Test Bank for Introduction to Genetic Analysis, 12th Edition Griffiths, Doebley, David

Introduction to Genetic Analysis, 12th Edition By Anthony Griffiths , John Doebley , Catherine Peichel, David Wassarman

Name:

Class:

Date:

Chapter 01: The Genetics Revolution Multiple Choice 1. The early 1900s was an important period for genetics due to which of the following major events? a. the rediscovery of Gregor Mendel's scientific findings b. Watson and Crick solving the structure of DNA c. Walter Sutton and Theodore Boveri hypothesizing that chromosomes are the hereditary elements d. the rediscovery of Gregor Mendel's scientific findings and Walter Sutton and Theodore Boveri hypothesizing that chromosomes are the hereditary elements e. All of the answer options are correct.

ANSWER: e 2. A sample of normal double-stranded DNA was found to have a guanine content of 18%. What is the expected proportion of adenine? a. 9% b. 32% c. 36% d. 68% e. 82% ANSWER: b 3. In one strand of DNA, the nucleotide sequence is 5'-ATGC-3'. The complementary sequence in the other strand must be a. 3'-ATGC-5'. b. 3'-TACG-5'. c. 5'-ATCG-3'. d. 5'-CGTA-3'. e. 5'-TACG-3'.

ANSWER: b 4. How many different DNA molecules that are eight-nucleotide-pairs long are theoretically possible? a. 24 b. 32 c. 64 d. 256 e. 65,536 ANSWER: e 5. Which of the following is/are TRUE about genes? a. Genes are located on chromosomes. b. Genes come in variants known as alleles. c. Genes usually encode protein products.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chapter 01: The Genetics Revolution d. All of the answer options are correct. e. None of the answer options is correct.

ANSWER: d 6. Wild cats (Felis silvestris) and common mice (Mus musculus) are diploid. In wild cats, 2n = 38, while in common mice, 2n = 40. Based on this information, we can conclude that wild-cat cells have a. less DNA than common-mouse cells. b. smaller genomes than common-mouse cells. c. fewer DNA molecules than common-mouse cells. d. fewer genes than common-mouse cells. e. fewer sets of chromosomes than common-mouse cells.

ANSWER: c 7. Which of the following is a component of DNA? a. alanine b. arginine c. cysteine d. guanine e. tyrosine ANSWER: d 8. Which of the following is/are TRUE of the DNA structure solved by Watson and Crick? a. It is a double-helical structure. b. Sugar–phosphate backbone is always toward the outside of the DNA. c. There are two hydrogen bonds between A and T and three hydrogen bonds between C and G. d. There are four types of nitrogenous bases. e. All of the answer options are correct.

ANSWER: e 9. Which of the following is a CORRECT representation of the central dogma? a. RNA → DNA → protein b. protein → DNA → RNA c. DNA → RNA → protein d. DNA → protein → DNA e. None of the answer options is correct.

ANSWER: c 10. You have come across a dog (named Cindy) that does not have a tail. Interestingly, all the puppies produced by this dog don't have a tail. If the lack of tail is caused by a genetic mutation, where has this mutation most likely taken place? a. in Cindy's gametes Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chapter 01: The Genetics Revolution b. in the cells that should normally have given rise to Cindy's tail c. in the cells that should normally have given rise to Cindy's and her puppies' tails d. in all of Cindy's cells (including her gametes) e. in a gamete of one of Cindy's parents ANSWER: e 11. Which of the following features makes a species suitable as a model organism? a. small organism b. short generation time c. small genome d. produce large number of offspring e. All of the answer options are correct.

ANSWER: e 12. Using molecular techniques, researchers have knocked out both copies of gene G in a series of genetically identical mouse embryos. These mice develop normally, except for their forelimbs, which are missing several small bones. What can be concluded from the results of this experiment? a. Gene G encodes a protein that is a crucial component of the forelimbs' small bones in mice. b. Gene G encodes a protein that is normally only present in the forelimb cells of developing mice. c. Gene G is necessary for proper development of the forelimbs' small bones in mice. d. Gene G is normally only present in the forelimb cells of developing mice. e. Gene G is normally only transcribed in the forelimb cells of developing mice.

ANSWER: c 13. Who originated the one-gene–one-enzyme hypothesis? a. Tatum and Beadle b. Gregor Mendel c. Watson and Crick d. Franklin and Wilkins e. Hershey and Chase ANSWER: a 14. What are alleles? a. gene variants b. enzymes c. regulatory elements d. de novo mutations e. quantitative trait loci ANSWER: a 15. Which enzyme cuts DNA at a specific location?

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chapter 01: The Genetics Revolution a. polymerase b. ligase c. nuclease d. allele e. ribosome ANSWER: c 16. Which type of mutation is a unique DNA variant that exists in a child but in neither of its parents? a. point mutation b. de novo mutation c. quantitative trait locus d. single nucleotide polymorphism e. dominant allele ANSWER: b 17. Which enzyme is responsible for DNA replication? a. polymerase b. ligase c. nuclease d. allele e. ribosome ANSWER: a 18. Which scientists offered the first compelling experimental evidence that genes are made of deoxyribonucleic acid (DNA)? a. Oswald Avery, Colin MacLeod, and Maclyn McCarty b. John Gurdon and Shinya Yamanaka c. François Jacob and Jacques Monod d. James Watson and Francis Crick e. Barbara McClintock and Erwin Chargoff ANSWER: a 19. The Central Dogma describes a. the hypothesis of how DNA is packaged into small molecules. b. the process by which RNA is processed within a cell. c. the flow of genetic information within cells from DNA to RNA to protein. d. how model organisms are used in experiments. e. the method of gene transfer between organisms.

ANSWER: c 20. The process of inserting foreign DNA molecules into the genomes of a recipient organism is called Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chapter 01: The Genetics Revolution a. replication. b. transformation. c. transcription. d. translation. e. ligation.

ANSWER: b Essay 21. Adenine and thymine are held together by two hydrogen bonds, while guanine and cytosine are held together by three hydrogen bonds. If you were to slowly heat a piece of DNA rich in GC base pairs—in order to denature it—would you expect the melting temperature to be higher or lower than a piece of DNA rich in AT base pairs?

ANSWER: The melting temperature would be higher for DNA rich in GC, owing to the three hydrogen bonds that must be broken in order for it to denature.

22. Arabidopsis thaliana is a diploid plant model organism with 2n = 10. a) How many copies of each gene does each Arabidopsis thaliana cell have? b) How many sets of chromosomes does the nucleus of an Arabidopsis thaliana leaf cell contain? c) How many pairs of homologous chromosomes does the nucleus of an Arabidopsis thaliana leaf cell contain?

ANSWER: a) two b) two c) five 23. Explain what it means to say that the genetic code is redundant. How does this redundancy help protect against mutations?

ANSWER: The genetic code is redundant because some of the amino acids are encoded by more than one triplet (codon). This protects against the effects of mutation since a change in the nucleotide base may not cause a different amino acid to be inserted.

24. Mutations are often viewed as negative events, and they are nearly always bad for an organism.

Paradoxically, without mutations there would be no evolution, and so they are essential. Explain how this is so.

ANSWER: Variation is introduced. So even though mutations are often viewed as negative events, all variation that we see around us originally came from mutations.

25. Describe the purpose and function of DNA polymerase, nuclease, and ligase.

ANSWER: DNA polymerase, nuclease, and ligase are tools for characterizing and manipulating DNA, RNA, and proteins. They each have a different function. DNA polymerase copies DNA, nucleases cut DNA molecules in specific locations or degrade an entire DNA molecule into single nucleotides, and ligases join two DNA molecules.

26. Why does the age of the father matter, while that of the mother seems to have no effect on the frequency of new point mutations?

ANSWER: Eggs are made prior to a woman's birth, while sperm production occurs throughout a man's life.

From the point of conception until formation of egg cells, there are about 23 rounds of cell division and DNA replication. Because egg formation occurs prior to birth, as a woman ages there is no chance for additional point mutations. By comparison, the cell divisions that produce sperm Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chapter 01: The Genetics Revolution continue throughout a man's life and with each cell division there is greater risk of introducing new point mutations.

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance Multiple Choice 1. If a plant of genotype A/a is selfed, and numerous offspring are scored, what proportion of the progeny is expected to have homozygous genotypes? a. 0 b. 25% c. 50% d. 75% e. 100% ANSWER: c 2. What is the maximum number of heterozygous genotypes that could be produced by monohybrid self? a. 1 b. 2 c. 3 d. 4 e. 6 ANSWER: a 3. A plant is heterozygous at three loci. How many different gamete genotypes can it theoretically produce with respect to these three loci? a. 2 b. 3 c. 4 d. 8 e. 16 ANSWER: d 4. In mountain rabbits, the EL-1 gene is located on chromosome 3. Four alleles of this gene have been identified in the population. With respect to EL-1, what is the maximum number of genotypes in the progeny of a SINGLE CROSS between two mountain rabbits? a. 1 b. 2 c. 3 d. 4 e. 6 ANSWER: d 5. A wild-type strain of haploid yeast is crossed to a mutant strain with phenotype d. What phenotypic ratios will be observed in the progeny? a. All wild type b. 75% wild type and 25% mutant (d) Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance c. 50% wild type and 50% mutant (d) d. 25% wild type and 75% mutant (d) e. All mutant (d) ANSWER: c 6. Mice (Mus musculus) have 40 chromosomes per diploid cell (2n = 40). How many double- stranded DNA molecules and how many chromosomes are there in a mouse cell that is in the G2 stage of the cell cycle? a. 40 DNA molecules and 20 chromosomes b. 40 DNA molecules and 40 chromosomes c. 40 DNA molecules and 80 chromosomes d. 80 DNA molecules and 40 chromosomes e. 80 DNA molecules and 80 chromosomes ANSWER: d 7. A mutation occurs in a germ cell of a pure-breeding, wild-type male mouse prior to DNA replication. The mutation is not corrected, and the cell undergoes DNA replication and a normal meiosis producing four gametes. How many of these gametes will carry the mutation? a. 1 b. 2 c. 3 d. 4 e. It is impossible to predict.

ANSWER: b 8. What is the mechanism that ensures Mendel's first law of segregation? a. formation of chiasmata b. formation of the kinetochore c. pairing of homologous chromosomes d. segregation of homologous chromosomes during meiosis I e. segregation of sister chromatids during meiosis II ANSWER: d 9. A laboratory mouse homozygous for an RFLP marker is mated to a wild mouse that is heterozygous for that marker. One of the heterozygous individuals resulting from this cross is mated back to the wild parent. What proportion of the offspring will have the same RFLP pattern as the original laboratory mouse? a. None of the offspring b. 1/4 c. 1/2 d. 3/4 e. All of the offspring ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance 10. The diagram below shows a part of the biochemical pathway responsible for fruit color in peppers (Caspicum annuum). Enzyme 1 is responsible for catalyzing the reaction that turns the colorless precursor into yellow pigment, whereas Enzyme 2 catalyzes the step that turns the yellow pigment into red pigment. A breeder crosses a pure-breeding plant that makes yellow peppers to a pure-breeding plant that makes red peppers. What proportion of the offspring will make red peppers?

a. All of the offspring b. 3/4 c. 1/2 d. 1/4 e. None of the offspring ANSWER: a 11. The wild-type eye color in the fruit fly Drosophila melanogaster is dark red, as a result of a mixture of bright red and brown pigments. Enzyme A is encoded by the a gene and is required to synthesize the bright red pigment. A lack of red pigment results in a somewhat brown eye color. You cross two fruit flies who are heterozygous for a recessive mutation that completely inactivates the a gene. What proportion of their offspring will have a recessive eye color phenotype? a. All of the offspring b. 3/4 c. 1/2 d. 1/4 e. None of the offspring ANSWER: d 12. In pet rabbits, brown coat color is recessive to black coat color. A black female rabbit gives birth to four black-coated and three brown-coated baby rabbits. What can be deduced about the genotype of the baby rabbits' father? a. He could be heterozygous black/brown or homozygous brown. b. He could be heterozygous black/brown or homozygous black. c. He must be heterozygous black/brown. d. He must be homozygous black. e. He must be homozygous brown.

ANSWER: a 13. "Dumpy" is a commonly used mutant phenotype in the nematode worm C. elegans. Two dumpy individuals are crossed to each other, and this cross produces 210 dumpy and 68 wild-type individuals. If one of the dumpy individuals used in this cross was mated with a wild type, what dumpy:wild-type ratio would we observe in the offspring? a. 0:1 Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance b. 1:0 c. 1:1 d. 1:3 e. 3:1 ANSWER: c 14. A female rabbit of phenotype c′ is crossed to a male rabbit with cch. The F1 is comprised of five rabbits with a c′ phenotype, two with cch phenotype, and three with c phenotype. Of the phenotypically c rabbits, two are females and are backcrossed to their father. This cross produces only rabbits with cch phenotype. These results suggest that a. c could be dominant or recessive to c′. b. c is dominant to c′ but recessive to cch. c. c is dominant to cch but recessive to c′. d. c is dominant to both c′ and cch. e. c is recessive to both c′ and cch.

ANSWER: e 15. A plant with small red flowers is crossed to a plant with large white flowers. The resulting F1 is comprised of 75 plants with small red flowers and 72 plants with small white flowers.

If flower color and flower size are controlled by a single gene each, what can be concluded from these results? a. Flower color is controlled by a sex-linked gene. b. Red color and small size are dominant to white color and large size, respectively. c. Small size is dominant to large size, but we cannot determine which color is dominant. d. We cannot determine which color and which size are dominant. e. White color and small size are dominant to red color and large size.

ANSWER: c 16. A dominant gene b+ is responsible for the wild-type body color of Drosophila; its recessive allele b produces black body color. A testcross of a heterozygous b+/b female by a black b/b male gave 52 black and 58 wild-type progeny. If a black female from these progeny were crossed with a wild-type brother, what phenotypic ratios would be expected in their offspring? a. All males will be wild type, and all females will be black. b. All progeny will be black. c. All progeny will be wild type. d. 75% will be wild type; 25% will be black. e. 50% will be wild type; 50% will be black.

ANSWER: e 17. A very common type of red–green color blindness in humans is caused by a mutation in a gene located on the X chromosome. Knowing that the mutant allele is recessive to the wild type, what is the probability that the Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance son of a woman whose father is color-blind is going to also be color-blind? a. 0% b. 25% c. 50% d. 75% e. 100% ANSWER: c 18. A phenotypically normal woman is heterozygous for the recessive Mendelian allele causing phenylketonuria, a disease caused by the inability to process phenylalanine in food. She is also heterozygous for a recessive X-linked allele causing red–green color blindness. What percentage of her eggs will carry the dominant allele that allows normal processing of phenylalanine and the X-linked recessive allele that causes color blindness? a. 0% b. 25% c. 50% d. 75% e. 100% ANSWER: b 19. A rare, curly winged mutant of Drosophila was found in nature. A mating of this fly with a true-breeding, normal laboratory stock produced progeny in the ratio 1 curly winged to 1 normal (both sexes had the same ratio). All curly winged progeny of this cross, mated with normal progeny of the same cross, again yielded 1 curly winged to 1 normal fly. When mated with one another, the curly winged progeny of the first cross yielded a progeny of 623 curly:323 normal. This ratio strongly suggests which of the following? a. Curly and normal are in the 3:1 ratio expected from intercrossing monohybrid genotypes for a recessive mutant allele (curly). b. Curly and normal are in the 3:1 ratio expected from intercrossing monohybrid genotypes for a dominant mutant allele (curly). c. The curly winged parent of the curly × curly cross is homozygous. d. Flies homozygous for the curly allele are lethal and never survive. e. The gene for curly is sex-linked.

ANSWER: d 20. A female Drosophila with the mutant phenotype a is crossed to a male who has the mutant phenotype b. In the resulting F1 generation all females are wild type and all males have the a mutant phenotype. Based on these results, we can conclude that the mode of inheritance of the phenotypes of interest is a. autosomal for a and X-linked for b. b. dominant for a and recessive for b. c. recessive for a and dominant for b. d. recessive for both a and b. e. X-linked for a and autosomal for b.

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance ANSWER: d 21. A recessive X-linked gene mutation is known to generate premature baldness in males but is without effect in women. If a heterozygous female marries an affected male, what proportion of all their children is expected to be prematurely bald? a. 1/4 b. 1/8 c. 1/16 d. 1/32 e. 1/216 ANSWER: a 22. You have three jars of gumballs. The first jar has 100 white gumballs and 25 green, the second jar has 50 white and 150 blue, and the third jar contains 500 white and 10 red.

If you randomly draw one gumball from each jar, what is the probability for ALL WHITE GUMBALLS? a. 0.196, or 19.6% b. 0.109, or 10.9% c. 0.056, or 5.6% d. 0.567, or 56.7% e. This is impossible (0% chance).

ANSWER: a 23. You have three jars of gumballs. The first jar has 100 white gumballs and 25 green, the second jar has 50 white and 150 blue, and the third jar contains 500 white and 10 red.

If you randomly draw one gumball from each jar, what is the probability for ALL WHITE OR ALL COLORED GUMBALLS? a. 0.199, or 19.9% b. 0.112, or 11.2% c. 0.058, or 5.8% d. 0.589, or 58.9% e. This is impossible (0% chance).

ANSWER: a 24. You have three jars of gumballs. The first jar has 100 white gumballs and 25 green, the second jar has 50 white and 150 blue, and the third jar contains 500 white and 10 red.

If you randomly draw one gumball from each jar, what is the probability for AT LEAST ONE WHITE GUMBALL? a. 0.997, or 99.7% b. 0.85, or 85% c. 0.69, or 69 % Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance d. 0.034, or 3.4% e. This is impossible (0% chance).

ANSWER: a 25. The following pedigree concerns the autosomal recessive disease phenylketonuria (PKU). The couple marked A and B are contemplating having a baby but are concerned about the baby having PKU. What is the probability of the first child having PKU? Unless you have evidence to the contrary, assume that a person marrying into the pedigree (i.e., not a descendant of the two parents at the top of the pedigree) is not a carrier.

The filled-in individuals have PKU.

a. 0 b. 1/12 c. 1/4 d. 3/4 e. 9/64 ANSWER: b 26. The following pedigree depicts the inheritance of a rare hereditary disease affecting muscles.

What is the MOST likely mode of inheritance of this disease? a. Autosomal dominant b. Autosomal recessive Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance c. X-linked dominant d. X-linked recessive e. Y-linked ANSWER: d 27. The following pedigree shows the inheritance of attached earlobes (black symbols) and unattached earlobes (white symbol). Both alternative phenotypes are quite common in human populations.

If the phenotypes are determined by alleles of one gene, then attached earlobes are inherited as a(n) a. autosomal dominant trait. b. autosomal recessive trait. c. dominant trait that could be either autosomal or X-linked. d. recessive trait that could be either autosomal or X-linked. e. X-linked dominant trait.

ANSWER: a 28. In the human pedigree shown below, black symbols indicate individuals suffering from a RARE genetic disease, whereas white symbols represent people who do not have the disease.

Based on the pedigree, what is the most likely mode of inheritance of this rare genetic disease?

a. Autosomal dominant b. Autosomal recessive c. X-linked dominant d. X-linked recessive e. Y-linked ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance 29. The following pedigree shows the inheritance of a mild, but very rare condition in Siberian Husky dogs. If individuals 1 and 2 are crossed, what is the probability that they will produce an affected pup?

a. 1/36 b. 1/16 c. 4/36 d. 4/16 e. 16/36 ANSWER: c 30. What is the probability that individual A is heterozygous with respect to the condition depicted in the pedigree?

a. 0% b. 25% c. 50% d. 75% e. 100% ANSWER: e 31. What is the most likely mode of inheritance of the exceptionally rare condition represented in the pedigree below, and why?

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance

a. Impossible to determine, because the condition is so rare. b. Recessive, because it is present in only one generation, but we do not have enough information to tell whether it is X-linked or autosomal. c. Recessive, because unaffected parents have an unaffected child, and autosomal, because there are more autosomes than there are X chromosomes. d. X-linked recessive, because this would require the smallest number of rare alleles in the pedigree. e. X-linked recessive, because it only affects a male, and his parents are unaffected.

ANSWER: d 32. A couple is both heterozygous for the autosomal recessive disease cystic fibrosis (CF). What is the probability that their first child will either be a boy or have CF? a. 6/8 b. 5/8 c. 3/8 d. 2/8 e. 1/8 ANSWER: b 33. Cystic fibrosis is an autosomal recessive condition. If the parents of a boy with cystic fibrosis have two more children, what is the probability that both of these children will be unaffected? a. 1/16 b. 3/16 c. 4/16 d. 9/16 e. 16/16 ANSWER: d 34. Which progeny phenotypic ratio is expected when a diploid monohybrid is selfed? a. 1:1 b. 1:2:1 c. 2:1:1 d. 3:1 Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance e. 1:0 ANSWER: d 35. Which progeny phenotypic ratio is expected in a diploid monohybrid testcross? a. 1:1 b. 1:2:1 c. 2:1:1 d. 3:1 e. 1:0 ANSWER: a 36. Which progeny phenotypic ratio is expected in a cross between a mutant and wild type in a haploid organism? a. 1:1 b. 1:2:1 c. 2:1:1 d. 3:1 e. 1:0 ANSWER: a 37. Which progeny phenotypic ratio is expected in a cross between a homozygous dominant and homozygous recessive (diploid)? a. 1:1 b. 1:2:1 c. 2:1:1 d. 3:1 e. 1:0 ANSWER: e 38. Which progeny phenotypic ratio is expected in a cross between mutant 1 and mutant 2 in a haploid organism? a. 1:1 b. 1:2:1 c. 2:1:1 d. 3:1 e. 1:0 ANSWER: a 39. Which pattern of inheritance best fits pedigree I shown below? Assume that individuals marrying into the family are homozygous for the wild-type allele.

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance

a. Autosomal dominant b. Autosomal recessive c. X-linked dominant d. X-linked recessive e. Y-linked ANSWER: e 40. Which pattern of inheritance best fits pedigree II shown below? Assume that individuals marrying into the family are homozygous for the wild-type allele.

a. Autosomal dominant b. Autosomal recessive c. X-linked dominant d. X-linked recessive e. Y-linked ANSWER: b 41. Which pattern of inheritance best fits pedigree III shown below? Assume that individuals marrying into the family are homozygous for the wild-type allele.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance

a. Autosomal dominant b. Autosomal recessive c. X-linked dominant d. X-linked recessive e. Y-linked ANSWER: a 42. Which pattern of inheritance best fits pedigree IV shown below? Assume that individuals marrying into the family are homozygous for the wild-type allele.

a. Autosomal dominant b. Autosomal recessive c. X-linked dominant d. X-linked recessive e. Y-linked ANSWER: d 43. What are leaky mutations? a. Mutations that cause a complete loss of function in the wild-type phenotype. b. Mutations that do not cause a complete loss of function because all wild-type function "leaks" into the mutant phenotype. c. Mutations that do not cause a complete loss of function in the wild-type phenotype because some Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance wild-type function "leaks" into the mutant phenotype. d. Mutations that cause a complete loss of function because all wild-type function "leaks" into the mutant phenotype. e. None of the above.

ANSWER: c 44. What is the phenotypic ratio of a cross between heterozygous dominant and wild-type parents? a. 1:1 b. 2:1 c. 3:1 d. 1:4 e. 1:2 ANSWER: a 45. An orange flower is crossed with a wild-type purple flower. All the F1 flowers are purple, and of 1760 F2 flowers sampled, 1320 are purple and 440 are orange. What is the phenotypic ratio of the F2 progeny? a. 2:1 b. 1:4 c. 4:1 d. 3:1 e. 1:3 ANSWER: d Essay 46. Mendel studied the inheritance of phenotypic characters determined by alleles of seven different genes. It is an interesting coincidence that the pea plant has seven pairs of chromosomes (n = 7). What is the probability that, by chance, Mendel's seven genes would each be located on a different chromosome? You may assume that the pea's chromosomes are all the same size.

ANSWER: 6!/76 = 6.12 * 10–3 Take each gene in turn. The probability is 1 that the first gene falls on a chromosome. The probability that the second gene falls on any of the remaining six chromosomes is 6/7, the next is 5/7, etc. The overall probability is the product of all these.

47. In Labrador retrievers, black color coat (B/–) is dominant to brown color coat (b/b). A breeder crosses two black individuals who have previously produced some brown puppies. If the cross produces six puppies: a. what is the probability that the first born will be brown? b. what is the probability that four of them will be brown and two will be black? c. what is the probability that at least one of them will be brown?

ANSWER: a. Both parents must be heterozygotes B/b because they have previously produced brown puppies.

The probability that they produce a brown puppy is therefore 1/4. b. Each pup has 3/4 chance of being black and 1/4 chance of being brown. The order in which the brown and black puppies are born does not matter, so there are 15 different permutations of 4 brown Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance + 2 black (5!). Hence, the probability is 15[(1/4)(1/4)(1/4)(1/4)(3/4)(3/4)] = 135/4096 = 3.3%. c. In this case, the only instance that does not satisfy the condition is the case in which all puppies are black. The probability of this event is (3/4)6 = 729/4096 = 17.8%. Therefore, the probability of obtaining at least one brown puppy is 1 – (729/4096) = 82.2%.

48. In a particular species of plants, flower color is dimorphic: some individuals have red flowers, whereas others have yellow flowers. If flower color is controlled by a single gene with two alleles (cred and cyellow): a. what would be the simplest way to determine which allele is dominant? b. what will be the genotypic ratio in the offspring of a cross between a monohybrid and a pure-breeding individual?

ANSWER: a. Cross a pure-breeding red to a pure-breeding yellow individual, and assess the phenotype of the monohybrid produced. If it makes red flowers, then cred is dominant; if it makes yellow flowers, then cyellow is dominant. b. 1:1; half of the offspring will be heterozygous, and half will be homozygous like the purebreeding parent.

49. Suppose that red flower color (RR or Rr) is dominant to white flower color (rr) in a petunia. A friend has a petunia plant with red flowers and wants to determine whether the plant is RR or Rr. a. What cross could you perform to help your friend determine the genotype of his petunia plant? b. How will this cross help you determine the genotype of your friend's red-flowered petunia? That is, how will the results from this cross differ if the red-flowered petunia is RR versus Rr? ANSWER: a. Perform a testcross (test the red petunia to a genotypically rr petunia). b. You will observe different segregation in the testcross progeny, depending on the genotype of the red petunia. If the red petunia is RR, then all testcross progeny will be red; if the red petunia is Rr, then 1/2 of the testcross progeny will be red (Rr) and 1/2 will be white (rr).

50. Suppose that a single gene controls fruit color in mango. Yellow fruit (Y) is dominant to red fruit (y).

Suppose a true-breeding yellow mango plant was crossed with a red-fruited plant, and the resulting F1 was selfed. The F2 segregated as expected. If one of the yellow-fruited plants was randomly selected and selfed, what is the probability that its progeny would segregate for fruit color? Explain your logic.

ANSWER: The F2 consists of 1/4 YY:1/2 Yy:1/4 yy. Thus, the yellow-fruited plant that was randomly picked could be either YY or Yy. There is a 1/3 chance that it was YY and 2/3 chance that it was Yy. If a YY plant was selected and selfed, the progeny would not segregate for fruit color. If a Yy plant was selected, the progeny would segregate for fruit color.

51. The wild-type flower color of a particular species of plant is blue. The diagram below shows a simplified version of the biochemical pathways responsible for the synthesis of the blue pigment. Suppose that gene A codes for Enzyme A and gene B for Enzyme B. A friend provides you with a pure-breeding plant that makes colorless (white) flowers. What genetic experiment(s) could you perform to determine whether your plant lacks Enzyme A or Enzyme B? (Suppose that you have access to any pure-breeding lines that you need.)

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance ANSWER: The "unknown" white mutant can be crossed to a pure-breeding mutant that lacks Enzyme 2 (genotype b/b); if the "unknown" mutant lacks Enzyme 2, then the entire F1 should make only white (colorless) flowers, but if the "unknown" mutant lacks Enzyme 1, then the F1 should inherit a functioning A allele from the b/b parent and a functioning B allele from the "unknown" (a/a) parent, and therefore make blue flowers.

ALTERNATIVELY: A heterozygous A/a can be produced by crossing a wild type to a pure line that lacks Enzyme A. This heterozygote can be crossed to the "unknown" mutant; if a 1:1 of blue:white is observed in the offspring, then our "unknown" mutant most likely lacks Enzyme A and is therefore a/a.

52. Yellow leaves on a plant can be caused by genetic mutations, viruses, or unfavorable environmental conditions. Suppose you find a plant that has yellow leaves, and you want to determine if the cause of the phenotype is a genetic mutation or an environmental stress. Design an experiment to differentiate between the different possibilities.

ANSWER: Cross the yellow plant with a normal plant. Self the resulting F1 and look for a consistent, predictable segregation pattern. For example, the presence of a 3 green:1 yellow segregation ratio would suggest that the yellow phenotype was caused by a recessive mutation.

53. Suppose that the length of a duck's tail is determined by a single autosomal gene with two alleles: L (long tail) and l (short tail). When a female duck with a long tail was backcrossed to her father, she produced three ducklings with a long tail and three with a short tail. a. What are the possible genotypes of the female duck and of her father? b. What is the most likely genotype of the female duck's father? (Justify your answer using probabilities.)

ANSWER: a. The presence of ducklings with the recessive phenotype among the offspring indicates that both the mother (the "female duck") and the male used in the cross carry the recessive l allele. The mother must be L/l as she has a long-tail phenotype. The father could be L/l or l/l. b. l/l is more likely. The probability of the cross L/l × l/l producing a 1:1 ratio within an offspring of six ducklings is [(1/2)6]*10 = ~15%, whereas the probability that the cross L/l × L/l produce a 1:1 ratio in an offspring of six ducklings is [(3/4)3(1/4)3]*10 = ~6.6%.

54. Loppins are fictitious (but useful) diploid invertebrates that produce large offspring and normally have long antennae. Short antennae mutants also exist. Unfortunately for the geneticists working on these organisms, the males' antennae do not fully develop until the loppin equivalence of "middle age." A female with short antennae is crossed to a young male, and all the females in their offspring have the short antennae mutant phenotype. A subset of these F1 females are crossed to a middle-aged male with short antennae, and all the females produced by these crosses have short antennae. However, all the crosses between the F1 females and their brothers produce both short antennae and long antennae loppins in a ratio of about 3:1. How can these results be explained? Provide the genotypes of as many individuals as possible.

ANSWER: The 3:1 ratio obtained in the cross between brothers and sisters suggests that short antennae (S) is dominant to long antennae (s), and that the F1 females and their brothers are heterozygous (S/s). The young male used in the original cross is probably homozygous for the long antennae allele (s/s); the middle-aged male with short antennae is probably homozygous (S/S, because all the progenies have short antennae). The cross between F1 females and the middle-aged male produces about 50% S/s Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance and 50% S/S individuals; the cross between the F1 females and their brothers produces about 25% S/S, 25% s/s, and 50% S/s, hence the observed phenotypic ratios.

55. Wild-type Drosophila melanogaster have a brown/gray body color. Mutants exist that have a yellow body color. Several crosses were performed between phenotypically wild type and yellow individuals, and the results of each cross are reported in the table below. Deduce the mode of inheritance of the yellow body phenotype and genotypes of the parents and offspring in the following crosses.

Progeny ––––––––––––––––––––––––––––––––––––––––––––––– Females Males Parents wild type yellow wild type yellow Female Male ––––––––––––––––––––––––––––––––––––––––––––––– a) wild type * yellow 198 0 203 0 b) yellow * yellow 0 156 0 145 c) yellow * wild type 210 0 0 190 d) wild type * yellow 102 98 99 97 ANSWER: All the offspring in cross (a) are wild type; yellow is recessive to wild type; let's define A as the dominant wild-type allele and a as the yellow mutant recessive. In all the progenies we have roughly equal numbers of males and females, which is what is expected. However, there is some sex bias, and reciprocal crosses give different results: all the sons of yellow females (homozygous a/a) are yellow; all the daughters of wild-type males are wild type; this suggests sex linkage.

In fact:

Parents Female a) XA/XA expected ratio

Progeny ––––––––––––––––––––––––––––––––––––––––––––– Females Males wild type yellow wild type yellow ––––––––––––––––––––––––––––––––––––––––––––– not possible XA/Xa Xa/Xa XA/Y all WT, as observed all WT, as observed

*

Male Xa/Y

b) Xa/Xa expected ratio

*

Xi/Y

not possible Xa/Xa all yellow, as observed

c) Xa/Xa expected ratio

*

XA/Y

not possible not possible Xa/Y XA/Xa all WT, as observed all yellow, as observed

d) XA/Xa expected ratio

*

Xa/Y

XA/Xa Xa/Xa 1:1, as observed

not possible Xa/Y all yellow, as observed

XA/Y Xa/Y 1:1, as observed

56. The black and yellow pigments in the coats of cats are controlled by an X-linked pair of alleles. Females heterozygous for these alleles have areas of black and areas of yellow in their coat (called tortoise-shell, or calico if there are also patches of white hair). a. A calico cat has a litter of eight kittens: one yellow male, two black males, two yellow females, and three Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance calico females. Assuming there is a single father for the litter, what is his probable color? b. A yellow cat has a litter of four kittens: one yellow and three calico. Assuming there is a single father for the litter, what is the probable sex of the yellow kitten? c. How would you prove that XO cats are phenotypically female? What female kitten colors (with respect to yellow, calico, and black) would you look for in which types of parental color crosses?

ANSWER: a. Yellow (genotype: yw; Y chromosome) where yw = yellow; yw+ = black b. Male. Since the father must be black (genotype yw+; Y chromosome), the only true yellow progeny cannot have received a color gene from the father. It must be male, and must have received its one X chromosome from its mother. c. Look for female kittens that fail to express an allele they should have inherited from their mothers: black female kittens from yellow mothers or yellow female kittens from black mothers.

These kittens should have an X chromosome from their fathers as usual; the fact that they show no alleles from their mothers may suggest they developed from eggs without an X chromosome and therefore that XO is female. Similarly, look for female kittens that fail to express a color that should have been inherited from their father. Female progeny of yellow tomcats should be either yellow or calico and of black tomcats, either black or calico, depending on the allele inherited from the mother. A black daughter of a yellow tomcat might come from a sperm lacking any sex chromosome. Chromosomal checks would be required on these unexpected progeny.

57. A young woman is worried about having a child because her mother's only sister had a son with Duchenne muscular dystrophy (DMD). The young woman has no brothers or sisters. (DMD is a rare X-linked recessive disorder.) a. Draw the relevant parts of the pedigree of the family described above. (Be sure to include the grandmother, the three women mentioned, and all their mates.) b. State the most likely genotype of everyone in the pedigree. c. Calculate the probability that the young woman's first child will have DMD. ANSWER: (a) pedigree and (b) genotypes

c. The grandmother must have been D/d. There is a 1/2 chance that the mother is D/d and, if so, a further 1/2 chance that the woman herself is D/d. If she is, 1/2 of her sons will have DMD. Since the probability of a son is also 1/2, the overall probability is 1/2 * 1/2 * 1/2 * 1/2 = 1/16.

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance 58. a. In families with four children, what proportion of the families will have at least one boy? b. In families with two girls and one boy, what fraction of the families will have the boy as the second child? c. In families with four children, what fraction of the families will have the gender order male-female-femalemale?

ANSWER: a. 0.9375, since 1 – Prob. of 4 girls, or 1 – (.5)4 = 1 – 0.0625. The frequency can be calculated more laboriously by expanding the binomial (p + q)4 = p4 + 4p3q + 6p2q2 + 4pq3 + q4 and calculating that 15/16 (0.9375) of the distribution has one boy. b. 1/3, because the frequencies of MFF, FMF, and FFM families are equal. c. Of four-child families, 6/16 have two boys and two girls; only 1/6 of such families will have the birth order MFFM. Therefore, 1/16 will have that particular birth order. The same answer can be derived as (0.5)4.

59. A man whose mother had cystic fibrosis (autosomal recessive) marries a phenotypically normal woman from outside the family, and the couple considers having a child. a. If the frequency of cystic fibrosis heterozygotes (carriers) in the general population is 1 in 25, what is the chance that the first child will have cystic fibrosis? b. If the first child does have cystic fibrosis, what is the probability that the second child will be normal?

ANSWER: a. The man must be a heterozygote, C/c. The probability that his wife is C/c is 1/25, and if they are both C/c, the probability of having an affected child is 1/4. Overall, the probability is (1/25)(1/4) = 1/100. b. The first child shows that both parents must have been C/c, so the probability that the next child will be normal is 3/4.

60. Consider the following pedigree of a rare autosomal recessive disease. Assume all people marrying into the pedigree do not carry the abnormal allele.

a. If individuals A and B have a child, what is the probability that the child will have the disease? b. If individuals C and D have a child, what is the probability that the child will have the disease? c. If the first child of C * D is normal, what is the probability that their second child will have the disease? d. If the first child of C * D has the disease, what is the probability that their second child will have the disease?

ANSWER: a. Choosing M for unaffected and m for the disorder, male B must be M/m, and female A has a 2/3 Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance chance of being M/m. The overall chance of an affected child is 1 * 2/3 * 1/4 = 1/6. b. If C's mother A is heterozygous, C stands a 1/2 chance of being heterozygous. D's mother must be heterozygous, and D stands a 1/2 chance of inheriting that heterozygosity. The overall chance of an affected child is 2/3 * 1/2 * 1 * 1/2 * 1/4 = 1/24. c. The probability is still 1/24. d. Now that we know individuals C and D must both be M/m, the chance of the second child being m/m is 1/4.

