Fundamental Molecular Biology, 2nd Edition , Lizabeth A Allison Test Bank

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Fundamental Molecular Biology, 2nd Edition By Lizabeth A. Allison

Email: richard@qwconsultancy.com

Fundamental Molecular Biology, 2nd Edition Instructor Test Bank Lizabeth A. Allison

Chapter 1 The Beginnings of Molecular Biology

Multiple Choice 1) In 1869 Friedrich Miescher isolated a substance called “nuclein” from pus from soiled bandages that was a(n) A. acidic, phosphorus-rich substance B. basic, phosphorus-rich substance C. amino acid-rich substance D. phospholipid-rich substance Answer: A 2) The basic principles of genetics attributed to Gregor Johann Mendel are: A. the laws of segregation and independent assortment B. the Punnett square C. the concept of dominant and recessive traits D. A and C E. A, B, and C Answer: D 3) The Augustinian monk, Gregor Johann Mendel, crossed yellow-seeded and green-seeded pea plants and then allowed the offspring to self-pollinate to produce an F2 generation. The results were as follows: 6018 yellow and 2002 green (8020 total). The allele for green seeds has what relationship to the allele for yellow seeds?

A. semidominant B. incompletely dominant C. recessive D. dominant Answer: C 4) Why is the allele for wrinkled seed shape in garden peas considered recessive?

A. The wrinkled allele is a rare mutant that is less common than the dominant allele.

B. The trait associated with the allele is not expressed in heterozygotes.

C. Individuals with the allele have overall lower fitness than that of individuals with the dominant allele.

D. The wrinkled allele “recedes” in the F2 generation when homozygous parents are crossed.

Answer: B

5) In 1928 Frederick Griffith discovered that Streptococcus pneumonia caused pneumonia in mice. In his experiments, mice were injected with different strains of treated and untreated bacteria. Which of the following is not likely to have occurred in the series of experiments by Griffith?

A. Mice injected with living S (pathogenic) strain die.

B. Mice injected with living R (nonpathogenic) strain live.

C. Mice injected with living R plus heat-killed S die.

D. Mice injected with living S plus heat-killed R live.

Answer: D 6) In 1928, Frederick Griffith injected living S (smooth) Streptococcus pneumonia into mice, and the mice died. When he injected living R (rough) Streptococcus pneumonia into mice, the mice lived. When he injected heat-killed S bacteria into mice, the mice lived. What was the result when he mixed heat-killed S bacteria with live R bacteria and injected this mixture into mice?

A. The mice died, and living bacteria that appeared smooth could be isolated from the dead mice.

B. The mice lived, and living bacteria that appeared smooth could be isolated from the living mice.

C. The mice died, indicating that the heat-killed S bacteria came back to life.

D. The mice lived and neither smooth nor rough-appearing bacteria could be isolated from the living mice.

Answer: A 7) In 1944, Oswald Avery, Colin MacLeod, and Maclyn McCarty used an in vitro agglutination assay to demonstrate that purified DNA was sufficient to cause transformation of cells, and that the transforming factor could be destroyed by A. ribonucleases but not by deoxyribonuclease or protease enzymes.

B. proteases but not by deoxyribonuclease or ribonuclease enzymes.

C. deoxyribonucleases but not by protease or ribonuclease enzymes.

D. both deoxyribonucleases and proteases but not by ribonuclease enzymes.

Answer: C

A → B → C → D enzyme a

enzyme b

enzyme c

8) According to Beadle and Tatum’s one-gene, one-enzyme hypothesis, how many different genes are apparently operating in the pathway shown above?

A. It cannot be determined from the pathway.

B. 1 C. 2 D. 3 Answer: D

9) Beadle and Tatum’s “one gene – one enzyme” hypothesis was later revised to the A. one gene – one polypeptide hypothesis.

B. one gene – one protein hypothesis.

C. one gene – one DNA molecule hypothesis.

D. one gene – one enzyme, one lipid, one RNA molecule, or one carbohydrate hypothesis.

Answer: A 10) In the 1950s, Hershey and Chase conducted a now famous experiment to determine whether DNA or protein carried the hereditary information in bacteriophage T2. What method did they used to selectively label the DNA and protein components of bacteriophage T2?

A. 35S to label the DNA and 32P to label the protein B. 35S to label the protein and 32P to label the DNA C. 35P to label the protein and 32S to label the DNA D. none of the above Answer: B 11) The classic experiment performed by Alfred Hershey and Martha Chase revealed: A. RNA was not the hereditary material.

B. Microbes could exchange genetic information.

C. In a bacteriophage infection, protein was transferred into the infected bacterial cell.

D. In a bacteriophage infection, DNA was transferred into the infected bacterial cell.

Answer: D 12) In 1953, James Watson and Francis Crick proposed A. the tetranucleotide structure of DNA. B. the one gene – one enzyme hypothesis.

C. the double helix as a model for the structure of DNA. D. that Griffith’s transforming principle was DNA. Answer: C 13) The primary principle of the “neutral” theory of natural selection is that A. evolution is primarily driven by purifying selection against deleterious mutations.

B. advantageous and deleterious mutations play an equal role in evolution.

C. genetic drift by random allele fixation plays a dominant role in evolution.

D. it is impossible to deduce whether observed allele frequencies are caused by positive or negative selection.

Answer: C 14) Neo-Darwinism did not recognize a significant role of _________________ variation in evolution.

A. advantageous B. deleterious C. adaptive D. neutral Answer: D

15) Which of the following theories suggests that mutations create a pool of genetic diversity that may undergo selection at a later time when conditions change?

A. Darwinism B. Neo-Darwinism C. Neutral Theory D. Nearly Neutral Theory Answer: D 16) Darwin’s original theory of evolution recognized ________________ variations.

A. advantageous, deleterious, and neutral B. advantageous, deleterious, and dominant C. dominant, recessive, and advantageous D. neutral, dominant, and recessive Answer: A

Short answer/analytical 17) Define the terms “in vivo” and “in vitro.”

Answer: Section 1.2, p. 6 18) The discovery by Avery, MacLeod, and McCarty that Griffith’s transforming principle was DNA came as a surprise to scientists at the time. Provide an explanation for why their finding was unexpected.

Answer: Section 1.2, p. p.8 19) Explain how technological advances allowed Alfred Hershey and Martha Chase to carry out their classic experiment in 1952.

Answer: Section 1.2, p. 11 20) Discuss scientific findings at the time that provided the necessary context for the proposed structure of DNA by Watson and Crick in 1953.

Answer: Section 1.2, p. 11-12 21) Sodium hydroxide denatures both proteins and nucleic acids, while phenol denatures proteins but not nucleic acids. In the transformation experiments performed by Griffith with Streptococcus pnemoniae, what result would be expected if an extract of S-strain bacteria was treated with phenol? What would be expected if it was treated with sodium hydroxide?

Answer: If an extract of S-strain bacteria was treated with phenol and then injected into mice along with live R-strain cells, one would expect the mice to die. Phenol would denature the proteins but not the nucleic acids in the bacterial extract. The S-strain bacterial DNA would be able to transform the R-strain bacteria. If the extract of S-strain bacteria was treated with sodium hydroxide and then injected into mice along with live R-strain cells, one would expect the mice to remain healthy. The extract would not promote transformation, since the DNA would be denatured.

22) When Hershey and Chase infected bacterial cells that had been grown in the presence of radioactive phosphorus with bacteriophage T2, both the phage RNA and DNA must also have incorporated the labeled phosphorus and yet the experimental results were not affected. Why not?

Answer: Even if both the phage RNA and DNA incorporated the labeled phosphorus, the experimental results would not be affected because only the phage DNA is packaged in the progeny particles that were isolated from the lysed bacteria. Either way, the experiment still demonstrated that the protein coat did not enter the host bacterium and therefore could not be providing the genetic information.

23) Four independent methionine auxotrophs of Neurospora were studied to determine which related compounds might substitute for their methionine requirement. In the table below “+” indicates growth on minimal medium supplemented with the indicated compound, and “-” indicates no growth.

Mutant strain Wild-type Mutant A Mutant B Mutant C Mutant D

Nothing added: O-acetyl (minimal medium) homoserine + + + -

Methionine Homocysteine Cystathione + + + + +

+ + + + -

+ + + -

Based on the results, give the order of these four compounds in the biosynthetic pathway of methionine.

Answer: The order of the four compounds in the biosynthetic pathway of methionine would be: O-acetyl homoserine cystathione homocysteine methionine.

Mutant B can grow on medium supplemented with both cystathione and homocysteine, but not when supplemented with O-acetyl homoserine alone, indicating that the block in the pathway must occur after homoserine, but before cystathione and homocysteine. Mutant C can grow on medium supplemented with homocysteine, but not with cystathione, indicating that the block in the pathway occurs after cystathione but before homocysteine. From this, the order of these four compounds can be deduced.

24) Explain how Tomoko Ohta’s “nearly neutral” theory of molecular evolution reconciled the differences between neutral evolutionary theory and selection evolutionary theory.

Answer: Section 1.5, p. 14-15

Fundamental Molecular Biology, 2nd Edition Instructor Test Bank Lizabeth A. Allison Chapter 2 The Structure of DNA Multiple Choice 1) Nucleic acids are polymers made up of which of the following monomers?

A. nucleotides B. sugars C. amino acids D. nitrogenous bases Answer: A 2) Which figure (I or II) shows the correct structure for a nucleic acid chain?

A. I B. II Answer: A 3) The phosphodiester linkage between adjacent nucleotides to form a DNA or RNA chain occurs by a _______________ reaction involving the removal of _____________ and pyrophosphate.

A. hydrolysis, H2O B. hydrolysis, CO2 C. condensation, H2O D. condensation, CO2 Answer: C 4) What is the difference between a ribonucleotide and a deoxyribonucleotide?

A. ribonucleotides contain a phosphate group.

B. ribonucleotides have a hydroxyl group on the 2′ carbon of their sugar subunit.

C. ribonucleotides contain a sugar with five carbon atoms.

D. ribonucleotides have a hydrogen atom on the 1′ carbon of their sugar subunit.

Answer: B

5) Which of the following nitrogenous bases is found in RNA, but not DNA? A. thymine B. adenine C. uracil D. guanine Answer: C 6) Which of the following nitrogenous bases are purines?

A. cytosine and uracil B. cytosine and thymine C. cytosine, uracil, and thymine D. adenine and guanine Answer: D 7) A nucleotide is made up of which of the following subunits?

A. a 5-carbon sugar, a phosphate group, and a nitrogenous base B. a 5-carbon sugar, a sulfate group, and a nitrogenous base C. a 5-carbon sugar and a nitrogenous base D. a 5-carbon sugar, an amino group, and an “R-group” Answer: A 8) What forms the “backbone” of a nucleic acid?

A. a chain of sugar and phosphate groups, linked through phosphodiester bonds B. purine and pyrimidine pairs, hydrogen-bonded to each other C. a chain of amino and carboxyl groups, linked via peptide bonds D. a double helix of antiparallel strands Answer: A 9) By convention, the sequence of bases in a nucleic acid is always written in which direction?

A. amino to carboxyl B. carboxyl to amino C. 3′ to 5′ D. 5′ to 3′ Answer: D 10) Edwin Chargaff found that the base composition of DNA, defined as the “percent G+C,” A. differs among species B. is constant in all cells of an organism within a species C. is the same among species D. both A and B Answer: D 11) Which of the following did Watson and Crick know when they were trying to determine the structure of DNA? A. The number of purines is always larger than the number of pyrimidines.

B. The number of purines is always the same as the number of pyrimidines.

C. The number of cytosines is always the same as the number of adenines.

D. The number of guanines is always the same as the number of thymines.

Answer: B

12) The abbreviation dNTP stands for A. nucleoside 5′-triphosphate.

B. nucleoside 3′-triphosphate.

C. deoxynucleoside 5′-triphosphate.

D. deoxynucleoside 3′-triphosphate.

Answer: C 13) In practice, the unit of length used for DNA is the kilobase pair, which corresponds to A. 10 base pairs B. 100 base pairs C. 1000 base pairs D. 1,000,000 base pairs Answer: C 14) In the two common Watson-Crick base pairs of DNA A. Adenine (A) is joined to thymine (T) by three hydrogen bonds, while guanine (G) is joined to cytosine (C) by two hydrogen bonds.

B. Adenine (A) is joined to thymine (T) by two hydrogen bonds, while guanine (G) is joined to cytosine (C) by three hydrogen bonds.

C. Adenine (A) is joined to cytosine (C) by three hydrogen bonds, while guanine (G) is joined to thymine (T) by two hydrogen bonds.

D. Adenine (A) is joined to cytosine (C) by two hydrogen bonds, while guanine (G) is joined to thymine (T) by three hydrogen bonds.

Answer: B 15) One DNA molecule varies from another by the A. composition of the sugar-phosphate backbone B. spacing of the major and minor grooves C. way in which the bases are attached to the sugars D. sequence of nitrogenous bases Answer: D 16) The __________________ carries a message (the base sequence of DNA) in a form that can be read by DNA-binding proteins; the bases form __________________ with the side chains of amino acids of proteins.

A. major groove, hydrogen bonds B. minor groove, hydrogen bonds C. major groove, peptide bonds D. minor groove, peptide bonds Answer: A 17) The predominant form of DNA in vivo is A. A-DNA B. Z-DNA C. B-DNA D. C-DNA Answer: C

18) Which statement about B-DNA is not true?

A. B-DNA occurs under conditions of relatively low humidity and relatively low salt.

B. B-DNA forms a right-handed helix.

C. B-DNA has 10.5 bases per turn of the helix.

D. B-DNA has a wide major groove of moderate depth.

Answer: A 19) Which statement is not true about Z-DNA? A. Z-DNA forms a left-handed helix B. Z-DNA is not present in cells C. Z-DNA may play a role in regulating gene expression D. At a B-Z junction, one base pair is extruded from the DNA helix Answer: B 20) Watson and Crick noted that DNA’s structure is interesting because it suggested a possible copying mechanism. What about DNA’s structure facilitates copying?

A. It has the same number of purines and pyrimidines.

B. The nitrogenous bases are located on the inside of the double helix.

C. The strands of the double helix are complementary.

D. DNA always goes from 5’ to 3’.

Answer: C 21) As DNA denatures its absorption of UV light increases, a phenomenon known as: A. melting B. reannealing C. thermal agitation D. hyperchromicity Answer: D 22) The melting temperature (Tm) of DNA is: A. the temperature at which half the bases in a double-stranded DNA sample have denatured.

B. the temperature at which half the bases in a single-stranded DNA sample have renatured.

C. the temperature at which half the phosphodiester bonds in a double-stranded DNA sample have broken.

D. the temperature at which all the bases in a double-stranded DNA sample have denatured.

Answer: A 23) When two strands of DNA from different sources are hybridized in the lab, what provides the chemical stability for holding the two strands of DNA in a double helix structure?

A. hydrophobic interactions (base-stacking) interactions and hydrogen bonds B. covalent bonds C. phosphodiester bonds D. hydrogen bonds only Answer: A 24) Lowering the salt concentration of a DNA solution A. promotes renaturation.

B. promotes denaturation.

C. promotes formation of hydrogen bonds.

D. removes cations that shield the positive charges on the two strands from each other.

Answer: B 25) You determine that a sequence of DNA contains numerous tandem repeats. What unusual DNA secondary structure would this sequence be most likely to form?

A. cruciform B. triple helix C. slipped structure D. tandem helix Answer: C 26) Friedreich’s ataxia is a rare inherited neurological disease caused by a A. trinucleotide repeat expansion in the first intron of the Friedrich’s ataxia gene B. formation of triple helix DNA in the Friedrich’s ataxia gene C. cruciform structures in the frataxin protein D. both A and B Answer: D 27) A Hoogsteen base pair differs from a Watson-Crick base pair by A. an altered pattern of hydrogen bonding in the Hoogsteen AT pair B. the Hoogsteen GC pair only forms two hydrogen bonds C. Hoogsteen GC pairs are not stable at neutral pH D. all of the above Answer: D 28) Consider a linear DNA molecule of 20 complete turns (or twists, T=20) with 10.5 base pairs (bp) per turn in solution. If the double helix is underwound by one full turn to the left and then the ends are sealed together, the result is a strained circle with 11.67 bp per turn, where L (linking number) =19 and T=19. If one negative supercoil is spontaneously introduced, the DNA circle will have the following characteristics: A. L=19, T=20, 11.6 bp/turn B. L=19, T=20, 10.5 bp/turn C. L=20, T=20, 10.5 bp/turn D. L=20, T=20, 11.6 bp/turn Answer: B 29) Topoisomerases are enzymes that A. relax supercoiled DNA B. denature double-stranded DNA C. synthesize a new strand of DNA from a single-stranded DNA template D. synthesize a strand of DNA from an RNA template Answer: A

30) Which statement is not true about DNA supercoiling?

A. Negative supercoiling puts energy into DNA. B. In bacteriophages, negative supercoiling is associated with decreased activity in replication and transcription.

C. Positive supercoiling occurs ahead of replication forks and transcription complexes.

D. Most DNA within both prokaryotic and eukaryotic cells exists in the negative supercoiled state.

Answer: B 31) As genome complexity increases, the time required for reannealing A. doesn’t change B. increases C. decreases D. either B or C depending on the genome Answer: B 32) Supercoiling is a form of DNA A. primary structure B. secondary structure C. tertiary structure D. quaternary structure Answer: C 33) Most sequence-specific DNA binding proteins interact with A. the minor groove of Z-DNA B. the major groove of B-DNA C. the backbone of A-DNA D. locally unwound DNA Answer: B 34) DNA loop domains A. are stabilized by proteins at loop junctions B. join B-DNA and Z-DNA segments C. are also known as DNA cruciforms D. form spontaneously Answer: A 35) The polarity of doubled-stranded DNA arises from A. negative charges on the backbone phosphates B. the presence of 5’ phosphates on one end and 3’ hydroxyls on the other C. base pairing D. the sequence of bases, which is strand-specific Answer: B 36) DNA-mediated heredity derives from which part of its nucleotide building blocks?

A. the sugars B. the phosphates C. the nitrogenous bases D. none of the above Answer: C

Short answer/analytical 37) Distinguish between the terms “nucleoside” and “nucleotide.”

Answer: Section 2.2, p. 18 and 20 38) Draw the general structure of a deoxynucleoside monophosphate. Show the sugar structure in detail and indicate the positions of attachment of the base and the phosphate. Also indicate the deoxy position.

Answer: Section 2.2, Fig. 2.3, p. 20 39) Starting with the dNTP shown below, draw a clearly labeled diagram showing where the linking bond is between two nucleotides in the same DNA chain. Name the bond. Label the 5′ and 3′ ends of the DNA chain.

P Base

Answer: Section 2.2, Fig. 2.3, p. 20 40) You have isolated a DNA molecule that is 12,200 base pairs in length. How many kilobase pairs is the DNA molecule?

Answer: Section 2.2, p. 21 41) Briefly describe the two main types of interactions that provide chemical stability to the DNA double helix.

Answer: Section 2.3, p. 22 42) Which DNA purine forms three H bonds with its partner in the other DNA strand? Which forms two H bonds?

Answer: Section 2.3, Fig. 2.4, p. 122; Section 2.2, p. 19 43) Which DNA pyrimidine forms three H bonds with its partner in the other DNA strand? Which forms two H bonds?

Answer: Section 2.3, Figure 2.4, p. 22; Section 2.2, p. 19 44) Explain what is meant by the following descriptions of the DNA double helix: “polarity in each strand,” and “the two strands are antiparallel.”

Answer: Section 2.2, p. 20-21; Section 2.3, p. 25 45) Which form of DNA is predominant in cells – A, B, or Z DNA? Do other forms ever occur under special circumstances? Explain your answers.

Answer: Section 2.3, p. 27-28

46) Use a drawing to illustrate the principles of denaturation, renaturation, and nucleic acid hybridization.

Answer: Section 2.3, Fig. 2.11, p. 29 47) Draw a typical DNA denaturation curve. Label the axes and point out the melting temperature (Tm). Indicate at what points all of the DNA is double-stranded and all of the DNA is single-stranded.

Answer: Section 2.3, Fig. 2.12, p. 30 48) Use a graph to illustrate the relationship between the G + C content of a DNA and its melting temperature. What is the explanation for this relationship?

Answer: Section 2.3, Figure 2.13, p. 29-30 49) Use a graph to illustrate the effect of lowering the salt concentration of a solution on the melting temperature (Tm) of a DNA sample. What is the explanation for this relationship?

Answer: Section 2.3, Figure 2.13, p. 29-30 50) Explain to a patient with Friedreich’s ataxia the underlying cause of his disease.

Answer: Section 2.4, p. 34, Disease Box 2.1 51) Discuss the significance of DNA supercoiling in vivo.

Answer: Section 2.5, p. 35-36 52) The DNA duplexes below are denatured and then allowed to reanneal. Which of the two molecules would have the highest Tm? Which of the two is least likely to re-form the original structure? Why?

(a) 5′-ATATCATATGATATGTA-3′ 3′-TATAGTATACTATACAT-5′ (b) 5′-CGGTACTCGTGCAGGT-3′ 3′-GCCATGAGCACGTCCA-5′ Answer Section 2.3, p. 29; Section 2.4, p. 31-32: The DNA molecule shown in (b) would have the highest melting temperature (Tm) because the sequence is GC-rich. The GC base pair has three hydrogen bonds to every two in an AT base pair. In addition, the base stacking (hydrophobic) interactions of GC base pairs with neighboring base pairs are more favorable energetically than interactions of AT base pairs with their adjacent base pairs. Thus, the higher the GC content in a given molecule of DNA, the higher the temperature required to denature the DNA. The DNA molecule shown in (a) would be least likely to reform the original structure because it is composed of tandem repeats of “ATAT.” These tandem repeats have the potential to form slipped structures, with compensating single-stranded loops in alternate strands.

53) When the base composition of DNA from a grasshopper was determined, 29% of the bases were found to be adenine. (a) What is the percentage of cytosine? (b) What is the % composition of each of the four nucleotides in the DNA sequence? (c) What is the [G] + [C] content?

Answer: (a) 21% cytosine (b) 29% adenine, 29% thymine, 21% guanosine, 21% cytosine (c) [G] + [C] content = 42% Answer: Section 2.2, p. 19. The answers were calculated following Chargaff’s rules: [A] = [T] and [G] = [C], and [A] + [G] = [T] + [C].

Since the percentage of adenine is 29% and [A] = [T], the percentage of thymine is also 29%, It follows then, that the [A] + [T] content is 29% + 29% = 58%. Accordingly, the [G + C] content must be 100%-58% = 42%. Thus, [C] = 21% and [G] = 21%.

54) Explain why positively supercoiling of genomic DNA might be advantageous for themophilic organisms.

Answer: Section 2.5, p. 36. Positive supercoiling makes it more difficult to unwind DNA, because doing do creates further positive supercoiling strain. For an organism that lives at high temperatures, positive supercoils would oppose the effect of heat, which would normally promote DNA melting (i.e., unwinding).

55) Explain how “hydrophobic bonding” stabilizes double-stranded DNA in aqueous solution.

Answer: Section 2.3, pp. 22-23. Nitrogenous bases contain nonpolar rings that are hydrophobic. Water tends to exclude these portions of dsDNA, which stack over one other by helical twist, to form a “hydrophobic core” analogous to the hydrophobic cores of proteins or the interior of a lipid bilayer. The hydrophilic sugar-phosphate backbone, on the other hand, freely interacts with water.

Fundamental Molecular Biology, 2nd Edition Instructor Test Bank Lizabeth A. Allison Chapter 3 The Versatility of RNA

Multiple Choice

1) What is the difference between a ribonucleotide and a deoxyribonucleotide?

A. ribonucleotides contain a phosphate group.

B. ribonucleotides have a hydroxyl group on the 2′ carbon of their sugar subunit.

C. ribonucleotides contain a sugar with five carbon atoms.

D. ribonucleotides have a hydrogen atom on the 1′ carbon of their sugar subunit.

Answer: B 2) Which of the following nitrogenous bases is found in RNA, but not DNA? A. thymine B. adenine C. uracil D. guanine Answer: C 3) Which of the following is not a difference between RNA and DNA? A. One is typically single stranded and the other is typically double stranded.

B. One contains uracil and the other does not.

C. One contains ribose sugar and the other contains deoxyribose sugar.

D. One is made of nucleotide monomers and the other one is not.

Answer: D 4) Hairpin loops, base-paired stems, and bulges make up what part of RNA’s structure?

A. primary B. secondary C. tertiary D. quaternary Answer: B 5) Base-paired stems in RNA A. form a B-type double helix B. form a Z-type double helix C. form an A-type double helix D. do not form a double helix Answer: C 6) A GU wobble is an example of A. a noncanonical base pair B. a base triple C. a Watson-Crick base pair D. a modified base Answer: A

7) Which of the following statements about tRNA is not true?

A. In general, tRNAs contain more than 50 modified bases, such as pseudouridine.

B. tRNA loops each have a separate function.

C. Base-paired stems are often involved in long-range interactions with other stems by coaxial stacking.

D. The three-dimensional structure of tRNA is referred to as a “cloverleaf.”

Answer: D 8) Pseudoknots and tetraloops make up what part of RNA’s structure?

A. primary B. secondary C. tertiary D. quaternary Answer: C 9) Which statement about a pseudoknot motif is not true?

A. Pseudoknot motifs have only been characterized in plant RNA viruses.

B. A highly conserved pseudoknot is required for telomerase activity.

C. Coaxial stacking of base-paired stems forms a quasicontinuous double helix.

D. A single-stranded loop base pairs with a complementary sequence outside this loop.

Answer: A 10) An A-minor motif is characterized by which of the following folding patterns?

A. Hydrogen bonding occurs between the 2′-OH of a ribose in one helix and the 2-oxygen of a pyrimidine base of the other helix between their respective minor groove surfaces.

B. Single-stranded adenosines make tertiary contacts with the minor grooves of RNA double helices by hydrogen bonding and van der Waals contact.

C. An asymmetric internal loop embedded in RNA double helices has a sharp bend in the phosphodiester backbone of the three-nucleotide bulge associated with this structure.

D. Base-stacking interactions promote and stabilize the stem-loop structure.

Answer: B 11) A particular RNA chain A. always folds into the same secondary structure.

B. can fold into many different secondary structures and still be functional.

C. can misfold into different secondary structures by incorrect base-pairing, but will still always form the same tertiary structure.

D. can misfold and become trapped in a misfolded structure in the absence of RNA chaperones.

Answer: D 12) There is a tremendous versatility of functional RNA products involved in a wide range of cellular processes. One of the key contributing factors to this versatility is A. the ability of RNA to form unique 3D structures that act similarly to proteins B. the ability of RNA to cleave phosphodiester bonds in other nucleic acids C. the ability of RNA to be copied into DNA D. the ability of RNA to serve as a messenger during protein synthesis Answer: A

13) The term ribozyme describes what property of RNA molecules?

A. they are single-stranded B. they can be catalytic C. their shape D. the formation of hairpin loops Answer: B 14) You are characterizing the properties of a self-splicing ribozyme and obtain the following results: in the presence of Mg2+, self-splicing occurs. However, in the absence of Mg2+, selfsplicing does not occur. Which of the following hypotheses does this observation support?

A. Mg2+ is a cofactor for the ribozyme.

B. Mg2+ is a substrate for the ribozyme.

C. Mg2+ helps maintain the tertiary or quaternary structure of the protein.

D. Mg2+ is not required for the activity of the ribozyme Answer: A 15) Which statement is true?

A. All proteins are enzymes.

B. All proteins are enzymes, and some RNA molecules also are catalytic.

C. All enzymes are proteins.

D. Most enzymes are proteins, but some RNA molecules also are catalytic.

Answer: D 16) Many ribozymes A. are autocatalytic B. are metalloenzymes C. cleave phosphodiester bonds D. all of the above Answer: D 17) Why do many researchers consider RNA to be the best candidate for the first life-form?

A. It is simple in structure.

B. It is capable of self-replication and catalysis.

C. It carries more information than any other molecule.

D. All of its nucleotide components have been created under laboratory conditions that mimic early Earth.

Answer: B 18) Which of the following is not an example of a tertiary RNA structure?

A. tetraloop B. kink-turn C. base-paired stem D. ribose zipper Answer: C 19) Which of the following is not a critical cellular role for non-coding RNAs? A. mRNA splicing B. Ribosome function C. DNA replication D. Regulation of gene expression

Answer: C 20) A retrovirus, such as HIV-1, A. uses non-coding RNAs to package its genome into a capsid B. stores its genome as an RNA sequence C. uses host cell ribozymes to process its genome D. converts its DNA genome to an RNA once inside the host cell Answer: B 21) Subviral RNA particles such as viroids and satellite RNAs differ from viruses in that A. subviral RNA particles usually do not encode proteins B. subviral RNA particles can only infect plant cells C. subviral RNA particles usually do not cause disease D. the infectious form of subviral RNA particles are usually encased by proteins Answer: A 22) Which of the following is a similarity between the catalytic mechanisms of self-splicing introns and DNA polymerase?

A. both use two metal ions in their active sites B. both bind to their substrates by base-pairing C. both use enzyme-RNA covalent intermediates D. both cleave phosphodiester bonds Answer: A

Short answer/analytical

23) Draw the general structure of a nucleoside monophosphate. Show the sugar structure in detail and indicate the positions of attachment of the base and the phosphate. Also indicate the location of the 2′-OH group.

Answer: Section 3.3, p. 42, Fig 3.2 24) Summarize the major types of secondary structure found in RNA. Include a labeled diagram.

Answer: Section 3.3, p. 42-43, Fig. 3.3 25) Give an example of a common noncanonical base pair in RNA and discuss the functional significance of this type of base pair.

Answer: Section 3.3, p. 42-43, Fig 3.3 26) Give an example of a common base modification in RNA and discuss the functional significance of base modifications.

Answer: Section 3.3, p. 45 and Fig. 3.6 27) The study of tRNA structure has provided many important insights into RNA structural motifs.

Discuss this statement.

Answer: Section 3.3, p. 44-46 28) Discuss why Francis Crick stated that “tRNA looks like Nature’s attempt to make RNA do the job of a protein.”

Answer: Section 3.3, p. 44-46 29) Give examples of the structural and functional importance of the 2′-OH group on the ribose sugar of RNA. Answer: Section 3.3, p. 49 (Ribose zipper motifs); Section 3.3, p. 47 (involvement of 2’-OH in tertiary structures in general); Section 3.3, p. 42 (implications of 2’-OH and secondary structure/A form helix) 30) You are characterizing a novel ribozyme. Describe an experiment to analyze the folding kinetics of the ribozyme.

Answer: Section 3.4, pp. 50-51, Fig 3.15 31) For many years the pathway of gene expression from DNA to functional product via an RNA intermediate emphasized proteins as the ultimate goal. Discuss how this view has changed.

Answer: Section 3.2, p. 40-42, Table 3.1 32) You are investigating a ribosomal RNA (rRNA) gene with one large intron and two short exons. Show the results of R looping experiments performed with: a. final rRNA product and single-stranded DNA b. final rRNA and double-stranded DNA c. rRNA precursor and single-stranded DNA d. rRNA precursor and double-stranded DNA Answer: Section 3.4, p. 52, Fig 3.17 33) Show where the 2′-OH group is located on the NTP below. What role does the 2′-OH play in a hammerhead ribozyme?

P Base

Answer: Section 3.4, p. 57 and Fig 3.2 34) You have discovered an RNA that you think is a self-splicing ribozyme. You also have evidence that it requires a pseudoknot for catalytic activity. Briefly outline an experiment you could perform to test these two hypotheses.

Answer: Section 3.3, p. 47-48; Section 3.3 p. 50-51, Fig. 3.15 35) Compare and contrast the mode of action of “ribozymes” and “enzymes.”

Answer: Section 3.4 p. 54-55 36) Make up an RNA sequence that will form a hairpin with a 9 bp stem and a 7 bp loop. Draw both the primary structure and the secondary structure.

Answer: Section 3.3, p. 42, Figure 3.4. One of the many possible sequences that could be drawn is shown below:

Primary structure: 5′-ACGUGCUCGAUCGACAGAACCUCGAUCGAGGCGCAA – 3′ Secondary structure: G A A A C C C

ACGUG

A-U G-C C-G U-A A-U G-C C-G U-A C-G

GCGCAA

37) What addition(s) would you need to make to the primary sequence in question 36 to allow pseudoknot formation?

Answer Section 3.3, p. 47-48 and Figure 3.8. A pseudoknot motif forms when a singlestranded loop base pairs with a complementary sequence outside this loop. The sequence in (1) would need to be modified to allow complementary base pairing between the loop and the sequence 3′ of the loop. One possible example is shown below: 38) You suspect that a tetraloop is critical for folding of a ribozyme into its active form. Describe an experiment to demonstrate whether the RNA folds into a similar tertiary structure when the tetraloop is deleted.

Answer Section 3.3 p. 49; Section 3.4 p. 50-51, Figure 3.15. Carry out a hydroxyl radical footprinting experiment, comparing the cleavage pattern of the RNA containing the tetraloop motif with the cleavage pattern of a mutant RNA lacking the tetraloop motif. Tertiary contacts within a folded RNA molecule result in local reductions in solvent accessibility. The hydroxyl radicals cannot react with the protected backbone sugar and hence there is a reduced cleavage of protected nucleotides, which can be visualized by electrophoresis. If the tetraloop motif is required for folding into the active form, then the mutant RNA would be predicted to be more accessible to solvent and to show enhanced cleavage.

39) You have discovered a small RNA involved in removal of a novel class of introns. Design an experiment to determine whether the small RNA functions as a catalytic RNA or RNP. Show sample positive results.

Answer: Section 3.4, p. 55 and Fig 3.18. Prepare a labeled intron-containing precursor RNA (pre-RNA) by in vitro transcription from a DNA template. Incubate the pre-RNA with the small RNA either in the presence or absence of its associated proteins. Separate the samples by polyacrylamide gel electrophoresis and visualize the results by autoradiography. If the small RNA functions as a catalytic RNA, then the RNA alone will remove the intron from the pre-RNA. If the small RNA functions as a catalytic RNP, then it will remove the intron from the pre-RNA

only when its associated proteins are present in the reaction. The proteins alone would not mediate intron removal in either case. Sample positive results are shown for the first scenario: 40) Explain how the discovery of ribozymes spawned the RNA World hypothesis.

Answer: Section 3.4, p.51-52. When it was discovered that RNAs can catalyze reactions, it was suddenly clear that RNAs could in principle be self-sufficient molecules, encoding themselves genetically by their base sequence, and catalyzing their own replication. Thus, biologists came to the idea that before the DNA-RNA-protein-based cells that exist today, there may have been simple RNA-only based cells containing self-replicating RNAs. 41) Compare and contrast small ribozymes and large ribozymes.

Answer: See Table 3.2.

Similarities: Both are RNA catalysts, and most in both categories cleave phosphodiester bonds.

Differences: Small ribozymes generate a free 5’OH and 2’,3’ cyclic phosphate, whereas large ribozymes generate a wider variety of products, which usually include a free 3’OH. Most large ribozymes are expressed widely in eukaryotes and prokaryotes, whereas most small ribozymes are found only in single cellular organisms.

42) The most effective therapies known for the treatment of HIV infection are inhibitors of RNAdependent DNA polymerases (also known as reverse transcriptases). Explain how this class of inhibitors would affect the virus’s activities.

Answer: Section 3.5, p. 59. Reverse transcriptase inhibitors would prevent the conversion of the viral RNA genome to DNA, which would in turn prevent integration of the viral genome into the host genome. (Reverse transcriptase inhibitors would not inhibit the entry of HIV into cells.)

43) How do eukaryotic RNA viruses differ from retroviruses?

Answer: Section 3.5, p. 59-62. Eukaryotic RNA viruses replicate their RNA without a DNA intermediate, whereas retroviruses copy their genomes into DNA, which is then copied again into RNA for the formation of new virus particles.

Fundamental Molecular Biology, 2nd Edition Instructor Test Bank Lizabeth A. Allison Chapter 4 Protein Structure and Folding

Multiple Choice

1) The following is a short segment of an mRNA molecule. The polypeptide it codes for is also shown: 5′ AUGGUGCUGAAG 3′: methionine-valine-leucine-lysine A mutation in the DNA occurs so that the fourth base (counting from the 5′ end) in the messenger RNA now reads A rather than G. What sequence of peptides will the mRNA now code for? (You do not need a copy of the genetic code to answer the question.)

A. methionine-lysine-leucine-lysine B. methionine-leucine-leucine-lysine C. methionine-valine-methionine-lysine D. methionine-methionine-leucine-lysine Answer: D 2) The genetic code is said to be “degenerate” because A. More than one type of tRNA can be charged with a particular amino acid.

B. A single tRNA specific for a particular amino acid may respond to multiple codons in an mRNA C. A single tRNA may couple a single mRNA codon to multiple amino acids.

D. A single codon in an mRNA can couple to the anticodons of many different types of tRNAs. Answer: B 3) Which statement about the genetic code is NOT correct?

A. Non-Watson-Crick base pairing can occur at the third position of codons and anticodons.

B. The sense strand of DNA serves as the template for RNA synthesis.

C. A particular amino acid may be coded for by more than one codon.

D. A DNA sequence is read in triplets.

Answer: B 4) _____________ different genetically-encoded amino acids are found in the proteins of cells; they are distinguished by___________________ .

A. 22; the location of their carboxyl group B. 20; the location of their amino group C. 20; the angles between the central carbon and their side chains, or R-groups D. 22; the composition of their side chains, or R-groups Answer: D

5) Which of the following are rare amino acids that are sometimes encoded by stop codons?

A. isoleucine and pyrroglutamine B. selenocysteine and pyrrolysine C. pseudouridine and D-alanine D. abrine and homocysteine Answer: B 6) The modified nucleotide inosine (I) found in tRNA can pair with A. the base A only B. any of the three bases, U, C, and A C. either U or A D. any of the four bases, U, C, G, and A Answer: B 7) The modified base 2-thiouracil found in tRNA can pair with A. the base A only B. any of the three bases, U, C, and A C. either A or G D. any of the four bases, U, C, G, and A Answer: A 8) The “wobble hypothesis” states that A. alternative (non-Watson-Crick) base-pairing is permitted between a codon and anticodon at the third position.

B. a peptide backbone can bend between the central carbon and the amino group’s N or the carboxyl group’s C. C. an allosteric regulator can induce a protein to “wobble” between two conformations.

