of the important pathogenic fungi exhibit the ability to grow in either yeast or mold form. The yeast-to-mold or mold-to-yeast transition is frequently associated with infection. 12. Answer: A Difficulty: 2 Section: Introduction to Pathogens and the Human Microbiota

Feedback: Since the mutation confers a health advantage, it is selected for in populations with a high incidence of malaria infection. 13. Answer: VPVV Difficulty: 2 Section: Introduction to Pathogens and the Human Microbiota Feedback: Viral genomes (made up of DNA or RNA) are packaged in a protein coat called a capsid. Some viruses are enclosed by a lipid bilayer membrane, or envelope. Viruses cause diseases such as polio, smallpox, and the common cold. 14. Answer: DABC Difficulty: 2 Section: Introduction to Pathogens and the Human Microbiota Feedback: In general, viral replication involves entry into the host cell, disassembly of the infectious virus particle, replication of the viral genome, transcription of viral genes, synthesis of viral proteins, assembly of these viral components into progeny virus particles, and finally release of progeny virions. 15. Answer: ENNE Difficulty: 3 Section: Introduction to Pathogens and the Human Microbiota Feedback: Nonenveloped viruses (such as adenoviruses) usually leave an infected cell by lysing it. For enveloped viruses (such as retroviruses), the nucleocapsid is enclosed within an envelope derived from cellular membranes, making these viruses more labile to heat, drying, detergent, and other harsh treatments. 16. Answer: C Difficulty: 1 Section: Cell Biology of Infection Feedback: Primary pathogens do not need to wait for a wound to gain access to their host. The human pathogen Yersinia pestis, for example, survives in the flea’s foregut, and uses the flea as a vector to spread from one mammalian host to another. 17. Answer: D Difficulty: 2 Section: Cell Biology of Infection Feedback: The wild-type bacterium blocks the flea gut, preventing the flea from digesting its blood meal, and leading to repetitive attempts at biting and dissemination of the infection. The mutant bacteria cannot block the flea gut and are transmitted less efficiently. 18. Answer: YNYN Difficulty: 2 Section: Cell Biology of Infection

Feedback: Whereas many epithelial barriers such as the skin and the lining of the mouth, vagina, and large intestine are densely populated by normal flora, others, including the lining of the lower lung and the bladder, are normally kept nearly sterile. 19. Answer: E Difficulty: 2 Section: Cell Biology of Infection Feedback: H. pylori uses multiple mechanisms to persist in the stomach, causing stomach ulcers and some stomach cancers. 20. Answer: EDCBA Difficulty: 2 Section: Cell Biology of Infection Feedback: Enteropathogenic E. coli (EPEC) uses a type III secretion system to deliver Tir into the host intestinal epithelial cell. Tir then inserts into the plasma membrane of the host cell, and its extracellular domain now binds to the bacterial surface protein intimin, triggering actin polymerization in the host cell. This results in the formation of a pedestal, pushing the bacterium up from the host-cell membrane, and thereby promoting bacterial movement along the cell surface. 21. Answer: FFFF Difficulty: 2 Section: Cell Biology of Infection Feedback: Some enveloped viruses such as human immunodeficiency virus (HIV) fuse with the host-cell plasma membrane for entry; others such as influenza A virus fuse with a host-cell membrane after endocytosis. Inhibition of endocytosis is therefore expected to mostly affect influenza virus entry. Nonenveloped viruses such as adenovirus enter the cell via endocytosis followed by endosomal membrane disruption. Individuals with a defective CCR5 gene are less susceptible to HIV infection, as the virus requires CCR5 to enter some host cells. Most viruses enter the host cell by endocytosis or macropinocytosis. 22. Answer: C Difficulty: 3 Section: Cell Biology of Infection Feedback: Loss-of-function mutations in the co-receptor CCR5 can result in less susceptibility to human immunodeficiency virus (HIV) infection. Additionally, inhibition of membrane fusion, inhibition of the co-receptor CXCR4, or saturation of the viral envelope surface protein with soluble CD4 decoys can help limit viral entry. HIV can fuse with the plasma membrane at neutral pH, and increasing endosomal pH does not inhibit HIV infection. 23. Answer: BBAA Difficulty: 2 Section: Cell Biology of Infection

Feedback: The zipper mechanism (B) depends on invasin proteins on the surface of the bacterium that bind to host-cell adhesion proteins, eventually resulting in the phagocytosis of the bacterium. The trigger mechanism (A) depends on the injection of effector proteins into the host cell, resulting in the engulfment of the bacterium by a process that resembles macropinocytosis. 24. Answer: B Difficulty: 2 Section: Cell Biology of Infection Feedback: The energy for the invasion of T. gondii comes from actin polymerization in the parasite rather than the host cell. 25. Answer: BCD Difficulty: 3 Section: Cell Biology of Infection Feedback: In contrast to intracellular bacterial pathogens such as S. enterica, intracellular eukaryotic parasites such as T. gondii actively invade host cells. The energy of active invasion seems to come from actin polymerization in the parasite rather than the host. It follows that interfering with actin dynamics in the parasite (but not the host) using CyD would inhibit internalization. CyD-resistant T. gondii (column C) can therefore invade both host-cell types whether or not CyD is present. Wild-type T. gondii (column B), in contrast, can efficiently infect the hosts only in the absence of the drug. Finally, wild-type S. enterica (column A) proliferates inside the hosts under all conditions except in the wild-type host (row D) in the presence of CyD. 26. Answer: A Difficulty: 3 Section: Cell Biology of Infection Feedback: The appearance of RFP-positive compartments shortly after infection represents the lysosome-dependent pathway. At this time, the pathogens invading via the lysosome-independent pathway are enclosed by a membrane derived from the host-cell plasma membrane, which is GFP-positive. 27. Answer: D Difficulty: 2 Section: Cell Biology of Infection Feedback: Listeriolysin O disrupts the phagosomal membrane and releases the bacterium into the host-cell cytosol. Despite its persisting secretion, this protein does not disrupt other cellular membranes since it is much less active at the neutral pH of the cytosol, and also because it is rapidly degraded by the host proteasomes. 28. Answer: 121 Difficulty: 2 Section: Cell Biology of Infection