61. Below is the pedigree of a family where some individuals are affected with a mild condition of the skin.

a. Based on the pedigree, what is the most likely mode of inheritance of this condition, and why? b. Indicate the respective genotypes of each individual represented. For individuals who could have two or more genotypes, calculate the relative probability of each possible genotype. c. What is the probability that individuals 1 and 2 will have an affected daughter?

ANSWER: a. Autosomal recessive; it is the only mode of inheritance whereby two unaffected parents can have an affected daughter (as is the case for I-1 and I-2 and their first child). b. A = WT; a = mild condition The affected individuals are a/a; all four individuals in generation I are A/a; the unaffected people in generation II have a probability of 2/3 of being A/a and 1/3 of being A/A; and for individual "2" a few more calculations are required: • • • •

If both her parents are A/A (probability of 1/9), then she's A/A. If one of her parents is A/A and the other A/a (probability of 4/9), then she has a 50% chance of being A/A and 50% chance of being A/a.

If both of her parents are carriers (probability of 4/9), then she has a 2/3 chance of being A/a and 1/3 chance of being A/A. Overall, her probability of being A/A is (1/9) + (1/2)(4/9) + (4/9)(1/3) = 13/27, and her probability of being A/a is (1/2)(4/9) + (4/9)(2/3) = 14/27.

c. (14/27)(1/2) = 7/27 62. In the late 1800s, Mendel defined two fundamental laws of transmission genetics; these were subsequently used to establish chromosome theory as scientists examined visible chromosomes in meiotic cells. Define these two laws, and diagram where in the process of meiosis these two processes actually occur.

ANSWER: Students can diagram meiosis, which is a healthy review of their understanding of the process, and identify within this process the observations of Mendel (working without knowledge of meiosis).

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chapter 02: Single-Gene Inheritance Mendel's first law focused upon the segregation of genetic determinants during meiosis. This is essentially the anaphase I-mediated process of reducing ploidy during meiosis I. Mendel's second law, independent assortment, occurs during meiosis I as homologous chromosomes are lined up and assorted to meiocytes. This process is distinctly random in each meiotic process and is key to genetic diversity within gamete production.

63. Describe two ways in which the principles of inheritance (the law of equal segregation) can be applied.

ANSWER: The principles of inheritance can be applied in the following scenarios: (1) when inferring genotypes from phenotypic ratios, and (2) when predicting phenotypic ratios from parents of known genotypes.

64. Why would a cross between a mutant and a wild type produce a 4:1 ratio of the F1 progeny?

ANSWER: A ratio of 4:1 in this instance may be because the phenotype relies on the interactions of multiple genes, or due to an environmental effect.

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chapter 03: Independent Assortment of Genes Multiple Choice 1. Mendel crossed Y/Y ; R/R (yellow wrinkled) peas with y/y ; r/r (green smooth) peas and selfed the F1 to obtain an F2. In the F2, what proportion of the yellow wrinkled individuals were pure-breeding? a. 1/9 b. 3/16 c. 1/4 d. 3/4 e. 9/16 ANSWER: a 2. Mendel crossed Y/Y ; R/R (yellow wrinkled) peas with y/y ; r/r (green smooth) peas and selfed the F1 to obtain an F2. What proportion of the F2 individuals was pure-breeding? a. 1/9 b. 3/16 c. 1/4 d. 3/4 e. 9/16 ANSWER: c 3. If genes assort independently, a testcrossed dihybrid characteristically produces progeny phenotypes in the ratio a. 1:1. b. 1:1:1:1. c. 1:2:1. d. 3:1. e. 9:3:3:1.

ANSWER: b 4. A fish of genotype a/a ; B/b is crossed with a fish whose genotype is A/a ; B/b. What proportion of the progeny will be heterozygous for at least one of the genes? (Assume independent assortment.) a. 1/8 b. 2/8 c. 4/8 d. 5/8 e. 6/8 ANSWER: e 5. In the offspring of a dihybrid self, what percentage of the individuals are themselves dihybrid? a. 6.25% b. 12.50% Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chapter 03: Independent Assortment of Genes c. 18.75% d. 25.00% e. 56.25% ANSWER: d 6. A leucine-requiring mutant strain of haploid yeast is crossed to a cysteine-requiring mutant strain. Assuming independent assortment, what proportion of the spores produced will be mutant? a. 1/16 b. 3/16 c. 1/4 d. 1/2 e. 3/4 ANSWER: e 7. In a haploid fungus similar to Neurospora, a red-colored mutant is crossed with an alanine-requiring mutant.

Assuming independent assortment, what proportion of the spores produced will be alanine-requiring? a. 1/16 b. 3/16 c. 1/4 d. 1/2 e. 3/4 ANSWER: d 8. Two pure-breeding mutant plants were crossed: One had small leaves (wild-type leaves are large), and the other made pink flowers (wild-type flowers are purple). All F1 individuals had small leaves and purple flowers.

Assuming independent assortment, what proportion of the F2 individuals are expected to be phenotypically wild type? a. 1/16 b. 3/16 c. 1/4 d. 9/16 e. 3/4 ANSWER: b 9. Mendel's Y/y ; R/r dihybrid pea plants were the F1 of the cross between a double homozygous dominant and a double homozygous recessive. If we testcrossed these dihybrids, what proportion of the offspring would be recombinant and phenotypically resemble the F1 dihybrid? a. 0% b. 25% c. 50% d. 75% Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chapter 03: Independent Assortment of Genes e. 100% ANSWER: a 10. The following is known about the inheritance of size and fur color in Holland lop rabbits: - Crosses between large individuals produce only large individuals.

- Crosses between dwarf individuals produce both large and dwarf rabbits in a ratio of 1:2. Such crosses also produce some very small kits (baby rabbits) that generally die within a few days.

- Crosses between brown rabbits produce only brown kits.

- Some crosses between black rabbits produce only black kits, whereas others produce both black and brown kits.

- The size and fur-color phenotypes segregate independently.

What are the expected phenotypes in the F1 of a cross between a dwarf rabbit that breeds true for brown fur color and a large rabbit that breeds true for black fur color? a. black dwarf only b. black dwarf and black large only c. brown dwarf and black dwarf only d. brown dwarf, brown large, black dwarf, and black large e. brown large only ANSWER: b 11. The following is known about the inheritance of size and fur color in Holland lop rabbits: - Crosses between large individuals produce only large individuals.

- Crosses between dwarf individuals produce both large and dwarf rabbits in a ratio of 1:2. Such crosses also produce some very small kits (baby rabbits) that generally die within a few days.

- Crosses between brown rabbits produce only brown kits.

- Some crosses between black rabbits produce only black kits, whereas others produce both black and brown kits.

- The size and fur-color phenotypes segregate independently.

A large black doe (female rabbit) gives birth to nine black kits; five are large, and four are dwarf. What are the possible phenotypes of the kits' father? a. black dwarf only b. black dwarf, brown dwarf, black large, or brown large c. black dwarf or black large only d. black dwarf or brown dwarf only e. black dwarf or brown large only ANSWER: d 12. In hogs, a dominant allele B results in a white belt around the body. At a separate locus, the dominant allele S causes fusion of the two parts of the normally cloven hoof resulting in a condition known as syndactyly. A belted syndactylous sow was crossed to an unbelted cloven-hoofed boar, and in the litter there were: 25% belted syndactylous Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chapter 03: Independent Assortment of Genes 25% belted cloven 25% unbelted syndactylous 25% unbelted cloven The genotypes of the parents can best be represented as which of the following? a. B/B ; S/S × b/b ; s/s b. B/b ; S/s × b/b ; S/S c. B/b ; S/s × B/B ; s/s d. b/b ; S/s × B/b ; s/s e. B/b ; S/s × b/b ; s/s ANSWER: e 13. A female Drosophila with singed bristles is crossed with a male with sepia (dark brown) eyes. In the F1, all females are wild type, and all males have singed bristles. F1 males and females are crossed together, and the following F2 phenotypic ratios are obtained.

748 wild-type females 753 wild-type males 742 singed females 747 singed males 249 sepia females 250 sepia males 244 singed, sepia females 246 singed, sepia males If sn and se represent the mutant alleles of the singed and sepia gene, respectively, and sn+ and se+ are their wild-type counterparts, what are the most likely genotypes of the original parents? a. X/X ; sn/sn ; se+/se+ and X/Y; sn/sn+ ; se/se b. X/X ; sn/sn ; se+/se

and

X/Y ; sn/sn+ ; se/se

c. Xsn/Xsn+ ; se+/se+

and

Xsn+/Y ; se/se

d. Xsn/Xsn ; se+/se+

and

Xsn+/Y ; se/se

e. Xsn/Xi ; se+/se

and

Xsn+/Y ; se/se

ANSWER: d 14. The pedigree below shows the inheritance patterns of two extremely rare, independent genetic conditions (represented with black and gray symbols, respectively). If individuals A and B have a child, what is the probability that it will have both conditions?

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chapter 03: Independent Assortment of Genes

a. 1/16 b. 1/9 c. 1/4 d. 1/3 e. 4/9 ANSWER: b 15. A pure line of guinea pigs with genotype A/A ; b/b is crossed with a pure line with genotype a/a ; B/B. The F1 male and female guinea pigs are crossed to each other to generate the F2. What proportion of the F2 individuals are the result of a recombinant egg being fertilized by a recombinant sperm? a. 1/16 b. 2/16 c. 4/16 d. 6/16 e. 10/16 ANSWER: c 16. If a plant of genotype A/a ; B/b ; C/c ; D/d is selfed and the genes assort independently, how many different genotypes will be found among the progeny? a. 4 b. 16 c. 24 d. 64 e. 81 ANSWER: a 17. If two mice of genotype F/f ; G/g ; H/h ; I/I ; J/j are repeatedly mated, how many different phenotypes will be found in the progeny? (Assume complete dominance for all genes.)

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chapter 03: Independent Assortment of Genes a. 15 b. 16 c. 32 d. 128 e. 256 ANSWER: c 18. In a trihybrid, how many different gamete genotypes can be produced with respect to the three genes of interest? a. 3 b. 4 c. 8 d. 16 e. 27 ANSWER: c 19. If an individual has 10 gene pairs, how many different gametes can be formed if 5 of the gene pairs are homozygous and the remaining 5 gene pairs are heterozygous? a. 20 b. 32 c. 60 d. 64 e. 1024 ANSWER: b 20. Two pure lines obtained from Mendel's experiments are crossed with each other. The first line makes green smooth seeds (peas), and the second line has white flowers. All F1 individuals have purple flowers and produce yellow wrinkled seeds. What proportion of the F2 plants should have purple flowers and make green wrinkled seeds? a. 1/64 b. 3/64 c. 1/16 d. 9/64 e. 3/16 ANSWER: b 21. In the cross between a female A/a ; B/b ; c/c ; D/d ; e/e and male A/a ; b/b ; C/c ; D/d ; e/e, what proportion of the progeny will be phenotypically identical to the female parent? (Assume independent assortment of all genes and complete dominance.) a. 1/16 b. 9/64 Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chapter 03: Independent Assortment of Genes c. 3/16 d. 9/16 e. 3/8 ANSWER: b 22. A corn plant of genotype A/a ; B/B ; C/c ; D/d ; E/e ; f/f ; G/g is selfed. How many different completely homozygous genotypes could be produced? a. 2 b. 4 c. 8 d. 16 e. 32 ANSWER: e 23. A corn plant of genotype A/a ; B/B ; C/c ; D/d ; E/e ; f/f ; G/g is selfed. What is the probability of producing a completely homozygous individual out of all possible offspring? a. 0.000 b. 0.008 c. 0.013 d. 0.031 e. 0.062 ANSWER: d 24. A trihybrid cross is a cross between two individuals who are heterozygous for three genes. Assuming independent assortment and complete dominance, what phenotypic ratios would be observed in F1? a. 27:9:9:9:3:3:3:1 b. 16:9:9:3:3:1 c. 12:9:3:1 d. 9:3:3:1 e. 1:1:1:1 ANSWER: a 25. Consider the following cross when examining five unlinked genes. Identify what proportion of progeny will be phenotypically identical to parent 2.

Parent 1: A/a ; B/b ; C/c ; D/d ; E/e × Parent 2: A/A ; b/b ; C/c ; D/d ; E/e a. 1/216 b. 27/128 c. 1/64 d. 1/16 e. 1/32 Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chapter 03: Independent Assortment of Genes ANSWER: b 26. Consider the following cross when examining five unlinked genes. Identify what proportion of the progeny will be genotypically identical to parent 2.

Parent 1: A/a ; B/b ; C/c ; D/d ; E/e × Parent 2: A/A ; b/b ; C/c ; D/d ; E/e a. 1/216 b. 27/128 c. 1/64 d. 1/16 e. 1/32 ANSWER: e 27. In a Drosophila cross, you mate a parental line displaying crossed eyes (recessive trait, autosomal) and corkscrew bristles (recessive trait, autosomal) by a wild-type fly. The sites of the two genes are known to be unlinked. The F1 flies are wild type. You self the F1 flies in the hope of regenerating the cross-eyed/corkscrew phenotype.

How many individual F2 flies would you need to analyze to have 95% confidence of finding at least one crosseyed/corkscrew fly? a. 16 flies b. 8 flies c. 32 flies d. 47 flies e. 216 flies ANSWER: d 28. You are studying meiosis in fungi with an organism in which the haploid chromosome number is 5 (n = 5).

You combine two haploid cells together (strain 1 by strain 2), and then activate meiosis to regenerate haploid cells. What are the odds that any individual haploid cell will have either the five chromosomes from strain 1 or the five chromosomes from strain 2? a. 1/64 b. 1/2 c. 1/16 d. 1/32 e. 1/8 ANSWER: c 29. Which of the following is/are TRUE with respect to haploid and diploid cells? a. Following DNA replication, haploid cells become diploid. b. In diploid cells, meiosis I results in the formation of two haploid cells. c. In haploid cells, condensed chromosomes have an "I" shape, while in diploid cells they have an "X" Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chapter 03: Independent Assortment of Genes shape. d. In haploid cells meiosis is simpler, owing to unpaired chromosomes. e. The process of meiosis is identical in haploid and diploid cells.

ANSWER: b 30. Meiosis I is about to start in a cell of a plant in which 2n = 18. At this stage, the cell has a. 9 chromosomes, and a total of 18 chromatids. b. 18 chromosomes, and a total of 18 chromatids. c. 18 chromosomes, and a total of 36 chromatids. d. 36 chromosomes, and a total of 36 chromatids. e. 36 chromosomes, and a total of 72 chromatids.

ANSWER: c 31. In a plant in which 2n = 24, what is the total number of chromosomes present at the end of interphase? a. 6 b. 12 c. 24 d. 48 e. 96 ANSWER: c 32. A male Drosophila melanogaster has the genotype A/a ; B/b ; C/c ; XD/Y. How many different sperm genotypes can it produce through meiosis of one single pre-gametic (2n) cell? a. 1 b. 2 c. 4 d. 8 e. 16 ANSWER: b 33. In a particular dihybrid, the two genes of interest, A and B, are located on different chromosomes. Which of the following diagrams does NOT represent a stage of meiosis in this organism? (Note: For each answer, only some chromatids may be represented.) a.

b.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chapter 03: Independent Assortment of Genes c.

d.

e.

ANSWER: e 34. What process occurring during meiosis ensures independent assortment? a. the arrangement of homologous chromosomes on the metaphase plate b. the fusion of male and female gametes c. the pairing of homologous chromosomes in prophase d. the replication of DNA and formation of sister chromatids e. the "splitting apart" of sister chromatids in anaphase ANSWER: a 35. In a diploid organism, 2n = 6, and there are two long, two intermediate, and two short chromosomes. What is the most accurate representation of a gamete resulting from meiosis in this organism? a. b. c. d. e.

ANSWER: b 36. Which diagram most accurately shows the arrangement of homologous chromosomes during the first metaphase of meiosis? a. b. c. d. e.

ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chapter 03: Independent Assortment of Genes 37. Which diagram most accurately shows the arrangement of homologous chromosomes during the first metaphase of mitosis? a. b. c. d. e.

ANSWER: b 38. Fruit color in a particular plant is controlled by a set of three QTLs (quantitative trait loci, or "polygenes") that work in an equal and additive manner. Each QTL has two alleles (i.e., A and a); each allele represented by a capital letter produces one dose of yellow pigment, while alleles represented with lowercase letters do not produce any pigment at all.

Assuming no effects by the environment, how many different shades of yellow fruit color are there? a. 3 b. 7 c. 9 d. 12 e. 64 ANSWER: b 39. Fruit color in a particular plant is controlled by a set of three QTLs (quantitative trait loci, or "polygenes") that work in an equal and additive manner. Each QTL has two alleles (i.e., A and a); each allele represented by a capital letter produces one dose of yellow pigment, while alleles represented with lowercase letters do not produce any pigment at all.

A trihybrid plant (A/a ; B/b ; C/c) is selfed. Assuming no effects of the environment, what proportion of the offspring will have the same fruit color phenotype as the dihybrid parent? a. 0.031 b. 0.125 c. 0.250 d. 0.312 e. 0.422 ANSWER: d 40. Fruit color in a particular plant is controlled by a set of three QTLs (quantitative trait loci, or "polygenes") that work in an equal and additive manner. Each QTL has two alleles (i.e., A and a); each allele represented by a Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chapter 03: Independent Assortment of Genes capital letter produces one dose of yellow pigment, while alleles represented with lowercase letters do not produce any pigment at all.

If a trihybrid plant (A/a ; B/b ; C/c) is testcrossed, what proportion of the offspring will have a fruit color phenotype that is different from both the trihybrid and the tester parents? (Assume no environmental effects.) a. 1/8 b. 1/4 c. 3/8 d. 1/2 e. 3/4 ANSWER: e 41. In animals, selfing is accomplished by mating animals of a. identical genotype. b. similar genotype. c. identical phenotype. d. similar phonotype.

ANSWER: a 42. You generated a plant that is heterozygous for three quantitative trait loci (QTL). In the heterozygous state, each locus has one allele that contributes one dose of trait (pigment) and another allele that does not contribute a dose to the trait; genotype: R1/r1; R2/r2; R3/r3. If you testcross this plant (by r1/r1; r2/r2; r3/r3, all "0-dose" alleles) what distribution of pigment doses would be observed in the progeny? a. 1/8, three doses; 3/8, two doses; 3/8, one dose; 1/8, zero doses b. 9/16, three doses; 3/16, two doses; 3/16, one dose; 1/16, zero doses c. 1/4, three doses; 1/4, two doses; 1/4, one dose; 1/4, zero doses d. 1/2, two doses; 1/2, zero doses e. One hundred percent will be pigmented, equally.

ANSWER: a 43. What statistical method is used to assess whether or not the number of observed individuals with certain phenotypes are an acceptable fit to an expected Mendelian ratio? a. the sum rule b. the product rule c. the sum and product rule combined d. chi-square test e. multiplication of probabilities ANSWER: d Essay 44. Why are homozygous pure lines an essential tool of genetics?

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chapter 03: Independent Assortment of Genes ANSWER: Homozygous pure lines are important research tools that allow geneticists to maintain a source of any given genotype. Members of a pure line interbreed over time and provide a constant source of the genotype for use in experiments. In addition, recessive alleles can be expressed only in pure lines.

45. In a species of Mexican mollies, fish with ragged fins and nonragged fins were found. When mollies of a true-breeding (i.e., pure-breeding) ragged strain were crossed with nonragged fish, the F1 all had ragged fins, and in the F2, three ragged appeared to one nonragged. Diagram this cross, letting R = ragged and r = nonragged, and give parental phenotypes, genotypes, and germ cells, as well as F2 genotypes and phenotypes.

ANSWER:

The F2 (from R/r intercrossed) is 1/4 R/R and 1/2 R/r (for a total of 3/4 ragged) and 1/4 r/r (nonragged).

46. Two albino strains of Neurospora crassa were isolated in two different laboratories. When either one was mated with wild type (orange), it produced 1/2 albino: 1/2 orange progeny. a) What can you now deduce about the genetics of each of these albino strains? b) Assume first that the two strains carry allelic mutations. Using appropriate symbols, diagram a cross of the two albino strains. Include the genotypes of the parents, a stick-and-circle diagram of the relevant chromosomes at metaphase I of meiosis, and the genotypes and ratios of the progeny. c) Now assume that the two strains carry mutations in different genes, and that the genes are on different chromosomes. Answer the same questions as above. Assume that there are no crossovers between either gene and the centromeres.

ANSWER: a) They are each single-gene mutations. Multiple-gene mutations would produce fewer wild-type progeny.

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chapter 03: Independent Assortment of Genes

47. An arginine-requiring strain of Neurospora is crossed with a wild-type (prototrophic) strain. a) In the first cross, half of the progeny are arginine-requiring and half are wild type. What does this tell you about the genetic basis of the arginine-requiring mutant? b) In the second, the progeny are 83 arginine-requiring and 13 wild type. The germination percentage of ascospores is 95–100%. Explain these results simply, and tell briefly how you would confirm your explanation.

ANSWER: a) The arginine mutant results from a mutation at one gene. Two alleles: mutant and wild type. b) In a haploid, failure to get 1:1 segregation for a character, without inviability of one class, makes one consider two- and three-gene models. If the arg parent were carrying two mutations in the same path, and they assorted independently, the cross (a/b × a+/b+) would yield four progeny types, with only one being able to grow without arginine. In the present case, three genes seem to be assorting (independently) since only 1/8 (13/100) are prototrophic.

48. A woman with tritanopia (a rare form of inherited blue-yellow color blindness) marries a man who also has tritanopia. Their first daughter has normal color vision but is diagnosed with PKU. a) Based on this information, what is the most likely mode of inheritance of tritanopia? b) Using symbols of your own invention, write the genotypes of the two parents and of their daughter. (Assume independent assortment.) c) What is the probability that the couple's next child is a son with tritanopia and PKU? (Assume independent assortment.)

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chapter 03: Independent Assortment of Genes ANSWER: a) Two affected parents have an unaffected child; this is indicative of a dominant condition. The unaffected child is female, which rules out the possibility of X-linked inheritance. The most likely mode of inheritance is autosomal dominant. b) PKU is inherited as an autosomal recessive (unaffected parents have an affected daughter). Let p = PKU, P = no PKU; T = tritanopia, t = normal color vision; X and Y represent the X and Y chromosomes, respectively.

Parents: X/X ; T/t ; P/p and X/Y ; T/ t ; P/p Daughter: X/X ; t/t ; p/p c) Prob(get Y) × Prob(get T) × Prob (get p) × Prob (get p) = (1/2)(3/4)(1/2)(1/2) = 3/32 49. Vanessa has obtained two true-breeding strains of mice, each homozygous for an independently discovered recessive mutation that prevents the formation of hair on the body. The discoverer of one of the mutant strains calls his mutation naked, and the other researcher calls her strain hairless. To determine whether the two mutations are simply alleles for the same gene, Vanessa crosses naked and hairless mice with each other. All the offspring are phenotypically wild type. After intercrossing these F1 mice, however, Vanessa observes 115 wildtype mice and 85 mutant mice in the F2. a) What is the most likely explanation for the segregation of wild-type mice and mutant mice in the F2? Are the naked and hairless mutations alleles for the same gene? b) Using symbols of your own choosing, indicate the genotypes of the parents, the F1, and each of the phenotypic classes of the F2 progeny. Describe any gene interactions in this set of crosses.

ANSWER: The naked and hairless mutations are not the same allele because complementation occurred and the F1 hybrids are phenotypically wild type. Naked and hairless are instead mutations of two different genes. To explain the phenotypic ratio in the F2, let us first adopt symbols for these mutations and their dominant wild-type alleles: n = naked mutation, N = wild-type allele, h = hairless mutation, H = wild-type allele The genotypes of the true-breeding parental strains are nnHH (naked) and NNhh (hairless). F1 hybrids produced by crossing these strains are therefore NnHh. When these hybrids are intercrossed, we expect several genotypes to appear in the offspring. Each recessive allele, however, when homozygous, prevents the formation of hair on the body.

Thus, only mice that are genotypically N–H– will develop hair; all the others—homozygous nn, or homozygous hh, or homozygous for both recessive alleles—will fail to develop body hair. We can predict the frequencies of the wild-type and mutant phenotypes if we assume that the naked and hairless genes assort independently.

This gives the following genotype and phenotype ratios:

Genotypes

Phenotypes

9 N–H– 3 N–hh 3 nnH– 1 nnhh

normal no hair no hair no hair

The ratio of normal mice to those with no hair is 9:7.

In a sample of 200 F2 progeny, we would expect 200 × 9/16 = 112 to be wild type and 200 × 7/16 = Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chapter 03: Independent Assortment of Genes 88 to be mutant. The observed frequencies of 115 wild-type mice and 85 mutant mice are close to these expected numbers, suggesting that the hypothesis of two independently assorting genes for body hair is, indeed, correct.

50. A tomato plant with large leaves and round fruits was crossed to a plant with small leaves and oblong fruits.

The F1 consisted of plants with large leaves; about half of these plants made round fruits and the other half made oblong fruits. Two F1 plants (one with round fruits and one with oblong fruits) were then analyzed through further crossing. The results were: Cross 1. F1 plant #1 with round fruits × parent with round fruits 2. F1 plant #1 with round fruits × parent with oblong fruits

3. F1 plant #1 with round fruits × F1 plant #2 with oblong fruits

Progeny 35 large leaves, round fruits 12 large leaves, oblong fruits 17 large leaves, round fruits 20 large leaves, oblong fruits 7 small leaves, round fruits 5 small leaves, oblong fruits 27 large leaves, round fruits 26 large leaves, oblong fruits

a) What are the genotypes of the two original parents? b) What are the expected phenotypic ratios in the progeny of a cross between "F1 plant #1" and a tester? c) What gamete genotypes can "F1 plant #2" produce?

Explain your reasoning behind each of your answers.

ANSWER: Since we don't know whether the original parents are pure-breeding or not, the easiest way to determine dominance and derive some of the genotypes is by looking at the results of the three crosses presented. From cross 1, we can tell that oblong fruit shape is recessive to round fruit shape because two plants with round fruits produce some individuals with oblong fruits. From cross 2, we can tell that small leaf size is recessive to large leaf size (same reasoning). a) Let R = round, r = oblong, L = large, and l = small.

- Parent with oblong fruits must be L/l ; r/r (phenotypically oblong fruits that can produce individuals with small leaves when crossed to the F1 plant with large leaves) - Parent with round fruits must be L/L ; R/r (produces only offspring with large leaves, has round fruits, but can produce offspring with small fruits) b) The F1 plant with round fruits must be L/l ; R/r (reasoning is as above), so in a testcross it would produce a phenotypic ratio of 1:1:1:1. c) From cross 3, we can tell that the F1 plant with oblong fruits must be L/L ; r/r, and therefore it can produce only one gamete genotype, L ; r.

51. Define Mendel's experimental evidence for diploid chromosome content in pea plants.

ANSWER: Mendel noticed that recessive traits were "hidden" in the F1 progeny of his monohybrid and dihybrid crosses. When the F1 plants were selfed, the recessive trait appeared again in the next generation of plants (F2). This was reproducible and followed a ratiometric pattern consistent with two genetic determinants (genes) being carried for each trait.

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chapter 03: Independent Assortment of Genes 52. Suppose red-striped laughing frogs are diploid and have 32 chromosomes in somatic cells. Answer the following questions as they pertain to these vanishingly rare but beautiful creatures. a) How many pairs of homologous chromosomes does a red-striped laughing frog have in its somatic cells? b) How many chromosomes are there in a haploid set? c) How many chromosomes are there in a mature sperm cell? d) How many chromosomes are there in a somatic cell just after mitosis? e) How many double-stranded DNA molecules are there in a somatic cell as mitosis is about to start? f) How many chromosomes are there in a cell prior to DNA replication?

ANSWER: a) 16 b) 16 c) 16 d) 32 e) 64 f) 32 53. Two haploid yeast cells, one wild type and one having a mutation blocking the synthesis of adenine, were mated to produce a diploid zygote. Through asexual reproduction (mitosis), this zygote gave rise to hundreds of identical cells, and all were wild type in phenotype. These (diploid) cells were then induced to undergo meiosis, and the progeny (a random sample of haploid meiotic products) consisted of 214 cells able to grow without adenine and 203 unable to grow without adenine. a) How many genes are involved? b) What are the dominance relationships of the alleles of the gene or genes?

ANSWER: This problem forces you to recognize the ability or inability to grow without adenine as alternative characters that segregate in a 1:1 ratio. These meet the criteria of allelism: ade+ and ade– are alleles of the ade gene. There is only one gene in the cross, and the ade+ allele is dominant; the diploid formed in the cross has the phenotype of the ade+ parent.

54. Several pairs of light brown hamsters with identical genotypes were mated. Altogether, the F1 was comprised of 143 individuals: 62 light brown, 41 brown, 38 beige (very, very light brown), 1 dark brown, and 1 white. What mode of inheritance of the phenotype "fur color" could most easily explain these results?

ANSWER: Polygenic inheritance. The different phenotypes can be considered as different "degrees of brownness" and could be explained by a system of two genes with two alleles each that work in an equal and additive manner. The parents would be A/a ; B/b. A cross between two such individuals would produce a ratio of 1:4:6:4:1 in terms of "doses" if we consider the dominant alleles are "functioning" and the recessive ones as "nonfunctioning," in terms of producing pigment.

55. If stem length in a certain species of crops were solely determined by polygenic inheritance, how many different genes would be required in this system to explain the existence of 15 different stem lengths ranging from 15 to 29 cm?

ANSWER: A system with seven genes and two alleles each work in an equal and additive manner. It can be helpful to start by thinking about how many genes are required for 3, 4, 5, etc., different phenotypes.

From there, one will notice that the number of possible phenotypes is always equal to 2 × the number of genes + 1.

56. In a perfectly controlled laboratory environment, a pure-breeding fish with a 23-mm tail was crossed to a Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chapter 03: Independent Assortment of Genes pure-breeding fish with a 15-mm tail. In the F1, all individuals had tails of an intermediate length: 19 mm. If tail length is a typical polygenic trait controlled entirely by the equal and additive effect of two unlinked genes, each with two alleles ("+" and "–"), what phenotypic ratios do you expect to find in the F2?

ANSWER: The F1 individuals have the "intermediate" phenotype and are dihybrid, so two "doses" correspond to 19 mm. On the other hand, 15 mm and 23 mm must be the result of "zero doses" and "four doses," respectively. In F2, we should see a 1:4:6:4:1 ratio of 23:21:19:17:15 mm.

57. The pedigree below shows the inheritance of a rare muscle condition in bats. What inheritance model best explains the peculiar pattern?

ANSWER: The condition is passed on only by affected mothers (all the offspring of affected females are themselves affected, but the offspring of affected males are not). This pattern could be explained by mitochondrial inheritance.

58. You just isolated a new strain of mutant mice, and preliminary matings suggest that the new mutant phenotype is not inherited autosomally. Circumstantial evidence seems to indicate that the mutant phenotype may be following either an X-linked dominant or a mitochondrial-type inheritance pattern. a) What informative crosses would you set up to distinguish between these two possibilities? b) How would you interpret your results?

ANSWER: a) affected female × unaffected male non-affected female × affected male b) If the phenotype is X-linked dominant, then in the F1 of the first cross, all individuals should be affected (males and females); and in the F1 of the second cross, all the females but none of the males should be affected.

If it's mitochondrial inheritance, then in the F1 of the first cross, all individuals should be equally affected (males and females); but in the F1 of the second cross, nobody should be affected.

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination Multiple Choice 1. A plant of genotype C/C ; d/d is crossed to c/c ; D/D, and the F1 is testcrossed. If the genes in question are unlinked, the percentage of double homozygous recessive individuals in the offspring of the testcross will be a. less than 25%. b. about 25%. c. more than 25%, but less than 50%. d. about 50%. e. more than 50% but less than 75%.

ANSWER: b 2. A plant of genotype C/C·d/d is crossed to c/c·D/D, and the F1 is testcrossed. If the genes in question are linked, the percentage of double homozygous recessive individuals in the offspring of the testcross will be a. less than 25%. b. about 25%. c. more than 25% but less than 50%. d. about 50%. e. more than 50% but less than 75%.

ANSWER: a 3. A sweet pea plant of genotype A/A·B/B is crossed to one that is a/a·b/b to produce a dihybrid F1. One cell of one F1 individual (genotype A/a·B/b) goes through meiosis and produces four gametes of the following genotypes: A·b, A·b, a·B and a·B. What can be concluded regarding the linkage relationship between the two genes? a. Genes A and B are linked; the result is indicative of linkage as the gametes are recombinants. b. Genes A and B are linked; the result is indicative of linkage as the gametes are the products of crossovers. c. Genes A and B could be linked or unlinked; the result is not indicative of either, because in both cases it's possible to obtain 100% recombinant gametes from one meiosis. d. Genes A and B could be linked or unlinked; the result is unusual, as we should never obtain more than 50% recombinants from one meiosis. e. Genes A and B are unlinked; the result is indicative of independent assortment as only two genotypes are represented among the gametes.

ANSWER: c 4. A diploid plant is a trihybrid for flower color (gene F), leaf size (gene L) and seed weight (gene S); its phenotype includes red flowers, large leaves, and heavy seeds. This plant is crossed to a tester plant (which has white flowers, small leaves, and light seeds). The progeny is as follows: 23 Red, heavy, large 25 Red, heavy, small 230 Red, light, large

232 White, heavy, large 228 White, heavy, small 25 White, light, large

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination 235 Red, light, small

26

White, light, small

What can be concluded about the linkage relationships of the three genes in question? a. Genes F, L, and S are linked. b. Genes F and L are linked, while S assorts independently. c. Genes F and S are linked, while L assorts independently. d. Genes S and L are linked, while F assorts independently. e. Genes F, L, and S are unlinked and assort independently.

ANSWER: c 5. In a diploid invertebrate, genes D and E are closely linked. Single crossovers between these two genes occur in only 1 out of 40 meioses, and multiple crossovers are never observed. If an individual has the genotype D e / d E, what percentage of its gametes are expected to be recombinant (either DE or de)? a. 1.25% b. 2.5% c. 4% d. 5% e. 20% ANSWER: a 6. A male mouse of genotype A/a·B/b is testcrossed multiple times. Overall, these testcrosses produce 11 mice of phenotype AB, 81 Ab, 77 aB, and 14 ab. Based on these data, the genotypes of the male mouse's parents are most likely a. A/a ; B/b and A/a ; B/b. b. A B / A B and a b / a b. c. A B / a b and A B / a b. d. A b / A b and a B / a B. e. A b / a b and A b / a b.

ANSWER: d 7. In an experimental plant, the genes t and q are linked. A pure-breeding t/t individual is crossed to a purebreeding q/q individual to obtain a dihybrid F1. Genotypically, the F1 individuals can be described as a. complementary dihybrids. b. crossover dihybrids. c. dihybrids in cis. d. recombinant dihybrids. e. dihybrids in trans.

ANSWER: e 8. Two genes, A and B, are linked. An individual of genotype A b / A b is crossed to one that is a B / a B. The F1 is testcrossed. Which of the following ratios most likely represents the phenotypic ratio observed in the progeny of this testcross?

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination a. 1:1:1:1 b. 1:4:4:1 c. 9:3:3:1 d. 51:24:24:1 e. 66:9:9:16 ANSWER: b 9. Two genes, A and B, are linked. An individual of genotype A b / A b is crossed to one that is a B / a B. The F1 is selfed and an F2 is obtained. Which of the following ratios most likely represents the phenotypic ratio observed in this F2? a. 1:1:1:1 b. 1:4:4:1 c. 9:3:3:1 d. 51:24:24:1 e. 66:9:9:16 ANSWER: d 10. In Drosophila, the genes crossveinless-c and Stubble are linked, about 7 map units apart on chromosome 3. cv-c is a recessive mutant allele of crossveinless-c (cv-c+ is wild type), while Sb is a dominant mutant allele of Stubble (Sb+ is wild type). A dihybrid female Drosophila with genotype cv-c Sb+ / cv-c+ Sb is testcrossed. The proportion of phenotypically wild-type individuals in the progeny of the testcross will be a. 0.035 b. 0.070 c. 0.350 d. 0.465 e. 0.930 ANSWER: d 11. In Drosophila, the two genes w and sn are X-linked and 25 map units apart. A female fly of genotype w+ sn+ / w sn is crossed to a male from a wild-type line. What percent of male progeny will be w+ sn? a. 0% b. 12.5% c. 25% d. 37.5% e. 50% ANSWER: b 12. The maize genes sh and bz are linked, 40 map units apart. If a plant sh+ bz / sh bz+ is selfed, what proportion of the progeny will be sh bz / sh bz? a. 0.04 Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination b. 0.10 c. 0.20 d. 0.40 e. 0.50 ANSWER: a 13. In maize, two plants that are both heterozygous for the recessive alleles a and b are crossed. What frequency of double-mutant progeny will appear if a and b are 7.2 map units apart, and both parents carry a and b in repulsion (trans)? a. 0.001296 b. 0.005184 c. 0.036 d. 0.0625 e. 0.072 ANSWER: a 14. The F, G, and H loci are linked in the order written. There are 30 map units between F and G, and 30 map units between G and H. If a plant F G H / f g h is testcrossed, what proportion of progeny plants will be f g h / f g h if there is no interference? a. 0.70 b. 0.30 c. 0.245 d. 0.21 e. 0.15 ANSWER: c 15. Out of 800 progeny of a three-point testcross, there were 16 double crossover recombinants, whereas 80 had been expected on the basis of no interference. The interference is a. 0.690 b. 0.310 c. 0.155 d. 0.80 e. 0.40 ANSWER: d 16. The linkage relationships among the four mouse genes, A, B, C, and D, are shown in the map below.