D. amino acids can spontaneously isomerize between the D- and L- forms.

Answer: A 9) The frequencies with which different codons are used vary significantly between different organisms and between proteins expressed at high or low levels within the same organism.

This is referred to as: A. stratification B. codon variegation C. heterologous codons D. codon bias Answer: D 10) At the pH found in cells (about 7.0), what happens to the amino group on an amino acid?

A. It acts as a base and gains a proton, giving it a positive charge.

B. It acts as a base and gains a proton, giving it a negative charge.

C. It acts as an acid and loses a proton, giving it a negative charge.

D. It neither gains nor loses a proton and does not have a charge.

Answer: A 11) At the pH found in cells (about 7.0), what happens to the carboxyl group on an amino acid?

A. It acts as a base and gains a proton, giving it a positive charge.

B. It acts as an acid and loses a proton, giving it a positive charge.

C. It acts as an acid and loses a proton, giving it a negative charge.

D. It neither gains nor loses a proton and does not have a charge.

Answer: C 12) Suppose you have discovered a new amino acid. Its R-group contains only carbon and hydrogen atoms. Predict the behavior of this amino acid.

A. It is hydrophilic.

B. It is hydrophobic.

C. It is polar.

D. Its R-group is positively charged at physiological pH. Answer: B 13) Amino acids are joined by a condensation reaction to form a polypeptide chain with a ___________ backbone and ___________ side chain.

A. variable, common B. variable, variable C. common, variable D. common, common Answer: C 14) The peptide bond acts as a partial double bond as a result of A. the acid-base properties of amino acids B. wobble base pairing C. the side chain or “R group” D. resonance Answer: D 15) The primary structure of a protein is A. the sequence of a chain of amino acids joined together by phosphodiester bonds.

B. the sequence of a chain of amino acids joined together by peptide bonds.

C. nearby amino acids in a chain joined together by hydrogen bonds.

D. the three-dimensional folding of amino acids in a chain.

Answer: B 16) The C-terminus of a protein corresponds to A. the R-group of the first amino acid B. the R-group of the last amino acid C. the amino group of the first amino acid D. the carboxyl group of the last amino acid Answer: D 17) You have just sequenced a new protein found in Tetrahymena and observe that sulfurcontaining cysteine residues occur at regular intervals in the sequence. Which hypothesis is supported by this observation?

A. This protein contains many alpha helices.

B. This protein contains many beta-pleated sheets.

C. This protein’s tertiary structure is stabilized by disulfide bridges.

D. This protein’s tertiary structure is stabilized by hydrogen bonding.

Answer: C

18) Which amino acid shown below, glutamine or valine, would be more likely to be found within the membrane region of a transmembrane protein, and less likely to be found in the intracellular region?

Valine Glutamine

A. Glutamine B. Valine Answer: B 19) You have discovered an enzyme that has three active sites. When you denature the enzyme and study its composition, you find that each active site occurs on a different polypeptide. Which of the following hypotheses does this observation support?

A. The enzyme is subject to allosteric regulation.

B. The enzyme requires a cofactor to function normally.

C. The structure of the enzyme is affected by temperature and pH. D. The enzyme has quaternary structure.

Answer: D 20) D- and L- amino acids A. contain different R groups B. are used in equal proportion in most proteins C. are enantiomers D. contain different numbers of backbone carbon atoms Answer: C

21) A particular polypeptide has a molecular weight of 40 kDa. The average molecular weight of an amino acid is 110. The polypeptide chain thus contains approximately how many amino acids?

A. 3.6 B. 364 C. 4400 D. 4,400,000 Answer: B

22) The Dalton unit, which is used to describe the mass of proteins, is defined as A. one atomic mass unit.

B. the atomic mass of glycine, the simplest amino acid.

C. the average atomic mass of the 20 common amino acids.

D. the precise atomic mass of a given protein.

Answer: A 23) An average molecular weight for a protein is A. 110 Daltons B. 1100 Daltons (1.1 kiloDaltons) C. 50,000 Daltons (50 kiloDaltons) D. 500,000 Daltons (500 kiloDaltons) Answer: C 24) Proline is a “helix-breaking” residue because A. its R-group is too large to be accommodated in the alpha helix interior B. its R-group is too hydrophobic to interact with water surrounding the alpha helix C. its backbone nitrogen can only form hydrogen bonds in the context of a pleated beta sheet D. cyclization prevents its backbone nitrogen from participating in hydrogen bonding that stabilizes the helix Answer: D 25) An alpha helix is stabilized by A. van der Waals interactions between R-group atoms in the helix interior B. ionic bonding between R-group atoms of residues located near one another C. hydrogen bonding between backbone atoms of residues located near one another D. peptide bonding between backbone atoms of non-adjacent residues Answer: C 26) A beta-pleated sheet is stabilized by A. hydrogen bonding between backbone atoms of residues located in adjacent beta strands B. van der Waals interactions between R-group atoms that extend perpendicularly from the plane of the sheet C. ionic bonding between R-group atoms of residues located near one another D. peptide bonding between backbone atoms of adjacent beta strands Answer: A 27) Short unstructured regions of a protein that connect secondary structural elements are called A. beta hooks B. turns C. barrels D. branch points Answer: B 28) In a beta-pleated sheet, A. adjacent polypeptide strands interact in an extended conformation B. beta strands stack at perpendicular angles C. R-groups of hydrophobic amino acids stabilize adjacent stretches of amino acid residues D. alpha helices align in parallel or antiparallel arrays Answer: A

29) Which of the following is NOT true about protein structure?

A. A protein’s tertiary structure is ultimately determined by its primary amino acid sequence.

B. An alpha helix is an example of tertiary structure.

C. Tertiary structure may be stabilized by hydrogen bonds, disulfide bonds, ionic bridges, and van der Waals interactions D. Enzymes such as lysozyme have tertiary structure.

Answer: B 30) A common structural element in fibrous proteins is A. transmembrane domain.

B. beta pleated sheet.

C. hydrophobic core.

D. coiled coil.

Answer: D 31) You have isolated a novel protein, characterized its structure and determined that it catalyzes the breakdown of a large substrate and has two binding sites. One of these is large, apparently the binding site for the large substrate; the other is small, possibly a binding site for a regulatory molecule. What do these findings suggest to you about the mode of action of this protein?

A. It is probably a structural protein that is found in tendons or bone.

B. It is probably an enzyme that is controlled through allosteric regulation.

C. It is probably an enzyme that works through autophosphorylation.

D. It is probably a transmembrane protein—like a G protein-coupled receptor.

Answer: B 32) Which of the following illustrates quaternary protein structure?

A. an allosteric ligand B. an actin monomer C. hemoglobin D. a metal-binding protein Answer: C 33) Which level(s) of protein structure could be disrupted by treatment with a reducing agent?

A. primary B. primary and secondary C. tertiary D. tertiary and quaternary Answer: D 34) An enzyme’s active site usually A. is on the outer surface of an enzyme.

B. is hydrophobic.

C. forms a cleft or pocket in the enzyme.

D. undergoes irreversible modification during catalysis.

Answer: C 35) The term induced fit refers to A. the coalescence of nonpolar amino acid residues in a protein’s amino acid sequence into a hydrophobic core.

B. a conformational change in an enzyme upon substrate binding.

C. the process of inducing a protein to fold into its lowest energy tertiary structure.

D. any specific protein-protein interaction.

Answer: B 36) Extrapolating from the example of lysozyme, which part of an enzyme do you suppose directly promotes catalysis?

A. the peptide backbone B. amino acid side chains C. alpha helices D. the hydrophobic core Answer: B 37) Which of the following is not true when comparing an uncatalyzed reaction to the same reaction with a catalyst?

A. The catalyzed reaction will be faster.

B. The catalyzed reaction will have a different G. C. The catalyzed reaction will have lower activation energy.

D. The catalyzed reaction will not consume any of the catalyst.

Answer: B 38) An enzyme that catalyzes phosphorylation of a substrate is called A. a kinase.

B. a transferase.

C. a phosphoprotein.

D. a glycoprotein.

Answer: A 39) Many steps in gene expression and cell signaling involve post-translational modification of proteins by A. reversible disulfide bridge formation.

B. irreversible disulfide bridge formation.

C. reversible phosphorylation.

D. irreversible phosphorylation.

Answer: C 40) Cyclin-dependent kinase activity is regulated by A. phosphorylation.

B. allosteric modification.

C. allolactose.

D. both A and B. Answer: D 41) Allosteric regulation always involves A. a change in protein shape or conformation.

B. a post-translational modification.

C. interaction with Hsp90.

D. the ubiquitin-proteasome system.

Answer: A

42) Which of the following is not true about molecular chaperones?

A. They decrease the efficiency of the overall process of protein folding.

B. Some of them enhance the rate of formation of disulfide bonds.

C. They reduce the probability of protein aggregation.

D. They aid in the destruction of misfolded proteins.

Answer: A 43) You determine that a protein has an attached polyubiquitin chain. It is likely that this protein will be A. targeted to the lysosome for degradation.

B. targeted to the 26S proteosome for degradation.

C. included in insoluble toxic deposits known as amyloids fibers.

D. degraded by E1, E2, and E3 enzymes.

Answer: B 44) Alzheimer’s disease, Huntington’s disease, and Creutzfeld-Jakob disease are all examples of: A. transmissible spongiform encephalopathies.

B. trinucleotide repeat disorders.

C. protein misfolding diseases.

D. diseases that are not related in any way.

Answer: C 45) A prion is an infectious protein A. with an excess of phosphate modifications.

B. associated with viruses.

C. that induces genetic changes (changes in DNA sequence).

D. with an altered conformation.

Answer: D

Short answer/analytical

46) Do UAG, UAA, and UGA always act as termination codons during protein synthesis?

Explain your answer.

Answer: Section 4.2, p. 69-70

47) Explain how a single base change in a gene could lead to premature termination of translation of the mRNA from that gene.

Answer: Section 4.2, p.68-70, Table 4.1 and Fig. 4.5 For example, a change in the first nucleotide of the codon GAG to T converts the codon for glutamic acid to a stop codon.

48) You are attempting to express a 50 kDa Tetrahymena protein from a cloned gene introduced into E. coli. When you purify the overexpressed protein you find that it is only 25 kDa in size. Discuss possible reasons for this result.

Answer: Section 4.2, p. 71

49) Draw the general structure of an amino acid.

Answer: Section 4.2, p. 67, Fig. 4.2 50) Draw a clearly-labeled diagram showing where the linking bond is between two amino acids in a polypeptide chain. Provide the name of the bond.

Answer: Section 4.2, p. 67-68, Fig. 4.3 51) Use a rough diagram to compare the structures of a protein alpha-helix and a beta-pleated sheet. For simplicity, show only the backbone of the polypeptide chain.

Answer: Section 4.3, p. 72-73, Fig. 4.8 52) What is meant by primary, secondary, tertiary, and quaternary structures of proteins?

Answer: Section 4.2-4.3, Fig. 4.8 53) Compare and contrast the overall shape and function of globular, fibrous, and membrane proteins. Give a specific example of a protein with each type of tertiary structure.

Answer: Section 4.3, p.74-76, Figs. 4.10, 4.11, 4.12.

54) You have determined the amino acid sequence of a new protein. Is it possible to determine the three-dimensional structure of the protein from its sequence? Why or why not?

Answer: Section 4.3, p. 76 55) Draw a graph comparing progress of a reaction with or without a catalyst. Label the axes, EA, and G. Answer: Section 4.4, p. 77-78, Fig. 4.14 56) Explain the “induced-fit” model for enzyme action.

Answer: Section 4.4, p. 77-79 57) What does a kinase do?

Answer: Section 4.4, p. 81 58) Describe a specific example of control of enzyme activity by post-translational modification and allosteric regulation.

Answer: Section 4.4, p. 81-83 59) Discuss the roles of molecular chaperones in protein folding and degradation.

Answer: Section 4.5, p.83, Figs. 4.20 and 4.21 60) Describe two possible fates for a misfolded protein in a cell.

Answer: Section 4.5, p. 83-36 Two of the more common outcomes are (1) “rescue” to a properly folded state by molecular chaperones and (2) ubiquitylation-mediated targeting to the proteasome for degradation. A third possibility is aggregation.

61) You treat cells with an inhibitor of the 26S proteasome. Predict the effect on cellular protein levels.

Answer: Section 4.5, p. 85-86 Cellular protein levels would increase because the proteasome carries out a large proportion of total protein turnover in the cell.

62) Use a diagram to outline the key steps in ubiquitin-mediated protein degradation.

Answer: Section 4.5, p. 85, Fig. 4.22 63) The sequence of a portion of a gene is presented below: 5′-AGCAATGCATGCATCGTTATGG-3′ 3′-TCGTTACGTACGTAGCAATACC-5′ (a) Assuming that transcription starts with the first T in the template strand of the DNA, and continues to the end, what would be the sequence of the transcribed mRNA? (b) Identify the initiation codon in this mRNA. (c) Would there be an effect on translation of changing the first C in the template strand to a G? If so, what effect? (d) Would there be an effect on translation of changing the third G in the template strand to a T? If so, what effect? (e) Would there be an effect on translation of changing the last A in the template strand to a C? If so, what effect? (f) Would there be an effect on translation if pyrrolysine were present?

Answer: Section 4.2 p. 69, Fig 4.5; Table 4.1; p.70 (a) 5′-AGCAAUGCAUGCAUCGUUAUGG-3′ (b) AUG (c) No. The first C is in the 5′-untranslated region (5′-UTR) of the mRNA. Changing it to a G would not have a direct effect on the translation; i.e., the amino acid sequence would be the same. (d) Yes. This would change GCA, which codes for alanine, to GAA, which codes for glutamine. (e) Yes. This would change UGG, which codes for tryptophan, to UAG, which is a stop codon. (f) Yes. The UAG codon in some instances can trigger incorporation of pyrrolysine (the 22nd amino acid) rather than termination.

64) Replacing an A with a T in a hemoglobin gene is associated with sickle-cell anemia in humans.

Normal: Sickle cell:

5′-ATGGTGCACCTGACTCCTGAGGAGAAGTCT-3′ 5′-ATGGTGCACCTGACTCCTGTGGAGAAGTCT-3′

(a) What is the nucleotide sequence of the normal and sickle-cell hemoglobin mRNA? (b) What is the amino acid sequence in this part of the polypeptide chain, and what is the amino acid replacement that results in sickle-cell hemoglobin? (c) Why might this amino acid substitution make a difference in protein structure?

Answer: Section 4.2 p. 69, Fig 4.5; Table 4.1; Section 4.1, Fig. 4.1, p. 66 (a) Normal: AUGGUGCACCUGACUCCUGAGGAGAAGUCU Sickle cell:

AUGGUGCACCUGACUCCUGUGGAGAAGUCU

(b) (Met) Val-His-Leu-Thr-Pro-Glu-Glu-Lys-Ser Replacement of the glutamine by valine (codons underlined) leads to the sickle cell form.

(c) Glutamine is a polar amino acid. Valine is nonpolar. Their differing side chains have distinct properties, leading to changes in secondary and tertiary interactions.

65) You determine the structure of a protein of unknown function. The protein adopts a filamentous coiled-coil of two alpha-helices. Is the protein likely to be an enzyme? Explain.

Answer: Section 4.3, pp. 75-76, Fig. 4.11. A protein that is a filamentous coiled coil of two alpha-helices is likely to be a fibrous, structural protein, like keratin. Enzymes are generally globular proteins that fold up into a tertiary structure with an active site within a deep pocket between folded regions.

66) Provide an explanation for why you should politely decline a serving of brains at a party hosted by cannibals.

Answer: Section 4.6, pp. 86-88, Disease Box 4.1. You should politely decline a serving of brains at a party hosted by cannibals to avoid consumption of infectious prions, the causative agent of “Kuru” a type of transmissible spongiform encephalopathy. This progressive neurodegenerative disease is characterized by sponge-like holes in the brain, dementia, and loss of muscle control or voluntary movement. There is no cure yet. Once symptoms appear, death results in 6-12 months. The prion protein (PrpSc) has the same amino acid sequence as a normal host cell protein (PrPC), but is misfolded. Prions propagate through a chain reaction in which the host protein PrPC is post-translationally misfolded to form new prions. Conversion of the normal cellular prion protein into an abnormally folded one leads to the formation of aggregates and amyloid plaques in the brain.

Fundamental Molecular Biology, 2nd Edition Instructor Test Bank Lizabeth A. Allison Chapter 5 Genome Organization and Evolution

Multiple Choice 1) The term genome is used interchangeably with all of the following terms except: A. chromosomal DNA B. nuclear DNA C. plasmid DNA D. genomic DNA Answer: C 2) How many base pairs of DNA are in the human genome?

A. 3 X 109 B. 3 X 108 C. 3 X 107 D. 3 X 106 Answer: A 3) Eukaryotic cells must fit approximately _____ of unpacked DNA into the spherical nucleus, which is a less than _________ diameter space.

A. 2 millimeters (mm), 10 mm B. 2 mm, 0.010 mm C. 2 meters (m), 10 mm D. 2 m, 0.010 mm Answer: D 4) Which of the following properties do you predict to be most critical to histone association with DNA? A. Histones are small proteins.

B. Histones are positively charged.

C. Histones are highly conserved.

D. There are at least five different histone proteins in every eukaryotic cell.

Answer: B 5) The term nucleosome means A. the core octamer of histones plus 146 base pairs of DNA. B. the core octamer of histones plus the linker histone and approximately 180 base pairs of DNA. C. the core octamer of histones.

D. an octamer of histone H1 proteins.

Answer: B 6) The core octamer is composed of A. two tetramers of histones H3 and H4 B. one each of 8 different types of histones.

C. a tetramer of histones H3 and H4, and two dimers of histones H2A and H2B. D. the linker histone H1

Answer: C 7) The __________________ are the site of many post-translational modifications of the histone proteins.

A. conserved wing helix folds B. charged tails at the N-terminal end C. charged tails at the C-terminal end D. extended histone-fold domain at the C-terminal end Answer: B 8) Brief digestion of eukaryotic chromatin with micrococcal nuclease gives a visible ladder of bands on an agarose gel separated from each other by multiples of 180-200 bp. The fragment size represents A. DNA that was wrapped around the core histone octamer.

B. linker DNA. C. the different histone proteins.

D. multiples of single nucleosomal repeat lengths of DNA. Answer: D 9) DNA wrapped around the histone core octamer appears as “beads-on-a-string” by electron microscopy. Upon the addition of salt or when observed in situ, nucleosomal arrays further compact to form: A. a 10 nm helical solenoid.

B. a 10 nm fiber.

C. a 30 nm fiber with a zig-zag ribbon structure.

D. a 30 nm loop domain Answer: C 10) A eukaryotic chromosome consists of A. a single molecule of double-stranded DNA. B. a single molecule of single-stranded DNA. C. multiple intertwined molecules of single-stranded DNA. D. multiple intertwined molecules of double-stranded DNA. Answer: A 11) What statement is not true about plasmid DNA? A. Plasmid DNA is extrachromosomal B. Plasmid DNA is only found in bacteria C. Plasmid DNA is self-replicating D. Most plasmids are small, double-stranded circles.

Answer: B 12) What statement is not true about bacteriophages?

A. Bacteriophages are viruses that infect bacteria.

B. Bacteriophages are viruses that infect both bacterial and mammalian cells.

C. Bacteriophages have been widely used as a tool for molecular biology research.

D. The bacteriophage genome usually consists of a single DNA molecule.

Answer: B

13) Electron micrographs of chromatin formation in simian virus 40 (SV40) have revealed that A. the virus uses host cell histone H1 to form nucleosomal-like particles.

B. the virus does not make use of host cell histones to package its genome.

C. the circular double-stranded viral genome is organized into several minichromosomes.

D. the circular double-stranded viral genome is organized into a single minichromosome.

Answer: D 14) What statement is not true about chloroplasts and mitochondria?

A. They both contain their own genetic information that is inherited independently of the nuclear genome.

B. Mitochondrial and chloroplast genomes are either circular or linear.

C. They both are thought to have originated from cyanobacteria.

D. The organelles are only contributed from the maternal gamete.

Answer: C 15) The location of the centromere within a chromosome is determined by A. the presence of the histone variant CenH3.

B. the presence of an evolutionarily conserved centromere-determining sequence.

C. the presence of linker histone.

D. the point of equidistance between the two chromosome ends.

Answer: A 16) You use two different fluorescently labeled antibodies to detect CenH3 and a telomerebinding protein in mouse cells. You find that the two signals are so close that they almost overlap at one of the two ends of each chromosome. From this observation, you conclude that A. mouse chromosomes are acrocentric.

B. mouse chromosomes are metacentric.

C. mouse chromosomes have non-functional centromeres.

D. mouse chromosomes have one non-functional telomere per chromosome.

Answer: A 17) Which of the following statements about sex chromosomes is not true?

A. All sexually reproducing animal species specify gender by sex chromosomes.

B. Sex chromosomes contain a centromere and a telomere at each end.

C. Sex chromosomes are also called autosomes.

D. In humans, the Y chromosome is always inherited from the male parent.

Answer: C 18) Heterochromatin is A. condensed, transcriptionally active chromatin.

B. condensed, transcriptionally silent chromatin.

C. decondensed, transcriptionally active chromatin.

D. decondensed, transcriptionally silent chromatin.

Answer: B 19) Which of the following occupies the largest portion of the human genome?

A. protein-coding sequences B. introns C. repeat sequences D. gene regulatory sequences Answer: C

20) Which of the following accounts for the interspersed locations of LINEs and SINEs in eukaryotic genomes?

A. They arise spontaneously.

B. Selective evolutionary pressure prevents the persistence of local spots of concentrated LINEs or SINEs. C. LINEs and SINEs are transposable elements that can insert at random positions in the genome.

D. LINEs and SINEs result from random integrations of retroviral DNA. Answer: C 21) Satellite DNA is a type of A. transposon.

B. centromere-determining sequence.

C. gene coding sequence.

D. tandem repeat.

Answer: D 22) Most archael genomes A. contain vast non-coding repeats.

B. are packaged into nucleoids.

C. are maintained as multiple linear chromosomes.

D. encode histone-like proteins.

Answer: D

Short answer/analytical 23) Describe the levels of packing involved in fitting 2 meters of DNA into the nucleus of a eukaryotic cell.

Answer: Section 5.3, p. 95-102, Fig. 5.5 24) Diagram a nucleosome as follows: (a) On a drawing of the histones without the DNA, show the general positions of all the histones. (b) On a separate drawing, show the path of DNA around the histones.

Answer: Section 5.3, p. 98-99, Fig. 5.8 25) The 10 nm chromatin fiber in eukaryotes has been described as "beads on a string." What are the "beads" (be precise and detailed in your use of terminology) and what is the "string"?

Answer: Section 5.3, p. 98 26) Compare and contrast key features that distinguish eukaryotic genomes from bacterial genomes.

Answer: Sections 5.2 and 5.6, p. 92-95, p. 109-110, Table 5.2 27) It has been said that the relationship between plasmids and host bacteria could be considered as either parasitic or symbiotic. Discuss this statement.

Answer: Section 5.6, p. 110 28) Discuss some of the reasons that DNA viruses have proved extremely useful systems for analysis of fundamental principles of molecular biology.

Answer: Section 5.6, p. 112-113

29) Leber’s hereditary optic neuropathy, a form of young adult blindness, is linked to a small inherited mutation in mitochondrial DNA. Discuss some of the reasons why individuals with this disease may have differences in the severity and kinds of symptoms they display.

Answer: Section 5.5, p. 107-108. Disease Box 5.1 30) Chloroplast DNA is usually depicted as a circular, double-stranded DNA molecule. Is this depiction correct? Explain your answer.

Answer: Section 5.5, p. 107 31) Although mitochondria in both plants and animals are thought to have derived from the same ancient prokaryote and carry out the same essential functions, the size of mtDNA varies dramatically between modern animal and plant species. How can these observations be reconciled?

Answer: Section 5.5, p. 108-109, Fig. 5.19 32) Brief digestion of eukaryotic chromatin with micrococcal nuclease gives DNA fragments ~200 bp long. You repeat the experiment, but incubate the samples for a longer period of time while you are in class. This longer digestion yields 146 bp fragments. Why? Answer: Section 5.3, p. 96; Fig 5.6. Treatment of chromatin with a low concentration of micrococcal nuclease will result in partial digestion of DNA, yielding samples that differ from each other by a single nucleosomal repeat length of DNA (multiples of ~180-200 base pairs).

Longer digestion will cleave all of the linker DNA in between the nucleosomes, leaving intact the DNA that was wrapped around the core histone octamer.

33) Do the 10 nm and 30 nm eukaryotic chromatin fibers exist in vivo? Discuss electron microscopic and biochemical evidence in support of your answer.

Answer: Section 5.3, pp. 98-99. The beads-on-a-string appearance of chromatin can be visualized by electron microscopy as a 10-11 nm fiber after low salt extraction. The linker histone is not required for this level of packing. Since isolation occurs under non-physiological conditions, the question remains as to whether the 10-11 nm fiber is present in vivo. Upon the addition of higher salt during extraction, or when observed in situ by electron microscopy, nucleosomal arrays form a compact fiber of approximately 30 nm in diameter. The classic solenoid model of six consecutive nucleosomes per turn is not seen at physiological salt concentrations. Studies under more physiological conditions now suggest that nucleosomes adopt a zig-zag ribbon structure that twists or supercoils.

34) Discuss the characteristics of polytene chromosomes that make them useful for the study of genes and gene expression.

Answer: Section 5.3, p. 103

Fundamental Molecular Biology, 2nd Edition Instructor Test Bank Lizabeth A. Allison Chapter 6 DNA Replication and Telomere Maintenance

Multiple Choice 1) DNA replication is said to be semiconservative because: A. half of the DNA in a new cell comes from one gamete and the other half from the other gamete.

B. the same process of DNA replication is used by all organisms.

C. the number of nucleotides within genes remains constant.

D. each new DNA molecule is composed of one old strand and one new strand.

Answer: D 2) DNA synthesis occurs in A. the 3′ to 5′ direction.

B. the 5′ to 3′ direction.

C. both directions.

D. different directions in different organisms.

Answer: B 3) Which of the following statements is correct?

A. DNA polymerase requires a primer to get started.

B. DNA polymerase can start synthesis de novo.

C. There is only one type of replicative DNA polymerase in mammalian cells.

D. DNA polymerase catalyzes the formation of a phosphodiester bond between the first 3′ hydroxyl group of the new dNTP and the 5′-phosphate of the last nucleotide in the newly synthesized strand.

Answer: A 4) Nucleotides used for DNA synthesis contain three phosphates. During catalysis, the 3’ hydroxyl of the primer attacks A. the phosphate closest to the deoxyribose, releasing the terminal two phosphates B. the central phosphate, releasing the terminal phosphate C. the terminal phosphate, incorporating all three phosphates into the DNA backbone D. the deoxyribose, releasing all three phosphates Answer: A 5) Discontinuous synthesis of DNA takes place on A. the lagging strand.

B. the leading strand.

C. both strands.

D. different strands in different organisms.

Answer: A

6) Phosphodiester bonds are formed between adjacent Okazaki fragments by A. DNA polymerase B. topoisomerase C. DNA ligase I D. telomerase Answer: C 7) During DNA replication, DNA ligase is most active on the lagging strand. This is because: A. The lagging strands contain more short DNA segments than the leading strand, and these short segments are ligated together with DNA ligase.

B. The lagging strand is synthesized in the 3′ to 5′ direction.

C. The lagging strand requires DNA ligase to couple the RNA primer to the Okazaki fragments.

D. The lagging strand is synthesized more slowly, and DNA ligase acts as a sliding clamp to speed up the DNA polymerase.

Answer: A 8) A laboratory technique called “nick translation,” in which nucleotides of a nicked dsDNA fragment are replaced with labeled nucleotides, takes advantage of which unique aspect of bacterial DNA Polymerase I? A. 5’ to 3’ polymerization activity B. 5’ to 3’ exonuclease activity C. 3’ to 5 exonuclease activity D. interaction with RNA Answer: B 9) The bacterial multi-protein macromolecular complex that simultaneously replicates the leading and lagging strands is called A. a replication factory B. replication complex A C. origin recognition complex (ORC) D. the replisome Answer: D 10) In an ATP-dependent reaction, Topoisomerase I A. makes a double-strand break in DNA B. uses the hydroxyl group of a serine to break the DNA backbone C. forms a covalent phosphotyrosine linkage with DNA D. uses the energy from the terminal phosphates released during replication to cleave a single strand of DNA Answer: C 11) Which enzyme creates a double-strand DNA break during catalysis?

A. DNA helicase B. topoisomerase II C. clamp loader D. the Klenow fragment Answer: B

12) Topoisomers can be physically separated by A. CsCl gradient ultracentrifugation, because they differ in molecular weight B. agarose gel electrophoresis, because they differ in shape C. sucrose density gradient centrifugation, because they differ in density D. electron microscopy, because they have different morphologies Answer: B 13) Which statement is not correct?

A. Eukaryotic chromosomal DNA replication occurs in “replication factories.”

B. Eukaryotic chromosomal DNA typically has only one origin of replication.

C. An origin of replication is a site on chromosomal DNA where a bidirectional pair of replication forks initiates.

D. Origin DNA sequences usually have many A-T base pairs.

Answer: B 14) Once bound to an origin of replication, the origin replication complex (ORC) recruits A. DNA polymerase/primase B. replication protein A C. topoisomerase I and II D. Cdc6, Cdt1, and the Mcm2-7 helicase complex Answer: D 15) Positive supercoils ahead of the replication fork are resolved by A. helicase B. DNA polymerase/primase C. topoisomerase I and/or II D. PCNA, the sliding clamp Answer: C 16) Replication licensing ensures that A. DNA replication occurs by a semiconservative mechanism.

B. DNA only replicates once per cell cycle.

C. only one origin of replication is active within a cell at a time.

D. DNA replication only occurs during the G2 phase of the cell cycle.

Answer: B 17) Pre-replication complex assembly A. occurs when cyclin-dependent kinase 1 (CDK1) activity is low B. occurs when cyclin-dependent kinase 1 (CDK1) activity is high C. is not regulated by cyclin-dependent kinase 1 (CDK1) activity D. occurs during the M phase of the cell cycle.

Answer: A 18) Enzymes that use the energy of ATP to unwind the DNA duplex are called A. DNA helicases B. DNA ligases C. DNA topoisomerases D. DNA kinases Answer: A

19) DNA polymerase  and DNA polymerase  have A. 5′ to 3′ polymerase activity only B. both 5′ to 3′ polymerase activity and 3′ to 5′ exonuclease activity C. both 3′ to 5′ polymerase activity and 5′ to 3′ exonuclease activity D. both 3′ to 5′ polymerase activity and 5′ to 3′ polymerase activity Answer: B 20) Nucleotide selectivity and proofreading by DNA polymerase depends on A. base-base hydrogen bonding.

B. the geometry of Watson-Crick base pairs.

C. levels of cyclin-dependent kinases D. Both A and B. Answer: D 21) A molecular biologist purified what she thought were all factors essential for DNA replication in order to set up an in vitro DNA replication system. When she tested the in vitro system, replication occurred. But, when she isolated the newly synthesized double-stranded DNA, denatured the DNA, and electrophoresed the single-stranded products, she noted that there were not only some long strands but also numerous short segments of DNA a few hundred nucleotides long. What was probably missing from the in vitro replication system she developed?

A. RNA primers B. Okazaki fragments C. DNA polymerase D. DNA ligase Answer: D 22) Phage X174 replication occurs by a A. a theta-shaped intermediate.

B. a rolling circle mechanism.

C. a strand coupled mechanism.

D. a strand displacement mechanism.

Answer: B 23) A hallmark feature of rolling circle replication is A. use of an alternative polymerase.

B. lack of semi-discontinuous synthesis on the lagging strand.

C. displacement of a strand to form a 5’ phosphate tail.

D. formation of a 3’OH-tyrosyl intermediate.

Answer: C 24) Chromatin assembly factor 1 (CAF-1) is thought to aid replication by A. initiating synthesis of histones.

B. removing histones ahead of the replication fork.

C. bringing histones to the replication fork.

D. phosphorylating nucleosomes behind the replication fork.

Answer: C

25) Two competing models to explain how nucleosomes are assembled behind replication forks differ by A. whether histones H3 and H4 are passed to daughter strands as dimers or tetramers.

B. whether histone octamers remain intact during the process.

C. whether CAF-1 is necessary for the process.

D. whether the leading or lagging strand binds to the parental histones.

Answer: A

26) At the ends of chromosomes in eukaryotes, a problem occurs upon repeated DNA replication of the __________ strand. __________ is responsible for providing a solution to this problem.

A. lagging; telomerase B. leading; replicase C. lagging; replicase D. leading; polymerase Answer: A 27) Telomerase directly elongates A. the 5′ end of the lagging template strand.

B. the 3′ end of the lagging template strand.

C. the 5’ end of the leading template strand.

D. the 3’ end of the leading template strand.

Answer: D 28) Telomerase is a ribonucleoprotein complex. The RNA component of the complex A. has reverse transcriptase activity B. provides the template for telomere repeat synthesis C. provides structural support only D. forms a t-loop Answer: B 29) Telomerase is usually active in all of the following cell types except for: A. single-celled eukaryotes B. human germ-line cells C. human fibroblast cells D. cancer cells Answer: C 30) The Hayflick limit is at least partly enforced by A. telomerase activation.

B. T-loop formation.

C. the TERC pseudoknot.

D. telomere shortening.

Answer: D 31) Which of the following is not a function of telomeres?

A. cellular clock B. Cdc6 regulation C. protection of chromosome ends D. shelterin binding

Answer: B 32) TRF1/TRF2 and POT1 form a complex called ______________ that __________________.

A. ORC; binds origins of replication B. replisome; coordinates replication of the leading and lagging strands C. telomerase; lengthens chromosome ends D. shelterin; protects telomeres Answer: D

Short answer/analytical

33) Describe and give the results of the classic experiment that showed that DNA replication is semiconservative. What are the names of the scientists who carried out this experiment?

Answer: Section 6.2, p. 118, Fig. 6.1 34) Describe electron microscopic evidence that DNA replication in E. coli is bidirectional.

Answer: Section 6.2, p. 120-121, Fig. 6.2 35) Draw a diagram showing the addition of a new dNTP to the daughter strand during DNA replication. The diagram should be detailed enough to illustrate what is meant by the statement “DNA synthesis occurs from 5′ to 3′ ”.

Answer: Section 6.3, p. 121, Fig. 6.3, p. 122 36) List the eukaryotic replicative DNA polymerases and their roles.

Answer: Section 6.5, p. 136, Table 6.2, p. 137 37) Why do you suppose eukaryotic cells “switch” from DNA polymerase alpha, which binds to primase and initiates DNA synthesis, to DNA polymerase delta/epsilon for elongation?

Answer: Section 6.5, p. 126. DNA Pol alpha has low processivity compared to DNA polymerase delta and epsilon.

38) Of the three DNA polymerase in E. coli, which is essential for DNA replication?

Answer: Section 6.4, p.125 DNA Polymerase III 39) Draw a bidirectional eukaryotic origin of replication. Label the 5′ and 3′ ends of each strand of the double helix. Label the lagging and leading strands at each replication fork. In which direction are new strands synthesized?

Answer: Section 6.3, Fig. 6.4, p. 123 40) Outline the steps of nuclear chromosome replication in eukaryotes, starting from origin firing and ending with telomere replication.

Answer: Section 6.5, p. 129-149, Fig. 6.10, Fig. 6.16, Fig. 6.17, Fig. 6.19, Fig 6. 20, Fig. 6.27 1. DNA melting at origin, 2. RNA primer synthesis, 3. DNA synthesis initiation, 4. Polymerase switching,

5. Continuous synthesis on leading strand and semi-discontinuous synthesis on lagging strand, 6. RNA primer removal and replacement with DNA, 7. Ligation at nicks, 8. Lagging strand template extension by telomerase, 9. “Filling in” of newly synthesized DNA on lagging strand.

41) Why are origins of replication often AT-rich?

Answer: Section 6.4, p. 125 Less energy is required to melt the two hydrogen bonds joining A with T, compared with the three hydrogen bonds joining guanine–cytosine (GC) base pairs. Thus it is easier to melt AT-rich sequences, which must occur to initiate DNA synthesis.

42) You are characterizing nuclear DNA replication in a previously unstudied single-celled eukaryote. Describe an experiment to test whether nuclear DNA replication occurs in replication factories from multiple origins. Show sample positive results.

Answer: Section 6.5, p. 131, Fig. 6.12 43) Define the term “replication licensing” and discuss its importance in DNA replication.

Answer: Section 6.5, p. 133-134, Fig. 6.15 44) List the major enzymes and proteins required for DNA replication in the order in which they act and provide a brief description of their function.

Answer: Section 6.3, p. 124, Table 6.1 45) Why are RNA primers required during DNA replication?

Answer: Section 6.3, p. 121 DNA polymerases cannot initiate synthesis de novo, whereas RNA polymerases can.

46) Describe an example of “polymerase switching.”

Answer: Section 6.5, p. 136, Fig. 6.17 47) What three critical steps are required for “maturation” of the newly synthesized lagging strand?

Answer: Section 6.5, p. 139 RNA primers must be removed, the resulting gap must be filled in by DNA polymerase, and these DNA-only Okazaki fragments must be ligated together.

48) What mode of interaction between DNA ligase I and DNA does X-ray crystallographic analysis suggest?

Answer: Section 6.5, p. 141, Fig. 6.21 49) Describe two opposing models for how parental histones are distributed to daughter strands behind the replication fork.

Answer: Section 6.5, p. 142, Fig. 6.22 50) Explain the molecular defect that causes a human form of dwarfism called cartilage-hair hypoplasia.

Answer: Disease Box 6.2, p. 145

51) Indicate whether the following enzymes normally require ATP for their functions. a) topoisomerase I b) topoisomerase II c) DNA ligase d) DNA helicase Answer: a) no; Section 6.4, p. 127 b) yes; Section 6.4, p. 128 c) yes; Section 6.5 p. 141 d) yes; Section 6.5, p. 136 52) Diagram the rolling circle replication mechanism used by phage X174.

Answer: Section 6.6, Fig. 6.23a 53) Xenopus oocytes contain amplified ribosomal DNA. Diagram the mechanism by which the extrachromosomal circles are produced.

Answer: Section 6.6, Fig. 6.23b 54) Choose one alternative mode of replication to describe (from among the rolling circle, strand displacement, and strand-coupled models) and compare and contrast this alternative mode with the conventional mode of replication for eukaryotic chromosomes.