Feedback: Pathogens that follow strategy 1 include all viruses, the protozoan Trypanosoma cruzi, and the bacteria Listeria monocytogenes and Shigella flexneri. Those that follow strategy 2 include the bacterium Legionella pneumophila. Those that follow strategy 3 include protozoans of the Leishmania genus and the bacterium Salmonella enterica. 29. Answer: BAC Difficulty: 2 Section: Cell Biology of Infection Feedback: The bacteria in this example are Mycobacterium tuberculosis (A), Salmonella enterica (B), and Legionella pneumophila (C). 30. Answer: LLHL Difficulty: 2 Section: Cell Biology of Infection Feedback: L. monocytogenes moves around the host cell’s cytosol by hijacking the actin cytoskeleton and nucleating new, short-lived actin filaments at one pole of the bacterial cell. This pushes the cell forward, possibly against the host cell’s plasma membrane. Viruses such as the herpes simplex virus, which infects neurons, rely on microtubule tracks and their associated motor proteins for transport. 31. Answer: C Difficulty: 2 Section: Cell Biology of Infection Feedback: Viruses with a negative [–] strand RNA genome rely on a pre-packaged RNAdependent RNA polymerase to replicate their genome. 32. Answer: FTFF Difficulty: 2 Section: Cell Biology of Infection Feedback: Even bacteria taken from natural soil are typically already resistant to a number of antibiotics widely used in clinical practice. Altering the critical enzyme that is targeted by an antibiotic is but one of the three general ways through which bacteria can achieve antibiotic resistance: the other two are modifying the antibiotic and ejecting it from the cell. Although high genomic plasticity in some viruses has complicated attempts to develop vaccines against them, several highly effective vaccines have been successfully used in immunization programs for viral infections.

MOLECULAR BIOLOGY OF THE CELL, SIXTH EDITION CHAPTER 24: THE INNATE AND ADAPTIVE IMMUNE SYSTEMS Copyright © 2015 by W.W. Norton & Company, Inc. 1. Indicate whether each of the following descriptions better applies to an adaptive (A) or innate (I) immune response. Your answer would be a four-letter string composed of letters A and I only, e.g. AAAA. ( ) It is found in invertebrate and vertebrate animals as well as in plants. ( ) It has a long-lasting memory. ( ) It is highly specific to the particular invading pathogen. ( ) It employs natural killer (NK) cells to induce apoptosis in infected host cells. 2.

Which of the following cell types is NOT phagocytic? A. Macrophage B. Monocyte C. Neutrophil D. Lymphocyte E. Dendritic cell

3. Blood-group compatibility is an important consideration in red blood cell transfusions that are administered in millions of liters worldwide every year. In the ABO blood-group system, individuals with AB blood type are the universal recipient (can accept blood from any donor), while those with O type are universal donors. If an incompatible transfusion is made between an A-type (donor) and a B-type (recipient) individual, for example, the anti-A antibodies present in the recipient’s plasma would rapidly destroy the transfused blood cells with the help of the complement system, with potentially serious consequences. But why do we produce antibodies against other blood-group antigens even without having been exposed to those foreign blood cells before? It has been suggested that these antibodies are generated in response to minor infections (in early life) with microbes of the normal flora of our bodies, and that, in addition to binding to the microbial antigens, these antibodies can cross-react with the similar A- and B-type carbohydrate antigens of the ABO blood-group system. If this is indeed the case, blood plasma from a “germ-free” individual would react with … A. blood of any ABO type. B. blood of no other ABO type.

C. blood from O-type individuals only. D. blood from non-O-type individuals only. E. blood from AB-type individuals only. 4. Which of the following are recognized by pattern recognition receptors as pathogenassociated molecular patterns? A. Double-stranded viral RNAs B. Formylmethionine-containing proteins C. Unmethylated CpG motifs D. Bacterial flagella components E. All of the above 5.

Which of the following groups of proteins are pattern recognition receptors? A. NOD-like receptors B. Toll-like receptors C. RIG-like receptors D. C-type lectin receptors E. All of the above

6. Gout patients have high uric acid levels in their blood and suffer from arthritis in their joints as a result of formation of intracellular and extracellular uric acid crystals. Would you expect treatment with glucocorticoids (G), which inhibit prostaglandin synthesis, or with tumor necrosis factor (TNF) to be normally used to treat such patients? 7. Indicate whether each of the following descriptions is shared (S) or not shared (N) between macrophages and neutrophils in destroying invading pathogens in vertebrates. Your answer would be a four-letter string composed of letters S and N only, e.g. SSSS. ( ) They use a respiratory burst and create toxic reactive oxygen species to kill engulfed pathogens. ( ) They are normally found in most tissues even before pathogen invasion. ( ) They are long-lived and normally survive long after activation. ( ) They secrete pro-inflammatory cytokines upon encountering invading pathogens. 8. Indicate true (T) and false (F) statements below regarding the complement system. Your answer would be a four-letter string composed of letters T and F only, e.g. TTFT.

( ) The early complement components form the membrane attack complex, whereas the late complement components cleave and activate C3. ( ) The classical pathway of complement activation is triggered by binding of a secreted PRR to mannose-containing glycoproteins or glycolipids on the surface of pathogens. ( ) Some pathogens such as Neisseria gonorrhoeae hide from the complement reaction via sialic acid camouflage. ( ) Once a membrane attack complex is formed, it can attack multiple pathogens before becoming inactivated. 9.