What proportion of the eggs produced by a female mouse with genotype A b / a B ; C D / c d will carry all the recessive alleles?

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination a. 0.064 b. 0.075 c. 0.150 d. 0.250 e. 0.255 ANSWER: a 17. The three genes, D, E, and F, are closely linked as indicated on the map below. In this region of the genome, there is no interference (the coefficient of coincidence is 1).

If a trihybrid with genotype A B C/a b c is testcrossed, what is the expected percentage of "parentals" within the progeny? a. 8.5% b. 15% c. 42.5% d. 85% e. 85.5% ANSWER: e 18. In maize, the genes W and D are so tightly linked that virtually no crossovers occur between them. A dihybrid W d / w D is testcrossed to w d / w d. The percentage of progeny with W– / D– phenotype will be a. 0 b. 25 c. 50 d. 75 e. 100 ANSWER: a 19. The mouse autosomal genes B and S are linked and 38 map units apart. Genotypes B S / B S and b s / b s are intercrossed, and the F1 is testcrossed to b s / b s. The proportion of B– / S–progeny will be a. 0.19 b. 0.31 c. 0.50 d. 0.38 e. 0.76 ANSWER: b 20. A dihybrid in cis is crossed to a dihybrid in trans (the two genes of interest, A and B, are linked). What proportion of the progeny is expected to be homozygous recessive with respect to the two genes of interest?

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination a. less than 6.25% b. more than 6.25%, but less than 18.75% c. about 18.75% d. more than 18.75%, but less than 37.5% e. more than 37.5% ANSWER: a 21. A corn plant known to be hybrid for three linked genes is testcrossed. The progeny phenotypes and frequencies are: +++ abc a++ +bc ++c ab+ a+c +b+

74 70 44 50 4 2 368 388

The coefficient of coincidence in this genomic region is a. 0.06 b. 0.15 c. 0.40 d. 0.44 e. 1.36 ANSWER: c 22. In Neurospora, a, b, and c are auxotrophic mutants (require particular compounds for growth). The mating a b+ × a+ b gives 5% prototrophs (grown on unsupplemented medium). The mating b+ c × b c+ gives 2.5% prototrophs, and the mating a+ c × a c+ gives 2.5% prototrophs. Assuming complete interference prevails in the chromosome segment in question, what percent of prototrophs would you expect in the mating a+ bc+ × ab+ c?

(The order of mutations is not necessarily a, b, c: It can be determined on the basis of the two-point crosses above.) a. 0% b. 0.0625% c. 0.125% d. 2.5% e. 5% ANSWER: d 23. In a haploid organism, the loci leu and arg are linked, 30 map units apart. In a cross of leu+ arg × leu arg+, what proportion of progeny will be leu arg?

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination a. 0 b. 0.05 c. 0.10 d. 0.15 e. 0.30 ANSWER: d 24. Ascospores from a cross leu-2 × ad-7 met-3 (all auxotrophic markers) are plated on minimal medium (no adenine, leucine, or methionine). If the three genes are unlinked, what proportion of ascospores is expected to grow into a colony on this medium? a. 0.005 b. 0.045 c. 0.125 d. 0.250 e. 0.500 ANSWER: c 25. Ascospores from a cross a × b c (all auxotrophic markers) are analyzed. If genes b and c are linked, 10 map units apart, but gene a is located on a different chromosome, what proportion of ascospores is expected to be a+ b+ c+? a. about 12.5% b. about 22.5% c. about 25% d. about 45% e. about 50% ANSWER: b 26. Ascospores from a cross of c × d e (all auxotrophic markers) are analyzed. If genes c and d are linked, 10 map units apart, but gene e is located on a different chromosome, what proportion of ascospores is expected to be c+ d+ e+? a. about 1% b. about 2.5% c. about 5% d. about 40% e. about 50% ANSWER: b 27. In a dihybrid female fruit fly with genotype Aa Bb, 88% of the meiosis there are no chiasmata between the A and B loci, and in 12% there is one. What is the expected frequency of recombinant gametes produced by this fly? a. 3% Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination b. 6% c. 12% d. 24% e. 50% ANSWER: b 28. An SSLP is located 2 map units away from the X-linked locus responsible for red-green color blindness.

Three alleles of this SSLP are known: "4 repeats," "5 repeats," and "7 repeats." A woman with normal color vision, but whose father is color-blind, marries a man with normal vision, and they have a son.

Family member Woman’s father Woman

SSLP genotype

Family member

SSLP genotype

“5 repeats”

Woman’s husband

“3 repeats”

“5 repeats”/“7 repeats” Couple’s child

“5 repeats”

Given the SSLP genotypes of the family members shown above, what is the probability that the couple's son is color-blind? a. 2% b. 50% c. 98% d. 99% e. 100% ANSWER: c 29. In loppins (small, diploid invertebrates), chocolate-dependent problem-solving ability is an autosomal recessive trait. You have been hired to map the position of the gene controlling this phenotype (called "lu") using two SSLP markers (called "A" and "B"). You first cross two pure-breeding loppins: lu+ A[3 repeats] B[5 repeats] / lu– A[3 repeats] B[5 repeats]× lu– A[6 repeats] B[8 repeats] / lu– A[6 repeats] B[8 repeats] The F1 is then crossed back to the lu– / lu– parent. Which of the following results would indicate that the lu gene is much closer to SSLP "A" than to SSLP "B"? a. lower proportion of A[3 repeats] B[5 repeats] / A[3 repeats] B[5 repeats] than A[3 repeats] B[8 repeats] / A[3 repeats] B[5 repeats] individuals b. lower proportion of lu+ A[3 repeats] / lu– A[6 repeats] than lu+ A[3 repeats] / lu– A[3 repeats] individuals c. lower proportion of lu+ A[3 repeats] / lu– A[6 repeats] than lu+ A[6 repeats] / lu– A[6 repeats] individuals d. lower proportion of lu+ A[6 repeats] / lu– A[3 repeats] than lu+ B[8 repeats] / lu– B[8 repeats] individuals e. lower proportion of lu+ A[6 repeats] / lu– A[6 repeats] than lu+ B[8 repeats] / lu– B[8 repeats] individuals ANSWER: e 30. Below is a pedigree for a very rare, late-onset genetic condition. Below the pedigree are represented the Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination RFLP banding patterns obtained for each individual in the family. The RFLP locus used is about 2 cM away from the gene responsible for the disease. Note that darker/thicker bands indicate homozygosity for that particular morph of the RFLP.

What is the probability that the child in generation III will develop the disease? a. 0% b. 0.4% c. 1% d. 2% e. 50% ANSWER: d 31. In a linear tetrad analysis, the second division segregation (MII) frequency of the cyh locus is 16%. The map distance from this locus to its centromere is a. 4 m.u. b. 8 m.u. c. 16 m.u. d. 32 m.u. e. 50 m.u.

ANSWER: b 32. A second-division segregation ascus is an indication of a. absence of linkage between centromere and locus. b. a crossover between centromere and locus. c. a four-strand crossover between centromere and locus. d. interference in the region between centromere and locus. e. a two-strand double crossover between centromere and locus.

ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination 33. The maximum second-division segregation frequency normally possible is a. 25% b. 33.3% c. 50% d. 66.7% e. 100% ANSWER: d 34. In an unordered ascus analysis of the Neurospora loci cot and fl, there were 42 tetratypes and 8 nonparental ditypes out of a total of 290 asci. These loci are linked at a distance of a. 2 m.u. b. 5 m.u. c. 10 m.u. d. 21 m.u. e. 25 m.u.

ANSWER: c 35. In unordered tetrad analyses of two linked loci, a nonparental ditype is an indication of a. no crossovers. b. single crossover. c. two-chromatid double crossover. d. three-chromatid double crossover. e. four-chromatid double crossover.

ANSWER: e 36. In the cross mt+ al+ × mt al, the mt locus shows an MII frequency of 12%, and the al locus shows an MII frequency of 10%. Most asci were parental ditypes. The order of loci is a. al, centromere, mt. b. al, mt, centromere. c. centromere, al, mt. d. centromere, mt, al. e. mt, centromere, al.

ANSWER: c 37. Neurospora loci d and f are located on the same chromosome. The d locus always segregates at MI, while the e locus segregates at MII at a frequency of 10%. Assuming complete interference, the expected frequency of parental ditypes in the progeny of a cross d+ e+ × d e is a. 30%. b. 40%. c. 45%.

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination d. 80%. e. 90%.

ANSWER: e 38. Neurospora loci d and f are located on the same chromosome. The d locus always segregates at MI, while the e locus segregates at MII at a frequency of 10%. Assuming complete interference, the distance between loci d and f is a. 5 map units. b. 10 map units. c. 20 map units. d. 40 map units. e. 45 map units.

ANSWER: a 39. A dihybrid Aa Bb female Drosophila is testcrossed with an aa bb male. The following offspring genotypes were obtained.

Aa Bb Aa bb aa Bb aa bb

99 83 86 92

A chi-square test was performed to determine whether the data supported the original hypothesis that the genes are unlinked. How many degrees of freedom should be used in this analysis? a. 1 b. 2 c. 3 d. 4 e. 5 ANSWER: c 40. Dihybrid Aa Bb female Drosophila were testcrossed with aa bb males. The following offspring genotypes were obtained.

Aa Bb Aa bb aa Bb aa bb

99 83 86 92

A chi-square test was performed to determine whether the data supported the original hypothesis that the genes are unlinked. What is the correct value of χ2? a. 0.24 b. 0.42 c. 0.47 Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination d. 1.67 e. 2.79 ANSWER: d 41. Using data from a testcross of a dihybrid, a χ2 value of 6.5 was obtained, with 3 degrees of freedom. What is the correct interpretation of this result? a. The p value is approximately 0.01 (1%), so the hypothesis of no linkage should be rejected. b. The p value is approximately 0.01 (1%), so the hypothesis of no linkage should not be rejected. c. The p value is approximately 0.1 (10%), so the hypothesis of no linkage should be rejected. d. The p value is approximately 0.1 (10%), so the hypothesis of no linkage should not be rejected.

ANSWER: d 42. In one short chromosome arm the average frequency of crossovers per meiosis was measured to be 0.5.

What proportion of meioses will have no crossovers at all in this region? a. 0.40 b. 0.52 c. 0.60 d. 0.74 e. 0.90 ANSWER: c 43. In a fungal cross of A B × a b (the loci are linked) in meiosis in which there has been a four-strand double crossover between the loci, the percent of meiotic products that will be recombinant for A and B for that ascus is a. 0 b. 25 c. 50 d. 75 e. 100 ANSWER: e 44. In a cross in which there is a mean crossover frequency of 0.9 between two marker loci, the recombinant frequency for those loci will be a. 5% b. 10% c. 30% d. 45% e. 90% ANSWER: c 45. The fungal cross a·b+ × a+·b yields 30 parental ditype, 27 nonparental ditype, and 18 tetratype asci. The two genes, a and b, are Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination a. linked, 36 map units apart (the RF being 36%). b. linked, 45 map units apart (the RF being 45%). c. unlinked, 90 map units apart (the RF being 45%). d. unlinked, 96 map units apart (the RF being 48%). e. unlinked, over 100 map units apart (the RF being 48%).

ANSWER: e 46. The molecular process that results in the formation of recombinant chromosomes is initiated by a. a single-stranded break in one of the chromatids. b. a double-stranded break in one of the chromatids. c. the formation of heteroduplex DNA. d. the invasion of single-stranded DNA. e. a Holliday junction.

ANSWER: b 47. Which of the following is one of the steps that occurs during the process of crossing over? a. mitosis b. meiosis c. translation d. transcription e. replication ANSWER: e 48. Holliday junctions are a. formed during the process of crossing over. b. the end products of the process of crossing over. c. a type of mismatched DNA. d. a type of mutation. e. a type of heteroduplex DNA. ANSWER: a 49. DNA in which there is a mismatched nucleotide pair in a gene under study is called a. a chiasma. b. segregated DNA. c. heteroduplex DNA. d. unlinked DNA. e. DNA breakage.

ANSWER: c 50. When two dominant, or wild-type, alleles are present on the same homolog, the arrangement is called a a. chiasma.

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination b. trans conformation. c. cis conformation. d. linked conformation. e. heteroduplex.

ANSWER: c 51. One genetic map unit (m.u.) is the distance between genes for which 1 product of meiosis in _____ is recombinant. a. 100 b. 75 c. 50 d. 25 e. 10 ANSWER: a Essay 52. Why is a genetic map an important tool for geneticists?

ANSWER: Gene maps are crucial for strain building in experimental and commercial applications. Also, knowing the position occupied by a gene provides a way of interpreting evolutionary mechanisms and discovering a gene's unknown function.

53. In the following cross, the genes are inherited independently except for C and D, which are tightly linked and show zero recombination: A/A ; b/b ; c/c ; D/D ; e/e ; F/F·a/a ; B/B ; C/C ; d/d ; E/E ; f/f. In the F2, what proportion of individuals will be a) pure breeding? b) heterozygous for all loci? c) homozygous recessive for all loci? d) A/a ; B/B ; c/c ; D/D ; E/e ; f/f? e) phenotypically like either parent? f) phenotypically unlike either parent? g) genotypically like either parent? h) genotypically unlike either parent?

ANSWER: The F1 is of the following constitution: A/a ; b/B ; c D / C d ; e/E ; F/f. The C and D loci behave like one locus. a) 1/2 × 1/2 × 1/2 × 1/2 × 1/2 = 1/32 b) 1/2 × 1/2 × 1/2 × 1/2 × 1/2 = 1/32 c) zero because of the trans arrangement of c and d d) 1/2 × 1/4 × 1/4 × 1/2 × 1/4 = 1/256 e) (3/4 × 1/4 × 1/4 × 1/4 × 3/4) + (1/4 × 3/4 × 1/4 × 3/4 × 1/4) = 18/1024 f) subtract answer for part (e) from 1; answer = 1006/1024 g) (1/4 × 1/4 × 1/4 × 1/4 × 1/4) + (1/4 × 1/4 × 1/4 × 1/4 × 1/4) = 2/1024 = 1/512 h) subtract answer for part (g) from 1; answer = 511/512 Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination 54. A fruit fly of genotype a+/a·b+/b (parent 1) is crossed to another fruit fly of genotype a/a·b/b (parent 2).

The progeny of this cross were: Genotype a+/a·b+/b a/a·b/b a+/a·b/b a/a·b+/b

Number of individuals 38 37 12 13

a) What gametes were produced by parent 1, and in what proportions? b) Do these proportions demonstrate independent assortment of the two genes? c) What can be deduced from these proportions? d) Demonstrate the arrangement of the genes on the chromosomes in parents 1 and 2 using the appropriate symbols. e) Draw one or more diagrams to explain the origin of the two rarest progeny genotypes.

ANSWER: a) The proportions were 38% a+; b+, 37% a ; b, 12% a+ ; b, and 13% a ; b+. b) No, because we expect 25% of each under independent assortment c) The two gene pairs must be linked; the %R (RF) = 12 + 13 = 25%, and the two loci are 25 map units apart on the same chromosome. d) a+ b+ / a b and a b / a b e) crossover at meiosis in parent 1 to give rise to a+ b and a b+ gametes 55. What is the specific convention used to symbolize linkage?

ANSWER: (1) Alleles on the same homolog have no punctuation between them. (2) A slash symbolically separates the two homologs. (3) Alleles are always written in the same order on each homolog. (4) Genes known to be on different chromosomes (unlinked genes) are shown separated by a semicolon.

(5) Genes of unknown linkage are shown separated by a dot.

56. In humans, the allele N causes an abnormal shape of the patella in the knee (n is the normal allele). A separate gene is concerned with finger length, and the allele B causes abnormally short fingers, whereas b gives normal length. A study focused on people who have both abnormal patellae and short fingers (they were most likely N/n·B/b in genotype), having inherited the N allele from one parent and the B allele from the other parent.

These N/n·B/b people mated with normal spouses, producing 40 progeny classified as follows: Normal Abnormal knees and fingers Abnormal knees only Abnormal fingers only

3 2 17 18

a) Draw the chromosomes of the N/n·B/b individuals, their parents, and their four types of children, showing the positions of the alleles. b) Explain why the four types of children were in the proportions shown.

ANSWER: a) The results clearly depart from independent assortment, so the loci must be linked. The cross must have been N b / n B x n b / n b and the gametes n b (3), N B (2), N b (17), and n B (18). The parents of the N b / n B individual were most likely: N b / n b and n B / n b.

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination b) Total % R (RF) = (3 + 2)/40 = 5/40 = 12.5%, or 12.5 map units.

57. Four possibly different albino strains of Neurospora were each crossed to wild type. All crosses resulted in 1/2 wild-type and 1/2 albino progeny. Crosses were made between the first strain and the other three with the following results: 1 × 2: 975 albino : 25 wild type 1 × 3: 1000 albino 1 × 4: 750 albino : 250 wild type Which mutations represent different genes and which genes are linked? How did you arrive at your conclusions?

ANSWER: Genes 1 and 2 are linked, since they yield less than 25% wild type. Twenty-five percent wild type is expected for independent assortment, the other recombinant (the double mutant) being among the albino progeny. The 25 wild types in the 1 × 2 cross are presumably accompanied by the albino, reciprocal class (double mutant) in approximately equal frequency. The true RF is thus about 50/1000, or 5%. The expectation of independent assortment is realized in the 1 × 4 cross. The 1 × 3 cross indicates, as far as one can tell, that the two mutations are allelic since they fail to yield any recombinants among a sizable progeny.

58. Draw a linkage map or maps consistent with the following recombination frequencies: a) A-B 30%, A-C 50%, B-C 50%, B-D 50%, A-D 50%, C-D 15% b) A-B 30%, A-C 50%, B-C 20%, B-D 35%, A-D 50%, C-D 15% ANSWER: a) It is most likely that A and B are linked and C and D are linked. Therefore, the map would be:

b) In this case, all four genes must be linked, and a consistent map is:

59. An experiment was done to determine the linkage relationship of three genes (a, b, and c) in Drosophila melanogaster. Homozygous females phenotypically a, c were crossed with homozygous males phenotypically b. The F1 females were all wild type in appearance, and the F1 males were all a, c. The F1 females and males were crossed to give the following F2 phenotypes and numbers.

a+c +b+ ab+ ++c abc +++ a++ +bc

Number of Males 242 238 157 163 59 61 38 42

Females 310 0 0 190 0 295 205 0

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination Total

1000

1000

a) Give the genotypes of the parents. b) Give the genotypes of the F1 flies, and show the genotypic arrangement in the female F1. c) Draw a genetic map showing the order of the genes and the distances between them. d) Calculate the interference (only if necessary; if not, say "I = 0").

ANSWER: The data clearly show sex linkage due to disproportionate representation of phenotypic classes between the sexes. a) P: ac+ × ++ b ac+ b)

F1

ac+ ×

Y a

c +

++b Y Loci order must be a c b because the most rare F2 classes of 38 and 42 flies each (due to D-CO) are a++ and +bc. c) a-b: Look for recombinants a b (157 + 59) and a+ b+ (163 + 61) Total = 440 RF = 440/1000 = 44% a-c: Look for recombinants a c+ (157 + 38) and a+ c (163 + 42) Total = 400 RF = 400/1000 = 0.40 or 40% b-c: Look for recombinants b c (59 + 42) and b+ c+ (61 + 38) Total = 200 RF = 200/1000 = 0.20 or 20% d) Expected D-CO = 0.4 × 0.2 × 1000 = 80 Observed D-CO = 80; therefore, Interference (I) = 0. The correct answer is "I = 0."

60. Three crosses involving three linked genes were performed, with the resulting "shorthand" designation of phenotypes indicated: R T / r tr t / r t:

43% R T,

43% r t,

7% R t,

7% r T

R S / r sr s / r s:

45% R S,

45% r s,

4.5% R s,

4.5% r S

T S / t st s / t s:

47.5% T S, 47.5% t s,

2.5% t S,

2.5% T s

What is the best map of these three genes?

ANSWER: 61. Three genes of corn, R, D, and Y, lie on chromosome 9. The map of the three genes, using map units is as follows:

In a testcross of a trihybrid plant (R D Y / r d y) arising from true-breeding R D Y / R D Y and r d y / r d y parents, how many R d Y / – d – plants would be expected in a progeny of 1000 if there were no interference? If Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination the coefficient of coincidence were 0.3, how would this change your answer?

ANSWER: First, if there is no interference, the number of double-crossover progeny can be predicted simply from the map distances given (10% and 20%). When converted to frequencies and multiplied, we have the expected number of double crossovers, 20 of 1000 progeny. Of these 20 double crossovers, 10 are R d Y (answer to the first question) and 10 are r D y. The 1000 progeny should have 10% progeny (or 100 in all) that have crossovers between R and D. Twenty of these are accounted for among the double crossovers, leaving 80 single crossovers, 40 R d y and 40 r D Y. By similar means, we find there are 180 single crossovers in the second region, 90 r d Y and 90 R D y. The remainder of the 1000 progeny will be parentals, R D Y and r d y.

62. Two albino strains of Neurospora are crossed. Of 200 progeny, 195 are albino and 5 are wild type (orange).

Each parent, when mated with wild type, gives 1/2 wild-type and 1/2 albino progeny. a) How can you tell that the two mutations involved here represent different genes? b) What is your best estimate of map distance between the two mutations, and how did you calculate it?

ANSWER: a) Wild-type progeny indicates recombination between two different genes. b) 5 map units: 5/200 wild type + 5/200 hidden double mutant = 10/200, or 5% recombination.

63. A cross was made between two Neurospora strains with genotypes + b and a +. The genes are linked. The genotypes of the spores in one ascus from this cross are listed in order, from left to right: + b ; + b ;+ + ; + + ; a b ; a b ; a + ; a + a) Is the ascus a parental ditype (PD), a tetratype (T), or a nonparental ditype (NPD)? b) Do the alleles of the a gene show first- or second-division segregation? c) Do the alleles of the b gene show first- or second-division segregation? d) Show how the above tetrad of spore pairs was derived by diagramming the chromosomes at pachytene of meiosis I. Indicate with a line between appropriate chromatids, the simplest crossover that can yield this tetrad. e) If recombination between a and b is 10% and 2/100 asci are NPD, how many tetratype asci would you expect to find? How many PD? f) Diagram how a tetrad with the same spore pairs, as shown above, would arise if the genes showed independent assortment.

ANSWER: a) T b) first-division segregation c) second-division segregation d)

e) 16T, 82PD [% recombination = (NPD + (1/2)T)(100)/Total] f)

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination

64. In a cross between an arginine-requiring, acetate-requiring strain of Neurospora and a wild-type strain (arg ac × + +) the following ordered tetrads of asci were found, with the numbers of each type indicated below the particular tetrad: I arg ac arg ac + + + + 69

II arg + arg + + ac + ac 0

III arg ac arg + + + + ac 10

IV arg ac + ac + + arg + 18

V arg ac + + + + arg ac 0

VI arg + + ac + ac arg + 1

VII arg + + ac + + arg ac 2

Total 100

a) Calculate the map distance between arg and its centromere. b) Calculate the map distance between ac and its centromere. c) What is the most likely map of the two genes in relation to each other and to the position of their respective centromere(s)?

ANSWER: a) 10.5% by 1/2 (MII for arg) b) 6.5% by 1/2 (MII for ac) c) Using the formula: RF = (1/2)T + NPD, one can see the genes are on opposite sides of the centromere, as the RF value would be much lower if they were on the same chromosomal arm.

65. In Neurospora, an Irish strain makes big as well as green-colored colonies (genotype b g). It is crossed to a wild-type strain, which makes small and orange-colored colonies (genotype: + +). The following ordered tetrads of asci were found, with the numbers of each type indicated below each tetrad.

I bg ++ +g b+ 4

II b+ +g b+ +g 1

III bg ++ bg ++ 24

IV bg +g b+ ++ 1

V bg b+ +g ++ 91

VI b+ b+ +g +g 2

VII bg bg ++ ++ 177

a) Classify the asci above (I–VII) according to type. (i) those that are PD (ii) those that are NPD Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination (iii) those that are T b) Calculate the percentage recombination between b and its centromere. c) Calculate the percentage recombination between g and its centromere. d) What is the most likely map of the two genes in relation to each other and to the position of their respective centromere(s)?

ANSWER: a) PD: III, VII NPD: II, VI T:I, IV, V b) 5% c) 20% d) The map distance between the two genes is 17% (using the formula (1/2)T + NPD). Therefore, the map is as follows:

66. In a fungal cross a b × + +, gene a is on chromosome 1 and gene b is on chromosome 6. It is known that the two loci are so closely linked to their centromeres that there is never any crossing over in either centromere to locus region. What proportion of linear tetrads will be a) MI MI PD? b) MI MI NPD? c) MI MII T? d) MII MI T? e) MII MII PD? f) MII MII NPD? g) MII MII T? ANSWER: Because there is no crossing over in either gene-to-centromere region, there can be no MII asci. a) 0.5 b) 0.5 c) through g) 0 67. In the fungus Neurospora, a strain that was auxotrophic for thiamine (mutant allele t) was crossed to a strain that was auxotrophic for methionine (mutant allele m). Linear asci were isolated, and they were classified in the following groups: Spore Pair 1 and 2 3 and 4 5 and 6 7 and 8

Ascus types t. + t + t. + t m + m + + + m + m 260 76

t + + m t + + m 4

t + + + t m + m 54

t m t m + + + + 1

t m + + t + + m 5

a) Determine the linkage relationships of these two genes to their centromere(s) and to each other. Specify distances in map units.

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination b) If 1000 randomly selected ascospores are plated on a plate of minimal medium, how many are expected to grow into colonies?

ANSWER: The distance between the genes is 17.1 m.u. (using the formula 1/2 T + NPD/total).

b) RF = 17.1%, and only 1/2 of these will be + + prototrophs; 8.6% or 86 out of 1000 will be prototrophic.

68. Dominant alleles at two different loci affect the tail of mice. These genes are linked and both are lethal in the embryo when homozygous. One of the tail genes is called brachyury (T), and these mice have short tails. A third gene, histocompatibility-2 (H-2), is linked to the tail genes and is concerned with tissue transplantation.

Mice that are H-2/+ will accept tissue grafts, unlike +/+. A cross was carried out between male T H-2 / + + and female + + / + + mice, and produced the following baby mice.

Brachy tail, accepts graft Normal tail, rejects graft Brachy tail, rejects graft Normal tail, accepts graft Total

105 115 14 16 250

Determine if the linkage between H-2 and T is significant using the chi-square test.

ANSWER: O E (O – E) (O – E)2 (O – E)2/E T H-2 / + + 105 62.5 42.5 1806.25 28.9 ++/++ 115 62.5 52.5 2756.25 44.1 T+ / + + 14 62.5 –48.5 2352.25 37.6 +H-2 / + + 16 62.5 –46.5 2162.25 34.6 2 Total 250 χ = 145.2 d.f. = number of categories –1 = 4 – 1 = 3 p < 0.0001, therefore the null hypothesis of absence of linkage is not supported.

69. Consider the following experimental results.

(M = mapping function and MU = M × 50) OBSERVED RF 18.5%

M 0.46

MU 23.1

6.4%

0.137

6.8

Based on these results, would you advocate the use of mapping functions? In which situations would you recommend using them?

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chapter 04: Mapping Eukaryote Chromosomes by Recombination ANSWER: Yes, you should recommend using mapping functions. The numbers in columns 1 and 3 in row 2 indicate that the calculated map unit distance is very similar to the observed RF, and so for short distances, use of RF is relatively accurate. For larger distances (as seen in row one), there is a much greater discrepancy between the distance calculated using the mapping function and the observed RF. Use the mapping function to calculate the corrected map distance between loci having a recombination frequency of 20% or greater.

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chapter 05: Gene Interaction Multiple Choice 1. The gene C codes for the membrane-associated protein C, responsible for signal transduction. A mutant allele of this gene, Cm, encodes a malfunctioning C protein, which has the same size as the wild type and associates with the plasma membrane like the wild type, but it has no function in signal. With respect to the expression of the C proteins, the wild-type allele is a. codominant to the Cm allele. b. dominant to the Cm allele. c. haploinsufficient. d. incompletely dominant to the Cm allele. e. recessive to the Cm allele.

ANSWER: a 2. In a flowering plant, gene A encodes an enzyme responsible for the presence of dots on the flowers' petals.

A1, A2, and A3are the three known alleles of this gene; A1 is the wild type, A2 is a null allele, while A3 has a mutation in the promoter region of the gene, which results in the synthesis of very little gene product. If the A gene is haploinsufficient, what is the predicted phenotypic ratio in the F1 of a cross between a wild type and a A2/A3 heterozygous? a. 100% of the F1 plants will have flowers with dots. b. 25% of the F1 plants will have flowers with dots, and 75% will have flowers with no dots. c. 50% of the F1 plants will have flowers with dots, and 50% will have flowers with no dots. d. 50% of the F1 plants will have flowers with a few dots, 25% will have flowers with normal amounts of dots, and 25% will have flowers with no dots. e. 100% of the F1 plants will have flowers with no dots.

ANSWER: e 3. In a wild-type fungus, protein E (encoded by the haplosufficient gene E) normally homodimerizes, and the EE dimer catalyzes a biochemical reaction necessary for the production of a dark pigment. ED represents a mutant, dominant, negative allele of gene E. What is the predicted phenotype of a fungus cell of genotype E+/ED, and why? a. mutant (no pigment production), as E is haplosufficient b. mutant (no pigment production), as no E-E dimers will form in the heterozygous c. mutant (no pigment production), as the mutant allele ED is dominant d. wild type (normal production of the dark pigment), as ED is a negative allele e. wild type (normal production of the dark pigment), as E is haplosufficient ANSWER: c 4. Crosses between two ducks with spread-out tail phenotype always produce ducklings with spread-out, normal, and pointy tails in a 2:1:1 ratio. In addition, normal-tailed ducks crossed to short-tailed ducks always Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chapter 05: Gene Interaction produce only normal-tailed individuals, while crosses between pointy-tailed and short-tailed individuals always produce only pointy-tailed ducklings. If the tail shape phenotype is controlled by a single locus, what is the expected phenotypic ratio in the progeny of the cross spread-out tail × short tail? a. 1:0 of spread-out : short b. 1:1 of normal : pointy c. 1:1 of spread-out : short d. 1:1:1:1 of spread-out : normal : pointy : short e. 3:1 of spread-out : short ANSWER: b 5. In the purple penguin, an allelic series occurs at the p locus on an autosome. All alleles affect the color of feathers: pd = dark-purple, pm = medium-purple, pl = light-purple, and pvl = very pale purple. The order of dominance is pd> pm> pl> pvl. If a pl/pvl female is crossed to a pd/pm male, the ratio of phenotypes expected among the baby penguins would be a. 2 dark : 1 light : 1 very pale. b. 2 dark : 1 medium : 1 light. c. 1 dark : 1 medium. d. 1 dark : 1 medium : 1 light : 1 very pale. e. 1 medium : 1 light.

ANSWER: c 6. In chickens, the dominant allele Cr produces the creeper phenotype (having extremely short legs). However, the creeper allele is lethal in the homozygous condition. If two creepers are mated, what proportion of the living progeny will be creepers? a. 1/4 b. 1/3 c. 1/2 d. 2/3 e. 3/4 ANSWER: d 7. When selfed, a particular plant always produces an F1 with a phenotypic ratio of 2:1. If this plant is testcrossed, what is the expected phenotypic ratio in the progeny? a. 1:0 b. 1:1 c. 1:1:1:1 d. 1:2:1 e. 3:1 ANSWER: b 8. In laboratory mice, the "short tail" phenotype is dominant to the wild-type ("long tail") phenotype. However, Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chapter 05: Gene Interaction crosses between any two short-tailed mice always produce mixtures of short- and long-tailed offspring.

Assuming that tail length is controlled by a single locus, a likely explanation for these results is that the "short tail" allele is a. a codominant allele. b. a dominant lethal allele. c. haploinsufficient. d. an incompletely dominant allele. e. a recessive lethal allele.

ANSWER: e 9. A 15:1 phenotypic ratio in a typical F2 data set would be consistent with _____________________ affecting a particular phenotype. a. incomplete dominance b. codominance c. alleles of two genes d. epistatic alleles e. the environment ANSWER: c 10. In the multiple-allele series that determines coat color in rabbits, c+ encodes agouti, cch encodes chinchilla, and ch encodes Himalayan. Dominance within this allelic series is c+ > cch > ch. In a cross of c+/ch × cch/ch, what proportion of progeny will be chinchilla? a. 0% b. 10% c. 25% d. 50% e. 100% ANSWER: c 11. For a (fictitious) haploid fungus, the starting point in the biosynthesis of the amino acid arginine is Compound X, which is always present in and absorbed from the environment. The arginine biosynthetic pathway is:

A mutant strain of genotype a– (lacking enzyme A) is crossed to a mutant strain of genotype c– (lacking enzyme C). What proportion of the progeny is expected to grow on medium supplemented with Compound Y? a. 0% b. 25% c. 50% d. 75% Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chapter 05: Gene Interaction e. 100% ANSWER: b 12. Two new mutant lines of a flowering plant have been obtained; one breeds true for blue flower color and the other breeds true for red flower color (wild-type flower color is purple). Consider the following crosses.

1. blue × purple → F1 all purple → F2 75% purple; 25% blue 2. red × purple → F1 all purple → F2 75% purple; 25% red 3. red × blue → F1 all purple If the purple flower color results from the mixture of red and blue pigments, and all the intermediate products in the pigment pathways are colorless, what phenotypic ratios do you expect in the F2 of cross 3? a. 3 purple : 1 white (colorless) b. 9 purple : 3 blue : 3 red : 1 white (colorless) c. 9 purple : 4 blue : 3 red d. 9 purple : 7 white (colorless) e. 12 purple : 3 blue : 1 white (colorless) ANSWER: b 13. In the fruit fly Drosophila melanogaster, red and yellow eye pigments are synthesized through the pterin pathway. These, plus the brown pigment synthesized through the ommochrome pathway, produce the dark red Drosophila eye color. A part of the pterin pathway (simplified) is shown below.

Consider the genes encoding Enzyme 1 (e1), Enzyme 2 (e2), and Enzyme 3 (e3), respectively. What are their predicted genetic interactions? a. e1 is recessively epistatic to e2, and e2 is recessively epistatic to e3. b. e1 is recessively epistatic to e3, and e3 is recessively epistatic to e2. c. e3 is recessively epistatic to e1, and e2 is recessively epistatic to e1. d. e1 is recessively epistatic to e2 and to e3. e. e2 is recessively epistatic to e1 and to e3.

ANSWER: d 14. Two new mutant lines of a flowering plant have been obtained; one breeds true for blue flower color and the Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chapter 05: Gene Interaction other breeds true for red flower color (wild-type flower color is purple). Consider the following crosses.

1. blue × purple → F1 all purple → F2 75% purple; 25% blue 2. red × purple → F1 all purple → F2 75% purple; 25% red 3. red × blue → F1 all purple If the red pigment is a precursor of the purple pigment, the blue pigment is a precursor of the red pigment, and a colorless compound is the precursor of the blue pigment in a linear biosynthetic pathway controlling flower color, what phenotypic ratios do you expect in the F2 of cross 3? a. 3 purple : 1 white (colorless) b. 9 purple : 3 blue : 3 red : 1 white (colorless) c. 9 purple : 4 blue : 3 red d. 9 purple : 7 white (colorless) e. 12 purple : 3 blue : 1 white (colorless) ANSWER: c 15. The diagram below shows a simplified version of the biochemical pathway responsible for fruit color in peppers. Assume that Enzyme 1 is encoded by gene A (a is a null allele), Enzyme 2 is encoded by gene B (b is a null allele), and Enzyme 3, which breaks down the chlorophyll present in the fruit, is encoded by gene C (c is a null allele). In the absence of Enzyme 3, the fruit takes a brown color in the presence of red pigment, but it remains green in the absence of red pigment.

Consider two genotypically different pure lines that make colorless peppers. If these two plants are crossed, what phenotypes should be observed in the F1? a. colorless only b. colorless and yellow only c. red and yellow only d. red only e. yellow only ANSWER: a 16. The diagram below shows a simplified version of the biochemical pathway responsible for fruit color in peppers. Assume that Enzyme 1 is encoded by gene A (a is a null allele), Enzyme 2 is encoded by gene B (b is a null allele), and Enzyme 3, which breaks down the chlorophyll present in the fruit, is encoded by gene C (c is a null allele). In the absence of Enzyme 3, the fruit takes a brown color in the presence of red pigment, but it Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chapter 05: Gene Interaction remains green in the absence of red pigment.

A pepper plant of genotype c/c is crossed to a plant of genotype b/b. The F1 is selfed to obtain an F2. What are the expected phenotypic ratios in this F2? a. 9 brown : 4 green : 1 yellow b. 9 brown : 3 green : 3 red : 1 yellow c. 3 brown : 1 green : 9 red : 3 yellow d. 1 brown : 9 green : 3 red : 3 yellow e. 1 green : 9 red : 4 yellow ANSWER: c 17. The diagram below shows a simplified version of the biochemical pathway responsible for fruit color in peppers. Assume that Enzyme 1 is encoded by gene A (a is a null allele), Enzyme 2 is encoded by gene B (b is a null allele), and Enzyme 3, which breaks down the chlorophyll present in the fruit, is encoded by gene C (c is a null allele). In the absence of Enzyme 3, the fruit takes a brown color in the presence of red pigment, but it remains green in the absence of red pigment.