Answer: Section 6.6, p. 143-146.

Answers should minimally indicate whether a “replication bubble” is formed or DNA is nicked and displaced, and whether synthesis on each strand is continuous or discontinuous.

55) What is the “end-replication problem?”

Answer: Section 6.7, p. 146, Fig. 6.25 56) Why do eukaryotes need telomeres but bacteria do not?

Answer: Section 6.7, p. 146. Bacterial genomes are circular. Therefore, bacterial replicative polymerases never encounter ends. By contrast, nearly all eukaryotic chromosomes are linear.

57) Why was Tetrahymena a good choice of organism in which to study telomerase?

Answer: Section 6.7, p. 147 Tetrahymena have more than 40,000 telomeres. By comparison, most other eukaryotes have less than 100 telomeres.

58) List at least two molecular characteristics of telomeres and relate these characteristics to functions.

Answer: Section 6.7, p. 146-150; Figs. 6.27 and 6.28 Possible answers include the presence of repeats, which allow extension by telomerase; binding to shelterin complex, which helps to protect chromosome ends and regulate telomerase activity; and formation of the t-loop structure, in which the 3’ end of a chromosome is protected.

59) Explain how the POT1 protein is thought to regulate in telomere length.

Answer: Section 6.7, Fig. 6.28, p. 150 60) What evidence supports the existence of t-loops in telomere structure?

Answer: Section 6.7, Fig. 6.27, p. 149

61) Describe an experiment (or observation) that you think has provided the most convincing evidence for a causal relation between telomere shortening and cellular senescence or aging.

Answer: Section 6.7, p. 151-154 62) Describe the molecular basis and primary (life-threatening) symptoms of the disease dyskeratosis congenita. Explain why the symptoms are not surprising in view of the molecular defect.

Answer: Section 6.7, Disease Box 6.3, p. 153-154 63) Gene therapy-mediated delivery of the telomerase RNA to cirrhotic liver has been performed in mice. What important risk associated with telomerase activation explains why this strategy not yet progressed to human trials?

Answer: Section 6.7, p. 154, Fig. 6.30 Telomerase activation increases the risk of tumor formation.

64) Illustrate and explain the results that Meselson and Stahl would have obtained if DNA replication were conservative.

Answer: Section 6.2, p. 118, Fig. 6.1. In conservative replication, one daughter molecule would consist of the original parent and the other daughter would be totally new DNA. One cell generation after transfer to 14N, there would be one heavy band and one light band. Two cell generations after transfer to 14N, there would still be a heavy band a light band; there would be no intermediate band, as there is in semiconservative replication.

Before transfer to 14N

One cell generation after transfer to 14N

14

N

15

N

15

N

65) The diagram below shows a replication fork in nuclear DNA. (a) Label the “leading strand” and “lagging strand” and indicate to which strand of DNA telomerase adds repeats. (b) Show on the drawing what happens next on each strand as more of the duplex DNA unwinds at the replication fork. Use arrows to show the direction of synthesis for each strand.

You do not need to show all the protein components of the replication machinery.

= new strand 3 5

Answer: Section 6.3, p. 123, Fig. 6.4; also below: Telomerase adds repeats here (3′ end of template for lagging strand) RNA primer

Lagging strand

Leading strand

66) You are studying a protein that you suspect functions to recruit other components of the licensing protein complex for eukaryotic DNA replication. Describe how you would assay the protein for this activity and show sample positive results.

Answer: Section 6.5, pp. 133-134, Fig. 6.14. Design an experiment similar to the one shown in Fig. 6.14. Incubate DNA containing an origin of replication (coupled to steptavidin-coated magnetic beads) with or without the protein of interest (protein X). Next, add cell extract depleted of this protein, or add purified licensing complex proteins. Collect the beads, wash, and analyze the associated proteins by immunoblotting. If components of the licensing protein complex (such as Mcm2-7 and Cdc6) are only present when the DNA was pre-incubated with the protein of interest, then this would indicate that the protein of interest functions to recruit other components of the licensing complex.

67) You are studying a mammalian DNA virus with a 200 kb double-stranded genome. Based on the size, you suspect that the genome has more than one origin of replication. Propose experiments to test your hypothesis and map the origins.

Answer: Options found in Section 6.5, Fig. 6.10; Section 6.5, Fig. 6.11, p. 129-131. To determine whether the viral genome has more than one origin, a number of different techniques could be used: Allow the viral DNA to replicate for one whole generation and a portion of the second in the presence of radioactive nucleotides. Visualize whether there is more than one bidirectional replication fork by autoradiography. Electron microscopy of replicating viral DNA also could be used to visualize the formation of more than one replication bubble. Replication origins could be mapped using the technique of two-dimensional agarose gel electrophoresis or by replication point (RIP) mapping to detect the precise site of replication initiation within the origins.

68) Cells you have been culturing usually undergo senescence after about 20 divisions. But, some cells have become immortal. You suspect that telomerase has been activated. Propose an experiment to assay for telomerase activity and show sample positive results.

Answer: Section 6.7, p. 147, Fig. 6.26. To test for telomerase activity, design an experiment similar to the one shown in Fig. 6.26. Prepare a cell-free extract from the immortal cells and incubate the extract with a synthetic oligonucleotide primer having four repeats of the mammalian telomere repeat sequence TTAGGG (input). After incubation in the presence of 32P-

labeled dGTP and unlabeled dNTPs, separate the products by electrophoresis and detect them by autoradiography. If telomerase activity is present, then the lane should show periodic extension of the telomere. As a control, include a lane with the Klenow fragment of E. coli DNA polymerase I. In this lane, no extension of the repeated sequence would be expected.

Fundamental Molecular Biology, 2nd Edition Instructor Test Bank Lizabeth A. Allison

Chapter 7 DNA Repair Pathways

Multiple Choice 1) An induced mutation occurs A. as a result of natural processes in cells.

B. as a result of interaction of DNA with an outside agent.

C. spontaneously.

D. as a result of DNA replication errors.

Answer: B 2) A transition mutation is one in which A. one pyrimidine base is replaced with another, or one purine base with another.

B. one pyrimidine base is replaced with a purine base or vice versa.

C. there is an A to T substitution.

D. there is a C to G substitution.

Answer: A 3) A transversion mutation is one in which A. one pyrimidine base is replaced with another, or one purine base with another.

B. one pyrimidine base is replaced with a purine base or vice versa.

C. there is a C to T substitution.

D. there is an A to G substitution.

Answer: A 4) A nonsense mutation is one in which: A. the altered codon codes for the same amino acid.

B. the altered codon codes for a different amino acid.

C. the mutant protein is shorter than normal.

D. insertion or deletion results in a shift in the reading frame.

Answer: C 5) A silent mutation is one in which A. the altered codon codes for the same amino acid.

B. the altered codon codes for a different amino acid.

C. the mutant protein is shorter than normal.

D. insertion or deletion results in a shift in the reading frame.

Answer: A 6) A missense mutation is one in which A. the altered codon codes for the same amino acid.

B. the altered codon codes for a different amino acid.

C. the mutant protein is shorter than normal.

D. insertion or deletion results in a shift in the reading frame.

Answer: B

7) A type of mutation that results in multiple adjacent amino acid changes in a polypeptide is probably caused by the following type of mutation: A. frameshift B. transversion C. transition D. missense Answer: A 8) Which mutagen does not covalently modify DNA? A. UV light B. base analogs C. intercalating agents D. ionizing radiation Answer: C 9) Which statement is not true about the deamination of cytosine?

A. Deamination can occur spontaneously from the action of water B. Deamination is an example of DNA damage resulting in a single base change.

C. Methylated cytosine is protected from deamination.

D. Deamination converts cytosine to uracil.

Answer: C 10) The most frequent ultraviolet (UV) light-induced lesions of DNA are A. oxidation of guanine to generate 8-oxoguanine.

B. formation of a cyclobutane ring between adjacent thymines, forming a thymine dimer.

C. induction of 5-bromouracil.

D. double-strand breaks.

Answer: B 11) Expansion of trinucleotide repeats can occur by A. non-homologous end joining during double strand break repair B. unequal crossing over during meiosis C. DNA slippage during replication D. B and C Answer: D 12) In human cells, pyrimidine dimers can be repaired by A. photoreactivation.

B. nucleotide excision repair.

C. mismatch repair.

D. both A and B. Answer: B 13) Translesion synthesis is mediated by A. error-prone DNA polymerases B. replicative DNA polymerases C. DNA glycosylase D. DNA photolyase Answer: A

14) Which DNA polymerase performs translesion synthesis past a thymine dimer by inserting two adenine residues?

A. DNA polymerase iota () B. DNA polymerase eta () C. Both A and B D. Neither A nor B; there are no error-free repair polymerases.

Answer: B 15) Which statement is not true about the enzyme methyltransferase?

A. A sulfhydryl group of a cysteine residue in the enzyme accepts a methyl group.

B. The enzyme mediates direct reversal of DNA damage.

C. After accepting a methyl group from methylguanine, the enzyme can be used again.

D. The enzyme is present in organisms ranging from E. coli to humans.

Answer: C 16) Steps in the base excision repair pathway occur in the following order: A. Damaged base removal, endonuclease cleavage, nucleotide excision, repair synthesis ligation.

B. Endonuclease cleavage, damaged base removal, nucleotide excision, repair synthesis, ligation.

C. Damaged base removal, endonuclease cleavage, ligation, nucleotide excision, repair synthesis.

D. Endonuclease cleavage, damaged base removal, nucleotide excision, ligation, repair synthesis.

Answer: A 17) The disease hereditary nonpolyposis colorectal cancer results from defects in which DNA repair pathway?

A. nucleotide excision repair B. base excision repair C. double-strand break repair D. mismatch repair Answer: D 18) In the mismatch repair pathway, DNA damage is recognized by A. RFC and PCNA B. exonuclease EXO1 C. MutS D. DNA polymerase Answer: C 19) The disease xeroderma pigmentosum (XP) is a rare autosomal recessive condition in humans. Individuals with this condition are extremely sensitive to UV light and nucleotide excision repair activity is reduced. When cell cultures from normal individuals and from XP patients are irradiated with various doses of UV light and cultured in radiolabeled thymidine…

A. cells from XP patients incorporate about the same amount of radiolabeled thymidine into their DNA as the cells from normal individuals.

B. cells from XP patients incorporate significantly less radiolabeled thymidine into their DNA. C. cells from XP patients incorporate significantly more radiolabeled thymidine into their DNA.

D. neither cells from XP patients nor cells from normal individuals will incorporate radiolabeled thymindine into their DNA after UV irradiation.

Answer: B 20) In mammalian cells double-strand breaks in DNA are primarily repaired by A. nucleotide excision repair.

B. translesion synthesis.

C. homologous recombination.

D. nonhomologous end-joining.

Answer: D 21) Which repair pathway requires a set of enzymes that recognize specific chemical modifications to DNA? A. nucleotide excision repair B. base excision repair C. mismatch repair D. non-homologous end joining Answer: B 22) Repair of a DNA double-strand break by homologous recombination is initiated by A. the Mre11-Rad50-Nbs1 (MRN) complex.

B. the resolvasome.

C. Rad52.

D. BRCA1.

Answer: A 23) Repair of a DNA double-strand break by nonhomologous end-joining is initiated by A. the endonuclease Artemis.

B. Ku70 and Ku80.

C. DNA polymerase.

D. DNA-dependent protein kinase catalytic subunit.

Answer: B 24) Homologous recombination repairs double-strand breaks by A. direct ligation of DNA ends.

B. DNA synthesis without a template.

C. retrieving genetic information from an undamaged homologous chromosome.

D. translesion synthesis.

Answer: C 25) Nonhomologous end-joing repairs double-strand breaks by A. direct ligation of DNA ends.

B. DNA synthesis without a template.

C. retrieving genetic information from an undamaged homologous chromosome.

D. translesion synthesis.

Answer: A

26) The Holliday junction is resolved into two duplexes by an enzyme complex called A. DNA ligase B. the resolvasome C. Rad54 D. topoisomerase Answer: B

Short answer/analytical 27) Explain why high-fidelity DNA replication is not always an advantage.

Answer: Section 7.2, p.160 Low-fidelity DNA replication is beneficial for the evolution of species and for generating diversity leading to increased survival when organisms are subjected to changing environments.

28) Diagram how a point mutation changing a GC base pair to a GA base pair can be permanently incorporated by DNA replication.

Answer: Section 7.2, Fig. 7.1, p. 161 29) Compare and contrast silent mutations, missense mutations, nonsense mutations, and frameshift mutations.

Answer: Section 7.2, p. 160-162, Fig. 7.2 30) In addition to deploying repair machinery, cells respond to global DNA damage by preventing entry into S phase. Provide a rationale for this observation.

Answer: Section 7.2, Fig. 7.1, p. 161. DNA damage can be transformed into permanent mutations by replication. By preventing entry into S phase when DNA damage is detected, cells give DNA repair machinery time to repair DNA damage, thereby preventing the damage from manifesting as mutations.

31) Draw the structure of a cyclobutane ring between adjacent thymines and explain how formation of thymine-thymine dimers distorts the double helix.

Answer: Section 7.2, p. 164, Fig. 7.4 Because covalent bonds form between thymines on the same strand, this disrupts the complementary base pairs that form the double helix. Thymine dimers thus distort the structure of the duplex DNA. 32) Compare and contrast the three major classes of DNA damage.

Answer: Section 7.2, p. 163-165 33) Explain why unequal crossing over and DNA slippage results cause the specific problem of trinucleotide repeat expansion rather than random errors. Use diagrams to illustrate your answer.

Answer: Section 7.2, Fig. 7.3, p. 163 DNA encoding trinucleotide repeats may be expanded because repeats can misalign with repeats on the homologous chromosome (in the case of meiosis) or can base-pair with misaligned repeats on the complementary strand (in the case of replication), resulting in changes in the number of repeats in a chromosome. The misalignment occurs because of the

repeat structure itself and therefore only creates errors in this context. Non-repeat sequences are not subject to this type of error because the sequence is too specific and diverse for misalignment to occur.

34) Define the term “translesion DNA synthesis” and briefly explain its mechanism.

Answer: Section 7.3, p. 166-167, Fig. 7.5 35) Explain why translesion synthesis is not truly a repair system.

Answer: Section 7.3, p. 167 The lesion is still present in the original parent strand of the DNA double helix after translesion synthesis.

36) What two enzymes catalyze direct reversal of DNA damage? Diagram/briefly explain the mechanisms they use. Are both repair pathways present in human cells?

Answer: Section 7.4, p. 167-169, Figs. 7.6 and 7.7 DNA photolyase (not present in humans) and DNA methyltransferase (present in humans) 37) Explain why repair of the methylated base O6-methylguanine by methyltransferases is considered very costly to a cell.

Answer: Section 7.4, p. 168-169 Once the methyltransferase has accepted a methyl group from the damaged guanine, the enzyme cannot be used again.

38) Exposure of DNA to an oxidizing agent leads to the formation of 8-oxoguanine (oxoG).

Name the pathway used to repair this damage and describe the two-step model based on X-ray crystallographic analysis for how the damage is first recognized.

Answer: Section 7.5, p. 169-170, Fig. 7.8 39) Compare and contrast base excision repair and nucleotide excision repair. Diagram both processes. For what types of damage is each primarily responsible?

Answer: Section 7.5, p. 169-170, 174-177, Figs. 7.9 and 7.12 40) What is the mismatch repair system and when is it used? Include the key players of the system and the concept of strand discrimination in your answer.

Answer: Section 7.5, p. 171-174, Fig. 7.10 Mismatch repair occurs in the final stages of DNA replication.

41) Describe the “molecular switch model” for MutS’s mode of action. Include in your answer the proposed role of ATP. Answer: Section 7.5, p. 172 The “molecular switch model” proposes that ATP-bound MutS forms a sliding clamp on mismatched DNA. MutSα is thought to diffuse either 5’ or 3’ for several thousand nucleotides along the DNA backbone. This model suggests that ATP hydrolysis is not needed for the activity of MutS but instead triggers its dissociation from the damage site when repair is complete.

42) What process is defective in people with hereditary non-polyposis colon cancer? Explain your answer.

Answer: Disease Box 7.1, p. 172 Mismatch repair is defective.

43) Predict the molecular and phenotypic consequences of loss of strand discrimination by the mismatch repair pathway.

Answer: Section 7.5, pp. 171-174. If the mismatch repair pathway could not discriminate between the parental and newly synthesized DNA strands, the pathway would “repair” the parental strand to match the misincorporated nucleotide in the newly synthesized strand 50% of the time. This would decrease the fidelity of replication and would increase the incidence of mutations, leading to susceptibility to certain cancers, such as hereditary nonpolyposis colorectal cancer.

44) What damage can be caused to DNA by UV irradiation? Describe the key steps of three different mechanisms by which this damage can be repaired. Do all three mechanisms repair the damage accurately? Are all mechanisms active in humans?

Answer: Section 7.2, p. 165 Section 7.3, p. 166, Fig. 7.5; Section 7.4, p. 167, Fig. 7.6; Section 7.5, p. 174, Fig. 7.12 UV irradiation predominantly induces the formation of pyrimidine dimers, but can also induce dimers between cytosine and thymine called pyrimidine (6,4)-pyrimidone photoproducts. Repair can occur by nucleotide excision repair (accurate), translesion DNA synthesis (usually not accurate), and by direct reversal via DNA photolyase (accurate). Humans do not express DNA photolyase.

45) How does transcription coupled repair (TCR) differ from global genome repair (GGR)?

Section 7.5, p. 176 The repair pathway responsible for recognizing lesions in the whole genome is called global genome repair, while the transcription-coupled repair pathway identifies lesions in the transcribed strand of active genes.

46) What DNA repair system is missing in xeroderma pigmentosum variant (XPV) patients?

What is the backup system for lesions missed by the nucleotide excision repair system in XP-V patients?

Answer: Disease Box 7.2, p. 175-176 47) Compare and contrast repair of double strand breaks by homologous recombination or nonhomologous end-joining. Which is used predominantly in mammalian cells?

Answer: Section 7.6, p. 177-181, Figs. 7.13 and 7.15 Homologous recombination repairs double-strand breaks by retrieving genetic information from an undamaged homologous chromosome, restoring sequence “perfectly” (except if the homologous chromosome contains a non-identical sequence).

Nonhomologous end-joining rejoins double-strand breaks via direct ligation of the DNA ends without any sequence homology requirements, and is almost always mutagenic because some sequence is lost during resection. Both pathway require resection of broken DNA ends and result in restoration of an uninterrupted DNA duplex.

48) Describe experiments that led to the characterization of the Holliday junction resolvasome.

Answer: Section 7.6, p. 180, Fig. 7.14 49) Outline the nonhomologous end-joining mechanism mammals used to repair double-strand DNA breaks. Show how this process can lead to loss of nucleotides at the repair site.

Answer: Section 7.6, p. 180-181, Fig. 7.15

50) What would be the effect on reading frame and gene function if: (a) Two nucleotides were inserted into the middle of an mRNA? (b) Three nucleotides were inserted into the middle of an mRNA? (c) One nucleotide was inserted into one codon and one subtracted from the next? (d) A transition mutation occurs from GA during DNA replication? What is the effect after a second round of DNA replication? (d) Exposure to an alkylating agent leads to the formation of O6-methylguanine? What is the effect after DNA replication?

Answer: Section 7.2, pp. 160-162 (a) If two nucleotides were inserted into the middle of an mRNA this would shift the reading frame and alter all of the amino acids downstream from the site of the mutation. (b) If three nucleotides were inserted, this would not result in a frameshift but the mRNA would encode one additional amino acid. Whether this additional amino acid has an effect on gene function would depend on the location of the amino acid insertion and whether it disrupted critical secondary or tertiary folding interactions, or interfered with interaction of the protein with another protein or nucleic acid, etc. (c) If one nucleotide were inserted into one codon and one subtracted from the next, this would not change the reading frame, but depending on whether the nucleotide substitutions were silent, missense, or nonsense mutations the function of the protein might be disrupted. (d) After a second round of DNA replication, the DNA molecule with the A-C mismatch would be replicated to yield one mutant molecule with an A-T base pair, and one wild type molecule with a G-C base pair. (e) O6-methylguanine often mispairs with thymine. So, if the damage were not removed by the base excision repair pathway or reversed by methyltransferase, this would result in change of a GC base pair into an AT base pair in one daughter molecule when the damaged DNA was replicated (as outlined in (d)). This nucleotide substitution would not affect reading frame, but depending on whether the nucleotide substitution was a silent, missense, or nonsense mutation the function of the protein might be disrupted.

51) A friend of yours with xeroderma pigmentosum seeks your advice about participating in a sun tanning competition in Florida during Spring Break. Provide appropriate advice.

Answer: Section 7.5, Disease Box 7.2, pp. 175-176. Xeroderma pigmentosum is a rare syndrome marked by unusually high sensitivity to UV light and a greatly increased risk of sunlight-induced cancer. UV light promotes DNA damage, most often in the form of thyminethymine (pyrimidine) dimers. Because covalent bonds form between thymines on the same strand, this disrupts complementary base pairs that form the double helix. T-T dimers distort the structure of the DNA and may impede transcription and replication by blocking the movement of the polymerases This type of damage is normally repaired by the nucleotide excision repair pathway, or bypassed by repair polymerases during translesion synthesis. One of the repair polymerases, DNA polymerase eta performs TLS past T-T dimers by inserting two adenine residues. This results in the lesion being bypassed in an error-free manner. In contrast, other repair polymerase such as DNA polymerase iota are highly error prone, resulting in the accumulation of mutations. Individuals with xeroderma pigmentosum either have a defect in one of the seven genes encoding proteins involved in the nucleotide excision repair pathway, or in

the gene encoding DNA polymerase eta. An impaired ability to repair damaged DNA leads to an increased mutation frequency that contributes to the increased risk of skin cancer. Do not participate in a sun tanning competition!

52) Draw a diagram of a Holliday junction during double-strand break repair. Starting with that diagram, illustrate branch migration and resolution. Is the resulting DNA duplex “repaired” to its original state?

Answer: See Fig. 7.13 for an illustration of branch migration and resolution. Homologous recombination repairs double-stranded breaks by retrieving genetic information from an undamaged homologous chromosome. When the two chromosomes are not exactly identical in sequence (i.e. they have alternative alleles), gene conversion may take place. In this case, the resulting DNA duplex would not be repaired to its original state, because the sequence of the repaired region would differ slightly from the original sequence.

53) You have isolated a novel protein factor you suspect is essential for efficient mismatch repair in mammalian cells. Design an experiment to test for repair activity in vitro. Show sample positive results.

Answer: Design an experiment using an in vitro system as shown in Fig. 7.11, where efficient mismatch repair regenerates a restriction endonuclease (HindIII) recognition site in a circular DNA of 5.0 kb that also contains an EcoRI recognition site. After incubation with other purified repair factors in the presence or absence of protein X (the protein suspected to be essential for efficient repair), samples are digested with both HindIII and EcoRI. Samples are then separated by agarose gel electrophoresis. Successful repair is indicated by two fragments on the agarose gel (3.0 and 2.0 kb), while unsuccessful repair yields a 5.0 kb linear fragment after a single cut with EcoRI.

+ protein X 5.0 kb 3.0 kb 2.0 kb

− protein X

Fundamental Molecular Biology, 2nd Edition Instructor Test Bank Lizabeth A. Allison Chapter 8 Recombinant DNA Technology and Molecular Cloning

Multiple Choice 1) Following the injection of a linear phage DNA into E. coli host cells, the phage genome circularizes by A. recombination at attachment (Att) sites.

B. joining of the attachment (Att) sites.

C. recombination at cohesive (cos) sites.

D. joining of the cohesive (cos) sites.

Answer: D 2) When E. coli host strain K-12 is infected by phage lambda K, the phage DNA is A. recognized as foreign and cleaved by a specific restriction endonuclease.

B. recognized as foreign and marked with a methylation pattern.

C. not recognized as foreign because it has the same methylation pattern as the E. coli genome.

D. not recognized as foreign because of the phage restriction and modification system.

Answer: C 3) Recombinant DNA technology became possible with the discovery of which of the following enzymes?

A. terminal deoxynucleotidyl transferase B. restriction endonucleases and DNA ligase C. DNA polymerase and DNA ligase D. reverse transcriptase Answer: B 4) Type II restriction endonucleases are useful in recombinant DNA research primarily because A. they cut both strands of DNA at a specific recognition site.

B. they can distinguish between genomic and cDNA. C. they can join two pieces of DNA by forming phosphodiester bonds.

D. they cleave DNA at a distance from the recognition site.

Answer: A 5) Which statement is not true about DNA binding and cleavage by type II restriction endonucleases?

A. the recognition process triggers a conformational change in both the enzyme and the DNA. B. the first contact with DNA is specific.

C. the target site is located by sliding, hopping, and jumping.

D. the enzyme interacts directly with the nitrogenous bases.

Answer: B

6) What is suggested by the fact that type II restriction endonuclease recognition sequences are usually palindromic?

A. A particular restriction endonuclease can cleave DNA at several different sites.

B. Restriction endonucleases must cleave DNA in a multistep process.

C. Restriction endonucleases function as dimers.

D. Restriction endonucleases recognize the sugar-phosphate backbone of DNA. Answer: C 7) Which statement is not true about DNA ligase?

A. It requires ATP as a cofactor.

B. It joins two pieces of DNA by forming phosphodiester bonds.

C. It can circularize a linear DNA with blunt ends.

D. It can only ligate fragments of DNA with sticky (complementary) ends.

Answer: D 8) In the context of recombinant DNA technology, plasmids are used as vectors. This means that plasmids A. allow for transfer of foreign DNA molecules into a host cell.

B. allow for infection and lysis of host cells.

C. serve as host cells for amplification of foreign DNA molecules.

D. function as artificial chromosomes in host cells.

Answer: A 9) Why is E. coli so widely used as a host organism in which to clone genes?

A. E. coli has a rapid growth rate.

B. Growing cultures of E. coli cells is relatively easy.

C. Many different E. coli strains with distinct and useful properties are available.

D. All of the above Answer: D 10) In recombinant DNA technology, the term competency refers to A. the ability of DNA ends to be ligated B. the ability of an enzyme to be activated by ATP C. the ability of cells to take up foreign DNA D. the ability of a medium to select for transformed cells Answer: C 11) Which of the following vectors can accommodate the largest DNA insert?

A. plasmid B. cosmid C. phage D. pUC18 Answer: B 12) You have ligated a foreign DNA fragment into the EcoRI site in the multiple cloning site of pUC18. You then transform host E. coli with the ligation mixture. Some of the bacterial colonies growing on the nutrient agar plate that contains ampicillin and X-gal are white and some are blue. Bacterial colonies that are blue A. contain recombinant plasmid DNA B. contain nonrecombinant plasmid DNA C. are untransformed bacteria

D. have hydrolyzed ampicillin to form a colored compound Answer: B 13) To separate acidic proteins from basic proteins in a crude cellular lysate you could use A. gel filtration chromatography B. antibody-affinity chromatography C. ion-exchange chromatography D. A and C Answer: C 14) In order to function, yeast artificial chromosome (YAC) vectors must include the following: A. an origin of replication B. a centromere and telomeres C. growth selectable markers D. all of the above Answer: D 15) In red-white selection for recombinant yeast artificial chromosome (YAC) vectors in yeast cells, a red colony contains A. recombinant YAC vector DNA. B. nonrecombinant YAC vector DNA. C. no YAC vector DNA. D. circular YAC vector DNA Answer: A 16) How does a genomic library differ from a gene clone?

A. A genomic library contains many different sequences that represent the entire genome of an organism; a gene clone contains one type of sequence.

B. A genomic library contains one type of sequence; a gene clone contains many different sequences that represent the entire genome of an organism.

C. A genomic library is much smaller than a gene clone.

D. A genomic library is sequence information stored in silico (i.e., in a computer); a gene clone is an actual sequence of DNA in a plasmid vector.

Answer: A 17) What aspect of a common recombinant DNA technology tool makes direct use of knowledge acquired from studies of retrovirus replication?

A. Using restriction endonucleases to cleave DNA B. Complementary DNA (cDNA) synthesis C. Introducing DNA into bacterial cells by transformation D. Using DNA ligase to join two pieces of DNA Answer: B 18) Reverse transcriptase synthesizes A. DNA from DNA template B. DNA from an RNA template C. RNA from a DNA template D. RNA from an RNA template Answer: B

19) A cDNA library contains A. DNA fragments that represent the entire set of genes in the genome of an organism.

B. DNA fragments that represent the coding regions of expressed genes from a specific tissue, cell type or developmental stage.

C. DNA fragments that represent the complete sequence of expressed genes from a specific tissue, cell type or developmental stage.

D. RNA fragments that represent the coding regions of expressed genes from a specific tissue, cell type or developmental stage.

Answer: B 20) How can an amino acid sequence be used to design a gene-specific hybridization probe?

A. Synthesize a portion of a polypeptide chain based on the amino acid sequence of the full protein, and use this polypeptide chain as a probe.

B. Purify a protein that contains this amino acid sequence, digest it with proteases that cleave at specific sites, and use one of these peptide fragments as a probe.

C. Synthesize all possible nucleotide sequences long enough to encode the amino acid sequence and thereby find one sequence that will encode the amino acid sequence.

D. Synthesize a set of nucleotide sequences that could encode the amino acid sequence, and this set of oligonucleotides as a probe.

Answer: D 21) What DNA sequence information is essential for the success of a polymerase chain reaction (PCR)?

A. The sequence of the ends of the DNA to be amplified must be known.

B. The complete sequence of the DNA to be amplified must be known.

C. The sequence of restriction endonuclease recognition sites in the DNA to be amplified must be known.

D. The sequence of restriction endonuclease recognition sites in the DNA to be amplified and in the plasmid where the amplified DNA fragment will be cloned must be known.

Answer: A 22) Which of the following sequences is in the correct order for one cycle of PCR? A. Denature DNA; primer extension, anneal primers.

B. Primer extension; anneal primers, denature DNA. C. Anneal primers; denature DNA; primer extension.

D. Denature DNA; anneal primers; primer extension.

Answer: D 23) In a single PCR cycle consisting of 30 seconds at 95C, 1 min at 55C, and 1 min at 72C, what is happening in the step run at 55C? A. The DNA to be amplified is being denatured.

B. Primers are being denatured.

C. DNA polymerase is extending new DNA from the primers.

D. Primers are annealing to the DNA to be amplified.

Answer: D 24) Which of the following enzymes is involved in the PCR process?

A, DNA ligase B. reverse transcriptase C. DNA polymerase D. DNA primase

Answer: C 25) Phosphorus-32 (32P) is commonly used for labeling A. dNTPs and NTPs B. amino acids C. nitrogenous bases D. chloramphenicol Answer: A 26) What does it mean when some bands on an autoradiograph are darker than others?

A. The concentration of the labeled macromolecule in that band is higher.

B. The concentration of the labeled macromolecule in that band is lower.

C. The experiment failed because the bands should be of the same intensity throughout the autoradiograph.

D. The darker bands represent larger labeled macromolecules.

Answer: A 27) To prepare a hybridization probe for screening a library, which labeling technique would be most desirable?

A. Klenow fill-in B. 5′ end-labeling C. random primed labeling D. 3′ end-labeling Answer: C 28) The appropriate hybridization temperature during library screening is calculated according to the G + C content and the percent homology of the probe to target. The more mismatches there are: A. the higher the melting temperature (Tm) will be.

B. the lower the melting temperature (Tm) will be.

C. the more stable the heteroduplex will be.

D. the more stringent the wash should be.

Answer: B 29) You have screened a library by nucleic acid hybridization with a probe for a particular gene sequence. In the final phase of library screening, you apply an X-ray film to the membrane.

The resulting autoradiogram has two black spots corresponding to two separate bacterial colonies. Which of the following explanations is the most likely?

A. The gene is large and fragmented between two different clones.

B. One of the spots must be an artifact because only one bacterial colony should contain the gene of interest.

C. The library screening was a failure because there should be blue and white colonies on the autoradiogram.

D. The temperature at which the high stringency washes was performed was too high.

Answer: A 30) What is the difference between a restriction endonuclease recognition site and a RFLP? A. Restriction endonuclease recognition sites are any sequence of DNA recognized by a restriction endonuclease; RFLPs are any sequence of RNA amplified by reverse transcriptase and PCR. B. Restriction endonuclease recognition sites produce DNA fragments that differ in size

when the source of the DNA is from different chromosomes; RFLPs are any sequence of RNA amplified by reverse transcriptase and PCR. C. Restriction endonuclease recognition sites produce DNA fragments that differ in size when the source of the DNA is from different chromosomes; RFLPs are any sequence of DNA recognized by a restriction endonuclease.

D. Restriction endonuclease recognition sites are any sequence of DNA recognized by a restriction endonuclease; RFLPs are DNA fragments produced by restriction endonuclease digestion that differ in size when DNA sequences differ between chromosomes or individuals.

Answer: D 31) The basic idea of using RFLPs as markers of genetic diseases is to A. calculate the frequency of a RFLP allele and a disease phenotype.

B. determine that a particular locus gives restriction endonuclease digestion fragments of different sizes.

C. establish close linkage between a RFLP allele and a disease-associated allele.

D. generate a genetic map using biochemical markers, such as the presence or absence or enzyme activity, and correlate these markers with the presence or absence of a disease.

Answer: C 32) What does it mean to say that a genetic marker and a disease gene are closely linked?

A. The genetic marker lies within the coding region for the disease gene.

B. The genetic marker and the disease gene are in close physical proximity on the same chromosome and tend to be inherited together.

C. The genetic marker and the disease gene are on different chromosomes.

D. The DNA sequence of the genetic marker and the DNA sequence of the disease gene are nearly identical.

Answer: B 33) When performing gel electrophoresis, why do molecular biologists add a “ladder” made up of nucleic acids or proteins of known size?

A. To help determine the size of each macromolecule after the separation has occurred.

B. To help macromolecules move through the gel more rapidly.

C. To make sure the dye included in the sample buffer doesn't interfere with the movement of the macromolecules that are being studied.

D. The macromolecules being studied would be invisible without the ladder.

Answer: A

34) Which diagram (I or II) shows the correct drawing of gel electrophoresis of DNA fragments of differing size?

I Positive electrode

II Larger DNA fragment

Positive electrode

Smaller DNA fragment Negative electrode

Smaller DNA fragment

Larger DNA fragment Negative electrode

A. I B. II Answer: B

35) To analyze the results of a DNA sequencing reaction, which type of gel should you prepare?

A. denaturing polyacrylamide gel B. agarose gel C. pulsed field gel D. nondenaturing polyacrylamide gel Answer: A 36) Which step in Southern blotting allows an investigator to identify a particular DNA fragment that is present within a complex mixture of fragments?

A. restriction endonuclease digestion of DNA B. agarose gel electrophoresis C. blotting D. hybridization with a labeled probe Answer: D

37) Pedigree analysis was used to determine whether a genetic marker is inherited with a disease trait (black circle). Results of Southern blot analysis of restriction endonucleasedigested DNA from members of the family are shown. What conclusions can be drawn from the data?

A. A restriction fragment length polymorphism (RFLP) is linked to this autosomal recessive trait.

B. A restriction fragment length polymorphism (RFLP) is linked to this autosomal dominant trait.

C. The disease is caused by expansion of a trinucleotide repeat.

D. No information can be gained from this analysis because researchers should have performed a Northern blot, not a Southern blot.

Answer: A 38) Pedigree analysis was used to determine whether a genetic marker is inherited with a disease trait (black symbols). Results of Southern blot analysis of restriction endonucleasedigested DNA from members of the family are shown. What conclusions can be drawn from the data?

A. A restriction fragment length polymorphism (RFLP) is linked to this autosomal recessive trait.

B. A restriction fragment length polymorphism (RFLP) is linked to this autosomal dominant trait.

C. The disease is caused by expansion of a trinucleotide repeat.

D. No information can be gained from this analysis because researchers should have performed a Northern blot, not a Southern blot.

Answer: B

39) Dideoxy DNA sequencing is based on chain termination with dideoxynucleotides. Why are normal deoxynucleotides also included in the reaction?

A. to act as a primer for chain termination by the dideoxynucleotides B. to provide a template for DNA polymerase C. to allow production of a range of synthesis products of different lengths.

D. to provide a 2′-OH required for incorporation of the dideoxynucleotides Answer: C 40) It is common to use ddNTPs (dideoxynucleoside triphosphates) for sequencing DNA because, when incorporated into a growing DNA polymer, the ddNTP acts as a chain terminator, allowing the researcher an opportunity to determine the last-added base to a given chain. From your knowledge of nucleotide structure and DNA synthesis, what is the most likely attribute of ddNTPs for DNA sequencing?

A. The ddNTPs lack a nitrogenous base.

B. The ddNTPs lack the sugar component.

C. The ddNTPs lack a 5′ carbon.

D. The ddNTPs lack a 3′ hydroxyl group.

Answer: D 41) The smallest fragment at the end of a DNA sequencing gel represents the A. 2′ end of the newly synthesized DNA. B. 3′ end of the newly synthesized DNA. C. 5′ end of the newly synthesized DNA. D. either end of newly synthesized DNA, depending on how the sample was loaded on the gel.

Answer: C 42) The sequence read from a DNA sequencing gel is A. the same sequence as the original strand.

B. complementary to the original strand.

C. the same polarity as the original strand.

D. both B and C. Answer: B 43) Automated DNA sequencing differs from manual DNA sequence in which way?

A. Instead of having to run four separate sequencing reactions, in automated sequencing the reactions can be combined into one tube.

B. ddNTPs are not used in automated DNA sequencing.

C. Automated DNA sequencing uses radioactive markers.

D. In automated DNA sequencing, DNA fragments are separated by sequence instead of by size.

Answer: A 44) Which of the following statements regarding 454 pyrosequencing is not true?

A. Each DNA fragment to be sequenced must be clonally amplified.

B. Dideoxy nucleotides are used to reversibly terminate chain elongation.

C. Nucleotide incorporation is detected as light generated by the luciferase enzyme.

D. Many sequencing reactions are carried in in parallel in a multi-well plate.

Answer: B

45) Pulse-field electrophoresis is useful for separating A. small single-stranded RNAs B. very large DNA molecules C. degenerate oligonucleotide probes D. DNA-RNA hybrid duplexes Answer: B 46) To read an old-fashioned Sanger sequencing gel in the 5’ to 3’ direction, on must read the gel A. from top to bottom.