A cell has been infected with an RNA virus. As a consequence, it may … A. reduce the expression of its surface class I MHC molecules. B. secrete type I interferons such as interferon-α. C. shut down most protein translation and destroy most of its RNA. D. be induced to die by apoptosis. E. All of the above.

10. Consider two receptors, one inhibitory and one activating, on the surface of natural killer (NK) cells: KIR receptors that interact with certain class I MHC proteins, and special Fc receptors that recognize the tail region of IgG antibodies. When NK cells are incubated with an immortalized B cell line derived by infection of human B cells with Epstein–Barr Virus, the B cells are killed by NK cells. Normal B cells, in contrast, are not efficiently killed by NK cells. In each of the following scenarios, indicate whether the described changes are expected to enhance (E) or suppress (S) the cytotoxicity (killing activity) of the NK cells in this system. Your answer would be a four-letter string composed of letters E and S only, e.g. SEEE. ( ) The cell line is transformed with human class I MHC genes, and the genes are expressed. ( ) An anti-CD23 IgG antibody is added to the cells. (CD23 is a known marker on the surface of the immortalized B cells.) ( ) The cells are treated with stibogluconate, an inhibitor of KIR receptor signaling. ( ) Type I interferons are added to the cells. 11.

How are natural killer (NK) cells different from cytotoxic T (TC) cells? A. NK cells induce apoptosis in their target cells, whereas TC cells are professional phagocytes. B. NK cells kill virus-infected cells, whereas TC cells kill cancerous cells.

C. NK cells kill cells with a high level of class I MHC protein expression, whereas killing by TC cells requires low expression levels. D. NK cells respond quickly to a virus infection, whereas the activation of TC cells to become cytotoxic is a slow process. E. NK cells are normally abundant in the tissue even before infection, whereas TC cells are mostly in the bloodstream. 12. Fill in the blank in the following paragraph regarding the innate and adaptive immune systems. Do not use abbreviations. “As key components of the innate immune response, … cells provide the link between the innate and adaptive immune responses. Upon exposure to pathogens in tissues, they engulf the microbe, become activated, and travel to nearby lymphoid organs where they present the processed antigens to the lymphocytes of the adaptive immune system.” 13. Polyclonal antibodies can be generated against almost any protein of interest by repeatedly injecting the purified protein into an animal (such as a rabbit) and collecting blood serum once the animal’s immune system has produced antibodies against the protein. In this process, the antigen is not injected alone; it is co-injected with a so-called adjuvant to “trick” the immune system. Additionally, the adjuvant used in the first injection is often different from that used in later “booster” injections. Freund’s complete adjuvant (C) contains a water–oil emulsion plus heat-killed Mycobacteria. In contrast, Freund’s incomplete adjuvant (I) lacks the Mycobacteria component. Which adjuvant (C or I) do you think is used in the booster injections? Which immune response—primary (P) or secondary (S)—is triggered after the first purified protein injection? Activation of which antigen-specific immune cells—memory (M) or effector (E)—is chiefly responsible for the “boosting” effect of the later injections? Write down your answer as a three-letter string, e.g. ISM. 14.

In contrast to a primary immune response, a secondary immune response generally … A. has a longer lag period. B. is stronger. C. involves induction of effector cells to become memory cells. D. takes longer to develop. E. All of the above.

15. Mice whose immune systems had previously rejected a tissue transplant from a certain strain of donor mouse received another transplant shortly afterward, either from the same donor strain or a new one. All of the recipient mice eventually rejected the transplant. However, one group did so more quickly than the other group. Do you think this group received the graft from the original mouse strain (O) or the new strain (N)? Write down O or N as your answer. 16. Indicate true (T) and false (F) statements below regarding lymphocytes in the adaptive immune system. Your answer would be a four-letter string composed of letters T and F only, e.g. TTFF. ( ) To induce an adaptive immune response, an invading pathogen must invade a peripheral lymphoid organ. ( ) Lymphocytes continuously circulate between central and peripheral lymphoid organs. ( ) Upon encountering its specific antigen in a peripheral lymphoid organ, a lymphocyte remains there to proliferate and differentiate into effector and memory cells. ( ) Memory B and T cells do not recirculate, but rather remain in the peripheral lymphoid organ in which they develop. 17. If mice are injected with antibodies against selectin homing receptors or against the integrin ligand I-CAM1 (which is expressed on the surface of endothelial cells), they show immune defects, such as higher susceptibility to bacterial infections. Which of the following properties of lymphocytes at endothelial cells would you expect to observe after injection with both antibodies? A. Normal lymphocyte rolling and normal lymphocyte adhesion B. Impaired lymphocyte rolling but normal lymphocyte adhesion C. Normal lymphocyte rolling but impaired lymphocyte adhesion D. Impaired lymphocyte rolling and impaired lymphocyte adhesion 18.

If heart surgery is necessary in a newborn, the thymus is sometimes removed to gain

better access to the heart. If a lymph node of such a patient is later studied in a microscope, one particular area of the node is found to have an unusually low number of cells. Looking at the drawing below, what is the name of this area?

19. Indicate whether each of the following descriptions better applies to the clonal anergy (A), clonal deletion (D), receptor editing (E), or clonal suppression (S) mechanism of immunological self-tolerance. Your answer would be a five-letter string composed of letters A, D, E, and S only, e.g. SADSS. ( ) It involves apoptosis of self-reactive cells. ( ) It is mostly limited to B cells. ( ) It involves regulatory T (Treg) cells. ( ) It involves DNA recombination. ( ) It is limited to central tolerance processes. 20. Mammals produce five major classes of immunoglobulins. Indicate whether each of the following descriptions better applies to the immunoglobulin A, D, E, G, or M class. Your answer would be a five-letter string composed of letters A, D, E, G, and M only, e.g. GMADE. ( ) It binds to cells that secrete histamine. ( ) It is the major antibody in secretions such as saliva. ( ) It is the major antibody circulating in the blood. ( ) It is found on the surface of naïve B cells along with IgM. ( ) It forms wheel-like pentamers involved in the primary immune response.