A pepper plant that produces red fruits was selfed, and the progeny obtained consisted of 176 plants that made red fruits, 78 that made colorless fruits, and 62 that made yellow fruits. If the original red fruit-producing plant was crossed to a tester of genotype a/a ; b/b ; c/c, what would be the expected phenotypic ratio in the offspring? a. 1 brown : 1 colorless : 1 red : 1 yellow b. 1 colorless : 1 green : 1 red : 1 yellow c. 1 colorless : 2 red : 1 yellow d. 1 colorless : 1 red : 2 yellow e. 2 colorless : 1 red : 1 yellow Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chapter 05: Gene Interaction ANSWER: e 18. The following noncomplementing E. coli mutants were tested for growth on four known precursors of thymine, compounds A–D. The data provided in the table reveal a simple linear biosynthetic pathway of the four precursors and the end product, thymine. In what order do the enzymes appear in this pathway (designate by their mutant number)?

Mutant 9 10 14 21

A + – + –

B – – + –

C + + + –

D – – – –

Thymine + + + +

a. 14, 9, 10, 21 b. 9, 10, 14, 21 c. 10, 9, 14, 21 d. 21, 14, 9, 10 e. 10, 21, 14, 9 ANSWER: a 19. Two new pure-breeding strains of mouse (strain 1 and strain 2) have been obtained. Crosses between strain 1 and wild type, as well as crosses between strain 2 and wild type and between strain 1 and strain 2, always produce 100% wild-type mice. What kind(s) of interactions can be deduced from these results? a. complementation only b. dominance only c. epistasis only d. complementation and dominance e. complementation, epistasis, and dominance ANSWER: d 20. Pet rabbits can have a variety of coat colors. Below are the results of a series of crosses performed with black-, blue-, and brown-coated pure lines.

Cross 1.

2.

3.

Phenotypes of parents Black × blue Black × brown Blue × brown

F1 phenotypes All black All black All black

If the F1 of cross number 3 is selfed, what is the expected proportion of black kits (baby rabbits) in the offspring? a. 25% b. more than 25%, but less than 50% c. 50% Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chapter 05: Gene Interaction d. more than 50%, but less than 75% e. 75% ANSWER: d 21. In swine, when a pure-breeding red is crossed to a pure-breeding white, the F1 are all red. However, the F2 shows 9/16 red, 1/16 white, and 6/16 as a new color—sandy. A particular red individual (Parent 1) is crossed to a particular sandy individual (Parent 2), and the progeny is comprised of four sandy, three red, and one white piglets. What are the MOST LIKELY genotypes of the two parents? a. A/A ; B/B and A/a ; B/b b. A/A ; B/b and A/a ; B/b c. A/A ; b/b and A/a ; B/b d. A/a ; B/b and a/a ; B/B e. A/a ; b/b and A/a ; B/b ANSWER: e 22. Remember that, in Labrador retrievers, B (black coat) is dominant to b (brown coat) and homozygosity for the recessive epistatic allele e always results in a yellow coat. A brown female is crossed to a yellow male, and their progeny consists of six black puppies. Interestingly, when one of the F1 females is crossed to her yellow father, she gives birth to seven puppies: three yellow, three black, and one brown. What are the MOST LIKELY genotypes of the original brown female and yellow male? a. b/b ; E/E (female) and B/B ; e/e (male) b. b/b ; E/E (female) and B/b ; e/e (male) c. b/b ; E/e(female) and B/B ; e/e (male) d. b/b ; E/e(female) and B/b ; e/e (male) e. b/b ; E/E (female) and b/b ; e/e (male) ANSWER: b 23. Drosophila eyes are normally red. Several purple-eyed strains have been isolated as spontaneous mutants, and the purple phenotype has been shown to be inherited as a Mendelian autosomal recessive in each case. To investigate allelism between these different purple mutations, two purple-eyed pure strains were crossed.

If the purple mutations are in different genes (i.e., they are not allelic), the phenotypic ratios in the F1 are expected to be a. 100% red. b. 75% red : 25% purple. c. 50% red : 50% purple. d. 25% red : 75% purple. e. 100% purple.

ANSWER: a 24. Drosophila eyes are normally red. Several purple-eyed strains have been isolated as spontaneous mutants, and the purple phenotype has been shown to be inherited as a Mendelian autosomal recessive in each case. To Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chapter 05: Gene Interaction investigate allelism between these different purple mutations, two purple-eyed pure strains were crossed.

If the purple mutations are in the same gene (i.e., they are allelic), the phenotypic ratios in the F1 are expected to be a. 100% red. b. 75% red : 25% purple. c. 50% red : 50% purple. d. 25% red : 75% purple. e. 100% purple.

ANSWER: e 25. In loppins (fictitious diploid invertebrates), ear shape is controlled by two genes with two alleles each; upright ears (F) are dominant to floppy ears (f), and double-pointed ears (P) are dominant to single-pointed ears (p). Moreover, due to genetic interactions, floppy ears are always single-pointed. A loppin with floppy ears is crossed with a loppin with upright single-pointed ears, and the phenotypic ratios in the offspring are 2 floppy : 1 double-pointed upright : 1 single-pointed upright. What are the genotypes of the parents? a. p/p ; f/f and p/p ; F/f b. P/p ; f/f and p/p ; F/F c. P/p ; f/f and p/p ; F/f d. P/P ; f/f and p/p ; F/F e. P/p ; F/f and p/p ; F/f ANSWER: c 26. A dihybrid plant is selfed, and the offspring shows a 13:3 phenotypic ratio for flower color. If the same dihybrid plant was testcrossed instead of being selfed, what would be the expected phenotypic ratio in the offspring? a. 1:0 b. 1:1 c. 1:1:1:1 d. 2:1:1 e. 3:1 ANSWER: e 27. In a certain breed of dog, the alleles B and b determine black and brown coats, respectively. However, the allele Q of an unlinked gene is epistatic to the B and b color alleles, resulting in a gray coat (q has no effect on color). If animals of genotype B/b ; Q/q are intercrossed, what phenotypic ratio is expected in the progeny? a. 1 black, 2 gray, 1 brown b. 3 black, 12, gray, 1 brown c. 4 black, 9 gray, 3 brown d. 9 black, 1 gray, 6 brown e. 9 black, 4 gray, 3 brown ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chapter 05: Gene Interaction 28. In Drosophila, the recessive alleles for brown and scarlet eyes (of two independent genes) interact so that b w / b w ; s t / s t is white. If a pure-breeding brown is crossed to a pure-breeding scarlet, what proportion of the F2 will be white? a. 1/16 b. 1/4 c. 7/16 d. 3/4 e. 13/16 ANSWER: a 29. In sweet peas, the allele C is needed for color expression (c results in white). The precise color expressed is determined by the alleles R (red) and r (blue). A cross between certain red and blue plants resulted in progeny as follows: 3/8 red, 3/8 blue, 1/4 white. What were the genotypes of the plants crossed? a. C/c ; R/r × C/C ; R/R b. C/c ; R/r × C/c ; R/r c. C/c ; R/r × C/c ; r/r d. C/c ; R/r × c/c ; R/r e. C/c ; r/r × C/C ; R/r ANSWER: c 30. Four pure lines of Guinea pigs (black, brown, gray, and tan) are used in a series of crosses. The results are as follows: Cross Mother Father F1 1.

Black Brown All black 2.

Black

Gray

All black

3.

Black

Tan

All black

4.

Brown

Tan

All brown

5.

Gray

Tan

All gray

6.

Tan

Black

??

F2 55 black females, 57 black males, 19 black females, 18 black males 75 black females, 38 black males, 36 gray males 53 black females, 26 black males, 25 gray males, 18 brown females, 9 brown males, 8 tan males 72 brown females, 37 brown males, 34 tan males 56 gray females, 57 gray males, 19 tan females, 18 tan males

What is the expected phenotypic ratio in the F1 of cross 6? a. 50% black females, 50% black males b. 50% black females, 50% gray males c. 50% black females, 50% tan males d. 50% brown females, 50% black males e. 50% tan females, 50% tan males Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chapter 05: Gene Interaction ANSWER: b 31. An A/a ; B/b dihybrid is testcrossed, and about 3/4 of the progeny phenotypically resembles the dihybrid parent, while 1/4 resembles the tester parent. If the dihybrid parent was selfed, what would be the expected phenotypic ratio in the progeny? a. 9:3:4 b. 9:7 c. 12:3:1 d. 13:3 e. 15:1 ANSWER: e 32. In Drosophila melanogaster, the dominant allele Cy results in curly wings. Interestingly, however, temperature also plays a significant role on the wing phenotype.

Genotype Cy+/Cy+(wild type) Cy+/Cy+(wild type) Cy/Cy+(mutant) Cy/Cy+(mutant)

Temperature at which raised 17°C 26°C 17°C 26°C

Proportion of individual with curly wings 0% 0% 15% 100%

Based on this information we can say that the a. expressivity and penetrance of the Cy allele are temperature-dependent. b. expressivity of the Cy allele is incomplete. c. expressivity of the Cy allele is temperature-dependent. d. penetrance of the Cy allele is incomplete. e. penetrance of the Cy allele is temperature-dependent.

ANSWER: e 33. In the nematode C. elegans, homozygosity for the e mutant allele causes an extreme "uncoordinated" phenotype, where the worm completely loses its ability to move. Examination of 100 individuals with genotype e/e reveals that 58 of them can't move at all, 45 show a very reduced ability to move, and the remaining 17 seem to have a completely wild-type phenotype with respect to movement ability. These observations suggest that e has a. high environmental components. b. incomplete expressivity. c. incomplete penetrance and variable expressivity. d. low expressivity and variable penetrance. e. low penetrance.

ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chapter 05: Gene Interaction 34. An organism that requires a particular nutrient for growth which the wild-type strain does not require is called a. auxotrophic mutant. b. prototrophic mutant. c. mutagenic. d. bradytroph. e. autotropic mutant.

ANSWER: a 35. The one-gene–one-polypeptide hypothesis is a. The theory that one enzyme is responsible for the cleavage of a single DNA strand. b. The theory that one enzyme is responsible for the cleavage of a single polypeptide. c. The theory that single-stranded DNA is responsible for the synthesis of a single polypeptide. d. The theory that a single gene is responsible for the synthesis of all polypeptides in Neurospora. e. The theory that each gene is responsible for the synthesis of a single polypeptide.

ANSWER: e 36. Two gametes, each carrying a mutant recessive allele for a different gene/enzyme in the adenine biosynthetic pathway, come together to form a diploid embryo. The individual derived from this embryo will display a a. lethal phenotype in media lacking adenine supplementation. b. wild-type phenotype, capable of synthesizing adenine. c. weak adenine requirement for survival (intermediate phenotype). d. profound defect in adenine synthesis as two enzymes are missing. e. embryonic lethality.

ANSWER: b 37. Which of the following is/are TRUE about functional RNA? a. Functional RNAs do not encode proteins and are active as RNA. b. Some functional RNAs encode proteins. c. Examples of functional RNA include mRNA, tRNA, and rRNA. d. Examples of functional RNA include tRNA, rRNA, snRNA, and mRNA. e. Functional RNAs do not encode proteins, and examples include tRNA, rRNA, and scRNA. ANSWER: e Essay 38. You are studying loppins (fictitious but useful invertebrates) and are interested in the inheritance of the colorful dots on their abdomen. You notice that some loppins have no dots, some have green dots, some have orange dots, and some have a mixture of orange and green dots. a) The first thing you notice is that it's relatively easy to obtain lines that breed true for the absence of dots, for Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chapter 05: Gene Interaction the presence of orange dots only, or for the presence of green dots only, but it seems impossible to get purebreeding loppins that have both green and orange dots. Every time that you cross two green-and-orange-dotted individuals, you obtain an F1 comprised of about 50% green-and-orange-dotted, 25% orange-dotted, and 25% green-dotted individuals.

What do you conclude? Explain your reasoning and use appropriate symbols to indicate the genotypes of the various individuals. b) How could you determine genetically whether or not the "no dots" phenotype is allelic to "green dots," "orange dots," and "orange-and-green dots"? Outline the crosses that you would set up as well as the expected results if "no dots" was allelic versus if it was not allelic to the other phenotypes.

ANSWER: a) Most likely, green (cG) and orange (cO) are codominant to each other so that it's possible to obtain homozygous cG/cG or cO/cO lines, but green-and-orange, having the genotype cO/cG, can't ever be homozygous (true breeding). In addition, when crossing two cO/cG individuals, we expect exactly 25% cG/cG, 25% cO/cO, and 50% cO/cG individuals, which is what is observed. b) Cross a pure-breeding "no dots" to a pure-breeding "orange": F1 should all be genotypically identical and either monohybrid or dihybrid. Cross brothers and sisters to obtain an F2: If the two phenotypes are allelic (i.e., we are working with only one gene), then the expected F2 phenotypic ratios are 3:1 or 1:2:1. If we are dealing with two genes, and the two phenotypes aren't allelic, then the F2 phenotypic ratios should be 9:3:3:1, or a variation on that theme.

The same procedure could/should be carried out with a pure-breeding "green" (with same expected results) and with "green-and-orange," where the expected phenotypic ratios would be typical "1 gene" ratios or typical "2 gene" ratios, although not typical "monohybrid" or "dihybrid" ratios.

39. A series of hamster crosses are performed to get an insight into the mode of inheritance of three coat colors: plain brown, plain beige, and patchy brown. Note that not all the parents were pure-breeding.

Cross Parents phenotypes 1.

2.

3.

4.

5.

6.

Plain brown × plain brown Plain beige × plain beige Patchy brown × patchy brown Plain brown × plain beige Patchy brown × plain brown Plain beige × patchy brown

7.

8.

Plain beige × plain beige Plain brown × plain brown

F1 phenotypes 14 plain brown 12 plain beige 6 patchy brown, 3 plain brown 13 plain brown 6 plain brown, 7 patchy brown 7 plain brown, 3 patchy brown, 3 plain beige 9 plain beige 10 plain brown, 3 plain beige

Propose a mode of inheritance for coat color in hamsters, using symbols of your own choosing to identify genes and alleles. Indicate relationships between alleles, and use the data above to support your explanation.

ANSWER: F1 ratios are indicative of one gene with multiple alleles. Plain brown seems to be dominant to plain beige, patchy brown seems dominant to plain brown, but the F1 ratios (e.g., cross 3) are compatible Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chapter 05: Gene Interaction with the idea that homozygosity for patchy brown is lethal. Some plain beige × plain beige crosses result in less numerous F1.

Possible explanation follows: B1 = plain brown B1/B1 = plain brown B1 > B3 B2 = patchy brown B2/B2 = lethal, but B2/B1 = patchy brown B3 = plain beige B3/B3 = plain beige, but B3/B2 = plain beige, too 40. In humans, the ABO blood type is determined mainly by four alleles of one gene, defined by the following allelic series: IA = IB> i. (IA is codominant to IB, and i/i results in blood type O). What matings can potentially produce children of any of the four blood types?

ANSWER: Matings where one parent is IA/i and the other is IB/i 41. Domestic rats can have black, silver, or light-gray patches on their necks. A pure-breeding silver female was crossed to a pure-breeding black male, and the F1 was comprised of black females and light-gray males. When F1 brothers and sisters were mated, the resulting F2 consisted of about 1/4 black females, 1/4 black males, 1/4 silver females, and 1/4 light-gray males. Propose a model for the mode of inheritance of patch color in domestic rats.

ANSWER: The gene controlling patch color is X-linked, and its "silver" allele is dose-sensitive so that B = black, b = silver, B > b, and b/b = silver, but b/Y chromosome = light gray.

42. On an illegal wolf ranch in Wyoming, cowboys breed gray wolves and sell them illegally. A mutation was identified, generating a beautiful gray "speckled" coat color. The speckled wolves were particularly popular with buyers. The breeders struggled to keep up with demand. They noticed that when two speckled wolves were bred, their pups always contained some normal gray and some speckled gray pups. In one typical season of interbreeding the speckled wolves, they cumulatively generated 82 speckled and 38 gray pups. State a clear genetic explanation for the pattern illustrated.

ANSWER: The speckled phenotype appears to be inherited in a dominant manner, with affected progeny being generated from affected parents. However, when two speckled are interbred, the offspring do not display a 3:1 ratio, and seem to follow a 2:1 ratio. The most likely explanation is that the allele is also a recessive lethal and that homozygous animals do not survive gestation.

43. The following example of gene regulation involves genes r+ and a+. Loss of either gene results in a similar phenotype. a) In complementation testing, would these genes appear as one gene or two distinct genes? Explain. b) What ratio might be generated by this example in a typical F2 data set where an F1 organism of phenotype r+/r ; a+/a had been selfed? Explain terms clearly.

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chapter 05: Gene Interaction

ANSWER: a) In complementation testing, an r/r ; a+/a+ would be crossed with an r+/r+; a/a organism. The parental lines would display similar traits and the F1 plant would be phenotypically normal, though heterozygous for the two genes. b) If an F1 that is heterozygous for these two loci were to be selfed, then 9/16 of the progeny would be wild type and 7/16 of the progeny would display the defective trait.

44. In a flowering plant, the a/a ; B/B genotype results in yellow flowers, and the A/A ; b/b genotype gives red flowers. Wild-type flowers are orange and are observed in plants with genotype A/A ; B/B. Two possibilities have been proposed for the biosynthetic pathways of flower color: 1. yellow pigment → red pigment → orange pigment 2. red pigment → yellow pigment → orange pigment Propose a genetic experiment that can determine which of the two possibilities is most likely correct, and explain how you would interpret your results.

ANSWER: Cross a red pure line to a yellow pure line to obtain a dihybrid A/a ; B/b, then self this dihybrid and inspect the F2. If possibility 1 is correct, then we expect a ratio of 9 orange : 4 yellow : 3 red. If possibility 2 is correct, we expect a ratio of 9 orange : 4 red : 3 yellow. Note: Since we are trying to differentiate between 9:3:4 and 9:4:3, it's essential that we have a large enough sample size.

45. Below is a simplified version of the biochemical pathway responsible for fruit color in peppers. Suppose that gene A encodes enzyme 1, gene B encodes enzyme 2, gene C encodes enzyme 3, and that each gene only has two alleles: completely functioning (capital letters) and null (lower case letters). Note that red plus chlorophyll is brown, but yellow plus chlorophyll is green.

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chapter 05: Gene Interaction a) A breeder crossed two pure lines of peppers and then selfed the resulting F1. The F2 was comprised of 351 plants with red peppers, 115 with yellow peppers and 154 with colorless (white) peppers. What must have been the genotypes and phenotypes of the two original pure lines? b) A pure-breeding plant with green fruits is crossed to a pure-breeding plant that makes yellow fruits. All the F1 plants make yellow fruits. The F1 is selfed, and in F2 we have 250 plants with yellow peppers, 84 with colorless peppers, and 112 with green peppers. What must have been the genotypes and phenotypes of the original pure lines?

ANSWER: a) The ratio in F2 is about 9:3:4 and the original pure lines were probably a/a ; B/B ; C/C (colorless) and A/A ; b/b ; C/C (yellow). If we carry through the genotypes in the crossing scheme, the expected ratios match the observed ones. b) Because of the absence of "brown" phenotype, we know that all the plants were probably b/b.

Therefore, we are working with two genes: A and C. The yellow pure line must have been A/A ; b/b ; C/C, and the green one must have been a/a ; b/b ; c/c since in F2 we obtain some colorless. If we carry through these genotypes in the crossing scheme, the expected ratios match the observed ones very closely.

46. Two mutant, colorless strains of the (fictitious) haploid fungus Cyanomyces are crossed. The progeny consists of 50% wild-type (blue) and 50% mutant (colorless) spores. a) How can this result be explained? b) Based on your explanation in part (a), what is the predicted result of a cross between a blue and a colorless strain?

ANSWER: Complementation → the two strains have mutations in different genes. For example, A ; b × a ; B and the 1:1 ratio is indicative of suppression, most likely "heterodimer suppression," where the combination of the two mutant alleles results in a wild-type phenotype. So a) progeny = 1/4 A ; B (wild type, blue), 1/4 A ; b (mutant, colorless), 1/4 a ; B (mutant colorless) and 1/4 a ; b (wild type, blue). b) Based on part (a), any blue × colorless cross should give 50% blue and 50% colorless. For example, A ; B × A ; b → 1/2 A ; B (blue) and 1/2 A ; b (colorless).

47. Suppose that pigmentation in rabbits is controlled by two genes (E and H). Each gene exhibits complete dominance, and the relationships between genotypes and phenotypes are listed below: Genotype E/− ; H/− e/e ; H/− E/− ; h/h e/e ; h/h

Phenotype albino brown albino albino

a) Two brown parents produce an albino baby rabbit. What are the genotypes of each of the parents and that of the kit (baby rabbit)? b) What phenotypic ratio would be expected among the progeny of an intercross between dihybrid rabbits?

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chapter 05: Gene Interaction c) In a litter of five baby rabbits from the above cross, what is the probability that all five of the baby rabbits will be albino?

ANSWER: a) parents: e/e ; H/h and e/e ; H/h kit: e/e ; h/h b) 13 albino : 3 brown c) (13/16)5 = 0.35 48. In Drosophila, two genes affecting body color are known. A mutant allele at one locus prevents the formation of granules on which pigment is laid down. An absence of these granules causes an albino. Another gene causes the color (the pigment involved) to be blood-red. True-breeding albino males were crossed to truebreeding blood-red females. All males and females in the resulting F1 generation were wild type in appearance.

Two F1 flies were intercrossed, producing a large F2 generation, with the following phenotypes.

Phenotypes Wild-type female Blood female Wild-type male Albino male Blood male Total

Observednumbers 39 8 13 17 3 80

Expected ratio ____ ____ ____ ____ ____ 16

a) What are the genotypes of the parents and the F1? b) Given the F1 genotypes in part (a), fill in the expected phenotypic ratio for the F2 in the spaces above. c) Use the chi-square test to determine how well the observed ratio fits the expected ratio.

ANSWER: a) Parents: male a/Y ; b+/b+ female a+/a+ ; b/b F1:

male a+/Y ; b+/b

female a+/a ; b+/b

b)

Phenotypes Observed numbers Expected ratio Wild-type female 39 6 Blood female 8 2 Wild-type male 13 4 Albino male 17 3 Blood male 3 1 Total 80 16 2 2 c) χ = 15.867; df = 4. At alpha = 0.05, the critical χ is 9.488; therefore, you can reject the null hypothesis that the observed values fit the expected model.

49. In wheat, a cross between red-kernel and white-kernel strains yielded F1 offspring with red kernels. When the F1 were intercrossed, the F2 plants had a ratio of 15 red-kernel : 1 white-kernel. A testcross of the red-kernel plants yielded 3 red-kernel : 1 white-kernel. a) What are the parental genotypes? b) What are the F2 genotypes and phenotypes? c) What conclusions can be made about the allelic and gene interactions?

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chapter 05: Gene Interaction ANSWER: a) A/A ; B/B (red)

a/a ; b/b (white)

b) 1 A/A ; B/B 2 A/a ; B/B 2 A/A ; B/b 4 A/a ; B/b

1 A/a ; b/b 2 A/a ; b/b 15 red 1 a/a ; B/B 2 a/a ; B/b 1 a/a ; b/b 1 white c) A is completely dominant to a, and B is completely dominant to b. Gene interaction: duplicate genes.

50. Snapdragons with red, normally shaped flowers are mated with plants with white, abnormally shaped flowers. In the F1, all the flowers are pink and have normal shape. The F1 intercross yields the following F2: 3/16 6/16 3/16 2/16 1/16 1/16

red, normal pink, normal white, normal pink, abnormal red, abnormal white, abnormal

a) What are the parental genotypes? b) What are the F2 genotypes and phenotypes? c) What conclusions can be made about the allelic and gene interactions?

ANSWER: a) A/A ; B/B (red, normal) a/a ; b/b (white, abnormal) b)

c) A (red) is incompletely dominant to a (white), and B is completely dominant to b. No gene interaction.

51. Certain recessive genes cause profound deafness, and individuals homozygous for such genes are occasionally found in high frequencies among extended families in small, isolated communities. A deaf man and a deaf woman from two different communities, each having deaf parents, had three children, all of whom had normal hearing. a) What are the parental genotypes? b) What conclusions can be made about the allelic and gene interactions?

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chapter 05: Gene Interaction ANSWER: a) a/a ; B/B ∞ A/A ; b/b b) A is completely dominant to a, and B is completely dominant to b. Gene interaction: complementary gene action.

52. Neurospora is a haploid, filamentous fungus normally having fluffy, orange masses of asexual spores called conidia. Two mutant strains, one having albino (white) conidia, and the other lacking conidia entirely (aconidial), were mated. Their progeny were as follows: 82 normal, 86 albino, and 166 aconidial. a) What are the parental genotypes? b) What are the progeny genotypes and phenotypes? c) What conclusions can be made about the gene interactions?

ANSWER: a) a ; B × A ; b b) 1/4 a ; b albino 1/4 A ; B normal 1/4 A ; b aconidial 1/4 a ; b aconidial c)

(tot = 50% aconidial)

b is epistatic to A and a.

53. The following four crosses all apply to diploid organisms. Assume that all the parents are true breeding.

You are given the phenotypes of the parental generation and the phenotypic ratio of the F2 generation. You must supply the following information: 1) genotypes of the parents 2) the phenotypic ratio expected for a testcross of the F1 3) a diagram of the chromosomes of the F1in cross (c) at metaphase I Define your symbols.

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chapter 05: Gene Interaction

Diagram the chromosomes at metaphase I. Give all possible arrangements of the bivalents. Assume there is no crossing over between genes and centromeres.

Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chapter 05: Gene Interaction ANSWER:

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chapter 05: Gene Interaction

54. Loppins can be purple or white, and this phenotype is inherited as a simple dominant/recessive trait with complete penetrance. In females, purple is dominant over white, while in males the opposite is true. If you cross a white female to a purple male, and then you "self" the F1 (i.e., you cross brothers and sisters) what phenotypic ratios would you expect in the F2? Assume that you always get a 1:1 ratio of males : females.

ANSWER: Let's define cP as the purple allele and cW as the white allele. White female must be cW/cW since white is recessive in females, and purple male must be cP/cP since purple is recessive in males. The F1 is therefore dihybrid cW/cP, with all the males being white and all the females being purple. If we have a 1:1 ratio of males to females, then the white : purple ratio is also 1:1.

55. If a green-eyed fly is mated with a homozygous wild-type black-eyed fly, a 1:1 ratio of green-eyed to blackeyed wild-type fly is always observed in the progeny. However, if two green-eyed flies are crossed with each other, the result is always 2:1 (green-eyed: black-eyed).

How can these ratios be explained?

ANSWER: a) The 1:1 ratio suggests that the green-eyed fly is always heterozygous for the green-eye allele and that the allele for green eyes is dominant over the wild type. b) The 2:1 ratio suggests the allele for green eyes is lethal when homozygous; hence, 1/4 of the progeny did not survive.

Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses Multiple Choice 1. A bacterium that is unable to grow on minimal medium is said to be a. allotrophic. b. autotrophic. c. auxotrophic. d. heterotrophic. e. prototrophic.

ANSWER: c 2. A gal– mutant a. can make its own galactose. b. cannot grow without galactose. c. cannot utilize galactose as a carbon source. d. can utilize galactose as a carbon source. e. is resistant to galactose.

ANSWER: d 3. A StrR mutant a. can grow in the presence of streptomycin. b. can make its own streptomycin. c. cannot grow in the presence of streptomycin. d. cannot grow without streptomycin. e. cannot make its own streptomycin.

ANSWER: a 4. A strain of E. coli has the genotype (and phenotype) leu– lac–. This notation means that this strain of bacteria is unable to a. metabolize the amino acid leucine and synthesize the sugar lactose. b. utilize the amino acid leucine and synthesize the sugar lactose. c. utilize both the amino acid leucine and the sugar lactose. d. synthesize the amino acid leucine and utilize the sugar lactose. e. synthesize both the amino acid leucine and the sugar lactose.

ANSWER: d 5. A strain of E. coli with the genotype arg– bio– gal– TetR is able to grow on minimal medium (containing glucose as a carbon source) supplemented with a. the amino acid arginine and the antibiotic tetracycline. b. the amino acid arginine and the vitamin biotin. c. the amino acid arginine and the sugar galactose.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses d. the vitamin biotin and the sugar galactose. e. the sugar galactose and the antibiotic tetracycline.

ANSWER: b 6. A bacterial strain is able to grow on rich medium (which contains all the amino acids, vitamins, and a variety of carbon sources), but not on minimal medium or on minimal medium supplemented with leucine. What could be the genotype(s) of this bacterial strain with respect to the arg and leu loci? a. arg+ leu– b. arg– leu+ c. arg+ leu– or arg– leu+ d. arg– leu– or arg+ leu– e. arg– leu– or arg– leu+ ANSWER: e 7. An E. coli culture is growing in (rich) liquid medium. A representative sample of this culture is plated on solid rich medium, and 950 colonies grow on this medium. A sample of bacteria from each of these colonies is then plated on minimal medium, and only 108 of these samples grow into a colony. What proportion of the bacteria in the original liquid culture are auxotrophic? a. 0.102 b. 0.114 c. 0.796 d. 0.886 e. 0.898 ANSWER: d 8. An E. coli colony grew on minimal medium supplemented with arginine and leucine. However, bacteria from this colony are unable to grow and form colonies on minimal medium supplemented with arginine and methionine. What is the genotype of the bacteria in this E. coli colony? a. arg+ leu+ met– b. arg+ leu– met+ c. arg+ leu– met– d. arg– leu+ met+ e. arg– leu– met– ANSWER: b 9. The table below summarizes the results of a replica-plating experiment where a mixture of 347 E. coli strains of different genotypes were tested for their ability to grow on a variety of media.

Medium complete medium (contains all amino acids, Copyright Macmillan Learning. Powered by Cognero.

Number of colonies that grew 347 Page 2

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses vitamins, and a variety of carbon sources) minimal medium (glucose as a carbon source) minimal medium + ampicillin (an antibiotic) minimal medium + methionine (an amino acid) minimal medium + arginine (an amino acid) minimal medium + methionine + ampicillin minimal medium + arginine + ampicillin

261 51 303 301 57 64

Based on the data presented in the table above, how many colonies are wild type (met+ arg+ AmpS)? a. 51 b. 210 c. 261 d. 296 e. 347 ANSWER: b 10. The table below summarizes the results of a replica-plating experiment where a mixture of 347 E. coli strains of different genotypes were tested for their ability to grow on a variety of media.

Medium complete medium (contains all amino acids, vitamins, and a variety of carbon sources) minimal medium (glucose as a carbon source) minimal medium + ampicillin (an antibiotic) minimal medium + methionine (an amino acid) minimal medium + arginine (an amino acid) minimal medium + methionine + ampicillin minimal medium + arginine + ampicillin

Number of colonies that grew 347 261 51 303 301 57 64

Based on the data presented in the table above, how many colonies are met– arg+ AmpS? a. 36 b. 44 c. 246 d. 252 e. 296 ANSWER: a 11. The table below summarizes the results of a replica-plating experiment where a mixture of 347 E. coli strains of different genotypes were tested for their ability to grow on a variety of media.

Medium complete medium (contains all amino acids, vitamins, and a variety of carbon sources) minimal medium (glucose as a carbon source) Copyright Macmillan Learning. Powered by Cognero.

Number of colonies that grew 347 261 Page 3

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses minimal medium + ampicillin (an antibiotic) minimal medium + methionine (an amino acid) minimal medium + arginine (an amino acid) minimal medium + methionine + ampicillin minimal medium + arginine + ampicillin

51 303 301 57 64

Based on the data presented in the table above, how many colonies are met– arg–? a. 2 b. 4 c. 36 d. 82 e. 86 ANSWER: b 12. Joshua Lederberg and Edward Tatum discovered a sex-like process in bacteria using a. high-powered microscopy in combination with antibiotic resistant genes. b. populations of cells with complementing auxotrophs, and the measurement of the generation of prototrophs during microbial mixing. c. mutations in bacteria that inhibit the mating process. d. comparison of bacterial behavior to yeast behavior. e. chromosome sequencing and analysis.

ANSWER: b 13. Bernard Davis tested the "cross-feeding" interpretation of some data that showed the phenotype of one microbe as capable of being changed by another microbe. His contribution can be summarized as a. using one strain of microbes as food for another, and then evaluating the impact upon microbial phenotype. b. combining microbes with complementing auxotrophs in a single tube and then analyzing the cells of gene exchange. c. placing a barrier between bacterial strains with complementing auxotrophs to assess whether contact between cells was required for genetic exchange. d. using fungi to feed bacterial strains with unusual auxotrophic mutations. e. Davis was not a significant contributor to this research question.

ANSWER: c 14. An F+ microbial strain would best be described as a. any bacterium that readily takes up naked DNA through its membrane. b. incapable of conjugation due to inhibitory genes on the F plasmid. c. a bacterial strain with specific auxotrophic characteristics. d. a bacterial strain that carries a phage with membrane fusion competency. e. a bacterium with a plasmid that confers mating competency (or fertility).

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses ANSWER: e 15. An Hfr microbial strain would best be described as a. any bacterium that readily takes up naked DNA through its membrane. b. a bacterial strain with specific auxotrophic characteristics. c. a bacterium containing a chromosomally integrated plasmid conferring mating competency. d. a bacterium carrying an Hfr phage with membrane fusion competency. e. a bacterium that lacks the histidine frequenase gene.

ANSWER: c 16. If a met– thr– Hfr strain is mated with an F– of genotype leu– thi–, prototrophic recombinants can be detected by plating the mixture on minimal a. medium. b. medium supplemented with leucine and methionine. c. medium supplemented with leucine and thiamine. d. medium supplemented with methionine, threonine, leucine, and thiamine. e. medium supplemented with threonine and thiamine.

ANSWER: a 17. An Hfr strain of E. coli with the genotype gly+ aziR StrS is mated with an F– strain of E. coli of genotype gly– aziS StrR. Gly refers to the amino acid glycine, azi refers to sodium azide, and Str refers to the antibiotic streptomycin. Conjugation occurs and the progeny are screened on a selective medium to detect recombinants.

If you wanted to select for the F– recombinant genotype gly+ aziR StrR, you should use a minimal medium containing a. glycine. b. glycine, sodium azide, and streptomycin. c. glycine and streptomycin. d. sodium azide. e. streptomycin and sodium azide.

ANSWER: e 18. Two bacterial strains were obtained with the following genotypes: Hfr: F–:

ala– leu+ aziS StrS ala+ leu– aziR StrR

After an uninterrupted conjugation, you want to select F– recombinants that are ala+ leu+. Which of the following media would you use for this selection? a. complete medium containing streptomycin b. minimal medium containing streptomycin c. minimal medium containing leucine and streptomycin Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses d. minimal medium containing sodium azide and leucine e. minimal medium containing alanine, leucine, and streptomycin ANSWER: b 19. Which of the following are characteristics of an E. coli strain of genotype F– gal+ lac –pur+ ile– strr? a. It can act as a donor in conjugation experiments. b. It can form a pilus. c. It can grow on minimal media where lactose is the only carbon source. d. It is auxotrophic for lactose. e. It is susceptible to streptomycin.

ANSWER: d 20. A cross is made between an Hfr strain that is StrS a+ b+ d+ in genotype and an F– strain that is StrR a– b– d– in genotype. Interrupted-mating studies show that b+ enters the recipient strain last, and that the Str locus is very far away from b+, so it never enters the recipient strain. The b+ recombinants are then tested for the presence of the a+ and d+ alleles. The following data were obtained: a+ b+ d+ a– b+ d+ a+ b+ d– a– b+ d–

326 2 14 58 400

What is the gene order? a. a d b Str b. b a d Str c. b d a Str d. d a b Str e. d b a Str ANSWER: d 21. A cross is made between an Hfr strain that is StrS a+ b+ d+ in genotype and an F– strain that is StrR a– b– d– in genotype. Interrupted-mating studies show that b+ enters the recipient strain last, and that the Str locus is very far away from b+, so it never enters the recipient strain. The b+ recombinants are then tested for the presence of the a+ and d+ alleles. The following data were obtained: a+ b+ d+ 326 a– b+ d+ 2 a+ b+ d– 14 a– b+ d– 58 Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses 400 What is the map distance between markers d and b? a. 4 map units b. 18 map units c. 19 map units d. 36 map units e. 38 map units ANSWER: c 22. An interrupted conjugation experiment is conducted using the following strains: Hfr StrS x+ y+ z+ and F– StrR x– y– z–. Samples of exconjugants are tested at regular intervals for their ability to grow on minimal medium (MM) containing streptomycin plus compound X, Y, or Z. The table below summarizes the number of colonies obtained on the different media when conjugation was interrupted after 10, 15, 20, or 25 minutes.

Time of interruption 10 min 15 min 20 min 25 min

Media used MM+Str+X 0 0 15 250

MM+Str+Y 0 10 250 250

MM+Str+Z 0 0 0 10

Based on these results, what is the order of entry of the four markers in question? a. Str x y z b. Str y x z c. Str z x y d. y x z Str e. z x y Str ANSWER: b 23. From one F+ strain, three distinct Hfr strains were derived. The first three markers transferred during an Hfr × F– cross (different for each of the three Hfr) are Hfr 1: . . . D A F→ Hfr 2: . . . E B F→ Hfr 3: . . . E C D→ Which of the following must be the order of the genes on the bacterial chromosomal circle? (A is shown at both ends to represent circularity. Assume that the Hfr picks up all intermediates between any two represented genes.) a. A D C E B F A b. A B C D F E A c. A C D F E B A Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses d. A E F B C D A e. A F B D E C A ANSWER: a 24. To demonstrate transformation of bacteria, one could a. extract DNA from an auxotroph and add it to prototrophic cells. b. extract DNA from arg– cells and add it to arg+ cells. c. extract DNA from arg+ cells and add it to arg– cells. d. extract DNA from StrS cells and add it to StrR cells. e. mix the DNA from arg+ and arg– cells to allow recombination.