B. from bottom to top.

C. from left to right.

D. from right to left.

Answer: B 47) The polymerase chain reaction does not require A. dNTPs B. DNA ligase C. DNA polymerase D. primers Answer: B

Short answer/analytical 48) Define the term “restriction endonuclease” and explain the function of restriction endonucleases in bacteria.

Answer: Section 8.2, p. 186-187 49) Describe some of the key experiments and discoveries leading to the development of recombinant DNA technology and the first cloning experiments.

Answer: Section 8.2, p. 186-189 50) When restriction endonucleases first bind to DNA they random walk, slide, hop, and jump.

Why don’t they immediately get on with their business of cutting DNA? Answer: Section 8.3, p. 191-192, Fig. 8.4 51) Why is sulfur-35 (35S) used to label proteins but not nucleic acids?

Answer: Section 8.5, Tool Box 8.4. The amino acids cysteine and methionine contain sulfur and thus can incorporate a 35S atom into a protein, whereas nucleic acids do not contain sulfur.

52) Explain the terms “random walk” and “coupling” in the context of the mode of action of restriction endonucleases.

Answer: Section 8.3, p. 186-89, p. 191-192, Fig. 8.4 53) Describe X-ray crystallographic evidence for the mode of action of EcoRI. Answer: Focus Box 8.2, p. 193

54) You want to join together two pieces of DNA that have been cut with SmaI. Outline a strategy you could use to increase the efficiency of blunt end ligation.

Answer: Section 8.3, p. 194-195, Fig. 8.5 55) Why do foreign DNAs need to be inserted into vectors to clone them?

Answer: Section 8.4, p. 195 Without a vector, DNA will not be replicated in a host cell.

56) You want to clone a 1.5 kb cDNA. Which vectors discussed would be appropriate to use?

Which would be inappropriate? Explain your answer.

Answer: Section 8.4, p. 195-196, Table 8.2 57) You want to make a genomic library with DNA fragments averaging about 85 kb in length.

Which vectors discussed would be appropriate to use? Which would be inappropriate? Explain your answer.

Answer: Section 8.4, p. 195-196, Table 8.2 58) Describe the process of cloning a DNA fragment into the multiple cloning site of the vector pUC18. How would you screen for clones that contain an insert?

Answer: Section 8.4, p. 196-197, Fig. 8.6 59) Explain the underlying principle between blue-white colony screening.

Answer: Section 8.4, p. 1197-199, Figs. 8.7 and 8.8 60) Describe the process of cloning a DNA fragment into bacteriophage lambda. How would you screen for clones that contain an insert?

Answer: Section 8.4, p. 201, Fig. 8.9 61) Describe the process of cloning a DNA fragment into a YAC. How would you screen for clones that contain an insert?

Answer: Section 8.4, p. 201-203, Fig. 8.10 62) Compare and contrast genomic libraries with cDNA libraries.

Answer: Section 8.4, p. 206 63) Describe the principle of ion-exchange chromatography. Describe the behavior of a positively charged protein and a negatively charged protein in a column containing positively charged resin during sample loading and elution.

Answer: Tool Box 8.1, p. 199-200 64) Describe the principle of gel filtration chromatography. For two proteins of different size, illustrate in which fraction they will elute.

Answer: Tool Box 8.1, p. 199-200 65) Describe the principle of antibody-affinity chromatography. For one protein that is recognized by the antibody and a second protein that is not, illustrate at which point each protein will be released from the column.

Answer: Tool Box 8.1, p. 199-200 66) Diagram the process for creating a double-stranded cDNA. Answer: Tool Box 8.2, p. 203-204

67) Explain why DNA polymerase from a thermophilic bacterium is used in PCR instead of DNA polymerase I from E. coli.

Answer: Tool Box 8.3, p. 205-206 68) Outline the polymerase chain reaction (PCR) method for amplifying a given sequence of DNA. Answer: Tool Box 8.3, p. 205-206 69) Compare and contrast the principles of autoradiography and phosphorimaging.

Answer: Tool Box 8.4, p. 210 70) What are radioisotopes and why are they useful in molecular biology research?

Answer: Tool Box 8.4, p. 210-212 71) Define the term “probe” and describe two important characteristics of a probe.

Answer: Section 85, p. 207 The probe (1) must have a sequence similar (complementary) to that of the nucleic acid to be detected and (2) must be modified in a way to allow detection.

72) What are the relative advantages and disadvantages of end-labeling versus internal-labeling a probe?

Answer: Tool Box 8.5, p. 211-212 Internal labeling provides maximal signal from the probe, making detection more efficient, whereas end-labeling must be used for methods that require the definition one of the end of the probe (e.g., DNase I footprinting).

73) (a) Outline the steps that would be necessary to generate a cDNA library from all of the genes expressed by adenocarcinoma cells of a breast tumor. (b) How could tumor-specific transcripts be identified from this library? (Hint: think about how you could use non-cancerous breast epithelial cells as a comparison.)

Answer: (a) Toolbox 8.2, p. 203-204 (1) Isolate mRNAs from cells of the tumor line using oligo-dT beads.

(2) Convert the mRNAs to cDNAs using an oligo-dT primer and reverse transcriptase.

(3) Digest the RNAs using a ribonuclease.

(4) Carry out second strand synthesis with Klenow.

(5) Ligase an adaptor with “sticky” ends complementary to a restriction site in the multiple cloning site a plasmid vector.

(6) Digest a plasmid vector with the restriction enzyme to generate compatible “sticky” ends (7) Ligate the cDNA fragments to the plasmid. This is the library. (b) The answer to (b) requires students to think their way to the approach of “subtractive” analysis, in which the genes expressed by non-cancerous cells are “subtracted” from the genes expressed by cancer cells to reveal what is different about them.

(1) Transform the tumor cDNA library into E. coli cells, and transfer cells to a nitrocellulose membrane.

(2) Prepare a labeled cDNA library from normal breast epithelial cells and probe the nitrocellulose membrane with the labeled cDNA library.

(3) Colonies that are not recognized by the normal epithelial cell library cDNAs contain putative

cancer-specific genes. Isolate the plasmid from these colonies and sequence the inserts to identify the genes.

74) Compare and contrast degenerate probes and EST-based probes.

Answer: Section 8.5, p. 208-209 Both are used to detect DNA clones when a protein sequence is known. For degenerate probes, all possible oligonucleotide combinations that could encode a known amino acid sequence are synthesized (based on the degeneracy of the genetic code).

For EST-based probes, small fragments of RNA that correspond to the known protein sequence are used to generate a single oligonucleotide probe.

75) Diagram a strategy for screening a cDNA library.

Answer: Section 8.5, p. 212-214, Fig. 8.13 76) If you had an antibody to the protein encoded by an unknown gene of interest, what type of library could you screen to identify the sequence of the gene? Using a diagram, explain how this would work.

Answer: Section 8.5, p. 214 77) Show how to use restriction mapping to determine the orientation of a restriction fragment ligated into a restriction site in a vector.

Answer: Section 8.6, p. 214, and Fig. 8.14, p. 216 78) Diagram the process of Southern blotting and probing to detect a DNA of interest. What kinds of information can be obtained from a Southern blot?

Answer: Tool Box 8.7, p. 218-219 79) Compare and contrast the key characteristics of automated sequencing using the Sanger method and 454 pyrosequencing.

Answer: Section 8.7, p. 220-223, Figs. 8.16 and 8.18 80) You have attempted to ligate a 1.5 kb fragment of foreign DNA into the EcoRI site in the multiple cloning site of the 4.0 kb plasmid vector shown below: Ampicillin resistance gene

HindIII LacZ gene

EcoRI

SmaI Multiple cloning site

(a) After ligation you use the DNA in the ligation mixture to transform host bacteria. Why is it important to use host bacteria that are deficient for restriction-modification? (b) You screen the bacteria that supposedly have been transformed with recombinant plasmid DNA. Some of the bacterial colonies growing on the nutrient agar plate that contains ampicillin and X-gal are white and some are blue. Explain these results.

(c) To confirm the presence of the foreign DNA insert, you perform EcoRI restriction endonuclease digests on DNA extracted from bacterial colonies. On the diagram of an agarose gel shown below, draw the pattern of bands (label their size in kb) you would expect to see for EcoRI-digested recombinant plasmid and EcoRI-digested non-recombinant plasmid vector, after electrophoresis and staining of the gel with ethidium bromide.

Answer: (a) Section 8.2 p. 186. If the host bacteria had an intact restriction modification system, then the unmethylated plasmid DNA would be recognized as foreign DNA and cleaved by host restriction endonucleases. (b) Section 8.4, pp. 197-199, Fig. 8.7. Bacteria are plated on selective agar medium containing the antibiotic ampicillin and the substrate X-gal. If foreign DNA is inserted into the multiple cloning site, then the lacZ′ coding region is disrupted and the N-terminal portion of galactosidase is not produced. Since there is no functional -galactosidase in the bacteria, the substrate X-gal remains colorless, and the bacterial colony containing recombinant plasmid DNA appears white, thus allowing the direct identification of colonies carrying cloned DNA inserts. If there is no insertion of foreign DNA in the multiple cloning site, then the lacZ′ gene is intact and enzymatically active -galactosidase is produced. The bacterial colonies containing nonrecombinant plasmid DNA thus appear blue. (c) Section 8.6, Tool Box 8.6, pp. 215-216, Fig. 8.14, p. 216; also see below:

EcoRIdigested recombinant plasmid

EcoRIdigested nonrecombinant plasmid − electrode

4.0 kb

Direction of migration 1.5 kb

+ electrode

81) A chromotography column in which oligo-dT is linked to an inert substance is useful in separating eukaryotic mRNA from other RNA molecules. On what principle does this column operate?

Answer: See Section 8.5, p. 207; Section 8.4, Tool Box 8.1, pp. 199-200; Section 8.4, Tool Box 8.2, pp. 203-204. Most eukaryotic mRNAs are polyadenylated at the 3′ end to form a poly(A) tail. The poly(A)region can be used to selectively isolate mRNA from total RNA by the principles of complementary base-pairing and hybridization of nucleic acids. Under conditions of relatively high salt the poly(A) RNA is retained by formation of hydrogen bonds between

adenine and thymine, and the RNA lacking a poly(A) tail flows through. The poly(A) mRNA is eluted from the column in low salt buffer which promotes denaturation of the hybrid.

82) Starting with the nucleotide sequence of the human DNA ligase I gene, describe how you would search for a homologous gene in another organism whose genome has been sequenced, such as the pufferfish Tetraodon nigroviridis. Then, describe how you would obtain the protein and test it for ligase activity.

Answer: For various approaches, see Section 8.5, p. 207, p. 214, Section 8.4, Tool Box 8.3, p. 204; Section 8.4 pp. 195-201; Section 8.3, p. 194. There are many possible approaches. The nucleotide sequence of the human DNA ligase I gene could be used as a probe to screen a pufferfish cDNA library. Alternatively, computer database searches could be used to find the homologous pufferfish sequence. This sequence could be used to design primers to isolate the pufferfish coding region by polymerase chain reaction (PCR). The PCR product could be inserted into an appropriate plasmid expression vector, and the protein could be expressed by in vitro transcription and translation and then tested for ligase activity. If the protein has ligase activity, it should be able to join linear fragments of DNA. Whether two fragment of know size were joined together or remained as separate fragment could be analyzed by agarose gel electrophoresis.

83) You plan to use the polymerase chain reaction to amplify part of the DNA sequence shown below, using oligonucleotide primers that are hexamers matching the regions shown in red. (In practice, hexamers are too short for most purposes). State the sequence of the primer oligonucleotides that should be used, including their polarity (5′3′), and give the sequence of the DNA molecule that results from amplification.

5′-TAGGCATGCAATGGTAATTTTTCAGGAACCAGGGCCCTTAAGCCGTAGGCAT-3′ 3′-ATCGGTACGTTACCATTAAAAACTCCTTGGTCCCGGGAATTCGGCATCGGTA-5′ Answer: Section 8.4, Tool Box 8.3, pp. 205-206; Primers: 5’-CGGCTT-3’ and 5’-GCAATG-3’ Amplified sequence: 5′-GCAATGGTAATTTTTCAGGAACCAGGGCCCTTAAGCCG-3′ 3′-CGTTACCATTAAAAACTCCTTGGTCCCGGGAATTCGGC-5′ 84) The following is a physical map of a region you are mapping for RFLP analysis:

3 kb

1

1 kb

2 kb

2

3

4

The numbered vertical lines represent restriction sites recognized by SmaI. The red sites (2 and

3) are polymorphic, the others are not. You cut the DNA with SmaI, electrophorese the fragments, and Southern blot them to a membrane. You have a choice of two probes that recognize the DNA regions shown above: probe A, green line; probe B, blue line. (a) Explain which probe you would use for analysis and why the other choice would be unsuitable. (b) Give the sizes of bands you will detect in individuals homozygous for the following genotypes with respect to sites 2 and 3.

Haplotype A B C D

Site 2 Present Present Absent Absent

Site 3 Present Absent Present Absent

Answer: See Section 8.6 pp. 214-219, Fig 8.15; Tool Box 8.7, p. 218-219. (a) Use probe B because it has regions of complementarity to all potential restriction fragments.

Probe A would not hybridize to all of the restriction fragments. (b) Haplotype A: 3 kb, 2 kb, 1 kb Haplotype B: 3 kb Haplotype C: 5 kb, 1 kb Haplotype D: 6 kb 85) Complete the incomplete diagrams below to show the key structural difference between an NTP, dNTP, and ddNTP (e.g., CTP, dCTP, ddCTP).

P

P

P

Base

Base

Base

e

ee

e

(a) What do “d” and “dd” stand for? (b) Explain why ddNTPs are called "chain terminators" in DNA sequencing reactions.

Answer: Section 8.7, pp. 220-222, Fig. 8.16; also see below:

OH

OH

OH

H

H

H

(a) “d” stands for “deoxy” meaning that the sugar lacks a 2′-OH. “dd” stands for “dideoxy”

meaning that both the 3′-OH and 2′-OH are missing. (b) Because ddNTPs lack the 3′-OH group, they cannot form a phosphodiester bond with another nucleotide. Thus, each sequencing reaction proceeds until a “chain terminator” is added and stops DNA replication.

86) The nucleotide sequence of a DNA fragment was determined by the Sanger (dideoxy) DNA sequencing method. The data are shown below. What is the 5′ to 3′ sequence of the nucleotides in the original DNA fragment? dGTP dATP + ddATP dCTP dTTP

dGTP + ddGTP dATP dCTP dTTP

dGTP dATP dCTP + ddCTP dTTP

dGTP dATP dCTP dTTP + ddTTP

Answer: The nucleotide sequence of the original DNA fragment is: 5′-TAGCCAGACCTT-3′.

Section 8.7, pp. 220-222, Fig. 8.16

3′ 5′ A→T T→A C→G G→C G→C T→A C→G T→A G→C G→C A→T A→T 5′ 3′

87) The following DNA ends were generated by some common restriction enzymes.

The dashes indicate the presence of additional nucleotides. Which of these ends could be labeled using Klenow? (There may be more than one.) Explain your answer.

EcoRI:

5’ ----G 3’ ----CTTAA

SacI

5’ ----GAGCT 3’ ----C

BamHI:

5’ ----G 3’ ----CCTAG

SmaI:

5’ ----GGG 3’ ----CCC

Answer: See Section 8.5, Tool Box 8.5, p. 211-212. The ends generated by EcoRI and BamHI could be labeled with Klenow because these contain free 3’ hydroxyl groups and 3’ overhands that can serve as a template for Klenow-mediated addition of labeled nucleotides. In the case of the SacI- and SmaI-generated ends, there are no nucleotides in a 3’ overhang to template synthesis. Note that the BamHI-generated fragment could be labeled with 32P-dCTP (because it contains a G in it’s 3’ overhang), whereas the EcoRI fragment could not. Both could be labeled with 32P-dATP.

Fundamental Molecular Biology, 2nd Edition Instructor Test Bank Lizabeth A. Allison

Chapter 9 Tools for Analyzing Gene Expression Multiple Choice 1) During a transient transfection assay A. DNA is introduced into a eukaryotic cell and remains extrachromosomal.

B. DNA is introduced into a eukaryotic cell and is stably integrated into a chromosome.

C. DNA is introduced into bacterial cells for amplification.

D. DNA is taken up and packaged in viral capsids.

Answer: A 2) You want to study regulatory regions that control expression of a gene whose protein product is difficult to measure quantitatively. What method should you use to analyze gene expression?

A. Western blot with an antibody specific for the protein of interest B. Reporter gene assay C. DNAse I footprinting D. Coimmunoprecipitation assay Answer: B 3) The technique of in situ hybridization gives information about: A. which genes are being transcribed in a particular cell B. which genes are present in a particular cell C. which proteins are being degraded in a particular cell D. which enzymes are active in a particular cell Answer: A 4) Western blotting is the process that is used to identify A. DNA B. RNA C. protein D. DNA-protein interactions Answer: C 5) Northern blotting is the process that is used to identify A. DNA B. RNA C. protein D. DNA-protein interactions Answer: B 6) Southern blotting is the process that is used to identify A. DNA B. RNA C. protein D. DNA-protein interactions Answer: A

7) To map the start of transcription for a given mRNA, the most useful method would be A. RNase protection assay B. RT-PCR C. Northern blot D. in situ hybridization Answer: A 8) A coimmunoprecipitation assay is a process that is used to identify A. protein structure B. transcriptional activity C. DNA-protein interactions D. protein-protein interactions Answer: D 9) A chromatin immunoprecipitation assay is a process that is used to identify A. protein structure B. transcriptional activity C. DNA-protein interactions D. protein-protein interactions Answer: C 10) You have received from a scientific collaborator some transgenic mice that have had the gene encoding the BRCA1 gene knocked out. Which method would help you to confirm the genotype of these mice most directly?

A. Southern blot B. Northern blot C. Western blot D. Chromatin immunoprecipitation Answer: A 11) You are interested in an uncharacterized gene that contains an interesting combination of predicted protein domains. To determine in which mouse tissues it is expressed, the best method is A. DNase I footprinting B. transient transfection assay C. Northern blot D. coimmunoprecipitation Answer: C 12) You have identified a specific region in a virus that you suspect is bound by the host transcription factor Sp1. Which of the following methods could you use to test your hypothesis?

A. chromatin immunoprecipitation B. electromobility shift assay C. DNase I footprinting D. all of the above Answer: D

13) Which of the following reporter genes allows one to visualize a protein in living cells?

A. beta-galactosidase (-Gal) B. green fluorescent protein (GFP) C. chloramphenicol acetyltransferase (CAT) D. luciferase (Luc) Answer: B 14) To analyze whether a putative transcriptional enhancer activates transcription, one could use which of the following approaches?

A. attach the enhancer to a reporter gene in a recombinant construct; transiently transfect the construct into cells B. attach a histidine tag to the enhancer sequence in a recombinant construct; purify the complex using nickel–nitrilotriacetic acid (Ni-NTA) resin C. co-transfect the enhancer sequence with a plasmid encoding RNA polymerase; perform in vitro transcription D. use in vitro mutagenesis to disrupt the enhancer; immunoprecipitate the enhancer Answer: A 15) Which of the following is not a property of green fluorescent protein (GFP)?

A. GFP can be detected directly in living cells.

B. GFP requires addition of a cofactor for light production.

C. GFP contains an intrinsic peptide fluorophore.

D. Following excitation with ultraviolet light, GFP emits green light.

Answer: B 16) You are analyzing the tissue-specific expression of a poorly characterized gene in mice.

Which type of gene would serve as the best positive control for your experiment?

A. a known gene that exhibits temporal expression B. a known gene that exhibits spatial expression C. a known gene that exhibits constitutive expression D. another poorly characterized or unknown gene Answer: C 17) Production of recombinant proteins in bacteria involves the cloning of the cDNA encoding the desired protein into an “expression vector” that contains a(n) A. promoter B. enhancer C. intron D. start codon Answer: A 18) Which method would best help you to identify specific amino acid residues that are critical for the function of a protein?

A. promoter bashing B. yeast two-hybrid assay C. alanine scanning mutagenesis D. quantitative PCR Answer: C

19) Which component is not required for in vitro translation of a recombinant protein?

A. T7 RNA polymerase B. amino acids C. rabbit reticulocyte lysate D. DNA polymerase Answer: D 20) The distance moved by a protein into a gel is inversely proportional to the A. logarithm of molecular charge B. logarithm of molecular weight C. logarithm of molecular shape D. Both A and C Answer: B 21) At its isoelectric point A. a protein has a positive charge B. a protein has a negative charge C. a protein has no net charge D. whether a protein has a positive or negative charge depends on the particular protein Answer: C 22) During the process of two-dimensional PAGE A. the first dimension separates proteins by isoelectric point and the second dimension separates them by size.

B. the first dimension separates proteins by size and the second dimension separates them by isoelectric point.

C. the first dimension separates proteins by size and the second dimension separates them by shape.

D. the first dimension separates proteins by shape and the second dimension separates by size.

Answer: A 23) In an indirect immunofluorescence assay, which component is fluorescently labeled?

A. The antigen.

B. The primary antibody.

C. The secondary antibody.

D. Both B and C. Answer: C 24) Morpholino oligonucleotides have proved useful for A. RNA interference (RNAi) B. antisense-mediated inhibition of gene expression C. reverse transcriptase-PCR D. RNase protection assays Answer: B

25) Double-stranded RNAs are processed by the enzyme _______ to form small interfering RNAs (siRNAs). One strand of the siRNA is loaded into a silencing complex named _________. The siRNAs target mRNA for ___________ .

A. RISC, Dicer, degradation B. Dicer, RISC, degradation C. Dicer, RISC, splicing D. RISC, Dicer, spicing Answer: B 26) You want to analyze the structure of a large protein complex at extremely high magnification. What method would you use?

A. confocal microscopy B. fluorescence microscopy C. cryoelectron microscopy D. scanning electron microscopy Answer: C 27) You are testing protein X and protein Y in a yeast two-hybrid assay that uses lacZ as a reporter gene. Production of a blue product signifies that: A. protein X interacts with protein Y B. protein X does not interact with protein Y C. protein X is required for production of protein Y D. protein Y is required for production of protein X Answer: A 28) Which of the following has not been used as a model organism in molecular biology research?

A. Danio rerio B. Caenorhabditis elegans C. Arabidopsis D. Homo sapiens Answer: D 29) Which of the following is a key advantage of NMR spectroscopy over X-ray crystallography?

A. NMR spectroscopy can accommodate larger sample sizes B. NMR spectroscopy can be under dynamic conditions C. NMR spectroscopy provides higher resolution data D. Only NMR spectroscopy can be used to study DNA-protein complexes at high resolution Answer: B

Short answer/analytical 30) A nonscientist friend of yours asks how findings in a worm or fly can be relevant to human biology. Explain to your friend the importance of model organisms in molecular biology research.

Answer: Focus box 9.1, p. 228-229.

31) Explain the information gained from an in situ analysis differs from that gained from an in vitro analysis, using protein or RNA expression as an example to illustrate your answer.

Answer: Section 9.5, p. 244 or Section 9.6, p. 249. An in situ analysis allows one to qualitatively assess the presence and location of a molecule within a tissue. For example, the expression of an mRNA might be restricted to one cell type in a complex tissue. On the other hand, in vitro analysis allows one to make more precise measurements of a molecule. For example, one could quantitatively compare the amount of RNA between two tissues and measure the size of the mRNAs. 32) Compare and contrast transient and stable transfection assays.

Answer: Section 9.2, p. 230, Fig. 9.1 33) Why is green fluorescent protein widely used as a reporter gene for investigation of tissuespecific gene expression and cellular localization or proteins?

Answer: Section 9.3, p. 235-237, Figs. 9.5 and 9.6 34) Describe the structure of green fluorescent protein and discuss how this structure relates to the function of the protein.

Answer: Section 9.3, p. 235-237, Fig. 9.5 35) Diagram a cotransfection assay. What information can be obtained from this type of assay?

Answer: Section 9.3, p. 232-233, Fig. 9.3 36) Describe the use of a vector that produces fusion proteins with glutathione-S-transferase (GST) at one end. Show the protein purification scheme to illustrate the advantage of the GST tag.

Answer: Section 9.3, p. 234, Fig. 9.4 37) Explain the principle of site-directed mutagenesis, then describe a method to carry out this process.

Answer: Section 9.4, p. 242, Fig. 9.7 38) You want to generate “optical sections” of a specimen. To this end, would you use conventional fluorescence microscopy or confocal microscopy? Explain your answer.

Answer: Section 9.3, Tool Box 9.2, p. 239 39) What kinds of information can be obtained from a Northern blot? Compare this with the information obtained from a Southern blot.

Answer: Section 9.5, p. 244, Fig. 9.8; Section 8.6, Tool Box 8.7, pp. 218-219.

A Northern blot gives information on gene expression, including how much mRNA is present and the size of the transcript. A Southern blot gives information on genomic DNA. 40) Describe fluorescence in situ hybridization (FISH). When would you use this method, rather than Northern blotting?

Answer: Section 9.5, p. 244, Fig. 9.8 41) What is the difference between reverse transcriptase PCR (RT-PCR) and standard PCR? For what purpose would you use RT-PCR? Answer: Section 9.5, p. 245, Fig. 9.8 42) Design an experiment to quantify transcript levels for a particular mRNA using RNase protection assay (RPA). Show sample results.

Answer: Section 9.5, p. 244-245, Fig. 9.8

43) Explain the molecular basis of an RNase protection assay. How can this method be used to map the transcription start site for a gene?

Answer: Section 9.5, p. 244-245, Fig. 9.8 A DNA/RNA probe that spans the region of the start of transcription can be annealed to the mRNA, followed by treatment with a single-stranded specific nuclease. The resulting digested probe has a 3’ end that corresponds to the precise 5’ end of the mRNA. (The sequence at this position can be determined by directly sequencing the probe fragment, or it can be inferred from the size of the probe fragment.)

44) What kinds of information can be obtained from a Western blot? Diagram the key steps in the method.

Answer: Section 9.6, p. 246-249, Fig. 9.10 45) What is SDS? What are its functions in SDS-PAGE? Answer: Section 9.6, Tool Box 9.3, p. 246-247 46) Explain the molecular basis for protein separation by isoelectric focusing (the first step of 2D-PAGE).

Answer: Section 9.6, Tool Box 9.3, pp. 246-247 47) You have used a polyclonal primary antibody made in rabbit to detect your protein of interest on Western blot. Give an example of an appropriate secondary antibody to use for detection. Explain your choice.

Answer: Section 9.7, Tool Box 9.4, p. 250-251 48) Compare and contrast the steps in production of polyclonal and monoclonal antibodies.

Answer: Section 9.7, Tool Box 9.4, p. 250-251 49) Diagram the key steps in a “sandwich” ELISA. What information can be obtained from this type of assay?

Answer: Section 9.6, p. 249, Fig. 9.10 50) Compare and contrast the electrophoretic mobility shift assay (EMSA) and DNase I footprinting methods for assaying specific DNA-protein interactions. What information does DNase footprinting provide that EMSA does not?

Answer: Section 9.8, p. 253, Fig. 9.13 51) You want to confirm the interaction of a particular protein with a specific sequence of DNA within the context of a living cell. What technique would you use and why?

Answer: Section 9.8, p. 253-255, Fig. 9.13 One would use chromatin immunoprecipitation because protein-DNA complexes are isolated from cells.

52) You want to confirm the interaction of a particular protein with another protein at a precise location within a cell. What technique would you use and why?

Answer: Section 9.9, p. 257, Fig. 9.14 53) Diagram the key steps in a GST pull-down assay. Is the GST-tagged protein the “bait” or the “prey”?

Answer: Section 9.9, p. 255-257, Fig. 9.14

54) In a yeast two-hybrid assay that uses lacZ as a reporter gene, what does the production of a blue product signify?

Answer: Section 9.9, p. 255-257, Fig. 9.14 55) Compare and contrast the type of information that can be obtained from X-ray crystallography with nuclear magnetic resonance (NMR) spectroscopy.

Answer: Section 9.10, p. 257-259, Fig. 9.15 56) Compare and contrast the type of information that can be obtained from cryoelectron microscopy with atomic force microscopy.

Answer: Section 9.10, p. 259, Fig. 9.15 57) You have purified a protein. When you subject it to SDS-PAGE, two bands are seen.

Provide a possible explanation and describe how you could test your hypothesis.

Answer: Section 9.6, Tool Box 9.3, pp. 246-247. The protein may be composed of two polypeptide subunits. When subject to SDS-PAGE, the protein would be denatured and the subunits dissociated by treatment with SDS, and any disulfide linkages in the protein would be broken by the reducing agent in the sample buffer (either 2-mercaptoethanol or dithiothreitol).

To test this hypothesis, electrophoresis could be performed under nondenaturing and nonreducing conditions. If only one band was present, this would suggest a two-subunit complex.

58) Consider a Northern blot: (a) Which strand of a DNA probe will hybridize to mRNA on the blot: the anti-sense strand, the sense strand, or both strands? (b) On the diagram of the DNA-RNA hybrid shown below, which bonds form when the DNA probe hybridizes to the mRNA?

Answer: Section 9.5, p. 244. (a) The antisense strand of a DNA probe is complementary to the mRNA and will hybridize to the mRNA on the blot.

(b) The hydrogen bonds (circled) form between the complementary base pairs when the DNA probe hybridizes to the mRNA.

59) Consider a cotransfection assay with expression vectors for two putative transcription factors that you suspect work together to activate transcription, and with a regulatory region of interest attached to a CAT reporter gene. You also test two deletion () mutants of the regulatory region. (a) How could you confirm that the transcription factors are transcribed and translated in the transfected cells? (b) Interpret the results of the CAT assay below. The plus and minus symbols indicate that presence or absence of expression vectors for the listed components in the transfection assay.

Transcription factor A Transcription factor B Full-length regulatory region (30 bp) Deletion mutant 1 (1-15) Deletion mutant 2 (16-30)

− − + − −

+ − + − −

− + + − −

+ + + − −

+ + − + −

+ + − − +

(c) How would you demonstrate protein-protein interaction between the two transcription factors in vivo and in vitro? Show sample positive results.

(d) Describe protocols to further characterize direct binding of the proteins to DNA in vitro and in vivo. Show sample results.

Answer: (a) Section 9.6, p. 247-249, Fig. 9.10. To confirm that the transcription factors are transcribed and translated in the transfected cells, you could perform a Western blot and probe it with antibodies specific for each transcription factor. (b) Section 9.3, p. 232, Fig. 9.2. In the absence of transcription factor A and transcription factor B there is no transcription of the CAT reporter gene. Thus, the enzyme chloramphenicol acetyltransferase (CAT) is not produced and the [14C]-chloramphenicol remains unacetylated (lower band on the autoradiogram). Likewise, there is no transcription when only one of the transcription factors is present. However, in the presence of both transcription factor A and transcription factor B, and the full-length regulatory region, there is transcription of the CAT reporter gene. The CAT enzyme is produced and some of the [14C]-chloramphenicol is acetylated (upper band). Deletion of the first 15 nucleotides of the regulatory region (Deletion mutant 1, 1-15) does not effect transcription, whereas deletion of the last 15 nucleotides (Deletion mutant 2, 16-30) abolishes transcription of the CAT reporter gene. These results suggest that transcription factor A and transcription factor B form a heterodimer that binds to nucleotides 16-30 of the regulatory region and thereby enhances transcription of the CAT reporter gene. (c) Section 9.9, p. 255-257, Fig. 9.14. A number of different methods could be used. For example, a GST-pull down assay could be used to show that the two transcription factors directly interact in vitro. A coimmunoprecipitation assay or yeast two-hybrid assay could be used to test for direct interaction in vivo. Sample positive results should appear similar to the results depicted in Fig. 9.14.. (d) Section 9.8, pp. 253-255, Fig. 9.13. A number of different methods could be used. Direct binding of the proteins to DNA in vitro could be tested by using an electrophoretic mobility shift assay (EMSA) or by DNase I footprinting. DNA-protein interactions in vivo could be tested using a chromatin immunoprecipitation (ChIP) assay. Sample positive results should appear similar to the results depicted in Fig. 9.13.

60) You have transfected siRNA against transcription factor A from Q. 59 into cells that normally express the transcription factor. (a) What would be the effect on CAT reporter gene expression? (b) How could you determine whether RNAi acts at the level of mRNA or translation in this case?

Answer: (a) Section 9.7, p. 252; Section 9.3, p. 232, Fig. 9.2. The siRNA would knockdown expression of transcription factor A, which is required for transcription of the CAT reporter gene.

Depending on the level of knockdown, CAT reporter gene expression would be expected to be significantly decreased. (b) Section 9.7, p. 252; Section 9.5, pp. 244-245; Section 9.6, p. 247-249, Fig. 9.10.

Quantitative reverse transcription (RT)-PCR or a Northern blot could be used to determine whether there is a decrease in transcription factor A mRNA levels. A Western blot could be used to determine whether there is a decrease in transcription factor A protein levels. If there was a decrease in mRNA levels and a corresponding decrease in protein levels, this would suggest that RNAi acts at the level of mRNA degradation. If there was no change in mRNA levels, but a decrease in the amount of protein present, this would suggest that RNAi acts at the level of translational repression.

Fundamental Molecular Biology, 2nd Edition Instructor Test Bank Lizabeth A. Allison

Chapter 10 Transcription in Bacteria Multiple Choice 1) A major difference between transcription in bacteria and eukaryotes is that A. transcription and translation are uncoupled in bacteria.

B. transcription and translation are coupled in bacteria.

C. there is no difference; transcription and translation are coupled in both bacteria and eukaryotes.

D. there are many additional levels of control of gene expression in bacteria.

Answer: B 2)Transcription is the process of A. formation of an RNA transcript complementary to the DNA template strand.

B. formation of an RNA transcript complementary to the DNA non-template strand.

C. formation of RNA transcripts complementary to both strands of DNA. D. formation of an RNA transcript complementary to the DNA sense (coding) strand.

Answer: A 3) Transcription proceeds in the A. 5′→ 3′ and 3′→ 5′ direction B. 3′→ 5′ direction only C. 5′→ 3′ direction only D. none of the above; 5′ and 3′ are only important for DNA replication Answer: C 4) RNA polymerase binds to a specific region on DNA called a(n) A. origin B. promoter C. sigma factor D. operon Answer: B 5) Which of the following statements is correct?

A. DNA polymerase requires a primer to get started.

B. RNA polymerase requires a primer to get started.

C. Both DNA and RNA polymerase require primers to get started.

D. Both DNA and RNA polymerase can start synthesis de novo.

Answer: A 6) Promoter strength A. refers to how rapidly transcript elongation occurs.

B. refers to the relative frequency of transcription initiation.

C. depends on the affinity of RNA polymerase for the promoter region.

D. both B and C Answer: D

7) Which of the following is almost always associated with high levels of transcription?

A. use of an alternative sigma factor B. the presence of an UP element in a promoter C. the presence of a Rho-independent terminator D. the linkage of genes in an operon Answer: B 8) Which statement is not true about the bacterial sigma factor?

A. Sigma is a regulatory protein, which is required for the initiation of transcription.

B. Sigma has the ability to join ribonucleotides (NTPs) by phosphodiester bonds using DNA as a template.

C. Sigma properly orients the RNA polymerase complex for transcription at the gene start site.

D. After initiation is complete, sigma does not dissociate completely; some domains are displaced and mRNA synthesis continues.

Answer: B 9) The TATA box (–10 or Pribnow box) has all the following characteristics except: A. It is both transcribed and translated.

B. It is upstream from most genes.

C. It is part of the binding site for the RNA polymerase.

D. It is part of a consensus (common) sequence of the promoter.

Answer: A 10) Rho-independent terminators have all of the following characteristics except: A. the consensus sequence is an inverted repeat B. stem-loop structures can form within the mRNA C. the repeat sequence is following by 7 to 8 uracil-containing nucleotides D. sigma factor binding to the terminator triggers transcript release Answer: D 11) Which statement best describes the role of ATP hydrolysis by the transcriptional terminator Rho? A. ATP hydrolysis induces a conformational change in the stem-loop structure of the terminator sequence.

B. ATP is used strictly to load Rho onto the mRNA. C. ATP fuels the travel of Rho along the RNA and dissociation of the RNA-DNA duplex.

D. ATP fuels the travel of Rho in the direction opposite of transcription, resulting in a headon collision between RNA polymerase and Rho. Answer: C 12) Which statement regarding bacterial Rho is not true?

A. Rho is a helicase.

B. Rho binds a specific C-rich sequence called a rut site.

C. Rho-dependent transcriptional termination depends on the ability of Rho to access the mRNA. D. Rho rides along with RNA polymerase until a termination sequence is encountered.

Answer: D

13) RNA polymerase forms a phosphodiester bond between A. two dNTPs (deoxyribonucleotide triphosphates).

B. two rNTPs (ribonucleotide triphosphates).

C. a phosphodiester bond between a dNTP and an rNTP. D. B and C. Answer: B 14) RNA polymerase has a _______ error rate because it has _______________ mechanism.

A. high, no proofreading B. low, proofreading C. high, proofreading D. low, no proofreading Answer: B 15) In which transcriptional complex(es) is the DNA duplex intact (not unwound)?

A. the closed complex B. the open complex C. the elongating complex D. A and B Answer: A 16) “Promoter clearance” refers to a step of transcription initiation in which A. activators and repressors dissociate from the promoter and sigma factor fully dissociates from RNA polymerase.

B. RNA polymerase unwinds the DNA at the start site of transcription.

C. RNA polymerase pauses before joining the first two nucleotides of the mRNA. D. a small RNA has been synthesized and some domains of sigma factor dissociate from RNA polymerase.

Answer: D 17) During transcription elongation, RNA polymerase A. generates positive supercoils (overwinding) ahead of itself.

B. generates positive supercoils (overwinding) behind itself.

C. prevents supercoiling by tracking around the DNA helix.

D. prevents supercoiling by generating nicks in the DNA helix.

Answer: A 18) The Jacob-Monod operon model A. led to the discovery of mRNA. B. proposed the existence of a repressor protein.

C. both A and B D. was later proved to be wrong.

Answer: C 19) Full induction of the lac operon occurs when: A. lactose levels are low and glucose levels are low B. lactose levels are low and glucose levels are high C. lactose levels are high and glucose levels are low D. lactose levels are high and glucose levels are high.

Answer: C

20) A compound known as X-gal is widely used in molecular biology research. When wild-type E. coli is grown on an agar plate containing X-gal, the bacterial colonies turn blue. In contrast, when lacZ mutants are grown on an agar plate containing X-gal, the bacterial colonies remain their normal white color. What is X-gal likely to be?