21. Indicate whether each of the following statements is true (T) or false (F) regarding all the cell-surface immunoglobulin molecules produced by a single, mature, naïve B cell. ( ) They all have the same heavy chain. ( ) They all have the same light chain. ( ) They all have the same antigen-binding site. 22.

Where are the hypervariable regions located in a typical antibody molecule? A. Only in the N-terminal region of the light chain B. Only in the C-terminal region of the light chain C. Only in the N-terminal region of the heavy chain D. Only in the C-terminal region of the heavy chain E. In the N-terminal region of both the light and the heavy chain

23. You have prepared glass beads coated with rabbit anti-Ig antibodies that were isolated from rabbits that were immunized with mouse immunoglobulin (Ig). If you incubate a sample of mononucleated cells from mouse spleen with these beads, which of the following cell types do you think would stick to the beads? A. Only macrophages B. Only B cells C. Only cytotoxic T cells D. B cells and cytotoxic T cells E. B cells and macrophages 24. Antigens and antibodies can form large, insoluble, cross-linked aggregates that precipitate from solution. The maximal amount of precipitant is observed at a certain antigen/antibody ratio; adding either excess antigen or excess antibody can prevent (or even reverse) the precipitation. Consider an antigen with multiple identical antigenic determinants that can be recognized by either bivalent or decavalent (i.e. pentameric) antibodies with binding sites that recognize the determinants. The amount of precipitation in the presence of a constant antigen concentration is plotted as a function of antibody concentration in the following qualitative graph. Which curve (1 or 2) in the graph do you think corresponds to the bivalent antibody? Would you expect to see such a bell-shaped curve if the antibodies were monovalent?

Amount of precipitation

1 2

Total molar antibody concentration (monomer or pentamer) A. Curve 1; yes B. Curve 2; yes C. Curve 1; no D. Curve 2; no E. Curve 2; only if the antigen was also monovalent 25. In the schematic drawing of an antibody molecule below, indicate which letter (A to E) corresponds to each of the following features. Your answer would be a five-letter string composed of letters A to E only, e.g. DAECB. D

A C

( ) N-terminus ( ) C-terminus ( ) Light chain ( ) Antigen-binding site ( ) Disulfide bond 26. After having cereal and nut mix for breakfast, your friend realizes that the mix contains a nut to which she is allergic. Soon afterward, symptoms of an allergic reaction start to appear, and she is going to take an antihistamine pill to alleviate them. During the allergic reaction in her body, … A. mast cells synthesize and secrete IgE antibodies into the blood. B. the allergen molecules bind to IgE antibodies that are bound to Fc receptors on the surface of basophils and mast cells. C. histamine and other amines are recognized by IgE antibodies. D. histamine binds to mast cells and induces IgE production. E. All of the above. 27.

How many hypervariable loops are there in a bivalent IgG antibody? Write down your

answer in digits, e.g. 5. 28. Indicate whether each of the following normally occurs before (B) or after (A) antigen stimulation of B cells. Your answer would be a four-letter string composed of letters A and B only, e.g. AAAA. ( ) V(D)J recombination ( ) Class switch recombination ( ) Somatic hypermutation ( ) Junctional diversification 29. Indicate whether each of the following descriptions better applies to class switching (C), somatic hypermutation (S), or both (B). Your answer would be a four-letter string composed of letters B, C, and S only, e.g. SSSB. ( ) It is dependent on activation-induced cytidine deaminase. ( ) It involves sequence changes in the CH domains only. ( ) It increases the affinity of antibody–antigen binding. ( ) It takes place in germinal centers after antigen stimulation.

30. Consider the following three diseases associated with activation-induced deaminase (AID) and indicate whether you think each of them is accompanied by a higher (H) or lower (L) AID activity in the affected cells compared to normal cells. Your answer would be a three-letter string composed of letters H and L only, e.g. LLL. ( ) Burkitt’s lymphoma is a cancer of B lymphocytes originating from the germinal center. In this cancer, chromosomal translocations are observed between the immunoglobulin heavy-chain genes and proto-oncogenes (genes whose overexpression can promote cancer), resulting in aberrant expression of the latter. ( ) Diffuse large B cell lymphoma is a cancer of B lymphocytes in which, in addition to chromosomal translocations, high mutation rates are observed in many genes including proto-oncogenes. ( ) Hyper IgM syndrome is a genetic immunodeficiency syndrome in which IgM antibodies are produced by B cells at high levels and for extended periods, while levels of IgA, IgE, and IgG antibodies are abnormally reduced. 31. Consider two solution chambers of equal volume divided by a dialysis membrane that is impermeable to antibody molecules but permeable to small antigens. Starting with an antigen concentration of 4.8 nM in both chambers, you add its specific IgG antibody to one of the chambers at 1 nM concentration. You then measure the concentration of the antigen in the other chamber after equilibrium is reached, which turns out to be about 4 nM. What is the association constant (Ka) for the binding of the antigen to each antigen-binding site of IgG? Assume independent antigen-binding sites in each antibody molecule. Write down your answer in liters per mole, in scientific notation with one decimal place, e.g. 5.0 × 102 liters/mole). 32. Indicate whether each of the following descriptions better applies to B cells (B) or T cells (T) of the adaptive immune system. Your answer would be a four-letter string composed of letters B and T only, e.g. BTTT. ( ) Their effector cells act mainly at short range. ( ) Their receptors recognize protein fragments that have been processed in antigenpresenting cells. ( ) They do NOT normally express CD8 protein on their surface. ( ) They can become “killer” cells. 33. Which of the following better describes cross-presentation of protein antigens by professional antigen-presenting cells to naïve TC cells? A. Presentation of intracellular antigens by class I MHC proteins