ANSWER: c 25. To demonstrate linkage of two markers A and B by transformation, one needs to demonstrate that the frequency of transformation by a. A and B is greater than the product of their individual transformation frequencies. b. A and B is greater than the sum of their individual transformation frequencies. c. A and B is less than the product of their individual transformation frequencies. d. A and B is less than the sum of their individual transformation frequencies. e. A is equal to the frequency of transformation by B. ANSWER: a 26. If two markers are closely linked, they will show a. a high frequency of recombination and a high frequency of co-transformation. b. a high frequency of recombination and a low frequency of co-transformation. c. a low frequency of recombination and a high frequency of co-transformation. d. a low frequency of recombination and a low frequency of co-transformation. e. the same frequency of recombination and frequency of co-transformation.

ANSWER: c 27. The most commonly used phage characters in the study of phage inheritance are a. phage morphology and host morphology. b. phage morphology and host range. c. plaque morphology and host morphology. d. plaque morphology and host range. e. plaque morphology and phage morphology.

ANSWER: d 28. A permissive bacterial strain is simultaneously infected with two strains of T4-like phage. One phage strain has the genotype a– b+, while the other is a+ b–. If the map distance between markers a and b is about 10 map units, what proportion of the progeny phages are expected to have the genotype a+ b+?

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses a. about 1/20 b. about 1/10 c. about 4/10 d. about 4/5 e. about 9/10 ANSWER: a 29. A double infection experiment is conducted to determine the distance between two markers in the bacteriophage RB49, a phage similar to T2 and T4. A mutation in the first marker (F13–) results in the inability to grow on E. coli strain B178, while a mutation in the second marker (P6–) prevents the phage from growing at high temperature. A permissive E. coli strain is infected with both F13– and P6– phages, and the lysate is analyzed. The analysis showed that 1002 plaques were formed on a lawn of permissive E. coli at 37°C, and out of these 1002, 49 also formed plaques on E. coli B178 at high temperature. What is the predicted frequency of phages with genotype F13+ P6– in the lysate? a. 5% b. 45% c. 47.5% d. 90% e. 95% ANSWER: b 30. A bacterial chromosome has four markers, A, B, C, and D, evenly spaced throughout the circle. A generalized transducing phage will a. always transduce at least two of the four markers. b. never transduce more than one marker. c. transduce any of the markers by different transduction events. d. transduce only one specific marker such as A. e. pick up all the markers in one particle.

ANSWER: c 31. If two bacterial genes are very closely linked (less than one map unit apart), then their frequency of cotransduction will be a. less than 1%. b. more than 1%, but no more than 25%. c. between 25% and 50%. d. between 50% and 75%. e. very high, almost 100%.

ANSWER: e 32. The three bacterial markers a, b, and c are linked, but their order is unknown. A cotransduction experiment is performed and the following results are obtained.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses - Markers a and b are co-transduced at a frequency of 2%.

- Markers a and c are co-transduced at a frequency of 20%.

- Markers b and c are co-transduced at a frequency of 65%.

What can be concluded about the arrangement of the three markers on the bacterial chromosome? a. The order is a-b-c, and b is closer to a than to c. b. The order is a-b-c, and b is closer to c than to a. c. The order is a-c-b, and c is closer to b than to a. d. The order is b-a-c, and a is closer to b than to c. e. The order is b-c-a, and c is closer to a than to b.

ANSWER: c 33. What are the three main ways by which bacteria exchange genes? a. Conjugation, transformation, and translocation. b. Transformation, transduction, and translocation. c. Conjugation, transformation, and transduction. d. Conjugation, translocation, and transduction. e. Conjugation, transmission, and transduction.

ANSWER: c 34. Bacteria that are capable of taking up DNA are a. plasmids. b. transformants. c. incompetent. d. competent. e. none None of the above.

ANSWER: d Subjective Short Answer 35. The following table contains statements that may apply to conjugation, transformation, and/or transduction.

For each blank space, insert a checkmark if that statement applies to that mode of DNA exchange.

Transformation Conjugation

Transduction

Bacteriophages are required for gene transfer.

DNA is acquired directly from the environment.

DNA is transferred via an F plasmid.

The movement of DNA is unidirectional from F+ to F–.

The DNA is replicated via rolling circle replication prior to the transfer.

The transferred DNA must recombine and replace the Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses DNA in the chromosome in order to be maintained.

It can be used to study the dominance relationship between alleles via the creation of partial diploids.

ANSWER:

Transformation Conjugation Bacteriophages are required for gene transfer.

DNA is acquired directly from the environment.

DNA is transferred via an F plasmid.

The movement of DNA is unidirectional from F+ to F–.

The DNA is replicated via rolling circle replication prior to the transfer.

The transferred DNA must recombine and replace the DNA in the chromosome in order to be maintained.

It can be used to study the dominance relationship between alleles via the creation of partial diploids.

Transduction √

√ √ √ √ √

( ) - Hfr = yes; F = no

√

√(F')

Essay 36. The following table contains statements pertaining to Hfr, F+, F′, and/or F– strains of bacteria. Indicate in respective boxes if the statements apply to each category by inserting a checkmark (√ ). + F– Statement Hfr F F' a) Contains genes that regulate conjugation and the transfer of DNA from one cell to another. b) Always acts as the recipient in conjugation. c) F factor is integrated into the host chromosome. d) Cannot form a pilus. e) F factor is in the plasmid form carrying a segment of chromosomal DNA. ANSWER: Statement a) Contains genes that regulate conjugation and the transfer of DNA from one cell to another. b) Always acts as the recipient in conjugation. c) F factor is integrated into the host chromosome. d) Cannot form a pilus. e) F factor is in the plasmid form carrying a segment of chromosomal DNA.

Hfr

F+

√

√

F–

F' √

√ √ √ √

37. Auxotrophic mutations have been extremely useful to study a number of processes in E. coli. a) Briefly define the term "auxotrophic." b) Suppose you were given two unlabeled tubes, one that contains a methionine auxotroph and the other that contains a leucine auxotroph. Design an experiment to determine which auxotroph is found in each tube.

ANSWER: a) It is a cell that is unable to synthesize essential nutrients and therefore cannot grow unless the Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses appropriate nutrient(s) are added to the medium. b) Experiment: Grow each sample on (1) minimal media (negative control), (2) complete media (positive control), (3) minimal + methionine, and (4) minimal + leucine and observe growth on the various media.

Growth = + No growth = – Medium WT/prototroph Contents: tube A met+ leu+ minimal – – complete + + minimal + methionine + + minimal + leucine + –

Contents: tube B – + – +

Therefore, tube A contains the met auxotroph, whereas tube B contains the leu auxotroph.

38. An E. coli F– strain has the following genotype: S– U– R– I– E– C–. Three different Hfr strains, all carrying S+ U+ R+ I+ E+ and C+ markers and all having lambda phages integrated at the same specific site in their chromosomes, are mated with the F strain in separate matings. The interrupted-mating results are given below.

The numbers indicate time (in minutes) when different donor markers appeared in F– cells after conjugation began. Assume that the E. coli map consists of 100 minutes, and gene C is mapped at 80 minutes.

Markers R I U E C S

Hfr 1 10 40 25 – – 55

Hfr 2 20 – 5 60 45 –

Hfr 3 – 5 – – – 20

The positions of Hfr 2, the gene C, and the polarity of Hfr 2 are given in the accompanying map.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses

a) Give the gene order, starting from C and going clockwise. b) What is the location of the origin of Hfr 1? Hfr 3? c) What is the location of the I marker? The S marker? d) What is the location of the lambda phage in these strains?

ANSWER: a) CRUISE (if counterclockwise, CESIUR) b) 95 minutes (Hfr 1); 30 minutes (Hfr 3) c) 35 minutes (I); 50 minutes (S) d) between S and E (50 to 65 minutes) 39. At time zero, an Hfr strain (strain 1) was mixed with an F– strain, and at various times after mixing, samples were removed and agitated to separate conjugating cells. The cross may be written as: Hfr 1: a+ b+ c+ d+ e+ f+ g+ h+ StrS F: a– b– c– d– e– f– g– h– StrR No order is implied in the above listing. The samples were then plated onto selective media to measure the frequency of h+ StrR recombinants that had received certain genes from the Hfr cell. The following graph shows the percentage of recombinants against time in minutes.

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses

a) Draw a linear map of the Hfr chromosome indicating the (1) point of insertion (origin).

(2) order of the genes a+, b+, e+, g+, and h+.

(3) shortest distance between consecutive genes on the chromosome in minutes. b) An additional four Hfr strains (strains 2–5) were obtained. Each carried the wild-type alleles of the genes listed below. They were mated individually to the same F– strain shown above (carrying the recessive alleles of all genes). With interrupted mating, the times of first appearance in minutes of individual Hfr markers in exconjugants were determined, as shown in the following chart.

Hfr marker a+ b+ c+ d+ e+ f+ g+

Minutes Hfr 1 Hfr 2 2 12 42

Hfr 3 11 28

Hfr 4 20 10 35

88

Hfr 5

5 50 35

Draw the genetic map showing the (1) positions of genes a through h.

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses (2) origins of the Hfrs (including strain 1).

(3) orientation of the origins.

(4) shortest distance between consecutive genes or origins.

ANSWER: a)

b)

Numbers on the outside of the circle are distances between genes. Numbers on the inside are distances between origins and adjacent genes. The above map is not to scale.

40. Strain A of E. coli is auxotrophic for methionine, while strain B is auxotrophic for lysine. Both strains are F– . You want to produce a strain that is prototrophic. Design an experiment to produce the desired prototroph.

Include in your experimental design your screening method for detecting the prototroph.

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses ANSWER: Since both strains are F–, a conjugation experiment cannot be conducted. Instead, a transformation or transduction experiment should be designed. To do a transformation experiment, take DNA from Strain A and mix it with Strain B bacteria (or vice versa). The goal is to combine the functional met+ gene from strain B and the functional lys+ gene from strain A via transformation and recombination.

Some of the transformants will contain the desired combination of met+ and lys+. Next, grow the transformed bacteria on minimal media. Any bacteria that can grow on minimal media are prototrophs.

41. In a transformation experiment, DNA of a p+ q+ strain was used to transform a recipient strain that is auxotrophic for these markers. The number of each class of transformants is shown below: Class Genotype Number 1 405 p+ q+ 2 300 p+ q– p– q+

3

195

a) If the total number of transformants isolated was 900, what is the co-transformation frequency for the p and q loci? b) Use diagrams to show how each class of transformants were obtained starting with the recipient auxotroph.

ANSWER: a) co-transformation frequency = (p+ q+)/p+ = 405/(405 + 300) = 0.574 b)

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses

42. A bacterium with the genotype met+ his+ lys– arg+ has been transformed with DNA from a second bacterium that has the genotype met+ his– lys+ arg–.

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses

Following the transformation, scientists streaked a sample of a transformed bacterium onto several different plates that contain different nutritional supplements in order to determine the transformant's genotype. Using the results of the experiments (shown below), determine what crossovers occurred to produce the bacterium. On the above diagram, draw the crossovers that are necessary to produce a bacterium that fits the observed growth profile.

ANSWER: The bacteria must be met+ his– lys– arg+, so a crossover must have occurred between the met and his genes and a second one between the his and lys genes, as shown below.

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses

43. Two labs calculated cotransduction frequencies for the genes A and B. Lab 1 calculated 0.63, and lab 2 calculated 0.47. Which lab reported the genes to be closer together?

ANSWER: Lab 1. A higher cotransduction frequency indicates a smaller distance between the markers because fewer crossovers occurred to break up the two genes.

44. In a generalized transduction experiment, donor E. coli cells have the genotype a+ c+ b–, and recipient cells have the genotype a– c– b+. Pl-mediated transductants for a+ were selected, and their total genotypes were determined, with the following results: Genotype a+ c+ a+ c– a+ c+ a+ c–

Number of progeny 95 b – 205 b 5 b+ 195 b+ 500 –

a) What is the order of the three genes? b) What is the cotransduction frequency for the following: a and c? a and b? b and c?

ANSWER: a) The donor is a+ c+ b–, the recipient is a– c– b+, and the rare class is a+ b+ c+. Since the b allele is not transferred but the other two are, b must be the middle gene. The order is a b c. b) cotransduction frequency for a and c = (95 + 5)/500 = 0.2 or 20%, and for a and b = (5 + Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses 195)/500 = 0.4 or 40%.

The frequency for b and c cannot be calculated because they are not selected markers.

45. A transduction experiment was conducted using a prototrophic donor and a recipient that was auxotrophic for lys, arg, and trp. Selection was done for trp+ transductants. The trp+ transductants could be categorized into the following classes: Genotype lys+ arg– trp+

Number 10

lys– arg+ trp+

13

lys– arg– trp+

82

lys+ arg+ trp+

1

a) Which marker is closest to trp? Explain your reasoning. b) What is the order of the three markers? Explain your reasoning.

ANSWER: a) arg. arg was co-transduced with trp at a higher frequency than was lys. b) arg trp lys (i.e., trp is in the middle). We can predict this because the frequency of triple transductants is much lower than either of the double transductants. If arg+ was in the middle, then the frequency of lys+ arg+ trp+ would be higher than lys+ arg– trp+ (because fewer crossovers would be needed). This would be true, also, if lys+ was in the middle.

46. A new bacteriophage of E. coli was isolated from the steam tunnels of New York City. Six mutant strains (a, b, c, d, e, f) were derived from it, each having a different single point mutation for the genes A, B, C, D, E, and F. Strain B D E was mated with strain b d e. The order of genes given below does not necessarily reflect the order of the genes on the chromosome. The following progeny were observed:

a) What are the genotypes of the double crossovers, showing gene order? b) Give the percent recombination between each pair of genes. c) Draw a map showing the positions of the B, D, and E genes, with map distances between genes, where appropriate. d) (1) Give the formula you would use to calculate the coefficient of coincidence.

(2) Calculate the coefficient of coincidence. e) Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses (1) Give the formula you would use to calculate interference.

(2) Calculate interference and specify whether the value is positive or negative.

ANSWER: a) B e D and b E d b) b, d: 25%; b, e: 20%; d, e: 5% c) 20 5 B–––E–––D (Distance from B to E: 20 m.u.; distance from E to D: 5 m.u.) d) (1) c.o.c. = observed double crossovers/expected DCOs (2) c.o.c. = 0.009/(0.2 0.005) = 0.9 e) (1) I = 1 – c.o.c.

(2) I = 1 – 0.9 = 0.1 47. A mixed infection of an E. coli strain was performed using the h– r+ and h+ r– genotypes of the T2 phage.

Progeny phage were collected after lysis and plated on two E. coli strains to obtain plaque characters important for detection of parental and recombinant types. h+ can infect only one strain of E. coli h– can infect both strains r+ shows slower lysis, forms small plaques r– shows rapid lysis In the table below, for each observed phenotype, give the genotype that would produce that phenotype, and indicate if the phenotype is parental or recombinant type. Calculate the map distance between the h and r genes using the information provided.

Plaque phenotype cloudy, large clear, small cloudy, small clear, large

Number of plaques 600 480 180 180

Plaque genotype

Parental/recombinant

ANSWER: RF% = (recombinants/total) × 100 = map distance Plaque phenotype cloudy, large clear, small cloudy, small clear, large

Number of plaques 600 480 180 180

Plaque genotype h+ r– h– r+ h+ r+ h– r–

Parental/recombinant parental parental recombinant recombinant

map distance = (360/1080) × 100 = 33.33 m.u.

48. How does conjugation differ from bacterial transformation?

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses ANSWER: In conjugation, DNA is transferred from one living cell to another through close contact, whereas in transformation, isolated pieces of external DNA are taken up by a cell through the cell wall and plasma membrane.

49. How are restriction enzymes used in the laboratory?

ANSWER: In the laboratory, restriction enzymes are used to cut DNA into smaller fragments that can be used to map DNA fragments and genomes.

50. Describe a method that is based on palindromic repeats, which can be used to edit the genome of eukaryotic cells.

ANSWER: The clustered regularly interspaced short palindromic repeats CRISPR system can be used to edit the genome of eukaryotic cells. DNA constructs comprising CRISPR associated protein 9 (Cas9) plus a guide RNA, homologous to a target gene to be modified, are inserted into the eukaryotic cell.

The guide RNA finds the target gene by base homology and Cas9 cuts it, resulting in a doublestrand break. The eukaryotic cellular repair mechanisms then take over and mend the break. By manipulating the nucleotide sequence of the guide RNA, the CRISPR/Cas9 system can be designed to target any DNA sequence for cleavage.

51. Bacterial cells have a variety of restriction enzymes that can degrade the DNA of an attacking virus. How do these bacterial cells protect their own DNA from being degraded by the very restriction enzymes that are meant to help them?

ANSWER: The bacterial DNA has been methylated at certain locations by special enzymes. This methylation occurs within the recognition sequences, and foreign DNA—not methylated—is recognized as foreign and destroyed.

52. An extraterrestrial scientist is trying to obtain a genetic map of a newly discovered bacterium using standard techniques such as interrupted conjugation and Hfr mapping. She is particularly interested in three loci (ata, tic, and lan) encoding proteins that are required for the synthesis of atanine, ticovine, and lancytrine (three essential extraterrestrial amino acids), respectively. With the help of interrupted-conjugation experiments, the scientist determines that ata is the last marker to enter the recipient bacterium.

She now sets up a cross between a wild-type Hfr strain that is tetracycline sensitive and an ata– tic– lan– F– strain that is tetracycline resistant. A total of 5000 exconjugants are recovered on minimal medium plus tetracycline, ticovine, and lancytrine, and subsequently replicated onto a variety of media.

The results are as follows: - minimal medium and tetracycline: 4704 colonies grow; - minimal medium plus lancytrine and tetracycline: 4708 colonies grow; - minimal medium plus tetracycline and ticovine: 4900 colonies grow. a) Draw a genetic map of these three loci (including all the map distances). b) Why is the scientist using tetracycline on her plates (what is the purpose)? c) Why is she recovering the exconjugants on plates that lack atanine?

ANSWER: a) 4704 ata+ lan+ tic+ 4 ata+ lan– tic+ Copyright Macmillan Learning. Powered by Cognero.

Page 22

Name:

Class:

Date:

Chapter 06: The Genetics of Bacteria and Their Viruses ata+ lan+ tic– ata+ lan– tic–

196 5000 − 4794 – 4 – 196 = 96

lan s in the middle, ata to lan2 m.u.; lan to tic4 m.u. b) to kill off the Hfr cells c) to select for the last marker (ensuring that all three markers have entered the recipient)

Copyright Macmillan Learning. Powered by Cognero.

Page 23

Name:

Class:

Date:

Chapter 07: DNA: Structure and Replication Multiple Choice 1. In the classic experiment by Griffith, evidence of the action of a hereditary biomolecule was identified by a. growth of the bacterial cells. b. transformation (phenotypic change) of the R strain by S strain biomolecules. c. rapid death of mice following injection with bacteria. d. isolation of pure DNA from the S strain cells. e. the ability of heat to destroy deadly biomolecules.

ANSWER: b 2. Mendel identified evidence for the location of genes on separate structures (chromosomes) through his a. electrophoresis experiments. b. purification of chromosomes from pea plants. c. identification of dominant and recessive alleles of genes. d. identification of independent assortment of distinct genes during meiosis. e. use of high-resolution microscopic techniques.

ANSWER: d 3. Smooth (S) and rough (R) strains of Streptococcus pneumonia are distinct because of their ability to cause illness (death in rodents). This trait is controlled by genes that regulate: a. production of a protective polysaccharide coat (capsule) around each cell. b. production of lethal toxins. c. speed of bacterial growth. d. metabolic rate of microbial glycolysis. e. bacterial ability to penetrate through the membrane of animal cells.

ANSWER: a 4. Evidence of transformation of R strain cells by biomolecules from S strain cells includes which one of these choices? a. R strain cells begin to produce high levels of neurotoxin. b. Transformed R strain cells gain the ability to kill mice. c. Isolation of smooth-looking cells from animals infected with R strain cells that had been incubated with heat-killed S strain cells. d. R strain cells produce neurotoxin and gain the ability to kill mice. e. R strain cells gain the ability to kill mice, and smooth-looking cells were isolated from animals infected with R strain cells previously incubated with heat-killed S strain cells.

ANSWER: e 5. Oswald Avery and colleagues strengthened scientific support that DNA was the transforming factor by replicating the Griffith experiment with some important differences in experimental design. The key difference was: a. purifying S strain DNA and RNA from other biomolecules.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chapter 07: DNA: Structure and Replication b. systematically eliminating the impact of classes of S strain biomolecules using enzymatic digestion before mixing with R strain live cells. c. injecting S strain DNA directly into R strain cells and observing the change in phenotype. d. Avery used Escherichia coli rather than Streptococcus as it is easier to transform. e. All of the answer options are correct.

ANSWER: b 6. Oswald Avery and colleagues strengthened scientific support that DNA was the transforming factor by replicating the Griffith experiment with some important differences in experimental design. They found that the enzyme ______________ was effective at destroying the transforming capacity of S strain biomolecules. a. protease (protein destruction) b. RNase (RNA destruction) c. DNase (DNA destruction) d. polysaccharide-destroying enzymes e. lipase (lipid destruction) ANSWER: c 7. Based on its association with chromosomes, this was once considered the hereditary molecule before the discovery of DNA. a. phospholipids b. proteins c. glycogen d. cholesterol e. None of the answer options is correct.

ANSWER: b 8. Alfred Hershey and Martha Chase examined transformation using bacteriophage and bacterial cells. In this experiment, how is transformation (altered phenotype) displayed? a. The bacteria are transformed (by phage DNA) into virus synthesizing cells. b. The bacteria are transformed from live to dead by the virus. c. The bacteria are transformed from rough to smooth by the virus. d. The bacteria are transformed into a virus. e. Transformation is not exhibited by a virus-infected bacterial cell.

ANSWER: a 9. Alfred Hershey and Martha Chase examined transformation using bacteriophage and bacterial cells. In these experiments, they used radioactivity to label nucleic acids (in this case DNA) and proteins. What do radioactive sulfur and phosphate specifically label? a. Sulfur labels DNA, and phosphate labels protein. b. Sulfur labels both protein and DNA. c. Phosphate labels both protein and DNA. d. Sulfur labels protein, and phosphate labels DNA. Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chapter 07: DNA: Structure and Replication e. None of the answer options is correct.

ANSWER: d 10. Alfred Hershey and Martha Chase examined transformation using bacteriophage and bacterial cells. If phage are labeled with radioactive sulfur and allowed to infect bacterial cells, the radioactive sulfur will be localized to: a. the inside of infected cells (in phage DNA). b. the outside of infected cells (in phage ghosts). c. newly synthesized phage viruses in the host bacterial cell. d. No radioactivity will remain after infection. e. The sulfur will be metabolically consumed, and therefore the radioactivity will be destroyed.

ANSWER: b 11. Alfred Hershey and Martha Chase examined transformation using bacteriophage and bacterial cells. If phage are labeled with radioactive phosphate and allowed to infect bacterial cells, the radioactive phosphate will be localized to: a. the inside of infected cells (in phage DNA). b. the outside of infected cells (in phage ghosts). c. newly synthesized phage viruses in the host bacterial cell. d. No radioactivity will remain after infection. e. The phosphate will be metabolically consumed, and therefore the radioactivity will be destroyed.

ANSWER: a 12. Chargaff's rules do NOT hold for which of the following genome types? a. yeast b. bacteria c. single-stranded DNA virus d. invertebrates e. Archaeobacteria ANSWER: c 13. Of the three key building blocks of DNA, which type(s) of building block is/are negatively charged and oriented on the outside of the double helical structure? a. phosphate b. deoxyribose c. nitrogenous bases d. both phosphate and nitrogenous bases e. All of the answer options are correct.

ANSWER: a 14. Of the three key building blocks of DNA, which type(s) of building block is/are structurally different in RNA molecules?

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chapter 07: DNA: Structure and Replication a. Phosphate b. Deoxyribose c. Nitrogenous bases d. Both deoxyribose and nitrogenous bases e. All of the answer options are correct.

ANSWER: d 15. Which statement BEST describes the arrangement of components in a DNA molecule? a. Nucleotides are located toward the outside of the strands and phosphates toward the interior. b. Nucleotides are located toward the inside of the strands and the phosphates toward the outside. c. Nucleotides, as well as the phosphates, are located toward the inside of the strands. d. Nucleotides, as well as the phosphates, are located toward the outside of the strands. e. None of the above.

ANSWER: b 16. Which statement below BEST describes the situation between nucleotides on opposite strands in a DNA molecule? a. A-T and C-G bonding between opposite strands involves two hydrogen bonds for either pair. b. A-T and C-G bonding between opposite strands involves three hydrogen bonds for either pair. c. A-T bonding between opposite strands involves two hydrogen bonds, whereas G-C bonding between strands involves three hydrogen bonds. d. A-T bonding between opposite strands involves three hydrogen bonds, whereas G-C bonding between strands involves two hydrogen bonds.

ANSWER: c 17. What structural feature of DNA suggests a possible method of its replication? a. The nucleotides between nucleotide strands are complementarily paired. b. The hydrogen bonds between A-T and G-C are the same on all DNA molecules. c. The DNA strands run antiparallel. d. The phosphate bonds in the backbone of all DNA molecules.

ANSWER: a 18. The following data were obtained from three organisms: an RNA virus, a DNA virus, and a wombat (an Australian mammal). Which is likely the sample from the RNA virus?

Sample Adenine Cytosine Guanine Thymine Uracil

(a) 28.0 22.0 22.0 0.0 28.0

Copyright Macmillan Learning. Powered by Cognero.

(b) 21.0 29.0 29.0 21.0 0.0

(c) 27.0 24.0 26.0 23.0 0.0

Page 4

Name:

Class:

Date:

Chapter 07: DNA: Structure and Replication a. (a) b. (b) c. (c) d. None of the answer options is correct.

ANSWER: a 19. When comparing the three key models of DNA replication, the model that included the synthesis of a brand new double-stranded DNA molecule from an original molecule was named a. dispersive replication. b. conservative replication. c. semiconservative replication. d. liberal replication. e. None of the answer options is correct.

ANSWER: b 20. When comparing the three key models of DNA replication, the model that included the separation of the two strands of the original DNA (template) and using those strands as templates to synthesize two new DNA strands is called a. dispersive replication. b. conservative replication. c. semiconservative replication. d. liberal replication. e. None of the answer options is correct.

ANSWER: c 21. The Meselson-Stahl experiment made clear predictions regarding experimental outcomes if dispersive, conservative, or semiconservative DNA replication was occurring in their Escherichia coli cells. This experiment enables the detection of "new" and "old" DNA by assessing the _______________ of DNA molecules in the cells. a. intensity of radioactive labeling b. density c. charge distribution along the double helix d. number of mutations e. size (length) ANSWER: b 22. The Meselson-Stahl experiment made clear predictions regarding experimental outcomes if dispersive, conservative, or semiconservative DNA replication was occurring in their Escherichia coli cells. To begin the experiment, cells were grown for a period of time in media containing __________ to label the chromosome. a. heavy nitrogen (15N) b. radioactive phosphate (32P) Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chapter 07: DNA: Structure and Replication c. chemically modified nitrogenous bases d. only sulfur and no nitrogen e. None of the answer options is correct.

ANSWER: a 23. If Escherichia coli, grown for a period of time in 15N, is transferred to 14N for one generation of DNA replication, the resulting DNA should have 14N added to all "new" DNA. If conservative replication is occurring, the 14N-containing "new" DNA will compose a. one strand of all bacterial chromosomes. b. both strands of DNA in half of all bacterial chromosomes. c. regions dispersed throughout all bacterial DNA. d. none of the new DNA; it will only be found in the old DNA. e. All of the answer options are correct.

ANSWER: b 24. If Escherichia coli, grown for a period of time in 15N, is transferred to 14N for one generation of DNA replication, the resulting DNA should have 14N added to all "new" DNA. If semiconservative replication is occurring, the 14N-containing "new" DNA will compose a. one strand of all bacterial chromosomes. b. both strands of DNA in half of all bacterial chromosomes. c. regions dispersed throughout all bacterial DNA. d. none of the new DNA; it will be found in the old DNA. e. None of the answer options is correct.

ANSWER: a 25. Linus Pauling proposed at one time that the DNA molecule was, in fact, a triple helix. If so, how would this situation affect cellular division? a. The triple-helix DNA could be replicated and passed on to daughter cells as it normally would. b. The triple-helix DNA could be replicated normally in mitosis but not in meiosis. c. Sperm and eggs produced by triple-helix DNA would be more fertile. d. The fact that the DNA had three nucleotide strands would make it difficult to replicate the DNA and reassemble the copies.

ANSWER: d 26. Examine Figure 7-16, recalling that DNA synthesis by DNA polymerases always occurs in the 5'-to-3' direction. The predominant location of small Okazaki fragments during DNA replication occurs at the a. replication fork. b. leading strand of replication. c. lagging strand of replication. d. site of primer synthesis by RNA primase.

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chapter 07: DNA: Structure and Replication e. major groove.

ANSWER: c 27. The replisome contains a protein subunit responsible for unwinding the double helix to enable DNA replication. This subunit/enzyme is named a. pol III holoenzyme. b. the beta clamp. c. primase. d. ligase. e. helicase.

ANSWER: e 28. The replisome contains a protein subunit responsible for attaching free ends of DNA on the newly formed strand. This subunit/enzyme is named a. pol III holoenzyme. b. the beta clamp. c. primase. d. ligase. e. helicase.

ANSWER: d 29. Topoisomerase and helicase have distinct functions that include which of the following? a. Topoisomerase is responsible for unwinding the double helix (separating strands). b. Helicase is responsible for unwinding the double helix (separating strands). c. Helicase relieves supercoiling that occurs in front of the replication fork. d. The enzymes have nearly identical activities but are located at different sites during DNA replication. e. These enzymes are nearly identical but have been given distinct names.

ANSWER: b 30. The complexity of lagging strand replication is necessary because a. there is room for only a single DNA polymerase III enzyme in the replisome. b. the helicase can only unwind double-helical DNA slowly. c. as polymerization occurs only in the 5'-to-3' direction, the lagging strand must be synthesized in consecutive small fragments. d. the RNA primase works better on the leading strand DNA. e. DNA polymerase I is more often associated with the lagging strand.

ANSWER: c 31. Initiation of replication occurs at an "origin of replication" site that typically includes an AT-rich region.

Initiation benefits from these AT-rich regions because a. adenine-thymine pairs are held together by two H-bonds, making them easier to separate during unwinding.

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chapter 07: DNA: Structure and Replication b. the AT-rich region recruits DNA polymerase to begin the process of DNA replication. c. the AT-rich region recruits topoisomerase to begin the process of DNA replication. d. GC-rich regions are impossible to replicate because of their strong H-bonds. e. None of the answer options is correct.

ANSWER: a 32. Experiments on chromosome structure and function have shown that, in eukaryotic chromatids, a. there are at least 50 DNA molecules per chromatid. b. there is more than one DNA replicating point per chromatid. c. DNA molecules duplicate conservatively rather than semiconservatively. d. DNA segregation to daughter chromatids occurs in a dispersive pattern. e. most DNA synthesis occurs during the M (mitosis) phase of the cell cycle.

ANSWER: b 33. In eukaryotic DNA replication, re-association of histones with newly formed DNA is accomplished by a. spontaneous association of histones with DNA. b. chromatin assembly factor (CAF-1) attaching to the primase enzyme. c. histones only associating with new DNA after the chromosome is completely replicated. d. histones not releasing DNA during replication. e. chromatin assembly factor 1 (CAF-1) and histones binding to the sliding clamp structure.

ANSWER: e 34. Cell-cycle progress enables the initiation of genome replication by a. activating DNA polymerase III activity directly. b. providing Cdc6 and Cdt1, which aid assembly of initiation components at the origin of replication. c. activating histone degradation near the origin of replication. d. activating expression of the ORC protein and helicase proteins. e. All of the answer options are correct.

ANSWER: b 35. When replicating the end of a chromosome, the lagging strand cannot copy the last ~10 nucleotides at the end of the chromosome. As a result, chromosomes contain telomere sequences at their ends, which are defined as a. special DNA sequences that do not require priming for replication. b. noncoding, repetitive sequences that can be copied independent of the replisome. c. noncoding, repetitive sequences that are not replicated during DNA replication. d. protein-based structures at the ends of chromosomes that protect the chromosome end. e. All of the answer options are correct.

ANSWER: b 36. Telomerase activity relies on ________________ for appropriate priming.

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chapter 07: DNA: Structure and Replication a. RNA primase b. a short, telomeric RNA sequence that is carried within its structure c. a primosome subunit associated with the telomere structure d. a unique protein-based primer e. DNA polymerase III holoenzyme ANSWER: b 37. Which of the following conditions is/are associated with abnormal telomerase activity resulting in premature aging? a. Down syndrome b. Patau syndrome c. Cri-du chat syndrome d. Werner syndrome ANSWER: d 38. Primase and telomerase enzymes are both considered types of a. DNA polymerases. b. retroviruses. c. transcription factors. d. reverse transcriptases.

ANSWER: d 39. The complex of proteins that coordinates reactions necessary for replication of DNA is the a. nucleosome. b. replication complex. c. polymerase. d. holoenzyme. e. replisome.

ANSWER: e 40. The complex of proteins that coordinates reactions necessary for replication of DNA is the a. G1 b. G2 c. S d. DNA is replicated in all cell cycles.

ANSWER: c 41. Why does DNA with a high G + C content require higher temperatures to melt? a. G–C base pairs have three hydrogen bonds. b. The pyrimidines in G–C stabilize the base pairs. c. The purines in G–C stabilize the base pairs.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chapter 07: DNA: Structure and Replication d. G–C base pairs will form major grooves preventing DNA unwinding. e. G–C base pairs form more minor grooves so DNA doesn’t unwind.

ANSWER: a Essay 42. Imagine it is the early 1900s and the nature of genetic material is not yet known. You believe that proteins (polypeptides) are the most likely candidates for storing heritable information, rather than DNA (which you regard as some kind of macromolecule used to store hydrocarbons). Make an argument supporting this position.

ANSWER: DNA and protein are found in chromosomes. Therefore, one of them must contain genetic information (as determined cytogenetically). At first blush, DNA seems too simple of a choice, only four bases from which to choose. Proteins have 20 amino acids from which to choose (though at the time, fewer than this number were known). Early methods of DNA extraction led researchers to believe DNA was a small, disorganized biomolecule.

43. Erwin Chargaff provided experimental data (Chargaff's rules) that helped Watson and Crick to gain important insight into the structure of DNA. What did Chargaff determine, and what did it mean to Watson?

ANSWER: Chargaff found that bases extracted from the nuclei of various organisms had different ratios between the bases, but in every case, there was a 1:1 ratio between guanine and cytosine and a 1:1 ratio between thymine and adenine. This proportionality suggested to Watson that there is some kind of link between A and T, and also between G and C. This turned out to be hydrogen-bond associations, and once he determined the ionization form of the bases in biological systems, he deduced that the double helix was constructed of complimentary molecules.

44. A-form DNA has very little external exposure of its nucleotide bases as compared to the B-form. B-DNA is the biologically significant form. Why do you think this is so?

ANSWER: Looking at the diagram, you can see that the major groove in A-form DNA is very narrow and deep.

B-form, however, is more open and provides easier access to the center of the DNA double helix.

Thus, special sequences inside are visible to the nuclear machinery.

Inside the nucleus, the order of nucleotides are important not only for constructing RNA containing essential information but also for most housekeeping functions, as well as identification of the genes themselves.

DNA-binding proteins are usually sequence specific; they identify certain nucleotide identities in order to bind. The origin of replication, for example, is a specific sequence that allows assembly of many proteins that initiate replication at that particular spot.

45. Imagine it is the year 2050. Because the Martian landers discovered that liquid water was present on Mars, another probe was sent to look for signs of life. Cells containing double-stranded DNA were found, and using an automated variant of the Meselson–Stahl experiment, the original DNA was labeled with 14N and then transferred to 15N (note that this is the reverse of the original M–S experiment). In the low temperatures of Mars, the replication machinery works more slowly, so each cell division takes 60 hours.

Assume you do not know if replication machinery uses a dispersive, conservative, or semiconservative approach. Predict what the centrifuge tube patterns should be at 60, 120, 180, and 240 hours.

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chapter 07: DNA: Structure and Replication ANSWER: The amount of DNA increases with time, as more "heavy" nitrogen accumulates in the molecules.

See the diagram below.

46. The antiparallel organization of DNA requires that replication be performed by a "machine of proteins."

What are the roles of proteins/enzymes associated with the leading strand? What enzymatic processes are predominantly associated with the lagging strand?

ANSWER: 1. needs to be unwound (helicase) 2. needs to have pieces remain single stranded (SSB) 3. needs to stay processive (PCNA/sliding clamp) 4. needs to have enzymes held in place 5. on lagging strand, need ligase, more primases, etc.

47. Telomeres regulate the replication of the ends of chromosomes in eukaryotes. Why is this structure implicated in human aging?

ANSWER: Telomeres are structures at the ends of eukaryotic chromosomes that contain tandem DNA sequences added to the 3' ends by the enzyme telomerase. Telomeres stabilize chromosomes by preventing the loss of genomic information after each round of DNA replication. Human somatic cells contain little or no telomerase, and those that do contain it get progressively shorter and enter premature senescence. People with diseases of premature aging, such as Werner syndrome and dyskeratosis congenital, have shorter telomeres than healthy people.