A. a compound chemically similar to glucose B. a compound chemically similar to galactose C. a compound chemically similar to lactose D. a compound that cannot be transported into lacZ mutants Answer: C 21) The real inducer of the lac operon is A. lactose B. allolactose C. IPTG D. galactose Answer: B 22) CAP is said to be responsible for positive regulation of the lac operon because: A. CAP binds cAMP. B. CAP binds the CAP binding site.

C. CAP prevents binding of the repressor to the operator.

D. CAP bound to the CAP binding site increases the frequency of transcription initiation by RNA polymerase.

Answer: D 23) You have discovered a bacterial operon involved in the synthesis of vitamin E. This operon is regulated by a repressor protein that binds to an operator sequence. Vitamin E is the allosteric effector of the repressor; i.e., the molecule that binds to the repressor to regulate its activity. Predict how vitamin E will influence repressor activity.

A. When vitamin E binds to the repressor, the repressor binds to the operator.

B. When vitamin E binds to the repressor, the repressor cannot bind to the operator.

C. When vitamin E binds to the repressor, the repressor can bind to RNA polymerase.

D. When vitamin E binds to the repressor, the repressor cannot bind to RNA polymerase.

Answer: A 24) How does a regulatory protein with a helix-turn-helix motif recognize a specific DNA sequence?

A. The regulatory protein opens up the two strands of DNA and detects functional groups on the bases on one of the strands.

B. The regulatory protein opens up the two strands of DNA and detects functional groups on the bases on both strands.

C. The regulatory protein detects functional groups on the base pairs that project into the major grooves of double-stranded DNA. D. The regulatory protein detects specific sequences in the sugar-phosphate backbone of the DNA double helix.

Answer: C

25) In the absence of arabinose, the regulatory protein AraC A. forms a dimer, binds to two regulatory sequences near the promoter, and activates transcription of the arabinose operon structural genes.

B. forms a dimer, binds to two distant sites, and forms a DNA loop that blocks access of RNA polymerase to the arabinose operon promoter.

C. binds to the arabinose operator and represses transcription of the arabinose operon structural genes.

D. binds to the arabinose operator and represses transcription of the arabinose operon structural genes.

Answer: B 26) Regulation of the tryptophan operon occurs through A. a tryptophan-activated repressor that binds to operator sites B. transcriptional attenuation by differential folding of RNA C. a ribozyme riboswitch D. both A and B Answer: D 27) A riboswitch regulates gene expression A. by influencing transcriptional initiation.

B. by influencing whether transcription continues to the end of a gene or operon.

C. by influencing whether a mRNA is translated.

D. B or C Answer: D 28) Which part of a riboswitch-regulated mRNA binds to a regulatory metabolite molecule?

A. the -35 element B. the aptamer C. the expression platform D. the terminator sequence Answer: B 29) A riboswitch is a domain within certain mRNAs that has all of the following characteristics, except: A. Riboswitches act as switchable “on-off” elements that selectively bind metabolites and control gene expression.

B. All riboswitches are ribozymes.

C. Riboswitches have two structural domains: an aptamer and an expression platform.

D. Riboswitches have the potential to form alternative anti-terminator and terminator hairpins.

Answer: B 30) Alternative sigma factors A. induce specific groups of functionally related genes.

B. replace the “standard” sigma factor 70 when cells are actively growing in optimal conditions.

C. bind to riboswitches.

D. are secreted by cells for quorum sensing.

Answer: A

31) Quorum sensing is a mechanism by which A. bacteria “count” the numbers of mRNA molecules that are made for a specific gene.

B. bacteria adjust the stability of an mRNA based on the presence of a metabolite.

C. bacteria change their transcriptional output in a manner that depends on their population density.

D. bacteria adjust total transcription levels in proportion to glucose levels.

Answer: C

Short answer/analytical 32) Are transcription and translation “coupled” or “uncoupled” in bacteria? Explain your answer.

Answer: Section 10.2, p. 264, Fig. 10.1 33) Diagram a typical bacterial promoter. Exact sequences are not necessary.

Answer: Section 10.2, p. 265, Fig. 10.2 34) Draw a rough sketch of the structure of the bacterial RNA polymerase holoenzyme based on X-ray crystallography. Point out the position of the active site.

Answer: Section 10.2, p. 265-266, Fig. 10.3 and 10.4 35) Design an experiment to demonstrate that the RNA polymerase holoenzyme binds DNA more tightly compared with the core polymerase. Show sample positive results.

Answer: Section 10.2, p. 266-267, Fig. 10.5 36) Which DNA strand is part of the DNA-RNA hybrid in the RNA polymerase “open complex,” the template or nontemplate strand? Explain your answer.

Answer: Section 10.2, p. 267 37) Diagram the difference between a closed and open promoter complex.

Answer: Section 10.2, p. 268-269, Fig. 10.6 38) Diagram the three-step transcription initiation process in E. coli.

Answer: Section 10.2, p. 268, Fig. 10.6 39) Diagram the key features of the bacterial elongation complex. Include a rough sketch showing the direction of RNA synthesis and where a new phosphodiester bond forms.

Answer: Section 10.2, p. 269, Fig. 10.7 40) Compare and contrast Rho-dependent and Rho-independent termination.

Answer: Section 10.2, p. 272-273, Fig. 10.9 41) What different catalytic activity does RNA polymerase have while “elongating” versus “backtracking?”

Answer: Section 10.2, p. 271-272, Fig. 10.8 42) Describe an experiment that showed that RNA polymerase can rotate DNA. Did the results of this experiment determine whether RNA polymerase rotates around DNA or vice versa in vivo?

Answer: Focus Box 10.1, p. 270

43) Transcription has a significant local effect on DNA structure (i.e., whether underwound or overwound). Discuss these structural changes and the enzymes that are required to resolve the situation.

Answer: Focus Box 10.1, p. 270 44) Diagram Jacob and Monod’s operon model. Discuss some of the testable hypotheses put forth by this model.

Answer: Section 10.3, p. 273-274, Fig. 10.10 45) Describe and give the results of an experiment that shows that the lac operator is the site of Lac repressor binding.

Answer: Section 10.3, p. 274-275, Fig. 10.11 46) Draw diagrams of the lac operon that illustrate repressed transcription, basal transcription, and activated transcription.

Answer: Section 10.3, p. 275-278, Figs. 10.12 and 10.14 47) “When glucose is abundant, bacteria use it exclusively as their food source, even when other sugars are present.” Discuss this statement in the context of how the presence of glucose effects induction of the lac operon.

Answer: Section 10.3, p. 278, Fig. 10.14 48) Draw a rough sketch of a helix-turn-helix motif and show the site of interaction with the DNA double helix.

Answer: Section 10.4, p. 279-280, Fig. 10.15 49) For bacterial operon induction, describe a specific example of the following: “The repressor and activators are DNA binding proteins that undergo allosteric modification.”

Answer: Section 10.4, p. 279-281, Fig. 10.15 and 10.16 50) For bacterial operon induction, describe a specific example of the following: “DNA looping is a mechanism used in gene regulation.”

Answer: Section 10.4, p. 281-282, Figs. 10.18 and 10.19 51) Present a model to explain attenuation in the trp operon in E. coli. Is attenuation the main regulatory mechanism? Explain your answer.

Answer: Section 10.5, p. 283-284, Fig. 10.20 52) Define the term “riboswitch.” Regulation of gene expression by riboswitches is common in bacteria but not in eukaryotes. Offer reasons why this is the case.

Answer: Section 10.5, p. 284-286 53) You have discovered a novel riboswitch that you think acts as a metabolite-responsive ribozyme. Design an experiment to demonstrate self-cleaving activity and metabolite specificity.

Show sample positive results.

Answer: Section 10.5, p. 285-286, Fig. 10.22 54) A particular sequence containing six base pairs is located in ten different organisms. The observed sequences are: 5′-ACGCAC-3′, ATACAC, GTGCAC, ACGCAC, ATACAC, ATGTAT, ATGCGC, ACGCAT, GTGCAT, and ATGCGC. What is the consensus sequence?

Answer: Section 10.2, p. 265 ACGCAC ATACAC GTGCAC ACGCAC ATACAC ATGTAT ATGCGC ACGCAT GTGCAT ATGCGC The consensus sequence is: ATGCAC If there is some variation in the sequence but certain nucleotides are present at high frequency (in 70-80% of the organisms analyzed); those nucleotides make up a consensus sequence.

55) Draw a diagram of a prokaryotic gene being transcribed and translated. Show the nascent mRNA with ribosomes attached. With an arrow, indicate the direction of transcription.

Answer: Section 10.2, p. 264, Fig. 10.1; also see below:

Direction of transcription

56) Show how you could use DNase I footprinting to demonstrate that a sigma () factor is required for specific binding of RNA polymerase to a bacterial gene promoter.

Answer: See Fig. 9.13c, p. 254. End-label a DNA fragment that contains a bacterial promoter.

Prepare three samples: DNA alone, DNA + RNA polymerase core enzyme, DNA + RNA polymerase core enzyme + sigma factor. Treat all three samples with deoxyribonuclease I (DNase I). Analyze samples by denaturing polyacrylamide gel electrophoresis and

autoradiography.. If the sigma factor is required for specific binding to the promoter, then the only lane in which the promoter sequence is protected from DNase digestion will be the one in which sigma is included. This lane will show a “footprint” over the promoter sequence.

57) Consider E. coli cells, each having one of the following mutations: (a) A mutant lac operator sequence that cannot bind lac repressor (b) A mutant lac repressor that cannot bind to the lac operator (c) A mutant lac repressor that cannot bind to allolactose (d) A mutant lac promoter that cannot bind CAP plus cAMP What effect would each mutation have on the function of the lac operon (assuming no glucose is present)?

Answer: Section 10.3, p. 273-279, Fig. 10.12 (a) A mutant lac operator sequence that cannot bind Lac repressor would result in constitutive expression of the lac operon. In both the presence and absence of lactose, -galactosidase and the other enzymes required for lactose metabolism would be produced. (b) A mutant Lac repressor that cannot bind to the lac operator would have the same effect as the mutant lac operator. There would be constitutive expression of the lac operon in both the presence and absence of lactose. (c) Allolactose is the real inducer that binds the Lac repressor and causes it to dissociate from the lac operator. A mutant Lac repressor that cannot bind allolactose would remain bound to the lac operator in both the presence and absence of allolactose, thereby inhibiting transcription of the lac operon. The lac operon would not be induced. (d) A mutant lac promoter that cannot bind CAP plus cAMP would not be capable of activated transcription in the presence of lactose, because CAP plays a role in recruiting RNA polymerase to the promoter. Although the Lac repressor would be released from the lac operator, allowing access of RNA polymerase to the promoter, in the absence of CAP RNA polymerase binds the promoter very poorly. Thus, only basal (low levels) of transcription would occur in the presence of lactose.

58) You are studying a new operon in bacteria involved in tyrosine biosynthesis. (a) You sequence the operon and discover that it contains a short open reading frame at the 5′ end of the operon that contains two codons for tyrosine. What prediction would you make about this leader sequence, the RNA transcript, and the peptide that it encodes? (b) How would you predict this operon is regulated; i.e., is it inducible or repressible by tyrosine?

Why? (c) Would this kind of regulation work in a eukaryotic cell? Why or why not?

Answer: Section 10.5, p. 283-284 (a) The following prediction could be made: When the amino acid tyrosine is abundant in the bacterial cell, the ribosome translating the short open reading frame at the 5′ end of the operon (leader sequence) does not stall at the tandem tyrosine codons in the leader sequence and rapidly reaches the leader sequence stop codon. A terminator hairpin forms in the RNA transcript, resulting in the termination of transcription. The structural genes for enzymes involved in tyrosine biosynthesis are not transcribed. When tyrosine is deficient in the bacterial cell, a deficiency in charged tRNATyr stalls the translating ribosome at one of the two tandem tyrosine codons in the leader sequence. This stalling allows an antiterminator hairpin to form in the RNA transcript, which prevents terminator formation. Transcription of the structural gene

coding regions takes place, and the enzymes required for tyrosine biosynthesis are translated. (b) The operon would be predicted to be repressible by tyrosine as described above. When tyrosine is abundant in the bacterial cell, the enzymes for tyrosine biosynthesis are not needed. (c) This kind of regulation in which translation of a leader peptide induces terminator hairpin formation in the RNA transcript would not work in a eukaryotic cell because transcription and translation are uncoupled (transcription occurs in the nucleus and translation occurs in the cytoplasm).

59) You are studying a repressor protein that you suspect forms a DNA loop between two operator sites, one located very near the promoter and the other located at a distance site upstream. Describe an experiment to test your hypothesis.

Answer: Section 10.4, p. 282. Perform a helical-twist experiment. Introduce half a turn of the DNA helix to the operon sequence anywhere between the upstream operator site and the operator site near the promoter. The added half-turn would rotate one of the two sites to which the repressor protein binds to the opposite side of the DNA, thereby interfering with proteinprotein interactions and hindering DNA loop formation. The introduction of half-rotations would be predicted to decrease repression of the operon, whereas the introduction of integral numbers of rotation that maintained the repressor binding sites on the same side of the DNA would not interfere with repression.

Fundamental Molecular Biology, 2nd Edition Instructor Test Bank Lizabeth A. Allison

Chapter 11 Transcription in Eukaryotes

Multiple Choice 1) Transcription of protein-coding genes in eukaryotes is carried out by A. RNA polymerase I B. RNA polymerase II C. RNA polymerase III D. RNA polymerase; there is only one type of RNA polymerase in eukaryotes.

Answer: B 2) Gene regulatory elements are A. trans-acting transcription factors B. cis-acting transcription factors C. trans-acting DNA sequences D. cis-acting DNA sequences Answer: D 3) Which of the following is involved in transcription in prokaryotes, but not in eukaryotic transcription?

A. regulatory proteins B. RNA polymerase C. promoters D. operators Answer: D 4) In order to be highly transcribed, a eukaryotic protein-coding gene may A. be transcribed by RNA Polymerase III B. leave its chromosome territory to associate with a transcription factory C. become dimethylated on Lys9 of histone H3 D. become covalently attached to the nuclear matrix Answer: B 5) Eukaryotic core promoter elements have all of the following characteristics, except: A. They serve as the recruitment site for RNA polymerase II. B. They serve as the recognition site for general transcription factors.

C. They all contain a TATA box.

D. They become nonfunctional when moved even a short distance from the start of transcription.

Answer: C 6) What types of proteins bind to promoter-proximal elements?

A. the TATA-binding proteins (TBP) B. general transcription factors plus RNA polymerase C. general transcription factors D. regulatory transcription factors

Answer: D 7) The primary difference between an enhancer and a promoter-proximal element is that: A. Enhancers are transcription factors; promoter-proximal elements are DNA sequences.

B. Enhancers enhance transcription; promoter proximal-elements inhibit transcription.

C. Enhancers are part of the core promoter; promoter-proximal elements are regulatory sequences distinct from the core promoter.

D. Enhancers are at considerable distances from the core promoter; promoter-proximal elements are close to the core promoter.

Answer: D 8) Which of the following statements about silencers is correct?

A. They contain a consensus sequence called a TATA box.

B. They are found in a variety of locations and are functional in any orientation.

C. They are located only in introns.

D. They are located only in 5′-flanking regions.

Answer: B 9) If the DNA double helix were inflexible, which of the following forms of transcriptional regulation would be most strongly affected?

A. The ability of the TATA-binding protein (TBP) to recognize the promoter.

B. The ability of RNA polymerase II to associate with the general transcription factors.

C. The ability of RNA polymerase II to initiate transcription.

D. The ability of enhancers to influence transcription.

Answer: D 10) A long-range regulatory element that marks the border between regions of heterochromatin and euchromatin is called a(n) A. enhancer B. insulator C. locus control region D. matrix attachment region Answer: B 11) Which of the following would not likely improve the expression of a transgene that integrated into a heterochromatic region of the genome?

A. inclusion of matrix attachment regions (MARs) B. inclusion of insulator elements C. inclusion of a strong promoter D. inclusion of a G-less cassette Answer: D 12) Patients with “Hispanic thalassemia” have a deletion of the ________ of the gene cluster that results in silencing of the genes.

A. locus control region B. coding region C. matrix attachment region D. promoter Answer: A

eta-globin

13) Which regulatory sequences must maintain their orientation with respect to the gene transcription start site to function?

A. promoter and silencer B. promoter and enhancer C promoter and locus control region D. silencer and enhancer Answer: C 14) What is the key property of DNase I that makes it useful for determining whether chromatin is in a closed (tightly condensed) or open (loosely packed) configuration?

A. DNase I is an enzyme.

B. DNase I will digest DNAs from all species equally effectively.

C. DNase I preferentially digests DNA not associated with protein.

D. DNase I cuts at specific DNA recognition sequences.

Answer: C 15) Imagine you’re assaying the DNase I sensitivity of the promoter regions of the beta–globin and vitellogenin genes in chicken liver, instead of in chick embryo erythroblasts. In this case you expect to find that: A. The beta–globin and vitellogenin promoters are equally sensitive to DNase I treatment.

B. The beta–globin and vitellogenin promoters are equally resistant to DNase I treatment.

C. The beta–globin promoter is much more sensitive to DNase I treatment.

D. The vitellogenin promoter is much more sensitive to DNase I treatment.

Answer: D

16) Small eukaryotic proteins that are known to add specificity to polymerase-dependent RNA synthesis are known as A. general transcription factors B. mediators C. histones D. elongation factors Answer: A 17) The unwinding of DNA during the initiation of transcription is mediated by the helicase activity of A. TFIID B. THIIE C. TFIIF D. TFIIH Answer: D 18) Reinitiation of transcription requires A. phosphorylation of the RNA polymerase II (RNA pol II) C-terminal domain (CTD).

B. dephosphorylation of the RNA pol II CTD. C. acetylation of the RNA pol II CTD. D. deacetylation of the RNA pol II CTD. Answer: B

19) TAFs and TBP are together known as A. TFIID B. TFIIE C. TFIIF D. TFIIH Answer: A 20) RNA polymerase II is only responsive to the presence of transcriptional activators in the presence of which protein complex?

A. spliceosome B. enhancer C. Mediator D. SWI/SNF Answer: C 21) In an in vitro transcription assay, a G-less cassette could be used to do which of the following?

A. generate an RNA product of defined length B. pause RNA polymerase at a defined site in a template C. induce RNA polymerase backtracking D. A and B Answer: D 22) Fundamentally, what makes one cell different from another in a multicellular eukaryote?

A. The different cells contain different sets of enhancers and promoter-proximal elements.

B. The different cells contain different sets of transcription factors.

C. The different cells contain different sets of cell-type-specific genes.

D. Answers A and B apply.

Answer: B 23) The basic leucine zipper (bZIP) motif has all of the following characteristics, except: A. it is a DNA-binding domain that directly contacts DNA in the major groove B. is a stretch of amino acids that fold into a long alpha-helix with leucines in every seventh position C. it facilitates dimerization of two similar polypeptide chains D. it is not as common as the zinc finger motif Answer: A 24) Which is not a common DNA binding motif?

A. chromodomain B. zinc finger C. helix-turn-helix D. basic helix-loop-helix Answer: A 25) You create a chimeric protein that contains the DNA binding domain of the Fork head transcription factor and the transactivation domain of the Sp1 transcription factor. Which of the following predictions would you make concerning the activity of your chimeric protein?

A. It would activate the transcription of genes normally activated by Fork head.

B. It would activate the transcription of genes normally activated by Sp1.

C. It would activate the transcription of genes normally activated by both Fork head and Sp1.

D. It would not activate the transcription of any genes.

Answer: A 26) Which assay would allow you to determine whether two transcription factors with nearby binding sites in a promoter sequence bind to the sequence synergistically?

A. in vitro run-off assay B. fluorescence recovery after photobleaching (FRAP) C. X-ray crystallography D. electromobility shift assay (EMSA) Answer: D 27) Which of the following is not true of Hox family genes?

A. Each Hox gene has a 180 base pair sequence called the homeobox.

B. In both fruit flies and mammals, the order of the Hox genes on the chromosome correlates with where they are expressed in embryos.

C. In fruit flies the order of Hox genes on the chromosome correlates with where they are expressed in embryos; in mammals there is no correlation.

D. The expression of the Hox genes is sequential, moving in order along the chromosome.

Answer: C 28) Histone acetyl transferases exert their effect on gene activity at least in part by: A. neutralizing positive charges on lysines of histones B. increasing the negative charge on glutamic acids of histones C. modifying the base sequence of the promoter D. adding bulky methyl groups to lysines and arginines of histones Answer: A 29) Which of the following is not true of regulatory proteins that are classified as coactivators?

A. Many coactivators function as chromatin modification complexes.

B. Many coactivators function as chromatin remodeling complexes.

C. All coactivators increase transcriptional activity.

D. All coactivators bind DNA directly.

Answer: D 30) Which statement best describes the key difference between the “enhanceosome” and “hit and run” models for transcriptional activation?

A. The “enhanceosome” model requires the involvement of both promoter proximal elements and the core promoter, whereas the “hit and run” model only involves promoter proximal elements.

B. The “enhanceosome” model requires that a stable transcriptional activation complex assembles in an ordered fashion, whereas the “hit and run” model suggests that such complexes are formed stochastically.

C. The “enhanceosome” model requires synergy between multiple transcriptional (co)activators, whereas the “hit and run” model allows that each transcriptional regulatory protein may function independently.

D. The “enhanceosome” model only applies to genes that undergo developmental regulation, whereas the “hit and run” model only applies to genes that undergo dynamic regulation even in a terminally differentiated cell.

Answer: B

31) Chromatin remodeling by the SWR1 family results in: A. nucleosome sliding B. replacement of a core histone with a variant histone C. nucleosome displacement D. remodeled nucleosomes Answer: B 32) Which is the correct order of recruitment of transcriptional regulatory proteins to a gene promoter?

A. SWI/SNF, histone acetyltransferase (HAT) complex, preinitiation complex B. HAT complex, SWI/SNF, preinitiation complex C. preinitiation complex, SWI/SNF, HAT complex D. the order of recruitment is gene-specific.

Answer: D 33) Which protein or protein complex helps RNA polymerase to traverse nucleosomes?

A. FACT B. SWI/SNF C. Sonic hedgehog D. Polycomb Answer: A 34) Which method was most instrumental in the development of the current model for transcript elongation by RNA Polymerase II? A. in vitro run-off assay B. fluorescence recovery after photobleaching (FRAP) C. X-ray crystallography D. electromobility shift assay (EMSA) Answer: C 35) Familial dysautonomia is a rare autosomal recessive disorder characterized by a mutation in a gene the plays a role in A. promoter clearance B. transcript elongation C. proofreading and backtracking D. nucleotide translocation Answer: B 36) Which of the following factors plays a role in RNA cleavage during polymerase backtracking?

A. TFIIS B. FACT C. HDAC D. Elongator Answer: A 37) Which of following events is not true about the nuclear import of proteins?

A. The nuclear localization sequence of the protein to be imported binds to importin.

B. The nuclear localization sequence of the protein is removed once the protein enters the nucleus.

C. The energy for import is provided by the small GTPase Ran.

D. Import may occur against a concentration gradient.

Answer: B 38) Nuclear import and export of proteins A. occurs by diffusion through the nuclear envelope.

B. is mediated by structures embedded in the nuclear envelope called nuclear pore complexes.

C. is mediated by structures embedded in the nuclear envelope called nuclear localization sequences.

D. occurs by active transport through phospholipid-lined channels in the nuclear envelope.

Answer: B 39) During nuclear transport, RanGTP A. causes disassembly of import complexes, but is required for the assembly of export complexes.

B. causes assembly of import complexes, but is required for the disassembly of export complexes.

C. is present in equal concentrations in the nucleus and cytoplasm.

D. is required to provide energy for cargo recognition and docking.

Answer: A

40) RanGAP stimulates the conversion of A. GTP to GDP B. GDP to GTP C. GDP to GMP D. GMP to GDP Answer: A 41) Steroid hormones, such as glucocorticoids, bind to receptors inside the cell. The hormonereceptor complex is transported into the nucleus, where it can directly affect gene expression. To get from the location where the receptor binds the hormone to its site of action: A. the receptor-hormone dimer must dissociate to form a monomer.

B. the receptor-hormone complex must become water soluble by binding to a carrier molecule.

C. the receptor-hormone complex must be transported through the nuclear pore complex.

D. the receptor-hormone complex must be activated by a signaling cascade.

Answer: C

Short answer/analytical 42) List the major types of RNA polymerases in eukaryotes and the types of genes they are responsible for transcribing.

Answer: Section 11.2, p. 293-294 43) How has in situ hybridization (FISH) evidence supported the chromosome territories and transcription factories hypotheses?

Answer: Section 11.2, p. 293, Fig. 11.2

44) Diagram the structure of a “typical” eukaryotic protein-coding gene, including all potential regulatory regions. Indicate where the RNA polymerase II preinitiation complex interacts.

Answer: Section 11.3, Fig. 11.3 45) Draw a diagram of a RNA polymerase II promoter, showing all of the types of elements it could have. Exact sequences are not necessary.

Answer: Section 11.3, p. 296-298 Table 11.2 46) You are investigating a new type of promoter, but you can find no familiar sequences.

Design an experiment to locate the promoter sequences, and show sample results.

Answer: Section 11.3, p. 297, Fig. 11.4 47) Compare and contrast the key characteristics of proximal promoter elements, enhancers, and locus control regions.

Answer: Section 11.3, p. 298-302 48) You have made a transgenic mouse using a cDNA coding for a human protein under control of a heterologous promoter. In most of the transgenic mice, you observe positiondependent expression of the human protein. Define the term “position-dependent” and give examples of regulatory regions that you could link to the promoter-human cDNA construct that might confer position-independent expression. Explain your choice of regulatory regions.

Answer: Section 11.3, Focus Box 11.1, p. 299-300 49) Imagine you’re assaying the DNase I sensitivity of the promoter regions of the beta–globin and vitellogenin genes in chicken liver, instead of in chick embryo erythroblasts. Show sample results from assay.

Answer: Section 11.3, Disease Box 11.1, p. 302-303 50) Describe a model for transcriptional regulation by matrix attachment regions (MARs).

Answer: Section 11.3, p. 303-304, Fig. 11.7, Focus Box 11.2 51) Explain the underlying genetic defect leading to Hutchinson-Gilford progeria.

Answer: Section 11.2, Focus Box 11.2, p. 304-306 52) Draw a rough diagram of the structure of yeast RNA polymerase II. Show where the DNA lies and show the location of the active site. What is the structure of the CTD? Answer: Section 11.4, p.307-308, Fig. 11.9 53) Does RNA polymerase II bind directly to the core promoter by itself to initiate transcription?

Explain your answer.

Answer: No, Section 11.4, p. 306-311, Fig. 11.8 54) Promoter clearance requires what post-translational modification of RNA polymerase II? Where does this post-translational modification occur?

Answer: Section 11.4, p. 314 55) List in order the proteins that assemble to form an RNA polymerase II preinitiation complex.

Answer: Section 11.4, p. 309, Fig. 11.8

56) What shape does TBP have? What is the geometry of the interaction between TBP, the TATA box, and the DNA double helix? Does TBP interact with the major or minor groove?

Answer: Section 11.4, p. 309-310, Fig. 11.10 57) Describe and give results of an experiment that shows that TFIIH phosphorylates the CTD of RNA polymerase II. Answer: Section 11.4, p. 310, Fig. 11.11 58) Describe the G-less cassette transcription assay that led to the discovery of Mediator.

Include a graph illustrating the effect of Mediator on activated versus basal transcription Answer: Section 11.4, p. 312, Fig. 11.13 59) Explain why transcription factors are called “modular” proteins. How is this exploited in a yeast two-hybrid assay?

Answer: Section 11.5, p. 315-316, Fig. 9.14 60) Compare and contrast four different classes of DNA-binding domains found in eukaryotic transcription factors.

Answer: Section 11.5, p. 316-323 61) Describe the structure and function of the homeodomain. What other protein domain does it most resemble?

Answer: Section 11.5, p. 317, Focus Box 11.3 62) Explain what is meant by the term “colinear expression” as it describes Hox genes in Drosophila.

Answer: Section 11.5, p. 317, Focus Box 11.3 63) Describe a model for how Polycomb group proteins silence homeobox genes.

Answer: Section 11.5, p. 317, Focus Box 11.3 64) Draw a detailed diagram of a zinc finger. Point out the DNA-binding motif of the finger.

Answer: Section 11.5, Fig. 11.17, p. 317-319 65) What does “Sonic the Hedgehog” have to do with zinc fingers?

Answer: Section 11.5, Disease Box 11.2, p. 320-321 66) Explain how the disease Greig cephalopolysyndactyly syndrome is related to the zinc-finger DNA binding domain.

Answer: Section 11.5, Disease Box 11.2, p. 320-321 67) Draw a detailed diagram of a basic leucine zipper (bZIP). Illustrate the interaction of this motif with another polypeptide and with DNA. Point out the leucine zipper and the basic DNAbinding domain.

Answer: Section 11.5, p. 319-323, Fig. 11.18 68) Describe a common motif found in transactivation domains.

Answer: Section 11.5, p. 323-324 69) Do coactivators and corepressors bind DNA directly? Explain your answer: Answer: No, Section 11.6, p. 324-325

70) Could the Polycomb-group protein complexes PRC1 and PRC2 be considered corepressors? Why or why not?

Answer: Section 11.5, Focus Box 11.3, p. 317-319 Formally, PRC1 and PRC2 could only be considered co-repressors if they are recruited or activated by sequence-specific DNA binding factors, which might depend on the context (both gene locus and organism). However, the role of PRC1 and PRC2 in generating a repressive chromatin state is very much in line with the function of many transcriptional corepressors.

71) Compare and contrast the structure and function of six post-translational modifications of histone N-terminal tails.

Answer: Section 11.6, p. 325-328, Fig. 11.20 72) What roles are histone acetyltransferases (HATs), histone deacetylases (HDACs), histone methyltransferases (HMTs) thought to play in activation and repression of transcription?

Answer: Section 11.6, p. 325-328 73) Define the “histone code” hypothesis and discuss evidence for and against it.

Answer: Section 11.6, Focus Box 11.4, p. 326-327 74) Explain how the same type of chemical modification of nucleosomes, e.g., methylation, can promote or repress transcription.

Answer: Section 11.6, p. 327 75) Discuss the role of the variant linker histone H1b in muscle progenitor cell differentiation.

Answer: Section 11.6, p. 328, Fig. 11.21 76) Describe four different ways in which chromatin structure can be altered by chromatin remodeling complexes. What role does chromatin remodeling play in activation and repression of gene transcription?

Answer: Section 11.6, p. 328-331, Figs. 11.22 and 11.23 77) Which is recruited first to a gene, the preinitiation complex, histone acetyltransferases (HATs), or SWI/SNF? Explain your answer.

Answer: Section 11.7, p. 331-333, Fig. 11.24 The order of recruitment depends on the chromatin structure of the gene promoter, the phase of the cell cycle, and many other factors.

78) Compare and contrast the “enhanceosome” and the “hit and run” model for transcription complex assembly.

Answer: Section 11.7, p. 331-334, Fig. 11.25 79) Explain how fluorescence recovery after photobleaching (FRAP) experiments may differentially support either the “hit and run” model or the “enhanceosome” models of transcriptional activation.

Answer: Section 11.7, p. 333, Fig. 11.26 80) Describe X-ray crystallographic evidence for a four step cycle of RNA synthesis during RNA polymerase II transcription.

Answer: Section 11.8, p. 334-335, Fig. 11.27

81) RNA polymerase has a single active site that switches between RNA synthesis and cleavage. Describe a model for this “tunable” RNA polymerase active site. How does this compare with the structure and function of DNA polymerase active sites?

Answer: Section 11.8, p. 335-336, Fig. 11.28 82) What is the proposed function of TFIIS in RNA polymerase II-mediate transcription?

Answer: Section 11.8, p. 336, Fig. 11.29 83) After initiation of transcription, how does RNA polymerase II move through nucleosomes?

Answer: Section 11.8, p. 336-337, Fig. 11.30 84) You have discovered a novel factor that you think is involved in facilitating transcription through nucleosome arrays. You propose that the factor displaces a histone dimer from the core octamer, leaving a core hexamer on the DNA. Design an experiment to test your hypothesis and show sample positive results.

Answer: Section 11.8, p. 337, Figs. 11.30 and 11.31 85) Draw a diagram illustrating the three key steps in the nuclear import of proteins. Which step requires the conversion of GTP to GDP? Answer: Section 11.9, p. 339-340, Fig. 11.32 86) Draw a rough sketch of the structure of a nuclear pore complex.

Answer: Section 11.9, Focus Box 11.5, p. 342-343 87) Discuss how Ran functions as a “molecular switch” in both nuclear import and export.

Answer: Section 11.9, p. 344, Figs. 11.32 and 11.33 88) Describe two models for translocation of cargo through the nuclear pore complex.

Answer: Section 11.9, p.343-344, Fig. 11.34 89) Assume that you find in Q. 90 that the zinc finger protein is sometimes in the nucleus and sometimes in the cytoplasm. Describe one possible mechanism for cytoplasmic retention and subsequent signal-mediated import of this transcription factor.

Answer: Section 11.10, p. 345-347, Figs. 11.35 and 11.36 90) Diagram the key steps in the regulated nuclear import of NF-B. Answer: Section 11.10, p. 345-347, Fig. 11.35 91) Diagram the key steps in the regulated nuclear import of the glucocorticoid receptor (GR).

Answer: Section 11.10, p. 347, Fig. 11.36 92) You suspect that a sequence upstream of a transcriptional start site is acting as an enhancer and not as a promoter. Describe an experiment you would run to test your hypothesis.

Predict the results.

Answer: Section 11.3, p. 296-300, Section 9.2, p. 230. Promoter elements increase the frequency of initiation only when positioned near the transcriptional start site. In contrast, an enhancer can function upstream, downstream, within an intron, at a distance, and in either orientation relative to the promoter. Various plasmid constructs could be made with the enhancer in different locations relative to a viral promoter driving expression of a reporter gene.

These constructs could then be used in transient transfection assays to test their ability to

enhance transcription of the reporter gene.

93) A team of molecular biologists has made transgenic mice using a viral promoter and a cDNA coding for a human protein called p45. In most of the transgenic mice, they observe position-dependent expression of the human gene. They seek your advice. Suggest regulatory regions that you could link to the promoter- p45 DNA construct that might confer positionindependent expression. Explain your choice of regulatory regions.

Answer: Section 11.3, p. 298-302. Regulatory regions that could be linked to the promoter-p45 DNA construct that might confer position-independent expression include an enhancer, an insulator, a locus control region, or matrix attachment regions. An enhancer might contribute to tissue-specific expression, insulators flanking the transgene could protect the transgene from the effects of nearby enhancers or silencers, or act as a chromatin boundary marker, preventing the spread of heterochromatin. A locus control region (LCR) upstream of the promoter could help organize and maintain a functional domain of active chromatin. The association of matrix attachment regions (MARs) with the nuclear architecture allows formation of an independent DNA loop domain that can adopt an altered chromatin structure distinct from the structure of neighboring chromatin. By including these long-range regulatory elements, position-independent expression units could be established, regardless of where the transgene randomly integrated into chromatin.

94) You have purified a transcription factor that has a leucine-rich region. You perform an electrophoretic mobility shift assay (EMSA) using a double-stranded oligonucleotide that you know from other studies contains the site recognized by this transcription factor in vivo.

However, the transcription factor does not bind to the labeled oligonucleotide in your EMSA. Provide an explanation for this result.

Answer: Section 11.5, p. 319-322. The transcription factor is likely to contain a leucine zipper motif. To bind DNA, the transcription factor must form a heterodimer with another polypeptide that also contains a leucine zipper motif. The joining of the two leucine zipper-containing polypeptides would result in correct positioning of two adjacent basic DNA-binding domains in the dimeric complex. If the dimerization partner was included in the EMSA, the heterodimer would then be able to bind the labeled oligonucleotide.

95) You are studying a new class of eukaryotic promoters recognized by a novel RNA polymerase. You discover two general transcription factors that are required for transcription of these promoters. You suspect that one has helicase activity and that the other is required to recruit the helicase and the RNA polymerase to the promoter. Describe experiments you would perform to test your hypothesis. Provide sample results of your experiments.

Answer: Section 11.4, Fig 11.12, p. 311; Section 9.8, p. 253. Perform an assay for helicase activity. Hybridize a labeled short piece of single-stranded DNA to a complementary region in a much larger unlabeled single-stranded DNA. Incubate this substrate with or without the transcription factor in the presence and absence of ATP. DNA helicase activity would unwind the partial duplex and release the labeled shorter DNA from the larger piece of DNA. The shorter fragment DNA could be distinguished from the hybrid by gel electrophoresis and autoradiography. Sample results would be similar to the experiment depicted in Fig. 11.12.

EMSA could be used to as a pilot study to test whether the other transcription factor is required for recruitment of the helicase and the RNA polymerase to the promoter. Incubate an endlabeled DNA fragment containing a promoter with various combinations of no protein, the helicase, transcription factor, and RNA polymerase to determine whether the factors can

individually bind the promoter (visualized as a shift in mobility of the complex), or whether the transcription factor must be present for binding of the polymerase and the helicase. Sample results would be similar in format to those shown in the next question.

96) You include an inhibitor of the protein kinase activity of TFIIH in an in vitro transcription assay. What step in transcription would you expect to see blocked? Describe an experiment you would run to test your hypothesis. Predict the results.

Answer: Section 11.4, p. 310-311, Fig. 11.13. The C-terminal domain (CTD) of RNA polymerase II undergoes dynamic phosphorylation during the transcription cycle. Transcription initiation requires an unphosphorylated CTD. Promoter clearance and elongation of the transcript requires a phosphorylated CTD. If the protein kinase activity of TFIIH was inhibited, elongation would be blocked. A G-less cassette assay could be used to test for the synthesis of a transcript of predictable size on a polyacrylamide gel (see Fig. 11.13). If transcript elongation is blocked by the inhibitor, there should be a significant reduction in the amount of labeled, fulllength transcript synthesized during the assay.

97) Assume that you are studying a transcription factor (TF). You have constructed a plasmid vector for expression of GFP-tagged TF. You discover that TF is sometimes in the nucleus and sometimes in the cytoplasm. Describe a possible mechanism for its cytoplasmic retention and subsequent signal-mediated import. Design an experiment to test your hypothesis and provide sample results.