B. Presentation of intracellular antigens by class II MHC proteins C. Presentation of extracellular antigens by class I MHC proteins D. Presentation of extracellular antigens by class II MHC proteins 34. Normally, naïve cytotoxic and helper T cells interact for the first time with foreign antigens presented by dendritic cells in … A. the thymus. B. the bloodstream. C. peripheral lymphoid organs. D. the bone marrow. E. inflamed tissues. 35. Indicate whether each of the following descriptions better applies to B cell receptors (B) or T cell receptors (T). Your answer would be a four-letter string composed of letters B and T only, e.g. BTTT. ( ) They can also be produced as secreted antibodies with the same antigen specificity. ( ) They diversify by class switching and somatic hypermutation. ( ) They often exist as heterodimers. ( ) They have a relatively low antigen-binding affinity. 36. Indicate true (T) and false (F) statements below regarding the proteins of our adaptive immune system. Your answer would be a four-letter string composed of letters T and F only, e.g. TTFT. ( ) Both B cell receptors and T cell receptors have domains with an immunoglobulin fold, which consists of a β-sandwich and an intrachain disulfide bond. ( ) MHC proteins are the most polymorphic among human proteins. ( ) MHC proteins and CD4 and CD8 co-receptors are all members of the Ig superfamily. ( ) Both B cell receptors and T cell receptors are stably associated with invariant transmembrane protein chains that are required for activating intracellular signaling pathways. 37. Indicate whether each of the following descriptions better applies to class I (1) or class II (2) MHC proteins of vertebrates. Your answer would be a four-digit number composed of digits 1 and 2 only, e.g. 2221. ( ) They mainly present intracellular peptide fragments. ( ) They are found on most nucleated cells.

( ) They present antigens mainly to helper or regulatory T cells. ( ) They are recognized by CD4 co-receptors on helper and regulatory T cells. 38. In antigen presentation to helper T cells (TH cells), both the TCR and co-receptor proteins… A. are expressed on the dendritic cell. B. bind to the class II MHC protein. C. bind to the presented peptide antigen. D. All of the above. 39.

T cell receptors that do not interact at all with self-peptide–MHC complexes during T cell

development in the thymus normally undergo … A. positive selection. B. negative selection. C. receptor editing. D. clonal suppression. E. “death by neglect.” 40.

The transcription regulator AIRE (autoimmune regulator) plays a key role in developing

immunological self-tolerance by promoting the production of various cellular proteins that are mainly expressed outside the thymus. Which of the following cells have the highest level of AIRE? A. Double-negative thymocytes B. Double-positive thymocytes C. Lymphoid follicle cells in lymph nodes D. Thymus epithelial cells E. Peyer’s patch cells 41. In the following schematic diagram, which curve (A to D) do you think better shows the qualitative relationship between the affinity of a developing T cell’s TCR for self-peptide–MHC complexes in the thymus and the survival chance of the cell?

Developing T cell Survival

A B C D Affinity of TCR binding to self peptide in complex with MHC

42. Sort the following events to reflect the order in which they occur in the presentation of viral peptides to cytotoxic T cells by a virus-infected cell. Your answer would be a four-letter string composed of letters A to D only, e.g. BDCA. (A) Peptide transport into the ER lumen (B) Partial proteasomal degradation of the viral protein into peptide fragments (C) Peptide binding to class I MHC protein in the ER lumen (D) The appearance of viral protein in the cytosol 43. Different mice have different allelic variants of class I MHC genes. Therefore, cytotoxic T cell (TC cell) clones can be generated by culturing lymphocytes harboring MHC allele A in the presence of nondividing nucleated cells harboring MHC allele B (and vice versa). These T cells can normally kill cells harboring allele B by inducing them to undergo apoptosis. Imagine two strains of mice with MHC allele A that are either wild type (WT) for the gene encoding perforin (A1) or are mutant and lack the gene (A2). Also imagine two strains of mice with MHC allele B that are either WT for the gene encoding Fas (B1) or are mutant and lack the gene (B2). Cytotoxic T cells derived from the A1 strain can induce apoptosis in both B1 and B2 cells. However, those derived from the A2 strain are only able to induce apoptosis in B1 cells and not B2 cells. Indicate whether each of the following statements is (Y) or is not (N) consistent with these observations. Your answer would be a four-letter string composed of letters Y and N only, e.g. NYNY. ( ) The perforin–granzyme pathway is the only way through which TC cells induce apoptosis in mouse target cells. ( ) The activation of the Fas–FasL pathway is sufficient to induce apoptosis by TC cells.

( ) There are other major pathways (not dependent on perforin or Fas) through which TC cells induce apoptosis in mouse target cells. ( ) Both pathways (Fas–FasL and perforin–granzyme) are required simultaneously for the induction of apoptosis by TC cells in mouse target cells. 44. The schematic drawing below summarizes the differentiation of naïve helper T cells (TH cells) into various effector T cells, which occurs in a peripheral lymphoid organ. Indicate which effector cell (A to E) in the drawing is better described by each of the following descriptions. Your answer would be a four-letter string composed of letters A to E only, e.g. DEAE.

( ) It can also develop from thymocytes in the thymus, as well as from naïve TH cells.