48. Telomeres have been a recent focus in experiments regarding both aging and cloning. For example, it has Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chapter 07: DNA: Structure and Replication been shown that in the absence of telomerase, the chromosome becomes shorter and shorter after each cell division. When chromosomes reach a certain length, the cell may cease to divide and die. Hence, maintaining telomerase action could be a "fountain of youth" for a cell.

Excess telomerase activity is associated with cancer cells, providing unlimited numbers of cell divisions (immortality) to these cells and presenting a dangerous threat to the organism.

Find a recent news or research article on telomeres, and discuss what the experiment tells you about the role of telomerase in the cell. Could activation of this enzyme in humans be beneficial with regard to age-related disease?

ANSWER: Responses to this question will vary depending on the article selected. This question would work well with a take-home exam or a class assignment. Telomerase activity has been the focus of aging research for over 20 years, as scientists examine the role chromosome ends play in this fascinating process. Some recent papers have reported success in extending life span by manipulating the activity of telomerase. Cancer cells often gain unlimited cell division cycles through activation of telomerase during the process of becoming cancerous. Student investigation into this topic will provide some interesting insight into numerous characteristics of genetics and chromosomes.

49. An epidemic disease affecting sheep grazing near the hot springs of Thermopolis, Wyoming, was found to be due to a new virus. This virus could infect sheep kidney cells cultured in vitro. The virus appeared to contain four chemically defined biomolecules, which we will call W, X, Y, and Z. Investigators wished to determine which of these components carried the genetic information of the virus.

An experiment was conducted in which one of the four components was radioactively labeled in different batches of kidney cells cultured and infected with the virus. Thus, four batches of labeled virus were generated (labeled W, X, Y, or Z).

Radioactive virus from each of these batches was allowed to attach to nonradioactive kidney cells. The cells were centrifuged down to remove the unattached viruses. The cells were then briefly exposed to a vigorous agitation to release viral particles on the cell surface, and again centrifuged. The supernatant (containing shaved-off viral parts) and the pelleted infected cells were examined for radioactivity, with the following results: % Radioactivity Labeled component: Supernatant Pelleted cells

W 100 0

X 0 100

Y 0 100

Z 0 100

a. On the basis of these results, which of these components does NOT carry genetic information? Why? Other experiments revealed that one of these purified biomolecules was contaminating sheep DNA, and the two remaining components did appear to be derived from a virus. The nitrogenous base compositions were as follows: Labeled component: %A

X 21

Copyright Macmillan Learning. Powered by Cognero.

Y 26

Z 28 Page 12

Name:

Class:

Date:

Chapter 07: DNA: Structure and Replication %G %C %T %U

21 32 0 26

32 21 21 0

22 22 28 0

b. What can you say about the type of nucleic acid found in each of these three components? c. Can you determine which of these purified biomolecules is most likely sheep DNA and which are more likely to belong to the virus? Explain your response.

ANSWER: a. As biomolecule W was found in the supernatant and not within the infected cells, it is most likely not carrying genetic information within its structure. The data are consistent with W being some type of structural component of the virus outer coat. b. Component X is RNA, as it contains uracil rather than thymine as a nitrogenous base, while Y and Z contain nitrogenous bases consistent with DNA. c. Sheep DNA is double-stranded, and thus should be consistent with Chargaff's rules (A=T, G=C).

The biomolecule Z appears to be double-stranded DNA, while Y is most likely single-stranded DNA and potentially a viral genome.

50. The number of proteins in the bacterial replisome is 13, while the number in a eukaryotic cell is 27. What are important protein components of the replisome in general? Why might a eukaryotic cell have more components?

ANSWER: Important components of the replisome are: Polymerase III holoenzyme: responsible for the addition of nucleotides.

Sliding clamp: keeps pol III attached to the DNA molecule.

Primase: enzyme that synthesizes the RNA primer.

Helicase: disrupts the hydrogen bonds that hold the two strands of the double helix together.

Topoiomerase: relaxes supercoiled DNA by breaking a single DNA strand or both strands, allowing DNA to rotate into a relaxed molecule.

Single-stranded binding protein: binds to single-stranded DNA and prevents the duplex from reforming.

One reason for the added complexity of the eukaryotic replisome is the higher complexity of the eukaryotic template. Unlike the bacterial chromosome, eukaryotic chromosomes exist in the nucleus as chromatin; thus, the replisome has to not only copy the parental strands but also disassemble the nucleosomes in the parental strands and reassemble them in daughter molecules.

51. Describe the structural components of a nucleotide.

ANSWER: Nucleotides are comprised of a phosphate, sugar, and nitrogenous bases. The bases adenine and guanine are purines with a double ring structure. The bases cytosine and thymine are pyrimidines with a single ring structure. The deoxyribose sugar is similar to ribose; however, deoxyribose has a hydrogen atom at the 2′ position and ribose has a hydroxyl group.

52. What overarching principles about hereditary material guided Watson and Crick's determination of DNA structure?

ANSWER: (1) Structural features of DNA must support accurate replication of genetic material. (2) Structural Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chapter 07: DNA: Structure and Replication features of DNA must have informational content for protein production. (3) Genetic material must be able to change to enable evolutionary selection.

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chapter 08: RNA: Transcription, Processing, and Decay Multiple Choice 1. Which of the following is/are TRUE for RNA compared to DNA? a. RNA has ribose sugar in its nucleotides, rather than the deoxyribose found in DNA. b. RNA is usually single stranded and can make more complex three-dimensional molecular shapes than double-stranded DNA. c. RNA contains the bases A, G, C, and U, whereas DNA contain the bases A, G, C, and T. d. RNA can catalyze biological reactions, but DNA cannot. e. All of the answer options are correct.

ANSWER: e 2. The spliceosome functions to a. insert introns into mRNA sequences before translation. b. create noncoding sequences in genes to enhance gene stability. c. remove noncoding introns from transcribed RNA. d. control translation, ensuring that only exons are translated. e. inhibit transcription of noncoding DNA regions.

ANSWER: c 3. Template strand DNA and encoded RNA are a. complementary of one another with antiparallel orientation. b. complementary of one another but share the same 5′-to-3′ orientation. c. identical sequences with the exception of U substituted for T. d. different with regard to the inclusion of introns. e. interconverted via nucleic acid remodeling.

ANSWER: a 4. In what cellular compartment are introns removed from pre-mRNA to make mature mRNA? a. cytoplasm b. endoplasmic reticulum c. nucleus d. mitochondria e. Golgi apparatus ANSWER: c 5. The experimental value of a "pulse–chase" cellular labeling experiment using radioactive uracil is that a. all RNAs within a cell are labeled and easily detected. b. the radioactivity allows for easy purification of cellular nucleic acids. c. all RNAs created during the pulse are labeled and detectable, revealing their stability and localization. d. the spliceosome function is easily seen with radiolabeled RNAs in the cellular cytosol.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chapter 08: RNA: Transcription, Processing, and Decay e. cells adapt to the pulse of radioactivity, revealing nucleic acid repair mechanisms.

ANSWER: c 6. Which of the following is evidence that RNA was a message-carrying intermediary between DNA and protein? a. A hydroxyl group is present on ribose. b. RNA is single stranded and thus cannot be copied by semiconservative replication in a manner similar to DNA. c. RNA structure includes a molecular code, proving that it carries a genetic message. d. RNA is produced in the nucleus (with DNA) and then migrates to the cytoplasm, the location of protein synthesis. e. Pulse–chase experiments revealed RNA to be exclusively localized to the nucleus.

ANSWER: d 7. Pulse–chase experiments provided evidence that a. RNA is made of ribose sugars rather than deoxyribose. b. RNA is an information-transfer intermediary between DNA and protein. c. many viruses have RNA genomes. d. RNA is capable of catalyzing biological reactions. e. RNA is synthesized in short pulses of transcriptional activity.

ANSWER: b 8. If the DNA template 5′-ATGCATGC-3′ were transcribed to RNA, the RNA would read a. 3′ TACGTACG 5′ b. 5′ AUGCAUGC 3′ c. 5′ UACGUACG 5′ d. 3′ UACGUACG 5′ e. 5′ ATGCATGC 3′ ANSWER: d 9. The spliceosome includes both protein and a functional type of RNA known as a. transfer RNA (tRNA). b. small nuclear RNAs (snRNAs). c. micro RNAs (miRNAs). d. small interfering RNAs (siRNAs). e. ribosomal RNAs (rRNAs).

ANSWER: b 10. Which of the following mRNA codons would form a codon-anticodon base-pairing interaction with the 3′UAG-5′ tRNA anticodon? a. 3′-ATC-5′ Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chapter 08: RNA: Transcription, Processing, and Decay b. 5′-GAU-3′ c. 5′-ATC-3′ d. 3′-AUC-5′ e. 5′-AUC-3′ ANSWER: e 11. RNA synthesis is always 5′ to 3′ because a. the unwinding of the double-stranded DNA can only move one direction. b. nucleotides can only be added to an available 3′-OH group on the transcript terminus. c. nitrogenous bases cannot pair up in the 3′-to-5′ direction. d. the structure of ATP restricts 3′-to-5′ polymerization into RNA. e. RNA synthesis can move in the 3′-to-5′ direction.

ANSWER: b 12. The role of tRNA is to a. serve as an intermediate in the decoding of genes. b. act as transporters bringing amino acids to the site of protein synthesis. c. serve as general translational components of the ribosome. d. facilitate splicing of pre-messenger RNAs. e. facilitate protein trafficking in protein secretion.

ANSWER: b 13. In a chromosome, which of the following is TRUE? a. RNAs of different genes can be transcribed off either DNA strand, but always 5′ to 3′. b. RNAs of different genes can be transcribed off either DNA strand, but always 3′ to 5′. c. The RNAs of all genes are synthesized 5′ to 3′ off the same DNA strand. d. The RNAs of all genes are synthesized 3′ to 5′ off the same DNA strand. e. Different genes can be transcribed off either strand, some in the 5′-to-3′ direction and some in the 3′to-5′ direction.

ANSWER: a 14. Which of the following acts before the others? a. tRNA alignment with mRNA b. aminoacyl-tRNA synthetase c. RNA polymerase d. ribosome movement to the next codon e. amino acid chain elongation ANSWER: c 15. RNA polymerase can transcribe in different directions on a chromosome. This is enabled because a. genes and their associated promoters can be oriented in either direction along a chromosome.

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chapter 08: RNA: Transcription, Processing, and Decay b. RNA polymerase can transcribe from either strand of DNA, synthesizing RNA in the 5′-to-3′ direction. c. the chemistry of RNA synthesis is identical regardless of which strand is transcribed. d. the 3′ end of a gene is always the template strand. e. All of the answer options are correct.

ANSWER: e 16. The -10 and -35 boxes found in most bacterial promoter elements were discovered by a. comparing the DNA sequences in front of highly expressed genes and determining nucleic acids that were common (consensus). b. the finding that RNA polymerase begins translation when a 10-nucleotide followed by a 35nucleotide repeat of guanine ("G") are encountered. c. comparing bacterial promoters to known eukaryotic promoters. d. examining data from the human genome project. e. Robert Koch as he characterized the microbe that causes tuberculosis.

ANSWER: a 17. The -10 and the -35 boxes found in bacterial gene promoters function to a. recruit DNA polymerase activity for DNA replication before cell division. b. activate transcription at the -35 region of a gene. c. orient RNA polymerase at a gene's transcription start site. d. identify the boundaries between introns and exons. e. identify the boundaries for DNA unwinding during transcription.

ANSWER: c 18. The sigma factor protein's role in transcription in E. coli includes which of the following? a. forms part of the core enzyme required for transcription initiation b. helps the holoenzyme to bind to the promoter c. contributes to the proof-reading activity of RNA polymerase d. plays a role in transcription termination e. All of the answer options are correct.

ANSWER: b 19. Why does E. coli have several different sigma factors? a. They allow different RNA polymerases to bind to the promoters. b. They allow the different subunits of the RNA polymerase holoenzyme to bind to each other. c. There is no good reason. They all perform the same function. d. One is needed to transcribe mRNA. A second is needed to transcribe tRNA. And a third is needed to transcribe rRNA. e. They allow RNA polymerase to recognize and bind to a different subset of promoters.

ANSWER: e Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chapter 08: RNA: Transcription, Processing, and Decay 20. Why is transcription in eukaryotes more complex than in prokaryotes? a. Compared to a few thousand genes in prokaryotes, eukaryotes have larger genomes and tens of thousands of genes to be transcribed. b. Transcription and translation takes place in the same cellular compartment in prokaryotes but in different compartments in eukaryotes. c. The template for transcription in prokaryotes is simple and exists as naked DNA, whereas the eukaryotic template is organized into a highly condensed chromatin structure. d. Eukaryotes have larger genomes and more genes than prokaryotes, and their genomes are organized into a highly condensed chromatin structure. e. All of the answer options are correct.

ANSWER: d 21. Which of the following statement/s about RNA pol II is/are FALSE? a. RNA pol II transcribes protein-encoding genes. b. General transcription factors (GTFs) help RNA pol II to position at the promoter start to initiate transcription. c. RNA pol II adds new nucleotides to the 3′ end of the transcript. d. The CTD domain of RNA pol II coordinates cotranscriptional modifications. e. RNA pol II reads the template DNA strand in a 5′-to-3′ direction during transcription.

ANSWER: d 22. In eukaryotic species, three separate RNA polymerases transcribe different categories of genes. RNA polymerase III is known to primarily transcribe a. messenger RNAs. b. protein-encoding genes. c. most ribosomal RNAs. d. transfer RNAs and small nuclear RNAs. e. genes responsible for DNA replication.

ANSWER: d 23. In eukaryotic species, three separate RNA polymerases transcribe different categories of genes. RNA polymerase I is known to primarily transcribe a. messenger RNAs. b. protein-encoding genes. c. most ribosomal RNAs. d. transfer RNAs and small nuclear RNAs. e. genes responsible for DNA replication.

ANSWER: c 24. A key characteristic of bacterial RNAs that is NOT observed with eukaryotic RNAs is that a. transcription generates RNAs that contain both introns and exons.

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chapter 08: RNA: Transcription, Processing, and Decay b. transcription can occur in the same cellular region as translation. c. a prerequisite for translation initiation is RNA processing. d. the genetic code used by bacteria is different from other organisms. e. bacterial RNAs are generated as double stranded.

ANSWER: b 25. For protein-encoding genes, six general transcription factors (GTFs) function to a. identify a gene's promoter, facilitating RNA polymerase II binding. b. terminate transcription at the end of an open reading frame. c. assemble the RNA polymerase subunits. d. arrest DNA replication. e. recruit the ribosome to newly synthesized RNAs. ANSWER: a 26. The carboxy-terminal domain of RNA polymerase II plays a key role in a. capping of the 5′ end of a new transcript. b. recruiting capping enzymes to the RNA polymerase enzyme. c. addition of a poly(A) tail at the 3′ end of a transcript. d. splicing of introns out of RNA transcripts. e. All of the answer options are correct.

ANSWER: e 27. What is the function of the TATA-binding protein? a. aids in the removal of introns from eukaryotic pre-mRNA b. allows prokaryotic RNA polymerase to bind to the promoter of genes c. allows eukaryotic RNA polymerase II to bind to the promoter of genes d. helps termination factors bind and terminate transcription. e. All of the answer options are correct.

ANSWER: c 28. The functional equivalent of the TATA-binding protein in prokaryotes is a. a sigma subunit. b. a holoenzyme. c. the Rho factor. d. TFIID. e. hairpin loops.

ANSWER: a 29. Which of the following is/are role(s) of the 5' cap? a. The cap helps the RNA polymerase find the promoter and initiate transcription. b. The cap plays a role in the removal of introns.

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chapter 08: RNA: Transcription, Processing, and Decay c. The cap acts as a binding site for the ribosome. d. The cap protects the RNA from degradation. e. The cap acts as a binding site for the ribosome and protects the RNA from degradation.

ANSWER: e 30. The initial discovery of the spliceosome resulted from experiments focused on a. autoimmunity in Lupus patients. b. the isolation of large RNA/protein complexes from cells. c. careful analysis of electron microscopy images from the cell nucleus. d. RNAs from plant cells. e. the cell biology of Huntington's disease.

ANSWER: a 31. The conserved adenine nucleotide in introns serves as the a. site for intron recognition for the spliceosome. b. branch point for the formation of the intronic "lariat." c. key point for ribosome assembly and initiation of translation. d. initial site of intronic RNA digestion and removal. e. molecular signal for RNA splicing.

ANSWER: b 32. Following splicing, the conserved adenosine nucleotide within the intron displays an unusual array of bonds, including a. a phosphate attached to the 4-carbon of the adenine nitrogenous base. b. phosphodiester bonds attached at three places on the ribose (2′, 3′, 5′). c. additional methyl (–CH3) groups added to the adenine. d. a DNA strand attached to the spliced RNA. e. attachment to an arginine amino acid.

ANSWER: b 33. The role of most microRNAs within a eukaryotic cell is to a. regulate the splicing of primary transcripts to mRNAs. b. bind with other RNAs to stabilize their secondary structure. c. associate with ribosomal proteins to facilitate translation. d. regulate RNA polymerase activity in the nucleus. e. repress the expression of genes by destroying mRNAs. ANSWER: e 34. siRNAs are distinct from microRNAs (miRNAs) in that siRNAs a. silence their own expression. b. are smaller than microRNAs. Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chapter 08: RNA: Transcription, Processing, and Decay c. are derived from rRNA sequences. d. silence gene expression. e. are made of double-stranded DNA. ANSWER: a 35. How is the half-life of an mRNA experimentally determined? a. by inhibiting all RNA polymerase I transcription, before measuring how long it takes for half of the existing mRNA molecules to be degraded b. by activating RNA polymerase II transcription, and measuring how long it takes for half of the existing mRNA molecules to be degraded c. by activating RNA polymerase I transcription, and measuring how long it takes for all of the existing mRNA molecules to be degraded d. by inhibiting all RNA polymerase II transcription, and then measuring how long it takes for all of the existing mRNA molecules to be degraded e. by inhibiting all RNA polymerase II transcription, before measuring how long it takes for half of the existing mRNA molecules to be degraded ANSWER: a 36. mRNA decay typically occurs in the a. cytoplasm. b. nucleus. c. mitochondria. d. chromatin. e. ribosome.

ANSWER: a Subjective Short Answer 37. The following table contains a list of statements that may apply to mRNA, tRNA, snRNA, or rRNA. For each class of RNA molecule, put a checkmark if the statement applies to that type of RNA. mRNA

tRNA

snRNA

rRNA

mRNA

tRNA

snRNA

Contains a 5UTR Translated into protein Contains an anticodon Classified as functional RNA Transcribed in the 5-to-3′ direction Major component of ribosomes Component of spliceosomes Transcribed by RNA polymerase II in eukaryotes Have a functional role in translation ANSWER: Copyright Macmillan Learning. Powered by Cognero.

rRNA Page 8

Name:

Class:

Date:

Chapter 08: RNA: Transcription, Processing, and Decay Contains a 5UTR Translated into protein Contains an anticodon Classified as functional RNA Transcribed in the 5-to-3′ direction Major component of ribosomes Component of spliceosomes Transcribed by RNA polymerase II in eukaryotes Have a functional role in translation

√ √ √ √ √

√

√ (√)

√ √

√ √ √

√ (√) √

√

38. The following table contains a list of statements that apply to replication, transcription, both, or neither. In each empty box, put a checkmark if that statement applies to replication or transcription.

Replication

Transcription

The new strand is made 5to 3.

The new strand is made 3to 5.

The new strand is identical to the template strand, with the exception of U replacing T. The new strand is complementary to the template strand.

The template strand is RNA. The product is DNA. The product is RNA. An RNA primer is required to initiate synthesis.

Synthesis of the new strand is initiated at a promoter.

The process is done only during the S phase of the cell cycle.

The process is done only during the G1 phase of the cell cycle.

ANSWER:

Replication √

The new strand is made 5to 3.

The new strand is made 3to 5.

The new strand is identical to the template strand, with the exception of U replacing T. The new strand is complementary to the template strand. √ The template strand is RNA. The product is DNA. √ The product is RNA. An RNA primer is required to initiate synthesis. √ Synthesis of the new strand is initiated at a promoter.

The process is done only during the S phase of the cell √ cycle.

Copyright Macmillan Learning. Powered by Cognero.

Transcription √

√

√ √

Page 9

Name:

Class:

Date:

Chapter 08: RNA: Transcription, Processing, and Decay The process is done only during the G1 phase of the cell cycle.

Essay Shown below is the structure of a Drosophila gene, divided into 10 segments designated A–J. The gene contains three exons, two introns, a promoter, and a site in I for poly(A) addition.

39. What segment or segments of the gene will be represented in the initial RNA transcript? List the appropriate letter or letters.

ANSWER: D, E, F, G, H, I 40. What segment or segments of the gene will be found in the completely processed transcript?

ANSWER: D, F, H, I 41. What segment or segments of the gene in the processed transcript will have additional nucleotides added to them?

ANSWER: D, I, (C – 5′ untranslated region) 42. What segment or segments of the gene contain the CCAAT box and TATA box elements?

ANSWER: C, (B) 43. What segment or segments of the gene will possess the translation initiation codon?

ANSWER: D 44. In the silk moth, a single gene encodes the protein that is the major constituent of the cocoon. a) The mRNA encoding this protein was isolated from silk moth larvae and hybridized with silk moth DNA. The following "electron micrograph" was obtained.

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chapter 08: RNA: Transcription, Processing, and Decay

(1) Explain the loops in the DNA strand.

(2) Show on the diagram which side of the mRNA is its 5′ end and which is its 3′ end. Explain. b) A mutant silk moth was isolated that cannot make the cocoon protein. The mutant makes mRNA encoding the protein at the same concentration as the wild type, but the mRNA molecules are slightly longer than usual.

In addition, an electron micrograph of the mRNA-DNA hybridization with mRNA from the mutant organism shows the following:

Suggest a possible explanation for the nature of the mutation.

ANSWER: a) (1) The loops are introns.

(2)

b) A mutation in the second intron prevents splicing, and no functional protein is made. Evidence is Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chapter 08: RNA: Transcription, Processing, and Decay that the mRNA and the DNA hybridize over the exon 2 region without looping. No intron removal has occurred at that site.

45. Consider the following piece of messenger RNA: 5′-AUGGAGUCGUUAAUUAAACCGGUGCGGAUCGUAUUUAGUCCCCAC-3′ a) Draw both strands of the segment of DNA from which this mRNA was transcribed. b) State which of the DNA strands the RNA synthesizing enzyme used as template for the transcription process.

ANSWER: a) 3-TACCTC............GGGGTG-5 5-ATGGAG............CCCCAC-3 b) the top strand 3 to 5 46. Pulse–chase experiments were vital for showing that RNA acts as an information-transfer intermediary between DNA and protein. Describe the basic steps of a pulse–chase experiment, and explain in your own words how the results suggested that RNA is the intermediate.

ANSWER: Cells are pulsed with radioactive uracil (pulse) and then moved to medium containing unlabeled uracil (chase). When this type of experiment is done, it is observed that the label is initially localized to the nucleus but moves into the cytoplasm with time. At the time of the experiment, it was known that DNA is located in the nucleus and proteins are made in the cytoplasm, so there must be some sort of messenger that moves from the nucleus to the cytoplasm. Thus, these experiments provided evidence that RNA is that messenger.

47. rRNA and tRNA exist in both eukaryotes and prokaryotes. However, snRNAs exist only in one of these types of organisms. Do snRNAs exist in prokaryotes or eukaryotes? Explain why they are found only in that type of organism.

ANSWER: Eukaryotes. snRNAs play a role in processing pre-mRNA. Because prokaryotes do not process mRNA, they do not have any of the snRNAs required to do such processing.

48. A segment of DNA from Drosophila melanogaster has the sequence: 5′ TCA AGC TTA AGA AGG CAT TTT 3′ a) Assuming that this is the template strand, what is the sequence of the encoded mRNA (make sure to denote 5′ and 3′ ends of the transcript)? b) If this sequence is the coding strand, what is the sequence of the encoded mRNA? ANSWER: a) 3′ AGU UCG AAU UCU UCC GUA AAA 5′ b) 5′ UCA AGC UUA AGA AGG CAU UUU 3′ 49. Suppose that you have sequenced five genes, including the putative promoter region of each gene, that are involved in the cold acclimation process in Arabidopsis. Each of the five genes is expressed only when the temperature falls below a certain temperature. You know that a specific transcription factor (CBF1) binds somewhere between –200 and –250 of these genes to initiate transcription. How could you use the sequences to determine the DNA sequence motif that is bound by CBF1?

ANSWER: Align the five sequences and look for a consensus sequence. This could be the desired sequence. To further test this hypothesis, mutate the sequence and see if the mutation prevents CBF1 from binding to it and initiating transcription under the inducing condition (cold temperature).

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chapter 08: RNA: Transcription, Processing, and Decay 50. The following sequence was identified at the 3′ end of a transcribed RNA in prokaryotes. a) What is the name of the structure shown below? b) What is its function in transcription? c) What is the probable base composition of the region marked by the A arrow? d) What is the probable base composition of the region marked by the B arrow? e) In the original gene encoding this RNA sequence, the B region was mutated to GGCCGGTGC. How might this mutation of B affect transcription of this gene?

ANSWER: a) hairpin loop b) terminates transcription in prokaryotes c) G/C rich d) string of U e) Transcription might not be terminated because there will be a bunch of triple bonds rather than double bonds between the template DNA and the RNA transcript. Normally, this region contains poly(A) on the DNA and poly(U) on the transcript. The enriched G/C interactions could perturb transcription termination by stabilizing interaction between the DNA and RNA, allowing transcription to continue.

51. Name the three major modifications of mRNA in eukaryotes before it is transported to the cytoplasm. In addition, tell why each of the modifications is necessary.

ANSWER: 1) Add 5′ cap to protect against RNA degradation and to act as a binding site for the initiation of translation.

2) Add poly(A) tail to the 3′ end of the transcript to protect against degradation and aid in the transfer of the mRNA to the cytoplasm.

3) Remove introns to generate an intact open reading frame for translation.

52. A major difference between eukaryotes and prokaryotes is that eukaryotes have a nucleus, whereas prokaryotes do not. Discuss the impact of having a nucleus on the creation of mature mRNA. ANSWER: Because prokaryotes do not have a nucleus, the translational machinery is able to attach to the mRNA as it is being synthesized. Thus, there is no time to process the mRNA. In eukaryotes, the translational machinery is not present in the nucleus. Thus, there is time for the cell to process the mRNA before it is transported from the nucleus to the cytoplasm and translated into protein.

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chapter 08: RNA: Transcription, Processing, and Decay 53. You generated a eukaryotic cell line that expresses an RNA polymerase II subunit that is missing the repeated sequences (seven amino acids per repeat) near the C-terminus. These repeated sequences are known to be important in the regulation of gene expression in eukaryotes. Define how expression of this RNA polymerase might affect the cell line. Define how these repeated sequences are regulated within a normal RNA polymerase II enzyme.

ANSWER: The C-terminal domain region contains repeats of a sequence of seven amino acids that serves as an assembly site for capping enzymes (5′ cap addition to RNA), for splicing machinery, and for polyadenylation enzymes—poly(A) added to the 3′ end of the transcript. The distinct sets of machinery are recruited by differential phosphorylation of specific amino acids within the seven amino acid repeat sequence. These are key processes in eukaryotic gene expression. A cell line expressing a defective RNA polymerase II lacking these sequences would generate improperly processed RNAs. 54. What is the function of RNAi in plants?

ANSWER: RNAi serves as an antiviral defense system in plants.

55. Describe how researchers have taken advantage of endogenous RNAi. ANSWER: Researchers use RNAi to study the specific function of a gene. They are able to silence the expression of a specific gene by introducing into cells dsRNA that is identical in sequence to the target gene. This results in the knockdown of the expression of that gene, and the ability to analyze the phenotype of an organism/cell in the absence of the gene of interest.

56. You have recently discovered a new species of fungi. How would you perform a genome-wide screen of the genes involved in cellular processes?

ANSWER: By using dsRNA to selectively silence genes, libraries of dsRNAs that target each of the proteincoding genes in the fungus can be generated. The genes involved in cellular processes would produce an altered phenotype when in injected with dsRNAs.

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chapter 09: Proteins and Their Synthesis Multiple Choice 1. Formation of the peptide bond could be described as a. a spontaneous reaction between two peptides, forming a covalent bond. b. a ribosome-catalyzed process occurring in the nucleus. c. a tRNA catalyzed process. d. a ribosome-catalyzed dehydration reaction (H2O released). e. transcription, a key process in gene expression.

ANSWER: d 2. The three-dimensional structure of an active hemoglobin (heterotetramer) structure would be referred to as its __________ structure. a. primary b. secondary c. tertiary d. quaternary e. quinternary ANSWER: d 3. Fibrous proteins contain a structure that is a. rounded and compact. b. globular. c. only comprised of secondary structure. d. linear/extended. e. random and variable.

ANSWER: d 4. Peptide bonds form between amino acids by the removal of a. nitrogen. b. a carboxyl group. c. an amino group. d. carbon. e. water.

ANSWER: e 5. Individual polypeptides in complexes are called a. primary structures. b. domains. c. subunits. d. globular proteins. e. α-helices.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chapter 09: Proteins and Their Synthesis ANSWER: c 6. The alpha-helix structure is stabilized by a. covalent bonds between adjacent functional groups. b. hydrogen bonds along the peptide backbone. c. association with helicase stabilizing enzymes. d. noncovalent interactions with water. e. cationic elements found in water solutions.

ANSWER: b 7. Which of the following constitutes the primary structure of a protein? a. the folding of a polypeptide chain b. the linear sequence of amino acids in a polypeptide chain c. the polypeptide chains stacked on top of each other d. a pleated sheet e. several polypeptide subunits ANSWER: b 8. How many nucleotides would be expected for a gene coding for a protein with 300 amino acids? a. 300 b. 100 c. 600 d. 1200 e. 900 ANSWER: e 9. A biochemical pathway making flower pigments shows the following sequential color conversions (each arrow represents an enzyme-catalyzed step).

Colorless  A

yellow  B

blue  red C

A plant is homozygous for mutations in enzyme "B," inactivating the active site for this enzyme. The resulting flowers will be a. orange. b. blue. c. yellow. d. red. e. white.

ANSWER: c 10. While one gene usually specifies one enzyme, which of the following is/are NOT true?

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chapter 09: Proteins and Their Synthesis a. One gene can specify a single enzyme if that enzyme contains a single type of polypeptide chain. b. One gene can specify parts of two enzymes if the enzymes have one type of polypeptide chain in common. c. One enzyme may be determined by two genes if the enzyme has two different types of polypeptide chains. d. One mutation may cause multiple nutritional requirements if it interferes with the synthesis of an intermediate common to several pathways. e. Two mutations in the same gene can never have different effects, even when studied in great biochemical detail.

ANSWER: e 11. In the 1940s, George Beadle and Edward Tatum developed the one-gene–one-polypeptide hypothesis by mutating the genes of Neurospora crassa (bread mold). They identified mutants that were unable to synthesize amino acids, such as arginine. When multiple mutant strains were identified for a single trait (adenine synthesis deficient), they showed that a. all the mutants encoded the same single enzyme. b. genes were transcribed before being translated. c. each step in the biochemical pathway was mediated by an enzyme encoded by a single gene. d. the biosynthesis of arginine could occur, even with mutated enzymes. e. separate genes/enzymes performed identical steps in the pathway.

ANSWER: c 12. We know that DNA and RNA (each with four nucleotide components) both use a three-nucleotide genetic code and 64 codons (43 = 64). Knowing that the minimum number of codons for the genetic code is 21 (20 amino acids, 1 stop codon), what codon size would be required if only three nucleotides were present in the genome? a. 2 nucleotides per codon b. 3 nucleotides per codon c. 4 nucleotides per codon d. 5 nucleotides per codon e. 6 nucleotides per codon ANSWER: b 13. Five nutritional mutants in Neurospora were independently isolated. They all require compound F to grow.

Intermediate compounds in the biosynthesis of compound F are known and tested for their ability to support the growth of each mutant. The results are given in the table below, where "+" indicates growth and "0" indicates no growth.

Mutants 1 2

Compounds A B C D E F –––––––––––––––––––––––––––––––––––– 0 0 0 + 0 + 0 + 0 + 0 +

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chapter 09: Proteins and Their Synthesis 3 4 5

0 0 +

0 + +

0 + +

0 + +

0 0 0

+ + +

Assuming a linear pathway, what is the order of the compounds A through

F? a. EABCDF b. ECBDAF c. ECDBAF d. EACBDF e. EBACDF ANSWER: d 14. The use of proflavin-induced mutations (like FCO) in the rII genome was found to experimentally support a three-nucleotide explanation for the genetic code because a. an insertion mutation could be suppressed by a deletion mutation. b. the rII genes are easier to mutate than other genes. c. proflavin was an inexpensive mutagenic compound. d. a gene with three insertions or three deletions could sometimes provide a functional protein. e. the rII gene encoded a protein with exactly the correct number of amino acids, as would be explained by a three-nucleotide genetic code.

ANSWER: d 15. Degeneracy in the genetic code is best illustrated by a. stop codons. b. threonine codons. c. tryptophan codons. d. methionine codons. e. operator sequences.

ANSWER: b 16. In vitro (in a test tube) translation of a synthetic RNA with the repeating sequence (AGA)20 would produce which of the following? a. polypeptides containing arginine, polypeptides containing glutamic acid, and polypeptides containing lysine b. polypeptides containing arginine c. polypeptides containing a mixture of amino acids (mostly arginine, glutamic acid, and lysine) d. no translation product because the translation start site was not included on the RNA e. polypeptides starting with methionine, followed by arginine additions ANSWER: a 17. The amber mutations initially discovered by Sydney Brenner and colleagues was eventually shown to include a(n) Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chapter 09: Proteins and Their Synthesis a. unusual amino acid that generated an amber color in the affected bacteria. b. premature stop codon within the gene's open reading frame. c. insertion mutation that creates a translational frameshift. d. previously unknown signal for speeding the translation process. e. deletion mutation that creates a translational frameshift.

ANSWER: b 18. Which could be an anticodon for the amino acid isoleucine? a. UAA b. UAG c. UAU d. UAA and UAG e. UAA, UAG, and UAU ANSWER: e 19. DNA probes are often synthesized based on knowledge of the protein produced by a gene. What might be a potential problem of synthesizing DNA probes in this manner? a. The deduced DNA probe would be missing introns associated with prokaryotic genes. b. Because codons in DNA are redundant, knowing which codon to assign an amino acid may be difficult. c. The deduced DNA probe would be missing introns associated with eukaryotic genes. d. Because codons in DNA are redundant, knowing which codon to assign an amino acid may be difficult, as well as the deduced DNA probe would be missing introns associated with eukaryotic genes. e. None of the answer options is correct.

ANSWER: d 20. Francis Crick proposed "adaptor" molecules to serve as an intermediary between mRNA and the synthesis of protein. Key to any model of how RNA is "read" in the process of protein synthesis would be a consideration of how a. amino acids are selected for addition to the new protein. b. the RNA code was read by the translation enzyme. c. peptide bonds are formed. d. one codon could select a specific amino acid. e. All of the answer options are correct.

ANSWER: e 21. In Neurospora, a linear biochemical pathway synthesizes an amino acid Z. E1, E2, and E3 are enzymes that catalyze the three reactions.

E1 E2 E3 WXYZ Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chapter 09: Proteins and Their Synthesis Null mutants for the enzyme E2 gene will grow on minimal medium supplemented with compound(s) a. W or X. b. Y or Z. c. W or X or Y or Z. d. Z only. e. W only.

ANSWER: b 22. The DNA sense strand for a particular amino acid is 5′-ATG-3′. What RNA sequence would be transcribed for this codon, what tRNA anticodon would recognize it, and what amino acid would be added in response to this codon? a. 5′-CAU-3′ (RNA sense); 3′-GUA-5′ (tRNA anticodon); histidine b. 3′-AUG-5′ (RNA sense); 3′-UAC-5′ (tRNA anticodon); valine c. 5′-CAU-3′ (RNA sense); 3′-AUG-5′ (tRNA anticodon); histidine d. 5′-UUU-3′ (RNA sense); 3′-AAA-5′ (tRNA anticodon); phenylalanine e. 5′-AUG-3′ (RNA sense); 3′-UAC-5′ (tRNA anticodon); methionine ANSWER: a 23. The aminoacyl-tRNA synthetase enzymes are responsible for a. synthesizing tRNA molecules in the nucleus from rRNA genes. b. adding amino acids to appropriate tRNAs (charging the tRNA). c. regulating the process of amino acid synthesis. d. catalyzing the enzymatic step of peptide bond formation during translation. e. matching each tRNA with the appropriate mRNA codon.

ANSWER: b 24. The "wobble" base is less important than the other two nucleotides in a codon and is found a. at the 5′ end of the RNA codon's sense strand. b. at the 3′ end of the mRNA codon's antisense strand. c. at the 3′ end of the tRNA anticodon. d. at the 5′ end of the tRNA anticodon. e. within a tRNA hairpin loop structure.

ANSWER: d 25. The "wobble" base is a tRNA site where a. inosine nucleotide can sometimes be found. b. mutations often create severe mutations in the encoded protein. c. the amino acid is closely located to facilitate its addition to the new protein. d. aminoacyl-tRNAs read the tRNA specificity. e. tRNAs display structural instability.

ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chapter 09: Proteins and Their Synthesis 26. Evidence that amino acids were "illiterate," or passively added to the newly synthesized protein at the direction of the tRNA to which they are attached, was derived from experiments with a. nickel hydride, altering the amino acids attached to charged tRNAs to determine if they were still utilized by the ribosome. b. proflavin-based mutations on subsets of anticodons to determine if the mutations affect the charging of tRNAs. c. abnormal forms of the aminoacyl-tRNA transferase enzymes. d. chemically synthesized, novel amino acid structures. e. eukaryotic ribosomes combined with prokaryotic tRNAs. ANSWER: a 27. How does a nonsense suppressor mutation prevent amber mutants from terminating their polypeptides prematurely? a. The mutation turns the amber codon back into a wild-type codon. b. The mutation alters a tRNA so that it reads the amber codon and inserts an amino acid. c. The mutation alters the release factors that would halt synthesis. d. The mutation results in a wobble that allows synthesis to continue. e. The mutation replaces the amber codon with an ochre codon.

ANSWER: b 28. A tRNA with the anticodon 3′-ACC-5′ would carry the amino acid a. phenylalanine. b. tyrosine. c. serine. d. threonine. e. tryptophan.

ANSWER: e 29. The anticodon on the tRNA molecule a. binds to the mRNA in a complementary fashion. b. is oriented and written in the 5′-to-3′ direction. c. is a catalytic part of protein synthesis. d. is the same for all tRNA molecules. e. contains amino-acyl-tRNA synthetase.

ANSWER: a 30. The ribosome is the primary site of a. oxidative phosphorylation. b. protein packaging. c. protein synthesis.

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chapter 09: Proteins and Their Synthesis d. cellular respiration. e. amino acid storage.

ANSWER: c 31. All of the following are associated with ribosomes EXCEPT a. decoding center. b. peptidyltransferase center. c. exit site. d. groove for RNA binding. e. groove for DNA binding.

ANSWER: e 32. The ribosome in eukaryotes can be described as being composed of a. rRNAs and over 50 different proteins. b. a 50S subunit and a 30S subunit, forming a 70S ribosome. c. proteins in quaternary structure, binding tRNAs within the nucleus. d. a large enzyme complex that is found only on the endoplasmic reticulum membrane. e. rRNAs, protein subunits, ATP, and aminoacyl-tRNA synthetase.

ANSWER: a 33. Which of the following are types of post-translational processing that will NOT occur in a prokaryote? a. addition of a phosphate group b. removal of a phosphate group c. addition of a co-factor d. cleavage of a mitochondrial targeting sequence e. None of the answer options is correct.

ANSWER: d 34. The A site, P site, and E site each control ______________ (in order) during translation. a. binding incoming tRNAs (A), retention of the peptide chain during elongation (P), exit of deacylated tRNAs (E) b. activation of tRNAs (A), protease-based release of the protein product (P), exit of empty rRNAs (E) c. activation of ribosome activity (A), processing of amino acids (P), elongation of the peptide chain (E) d. amino acid bond transfer (A), protein processing (P), and elongation of the peptide chain (E) e. binding incoming tRNAs (A), retention of the peptide chain during elongation (P), exit of deacylated tRNAs (E) as well as activation of ribosome activity (A), processing of amino acids (P), elongation of the peptide chain (E) ANSWER: a 35. In bacteria, the Shine–Dalgarno sequence is found on the mRNA and is recognized by the ________________________ to reveal __________________________.

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chapter 09: Proteins and Their Synthesis a. the 16S rRNA; the translation stop codon b. the 30S subunit; the translation start codon c. initiator tRNA; the translation start codon d. ribosome A site; the translation start codon e. amber tRNA; the translation stop codon ANSWER: b 36. During the initiation of prokaryotic translation, a defect in IF3 function would generate a. inefficient association of the 30S and 50S subunits. b. proteolysis of the 50S subunit. c. an inability of tRNA and mRNA binding to the 30S subunit. d. premature association of the 30S and 50S subunits. e. normal progression of translation (no defect).

ANSWER: d 37. Signal peptidase activity is responsible for a. directing defective proteins to the ubiquitin-proteasome pathway. b. adding a signal phosphate group to some proteins, regulating their activity. c. removing a hydrophobic secretion signal from the N-terminus of a protein. d. adding a signal glycosylation modification to a new protein. e. destroying peptide bonds in defective proteins.

ANSWER: c Subjective Short Answer 38. Which codons are susceptible to being converted to a "UAG" stop codon with only a single nucleotide mutation?

ANSWER: UGG (Trp), UCG (Ser), GAG (Glu), AAG (Lys), CAG (Gln), UAU/UAC (Tyr) Essay 39. Draw the general formula for an amino acid. What is it that distinguishes the 20 amino acids known to exist in proteins?

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chapter 09: Proteins and Their Synthesis ANSWER:

Each of the common 20 amino acids has a different R group that gives the amino acid its unique properties.

40. Distinguish between tRNAs and rRNAs. What does it mean for an RNA to be functional?

ANSWER: Functional RNAs are active as RNAs. They are never translated into proteins. The two main classes are: (1) tRNA, which functions to translate the three-nucleotide codon in the mRNA into the corresponding amino acid and which is brought to the ribosome in the process of translation; and (2) rRNA, which functions as the major component of ribosomes—large macromolecular complexes that assemble amino acids to form the protein whose sequence is encoded in a specific mRNA. 41. NASA scientists brought back from Mars a molecule they think is similar to DNA. Just like DNA, it is transcribed into an mRNA-like molecule. Unlike mRNA, though, it appears to use only three bases (A, C, and T). Assuming that proteins were still made from this mRNA molecule and it was thought that 30 amino acids existed for this organism, what is the minimum number of bases likely to be in a codon? Explain your answer.

ANSWER: With three bases to use in the code, the codons must be larger to accommodate a minimum of 30 amino acids (plus at least one stop codon). If the codon were three nucleotides in length, only 27 "words" would be in the code (33). If the codon were four nucleotides in length, the three nitrogenous bases would provide 81 codons (34).

42. Describe how it was determined that the genetic code was not overlapping. What evidence was used to support this?

ANSWER: Analyses of mutant RNAs showed that a single nucleotide change affects only one amino acid in the encoded protein. This was predictive of a nonoverlapping code. An overlapping code would predict that a single base change would alter as many as three amino acids at adjacent positions in the protein.

43. Crick, Brenner, and colleagues conducted experiments in 1961 using mutants in the rII locus of the T4 phage. They were able to show that certain mutations could be reversed if they first consisted of an insertion and then a deletion (or first a deletion and then an addition). Explain why—in Crick and colleagues' experiments—the "revertant" phenotypes did not look exactly like the wild types.

ANSWER: The "revertants" did not look just like the wild type because often the reading frame was restored a small distance away from the first mutation. For the region between the two mutations, the ribosome Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chapter 09: Proteins and Their Synthesis would be reading out of reading frame and would be adding amino acids that are inappropriate (incorrect) for that protein. The resulting protein would have an abnormal amino acid composition, with a number of amino acid substitutions between the mutations.

44. A gene makes a polypeptide 30 amino acids long containing an alternating sequence of phenylalanine and tyrosine. Assuming Phe = UUU and Tyr = UAU in mRNA, what are the sequences of nucleotides corresponding to this sequence in the following? a) the mRNA strand (sense) b) the non-template DNA strand (antisense or noncoding strand) c) the template DNA strand (sense or coding stand) d) the tRNA anticodon e) Do you think that the sequence is the only possible one for these amino acids? Assume that the wobble position for each of these codons can be changed to the base cytosine. Answer questions (a) through (d) again with this new information.

ANSWER: a) 5′-UUU UAU UUU UAU UUU . . . etc. b) 5′-TTT TAT TTT TAT TTT TAT . . . etc. c) 3′-AAA ATA AAA ATA AAA ATA . . . etc. d) Phe anticodon-3′-AAA, Tyr anticodon-3′-AUA e) Phe-also coded by UUC, Tyr-also coded by UAC a-2) 5′-UUC UAC UUC UAC UUC UAC . . . etc. b-2) 5′-TTC TAC TTC TAC TTC TAC . . . etc. c-3) 3′-AAG ATG AAG ATG AAG . . . etc. d-3) Phe anticodon-3′-AAG, Tyr anticodon-3′-AUG 45. Why is the attachment of an amino acid to the correct tRNA considered to be such an important step in protein synthesis?

ANSWER: The results of experiments have shown that the anticodon of a tRNA charged with an incorrect amino acid still interacts with the complementary codon in mRNA, thus causing the wrong amino acid to be incorporated into the growing polypeptide chain. No proofreading or editing mechanism exists in the ribosome to remove these incorrect amino acids.

46. Describe the types of RNAs participating in translation.

ANSWER: 1) Messenger RNA (mRNA) is encoded by distinct nuclear genes and carries the codon sequence to define synthesis of a functional protein.

2) Transfer RNAs (tRNA) carry an anticodon that interacts with a specific mRNA codon. tRNAs are charged with a specific amino acid, depending on the identity of the tRNA and its anticodon sequence.

3) Ribosomal RNAs (rRNA) are structural components of ribosomes and are a participant in the catalysis of peptide bond formation.

47. You believe you created a protein translation machine that operates outside of the cell. If you combine the "translation machine" with an mRNA and charged tRNAs, a protein product is generated. However, as you do additional experiments, you find that though the correct amino acids are joined together, the protein is nonfunctional. What might be missing in your system, causing these synthesized proteins to lack activity?

ANSWER: Proteins depend on three-dimensional structure in order to gain full activity. The folding of new proteins requires chaperone proteins that facilitate this process, and they often also require the Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chapter 09: Proteins and Their Synthesis addition of sugar residues (glycosylation) in order to complete the complex process of protein folding. Further, if the proteins are eukaryotic, they may require other post-translational modifications before they can become active.

48. Many antibiotics target prokaryotic ribosomal function, blocking translation and causing rapid death in susceptible microbial populations. You have been assigned to a structure-based drug design project for a biotechnology company. In order for an antibiotic to be effective, it must be able to block an essential ribosomal function, and the drug must have access to the portion of the enzyme or enzyme complex being targeted. What processes/enzymes might you target in your drug design?

ANSWER: This is an open-ended question, and there will be varied responses. Potential processes on the ribosome might be the entry site (A), where tRNAs enter the ribosome, the exit site (E), the peptidyl-transferase center, or even particular aminoacyl-transferase enzymes. Arrest of any of these processes would also arrest translation.

49. What is post-translational processing? Why is it important for protein function?

ANSWER: Some proteins cannot function until they are modified by the addition of certain molecules, such as phosphate or carbohydrate groups. In addition, some proteins do not function until they have been transported to a particular cellular compartment. Proteins destined for the nucleus contain nuclear localization sequences, whereas those destined for secretion, insertion into membranes, or transport into organelles contain signal sequences at their amino terminal end. These signal sequences are cleaved during targeting and are not part of the mature protein.

50. Describe the difference between translation in bacteria and translation in eukaryotes.

ANSWER: The major feature that difference in translation between bacteria and eukaryotes is the location of transcription and translation. In bacteria, translation and transcription occur in the same compartment. Translation of an RNA begins at its 5′ end while the rest of the mRNA is still being transcribed. In eukaryotes, mRNAs are exported from the nucleus for translation by ribosomes that reside in the cytoplasm. In contrast, transcription and translation are coupled in bacteria: translation of an RNA begins at its 5′ end while the rest of the mRNA is still being transcribed.

51. Describe the four levels of protein structure.

ANSWER: Primary structure is the sequence of amino acids. Secondary structure is the shape of a region of amino acids, such as α-helices or β-sheets. Tertiary structure is the three-dimensional shape of a whole polypeptide, and quaternary structure is the assembly of multiple polypeptides into a protein complex.

52. What are the four important structural features of tRNAs? ANSWER: The important structure features are (1) the sequence CCA at the 3′ end, (2) modified nucleotides such as dihydrouridine, pseudouridine, and inosine, (3) an overall inverted L shape, and (4) an anticodon.

53.

You found a mouse gene sequence containing an open reading frame that should produce a 43 kDa protein. You express this protein in cultured mouse cells by transforming the cells with a plasmid bearing your gene of interest. When you isolate proteins from these cells and probe for your protein of interest, you find that the detected protein is 49 kDa. Why might your protein be larger than expected when analyzed by immunoblot?

ANSWER: Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chapter 09: Proteins and Their Synthesis In eukaryotic cells, post-translational modification of proteins is common, especially with proteins that are to be localized within a cellular membrane, organelle, or proteins that are to be secreted. Post-translational modification can have a significant impact on a protein’s overall mass or mobility during electrophoresis.

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation Multiple Choice 1. Which of the following sequences includes a clear eight-base-pair palindrome? a. 5′-CCGATCGATCCC-3′ b. 5′-TGGGGTTTT G-3′ c. 5′-GGGGAAAA-3′ d. 5′-GATCCTAG-3′ e. 5′-AACCAACCAA-3′ ANSWER: a 2. In recombinant DNA technology, DNA is most often cut using a. DNA ligase. b. DNA polymerase. c. DNA gyrase. d. restriction endonucleases. e. terminal transferase.

ANSWER: d 3. A linear DNA molecule has n target sites for restriction enzyme EcoRI. How many fragments will be produced after complete digestion? a. n – 1 b. n c. n + 1 d. 2n – 1 e. 2n + 2 ANSWER: c 4. A circular DNA molecule has n target sites for restriction enzyme EcoRI. How many fragments will be produced after complete digestion? a. n – 1 b. n c. n + 1 d. 2n – 1 e. 2n + 2 ANSWER: b 5. Recombinant DNA techniques typically require the action of a. DNA polymerase and phosphatase. b. restriction enzymes and DNA ligase. c. RNA polymerase and RNA primase. d. reverse transcriptase.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation e. DNA helicase.

ANSWER: b 6. Which of the following statements about PCR is FALSE? a. PCR reaction contains DNA template, primers, deoxyribonucleotide triphosphates, and heat-resistant DNA polymerase. b. Each PCR cycle includes three steps: denaturation, annealing, and extension. c. After each PCR cycle, the amount of DNA increases in a linear manner. d. One limitation in PCR is that it requires prior knowledge of the sequence or at least part of the sequence to be amplified. e. PCR is a very sensitive technique and can amplify target sequences present in extremely low copy numbers.

ANSWER: c 7. The polymerase chain reaction requires ssDNA primers that a. anneal to the same strand of template DNA, though at distant sites. b. anneal to opposite strands of template DNA at distant sites, with their 3¢ ends directed toward each other. c. anneal to opposite strands of template DNA at distant sites, with their 5¢ ends directed toward each other. d. anneal to each other to prime DNA polymerization. e. hybridize only to DNA within the open reading frame of a selected gene.

ANSWER: b 8. The intermediate temperature cycle (72°C) in the polymerase chain reaction enables a. hybridization between the template DNA and the PCR primers. b. denaturation of the template DNA to enable primer access to matching sequence. c. activation of the primers to being the amplification procedure. d. activity of the thermostable DNA polymerase enzyme. e. a reduction of contamination in the amplification reaction.

ANSWER: d 9. Which of the following is NOT true of a cDNA clone? a. cDNAs are generally copies of mRNA from expressed genes. b. Reverse transcriptase activity is required for cDNA cloning. c. cDNA clones lack exons, as introns are spliced together before cloning. d. cDNAs are typically smaller than the chromosomal sequence for a particular gene. e. cDNAs can be fused to a promoter to enable expression of the encoded gene.

ANSWER: c 10. Identification of mRNA and initiation of priming for cDNA synthesis is accomplished by a. purifying only cytosolic RNAs before initiating the process.

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation b. coupling cDNA synthesis to exon splicing. c. detecting the 5′ cap sequence to initiation cDNA synthesis. d. use of oligo-dT to prime cDNA synthesis from the poly(A) tail. e. column purification of mRNA sequences.

ANSWER: d 11. Which of the following about plasmid vectors is/are TRUE? a. Plasmid vectors must have convenient restriction sites/polylinker sites. b. Plasmid vectors must have an origin of replication. c. Plasmid vectors must carry a selectable marker (drug resistance) for selection. d. Plasmid vectors can accept DNA inserts as large as 10 kb to 15 kb. e. All of the answer options are correct.

ANSWER: e 12. Perhaps the most common use of plasmid vectors is to a. enable the amplification of cloned DNA within a bacterial host cell. b. isolate viral DNA from infected cells. c. insert restriction enzyme cut sites into cDNAs. d. generate recombinant proteins. e. manipulate the cell biology of bacterial cells.

ANSWER: a 13. A plasmid vector has a gene for erythromycin resistance (EryR) and a gene for ampicillin resistance (AmpR).

The Amp gene is cut with restriction enzyme, and donor DNA treated with the same enzyme is added. What genotype of cells needs to be selected to show evidence of transformation? a. AmpR EryR b. AmpR EryS c. AmpS EryR d. AmpS EryS e. None of the answer options is correct.

ANSWER: c 14. Assume that a cosmid will carry inserts of about 50 kb and that cosmids are used to clone a 3 Mb (megabase) genome. Assuming you are particularly lucky and have no duplication in your library, what is the smallest number of cosmid clones you would need for a "complete" genomic library? a. 3000 clones b. 600 clones c. 300 clones d. 60 clones e. 30 clones Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation ANSWER: d 15. A radioactive probe is generated using the actin gene from yeast. This probe is used in a Southern blot analysis of EcoRI digested genomic DNA from the ciliated protozoan Tetrahymena thermophila. The autoradiogram shows a single-labeled band of 4 kb in size. This means that the Tetrahymena actin gene is a. 4 kb in size. b. the same size as the yeast gene. c. present in one copy in the genome. d. present in four copies in the genome. e. identical in sequence with the yeast gene.

ANSWER: c 16. A wild-type Aspergillus strain is transformed with a plasmid carrying a hygromycin resistance allele, and cells are plated on hygromycin. One resistant colony showed an aberrant type of aerial hyphae. When crossed to wild type, the progeny were 1/2 normal hyphae, Hyg sensitive, and 1/2 aberrant hyphae, Hyg resistant. The probable explanation is that a. the plasmid was inserted in a gene for normal hyphal development. b. the plasmid interfered with hyphal development. c. a mutation arose in a gene for hyphal development on the plasmid. d. a mutation arose in a gene for hyphal development on a recipient chromosome. e. the recipient was a heterokaryon carrying some mutant nuclei.

ANSWER: a 17. The cDNA for a eukaryotic gene B is 900 nucleotide pairs long. A cDNA clone is used to isolate a genomic clone of gene B, and the gene is sequenced. From start to stop codon, the gene is found to be 1800 nucleotide pairs long. The most probable reason for the discrepancy is that a. the mRNA broke during cDNA synthesis. b. the gene is present as a tandem duplication. c. the gene has 900 nucleotide pairs of introns. d. the genomic clone is not really gene B, just a related gene. e. there was a sequencing error.

ANSWER: c 18. Gel electrophoresis can be used to separate DNA or RNA molecules based on their a. charge. b. molecular weight. c. solubility in buffer. d. charge and molecular weight. e. All of the answer options are correct.

ANSWER: b 19. Which of the following about the Sanger sequencing method is TRUE? Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation a. Sanger sequencing uses dideoxynucleotide triphosphates to terminate DNA polymerization. b. Dideoxynucleotides lack 2′- and 3′-OH groups in the ribose sugar. c. All four deoxynucleotide triphosphates are used in a Sanger sequencing reaction. d. Fluorescent dyes can be used to automate Sanger sequencing. e. All of the answer options are correct.

ANSWER: e 20. In the Sanger sequencing method, the use of dideoxy adenosine triphosphate stops nucleotide polymerization opposite a. A's in the template strand. b. T's in the template strand. c. G's in the template strand. d. C's in the template strand. e. any base selected randomly in the template strand.

ANSWER: b 21. This vector is comprised of bacteriophage genome components and can carry cloned DNA up to 15 kb in size. a. pUC19 b. lambda c. fosmid d. bacterial artificial chromosome e. T-vector ANSWER: b 22. The value of the beta-galactosidase gene in cloning could best be described as the gene that a. provides valuable restriction enzyme cut sites. b. "reports" if a recombinant product is formed. c. confers resistance to antibiotics. d. facilitates easier transfer of the DNA into bacterial cells. e. encodes a special DNA ligase enzyme.

ANSWER: b 23. Bacterial artificial chromosomes are comprised of a. components of the bacterial F plasmid and can carry large inserts (200,000 bp). b. fosmid DNA sequences, enabling the transfer of DNA between bacteria. c. viral lambda DNA sequences, enabling the transfer of DNA between bacteria. d. a bacterial chromosome, with some artificial components. e. recombinant proteins isolated from specific phage viruses.

ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation 24. In the study of genetics, a "library" denotes a. the information carried in a digital database, such as GenBank. b. a collection of DNA fragments isolated from a particular group of cells (or tissue) representing the genetic content of those cells. c. a complex collection of restriction enzymes gathered for the purpose of restriction enzyme mapping. d. all the proteins present in a particular cell type, collected by protein extraction. e. a collection of books that explore the topic of genetics.

ANSWER: b 25. What type of library would be most valuable in the isolation of promoter sequences found in front of a gene's transcriptional start site? a. genomic DNA library b. cDNA library c. miRNA library d. mutant genomic DNA library e. repetitive DNA library ANSWER: a 26. Foreign DNA can be introduced into a cell using which of the following method(s)? a. transformation b. biolistic particle delivery c. micro injection d. electroporation e. All of the answer options are correct.

ANSWER: e 27. In Southern and Northern blotting, the probe being used to analyze DNA or RNA identifies the target sequence via a. attaching to the gene target via antibody-like protein-to-protein interactions. b. a DNA ligase-mediated attachment to the target DNA or RNA fragments. c. annealing between complementary sequences on the probe and the gene target. d. an enzymatic step that includes crosslinking between similar nucleic acid sequences. e. amplification during the polymerase chain reaction.

ANSWER: c 28. Transgenic plants can be generated using T-DNA plasmid carrying a gene of interest. To get the DNA into the plant cells, the researchers a. use a syringe needle to inject the DNA into pollen before fertilization. b. use a syringe needle to inject the DNA into groups of cells in the plant's root tissue. c. co-cultivate bacteria with T-DNA and plant cells, resulting in DNA transfer. d. add a mild detergent to cultures of plant cells to open holes in the cell wall.

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation e. remove the plant cell's normal DNA and replace it with the T-DNA. ANSWER: c 29. Detection of a transgenic plant can be accomplished in a number of different ways, but often the initially transformed cells (in tissue culture) are enriched for the presence of a transgene by a. screening groups of cells using the polymerase chain reaction. b. examining if the phenotype of the cells has been altered. c. placing the potentially transformed cells in the presence of a drug that inhibits the division of nontransgenic cells. d. microscopically examining the cells and isolating those that have "extra" DNA. e. None of the above. Transgenesis cannot be selected for in tissue culture.

ANSWER: c 30. Homologous recombination in yeast facilitates the a. segregation of genes during meiosis. b. targeted replacement of a gene in a living yeast cell. c. amplification of homologous yeast genes. d. independent assortment of separate gene alleles. e. expression of similar gene sequences via one promoter.

ANSWER: b 31. Embryonic stem (ES) cells have the unique ability to a. form a mouse pup without any additional cells. b. combine with other stem cells to generate a chimeric mouse pup. c. grow in the presence of gancyclovir. d. eliminate specific genes in response to chemical cues. e. divide more rapidly than other cell types.

ANSWER: b 32. A transgenic mouse is different from a knockout mouse in that a. transgenic mice have DNA added to their genome. b. transgenic mice do not have DNA added to their genome. c. the transgene is inserted at a random (ectopic) site in the genome. d. a knockout mouse is more healthy than a transgenic mouse. e. the knockout mouse is sterile.

ANSWER: c 33. What is the key difference between Southern and Northern blotting? a. In Southern blotting, DNA is transferred to the membrane, while in Northern blotting, RNA is transferred to the membrane. b. In Southern blotting, RNA is transferred to the membrane, while in Northern blotting, DNA is Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation transferred to the membrane. c. In Southern blotting, an antibody probe is used, while in Northern blotting, the probe is a radioactive particle. d. In Southern blotting, a radioactive probe is used, while in Northern blotting, the probe is an antibody. e. In Southern blotting, DNA is transferred to the membrane, while in Northern blotting, proteins are transferred to the membrane.

ANSWER: a 34. What experimental technique detects RNA and DNA in vivo through hybridization with radioactive nucleic acid probes, followed by autoradiography? a. immunofluorescence b. fluorescence in situ hybridization c. in situ hybridization d. Western blot e. Northern blot ANSWER: c 35. Which of the following is NOT required for a PCR reaction? a. DNA polymerase b. DNA primers c. deoxyribonucleotides d. template DNA e. helicase ANSWER: e Essay 36. Suppose you cut two different DNAs, one with BamH1:

5′-GGATCC-3′ 3′-CCTAGG-5′

and one with BglII:

5′-AGATCT-3′ 3′-TCTAG A-5′

then ligate them together through their compatible sticky ends. Once joined, could you separate these two DNAs again with either restriction enzyme? Why or why not? (Each of these enzymes is cut between the first two nucleotides, at the 5′ end of each strand of the palindrome.)

ANSWER: Neither enzyme cut site would be regenerated following ligation. After ligating the two DNAs together, the site (GGATCT or AGATCC) is a hybrid BamH1-like on one end and BglII-like on the other. Therefore, neither enzyme will recognize this site.

37. Restriction endonucleases BamH1, NheI, XbaI, and EcoRI make sequence-specific cuts in DNA at the following sequences: Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation BamHI: NheI: XbaI: EcoRI:

5′-GGATCC-3′ 5′-GCTAGC-3′ 5′-TCTAGA-3′ 5′-GAATTC-3′

All these enzymes break a phosphodiester bond between the first and second nucleotide (counting from the 5′ end), leaving a 5′ P and a 3′-OH. Could ligase covalently join two ends produced by the following enzymes? If so, would ligation regenerate one of the above restriction sites?

(1) two BamH1 ends (2) two XbaI ends (3) a BamH1 and an EcoRI end (4) an XbaI and an NheI end (5) a BamH1 and an NheI end ANSWER: Some cohesive ends can form perfect Watson-Crick pairs, and therefore can be ligated; others cannot. Ligations (1) and (2), each involving two ends created by a single restriction enzyme (BamH1 and XbaI, respectively), regenerate a restriction site for the corresponding enzyme.

Ligation (4) is successful because the cohesive ends are compatible, but ligation creates a "hybrid" hexanucleotide sequence that is not a restriction site for either XbaI or NheI. Ligations (3) and (5) cannot occur.

38. A 3-kb supercoiled circular plasmid contains the restriction sites shown below. N and X represent NheI and XbaI sites, respectively; A through F mark six points on the plasmid. A preparation of plasmid DNA was separately digested to completion with XbaI alone, NheI alone, and a mixture of the two enzymes. The products were separated by agarose gel electrophoresis. Draw the predicted appearance of the gel, assuming that you ran the following in the various lanes.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation

Lane M (size markers)—a mixture of known markers increasing in size from 1 kb to 6 kb, in steps of 1 kb Lane 1—sample from the original plasmid DNA preparation Lane 2—products of complete digestion with XbaI alone Lane 3—products of complete digestion with NheI alone Lane 4—products of complete digestion with XbaI plus NheI

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation ANSWER:

39. Human insulin can be made purer, and at a lower cost, using recombinant DNA technology. What sort of DNA clone might be ligated into a bacterial expression plasmid to be used in this strategy?

ANSWER: A cDNA clone, placed into a plasmid with a promoter that is active in bacterial cells, would generate the intact protein. Unlike genomic DNA, the cDNA would lack introns and would contain appropriate splicing of the exons.

40. You purified genomic DNA from tissue isolated from a population of roundworms (C. elegans). You wish to analyze this DNA, so you digest a portion of this DNA with the restriction enzyme EcoRI, and another portion of the DNA is digested with SalI. You make a gel made of agarose, with wells for loading your DNA. You place undigested DNA in well "A," the EcoRI-digested DNA in well "B," and the SalI-digested DNA in well "C." An electric field is generated by placing an anode (+ electrode) at the base of the gel, and a cathode (– electrode) at the top of the gel near the wells. The DNA migrates toward the anode. a)

Describe the movement of the DNA along the gel. How is an electric field able to generate movement of the DNA fragments?

b)

How can nucleic acids be visualized in an agarose gel?

c)

How would you expect lanes A, B, and C to appear?

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation d) Can you analyze the size of fragments that contain a particular gene?

ANSWER: a) DNA molecules contain a negative charge and so migrate toward the anode. The speed of the movement of a fragment depends on the size. Smaller pieces move faster as they are able to move through the gaps in the agarose more quickly. Pieces of similar size migrate at similar rates. b) Certain dyes attach selectively to nucleic acids. A commonly used dye (ethidium bromide) will fluoresce if stimulated with ultraviolet wavelengths of light. Thus, the DNA in our gel can be seen by adding ethidium bromide to the gel, and then exposing the gel to ultraviolet light. c) Undigested DNA (A) will contain large fragments of DNA, while the digested DNA (B, C) will exhibit a range of DNA fragment sizes, from small to large. The digested DNA would appear as a "smear" of DNA after the gel has been run. d) To examine a particular gene, you first would need a probe that complements the DNA sequences found in the genomic DNA. You would transfer the DNA from the gel to a paper known as nitrocellulose or positively charged nylon (binds DNA efficiently). The probe is labeled with radioactivity and then washed over the paper containing the transferred DNA. The probe will stick only to the gene of interest, likely showing different-sized bands for the two different enzyme digests.

41. The autoradiogram shown was generated by the dideoxy sequencing method. Define the DNA sequence of the template strand that was used in this sequencing reaction, noting the 5′ and 3′ ends of this template strand.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation

ANSWER: Reading the gel bottom to top reveals a 5′-to-3′ sequence of the synthesized strand, and the template strand is: 3′-TTTAGCGAATGCCCGATCGACCGTTTAGAT-5′ 42. In a certain community, a mutant allele for Tay-Sachs disease is prevalent. This mutation removes a Hind III restriction site in the gene sequence. A probe is available for this region, with the following homology:

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation

Two couples in this community are expecting children. Each couple has a DNA test to determine if their child will have Tay-Sachs. The results of the Southern blot, probed for the region described above, is shown below.

Using these results, how would you counsel the couples?

ANSWER: The first couple (M1, F1) are both heterozygous for the Tay-Sachs mutant allele and, thus, are phenotypically normal but carry the disease allele. Their child (Child 1) has inherited the mutant allele from each parent and will be affected by the disease. The second couple (M2, F2) are both heterozygous for the Tay-Sachs mutant allele and, thus, are phenotypically normal but carry the disease allele. Their child (Child 2) has inherited one mutant allele and one normal allele. This child will carry the disease allele but will be unaffected by the disease.

43. The restriction enzyme ClaI recognizes the sequence AT/CGAT. It cleaves the phosphodiester backbone between the T and the C bases as shown by the /. The restriction enzyme TaqI recognizes the sequence T/CGA. It cleaves the phosphodiester backbone between the T and the C bases as shown by the /. a) Indicate the ends of the double-stranded DNA molecule left after cleavage by ClaI. Show the sequence, polarity, and overhang, if any. b) On average, how many sites would TaqI recognize in a random sequence of 1200 base pairs? c) If a TaqI-digested DNA end is ligated to a ClaI-digested DNA end, will ClaI cleave the newly ligated DNA? Will TaqI cleave the newly ligated DNA? ANSWER: a) Overhangs are underlined.

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation ClaI 5′-AT CGAT-3′ 3′-TAGC TA-5′ b) TaqI is a 4-cutter enzyme, that on average would cut a random DNA sequence every 44 or 256 bases. Thus, it would cut a 1200 bp fragment ~4 times. ClaI is a 6-cutter enzyme and would cut, on average, every 46 or 4096 bases. c) TaqI would cut the newly ligated DNA. ClaI is possible, but it would depend on the DNA sequence adjacent to the TaqI cut site.

44. A new restriction enzyme is discovered that recognizes an 8-base restriction sequence. About how many fragments of the Wombat genome (approximately 4.2 × 108 in size) would you expect if you digested it with this enzyme?

ANSWER: An 8-base recognition site will appear on average once in about 48, or 65,536 bp. If the wombat genome is 420,000 kb in size, it would be cut into about 6409 fragments, although they would range greatly in size.

45. A chimeric plasmid contains vector DNA ligated to a fragment of chromosomal mouse DNA that is suspected of containing the obese gene. As a first step, a restriction map of the DNA clone was constructed. The DNA was digested with two restriction enzymes, BopI and ZapII, with the following results: BopI 8.0 kb

ZapII 2.0 kb 2.5 kb 3.5 kb

BopI +ZapII 1.0 kb 1.5 kb 2.0 kb 3.5 kb

a) From these data, answer the following questions.

(1) How large is the entire DNA clone?

(2) Is the clone linear or circular?

(3) Draw a restriction map consistent with these data. Include all BopI and ZapII sites and the physical distances between them. b) To better identify the location of the mouse chromosomal DNA on this chimeric plasmid, mouse genomic DNA was digested with ZapII, run on a gel, and analyzed using the Southern blot technique. The Southern blot was probed with radiolabeled Zap II DNA fragments from the plasmid DNA. Three lanes each containing the same amount of Zap II-digested chromosomal DNA were each probed with a different fragment from the plasmid. Autoradiograms showed the following:

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation

What is the maximum amount of mouse DNA present in the clone? Why didn't the 2.0-kb ZapII fragment hybridize to the mouse DNA? ANSWER: a) (1) 8.0 kb (2) A circular plasmid is the most likely scenario, as BopI clearly cuts the DNA one time (cuts the 2.5-kb ZapII band in the double digest).

(3)

b) The Southern blot revealed that only the 2.5-kb and the 3.5-kb Zap II fragments contain mouse Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation DNA because only they bind to mouse DNA in the Southern blot. The maximum amount of mouse chromosomal DNA in this plasmid is 6 kb.

46. A purified DNA preparation of a certain plasmid is digested to completion with BamH1 restriction endonuclease. In separate reactions, the same preparation was digested to completion with EcoRI and with a mixture of EcoRI and BamH1, respectively. The diagram below shows the appearance of the original molecules and the digestion products in the three digestion mixtures, after separation by electrophoresis in an agarose gel.

Lane M—linear DNA markers of the indicated length in kb Lane 1—sample of original plasmid DNA preparation Lane 2—sample of the products of BamH1 digestion Lane 3—sample of the products of EcoRI digestion Lane 4—sample of the products of digestion of BamH1 + EcoRI

Answer and justify the following: a) What was the original DNA molecule's length? b) Was the original plasmid DNA circular or linear? c) How many restriction sites for EcoRI and BamH1 did the original plasmid have? d) Draw the original plasmid, using B and E to indicate the location of the BamH1 and EcoRI restriction sites.

Indicate the number of kb between sites.

ANSWER: a) The original plasmid was 6 kb long since the linear fragments produced after digestion add up to 6 kb in lanes 3–5. b) The original plasmid could not be linear, or else it would have migrated with the same mobility as Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation the 6-kb linear marker. It must have been circular. c) The original circular plasmid had one restriction site for BamH1 (producing a full-length linear molecule when cut) and two sites for EcoRI (two fragments produced). d) A map of the restriction sites is shown. The sites were placed according to the following reasoning: The two EcoRI sites define two segments, 2 kb and 4 kb long (lane 4 of the gel). The single BamH1 site must be located within the 4-kb EcoRI segment since the 2-kb fragment shows up in both lanes 4 and 5, but the 4-kb fragment in lane 4 is missing from lane 5. BamH1 thus cuts the 4kb EcoRI fragment into two fragments, 3 kb and 1 kb, respectively. Locating the sites, as in the map above, accounts for all the observations.

47. You constructed a cDNA library in a lambda vector using mRNAs from a pea plant. You want to identify a rubisco (a protein involved in photosynthesis) clone among the 50,000 clones in the library. You have a rubisco cDNA from tobacco to use as a probe. When hybridizing the probe to the plaques in the library, would you use higher, lower, or the same temperature as for hybridizing the tobacco DNA to a tobacco rubisco clone? Why? ANSWER: Because the tobacco probe and the pea gene will almost certainly have different sequences, you should use a lower temperature (lower stringency) so that a stable hybrid can form between the two DNAs of similar, but not identical, sequence.

48. A certain animal virus contains a circular DNA that has five recognition sites for the restriction enzyme EcoRI. Infected cells make a large quantity of the viral surface protein. A cDNA corresponding to the mRNA Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation for this surface protein has been cloned. How would you use a Southern blot to determine which EcoRI fragment(s) contain the gene for the viral surface protein?

ANSWER: Cut the viral DNA with EcoRI, electrophorese the resulting five fragments, and Southern blot the fragments. Hybridize the Southern blot to a radioactive probe made from the cloned cDNA, and autoradiograph to see which of the five DNA fragments "light(s) up."

49. The drawing below depicts a restriction map of a segment of a human chromosome. The cut sites shown (1 through 6) are known to be polymorphic and can be present or absent depending on the genetic makeup of a specific individual's chromosomes. E identifies EcoRI sites, and P identifies Pst1 sites. The line below the map identifies the region of homology between the DNA and a probe used in Southern blot analyses to characterize this region.

a) Individual 1 has restriction sites 1 through 6 (all) on both copies of this chromosome. If DNA from this individual was digested with PstI, what are the sizes of the fragments that would be detected via Southern blot using the identified probe? b) DNA from Individual 1 is digested with both EcoRI and PstI and then analyzed by Southern blot. What bands are revealed? c) Individual 2 is homozygous for a polymorphism that eliminates site 3. If DNA from this individual was digested with both PstI and EcoRI, what are the sizes of the fragments that would be detected via Southern blot? d) Individual 3 has not previously been characterized over this chromosomal locus. The research team digests the DNA with PstI and EcoRI and generates the following banding pattern. What is the likely genotype of this individual?