Answer: Section 11.10, p. 346. The transcription factor may be sequestered in the cytoplasm by binding to an inhibitor protein. Transcription factor nuclear import may be regulated by a signal transduction pathway, similar to signal-mediated nuclear import of NF-B or the glucocorticoid receptor. Transient transfection assays and fluorescence microscopy could be used to track the movement of the GFP-tagged TF within cells in the presence or absence of the suspected extracellular signal (ligand).

Fundamental Molecular Biology, 2nd Edition Instructor Test Bank Lizabeth A. Allison

Chapter 12 Epigenetic Mechanisms of Gene Regulation Multiple Choice 1) As a general rule, cytosine DNA methylation marks genes for A. activation.

B. silencing.

C. mutation.

D. programmed gene rearrangements.

Answer: B 2) CpG islands have all of the following characteristics, except: A. They are found near gene promoters.

B. They are protected from spontaneous deamination.

C. They are typically highly methylated.

D. They occur in clusters near the 5′ end of genes.

Answer: C 3) The dinucleotide CG is underrepresented in the human genome because A. it is highly deleterious in most genomic locations due to methylation.

B. methylcytosine is highly susceptible to spontaneous deamination, resulting in the TG dinucleotide.

C. methylcytosine is read as an thymine during replication, resulting in the CA dinucleotide.

D. methyltransferases cannot methylate cytosines when a guanine is the next nucleotide.

Answer: B 4) During DNA replication, methylation is A. maintained by a semiconservative process.

B. erased and then a maintenance methylase adds back methyl groups to both the template and the newly synthesized strand.

C. erased and then a maintenance methylase adds back methyl groups to the template strand.

D. not maintained; replication results in hemimethylated progeny DNA. Answer: A 5) When the allele choice is determined by its parent-of-origin, monoallelic gene expression is referred to A. nutritional legacy.

B. transposition.

C. gene switching.

D. imprinting.

Answer: D 6) For imprinted genes, selection of the active allele A. does not depend on the parent-of-origin B. can either be random or nonrandom depending on the particular gene C. is random

D. is nonrandom Answer: D 7) Which statement is not true about imprinted genes?

A. Imprinting is maintained in the primordial germ cells.

B. Nearly all imprinted genes are organized in clusters in the genome.

C. Imprinting occurs in mammals, but no other vertebrates studied so far.

D. Imprinting affects a small subset of genes and results in the expression of those genes from only one of the two parental chromosomes.

Answer: A 8) Which of the following is not a proposed mechanism for monoallelic expression of imprinted genes?

A. Altered chromatin structure in the gene promoter.

B. Differential expression of an antisense RNA transcript.

C. Programmed gene rearrangements.

D. Blocking of an enhancer by an insulator.

Answer: C 9) Insulin-like growth factor 2 receptor (Igf2r) gene is expressed exclusively from the maternal allele on chromosome 17 due to A. differential methylation of an Igf2r intron.

B. differential expression of an antisense RNA transcript.

C. transposition of the maternal copy of Igf2r.

D. methylation of an imprinting control region.

Answer: B 10) Imprinted genes are regulated by imprinting control regions (ICRs). Within these regions there are allele-specific differences in A. DNA methylation B. histone modification C. the location of CpG islands D. Answers A and B apply Answer: D 11) X chromosome inactivation in humans is A. random B. nonrandom C. imprinted D. preferential for the paternal X chromosome Answer: A 12) XIST transcript levels are upregulated A. on the active X chromosome B. on the inactive X chromosome C. on an autosome D. by the expression of Tsix Answer: B

13) X chromosome inactivation is characterized by a series of epigenetic chromatin modifications, including all of the following except for: A. DNA methylation B. histone hyperacetylation C. enrichment of variant histone macroH2A D. coating with XIST RNA Answer: B 14) Transposable elements A. have the ability to move from one location to another in a genome.

B. occur very rarely in humans.

C. are small, extrachromosomal circles of DNA that have the ability to self-replicate.

D. are found only in bacteria and plants.

Answer: A 15) Transposable elements were first discovered while studying chromosomal breakage events in maize by A. Mary Lyon B. Gregor Johann Mendel C. James Watson and Francis Crick D. Barbara McClintock Answer: D 16) The wrinkled seed (rr) character of the garden pea described by Mendel is caused by A. insertion of a transposable element B. genomic imprinting C. X chromosome inactivation D. a nonsense mutation Answer: A 17) DNA transposons move by a(n) A. “copy and paste” mechanism.

B. “cut and paste” mechanism.

C. “Activator (Ac)-Dissociation (Ds)” mechanism.

D. “random walk” mechanism.

Answer: B 18) Retrotransposons move by a(n) A. “copy and paste” mechanism.

B. “cut and paste” mechanism.

C. “Activator (Ac)-Dissociation (Ds)” mechanism.

D. “random walk” mechanism.

Answer: A 19) Alu elements are A. active long interspersed nuclear elements (LINES) in the human genome B. inactive long interspersed nuclear elements (LINES) in the human genome C. active short interspersed nuclear elements (SINES) in the human genome D. inactive short interspersed nuclear elements (SINES) in the human genome Answer: C

20) About 21% of the human genome is composed of members of the A. Ac and Ds elements B. L1 family of long interspersed nuclear elements (LINES) C. Tc1/mariner superfamily elements D. P elements Answer: B 21) A nonautonomous transposon relies on other transposons for transposition because it A. does not contain transposase recognition sequences.

B. does not encodes transposase.

C. does not undergo methylation.

D. all of the above.

Answer: B 22) When a pregnant female of a particular strain of yellow agouti mice with the agouti viable yellow (Avy) allele was fed a diet rich in folic acid, the mouse give birth to offspring with mostly brown fur indicating that A. an LTR retrotransposon was hypermethylated and moved from one chromosome to another, thereby turning on expression of the brown fur gene B. an LTR retrotransposon was hypermethylated and moved from one chromosome to another, thereby shutting off expression of the agouti gene C. the LTR retrotransposon promoter region was hypomethylated, thereby shutting off expression of the downstream agouti gene.

D. the retrotransposon promoter region was hypermethylated, thereby shutting off expression of the downstream agouti gene.

Answer: D 23) A haploid yeast cell of mating-type that has budded and divided at least once A. can switch to the opposite mating type.

B. must bud and divide again before switching to the opposite mating type.

C. will form a diploid / spore.

D. will stop budding and dividing.

Answer: A 24) Suppose that prior to mating-type switching in a haploid yeast cell, MAT is expressed in the active locus. During mating-type switching by homologous recombination A. the MAT genes in the active locus switch places with HMRa genes at a silent, donor locus B. the MAT genes in the active locus are degraded and replaced with HMRa genes from a silent, donor locus.

C. the MAT genes in the active locus are returned to the HML silent locus; HMRa genes from a silent, donor locus are moved to the active MAT locus.

D. the MAT locus is inactivated and HMRa genes are activated by an epigenetic mechanism.

Answer: B 25) DNA methylation of transposable elements is thought to function A. as a host defense mechanism.

B. as a mechanism to counteract silencing of the transposable element.

B. to promote excision of transposable element sequences.

C. to promote insertion of new sequences into the transposable elements.

Answer: A 26) In plants, heterochromatinization of repeat sequences may be mediated by A. histone hyperacetylation.

B. formation of nuclear pore-associated chromatin loops.

C. riboswitch-containing mRNAs. D. RNA-directed DNA methylation.

Answer: D 27) During mating-type switching in haploid yeast cells, in the “off” state the recombination enhancer A. represses all mating-type switching B. represses mating-type switching in daughter cells C. represses recombination between MAT and HMLα D. represses recombination between MAT and HMRa Answer: C 28) Antigen switching in trypanosomes occurs by A. gene conversion B. reciprocal recombination C. switching the active expression site D. all of the above answers apply Answer: D 29) Which of the following statements is not true about variable surface glycoprotein (VSG) gene expression in the trypanosome, Trypanosoma brucei?

A. There are over 1000 different VSG genes, but only one is expressed at a time.

B. There is a single telomeric expression site.

C. VSG gene transcription is localized in the “expression site body.”

D. VSG switching occurs predominantly by DNA homologous recombination.

Answer: B 30) During development, millions of B cell types are produced. Each B cell has a unique antigen receptor generated from different V, D, and J gene segments by a series of A. alternative splicing events during B-cell differentiation B. multiple gene mutation events during B-cell differentiation C. site-specific recombination events during B-cell differentiation D. nucleotide excision repair events during B-cell differentiation Answer: C 31) During V(D)J recombination there is epigenetic control of the initial selection of the allele to be rearranged. This involves A. progressive histone hyperacetylation of the heavy chain allele to be rearranged B. differential methylation of the two light chain alleles C. early replication of the light chain allele that undergoes rearrangement D. all of the above answers apply Answer: D 32) Which of the following does not result from allelic exclusion?

A. transgenerational effects of diet B. mating type switching in yeast

C. antigen switching in trypanosomes D. unique immunoglobin gene expression in lymphocytes Answer: A 33) Which does not constitute a major epigenetic mechanism involved in mammalian X inactivation?

A. cytosine methylation B. histone methylation C. association with Xist RNA D. gene rearrangement Answer: D

Short answer/analytical 34) Draw the structure of the normal base cytosine compared with 5-methyl-cytosine. Which base is more likely to undergo spontaneous deamination to thymine? Explain your answer.

Answer: Section 12.2, Fig. 12.1, p. 355 35) The CG dinucleotide is often denoted as “CpG.” What does the “p” represent?

Answer: Section 12.2, p. 355 The p denotes phosphate.

36) You hypothesize that methylation of a gene of interest is correlated with decreased gene expression. Design an experiment to test your hypothesis and show sample positive results.

Answer: Section 12.2, p. 356-357, Fig. 12.2 37) You prepare two samples of genomic DNA and digest one with HpaII and the other with MspI. After digestion you separate the DNA samples by agarose gel electrophoresis. Sketch the pattern of bands you would see and indicate the location of any CpG islands.

Answer: Section 12.2, p. 358, Fig. 12.3 38) Discuss the connection between DNA methylation and fragile X mental retardation.

Answer: Section 12.2, Disease Box 12.2, p. 357-358 39) Diagram how genomic imprinting is “reset” in the germline and then maintained throughout development.

Answer: Section 12.3, p. 364-365, Fig. 12.4 40) Compare and contrast three general mechanisms for ensuring monoallelic expression of maternally or paternally imprinted genes.

Answer: Section 12.3, p. 365-367, Fig. 12.5 41) Describe how methylation analysis can be used for clinical diagnosis of Prader-Willi syndrome (PWS) and Angelman syndrome (AS).

Answer: Section 12.3, Disease Box 12.3, p. 362-364 42) A child shows symptoms of Angelman syndrome (AS), but their methylation pattern on chromosome 15 is normal. Explain how AS may have arisen in this case.

Answer: Section 12.3, Disease Box 12.3, p. 362-364

43) You have performed a bisulfite-PCR assay for distinguishing normal cytosine from 5-methyl cytosine. The starting sequence is: CGCGTAGCC. After sodium bisulfite treatment, PCR, and DNA sequencing, you obtain the following sequence: CGCGTAGTT. Which of the cytosines were methylated in the starting sequence? Explain your answer.

Answer: Section 12.3, Disease Box 12.3, Fig. 1, p. 362-363 The last two cytosines were methylated.

44) H19 is a maternally expressed imprinted gene. What evidence is suggestive that the H19 transcript is a protein-coding mRNA. What evidence favors its role as a regulatory RNA? Answer: Section 12.3, p. 367 45) Diagram the mode of regulation of the insulin-like growth factor 2 (Igf2)-H19 locus.

Discuss some of the known consequences of loss of imprinting of Igf2.

Answer: Section 12.3, p. 368, Fig. 12.5 46) Discuss evidence for and against the “conflict hypothesis” for the origins of genomic imprinting.

Answer: Section 12.3, p. 368 47) Compare and contrast X-chromosome inactivation with imprinting of a paternal or maternal allele of a gene.

Answer: Section 12.3, p. 360, Table 12.1, Section 12.4 The mechanism that determines which X chromosome undergoes inactivation in an embryo is random, whereas imprinting occurs in a manner dependent on the parent-of-origin. Both choices are epigenetically maintained throughout the lifetime of an organism.

48) Use a diagram to compare and contrast X chromosome inactivation in marsupials and placental mammals.

Answer: Section 12.4, p. 369, Fig. 12.6 49) Is XIST transcribed from the active or inactive X chromosome? Explain your answer.

Answer: Section 12.4, p. 370, Fig. 12.7 50) Discuss some of the implications of the findings that about 15% of genes on the inactive X chromosome escape inactivation and that in an additional 10% of genes the level of expression differs from woman to woman.

Answer: Section 12.4, p. 370-371, Fig. 12.8 51) Describe changes in gene expression that are likely to occur upon transposition of the Ds mobile genetic element in maize to a site within a kernel color gene.

Answer: Section 12.5, p. 372-373, Fig. 12.9 52) Diagram the mechanism by which DNA transposons move.

Answer: Section 12.5, p. 376-377, Fig. 12.11 53) Diagram the mechanism by which retrotransposons move.

Answer: Section 12.5, p. 377, Fig. 12.12 54) Describe a specific example of an active Alu element in humans and describe a diagnostic test for the presence of the Alu element.

Answer: Section 12.5, Disease Box 12.4, p. 375-376

55) Define the terms “LINE” and “SINE”. These elements are widespread in the human genome.

Are any still active? Explain your answer, giving specific examples.

Answer: Section 12.5, p. 378-379, Disease Box 12.4 56) You have discovered a new species of plant with flowers that are either purple or white.

Assume that you have enough sequence information for the purple pigment gene to design primers for PCR amplification of a specific region of the gene. You extract genomic DNA from purple flowers and white flowers and perform PCR. After gel electrophoresis of the PCR products, you find that you have amplified a 200 bp fragment from the purple flower. In contrast, using the same PCR primers, you amplify a 500 bp fragment from the white flower. Provide an explanation for these results.

Answer: Section 12.5, p. 379-380, Fig. 12.13 57) Discuss the connection between DNA methylation and the statement that “poor nutrition predisposes cells of an organism to cancer.”

Answer: Section 12.6, p.381-382 58) DNA methylation may serve to defend the genome. Give an example of epigenetic control of an active LTR retrotransposon in mouse.

Answer: Section 12.6, p. 381-382, Fig. 12.15 59) DNA methylation may serve to defend the genome from transposable elements. Other than methylation, what is another mechanism for epigenetic control of transposable elements?

Answer: Section 12.6, p. 380-381, Fig. 12.14 60) Diagram the process of gene conversion in yeast during switching from mating-type alpha to mating-type a.

Answer: Section 12.7, p. 383-386, Fig. 12.17 61) Describe the key characteristics of variant surface glycoprotein (VSG) gene expression in trypanosomes.

Answer: Section 12.7, p. 387-392, Disease Box 12.5 62) What disease is caused by African trypanosomes and why is this disease so difficult to treat?

Answer: Section 12.7, p. 387-392, Disease Box 12.5 63) Describe an experiment to demonstrate that RNA polymerase I colocalizes with active expression site sequences in the expression site body (ESB) of Trypanosoma brucei .

Answer: Section 12.7, Fig. 12.21, p. 392-393 64) Draw a sketch of an antibody showing the light and heavy chains and the antigen binding sites.

Answer: Section 12.7, Fig. 12.23, p. 395 65) Explain how thousands of immunoglobulin genes can give rise to many millions of antibody proteins.

Answer: Section 12.7, p. 395-398, Fig. 12.25 66) Diagram the rearrangement that occurs during B lymphocyte maturation at the

immunoglobulin light-chain locus.

Answer: Section 12.7, p. 395-398, Fig. 12.25 67) Present a model for the cleavage and rejoining of DNA strands at immunoglobulin lightchain locus recombination signal sequences.

Answer: Section 12.7, p. 396-397, Fig. 12.25 (inset) 68) Discuss evidence for the hypothesis that the V(D)J system evolved from a transposon.

Answer: Section 12.7, Focus Box 12.1, p. 394 69) Compare and contrast the mechanisms for allelic exclusion (both initiation and maintenance) of (1) the Igf2 gene in embryonic cells and (2) antibody-encoding genes in developing B cells.

Answer: Section 12.3, p.367, Fig.12.5, Section 12.7, p. 395-398, Fig. 12.26 70) During rearrangement of immunoglobulin genes there is epigenetic control of the initial selection of the allele to be rearranged, as well as maintenance of allelic exclusion. You determine that both light chain alleles in pre-B cells have hyperacetylated histones, but one allele is methylated and late-replicating and the other allele is unmethylated and early replicating. Which allele is likely to be subject to rearrangement? Explain your answer.

Answer: Section 12.7, p. 397-398, Fig. 12.26 71) You find that the maternal allele of the mouse MAMA gene is active, while the paternal allele is repressed. The gene encodes a growth factor. (a) Predict the effect on allelic expression of treating adult mice with 5-aza-deoxycytidine. (b) During the course of the experiment you find that mouse pups are born with completely yellow fur instead of the expected brown fur. Provide an explanation.

Answer: Section 12.2, p. 356-357, Section 12.5, p. 379-380, Fig. 12.13 (a) 5-aza-deoxycytidine is an analog of normal dCTP that can be incorporated into DNA during replication. However, it cannot be methylated because the 5-carbon of normal cytosine is replaced by nitrogen. Drug treatment leads to the gradual loss of methyl groups from cytosine.

Treating adult mice with 5-aza-deoxycytidine could lead to demethylation of the imprinting control region and thus a loss of imprinting. In this case, both the maternal and paternal alleles of MAMA would be expressed. (b) Demethylation may have activated a cryptic promoter within a retrotransposon and inappropriately activated the agouti viable yellow allele (e.g. see Fig. 12.13).

72) The figure below shows the results of methylation analysis of DNA from a “normal” child, a child with Prader-Willi Syndrome, and two children with Angelman Syndrome.

Southern blot of genomic DNA digested with HindIII and HpaII, probed with PW71B, a chromosome 15-specific probe.

Lane 1: “Normal” child Lane 2: Child with Prader-Willi Syndrome Lane 3: Child with Angelman Syndrome Lane 4: Child with Angelman Syndrome

6.0 kb

4.4 kb Lane Numbers

1

2

3

4

Interpret the results for each lane.

Answer: Section 12.3, Disease Box 12.3, p. 362-364. HindIII is a rare cutting enzyme and HpaII only cuts at a nonmethylated restriction site in the paternal chromosome (CCGG), and does not cut the methylated restriction site on the maternal chromosome. The probe is complementary to a DNA sequence located in between the two restriction sites (PW71B). A 4.4 kb HpaII-HindIII fragment from the paternal chromosome hybridizes with the probe on a Southern blot, but only a 6.0 kb HindIII-HindIII fragment from the maternal chromosome hybridizes with the probe.

Lane 1: A normal child has both the maternal and paternal fragments (6.0 and 4.4 kb bands).

Lane 2: A child with Prader-Willi Syndrome (PWS) is lacking a functional paternal contribution to the imprinted region due to biparental deletion, uniparental disomy, or ICR mutations, so only the maternal (6.0 kb) fragment is visible.

Lane 3: When Angelman Syndrome (AS) is due to biparental deletion, uniparental disomy, or ICR mutations, the child lacks a functional maternal contribution to the imprinted region. In this case, only the paternal (4.4 kb) fragment is visible.

Lane 4: When AS is due to a classical mutation in the UBE3A gene, the methylation pattern is identical to that of a normal child. Both the maternal and paternal fragments are visible.

73) On the inactive X chromosome many CpG islands are heavily methylated. Does this hold true for the CpG island upstream of the XIST gene? Why or why not?

Answer: Section 12.2, p.355, Section 12.4, p. 370. The CpG island upstream of the XIST gene would likely be unmethylated (hypomethylated) because the XIST gene is actively transcribed from the inactive X chromosome.

74) Assume you have two cell-free transposition systems that have all the enzymes necessary for transposition of a yeast Ty element and a vertebrate Tc1/mariner element, respectively.

What effect would the following inhibitors have on these two systems, and why? (a) Inhibitors of translation (b) Inhibitors of transcription (c) Inhibitors of double-stranded DNA replication (d) Inhibitors of reverse transcription Answer: Section 12.5, p. 373, Table 12.3 (a) Inhibitors of translation would block transposition of both the yeast Ty element (a retrotransposon) and the vertebrate Tc1/mariner element (a DNA transposon).

Retrotransposons code for an endonuclease and a reverse transcriptase, and transposons code for a transposase. These essential proteins would not be synthesized. (b) Inhibitors of transcription would block transposition of both the yeast Ty element (a retrotransposon) and the vertebrate Tc1/mariner element (a DNA transposon).

Retrotransposons code for an endonuclease and a reverse transcriptase, and transposons code for a transposase. These essential proteins could not be synthesized if the mRNAs were not transcribed. (c) Inhibitors of double-stranded DNA replication would block the final steps in transposition of both the yeast Ty element (a retrotransposon) and the vertebrate Tc1/mariner element (a DNA transposon). Both processes required DNA repair for completion. (d) Inhibitors of reverse transcription would only block transposition of the yeast Ty element (a retrotransposon). They transpose by a “copy and paste” mechanism which involves an RNA intermediate which must be copied into DNA by reverse transcription.

75) Molecular biologists have demonstrated that there is preferential switching to the opposite mating type in budding yeast. Switching from a→a (or →) with no change in mating type occurs rarely. Since the DNA sequence in the MATa locus would be the same as the donor HMRa, how would you show that a→a had occurred or not?

Answer: Section 12.7, p. 386. There are a number of possible strategies. One would be to use mutagenesis techniques to slightly alter the sequence of the donor HMRa locus in a strain of yeast. The movement and consequent expression of this altered allele could then be tracked using RT-PCR with primers specific for the altered allele.

76) In the most common mode of trypanosome antigen-switching, the VSG gene in the active expression site is degraded and replaced with the donor VSG gene. Why isn’t there progressive loss of VSG genes over time?

Answer: Section 12.7, p. 392, Fig. 12.20. There isn’t progressive loss of VSG genes over time because the most common mode of antigen switching involves gene conversion (duplicative transposition). In this mechanism, the VSG in the active expression site is degraded and replaced with a duplicated copy of a silent donor VSG by homologous recombination. Most likely the degraded VSG still has a silent copy present elsewhere in the trypanosome genome.

77) Is the genomic DNA in a bone marrow stem cell identical to the genomic DNA in a mature B cell? Why or why not?

Answer: Section 12.7, p. 393-396. The genomic DNA in a bone marrow stem cell would not be identical to the genomic DNA in a mature B cell. In the stem cells, there are multiple V, D, J, and C regions for the immunoglobulin genes. In the mature B cell, V, D, and J gene segments are rearranged to form one unique, functional coding region.

Fundamental Molecular Biology, 2nd Edition Instructor Test Bank Lizabeth A. Allison

Chapter 13 RNA Processing and Post-Transcriptional Gene Regulation

Multiple Choice 1) The ovalbumin gene is much longer than the mRNA produced by this gene. What causes this discrepancy?

A. The exons have been removed during mRNA processing.

B. The DNA represents a double-stranded structure and the RNA is single stranded.

C. There are more amino acids coded for by the DNA than the mRNA. D. The introns have been removed during mRNA processing.

Answer: D 2) RNA splicing is the process by which introns are removed from a primary RNA transcript at precisely defined splice points. Which statement is not true about introns?

A. Introns are that are joined together after splicing.

B. Introns are usually nonfunctional and are degraded, but some introns encode functional RNA products C. Introns are transcribed along with the exons.

D. Introns are larger on average than exons.

Answer: A 3) Some introns contain termination codons yet they do not interrupt the coding of a particular protein. Why? A. The translation machinery recognizes and skips over introns.

B. These triplets cause frameshift mutations, not termination.

C. More than one termination codon is needed to stop translation.

D. Splicing occurs before translation.

Answer: D 4) Group I intron self-splicing relies on A. An external guanosine nucleotide as a cofactor.

B. A bulged internal guanosine nucleotide.

C. An external adenosine nucleotide as a cofactor.

D. A bulged internal adenosine nucleotide.

Answer: A 5) The first nucleophilic attack in group II intron splicing is initiated by a A. An external guanosine nucleotide as a cofactor.

B. A bulged internal guanosine nucleotide.

C. An external adenosine nucleotide as a cofactor.

D. A bulged internal adenosine nucleotide.

Answer: D

6) The second transesterification step for group I and group II intron self-splicing is initiated by the A. 3′ end of the intron B. 5′ end of the intron C. 3′ end of the upstream exon D. 5′ end of the upstream exon Answer: C 7) Archael introns are A. self-splicing.

B. spliced by an endoribonuclease.

C. spliced by the spliceosome.

D. require an external guanosine nucleotide as a cofactor.

Answer: B 8) Which statement is true regarding eukaryotic nuclear tRNA genes?

A. There are no intron-containing nuclear tRNA genes in eukaryotes.

B. All eukaryotic nuclear tRNA genes contain at least one intron.

C. One of two classes of eukaryotic nuclear tRNA genes has introns.

D. Only yeast and plant nuclear tRNA genes have introns.

Answer: C 9) Select three posttranscriptional modifications often seen in the maturation of nuclear premRNA in eukaryotes.

A. 5′-capping; 3′-poly(A) tail addition; splicing B. 3′-capping; 5′-poly(A) tail addition; splicing C. 5′-capping, removal of exons; insertion of introns D. 3′-capping, insertion of introns; 5′-poly(A) tail addition Answer: A 10) During maturation of nuclear pre-mRNA in eukaryotes, capping takes place at the A. 5′ end of the primary RNA transcript.

B. 3′ end of the primary RNA transcript C. 2′ end of the primary RNA transcript D. At either the 5′ or 3′ end, depending on the particular primary RNA transcript Answer: A 11) The 7-methylguanosine cap structure, the ribose of guanosine is connected to the 5’ end of the mRNA by A. phosphates B. sulfates C. glycerol D. ethylene Answer: A 12) Which is not a function of the 7-methylguanosine cap?

A. protection from exonucleases B. promoting mRNA nuclear export C. stimulation of RNA polymerase elongation D. stimulation of translation Answer: C

13) The process of polyadenylation is accomplished by A. reverse transcriptase B. the CTD domain C. primase D. poly(A) polymerase Answer: D 14) Which is not required for efficient mRNA polyadenylation?

A. a polyadenylation sequence in the mRNA B. poly(A)-binding proteins C. Xrn2 exonuclease D. poly(A) polymerase Answer: C 15) To ensure that RNA polymerase is only released after it has transcribed a gene to its end, termination is mechanistically linked to A. 5’ capping B. excision of the lariat RNA C. alternative splicing D. 3’ end formation Answer: D 16) RNA polymerase is thought to be disengaged from the DNA template by A. the Xrn2 exonuclease B. cleavage and polyadenylation specific-factor (CPSF) C. poly(A) polymerase D. the spliceosome Answer: A 17) Nuclear pre-mRNA introns are A. self-splicing B. removed by an endoribonuclease C. removed by the spliceosome D. require an external guanosine nucleotide as a cofactor.

Answer: C 18) RNA splicing of nuclear pre-mRNA involves removal of A. introns from a primary RNA transcript by a series of two transesterification reactions.

B. exons from a primary RNA transcript by a series of two transesterification reactions.

C. introns from a primary RNA transcript by a series of polyadenylation reactions.

D. exons from a primary RNA transcript by a series of polyadenylation reactions.

Answer: A 19) The spliceosome is composed of A. U1, U2, U3, U4, and U5 snRNPs B. five different U-rich snRNAs and more than 200 proteins C. small ribosomal RNAs and associated proteins D. more than 200 proteins Answer: B

20) Which snRNPs are thought to contribute to the catalysis of pre-mRNA splicing?

A. U1 and U2 snRNPs B. U2 and U4 snRNPs C. U5 and U6 snRNPs D. U2 and U6 snRNPs Answer: D 21) The mechanism for nuclear pre-mRNA splicing is similar to the splicing mechanism of A. Group I introns B. Group II introns C. Archael introns D. nuclear tRNA introns Answer: B 22) It was once assumed that a count of the number of protein-coding genes in an organism would provide a reasonable estimate of the total number of gene products. This is now known to be incorrect, largely due to the discovery of widespread: A. chromatin modification control B. transcriptional initiation control C. alternative splicing D. translational control Answer: C 23) One way to detect alternative splicing of a given gene is to compare: A. the relative rate of transcription of the alternatively spliced gene with another gene B. the sequence of the alternatively spliced gene to that of another gene.

C. the size and sequence of different pre-mRNAs transcribed from this gene.

D. the size and sequence of mature mRNAs produced from this gene.

Answer: D 24) There is evidence supporting A. that the spliceosome assembles in a step-wise fashion as the splicing reaction occurs B. that the splicosome exists as a “pre-assembled” macromolecular complex C. that the order of snRNP binding to a pre-mRNA varies between specific transcripts D. A and B Answer: D 25) Which factors bind to pre-mRNA splice sites first?

A. U1 and U2AF B. U2 and a DEXH helicase C. U4/U6/U5 D. hnRNP proteins Answer: A 26) For some pre-mRNAs, splicing is regulated by A. RNA polymerase stalling B. splicing enhancer sequences C. localization of the pre-mRNA to subnuclear “speckles” D. transcription termination Answer: B

27) When an exon from one pre-mRNA joins to an exon from another pre-mRNA the process is called A. trans-splicing B. cis-splicing C. alternative splicing D. RNA editing Answer: A 28) RNA editing is a widespread mechanism for post-transcriptional modification of the base sequence of mRNA. In humans, RNA editing A. is mediated by guide RNAs and a multiprotein complex called the “editosome” that inserts or deletes uridines.

B. occurs by deamination of adenosine to inosine, or cytidine to uridine.

C. is mediated by small nucleolar RNAs and base modification enzymes.

D. is mediated by the U12-type intron splicing pathway.

Answer: B 29) RNA editing is a widespread mechanism for post-transcriptional modification of the base sequence of mRNA. In trypanosomes, RNA editing A. is mediated by guide RNAs and a multiprotein complex called the “editosome” that inserts or deletes uridines.

B. occurs by deamination of adenosine to inosine, or cytidine to uridine.

C. is mediated by small nucleolar RNAs and base modification enzymes.

D. is mediated by the U12-type intron splicing pathway.

Answer: A 30) The enzyme ADAR (adenosine deaminase acting on RNA) functions in A. RNA interference B. RNA editing C. alternative splicing D. formation of an alternative mRNA cap structure Answer: B 31) MicroRNAs are processed by the enzymes _______ and ______. One strand of the microRNA is loaded into a silencing complex named _________. The microRNAs target mRNA for ___________ or ___________.

A. RISC, Dicer, Drosha, mRNA degradation, translational repression B. Drosha, Dicer, RISC, mRNA degradation, translational repression C. Drosha, Dicer, RISC, alternative splicing, RNA editing D. RISC, Dicer, Drosha, alternative splicing, RNA editing Answer: B 32) RNA interference is the process of A. gene silencing at the transcriptional level.

B. gene silencing at the level of RNA splicing.

C. gene silencing at the post-transcriptional level.

D. gene silencing at the post-translational level.

Answer: C

33) RNA interference is thought to have evolved to A. downregulate cancer-causing genes B. promote transposition C. prevent spurious recombination D. defend against viruses Answer: D 34) The target mRNA for a particular miRNA is cleaved by A. RNA-directed RNA polymerase B. Dicer C. Drosha D. RISC Answer: D 35) Which component of the RNAi pathway confers inheritance of siRNA-induced silencing?

A. Dicer B. RISC (RNA induced silencing complex) C. RdRP (RNA-directed RNA polymerase) D. Drosha Answer: C 36) The targets of a microRNA are usually A. viral mRNAs B. bacterial dsRNAs C. tissue-specific genes D. “housekeeping” genes Answer: C 37) The nuclear exosome A. degrades aberrant mRNAs B. cleaves dsRNA into siRNAs C. promotes nuclear export of mature mRNA complexes D. facilitates spliceosome assembly Answer: A 38) Which aspect of an mRNA’s life cycle serves as a key mRNA quality control mechanism?

A. 3’ end cleavage B. transcription termination C. nuclear export D. splicing Answer: C 39) Nuclear basket proteins Mlp1 and Mlp2 act as A. nuclear exosomes B. a sorting filter for unspliced transcripts C. splicing factors D. miRNA exportins Answer: B

40) When mRNAs lacking a poly(A) tail compete with polyadenylated mRNAs for limiting translational machinery A. the polyadenylated mRNA is translated more efficiently.

B. the mRNA lacking a poly(A) tail is translated more efficiently.

C. the mRNAs are translated equally efficiently.

D. the efficiency of translation depends on the sequence of the mRNA, not the poly(A) tail.

Answer: A 41) Failure of the ribosome to remove an exon-exon junction complex (EJC) leads to A. degradation by nuclear exosomes B. translation of the mRNA C. storage of the mRNA in the cytoplasm D. nonsense-mediated mRNA decay Answer: D

Short answer/analytical 42) Describe the original observations that indicated the existence of “split genes.”

Answer: Section 13.2, p. 404-405. It was noticed that segments of viral DNA sequences were missing from the mature mRNA that spanned these segments. It was thus proposed that these sequences were removed from the RNA intermediate.

43) Define the terms “intron” and “exon.”

Answer: Section 13.2, p. 405, Fig. 13.2 44) What is a snRNP (“snurp”)?

Answer: Section 13.4, p. 415 45) Describe and show the results of an R-looping experiment that demonstrates that an intron is transcribed.

Answer: Section 13.2, p. 404-405, Fig. 13.1 46) Discuss how the discovery of intron-encoded small nucleolar RNAs (snoRNAs) has added to the challenge of cloning “genes”.

Answer: Section 13.2, p. 406, Focus Box 13.1 47) Draw a diagram of the steps involved in self-splicing of an RNA containing a group I intron.

Answer: Section 13.3, p. 407-408, Fig. 13.3 48) You are characterizing an RNA transcript that contains an intron. List the key features you would need to analyze to determine whether the RNA is self-splicing and, if it is self-splicing, to classify the intron as a group I or group II intron.

Answer: Section 13.3, p. 407-411 49) Diagram the lariat mechanism of group II intron splicing.

Answer: Section 13.3, p. 408-409, Fig. 13.4 50) What is unusual about the phosphodiester bond forming a lariat intron?

Answer: Section 13.3, p. 409

51) Compare and contrast the key features of “homing” and “retrohoming” by mobile group I and II introns, respectively.

Answer: Section 13.3, p. 409-410, Fig. 13.5 52) Diagram the Archael intron splicing pathway. Include a rough sketch of the conserved secondary structure of an archael intron.

Answer: Section 13.3, p. 410, Fig. 13.6 53) Diagram the tRNA splicing pathway of plants and fungi.

Answer: Section 13.3, p. 410-411, Fig. 13.7 54) a) Name the domain of RNA polymerase II that coordinates cotranscriptional processing of nuclear pre-mRNA. b) Name the post-translational modification that is required for function of this domain. c) Name the class of enzyme that carries out the post-translational modification in (b).

Answer: Section 13.4, p. 411-413 55) Diagram a model showing how nuclear pre-mRNA processing is coupled to transcription.

Answer: Section 13.4, p. 411-412, Fig. 13.8 56) Diagram the structure of the 7-methylguanosine cap structure with sufficient detail to show how the cap is added in the “wrong” direction to the initial nucleotide of the mRNA. Answer: Section 13.4, p. 413, Fig. 13.10 57) Describe and give the results of an experiment that showed that the carboxyl terminus of the RNA polymerase II carboxyl-terminal domain (CTD) is necessary and sufficient to enhance splicing.

Answer: Section 13.4, p. 411-412, Fig. 13.9 58) Diagram a two-step model for transcription termination in eukaryotes.

Answer: Section 13.4, p. 413-414, Fig. 13.11 59) Draw a diagram of the polyadenylation complex and the polyadenylation process.

Answer: Section 13.4, p. 413-414, Fig. 13.12 60) A friend of yours has been diagnosed with oculopharyngeal muscular dystrophy. Explain the genetic basis for this disease.

Answer: Section 13.4, Disease Box 13.1, p. 415 61) Describe the composition and structure of the spliceosome, based on cryoelectron microscopy and biochemical analysis. Relate the structure to function.

Answer: Section 13.5, p. 414-419, Figs. 13.13 and 13.14.

62) What are Sm proteins?

Answer: Section 13.4, p. 415-416, Fig. 13.14 63) Suggest how DEXH/D-box RNA helicases might facilitate splicing.

Answer: Section 13.4, p. 420

64) Draw a diagram of a pre-mRNA as it exists in a spliceosome just before the second step in splicing. Show the interactions with U2, U5, and U6 snRNPs. This scheme resembles the intermediate stage for splicing of what kind of self-splicing RNA? Answer: Section 13.4, p. 419-420, Fig. 13.15 65) Besides base-pairing with the pre-mRNA, U6 base-pairs with two snRNAs. Which ones are they?

Answer: Section 13.4, Fig. 13.15 U2 and U4 66) Describe and show the results of an experiment that demonstrates that U1 snRNP is the first snRNP to bind to the pre-mRNA. Answer: Section 13.4, Fig. 13.16 67) What process is thought to be defective in spinal muscular atrophy?

Answer: Section 13.4, Disease Box 13.2, p. 417-418 68) Describe and show the results of an experiment to demonstrate that the survival of motor neurons (SMN) protein colocalizes with coilin in Cajal bodies.

Answer: Section 13.4, Disease Box 13.2, p. 417-418 69) A friend of yours has been diagnosed retinistis pigmentosa. Explain the genetic basis for this disease.

Answer: Section 13.4, Disease Box 13.3, p. 422 70) What are SR proteins and where are they located within cells?

Answer: Section 13.4, p. 420, Fig. 13.17 71) Is the spliceosome a ribozyme? Describe experimental evidence for the catalytic activity of U2 and U6 snRNAs. Answer: Section 13.4, p. 421 Fig. 13.19 72) Define the term “alternative splicing.” Describe a specific example of alternative splicing.

Answer: Section 13.5, p. 423-426 73) Is alternative splicing within human genes common or uncommon? Explain your answer.

Answer: Section 13.5, p. 423 74) Use a diagram to compare and contrast discontinuous group II trans-splicing with spliced leader trans-splicing. Discuss how the discovery of discontinuous group II trans-splicing in particular has added to the challenge of cloning “genes”.