( ) It suppresses the development, activation, or function of most other immune cells. ( ) It produces IFNγ and activates macrophages. ( ) They reside in lymphoid follicles close to developing B cells. 45. Indicate whether each of the following cell-surface proteins is expressed mainly by B cells (B), dendritic cells (D), or T cells (T). Your answer would be a four-letter string composed of letters B, D, and T only, e.g. DBBB. ( ) Co-stimulatory protein B7, which is recognized by CD28 ( ) Invariant CD3 complex ( ) Inhibitory CTLA4 protein ( ) CD40 receptor, which recognizes the co-stimulatory protein CD40 ligand on a helper T cell 46. From an evolutionary perspective, which mating pattern confers a higher fitness to a mammalian population: preferred mating between pairs whose MHC genes are most similar (S) or between those whose MHC genes are most dissimilar (D)? Write down D or S as your answer. 47. A patient with persisting larger-than-normal lymph nodes is suspected to have a T cell lymphoma, a cancer of T lymphocytes. Abnormally enlarged lymph nodes are also found in local infections, when lymphocytes are activated to proliferate there. In order to distinguish between cancer and infection, you take tissue samples from the patient, as well as from a healthy individual, and perform a so-called clonality test. You extract DNA from the tissues and amplify a small chromosomal region that includes the T-cell-receptor β chain D–J junctions. Amplification is done by PCR, using primers that hybridize to specific DNA sequences flanking the junctions. You then analyze the size distributions of the PCR products by capillary electrophoresis, which makes it possible to distinguish between DNA molecules with small differences in size. Based on the results, presented in the following schematic graphs, you conclude that, unfortunately, it is likely that the patient has developed a T cell lymphoma. Which graph (1 or 2) represents the patient’s test results? Write down 1 or 2 as your answer.

Relative number of PCR products

2 100

200 Size of PCR product (nucleotide pairs)

Answers: 1. Answer: IAAI Difficulty: 2 Section: The Innate Immune System Feedback: The innate immune response constitutes the first line of defense, which employs general defense reactions and is short-lasting. Like the cytotoxic T cells of adaptive immunity, the NK cells of innate immunity can kill infected host cells by inducing apoptosis. 2. Answer: D Difficulty: 2 Section: The Innate Immune System Feedback: Lymphocytes (e.g. B and T cells of the adaptive immune system) are not phagocytes. 3. Answer: B Difficulty: 2 Section: The Innate Immune System Feedback: Such an individual would not produce the cross-reacting antibodies, and their plasma would therefore not react with blood from individuals of any ABO type. 4. Answer: E Difficulty: 1 Section: The Innate Immune System Feedback: Pattern recognition receptors (PRRs) recognize a variety of pathogenassociated antigens, and are of various classes. 5. Answer: E Difficulty: 1 Section: The Innate Immune System Feedback: Pattern recognition receptors (PRRs) recognize conserved features of pathogens. PRRs include Toll-like receptors, NOD-like receptors, RIG-like receptors, and C-type lectin receptors. 6. Answer: G Difficulty: 3 Section: The Innate Immune System Feedback: Prostaglandin is a pro-inflammatory signal molecule, and TNF is a proinflammatory cytokine. In gout, NOD-like receptor (NLR)-dependent inflammasome assembly leads to an inflammatory response, which can be suppressed by drugs such as glucocorticoids. 7. Answer: SNNS Difficulty: 1 Section: The Innate Immune System

Feedback: Neutrophils normally circulate in the bloodstream and are short-lived. They exit the blood and enter an infected tissue only when needed and are unable to survive the killing frenzy of the inflammatory response. 8. Answer: FFTF Difficulty: 2 Section: The Innate Immune System Feedback: Cleavage and activation of the pivotal complement component, C3, is a result of the activation of the early components (by antibodies in the classical pathway, or by lectin binding in the lectin pathway), and in turn activates the late components to form the membrane attack complex. The complex is assembled on a target membrane only and is not reusable as a whole. Most activated complement components are rapidly inactivated. 9. Answer: E Difficulty: 1 Section: The Innate Immune System Feedback: Type I interferons are produced in response to viral infection and help block viral replication in multiple ways, for example by shutting down most protein synthesis in the infected cell. If these measures fail to stop viral replication, the cell may even kill itself by apoptosis. Many viruses have evolved mechanisms to inhibit the expression of class I MHC protein on the surface of their host cells. 10. Answer: SEEE Difficulty: 1 Section: The Innate Immune System Feedback: The KIR receptors that bind to class I MHC proteins initiate inhibitory signals upon binding and therefore suppress NK cytotoxicity. Viral infection tends to decrease the expression of class I MHC proteins by host cells, which partly explains why the virusinfected B cells are susceptible to NK cytotoxicity, whereas normal B cells are not. Inhibition of the KIR receptor inhibitory signal by stibogluconate, therefore, enhances NK cell killing. Additionally, the Fc receptors that bind to the immunoglobulin G (IgG) Fc regions enhance NK cell killing of target cells coated with IgG antibodies. Finally, type I interferons enhance NK cell killing activity in general. 11. Answer: D Difficulty: 2 Section: The Innate Immune System Feedback: NK cells are components of the innate immune system, which responds quickly to infection; NK cells circulate in a partially active state and therefore can kill virally infected cells when they first encounter them. TC cells need to be activated in a slow process in a lymphoid organ before they can kill virus-infected cells. 12. Answer: dendritic Difficulty: 1 Section: The Innate Immune System

Feedback: Dendritic cells provide the link between our innate and adaptive immune systems. 13. Answer: IPM Difficulty: 2 Section: Overview of the Adaptive Immune System Feedback: The adjuvant usually enhances innate immune responses, which are required to activate the adaptive immune system to respond to an otherwise harmless foreign protein. The Mycobacteria component in the complete adjuvant performs this trick and elicits a strong primary immune response, but it is not required for later injections once memory T and B cells have developed. 14. Answer: B Difficulty: 2 Section: Overview of the Adaptive Immune System Feedback: A secondary immune response is different from a primary response in that it has a shorter lag time and is stronger and more efficient. This is because memory cells differentiated from naïve cells in the first exposure to the antigen are pre-existing at the time of the secondary immune response. 15. Answer: O Difficulty: 2 Section: Overview of the Adaptive Immune System Feedback: Rejection of a second graft from the original donor strain occurs more quickly, because memory B and/or T cells were produced in response to the first transplant. This results in a secondary adaptive immune response to the second transplant from the same mouse strain that is faster and stronger than the primary immune response elicited when the second transplant is from a different donor mouse strain. 16. Answer: FFTF Difficulty: 2 Section: Overview of the Adaptive Immune System Feedback: Pathogens or their products are carried to a peripheral lymphoid organ via the blood or lymph, usually by dendritic cells. The process of recirculation between peripheral lymphoid organs and the bloodstream greatly increases the chance that a lymphocyte will encounter its specific foreign antigen—usually a pathogen or one of its products. Both naïve and memory B and T cells take advantage of this recirculation to police the body; once activated in a peripheral lymphoid organ, they differentiate into effector cells and memory cells. 17. Answer: D Difficulty: 1 Section: Overview of the Adaptive Immune System Feedback: Selectins have a key role in the rolling step, which is followed by the integrindependent adhesion.

18. Answer: paracortex Difficulty: 2 Section: Overview of the Adaptive Immune System Feedback: T cells (which develop in the thymus) mostly occupy the paracortex, while B cells are found mainly in the lymphoid follicles. 19. Answer: DESEE Difficulty: 2 Section: Overview of the Adaptive Immune System Feedback: Receptor editing in B cells is a major self-tolerance mechanism that involves V(D)J recombination and occurs exclusively in a central lymphoid organ such as the bone marrow in adults. In clonal deletion, T and B cell clones that encounter self antigens are induced to die by apoptosis. In clonal suppression, Treg cells suppress the activity of self-reactive T and B cells. 20. Answer: EAGDM Difficulty: 2 Section: B Cells and Immunoglobulins Feedback: Each of the five major classes of immunoglobulins in mammals mediates a characteristic response. IgM and IgD are expressed as B cell receptors on the surface of naïve B cells. In its secreted form, IgM forms pentamers. The major antibody class in the blood is IgG. The principal antibody class in secretions such as saliva is IgA. IgE antibodies bind to Fc receptors on the surface of mast cells and basophils, which are involved in allergy. 21. Answer: FTT Difficulty: 1 Section: B Cells and Immunoglobulins Feedback: A single, mature, naïve B cell has IgM and IgD cell-surface immunoglobulin molecules with the same type of light chain and antigen-binding site. The IgM and IgD molecules have different heavy chains. 22. Answer: E Difficulty: 1 Section: B Cells and Immunoglobulins Feedback: The greatest diversity in the variable regions of light and heavy chains occurs in their hypervariable regions. The actual antigen binding site is formed by only about 510 amino acid residues in each hypervariable region. 23. Answer: E Difficulty: 1 Section: B Cells and Immunoglobulins Feedback: The mouse B cells will bind because the rabbit antibodies will recognize the cell-surface antigen receptors (B cell receptors; BCRs), which are Ig molecules.

Macrophages will bind because they have cell-surface Fc receptors that will recognize the Fc region of the rabbit anti-Ig antibodies. 24. Answer: D Difficulty: 3 Section: B Cells and Immunoglobulins Feedback: Monovalent antibodies would not be able to cross-link the antigen and therefore would not precipitate the antigen. With the same binding affinity of their antigen-binding sites, the pentameric IgM molecules would be more effective at crosslinking antigens than bivalent IgG molecules and so would precipitate the antigen at lower antibody concentrations (curve 1). The bell-shaped curve implies that both the antigen and antibody are multivalent. 25. Answer: DEBAC Difficulty: 1 Section: B Cells and Immunoglobulins Feedback: Please refer to Figure 24-23. 26. Answer: B Difficulty: 1 Section: B Cells and Immunoglobulins Feedback: IgE antibodies bind to Fc receptors on the surface of mast cells and basophils with high affinity and thereby act as allergen receptors on these cells. Binding of the allergen causes the cells to release histamine, which causes most of the symptoms of the allergic reaction. 27. Answer: 12 Difficulty: 2 Section: B Cells and Immunoglobulins Feedback: There are three hypervariable loops in each of the VH and VL domains, and each domain is present in two copies per IgG molecule. 28. Answer: BAAB Difficulty: 2 Section: B Cells and Immunoglobulins Feedback: Before they can make immunoglobulin (Ig) molecules and thereby bind antigen, B cells have to undergo V(D)J recombination, which is accompanied by junctional diversification. After stimulation with an antigen, B cells undergo somatic hypermutation (as part of their affinity maturation) and class switch recombination (as part of Ig class switching). 29. Answer: BCSB Difficulty: 2 Section: B Cells and Immunoglobulins

Feedback: Both class switching and somatic hypermutation depend on activation-induced deaminase (AID) and occur after antigen stimulation, which occurs mainly in germinal centers. 30. Answer: HHL Difficulty: 2 Section: B Cells and Immunoglobulins Feedback: If the mutagenic activity of AID is abnormally controlled and becomes excessive, it can induce hypermutation genome-wide and can also induce abnormal double-strand DNA breaks and thereby aberrant translocation events. Loss-of-function mutations in AID can lead to hyper IgM syndrome, because B cells cannot efficiently perform AID-dependent class switch recombination. 31. Answer: 1.0 × 109 liters/mole Difficulty: 3 Section: B Cells and Immunoglobulins Feedback: Free antigen concentration in both chambers is 4 nM at equilibrium, which means the concentration of bound antigen in the antibody-containing chamber is 1.6 nM, that is (4.8 nM – 4 nM) × 2, to account for the total antigen present in both chambers. This is also equal to the concentration of bound antigen-binding sites (not antibodies). Therefore, the concentration of unbound sites in this chamber is 0.4 nM, that is (1 nM × 2 – 1.6 nM). The association constant of antigen (Ag) binding to antigen-binding site (Abs) is thus calculated as: Ka = [Abs–Ag]eq/([Abs]eq[Ag]eq) = (1.6 nM)/(0.4 nM × 4 nM) = 1 liter/nmol = 1.0 × 109 liters/mole. 32. Answer: TTBT Difficulty: 2 Section: T Cells and MHC Proteins Feedback: In contrast to B cells, T cells act mainly at short range and recognize protein fragments bound to MHC molecules on the surface of antigen-presenting cells. Cytotoxic T cells, but not B cells, express the CD8 co-receptor. 33. Answer: C Difficulty: 1 Section: T Cells and MHC Proteins Feedback: Cross-presentation of foreign antigens by dendritic cells involves loading of class I MHC proteins with peptides derived from extracellular antigens. It enables special dendritic cells that are not infected by viruses to present viral antigens to cytotoxic T cells (TC cells). 34. Answer: C Difficulty: 1 Section: T Cells and MHC Proteins