Answer: Section 13.5, p. 426-427, Fig. 13.22 75) When the genome of the only characterized member of the kingdom Nanoarchaeota (Nanoarchaeum equitans) was sequenced, it appeared to be missing genes encoding several tRNAs, yet the tRNAs themselves were all present in the organism. Provide an explanation.

Answer: Section 13.5, p. 427-428, Fig. 13.22 76) Define the term “RNA editing” and explain why its discovery shook the central dogma of molecular biology.

Answer: Section 13.6, p. 428-429

77) Diagram the general mechanism for RNA editing in trypanosomes. You do not need to show a complete sequence for the pre-mRNA. Answer: Section 13.6, p. 428-429, Fig. 13.24 78) In what cellular compartment does editing take place in trypanosomes? Are all the components required for editing produced in this compartment? Explain your answer.

Answer: Section 13.6, p. 428-429, Fig. 13.24 79) Diagram the process of apolipoprotein B (CU editing) in humans.

Answer: Section 13.6, p. 433-434, Fig. 13.28 80) Besides CU editing, what other type of editing occurs in humans?

Answer: Section 13.6, p. 431 Adenosine to inosine editing 81) What unusual feature was revealed by X-ray crystallographic analysis of the structure of the adenosine deaminase ADAR? Answer: Section 13.6, p. 431, Fig. 13.27 82) Describe evidence for a model in which a defect in RNA editing contributes to the onset of amyotrophic lateral sclerosis.

Answer: Section 13.6, Disease Box 13.4, p. 432-433 83) You discover that a single gene can produce two different proteins. Provide possible explanations for your observation.

Answer: Sections 13.5 and 13.6, Fig. 13.20, p. 423 e.g., alternative splicing, RNA editing, alternative promoters 84) Describe some of the experiments that led to the discovery of miRNAs in the nematode worm, Caenorhabditis elegans.

Answer: Section 13.7, p. 438-439 85) Define the term RNA interference (RNAi). Compare and contrast key features of small interfering RNAs (siRNAs) and micro RNAs (miRNAs).

Answer: Section 13.7, p. 434-442, Figs. 13.29 and 13.33; see also Chapter 9, Section 9.7, p. 251 86) Diagram key steps in the pathway of miRNA biogenesis and post-transcriptional gene regulation.

Answer: Section 13.7, p. 439-441, Fig. 13.33 87) Do miRNAs target mRNA for degradation or translational inhibition or both? Explain your answer.

Answer: Section 13.7, p. 440-441, Fig. 13.33 88) RNA can complementarily base pair with other RNAs. Give specific examples of basepairing interactions that are critical for each of the following cellular processes: a) pre-mRNA intron splicing b) RNA editing in trypanosomes c) post-transcriptional gene regulation by microRNAs (miRNAs) Answer: see Sections 13.4, 13.6, 13.7

89) Describe the various “checkpoints” for RNA quality control in the nucleus.

Answer: Section 13.8, p. 442-443, Fig. 13.35 90) If the translational machinery is limiting, what is the likely fate of an RNA lacking a poly(A) tail: storage in the cytoplasm or translation?

Answer: Section 13.8, p. 444 91) Diagram the process of nonsense-mediated mRNA decay.

Answer: Section 13.8, p. 445, Fig. 13.36 92) You are investigating a gene with two large introns (intron 1 and 2) and three short exons (exons A, B, C). Show the results of R-looping experiments performed with: (a) mRNA with exons A, B, and C, and single-stranded DNA (b) pre-mRNA and single-stranded DNA (c) mRNA with exons A and C only, and single-stranded DNA A

1

B

2

C single-stranded DNA pre-mRNA mRNA

Answer:Section 13.2, p. 404-405, Figure 13.1 (a)

(b)

(c)

93) Sequence analysis of a novel cloned gene reveals that the exons are poorly conserved and the introns are highly conserved. What features would you look for in the RNA that might help in determining possible function of the intronic RNA? Answer: Section 13.2, Focus Box 13.1, p. 406; Section 13.7, p. 438 If the introns showed sequence complementarity to ribosomal RNA, they might contain the genes for small nucleolar RNAs (snoRNA). If the introns showed sequence complementarity to specific mRNAs, they might contain genes for microRNAs (miRNAs).

94) You have developed an in vitro assay for both splicing and polyadenylation. You generate in vitro radioactively labeled pre-mRNA substrates that either include a 5′ cap or lack the 5′ cap.

You incubate these labeled pre-mRNA substrates with mammalian cell nuclear extract and run out the products on a sequencing gel. You then distinguish the products based on their relative size in the gel. You get the following results where the number of pluses is related to the relative amount of radioactivity found in that band on the gel: Pre-mRNA RNA uncapped +++ RNA capped +

Spliced RNA + +++

Poly (A) + +++

Propose a hypothesis that explains these results.

Answer: Section 13.4, p. 413. The 5′ cap is essential for efficient splicing and polyadenylation.

95) You have discovered a minor class of rare introns that use a distinct splicing machinery involving snRNPs U110, U120, and U140. An experiment was performed to determine the order of assembly. A pre-mRNA was attached to agarose beads and incubated with purified splicing components. At time points from 0-20 minutes, RNA was extracted from the beads and analyzed by Northern blot with labeled probes specific for each RNA. In what order are the snRNPs assembled?

0

2

5

10

20

Time (min) U140 U120 pre-mRNA

U110

Answer: Section 13.4, Fig 13.16, p. 419. U120 snRNP joins the complex first, followed by U140 and U110.

96) You determine that a single gene encodes a longer protein in embryos and a shorter protein in the adult organism. Describe experiments to determine whether this results from alternative splicing or RNA editing. Provide sample positive results.

Answer: See Section 13.5; Section 13.6 If the longer and shorter proteins result from alternative splicing, then longer and shorter mRNAs would be expected to be present in embryonic cells compared with adult cells. RNA could be extracted from embryonic cells and adult cells and analyzed by Northern blot with a labeled probe specific for the mRNA. Two bands would be predicted, a slower moving band in embryos, and a faster moving band in adults. If the longer and shorter proteins resulted from RNA editing, then the mRNAs would be expected to be the same length, and there would be only one band present on the Northern blot. Most likely, there would be a single nucleotide change resulting in formation of a premature

stop codon in the mRNA encoding the shorter protein. A cDNA could be prepared by reverse transcription, followed by DNA sequencing to confirm this hypothesis.

97) You have cloned what you thought was the complete sequence of a gene. However, closer scrutiny reveals that the sequence lacks an entire exon normally found in other organisms.

Propose a hypothesis that explains these results.

Answer: Section 13.5, p. 426. The exon may be present in a different pre-RNA transcript, encoded by a separate gene. The exon would then be joined to the rest of the mRNA by transsplicing.

98) The results of the experiment shown in Fig. 13.34 suggest that miRNAs play a role in the tissue-specific suppression of certain genes during development. Propose an experiment to follow up on these studies. For example, when does suppression by miR-1 and miR-124 occur naturally?

Answer: See Section 13.7, p. 441-442. Examine the pattern of expression of miR-1 and miR-2 homologs in different stages of development, for example in Xenopus, Drosophila or C. elegans using in situ hybridization. Expression of the microRNA at a specific developmental stage will indicate that the microRNA may play a role in growth and/or development at that stage.

99) What is the likely fate of an RNA transcript if: (a) Alternative splicing produces a premature termination codon. (b) There is a mutation in a splice site that results in unspliced RNA. Answer: (a) Section 13.8, p. 445. If alternative splicing produces a premature termination codon, the mRNA will most likely be rapidly degraded by nonsense-mediated mRNA decay. Exon-junction complexes (EJCs) are assembled near exon-exon boundaries after mRNA splicing. When the first ribosome begins translating an mRNA, the EJCs are normally displaced. If a premature termination codon is present, this activates a surveillance complex. The surveillance complex interacts with the prematurely terminating ribosome and targets the mRNA for degradation. (b) Section 13.8, p. 442-443 If there is a mutation in a splice site that results in unspliced RNA, the unspliced RNA will most likely be retained in the nucleus and targeted for degradation by the nuclear exosome. mRNP surveillance also appears to occur at the nuclear periphery. Two filamentous proteins that are anchored at the nuclear basket of the nuclear pore complex in yeast (Mlp1p and Mlp2p) are proposed to function as a “sorting filter,” preferentially interacting with properly assembled mRNP particles. They may serve to retain unspliced transcripts within the nucleus, possibly via recognition of a component associated with the 5′ splice site.

Fundamental Molecular Biology, 2nd Edition Instructor Test Bank Lizabeth A. Allison

Chapter 14 The Mechanism of Translation

Multiple Choice 1) Which component is not directly involved in the process known as translation?

A. mRNA B. tRNA C. rRNA D. DNA Answer: D 2) Which statement is not true about the nucleolus?

A. The nucleolus is the site of rRNA synthesis.

B. The nucleolus is a membrane-bound organelle within the nucleus.

C. The nucleolus is the site of ribosome subunit assembly.

D. The nucleolus is composed of a fibrillar center, dense fibrillar component, and granular component.

Answer: B 3) Ribosomes carry out the process of protein synthesis in the ____________ after assembly in the _____________ within a special compartment called the ______________.

A. cytoplasm, cytoplasm, nucleolus B. nucleus, cytoplasm, nucleolus C. nucleolus, cytoplasm, nucleus D. cytoplasm, nucleus, nucleolus Answer: D 4) Which component of mammalian ribosomes accounts for 60% of the mass?

A. ribosomal RNA B. ribosomal protein C. messenger RNA D. transfer RNA Answer: A 5) In eukaryotes, ribosomal RNAs are transcribed by A. RNA polymerase I B. RNA polymerase II C. RNA polymerase III D. RNA polymerase IV Answer: A 6) What is an anticodon?

A. The base sequence of an mRNA that signals the termination of translation.

B. The base sequence of an mRNA that binds to a complementary codon in tRNA. C. The base sequence of a tRNA that binds to a complementary codon in mRNA. D. The base sequence of a tRNA that accepts an amino acid, via a reaction catalyzed by

aminoacyl-tRNA synthetase Answer: C 7) Aminoacyl-tRNA synthetases are enzymes that A. catalyze loading of an anticodon onto the 3′ end of a tRNA. B. catalyze loading of an amino acid onto the 3′ end of a tRNA. C. catalyze loading of an amino acid onto the anticodon loop of a tRNA. D. act as chaperones to fold tRNA into its functional L-shaped tertiary structure.

Answer: B 8) Transfer RNAs couple to amino acids by A. a high-energy ester bond B. a redox-sensitive disulfide bond C. hydrogen bonding interactions that mimic base-pairing D. a hydrogen bond between the tRNA 3’ OH and the amino group of the amino acid Answer: A 9) Which statement is not true about ribosomes: A. There are three tRNA binding sites on the ribosome, the A site, the P site, and the E site.

B. Ribosomal subunits are assembled in the nucleolus of the cell.

C. Peptide bond formation is catalyzed by a ribosomal protein.

D. Eukaryotic ribosomes are composed of a 40S and 60S subunit.

Answer: C 10) GTP hydrolysis is required for A. joining of the 40S and 60S subunits.

B. peptide bond formation.

C. start codon recognition and initiation factor release.

D. A and C. Answer: D 11) Hydrolysis of ATP indirectly fuels protein synthesis by A. generating a high energy bond between the amino acid and tRNA. B. loading the ternary complex onto the 40S subunit.

C. triggering translocation of the mRNA within the ribosome during elongation.

D. triggering dissociation of the large and small ribosomal subunits at termination so they can be recycled.

Answer: A 12) Which is not a member of the eukaryotic “ternary complex” of translation initiation?

A. Met-tRNA B. eIF2 C. mRNA D. GTP Answer: C

13) Below is a simplified diagram of several components of the translation machinery. The ribosome is moving from left to right in this figure. The arrow points to the:

A. 3′ end of the mRNA B. 5′ end of the mRNA C. 3′ end of the rRNA D. 5′ end of the rRNA Answer: A 14) The start codon for translation in eukaryotes is A. UUG B. GUG C. AUG D. all of the above Answer: C 15) Which of the following is/are required as an energy source for translation?

A. GTP only B. ATP only C. GTP and ATP D. No energy source is required.

Answer: C 16) Transfer RNAs are thought to improve the fidelity of mRNA translation by A. preventing translocation if an incorrect amino acid is present in the peptidyl transferase center.

B. increasing the rate of GTP hydrolysis by EF-Tu upon accurate codon-anticodon pairing.

C. activating proofreading activity of eEF2.

D. linking codon-anticodon mispairing in the P site to translation termination.

Answer: B 17) The stop codon is recognized by A. a special aminoacyl tRNA B. eRF1 and eRF2 C. GTP D. a molecular chaperone Answer: B

18) Which of the following does not need to occur for translation to terminate?

A. cleavage of the polypeptide from the tRNA in the P-site B. recognition of the stop codon by a release factor C. translocation by eEF2 D. binding of a tRNA to the A-site Answer: D 19) Nascent polypeptides exit the ribosome via A. the E (exit) site B. the mRNA exit channel C. a long tunnel D. a large cavity that allows protein folding before its release Answer: C 20) The crystal structure of the ribosome provided key new evidence that A. the mechanism of translation is fundamentally different among the three branches of life.

B. the mRNA is situated in a tunnel in the small ribosomal subunit.

C. ribosomal proteins facilitate structural transitions during translocation.

D. ribosomal RNAs catalyze peptide bond formation.

Answer: D 21) Which step of translation requires mRNA decoding by a protein?

A. Initiation B. Elongation C. Termination D. A and C Answer: C 22) The signal recognition particle (SRP) is a ribonucleoprotein complex that binds to ribosomes translating polypeptides that bear a signal sequence for targeting to A. the endoplasmic reticulum.

B. the nucleus.

C. the mitochondria.

D. the cytoplasm.

Answer: A 23) What is a molecular chaperone?

A. A protein that binds to ribosomes at the rough ER membrane.

B. A protein that post-translationally modifies a nascent polypeptide by adding or removing a phosphate group.

C. A protein component of the large ribosomal subunit that helps with correct peptide bond formation.

D. A protein that helps a nascent polypeptide chain to adopt its correct three-dimensional fold.

Answer: D

24) When the translation initiation factor eIF2 is phosphorylated in response to a sudden temperature increase or viral infection, translation can be greatly inhibited. In this case, phosphorylated eIF2 is under _______________ control, and the phosphorylation of eIF2 leads to global (general or widespread) _____________ control.

A. translational; post-translational B. post-translational; translational C. post-translational; transcriptional D. translational; transcriptional Answer: B 25) In response to unfavorable conditions such as hypoxia or osmotic shock, translation can be globally repressed by A. modification of mRNA 7-methylguanosine caps B. ribosome stalling via a conformation change in eEF2 C. phosphorylation of eIF2 D. puromycin synthesis Answer: C

Short answer/analytical 26) Give a brief explanation of the role of the following RNAs during protein synthesis: a. mRNA b. tRNA c. rRNA Answer: Section 14.2, p. 453; see also Chapter 3, Section 3.2, p. 40 27) Draw a rough sketch of a ribosome, showing the location of the large and small subunits and the E, P, and A sites.

Answer: Section 14.2, Fig. 14.3 28) Why do you suppose eukaryotes evolved a specialized subnuclear structure for ribosome biogenesis?

Answer: Section 14.2, p. 453-454. Ribosomes must be mass produced because they are required for the production of each and every protein that is produced by the cell, and their RNA content accounts for ~80% of total cellular RNA. For this reason, it is worthwhile for the cell to create a special location where ribosomal RNAs can be efficiently transcribed by a dedicated polymerase (RNA polymerase I) and assembled with ribosomal proteins into functional subunits.

29) Eukaryotic ribosomes misincorporate one amino acid for every 1,000-10,000, whereas the eukaryotic replication machinery misincorporates one nucleotide for every ~1x107, a number which goes down to ~1x109 following activity of the mismatch repair machinery. (a) Why can the cell tolerate a higher rate of error in translation than in replication? (b) Besides errors that occur in the ribosome, what is another potential source of error in protein synthesis?

Answer: (a) See Section 7.2, p. 160. Replication errors can result in a change in gene sequence, which in turn can change the amino acid sequence of every protein that is expressed from that gene in a cell and in all its progeny. In other words, replication errors are permanent.

By contrast, translation errors result in the production of a single mutant protein with no further consequences for the cell. (b) Section 14.3, p. 457. Errors in protein synthesis can also result from mis-charging of tRNAs by aminoacyl tRNA synthetases. For example, a tRNA with a tyrosine anticodon could be charged with a phenylalanine instead of tyrosine.

30) Mammalian ribosomes are composed of a 60S and a 40S subunit. What is “S”?

Answer: Section 14.2, p. 452 31) Which component of the nucleolus is thought to be the location of active ribosomal genes?

Answer: Section 14.2, p. 454 The dense fibrillar component is thought to be location of active rRNA genes.

32) An active ribosomal RNA gene has a “Christmas tree” structure as visualized by electron microscopy. What are the “branches” and what is the “trunk”?

Answer: Section 14.2, p. 454; see also the image on p. 451 The trunk is the rRNA gene (the DNA template) and the branches are rRNA gene products (RNA).

33) Diagram the pathway for assembly in the nucleolus, maturation, and nuclear export of ribosomal subunits in yeast.

Answer: Section 14.2, p. 455, Fig. 14.5 34) Diagram the two steps of aminoacyl-tRNA charging.

Answer: Section 14.3, p. 456-457, Fig. 14.6 35) Describe an in vitro assay to demonstrate that aminoacylation of tRNACUA with pyrrolysine requires a novel archael aminoacyl synthetase, PylS. Answer: Section 14.3, p. 457, Fig. 14.7 36) Aminoacyl-tRNAs display high fidelity, even for amino acids such as threonine and serine that have very similar side chains. Present a model for the proofreading activity of aminoacyltRNA synthetases based on recent X-ray crystallographic evidence.

Answer: Section 14.3, p. 457, Fig. 14.8 37) Draw a diagram to summarize the process of translation initiation in eukaryotes. Indicate at what points ATP or GTP hydrolysis are required.

Answer: Section 14.4, p. 458-461, Fig. 14.9 38) In eukaryotes, the mRNA to be translated is most likely folded into secondary and tertiary structures and coated with protective RNA-binding proteins. Discuss the importance of the 5′ cap and poly(A) tail for loading of the mRNA onto the 43S complex.

Answer: Section 14.4, p. 459, Fig. 14.10 39) Write the sequence of an ideal eukaryotic translation initiation site. Aside form the AUG, what are the most important positions?

Answer: Section 14.4, p. 461, Table 14.2 40) In order to recycle into its active form to bind another initiator Met-tRNA, eIF2-GDP must be converted to eIF2-GTP. Describe how this occurs and explain how this process differs from GTP hydrolysis.

Answer: Section 14.4, p. 461, Fig. 14.9 41) The younger sibling of a friend of yours has been diagnosed with a rare disease called leukoencephalopathy with vanishing white matter. Explain the genetic basis of this disease.

Answer: Section 14.4, Disease Box 14.1, p. 463

42) Explain (or draw a clearly labeled diagram) the sequence of events that occurs during translation in eukaryotic cells as a protein elongates by one amino acid. At each step, specify what is happening in the ribosome’s A site, P site, and E site. Indicate at what points GTP hydrolysis is required.

Answer: Section 14.5, p. 463-467, Fig. 14.11 43) Do stop codons always specify termination? Explain your answer.

Answer: Section 14.5, p. 463, Fig. 14.12 44) tRNA was originally thought to be a static “adaptor” during protein synthesis. Discuss recent evidence suggesting that tRNA plays an active role in initial selection and proofreading during translation.

Answer: Section 14.5, p. 463-466, Fig. 14.13 45) Draw the bond forming between the growing polypeptide chain and the amino acid attached to the tRNA in the A site. Name the bond.

Answer: Section 14.5, p. 466-467, Fig. 14.14 46) What is the function of and where is the “peptidyl transferase” region of the ribosome? Of what is this region primarily composed?

Answer: Section 14.5, p. 467-469, Fig. 14.14, Fig. 14.16 47) Diagram the cotranslation translocation pathway mediated by the signal recognition particle (SRP) from the ribosome to the endoplasmic reticulum.

Answer: Section 14.5, p. 470-471, Fig. 14.18 48) The E. coli trigger factor has been described as hunching over the exit of the ribosome tunnel “like a crouching dragon.” What is it doing?

Answer: Section 14.5, p. 471, Fig. 14.19 49) Does a special tRNA bind to the stop codon in the mRNA during termination?

Answer: Section 14.6, p. 471, Fig. 14.20 50) Diagram the mechanism of translational control by phosphorylation of eukaryotic initiation factor 2 (eIF2), starting with activation of protein kinase RNA (PKR).

Answer: Section 14.7, p. 472-475, Fig. 14.21, Fig. 14.22 51) Imagine you have treated reticulocytes from a heme-regulated inhibitor kinase (HRI) knockout mouse with hemin. Predict the effect of this treatment on eIF2 phosphorylation state and globin synthesis. Describe an experiment to test your predictions. Shown sample positive results.

Answer: Section 14.7, p. 475-476, Fig. 14.23 52) Describe a toeprint assay involving mammalian ribosomal subunits and a mRNA in a cellfree extract that contains all the factors necessary for translation. (a) What results would you expect to see with 40S ribosomal subunits alone? (b) With 60S subunits alone? (c) With both subunits and all amino acids except serine, which is required in the 15th position of the polypeptide?

Answer: Section 14.4, Tool Box 14.1, p. 462-463. In a toeprint assay, a radiolabeled

oligonucleotide primer is annealed to the mRNA template and is extended by reverse transcriptase to form a cDNA. On naked mRNA, extension proceeds to the 5′ end of the template. When ribosomes and their associated factors are bound at discrete positions on the RNA they inhibit primer extension, generating shorter cDNA fragments. cDNA products are analyzed on denaturing sequencing gels on which the full-length cDNA migrates most slowly. A ribosome with the AUG initiation codon in its P site blocks the movement of reverse transcriptase on the RNA template, causing reverse transcription to stop 15-17 nt downstream of the A of the AUG initiator. (a) With 40S ribosomal subunits, a toeprint would form 15-17 nt downstream of the A of the AUG initiator, caused by the assembly of 48S initiation complexes. (b) With 60S ribosomal subunits, there would be no toeprint. Extension would proceed to the 5′ end of the template because the 60S subunit cannot bind in the absence of the 40S ribosomal subunit. (c) With both subunits and all amino acids except serine, which is required in the 15th position of the polypeptide, the ribosome would stall when the codon for serine was in the A site. The toeprint would form 15-17 nt downstream of the first base of the codon for serine.

53) What would be the effect on the function of eIF2 of inhibiting its guanine exchange factor?

Would the outcome be different if eIF2 were phosphorylated?

Answer: Section 14.4, p. 459-461, Fig. 14.9. eIF2 would not be able to form a ternary complex with methionyl-tRNA if its guanine exchange factor (eIF2B) was inhibited. eIF2B mediates exchange of GDP for GTP in the eIF2 complex. Only the GTP-bound form of eIF2 can bind the initiator tRNA. The outcome would be the same if eIF2 was phosphorylated. Phosphorylation of eIF2 inhibits the exchange of GDP for GTP on the eIF2 complex, and thereby prevents formation of the ternary complex.

54) You are investigating a tRNA whose charging specificity appears to be affected by a U11:G24 to U11:A24 base pair in the D stem. Design an experiment using an in vitro reaction to show that changing this base-pair changes the charging specificity of the tRNA. Answer: Section 14.5, p. 565-466, Fig. 14.13. Measure rates of GTP hydrolysis by mixing programmed bacterial ribosomes with purified ternary complexes composed of EF-Tu, [32 P]GTP and with the wild-type or variant tRNA for tryptophan (Trp). Measure tRNA accommodation by assaying peptide bond formation. Mix programmed ribosomes containing f[35S]-Met-tRNAfMet in the P site with wild-type and mutant Trp-tRNATrp ternary complexes.

Quantify the amount of [35S]fMet-Trp dipeptide formed over time, on UGG (cognate codon), UGA or CGG programmed ribosomes. One possible outcome would be that the U11-A24 tRNATrp variant accelerates forward rates of GTPase activation and accommodation independent of correct codon:anticodon pairing. The mutation allows movement of tRNA into the peptidyl transferase center even on the stop codon UGA. 55) A skeptic thinks that only proteins can be enzymes. Provide X-ray crystallographic and biochemical evidence in support of the view that the ribosome is a ribozyme.

Answer: Section 14.5, p. 467-468, Figs. 14.15 and 14.16. Harry Noller’s group used the “fragment reaction,” which is a simple in vitro assay for peptidyl transferase activity, to test whether ribosomal RNA alone could catalyze formation of a model peptide bond. They treated bacterial 50S ribosomes with proteinase K and SDS. They showed that after protein extraction, the remaining 23S rRNA still had peptidyl transferase activity. Importantly, they also showed

that this catalytic activity was inhibited by treatment with ribonucleases and by known peptidyl transferase inhibitors. Their findings strongly suggested that 23S rRNA alone can catalyze peptide bond formation. Atomic resolution views of the large ribosomal subunit from the archaeon Haloarcula marismortui show that about two-thirds of the ribosome’s mass is composed of RNA. rRNA both creates a structural framework for the ribosome and, at the same time, forms the main features of its functional sites. The rRNA forms most of the intersubunit interface, the peptidyl transferase center, the decoding site, and the A and P sites. The ribosomal proteins are abundant on the exterior of the ribosome, but not in the active site. The ribosomal proteins follow the contours of the RNA. No protein is located within 15 Å of the site of catalysis.

Fundamental Molecular Biology, 2nd Edition Instructor Test Bank Lizabeth A. Allison

Chapter 15 Genetically Modified Organisms: Use in Basic and Applied Research

Multiple Choice 1) When making a transgenic mouse by microinjection of foreign DNA into fertilized eggs A. there is random integration of the transgene into the embryonic genome.

B. there is disruption or mutation of a targeted gene in the embryonic genome.

C. the resulting mice are genetically identical to the donor nucleus.

D. the integration event into the embryonic genome usually occurs by homologous recombination.

Answer: A 2) When making transgenic mice, to determine whether there has been stable integration of a transgene into the mouse chromosome, tail biopsies are taken from the mouse pups and tested for the presence of the transgene by A. Northern blot B. Southern blot C. Western blot D. Chromatin immunoprecipitation Answer: B 3) Mating mice from different founder lineages that carry identical transgenes will result in A. one transgene being expressed and the other silenced.

B. a heterozygote for the transgene.

C. a true homozygote in which independent segregation of the transgenic loci is predictable.

D. a homozygote in which segregation of the transgenic loci is not necessarily independent.

Answer: D 4) When making a mouse by gene targeting A. there is random integration of the transgene into the embryonic genome.

B. there is disruption or mutation of a targeted gene in the embryonic genome.

C. the resulting mice are genetically identical to the donor nucleus.

D. the integration event into the embryonic genome usually occurs by nonhomologous recombination.

Answer: B 5) Assume you have transfected embryonic stem cells with a gene-targeting vector containing the neomycin resistance gene as a positive selection marker and the thymidine kinase gene as a negative selection marker. You observe that some cells are resistant to neomycin but killed by ganciclovir. What was the outcome of gene targeting procedure?

A. No recombination.

B. Homologous recombination.

C. Nonhomologous recombination.

D. The transfection did not work.

Answer: C

6) You have received a lineage of “knockout mice” from a collaborator. What method would not be useful to confirm that the mice really do not express the targeted gene in all tissues of the body?

A. Northern blot B. Neomycin sensitivity assay C. Western blot D. Reverse transcriptase-PCR Answer: B 7) Gene knockouts often result in embryonic lethality. What strategy do investigators often use to make it possible to study a gene’s role later in development?

A. Gene knockout using an inducible tetracycline resistance gene.

B. Use of embryonic stem cells from a strain of mice with brown fur and a blastocyst from a strain of mice with black fur.

C. Implantation of embryos into a pseudopregnant female.

D. Inducible Cre/Lox system for site-specific recombination.

Answer: D 8) Creation of which of the following does not rely on homologous recombination into the genome?

A. transgenic mice B. knockout mice C. knockin mice D. conditional knockout mice Answer: A 9) Eduardo Kac the transgenic artist produced which of the following?

A. Mickey the GFP mouse B. A sculpture of the first transgenic rat-sized mouse C. Alba the GFP bunny D. A photo montage of pronuclear microinjection Answer: C 10) Which of the following correctly orders cells according to their phenotypic potential from greatest to least?

A. pluripotent, totipotent, terminally differentiated B. totipotent, terminally differentiated, pluripotent C. terminally differentiated, pluripotent, totipotent D. totipotent, pluripotent, terminally differentiated Answer: D 11) All of the following are true of stem cells except: A. They can be found in many developing organs.

B. They are terminally differentiated.

C. They have a high replicative capacity.

D. They may differentiate to replace cells lost to injury.

Answer: B 12) When cloning an organism by nuclear transfer A. there is random integration of a transgene into the recipient cell’s nuclear genome.

B. there is disruption or mutation of a targeted gene in the resulting organism.

C. the resulting organism is genetically identical to only the donor nucleus and not the recipient cell’s nucleus.

D. the cytoplasm of the donor cell is transferred to the recipient cell.

Answer: C 13) The term “reprogramming” refers to A. genetic modification of cells for the purpose of reproductive cloning B. altering gene expression to revert a somatic cell to a stem cell state C. artificial telomere elongation to effectively reduce cell age D. the inducible excision of a gene in a conditional knockout mouse Answer: B 14) The results of nuclear transplantation experiments in Xenopus laevis by John Gurdon and co-workers established the basic principle that A. cell differentiation depends on changes in the expression not content of the genome.

B. cell differentiation depends on changes in the content not expression of the genome.

C. cell differentiation depends on changes in both the expression and content of the genome.

D. cloning by nuclear transfer is only possible in amphibians.

Answer: A 15) When nuclear transfer is performed by cell fusion, the resulting cloned offspring A. will be genetically identical to the original adult cell nucleus but the mitochondrial DNA will come from the recipient egg.

B. will be genetically identical to the original adult cell nucleus but the mitochondrial DNA will come from both the donor adult cell cytoplasm and the recipient egg.

C. will be genetically identical to the original adult cell nucleus and the mitochondrial DNA will come from the donor adult cell cytoplasm.

D. will be a chimera of nuclear and mitochondrial DNA from the original adult cell and the recipient egg.

Answer: B 16) Which of the following is not a suspected reason for the inefficiency/poor outcomes of reproductive cloning?

A. inefficient reprogramming B. premature terminal differentiation in the developing embryo C. telomere length D. chromosome missegregation during early embryonic divisions Answer: B 17) “Pharming” is A. the growth of transgenic plants as food crops B. the use of Agrobacterium tumefaciens to transform animal cells C. the use of genetically modified organisms to produce pharmaceutical agents D. the use of plants or algae to produce biofuels Answer: C 18) What information did the study of crown gall disease provide that was critical for plant genetic engineering?

A. Understanding the role of plant hormones in growth promotion.

B. Knowledge of plant tumor-inducing genes.

C. Understanding the role of opine synthesis enzymes in tumor progression.

D. Discovery of a plasmid that introduces virulence genes into plants.

Answer: D 19) Plasmids are used as vectors in both plant and bacterial recombinant DNA technology.

However, there is a major difference in the fate of genes introduced into bacteria on most bacterial plasmids and into plants on Ti plasmids. What is this difference?

A. In bacteria, genes are stably expressed; in plants, gene expression is quickly lost.

B. Gene expression tends to decrease rapidly and unpredictably in bacteria; gene expression is much more stable in plants.

C. Bacterial plasmids are circular DNAs; Ti plasmid DNA is linear.

D. Bacterial plasmids and the genes they carry usually are not integrated into the chromosome; Ti plasmids and the genes they carry are integrated into the chromosome.

Answer: D 20) Which is not a method by which one can introduce foreign DNA into a plant cell?

A. calcium chloride treatment followed by heat shock B. infection with a specific bacterium C. microballistic “gene gun” D. electroporation Answer: A

Short answer/analytical 21) Describe the method for creating a transgenic mouse by pronuclear injection. What information can a transgenic mouse provide?

Section 15.2, p. 481-483, Fig. 15.2, Table 15.1 22) You are trying to make a transgenic mouse but there is very poor survivorship of the embryos after introduction of the foreign gene by pronuclear injection. Describe a method you could use to get around this problem.

Section 15.2, p. 485-486, Fig. 15.4 23) You have cloned the cDNA for human XPA. You now attempt to use the human XPA cDNA (linked to a promoter) to correct the genetic defect in mice that have the mouse equivalent of the human disease xeroderma pigmentosum, by making transgenic mice. Briefly explain how you would analyze the transgenic mice for: • Stable integration of the transgene into the mouse’s genomic DNA • Transcription of the transgene • Translation of the transgene • Ability of the XPA protein to correct the genetic defect in the transgenic mice.

Answer: Section 15.2, p. 482-85, Fig. 15.2, see also Chapter 7, Disease Box 7.2, p. 175 Southern blot hybridization; Northern blot, RT-PCR; Western or blot or gene activity (e.g., enzymatic) assay; Phenotypic analysis of the mice (e.g., skin pigmentation abnormalities) 24) Describe the method for creating a knockout mouse by gene targeting. Explain the importance of selectable marker genes in this procedure. What information can a knockout mouse provide?

Answer: Section 15.3, p. 486-490, Fig. 15.5, Table 15.1

25) You have created a strain of knockout mice. Describe an experiment to confirm that the mice lack expression of the targeted gene at the level of mRNA expression. Show sample positive results.

Answer: Section 15.3, p. 489-490, Fig. 15.6 26) Discuss some of the pros and cons of using genetically modified nonhuman primates in experimental medicine.

Answer: Section 15.4, p. 493-494 27) Describe a recently developed method for creating transgenic livestock by linker-based sperm-mediated gene transfer. Why don’t investigators just use pronuclear microinjection?

Answer: Section 15.4, p. 494, Fig. 15.9 28) Discuss some of the potential applications of “gene pharming.”

Answer: Section 15.4, p. 494 29) Describe amphibian nuclear transplantation experiments that established the basic principle that cell differentiation depends on changes in gene expression not changes in the content of the genome.

Answer: Section 15.5, p. 496-497, Fig. 15.10 30) Describe the method for cloning of mammals by nuclear transfer. What information can a cloned organism provide?

Answer: Section 15.5, p. 497-498, Fig. 15.12, Table 15.1 31) Less than 1% of all nuclear transfers from adult differentiated cells result in normalappearing cloned offspring. Give possible explanations for this lack of efficiency.

Answer: Section 15.5, p. 500, Fig. 15.13 32) Will a cloned calico cat have the exactly the same pattern of orange and black patches on her coat compare with her calico donor? Explain your answer.

Answer: Section 15.5, p. 501, Focus Box 15.4 33) Discuss some of the controversial applications of cloning by nuclear transfer.

Answer: Section 15.5, p. 502-506 34) What kind of vector would you use to insert a transgene into a plant such as tobacco?

Diagram the process you would use.

Answer: Section 15.6, p. 506-508, Fig. 15.17 35) Describe an experiment to determine whether the tomato you just bought from the grocery store is genetically modified. Show sample positive results.

Answer: Section 15.6, p. 506-508, Focus Box 15.5. (e.g. design primers specific for Ti plasmid sequences, extract DNA from tomatoes, use PCR to test for the presence of the Ti plasmid) 36) Design a Southern blot experiment to check a transgenic mouse’s DNA for integration of the GFP cDNA under control of a constitutive promoter. You many assume any array of restriction sites you wish in the target mouse chromosome and in the GFP transgene. (a) Show sample results for a successful and an unsuccessful integration. (b) Will mice from different founder lineages have the same site of integration?

(c) Describe techniques you would use to analyze the transgenic mice for transcription of the transgene and translation of the transgene.

Answer: (a) Section 15.4, Figure 15.9, p. 482-483; see also Section 8.6, Tool Box 8.7, p. 218-219.

Southern blot of transgenic mouse DNA hybridized with a GFP cDNA probe: input plasmid

unintegrated integrated plasmid (one plasmid recognition site for restriction endonuclease) − electrode

Direction of migration

+ electrode

(b) Section 15.2, p. 483. Mice from different founder lineages are not likely to have the same site of integration because the site of integration is random. (c) Section 15.2, p. 483, Figure 15.3. To analyze the mice for transcription of the transgene, a Northern blot, RT-PCR, or in situ hybridization could be used. To analyze the mice for translation of the transgene, a Western blot could be performed, or mice (whole animals or cells) could be viewed under UV light to look for GFP expression.

37) What would be the outcome if the procedure used to make gene-targeted (knock-out) mice was altered in the following ways (consider each situation separately). (a) The embryo-derived stem (ES) cells did not possess a gene responsible for visible phenotypic trait (such as agouti coat color) that differed from that of the cell in the recipient blastocyst. (b) The inserted “disrupting” gene (e.g. neoR) did not insert into the gene of interest. (c) Chemical transfection or electroporation failed. (d) The gene knock-out was embryonic lethal Answer: (a) Section 15.2, p. 482-483. If the ES cells did not possess a gene responsible for a visible phenotypic trait that differed from that of the cell in the recipient blastocyst, the outcome of the procedure would be the same, but the screening protocol would be much more labor intensive and cumbersome. Prior to breeding, all mice would have to be checked for their genotype by PCR or Southern blot analysis. (b) Section 15.3, p. 487-489. If the “disrupting gene” (e.g. neoR) did not insert into the gene of interest, all ES cells would be sensitive to neomycin and would be selected against in culture.

There would be no ES cells to inject into the recipient blastocyst.

(c) Section 15.3, p. 487-489. If the chemical transfection or electroporation procedure failed, all ES cells would be sensitive to neomycin and would be selected against in culture. (d) Section 15.3, p. 490. If the gene knockout was embryonic lethal, then it would not be possible to obtain homozygotes for the knockout since no mouse pups would be born from matings of heterozygous mice.

38) Consider the following hypothetical scenario: A young scientific genius, Dolly, is terminally ill with hereditary nonpolyposis colon cancer (HNPCC). When she dies, her parents feel that one of the most remarkable minds in science will die with her, and they feel they owe it to the world to not let this happen. The family travels to a secret lab on a small offshore island, which allegedly performs “reproductive cloning" (cloning by somatic cell nuclear transfer). Dolly’s parents hope to clone her from one of her skin cells. (a) If the lab is successful in cloning Dolly, can they guarantee that her clone will be as academically gifted as Dolly? Why or why not? (b) Would Dolly and her clone have identical DNA “fingerprints”? Why or why not? (c) Would Dolly and her clone share the same mitochondrial DNA? Why or why not? (d) Would you expect there to be a difference in the length of telomeres in Dolly's skin cells versus in the cells of her embryonic clone? Why or why not?