Feedback: Once exposed to a pathogen in an infected tissue, an immature dendritic cell becomes activated and migrates to a nearby peripheral lymphoid organ, where it presents processed antigen to naïve T cells. 35. Answer: BBTT Difficulty: 2 Section: T Cells and MHC Proteins Feedback: B cell receptors are tetramers made up of two identical heavy chains and two identical light chains; they can also be made in the form of soluble antibodies and can undergo affinity maturation by somatic hypermutation. T cell receptors are heterodimers; they do not undergo somatic hypermutation and affinity maturation and therefore have relatively low affinity for antigen. 36. Answer: TTTT Difficulty: 2 Section: T Cells and MHC Proteins Feedback: The immunoglobulin (Ig) superfamily includes a large number of important proteins of the immune system, such as antibodies, cell receptors (BCRs), T cell receptors (TCRs), co-receptors, and invariant chains associated with BCRs and TCRs. They all have Ig or Ig-like domains in their structure. Also bearing Ig domains, MHC proteins are highly polymorphic. 37. Answer: 1122 Difficulty: 2 Section: T Cells and MHC Proteins Feedback: Class I MHC proteins are found on the surface of most nucleated cells and mainly present intracellular antigens. Class II MHC molecules are present on special immune cells such as dendritic cells and present mainly extracellular antigens to helper or regulatory T cells with the help of CD4 co-receptors. 38. Answer: B Difficulty: 2 Section: T Cells and MHC Proteins Feedback: Note that T cell receptors (TCRs) bind to both the presented peptide and the polymorphic parts of the class II MHC protein, whereas the CD4 co-receptor binds only to the invariant part of the MHC protein. 39. Answer: E Difficulty: 2 Section: T Cells and MHC Proteins Feedback: Such receptors are useless without the ability to bind to self MHC proteins in complex with peptides. The thymocyte harboring the receptor fails to receive survival signals and dies. 40. Answer: D Difficulty: 2

Section: T Cells and MHC Proteins Feedback: A subset of epithelial cells in the adult thymus express AIRE, helping to induce self-tolerance to proteins mainly expressed outside the thymus. 41. Answer: C Difficulty: 3 Section: T Cells and MHC Proteins Feedback: Too low or too high an affinity would result in a reduction in survival. 42. Answer: DBAC Difficulty: 2 Section: T Cells and MHC Proteins Feedback: This form of antigen processing can be carried out by many types of host cells and uses class I MHC proteins. 43. Answer: NYNN Difficulty: 3 Section: T Cells and MHC Proteins Feedback: The ability of the A1 cytotoxic T cells to kill both B1 and B2 cells suggests that Fas in the target cell is not required to induce apoptosis; the perforin–granzyme pathway is sufficient. The ability of A2 cytotoxic T cells to kill B1 cells suggests that perforin in the TC cell is not essential either; activation of Fas is sufficient. Finally, the inability of A2 cytotoxic T cells to kill B2 cells suggests that at least one of these pathways (Fas–FasL or perforin–granzyme) should be functional for apoptosis to be induced and that there are no other major killing pathways in this cell system. 44. Answer: EEAC Difficulty: 2 Section: T Cells and MHC Proteins Feedback: Regulatory (suppressor) T cells can develop from thymocytes in the thymus as well as from naïve TH cells in peripheral tissues. TH1 cells secrete interferon-γ (IFNγ) which activates macrophages and induces B cells to switch the class of Ig they make. TFH cells reside in lymphoid follicles and secrete a variety of cytokines; they are especially important for the stimulation of Ig class switching and somatic hypermutation in B cells. 45. Answer: DTTB Difficulty: 1 Section: T Cells and MHC Proteins Feedback: Transmembrane B7 proteins are co-stimulatory signals on the surface of dendritic cells that are recognized by the co-receptor protein CD28 on the surface of naïve T cells. Similarly, the CD40 ligand on an effector helper T cell is recognized by CD40 receptors on B cells. T cells express the inhibitory protein CTLA4 on their surface to help regulate their activity. The CD3 complex associated with T cell receptors (TCRs) in T cells help relay the signal into the cell upon TCR stimulation by antigen binding. 46. Answer: D

Difficulty: 3 Section: T Cells and MHC Proteins Feedback: Mating between animals whose MHC genes are different provides the population with the advantage of increased MHC polymorphism and a broader capability for adaptive immune responses to pathogens. 47. Answer: 2 Difficulty: 3 Section: T Cells and MHC Proteins Feedback: V(D)J recombination in T lymphocytes creates deletions (or insertions) of various numbers of nucleotide pairs at the junctions of the V, D, and J segments. In response to an infection, a number of T lymphocytes (that recognize various antigenic determinants associated with the pathogen) proliferate to create effector and memory cells, leading to lymph-node enlargement. The different activated T cell clones normally differ in the exact number of deleted or inserted nucleotide pairs (graph 1). By contrast, cancer is a clonal disease; all malignant cells in a T cell lymphoma derive from a single original cell with its unique V(D)J junctions (graph 2).