Answer: Section 15.5, p. 500-502. (a) There is no guarantee that Dolly’s clone will be as academically gifted as Dolly because intelligence is a polygenic trait that depends on gene-gene and gene-environment interactions. (b) Dolly and her clone would most likely have identical DNA “fingerprints” since their nuclear DNA should be identical (c) Dolly’s clone would be a hybrid containing mitochondrial DNA (mtDNA) from the cytoplasm accompanying Dolly’s donor nucleus, and the mitochondria from the recipient egg. If Dolly provided the recipient egg, then they would have identical mitochondrial DNA. (d) This would depend on whether telomerase was reactivated in the embryonic clone. If it wasn’t, then the already short telomeres from Dolly’s skin cell nucleus would continue to get progressively shorter with each round of cell division and DNA replication. If telomerase was reactivated, then the telomeres would be extended and maintained, and would be longer than those in Dolly’s skin cells.

39) A team of molecular biologists seeks your advice regarding whether it would be more important to include regulatory elements such as insulators and matrix attachment regions in a gene construct for generating transgenic mice by gene-targeting in embryo-derived stem cells, or for generating transgenic mice by microinjection of foreign DNA into pronuclei of fertilized eggs? Provide advice and explain your rationale.

Answer: Section 15.3 p. 487-490; Section 15.2 p. 480-483; see also Section 11.3, p. 300301 and p. 303-304. It would be more important to included regulatory elements such as insulators and matrix attachment regions in a gene construct for generating transgenic mice by microinjection of foreign DNA into pronuclei of fertilized eggs. This is because in this technique, random integration of the transgene occurs and it could either integrate into euchromatin or heterochromatin, or could integrate into a region of DNA where it was inappropriately activated or repressed by an enhancer or silencer. Including insulators or MARs in the construct would promote formation of a position-independent chromatin domain. Gene targeting occurs by sitespecific homologous recombination, so these regulatory elements would not be necessary.

Fundamental Molecular Biology, 2nd Edition Instructor Test Bank Lizabeth A. Allison

Chapter 16 Genome Analysis: DNA Typing, Genomics, and Beyond

Multiple Choice 1) The majority of the human genome is A. protein-coding gene exons B. intergenic DNA C. genes and gene-related sequences D. interspersed elements Answer: B 2) What are minisatellites and microsatellites?

A. Small, extrachromosomal loops of DNA that are similar to plasmids.

B. Parts of viruses that have become integrated into the genome of an organism.

C. Incomplete or inactive remains of transposable elements in a host cell.

D. Short and simple repeated sequences in DNA. Answer: D 3) The variable number of short tandem repeats in the human genome primarily results from A. unequal crossing-over.

B. slippage during DNA replication.

C. alternative splicing.

D. movement of transposable elements.

Answer: B 4) The variable number of minisatellite repeats in the human genome primarily results from A. unequal crossing-over.

B. slippage during DNA replication.

C. alternative splicing.

D. movement of transposable elements.

Answer: A 5) Forensic DNA typing often involves recovery of very tiny amounts of DNA from a crime scene. What is a major concern in analyzing this DNA? A. There is rarely enough DNA to allow PCR amplification.

B. Even with these minute amounts of DNA, there still may be too much DNA for efficient PCR. C. It is important to avoid the introduction of contaminating human DNA into the sample.

D. Suspects may refuse to provide a sample of their DNA. Answer: C

6) In establishing paternity by DNA typing, how many short tandem repeat (STR) alleles in a child come from the father?

A. ¼ B. ½ C. ¾ D. none Answer: B 7) At approximately 2 PM; Ms. Smith, Ms. Taylor, and Ms. Jones each delivered a healthy baby girl at a local hospital. At 2:16 PM, the hospital's fire alarm sounded. Nurses and orderlies hurried to evacuate patients, and the three newborns were rushed to safety. After the danger had passed, the hospital staff discovered that in the confusion, they had forgotten which baby was which. Since the babies were moved before receiving their identification bracelets there was no easy way to identify them. The attending physician ordered that DNA typing be performed on the babies and their parents. Results of gel analysis are shown below. Your job is to decide to which set of parents Baby 1 belongs.

Smith

Taylor Jones

A. Baby 1 belongs to the Jones B. Baby 1 belongs to the Taylors C. Baby 1 belongs to the Smiths Answer: B

8) Multiplex PCR is A. the simultaneous amplification of many target DNA sequences of interest in one reaction by using more than one pair of primers.

B. the amplification of one target DNA sequence of interest in one reaction by using more than one pair of primers.

C. the amplification of many target DNA sequences of interest by performing multiple reactions with a different pair of primers in each reaction tube.

D. the amplification of one target DNA sequence of interest by performing multiple reactions with a different pair of primers in each reaction tube.

Answer: A 9) Following up on reports of a man who attacked an elderly lady (Victim Reference Sample) and spit in her face, the police went to the homes of Suspect 1 and Suspect 2, who each gave police a cheek swab. The electropherogram resulting from STR analysis is shown below. The Evidence Sample is spit from a handkerchief the elderly woman used to wipe her face. Your job is to decide which suspect committed the crime.

A. Suspect #1 committed the crime B. Suspect #2 committed the crime Answer: A 10) Which statement is not correct?

A. Mitochondrial DNA (mtDNA) typing usually involves PCR plus DNA sequencing.

B. mtDNA typing is often used for analysis of ancient DNA samples.

C. mtDNA typing can be used to trace paternal lineages.

D. mtDNA typing can be used to trace maternal lineages.

Answer: C

11) What distinguishes the field of genomics from genetics?

A. Genomics deals with genomes; genetics deals with gene products.

B. A goal of genomics is to understand how genes influence both genotype and phenotype; genetics only considers how genes influence phenotype.

C. Genomics is based almost solely on studies of human genome sequences; genetics is based on studies of organisms from bacteriophage to Arabidopsis.

D. Genomics is based on DNA sequence analysis and considers the sequence, function and interactions of many genes; genetics more often considers one or a small number of genes at a time and is generally involved in the passage of hereditary traits from one generation to the next.

Answer: D 12) All of the following apply to the discipline of bioinformatics except: A. Bioinformatics is a synonym for genomics.

B. The most commonly used bioinformatics tool is BLAST. C. Bioinformatics is an area of computer science devoted to to collecting, organizing and analyzing DNA and protein sequences.

D. Bioinformatics tools support high-throughput screening methods for both proteomic and genomic research programs Answer: A 13) What is the difference between an open reading frame (ORF) and a gene?

A. There is no difference.

B. An ORF is a potential gene identified by a potential protein-coding segment in DNA bordered by start and stop codons; a gene is a DNA segment known to produce a product.

C. An ORF is a known gene-containing DNA segment that is shorter than a full-length gene.

D. An ORF may or may not have introns; a gene must have introns.

Answer: B 14) Which best describes the logic behind shotgun sequencing?

A. Fragment the genome into short pieces. Sequence each piece. Use overlapping ends to assemble the pieces in the correct order.

B. Start with one end of each chromosome. Sequence straight through to the other end of the chromosome.

C. Use a variety of techniques to identify genes and open reading frames. Sequence these segments but not the noncoding and repeated sequences.

D. Fragment the genome into pieces. Map the location of each piece. Then sequence each piece.

Answer: A 15) What evidence do molecular biologists use to infer that a gene is part of a gene family?

A. Its sequence is almost exactly identical to that of a gene in another organism.

B. Its structure (meaning its pattern of exons and introns) is identical to that of a gene found in another species.

C. Its composition, in terms of percentage of A-T and G-C pairs, is unique.

D. Its sequence, structure, and composition are similar to those of another gene in the same genome.

Answer: D

16) What is a pseudogene?

A. A coding sequence that originated by DNA slippage during replication.

B. A gene whose sequence is similar to that of functioning genes but does not produce a functioning product.

C. A gene with more than two alleles present in a population.

D. An open reading frame (ORF) whose function has not yet been established.

Answer: B 17) An expectation that has been borne out by comparative genomic analysis is that: A. Increased degrees of complexity in multicellular organisms are closely correlated with increased genome sizes.

B. Increased genome size correlates with an increased number of genes.

C. All vertebrate genomes are roughly the same size.

D. Human and chimpanzee sequences are closely related.

Answer: D 18) Like DNA microarrays, Northern blots can be used to analyze levels of gene expression.

However, Northern blots are almost always prepared with a probe tagged with one label, whereas DNA microarrays almost always use two groups of probes, each one tagged with a different fluorescent label. What advantage is conferred by using two labels for the probes of microarrays?

A. The two labels increase the sensitivity of the microarray.

B. The two labels allow the analysis of the expression of more genes.

C. The two labels allow simultaneous comparison of gene expression between two samples.

D. The two labels allow simultaneous assay of levels of RNA and protein.

Answer: C 19) To begin the process of DNA microarray, the following cellular component is extracted from sample tissues: A. DNA B. RNA C. protein D. A and B Answer: B 20) In a comparative gene expression analysis between two samples labeled with Cy3 (green) and Cy5 (red), if the two samples contained equivalent amounts of mRNA for a particular gene, the spot on the array corresponding to that gene would appear A. red.

B. green C. brown D. yellow Answer: D 21) Which technique has almost eliminated the need to chemically sequence a polypeptide?

A. peptide mass fingerprinting using MALDI-TOF B. functional protein array C. DNA microarray D. whole-genome shotgun sequencing Answer: A

22) You want to analyze the entire proteome of a particular organelle. Which technique would you use?

A. DNA microarray B. tandem mass spectrometry C. Western blot D. analytical protein array Answer: B 23) Single nucleotide polymorphisms are A. mutations that cause genetic diseases B. the most common type of sequence variation in an individual C. variable number tandem repetitive sequences D. Answers A and B apply.

Answer: B 24) Which two types of genomic variation have been commonly analyzed in genome-wide association studies?

A. copy number variation and small deletions B. small deletions and single nucleotide polymorphisms C. small deletions and small insertions D. single nucleotide polymorphisms and copy number variation Answer: D 25) Genome-wide association studies are used for identifying the causes of A. monogenic (single gene defect) diseases B. diseases with multiple genetic components C. diseases that do not have a genetic component D. all of the above Answer: B

Short answer/analytical 26) Compare and contrast the techniques of AFLP and RAPD analysis in DNA typing.

Answer: Section 16.2, Focus Box 16.1, p. 512-513; p. 519-520, Fig. 16.5 27) About 60% of the human genome is composed of intergenic DNA. Describe two types of intergenic DNA that are used in DNA typing and how they are used.

Answer: Section 16.2, p. 514-518, Table 16.1, Figure 16.1, 16.2, 16.3 28) What type of analysis is currently the most widely used DNA typing procedure in forensic genetics – short tandem repeat (STR) analysis or minisatellite analysis? Which technique uses multiplex PCR? Which technique uses a Southern blot?

Answer: Section 16.2, p. 514-518, Table 16.1 29) “The power of DNA evidence lies in statistics.” Explain the meaning of this statement.

Answer: Section 16.2, p. 514, Table 16.2

30) When is mitochondrial DNA analysis considered the method of choice in forensic analysis?

Explain why it should not be used for paternity testing.

Answer: Section 16.2, p. 518-519 31) Compare and contrast the terms “bioinformatics,” “genomics,” and “proteomics.”

Answer: Section 16.3, p. 520-523 32) Compare and contrast the clone-by-clone sequencing strategy and the shotgun sequencing strategy for large genomes.

Answer: Section 16.4, p. 524-525, Fig. 16.7 33) When is a sequence considered a “rough draft” as opposed to a “finished” sequence? Does a genome sequence have to be 100% complete and accurate to be declared “finished”?

Answer: Section 16.4, p. 525 34) The Japanese pufferfish (Takifugu rubripes) genome is approximately eight times smaller than the human genome, but contains just as many protein-coding genes. How can that be?

Answer: Section 16.4, p. 525-526, Focus Box 16.3 35) What is the meaning of the term “synteny” in comparative genomics?

Answer: Section 16.4, Focus Box 16.3, p. 526-527 36) Is it accurate to call a cDNA a gene? Why or why not?

Answer: Section 16.4, p. 528-529 37) What is an open reading frame (ORF)? Write a DNA sequence containing a short ORF. Answer: Section 16.4, p. 528-529, See Section 4.2, p. 68-69, Table 4.1 38) Describe the process of making a DNA “chip.”

Answer: Section 16.5, p. 529 39) Describe a hypothetical experiment using a DNA microarray to measure the expression of viral genes at two stages of infection of human cells by the virus. Present sample results.

Answer: Section 16.5, p. 529-531, Fig. 16.9 40) Match each technique for high-throughput analysis of proteins with its possible application(s): Technique Analytical protein array Peptide mass fingerprinting using MALDI-TOF Shotgun proteomics using MS/MS Application Analysis of a single isolated protein from a protein gel Analysis of an entire proteome Analysis of ligand binding Answer: Section 16.5, p. 531-533, Figs. 16.10, 16.11, 16.12

41) What kinds of questions can be better addressed using a protein microarray rather than a DNA microarray?

Answer: Section 16.5, p. 529; p. 531. For example, a protein array could be used to identify a drug target or an antibody’s antigen, whereas a DNA microarray could not be used for these purposes.

42) Compare and contrast an analytical protein array with a functional array. What type of information can be obtained from each method?

Answer: Section 16.5, p. 531, Fig. 16.10 43) Diagram the process of peptide mass fingerprinting using MALDI-TOF. What type of information can be obtained from this method?

Answer: Section 16.5, p. 531-532, Fig. 16.11 44) Diagram the process of shotgun proteomics using tandem mass spectrometry. What type of information can be obtained from this method?

Answer: Section 16.5, p. 533, Fig. 16.12 45) Explain the key difference(s) between peptide mass fingerprinting using MALDI-TOF and shotgun proteomics using tandem mass spectrometry.

Answer: Section 16.5, p. 531-533 46) You have characterized the entire proteome of a particular organelle. What further analysis would you do to confirm localization of the identified proteins to this organelle in situ?

Answer: Section 16.5, Focus Box 16.4, p. 533-534 47) Write an example of a single nucleotide polymorphism. To be considered a SNP, how frequent must the polymorphism be in the population?

Answer: Section 16.6, p. 535 48) How could DNA microarrays be used in conducting a genome-wide association study to identify SNPs associated with, e.g., type II diabetes?

Answer: Section 16.5, p. 529-531, Fig. 16.9; Section 16.6, p. 535. One set of diseased patients and one set of control patients are genotyped at many genomic sites using DNA microarrays. The genomic DNA of each individual is amplified, labeled, and hybridized to an array containing probes that correspond to different haplotypes. The relative prevalence of different haplotypes between the diseased and control genotypes can then be analyzed statistically.

49) Variable number tandem repeats in the promoter region of the monoamine oxidase A gene have been linked to aggressive, impulsive, and violent behavior. A child has been classified as having the 3-repeat, low activity allele. Are they likely to show antisocial behavior as adults?

Does this example support the statement “DNA is destiny?”

Answer: Section 16.6, p. 538-539, Fig. 16.13, Fig. 16.14

50) A cigarette butt found at the scene of a violent crime is found to have a sufficient number of epithelial cells stuck to the paper for the DNA to be extracted and typed. Shown below are the results of typing for three probes (locus 1, locus 2, and locus 3) of the evidence (X) and 4 suspects (A through D). Which of the suspects can be excluded? Which cannot be excluded?

Can you identify the criminal? Explain your reasoning.

A

B

C

D

A

B

C

D

A

X

B

C

D

X

Locus 1

X

Locus 2

Locus 3

Answer: Section 16.2 p. 512-516, Figure 16.1, Figure 16.2. Suspects A, B, and D can be excluded because they do not have matching bands at all three loci analyzed. Suspect C cannot be excluded because the banding patterns match at all three loci. To increase the probability that the band patterns are not likely to be identical solely by chance, more loci should be analyzed. Currently, the FBI tests at least 13 different loci. More evidence would be required to charge Suspect C with the crime, since the cigarette butt only suggests that Suspect C was present at the crime site, not that he/she committed the violent crime. NOTE: I can’t see the data so I can’t tell what type of DNA analysis was done here - I’m assuming microsatellite in the reference above.

51) Based on the DNA fingerprinting (minisatellite analysis with a multilocus probe) results shown in the figure below, is Mr. X or Mr. Y the child’s father? Explain your answer.

Lane1: mother Lane 2: mother's child Lane 3: Mr.Y Lane 4: Mr. X

1

2

3

4

4

Answer: Section 16.2 p. 512-516, Figure 16.1, Figure 16.2. Mr. X is the child’s father. There is a simple Mendelian pattern of inheritance. For a child, 50% of the bands are derived from the mother and 50% are derived from the father. The bands that come from the mother can be identified; the remaining bands must come from the father, and these match the bands present in the lane representing Mr. X. 52) You separate a mixture of proteins by 2D-PAGE. From one protein spot, you obtain the following peptide sequence by mass spectrometry: EQAGGDATENFEDVGHSTDAR Use the BLAST tool to identify the unknown protein.

Answer: Section 16.3, p. 520, Fig. 16.6. Cytochrome b5 (Note that this question would require student access to a computer database during the exam.)

53) After sequencing the entire genome of a novel organism by the whole-genome shotgun approach, you use a computer program to search for start/stop signals for translation and homology searches based on similarity to cDNAs in the database. You predict approximately 20,000 genes. Is this an accurate representation of the total number of genes in the organism?

Why or why not?

Answer: Section 16.4, p. 525-526. Looking for start/stop signals for translation and similarity to cDNAs would only predict protein-coding genes. This would not be an accurate representation of the total number of genes in the organism because it would miss all the genes encoding functional RNAs (e.g. microRNAs, small nucleolar RNAs, ribosomal RNAs, tRNAs, small nuclear RNAs, etc.)

54) After SNP analysis of a region of the Apolipoprotein E gene, you determine that a patient carries the apoE4 allele. This allele is linked with Alzheimer’s disease. Counsel the patient regarding their prognosis.

Answer: Section 16.6, Disease Box 16.1, p. 535. Individuals who inherit at least one apoE 4 allele will have a greater chance or getting Alzheimer’s disease. However, inheritance of this allele is not an absolute indicator of disease development. Someone who has inherited two 4 alleles may never develop Alzheimer’s, while another one who has inherited two 2 alleles may.

The apoE gene is just one of several that have been linked to Alzheimer’s disease.

55) cDNA derived from normal cell mRNA was labeled with a green fluorescent tag (represented as black). cDNA derived from diseased cell mRNA was labeled with a red fluorescent tag (represented as white; the presence of both green and red tags is represented by stripes). The labeled cDNAs were mixed and hybridized to a prepared DNA microarray.

Results from a small section of the microarray grid are shown below. Which genes would you select for further analysis? Explain your answer.

Answer: Section 16.5, p. 529-531, Figure 16.9. Genes that are differentially expressed in diseased versus normal cells should be selected for further analysis. Any green spots (represented as black) indicate a gene, which is expressed at higher levels in a normal cell. Any red spots (represented as white) indicate a gene that is expressed at higher levels in a diseased cell. The presence or absence of expression of particular genes may be of equal importance in the progression of the disease.

Yellow spots (represented as stripes) indicate genes that are expressed at similar levels in both cell types and would not be selected for further analysis at this point.

56) In a study that involved 100 Caucasian males, you find a polymorphism in a gene for a neurotransmitter receptor. The “long allele” is linked with reckless behavior. What conclusions, if any, can you draw about the results of this study? Should you hold a press conference to announce the discovery of the “reckless” gene?

Answer: Section 16.6, p. 537-539, Figure 16.13. One can only conclude that the genetic cause of reckless behavior is closely linked with the neurotransmitter receptor, most likely residing in the same haplotype block. It is necessary to conduct further experiments to support a causal role for the long allele, and you should not hold a press conference announcing the neurotransmitter receptor as the “reckless” gene.

Fundamental Molecular Biology, 2nd Edition Instructor Test Bank Lizabeth A. Allison

Chapter 17 Medical Molecular Biology

Multiple Choice 1) Which statement is not true about cancer?

A. Cancer results from the accumulation of many genetic changes.

B. Gene mutations that increase the risk for developing cancer are always inherited.

C. Cancer cells grow in an uncontrolled and invasive way.

D. Cancer cells are immortal.

Answer: B 2) The process by which cancer cells travel from the tissue of origin to other parts of the body is called A. immortalization B. transformation C. metastasis D. carcinogenesis Answer: C 3) Can surgery successfully cure a patient of a cancer that has metastasized?

A. No; all the body cells of the patient are dividing uncontrollably.

B. Yes; surgery could remove all the patient’s cells with defective cell-cycle regulation.

C. No; cancer cells are no longer localized in one spot in the patient.

D. Yes; if the patient’s tumor is benign.

Answer: C 4) Cancer is a multi-step disease that requires A. at least four to eight genetic changes over the course of years B. mutation of the cancer gene C. at least 20 genetic changes over the course of years D. at least four to eight genetic changes over the course of days Answer: A 5) Chemical agents that cause cancer are known as A. transformers B. carcinogens C. tumor suppressors D. oncogenes Answer: B 6) What is a tumor suppressor gene?

A. A gene associated with tumor formation when its product does not function.

B. A gene associated with tumor formation when its product functions normally.

C. A gene that accelerates the cell cycle and leads to uncontrolled cell growth.

D. A gene that codes for a transcription factor involved in tumor formation.

Answer: A

7) What is an oncogene?

A. A gene associated with tumor formation when its product does not function.

B. A gene associated with tumor formation when it is inappropriately activated.

C. A gene that inhibits cell growth when its product functions normally.

D. A gene that codes for a transcription factor involved in tumor formation.

Answer: B 8) The c-src proto-oncogene encodes a membrane-associated, nonreceptor tyrosine kinase.

The v-src oncogene product differs from c-Src in what way?

A. Because of amino acid substitutions in the C-terminus, the v-Src protein remains in the active conformation and dephosphorylates target proteins.

B. Because of amino acid substitutions in the C-terminus, the v-Src protein remains in the active conformation, binds to target genes and activates their expression.

C. Because of amino acid substitutions in the C-terminus, the v-Src protein remains in the inactive conformation and cannot phosphorylate target proteins.

D. Because of amino acid substitutions in the C-terminus, the v-Src protein remains in the active conformation and phosphorylates target proteins.

Answer: D 9) The c-myc proto-oncogene encodes a A. helix-loop-helix transcription factor B. nonreceptor tyrosine kinase C. GTP-binding protein D. secreted growth factor Answer: A 10) What would be the effect on cell growth if the retinoblastoma protein (pRB) remained phosphorylated?

A. pRB would prevent E2F from binding to regulatory elements in S phase-specific genes, resulting in an inhibition of cell growth.

B. The E2F complex would continue to stimulate S phase-specific genes, resulting in an inhibition of cell growth.

C. The E2F complex would continue to stimulate S phase-specific genes, resulting in unrestrained cell growth.

D. Cell growth would be normal because pRB is a phosphoprotein.

Answer: C 11) What would be the consequence of inactivation of the E2F complex?

A. Cells would be unable to enter the S phase of the cell cycle.

B. Cells would immediately enter the S phase of the cell cycle.

C. Cells would rapidly proliferate and form a tumor.

D. The retinoblastoma protein (pRB) would become activated.

Answer: A 12) Regulatory proteins that serve to prevent a cell from entering the S phase of the cell cycle under conditions of DNA damage are also known as: A. cyclins B. cyclin-dependent kinases C. oncoproteins D. tumor suppressors Answer: D

13) Progression through the cell cycle is regulated by oscillations in the concentration of which factor?

A. cyclin-dependent kinases B. cyclins C. actin D. tubulin Answer: B 14) If DNA damage occurs early in G1 of the cell cycle, the tumor suppressor protein p53 A. activates the p21 gene which encodes a CDK inhibitor, thereby causing G1 arrest.

B. promotes apoptosis.

C. phosphorylates the retinoblastoma protein (pRB), thereby blocking the cell cycle.

D. is targeted for degradation by the proteasome.

Answer: A 15) In addition to promoting cell cycle arrest via cyclin/Cdk complexes in response to cytotoxic conditions such as DNA damage, the tumor suppressor p53 may A. initiate programmed cell death.

B. induce the degradation of oncoproteins C. stimulate translational arrest by phosphorylated eIF2.

D. cause cells to stop growing by limiting glycolysis.

Answer: A

16) Recent studies suggest that the expression pattern of a set of human _______ in cancer samples defines that cancer type better than microarray expression data from a set of 16,000 mRNAs. A. snRNAs B. snoRNAs C. miRNAs D. rRNAs Answer: C 17) The “Philadelphia chromosome” is a reciprocal translocation of chromosome 9 to 22, which causes chronic myelogenous leukemia. The chromosomal translocation fuses the genes encoding the tyrosine kinase ABL and BCR, an activator of ABL. The drug Gleevac has been shown to be effective in Phase II clinical trials. The drug works by: A. dephosphorylating ABL substrates B. blocking the active site of the ABL tyrosine kinase so that ATP cannot bind C. binding to BCR in a way that inhibits its ability to activate ABL D. degrading the ABL-BCR fusion protein Answer: B 18) Possible outcomes of DNA tumor virus infection are A. lysis, cell death, and release of progeny viruses in permissive cells.

B. transformation of nonpermissive cells.

C. interaction between viral-encoded proteins and host cell proteins that inhibit their normal tumor suppressor function.

D. all of the above Answer: D

19) Tumor viruses can cause cancer by A. encoding an oncogene B. activating proto-oncogenes C. activating retinoblastoma protein D. A and B Answer: D 20) Which statement is not true about retroviruses (RNA tumor viruses).

A. Study of the oncogenes carried by retroviruses has led to major advances in understanding of the molecular basis of cancer.

B. Most human cancers are the result of a retroviral infection.

C. When a retrovirus infects a cell, its RNA genome is converted to DNA by the viralencoded reverse transcriptase.

D. Occasionally retroviruses acquire a gene from the host that is normally involved in cellular growth control.

Answer: B 21) The vast majority of human cervical cancers are associated with high-risk human papilloma virus (HPV). HPV is a A. DNA tumor virus B. RNA tumor virus C. retrovirus D. phage Answer: A 22) Benzo(a)pyrene can induce formation of DNA adducts that interfere with replication and transcription. It acts by a A. promoter insertion mechanism B. enhancer insertion mechanism C. nongenotoxic mechanism D. genotoxic mechanism Answer: D 23) Which type of gene therapy has not yet been tried in the clinic?

A. germline gene therapy B. somatic cell gene therapy C. ex vivo gene therapy D. cancer gene therapy Answer: A 24) Which type of gene therapy involves genetic modification of gametes or embryos?

A. germline gene therapy B. somatic cell gene therapy C. ex vivo gene therapy D. cancer gene therapy Answer: A 25) Which gene might be useful in the gene therapy treatment of cancer?

A. E2F B. G1 cyclin

C. wild type proto-oncogenes D. p53 Answer: D 26) Currently, somatic cell gene therapy: A. has only been tested in cell culture and animal models B. is experimental and has never achieved any clinical success.

C. is experimental and has achieved only very limited clinical success.

D. is routine and has been used successfully to treat many diseases.

Answer: C 27) With respect to gene therapy vectors, which of the following is the key advantage of liposomes over viruses?

A. They can be incorporated into endosomes.

B. They do not cause a host immune response.

C. Unlike viral vectors, they allow DNA to enter the host cell nucleus.

D. They enter cells more efficiently than viruses.

Answer: B 28) To create a “safe” viral vector for delivery of genes into human cells, the virus must be engineered to: A. remove all viral coat proteins.

B. remove all viral genes, replacing them with the human genes to be delivered.

C. remove select viral genes and replace them with the human genes to be delivered.

D. remove the viral genome and coat proteins and replace them with plasmids carrying the human genes to be delivered.

Answer: C 29) The key step in the generation of “safe” (replication-incompetent) viral particles for gene therapy is A. creation of site-specific mutations in viral genes B. provision of viral genes on a - (psi-) DNA C. use of a packaging cell line that is not immortalized D. deletion of the viral reverse transcriptase gene Answer: B 30) Adenovirus vectors for gene therapy have all of the following characteristics except: A. they can infect both dividing and nondividing cells.

B. they have a larger capacity for foreign genes than other viral vectors.

C. they are very immunogenic D. they stably integrate into host chromosomes Answer: D 31) Retrovirus vectors for gene therapy have all of the following characteristics except: A. they can infect both dividing and nondividing cells.

B. they have a limited insert size.

C. they have generate only a mild immune response D. they stably integrate into host chromosomes Answer: A

32) Which mode of gene therapy accomplishes transgene integration into the target cell’s genome?

A. adenovirus-mediated B. retrovirus-mediated C. liposome-mediated D. A and B Answer: B 33) Which is a present concern regarding the use of retroviral vectors for gene therapy?

A. They will integrate their DNA into the genome instead of remaining extrachromosomal DNA. B. They will integrate their DNA into the genome in ways that disrupt or misregulate the expression of genes at or near the site of integration.

C. They will fail to introduce the therapeutic gene into any of the patient’s cells.

D. They will cause retroviral disease.

Answer: B 34) The most promising gene therapy clinical trial so far for cystic fibrosis used aerosol administration and A. liposome-mediated gene transfer B. adenovirus-mediated gene transfer C. adeno-associated virus-mediated gene transfer D. retrovirus-mediated gene transfer Answer: C 35) Current strategies for anti-HIV-1 gene therapy aim to do all of the following except: A. destroy every cell infected with HIV-1 B. reduce plasma viral load C. improve patient quality of life D. use ribozymes to cleave viral RNA Answer: A 36) Which class of diseases is least amenable to treatment by gene therapy?

A. monogenic loss-of-function metabolic diseases B. multigenic behavioral diseases C. viral diseases D. immunodeficiencies Answer: B

Short answer/analytical 37) Distinguish between “oncogenes,” “proto-oncogenes,” and “tumor suppressor genes.”

Answer: Section 17.2, p. 547 38) Does the “two hit” hypothesis explain the role of oncogenes and proto-oncogenes in cancer? Why or why not?

Answer: Section 17.2, p. 547; Disease Box 17.1, p. 553 No, because activated oncogenes/proto-oncogenes act dominantly to promote tumorigenesis by, for example, activating mitogenic signaling pathways; there is no need to activate the other allele to achieve its effect.

39) Draw a diagram illustrating how inappropriate activation of proto-oncogenes may be due to qualitative or quantitative changes.

Answer: Section 17.2, p. 547-549, Fig. 17.2 40) Describe the functional differences between c-Src and v-Src. Answer: Section 17.2, p. 549-550, Fig. 17.4 41) Describe the three major roles of c-Src in cancer cell metastasis.

Answer: Section 17.2, Focus Box 17.1 42) Explain why c-myc is referred to as a central “oncogenic switch.”

Answer: Section 17.2, p. 550-551, Fig. 17.5 43) Using retinoblastoma as an example, explain Knudson’s “two-hit” hypothesis.

Answer: Section 17.2, Disease Box 17.1, p. 553 44) What would be the effect on cell growth if the retinoblastoma protein (pRB) remained phosphorylated? Explain your answer.

Answer: Section 17.2, p.552-554, Fig. 17.6 45) After activation of wild-type p53 in response to DNA damage, what are two possible outcomes of p53 activity?

Answer: Section 17.2, p. 554-555, Fig. 17.8 46) Explain how deletions of both alleles of p53, or missense point mutations in one allele, can each lead to unrestrained cell growth.

Answer: Section 17.2, p. 555-556, Fig. 17.9 and Focus Box 17.2 47) Discuss recent evidence suggesting a role for microRNAs (miRNAs) in cancer progression.

Answer: Section 17.2, p. 557-558, Fig. 17.10 48) Explain the molecular mechanism for control of chronic myelogenous leukemia by imatinib mesylate (Gleevac).

Answer: Section 17.2, p. 560, Fig. 17.11 49) Diagram a model for human papilloma virus (HPV)-induced cervical cancer.

Answer: Section 17.2, p. 560-561, Disease Box 17.2 50) Diagram the two main mechanisms for transformation of cells by retrovirus infection.

Answer: Section 17.2, p. 562-564, Fig. 17.13 51) Compare and contrast genotoxic and nongenotoxic effects of carcinogens.

Answer: Section 17.2, p. 564-566, Fig. 17.14 52) Diagram the pathway of tumor promotion by dioxin through arylhydrocarbon receptormediated signal transduction.

Answer: Section 17.2, p. 564-566, Fig. 17.15

53) In gene therapy clinical trials using retrovirus and adenovirus vectors, the levels of gene expression have generally been disappointingly low or transient. Based on your knowledge of these vectors and of the regulation of eukaryotic gene expression, provide an explanation for why this is the case.

Answer: Section 17.3, p. 566-571 54) If you were attempting for the first time to develop a viral gene therapy for a monogenic disease, what factors would you need to consider in your selection of a vector? Explain your answer.

Answer: Section 17.3, Table 17.4, p. 568 Two essential factors to consider are: (1) the size of the therapeutic gene. This will dictate whether a vector with a large capacity (e.g., adenovirus) is needed; (2) the target cell type (cell type and whether it is actively dividing). This will potentially rule out vectors that cannot infect nondividing cells (e.g., neurons). Once these factors are considered, one would choose the vector with the longest persistence and lowest immunogenicity.

55) Compare and contrast in vivo and ex vivo somatic cell gene therapy approaches.

Answer: Section 17.3, p. 566, Fig. 17.16 56) Explain why a “packaging cell” step is used when preparing a retroviral vector for gene transfer.

Answer: Section 17.3, Focus Box 17.3, p. 569 57) Compare and contrast the strategies used for the treatment of ornithine transcarbamylase (OTC) deficiency, adenosine deaminase (ADA)-deficient severe combined immunodeficiency (SCID), and cystic fibrosis. Why are different treatment strategies and different vectors used for different genetic diseases?

Answer: Section 17.3, p. 570-574, Focus Box 17.4 58) Why were gene therapy trials for SCID-X1 recently suspended despite very encouraging initial results of clinical trials?

Answer: Section 17.3, p. 572-573, Fig. 17.19 59) Diagram the HIV-1 life cycle and indicate some of the popular targets for antiviral strategies.

Answer: Section 17.3, p. 575-577, Focus Box 17.5 60) Diagram the hypothesized action of a hairpin-derived catalytic antisense RNA against the HIV-1 long terminal repeat (LTR).

Answer: Section 17.3, p. 577, Figure 17.21 61) Can gene therapy be used for purposes other than treatment of “monogenic” diseases (e.g., cystic fibrosis)? Explain your answer.

Answer: Section 17.3, p. 571-578, Disease Box 17.4, p. 567 Examples of this include RNAi-based therapies, enhancement genetic engineering, cancer gene therapy, and HIV-1 gene therapy.

62) You suspect that a gene you have cloned is a proto-oncogene. Describe how you would test your hypothesis experimentally. Predict the results.

Answer: Section 17.2, p. 546-549. If the gene is a proto-oncogene then its inappropriate overexpression could lead to cellular transformation. The proto-oncogene cDNA could be cloned into an expression vector, introduced into cells in culture, and assayed for transforming activity.

Transformed cells will become rounded up, detach from the culture plate, and proliferate indefinitely in culture.

63) DNA microarray analysis reveals no expression of the pRB tumor suppressor gene in a bone cancer cell line. Would you expect this to be the only difference in expression pattern between normal and bone cancer cells. Why or why not?

Answer: Section 17.2, p. 552-554; Section 16.5, p. 529. No expression of the pRB tumor suppressor gene in a bone cancer cell line would be expected to be only one of many differences in expression pattern between normal and cancerous cells. Cancer is a multi-step disease and the emergence of a tumor cell requires the accumulation of at least 4-8 genetic changes.

64) Analysis of another cell line reveals normal expression levels of pRB and yet the cells exhibit unrestrained cell growth. Provide an explanation and suggest how to test your hypothesis.

Answer: Section 17.2, p. 552-554; Section 9.6, p. 246-247. Normal expression levels of pRB but unrestrained cell growth could indicate that pRB is being kept in the phosphorylated state by CDK2 catalytic activity. If pRB remains phosphorylated it cannot bind the E2F complex and the E2F complex continues to stimulate S-phase-specific genes. To test this hypothesis, pRB from cell line extracts could be analyzed by Western blot with antibodies specific for pRB. As controls, purified unphosphorylated pRB and phosphorylated pRB would be included on the blot. Phosphorylated pRB will show a shift in mobility and appear as a slower moving band compared with unphosphorylated pRB.

65) As they light up their cigarette, a friend of yours tells you that they have heard that benzo(a)pyrene in cigarette smoke is not really a carcinogen. Set them straight.

Answer: Section 17.2 p. 564. Benzo(a) pyrene is a “procarcinogen” in cigarette smoke. Within cells, it is converted through a series of enzyme-mediated reactions to the genotoxic carcinogen benzo(a)pyrene diol-epoxide (BPDE). In this carcinogenic form, it binds to DNA and is mutagenic, leading to skin, lung, and stomach cancer. Repair of the bulky BPDE adducts in DNA is inefficient and leads to the accumulation of mutations. For example, 60% of lung cancers shown inactivating mutations in the p53 tumor suppressor gene.

66) You are designing a somatic cell gene therapy protocol for spinal muscular atrophy, a genetic disorder that is caused by a defect in the Survival of Motor Neurons (SMN) gene. You have cloned the SMN cDNA. (a) What is the minimal cis-acting DNA regulatory element you would need to ligate to the SMN cDNA to get transcription of the cDNA in cells? (b) Assume that you decide to use retroviral-mediated gene transfer to introduce the SMN cDNA into the patient's target cells ex vivo. What are the advantages and disadvantages of this choice

of vector for gene delivery? Is there another vector that would work better?

Answer: (a) Section 11.3, p. 296. A promoter element would be the minimal requirement for transcription of the cDNA. (b) Section 17.3, p. 568-570, Table 17.4. The advantages are that retrovirus vectors enter cells efficiently, there are few immunity problems, and the therapeutic gene is stably integrated into the host chromosomes. The disadvantages are that it is hard to produce, an ex vivo approach is required, it has a limited insert size (~7 kb), infects only dividing cells, and integration may cause insertional mutations since it seems to preferentially insert near active genes. Since spinal muscular atrophy affects the motor neurons, the target cells may be difficult to manipulate ex vivo. Another possibility might be a herpes virus-mediated approach, since this virus targets nondividing nerve cells. Disadvantages include the possibility of an immune response.