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation

ANSWER: a) A Southern blot of PstI-digested DNA from Individual 1 and probed with the identified probe would yield bands of 4 kb and 7 kb. b) If DNA from Individual 1 is digested with EcoRI and PstI and then probed by Southern blot, bands of 2 kb, 4 kb, and 5 kb would be observed. c) Southern blot analysis of Individual 2, digested with EcoRI and PstI, would yield bands of 2 kb, 5 kb, and 7 kb. d) Individual 3 is likely heterozygous for cut site #4 (PstI).

50. When ligating pieces of DNA into a plasmid, a scientist will often transform the ligation products into competent bacteria to evaluate the success rate of the ligation (i.e., the number of plasmids that now have extra DNA ligated into their sequence). Define one quick way scientists can assess the success rate of a ligation just by looking at the bacterial colonies.

ANSWER: Interruption of a selectable marker, such as an antibiotic resistance gene or the beta-galactosidase gene (lacZ), enables a rapid screening of transgenic colonies for those containing recombinant plasmids.

51. Use the following data to generate a restriction map for the circular "pEPP1" plasmid.

PstI: 6.0 kb EcoRI: 6.0 kb SalI: 6.0 kb EcoRI + SalI: 2.0 kb, 4.0 kb EcoRI + PstI: 1.5 kb, 4.5 kb PstI + SalI: 2.5 kb, 3.5 kb Copyright Macmillan Learning. Powered by Cognero.

Page 20

Name:

Class:

Date:

Chapter 10: Gene Isolation and Manipulation PstI + SalI + EcoRI: 1.5 kb, 2.0 kb, 2.5 kb ANSWER:

52. cDNA synthesis begins with the purification of mRNA tissue or specific cells. Why is purification necessary, and how is it accomplished?

ANSWER: Purification is necessary because the majority of cellular RNA is comprised of rRNA and tRNA with mRNA accounting for only 5% of cellular RNA. Purification of mRNA from eukaryotic cells is done using affinity methods that target the unique features of mRNA relative to other types of RNA, such as the 5′-m7G cap and the 3′ poly(A) tail.

53. Describe the essential and nonessential features of plasmids for cloning.

ANSWER: The essential components of a plasmid include the origin of replication, a drug-resistance gene so that bacteria containing the plasmid can be identified, and a polylinker so that DNA can be inserted with restriction enzymes. Nonessential features of plasmids include sequence elements for the identification of plasmids that contain inserts, the constitutive or inducible expression of inserted genes, and the addition of an epitope tag onto a recombinant protein.

Copyright Macmillan Learning. Powered by Cognero.

Page 21

Name:

Class:

Date:

Chapter 11: Regulation of Gene Expression in Bacteria and Their Viruses Multiple Choice 1. The product of the regulator gene of the lac operon is a. the operator. b. the inducer. c. the repressor. d. the corepressor. e. β-galactosidase.

ANSWER: c 2. A prokaryotic operon is composed of a series of adjacent genes under the control of a. the same promoter. b. the same operator. c. an inducible promoter. d. the same operator and promoter. e. None of the answer options is correct.

ANSWER: d 3. The enzyme β-galactosidase can convert the disaccharide lactose into a. allolactose. b. glucose. c. allolactose and glucose. d. allolactose, glucose, and galactose. e. sucrose and glucose.

ANSWER: d 4. The Lac repressor protein controls expression of the lac operon by binding to the a. lac structural genes to repress expression. b. lacZ and lacY genes only to repress expression. c. lac operator site to repress expression. d. lac promoter site to repress expression. e. All of the answer options are correct.

ANSWER: c 5. The Lac repressor (encoded by LacI) binds to a. lactose and DNA. b. RNA polymerase. c. RNA polymerase and DNA. d. β-galactosidase, permease, and transacetylase. e. RNA and DNA. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chapter 11: Regulation of Gene Expression in Bacteria and Their Viruses 6. DNA-dependent RNA polymerase binds to the a. repressor gene. b. promoter. c. operator. d. permease gene. e. sugar lactose.

ANSWER: b 7. E. coli bacteria are placed into a medium containing both glucose and lactose. Which of the gene products (i.e., enzymes) listed below do you predict will be "turned on"? a. β-galactosidase b. lacI c. lacP d. permease e. None of the answer options is correct.

ANSWER: e 8. A null repressor mutation (I–) results in a. no transcription. b. inducible transcription. c. transcription but no translation. d. no translation. e. constitutive transcription.

ANSWER: e 9. A constitutive operator mutation (Oc) results in a. no transcription. b. inducible transcription. c. transcription but no translation. d. no translation. e. constitutive transcription.

ANSWER: e 10. A promoter mutation (P–) results in a. no transcription. b. inducible transcription. c. transcription but no translation. d. no translation. e. constitutive transcription.

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chapter 11: Regulation of Gene Expression in Bacteria and Their Viruses ANSWER: a 11. A super repressor mutation (IS) results in a. no transcription. b. inducible transcription. c. transcription but no translation. d. no translation. e. constitutive transcription.

ANSWER: a 12. A partial diploid of genotype I– P+ O+ Z+ / I+ P+ O+ Z– will show a. inducible production of repressor. b. inducible production of β-galactosidase. c. constitutive production of β-galactosidase. d. no production of β-galactosidase. e. constitutive production of lactose.

ANSWER: b 13. A partial diploid of genotype IS P+ O+ Z+ / I+ P+ O+ Z– will show a. inducible production of repressor. b. inducible production of β-galactosidase. c. constitutive production of β-galactosidase. d. no production of β-galactosidase. e. constitutive production of lactose.

ANSWER: d 14. A partial diploid of genotype I+ P+ O+ Z+ Y– / I+ P– O+ Z+ Y+ will show a. inducible production of β-galactosidase. b. inducible production of β-galactosidase and permease. c. constitutive production of β-galactosidase. d. constitutive production of β-galactosidase and permease. e. no β-galactosidase or acetylase production at all.

ANSWER: a 15. The repression of the transcription of lactose-metabolizing genes in the presence of glucose is an example of a. catabolite induction. b. anabolite induction. c. anabolite repression. d. catabolite repression. e. None of the answer options is correct.

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chapter 11: Regulation of Gene Expression in Bacteria and Their Viruses ANSWER: d 16. The lac operon is controlled by a. the Lac repressor. b. attenuation. c. catabolite activator protein (CAP). d. the Lac repressor and catabolite activator protein (CAP). e. the Lac repressor and attenuation.

ANSWER: d 17. Which of the following is an example of a cis-acting element? a. β-galactosidase b. operator site c. lacI repressor protein d. lactose e. permease ANSWER: b 18. Which of the following is an example of a trans-acting factor? a. β-galactosidase b. operator site c. lacI repressor protein d. lactose e. permease ANSWER: c 19. Which of the following is/are NOT part of the lac operon? a. lacZ b. lacI c. lacY d. lacA e. All (lacZ, lacI, lacY, and lacA) are part of the lac operon.

ANSWER: b 20. Which is the synthetic inducer Jacob and Monod used to examine the expression of β-galactosidase? a. allolactose b. isopropyl-β-D-thiogalactoside c. lactose d. albumin e. galactose ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chapter 11: Regulation of Gene Expression in Bacteria and Their Viruses 21. RNA polymerase binds to the promoter and transcribes the ara operon. In the presence of arabinose, both the CAP-cAMP complex and the AraC-arabinose complex must bind to a. araO. b. araI. c. araA. d. araB. e. araD. ANSWER: b 22. In the absence of arabinose, the AraC protein represses the ara operon by binding to: a. araI. b. araO. c. araI and araO. d. araI and araA. e. araO and araA. ANSWER: c 23. The order of the structural genes controlling the trp operon is a. trpE, trpD, trpC, trpB, trpA. b. trpA, trpC, trpB, trpD, trpE. c. trpE, trpD, trpC, trpA, trpB. d. trpA, trpB, trpC, trpD, trpE. e. None of the answer options is correct.

ANSWER: a 24. In the presence of the repressor molecule and free tryptophan, the trp operon is a. constitutively transcribed. b. derepressed. c. induced. d. repressed. e. transcribed but not translated.

ANSWER: d 25. In the presence of the repressor molecule and the absence of free tryptophan, the trp operon is a. constitutively transcribed. b. derepressed. c. induced. d. repressed. e. transcribed but not translated.

ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chapter 11: Regulation of Gene Expression in Bacteria and Their Viruses 26. Which of the following is/are NOT part of the trp operon? a. trpR b. trpE c. trpC d. trpA e. All (trpR, trpE, trpC, and trpA) are part of the trp operon.

ANSWER: a 27. The trp operon is controlled by a. the Trp repressor. b. attenuation. c. catabolite activator protein (CAP). d. the Trp repressor and catabolite activator protein (CAP). e. the Trp repressor and attenuation.

ANSWER: e 28. The attenuator region (AR) of the trp operon is located a. within the promoter region. b. between the promoter region and the operator site. c. within the operator site. d. between the operator site and the first structural gene. e. after the structural genes.

ANSWER: d 29. The regulation of operon gene expression by attenuator control is commonly found in a. glucose utilization pathways. b. lactose utilization pathways. c. amino acid biosynthetic pathways. d. fatty acid biosynthetic pathways. e. UV-protection responses.

ANSWER: c 30. In the presence of abundant tryptophan, the a. attenuator stem-loop structure forms, allowing transcription to continue. b. attenuator stem-loop structure forms, terminating transcription. c. preemptor stem-loop structure forms, allowing transcription to continue. d. preemptor stem-loop structure forms, terminating transcription. e. None of the answer options is correct.

ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chapter 11: Regulation of Gene Expression in Bacteria and Their Viruses 31. Under conditions where tryptophan is scarce, the a. attenuator stem-loop structure forms, allowing transcription to continue. b. attenuator stem-loop structure forms, terminating transcription. c. preemptor stem-loop structure forms, allowing transcription to continue. d. preemptor stem-loop structure forms, terminating transcription. e. None of the answer options is correct.

ANSWER: c 32. In the bacteriophage λ, the mutants cI, cII, and cIII form a. turbid plaques and are able to establish lysogeny. b. clear plaques and are unable to establish lysogeny. c. turbid plaques and are unable to establish lysogeny. d. clear plaques and are able to establish lysogeny. e. both clear and turbid plaques.

ANSWER: b 33. In the bacteriophage λ, the cI gene encodes the λ repressor, which a. represses lytic growth and promotes lysogeny. b. represses lysogeny and permits lytic growth. c. represses both lytic growth and lysogeny. d. promotes lytic growth. e. None of the answer options is correct.

ANSWER: a 34. In the bacteriophage λ, cro gene encodes the Cro repressor, which a. represses lytic growth and promotes lysogeny. b. represses lysogeny and permits lytic growth. c. represses both lytic growth and lysogeny. d. promotes lysogeny. e. None of the answer options is correct.

ANSWER: b 35. Which of the following DNA-binding proteins contain a helix-turn-helix domain? a. LacI repressor b. TrpR repressor c. AraC activator d. λ and Cro repressor e. All of the answer options are correct.

ANSWER: e 36. In the bacteriophage λ, the N gene encodes a(n) Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chapter 11: Regulation of Gene Expression in Bacteria and Their Viruses a. alpha repressor. b. positive regulator. c. Cro repressor. d. Lac repressor. e. negative regulator.

ANSWER: b 37. Any regulatory protein that acts by preventing transcription termination is called a(n) a. corepressor. b. coactivator. c. antiterminator. d. antipromoter. e. attenuator.

ANSWER: c 38. In Bacillus, the presence of alternate forms of which molecule is responsible for the control of large numbers of genes? a. DNA dependent RNA polymerase b. TFII c. σ factor d. Trp repressor e. Operator protein ANSWER: c 39. The specific conformation (shape) of DNA-binding regulatory proteins is such that these proteins can interact and bind to a. specific bases in the DNA through its major grooves. b. the phosphate backbone in the DNA. c. other transcription factors that are trans-acting elements. d. helicase to help induce transcription. e. specific bases in the DNA through its minor grooves.

ANSWER: a 40. Gene regulation in bacteria and bacteriophage utilize the following terminology. Twelve terms are given below (A–L). Select the letter of the correct term to match this statement: Phenotype with respect to β-galactosidase synthesis for the partial diploid I+ O+ Z+ / I– OC Z– a. temporal control b. positive control c. negative control d. inducible Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chapter 11: Regulation of Gene Expression in Bacteria and Their Viruses e. noninducible f. repressible g. constitutive h. operator i. repressor j. effector molecule k. feedback inhibition l. suppressor mutation ANSWER: d 41. Gene regulation in bacteria and bacteriophage utilize the following terminology. Twelve terms are given below (A–L). Select the letter of the correct term to match this statement: Phenotype of I+ O+ Z– / I+ OC Z+ a. temporal control b. positive control c. negative control d. inducible e. noninducible f. repressible g. constitutive h. operator i. repressor j. effector molecule k. feedback inhibition l. suppressor mutation ANSWER: g 42. Gene regulation in bacteria and bacteriophage utilize the following terminology. Twelve terms are given below (A–L). Select the letter of the correct term to match this statement: Phenotype of I+ O+ Z– / I– OC Z– a. temporal control b. positive control c. negative control d. inducible e. noninducible f. repressible g. constitutive h. operator Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chapter 11: Regulation of Gene Expression in Bacteria and Their Viruses i. repressor j. effector molecule k. feedback inhibition l. suppressor mutation ANSWER: e 43. Gene regulation in bacteria and bacteriophage utilize the following terminology. Twelve terms are given below (A–L). Select the letter of the correct term to match this statement: Phenotype with respect to the synthesis of tryptophan enzymes in the trp operon trp R+ OC S+ (trpR codes for repressor; S = structural gene) a. temporal control b. positive control c. negative control d. inducible e. noninducible f. repressible g. constitutive h. operator i. repressor j. effector molecule k. feedback inhibition l. suppressor mutation ANSWER: g 44. Gene regulation in bacteria and bacteriophage utilize the following terminology. Twelve terms are given below (A–L). Select the letter of the correct term to match this statement: Phenotype of trp R+ O+ S+ in the presence of tryptophan a. temporal control b. positive control c. negative control d. inducible e. noninducible f. repressible g. constitutive h. operator i. repressor j. effector molecule k. feedback inhibition Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chapter 11: Regulation of Gene Expression in Bacteria and Their Viruses l. suppressor mutation ANSWER: f 45. Gene regulation in bacteria and bacteriophage utilize the following terminology. Twelve terms are given below (A–L). Select the letter of the correct term to match this statement: Regulatory molecule must be present at a site in DNA (such as promoter) so that transcription occurs. a. temporal control b. positive control c. negative control d. inducible e. noninducible f. repressible g. constitutive h. operator i. repressor j. effector molecule k. feedback inhibition l. suppressor mutation ANSWER: b 46. Gene regulation in bacteria and bacteriophage utilize the following terminology. Twelve terms are given below (A–L). Select the letter of the correct term to match this statement: Small molecules that bind to regulatory molecule, such as repressor a. temporal control b. positive control c. negative control d. inducible e. noninducible f. repressible g. constitutive h. operator i. repressor j. effector molecule k. feedback inhibition l. suppressor mutation ANSWER: j 47. Gene regulation in bacteria and bacteriophage utilize the following terminology. Twelve terms are given below (A–L). Select the letter of the correct term to match this statement: Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chapter 11: Regulation of Gene Expression in Bacteria and Their Viruses Type of operons responding to catabolite repression (glucose effect) a. temporal control b. positive control c. negative control d. inducible e. noninducible f. repressible g. constitutive h. operator i. repressor j. effector molecule k. feedback inhibition l. suppressor mutation ANSWER: d 48. Gene regulation in bacteria and bacteriophage utilize the following terminology. Twelve terms are given below (A–L). Select the letter of the correct term to match this statement: Regulatory molecule that binds to operator region in DNA a. temporal control b. positive control c. negative control d. inducible e. noninducible f. repressible g. constitutive h. operator i. repressor j. effector molecule k. feedback inhibition l. suppressor mutation ANSWER: i 49. Alternative σ factors also play important roles a. in the regulation of alternative splicing. b. inducible mutations. c. in the virulence of human pathogens. d. at the lac promoter site to repress expression. e. all of the above.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chapter 11: Regulation of Gene Expression in Bacteria and Their Viruses ANSWER: c 50. On initiation of sporulation, σF is active in the _______ and σE is active in the _______ . a. forespore; mother cell. b. promoter; regulons. c. forespore; extracellular space. d. mother cell; forespore. e. None of the answer options is correct.

ANSWER: a 51. Which σ factors are active in the vegetative cells of Bacillus subtilis? a. σH and σG b. σA and σF c. σF and σE d. σG and σK e. σA and σH ANSWER: e Subjective Short Answer 52. Give the levels of β-galactosidase activity (high or low) expected for the following partial diploids for the lac operon.

Lactose absent +

+

+ –

+

–

Lactose present

+ +

a) I P O Z / I P O Z

b) I+ P+ O+ Z– / I– P+ O+ Z+ c) I+ P+ OC Z– / I+ P+ O+ Z+ d) I– P+ OC Z+ / I+ P+ O+ Z+ e) I+ P+ OC Z+ / IS P+ O+ Z+ ANSWER: a) I+ P+ O+ Z– / I+ P– O+ Z+ b) I+ P+ O+ Z– / I– P+ O+ Z+ c) I+ P+ OC Z– / I+ P+ O+ Z+ d) I– P+ OC Z+ / I+ P+ O+ Z+ e) I+ P+ OC Z+ / IS P+ O+ Z+

Lactose absent low low low high high

Lactose present low high high high high

53. Predict whether β-galactosidase will be produced by the following E. coli strains under the conditions noted.

The diploid genotypes represent F′ (lac) strains. Use + to indicate that the enzyme is synthesized at greater than basal levels, and 0 to indicate that the enzyme is not synthesized.

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chapter 11: Regulation of Gene Expression in Bacteria and Their Viruses Lactose absent +

+

Lactose present

+ +

a) I P O Z

b) I– P+ O+ Z+ c) I+ P+ OC Z+ d) I– P+ OC Z+ e) I– P+ OC Z– f) I+ P+ O+ Z– g) I+ P+ OC Z– / I+ P+ O+ Z+ h) I+ P+ O+ Z– / I+ P+ OC Z+ i) I+ P+ O+ Z– / I– P+ O+ Z+ j) I+ P+ O+ Z+ / I– P+ O+ Z– k) I+ P+ OC Z– / I– P+ O+ Z+ l) I+ P+ O+ Z– / I– P+ OC Z+ m) I– P+ O+ Z– / I– P+ OC Z+ n) I– P+ OC Z– / I– P+ O+ Z+ ANSWER: a) I+ P+ O+ Z+ b) I– P+ O+ Z+ c) I+ P+ OC Z+ d) I– P+ OC Z+ e) I– P+ OC Z– f) I+ P+ O+ Z– g) I+ P+ OC Z– / I+ P+ O+ Z+ h) I+ P+ O+ Z– / I+ P+ OC Z+ i) I+ P+ O+ Z– / I– P+ O+ Z+ j) I+ P+ O+ Z+ / I– P+ O+ Z– k) I+ P+ OC Z– / I– P+ O+ Z+ l) I+ P+ O+ Z– / I– P+ OC Z+ m) I– P+ O+ Z– / I– P+ OC Z+ n) I– P+ OC Z– / I– P+ O+ Z+

Lactose absent 0 + + + 0 0 0 + 0 0 0 + + +

Lactose present + + + + 0 0 + + + + + + + +

Essay 54. The lac operon is the classic "inducible system" for gene control, while the trp operon is the classic "repressible system." Arguably, the key difference between these control systems involves the manner in which the repressor proteins control transcription. Compare and contrast the LacI repressor protein with the TrpR repressor protein.

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chapter 11: Regulation of Gene Expression in Bacteria and Their Viruses ANSWER: Both repressor proteins bind to their respective operator sites. The LacI repressor is active in the "free" form and inactive when complexed with inducer (such as lactose, allolactose, or IPTG). The TrpR repressor is inactive in the "free" form and active when complexed with the co-repressor molecule tryptophan.

55. Consider an E. coli cell with the following mutations. What effect would each mutation have on the function of the lac operon (assuming no glucose is present)? a) a mutant lac operator that cannot bind repressor b) a mutant lac repressor that cannot bind the lac operator c) a mutant lac repressor that cannot bind to allolactose d) a mutant lac promoter that cannot bind CAP plus cAMP ANSWER: a) The lac operon would always be turned on because the repressor cannot turn it off by binding to the operator. b) Same as answer (a). c) The operon would be uninducible. The repressor would remain bound to the operator even in the presence of the inducer. d) The operon would be transcribed only weakly. Even in the presence of glucose, CAP plus cAMP would be unable to facilitate polymerase binding to the lac promoter.

56. A number of mutations affect the expression of the lactose operon in E. coli. Complete the table given below. Use + to indicate that the enzyme is synthesized at greater than basal levels, and 0 to indicate that enzyme is not synthesized.

Inducer (IPTG) absent -Galactosidase Permease a) I+ O+ Z+ Y+ b) IS O+ Z+ Y+ c) I– O+ Z+ Y+ d) I+ OC Z+ Y+ e) IS OC Z+ Y– f) I+ OC Z+ Y– / I– O+ Z– Y+ ANSWER:

Inducer (IPTG) present -Galactosidase Permease

Inducer (IPTG) absent -Galactosidase Permease + + + + 0 0 a) I O Z Y 0 0 b) IS O+ Z+ Y+ + + c) I– O+ Z+ Y+ + C + + + + d) I O Z Y + 0 e) IS OC Z+ Y– 0 f) I+ OC Z+ Y– / I– O+ Z– Y+ +

Inducer (IPTG) present -Galactosidase Permease + + 0 0 + + + + + 0 + +

57. Describe the σ factors that regulate sporulation in Bacillus subtilis.

ANSWER: During sporulation in B. subtilis, σF is activated in the forespore and activates a group of more than 40 genes. The σE factor activates genes in the mother cell, and σK and σG are activated in the mother Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chapter 11: Regulation of Gene Expression in Bacteria and Their Viruses cell and forespore, respectively.

58. Why do σ factors have different sequence-specific DNA-binding properties?

ANSWER: Characteristic sequences in the –35 and –11 regions of the promoter of genes allow specific σ factors to bind to and regulate different sets of genes.

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chapter 12: Regulation of Gene Expression in Eukaryotes Multiple Choice 1. In eukaryotes, transcriptional gene control is mediated by a. trans-acting factors binding to cis-acting elements. b. trans-acting factors failing to bind to cis-acting elements. c. repressor proteins binding to operator sites. d. metabolites binding to cis-acting elements. e. None of the answer options are correct.

ANSWER: a 2. The ________________ is a cis-acting element that binds RNA polymerase II. a. promoter b. enhancer c. promoter-proximal element d. GC-rich box e. All of the answer options are correct.

ANSWER: a 3. The ________________ is a cis-acting element that binds regulatory proteins. a. promoter b. operator c. promoter-proximal element d. TATA box e. All of the answer options are correct.

ANSWER: c 4. The ________________ is a cis-acting element that binds trans-acting DNA-binding proteins. a. promoter b. enhancer c. promoter-proximal element d. GC-rich box e. All of the answer options are correct.

ANSWER: e 5. The Gal4 protein has which of the following functional domains? a. a DNA-binding domain b. an activation domain c. a repressor domain d. a DNA-binding domain and an activation domain e. a DNA-binding domain and a repressor domain ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chapter 12: Regulation of Gene Expression in Eukaryotes 6. Which of the following GAL genes are located on the same chromosome? a. GAL1 and GAL2 b. GAL1 and GAL10 c. GAL2 and GAL10 d. GAL2 and GAL7 e. All of the answer options are correct.

ANSWER: b 7. In the presence of the sugar galactose, the expression levels of GAL1, GAL2, GAL7, and GAL10 are ______________ as compared to the absence of galactose. a. 1000-fold higher b. 1000-fold lower c. 10-fold higher d. 10-fold lower e. the same ANSWER: a 8. The GAL4 enhancers are located _____ of the genes they regulate and are known as ______. a. downstream; downstream activation sequences (DAS) b. downstream; downstream enhancer sequences (DES) c. upstream; upstream activation sequences (UAS) d. upstream; upstream enhancer elements (UES) e. None of the answer options is correct.

ANSWER: c 9. The "domain-swapping" experiment that grafts the Gal4 activation domain to the LexA DNA-binding domain generates a chimeric protein that will a. bind to the Gal4 site. b. bind to the LexA site. c. activate transcription of the Gal4 gene. d. activate transcription of the Gal7 gene. e. All of the answer options are correct.

ANSWER: b 10. Analyzing mutations in the GAL80 and GAL3 indicate that a. both GAL80 and GAL3 inhibit GAL gene expression. b. both GAL80 and GAL3 promote GAL gene expression. c. GAL80 promotes GAL gene expression, and GAL3 inhibits GAL gene expression. d. GAL80 inhibits GAL gene expression, and GAL3 promotes GAL gene expression. e. GAL80 and GAL3 are not involved in GAL expression.

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chapter 12: Regulation of Gene Expression in Eukaryotes ANSWER: d 11. GAL80 inhibits GAL gene expression by a. specifically binding to Gal4 and preventing DNA binding. b. specifically binding to Gal4 and preventing activation. c. specifically binding to Gal3 and preventing DNA binding. d. specifically binding to Gal3 and preventing activation. e. inducing an allosteric change in Gal3.

ANSWER: b 12. GAL3 promotes GAL gene expression by a. binding galactose only and subsequently binding to Gal80. b. binding ATP only, and subsequently binding to Gal80. c. binding galactose and ATP, and subsequently binding to Gal80. d. binding galactose and ATP, and subsequently binding to Gal1, Gal7, and Gal10. e. binding to the Gal upstream activation sequence (UAS).

ANSWER: c 13. In the bacterial lac operon, the physiologically regulated step is a. DNA binding by the transcriptional regulator. b. activity of the activation domain of the transcriptional regulator. c. feedback inhibition by the end product of the pathway. d. import of the inducer molecule. e. None of the answer options is correct.

ANSWER: a 14. In the eukaryotic GAL system, the physiologically regulated step is a. DNA binding by the transcriptional regulator. b. activity of the activation domain of the transcriptional regulator. c. feedback inhibition by the end product of the pathway. d. import of the inducer molecule. e. None of the answer options is correct.

ANSWER: b 15. In addition to yeast cells, Gal4 has been shown to activate transcription in a. insect cells. b. mouse cells. c. human cells. d. insect, mouse, and human cells. e. None of the answer options is correct.

ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chapter 12: Regulation of Gene Expression in Eukaryotes 16. The mating-type (MAT) locus in yeast has how many alleles? a. 1 b. 2 c. 3 d. 4 e. The precise number is unknown.

ANSWER: b 17. In yeast, which of these is NOT a function of MAT genes? a. controlling the expression of structural genes in a and α cells b. MATa locus encoding a regulatory protein (a1) that affects only diploid cells c. MATa locus encoding a regulatory protein (a1) that affects both haploid and diploid cells. d. MATα allele encoding a regulatory protein (α2) that represses expression of a specific genes e. None of the answer options is correct.

ANSWER: c 18. How many molecules of histones H2A, H2B, H3, and H4 (respectively) comprise a single nucleosome? a. one each (i.e., 1, 1, 1, 1) b. two each (i.e., 2, 2, 2, 2) c. three each (i.e., 3, 3, 3, 3) d. 1, 2, 2, 2 e. 2, 2, 2, 1 ANSWER: b 19. Which histone is considered the "linker" that compacts nucleosomes into higher-order structures? a. H1 b. H2A c. H2B d. H3 e. H4 ANSWER: a 20. Which of the following describes the histones associated with the nucleosomes of active genes? a. poor in acetyl groups (i.e., hypoacetylated) b. rich in acetyl groups (i.e., hyperacetylated) c. rich in phosphate groups (i.e., hyperphosphorylated) d. poor in phosphate groups (i.e., hypophosphorylated) e. None of the answer options is correct.

ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chapter 12: Regulation of Gene Expression in Eukaryotes 21. Which of the following describes the histones associated with the nucleosomes of inactive genes? a. poor in acetyl groups (i.e., hypoacetylated) b. rich in acetyl groups (i.e., hyperacetylated) c. rich in phosphate groups (i.e., hyperphosphorylated) d. poor in phosphate groups (i.e., hypophosphorylated) e. None of the answer options is correct.

ANSWER: a 22. The enzyme responsible for adding acetyl groups to histone proteins is called a. histone acetylase (HA). b. histone deacetylase (HDAC). c. histone acetylmethyltransferase (HAT). d. histone acetyltransferase (HAT). e. histone methyltransferase (HMT).

ANSWER: d 23. The enzyme responsible for removing acetyl groups from histone proteins is called a. histone acetylase (HA). b. histone deacetylase (HDAC). c. histone acetylmethyltransferase (HAT). d. histone acetyltransferase (HAT). e. histone methyltransferase (HMT).

ANSWER: b 24. The enzyme responsible for adding methyl groups to histone proteins is called a. histone acetylase (HA). b. histone deacetylase (HDAC). c. histone acetylmethyltransferase (HAT). d. histone acetyltransferase (HAT). e. histone methyltransferase (HMT).

ANSWER: e 25. Epigenetic inheritance is defined as the inheritance of a. nuclear DNA from one cell generation to the next. b. extra-chromosomal DNA from one cell generation to the next. c. mRNA from one cell generation to the next. d. chromatin states from one cell generation to the next. e. None of the answer options is correct.

ANSWER: d 26. What portion of a histone is typically modified by acetylation/deacetylation?

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chapter 12: Regulation of Gene Expression in Eukaryotes a. acidic amino acid residues in the "head" region of the histone b. basic amino acid residues in the "head" region of the histone c. basic amino acid residues in the "tail" region of the histone d. acidic amino acid residues in the "head" region and basic amino acid residues in the "tail" region e. None of the answer options is correct.

ANSWER: c 27. Methylation of H3 lysine residue 4 a. is associated with activation of gene expression. b. is associated with repression of gene expression. c. has no effect on gene expression. d. creates binding sites for other gene repressors. e. None of the answer options is correct.

ANSWER: a 28. What is an enhanceosome? a. a large complex that acts synergistically to activate translation b. a large complex that acts synergistically to activate transcription c. an aggregation of nucleosomes that activates transcription d. an aggregation of nucleosomes that activates translation e. a large complex that promotes epigenetic inheritance ANSWER: b 29. What is chromatin remodeling? a. the first stage of mitosis b. the changing of amino acid position c. the changing of nucleosome position d. the changing of nucleotide position e. None of the answer options is correct.

ANSWER: c 30. The regulatory element that evolved to prevent enhancers from promiscuously activating promoters is known as a. the first stage of mitosis. b. the changing of amino acid position. c. enhancer-blocking elements. d. the changing of nucleotide position. e. None of the answer options is correct.

ANSWER: c 31. Regions of a chromosome that are bundled in highly condensed chromatin are known as Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chapter 12: Regulation of Gene Expression in Eukaryotes a. homochromatin. b. heterochromatin. c. euchromatin. d. epichromatin. e. hyperchromatin.

ANSWER: b 32. Regions of a chromosome that are packaged in less-condensed chromatin are known as a. homochromatin. b. heterochromatin. c. euchromatin. d. epichromatin. e. hyperchromatin.

ANSWER: c 33. Compared to heterochromatin, euchromatin is a. rich in genes and comprised of densely packed nucleosomes. b. poor in genes and comprised of densely packed nucleosomes. c. rich in genes and comprised of loosely packed nucleosomes. d. poor in genes and comprised of loosely packed nucleosomes. e. rich in genes and completely lacking nucleosomes.

ANSWER: c 34. The phenomenon in D. melanogaster known as position-effect variegation (PEV) provides powerful evidence that a. DNA is the genetic material. b. post-transcriptional mechanisms regulate gene expression. c. transcriptional mechanisms regulate gene control. d. chromatin structure is unable to regulate gene expression. e. chromatin structure is able to regulate gene expression.

ANSWER: e 35. DNA elements that prevent the spreading of heterochromatin into adjacent euchromatic regions are called a. enhancer-blocking elements. b. euchromatin buffers. c. barrier insulators. d. barrier buffers. e. barrier-blocking elements.

ANSWER: c 36. In mice, the igf2 allele is expressed only if it is Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chapter 12: Regulation of Gene Expression in Eukaryotes a. inherited from the mother. b. inherited from the father. c. mutated in the mother. d. mutated in the father. e. None of the answer options is correct.

ANSWER: b 37. In mice, the H19 allele is expressed only if it is a. inherited from the mother. b. inherited from the father. c. mutated in the mother. d. mutated in the father. e. None of the answer options is correct.

ANSWER: a 38. Expression of a gene in a mammal can sometimes be associated with the parent from which the gene was inherited (mother or father). What term is used to describe the situation where an allele that is inherited from the father is expressed, but the homologous allele inherited from the mother is not? a. paternal imprinting b. maternal imprinting c. autosomal imprinting d. Y-linked inheritance e. X-linked inheritance ANSWER: b 39. Genomic imprinting in mice is mediated by the imprinting control region (ICR). In the maternal allele, the a. enhancer element is methylated and serves as a buffer-insulator. b. ICR is methylated, which allows H19 transcription. c. ICR is not methylated, which allows H19 transcription. d. ICR is not methylated, which allows Igf2 transcription. e. ICR is methylated, which allows Igf2 transcription.

ANSWER: c 40. Genomic imprinting in mice is mediated by the imprinting control region (ICR). In the paternal allele, the a. enhancer element is methylated and serves as a buffer-insulator. b. ICR is methylated, which allows H19 transcription. c. ICR is not methylated, which allows H19 transcription. d. ICR is not methylated, which allows Igf2 transcription. e. ICR is methylated, which allows Igf2 transcription.

ANSWER: e Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chapter 12: Regulation of Gene Expression in Eukaryotes 41. The sequences of ________ are among the most highly conserved in evolution. a. core histones b. linker histones c. histone acetyltransferases d. heterochromatin e. H1 histones ANSWER: a 42. Acetylation of lysines in histones by histone acetyltransferase a. blocks binding of transcription factors. b. blocks the binding site for bromodomains. c. results in closed access to chromatin. d. loosens interactions within and between nucleosomes. e. creates a binding site for coregulators.

ANSWER: d Essay 43. Describe the difference between transcriptional gene regulation and post-transcriptional gene regulation in eukaryotes.

ANSWER: In transcriptional gene regulation, regulatory proteins bind to specific DNA sequences outside of the protein-coding regions; the binding of these proteins modulates the rate of transcription and ultimately the expression of the related phenotype. In post-transcriptional gene regulation, mechanisms that occur after the generation of mRNA transcript modulate expression of the related phenotype; these mechanisms include mRNA stability, translation, protein modification and storage, and RNA interference.

44. Histones are essentially identical in sequence/structure in all eukaryotic organisms from yeast to plants to animals. What does this say about the biophysical properties of DNA packaging and the evolution of eukaryotic organisms?

ANSWER: Histones are basic proteins able to interact with DNA in a highly specific manner. DNA packaging into nucleosomes constrains the evolution of histones. The presence of histones in all eukaryotic organisms provides strong evidence that histone-mediated nucleosome packaging of DNA was a phenotype found in the last common ancestor of yeast, plants, and animals.

45. Some regulatory proteins do not directly bind DNA. What are they, and how do they regulate transcription without direct binding?

ANSWER: Coactivators and corepressors are regulatory proteins that do not directly bind DNA. Coactivators increase the amount of transcription through binding or modifying other regulatory factors.

Corepressors decrease transcription rates, and similar to coactivators, they work through binding or modifying other regulatory factors.

46. What functional domains are found in transcription factors?

ANSWER: Eukaryotic transcription factors have four types of functional domains. They have a DNA-binding Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chapter 12: Regulation of Gene Expression in Eukaryotes domain, an activation/repression domain, a domain for dimerization, and a domain for ligand binding.

47. Describe the major mechanisms that allow eukaryotic genes to have transcription states that range from silent to highly active.

ANSWER: Chromatin modification and chromatin remodeling are mechanisms that enable access of transcription machinery to DNA, which results in a wide range of transcription states. During chromatin modification, enzymes alter the chemical structure of amino acids in histones or nucleotides in DNA. The alteration affects recruitment of transcription factors, coregulators, and general transcription factors to chromatin. During chromatin remodeling, enzymes use energy from ATP hydrolysis to remodel nucleosomes which affects the accessibility of DNA to transcription factors, coregulators, and general transcription factors.

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chapter 13: The Genetic Control of Development Multiple Choice 1. A morphogen is a molecule that induces a. cells to undergo cellular division. b. cells to increase in size. c. various responses in surrounding tissue in a non-concentration-dependent manner. d. various responses in surrounding tissue in a concentration-dependent manner. e. None of the answer options is correct.

ANSWER: d 2. In the developing chick vertebral limb bud, the zone of polarizing activity (ZPA) organizes a pattern along the anteroposterior axis. Transplantation of the ZPA from a posterior to anterior position a. causes no change in development. b. induces extra digits with reverse polarity. c. induces extra digits with the same polarity. d. induces extra limbs. e. None of the answer options is correct.

ANSWER: b 3. In homeotic mutants, a. one body structure is missing. b. one body structure has been changed into another. c. one body structure has been duplicated. d. one body structure has significantly increased in size. e. the organism is infertile.

ANSWER: b 4. Drosophila melanogaster development proceeds in which of the following stages? a. egg–segmented embryo–pupa–larva–adult b. segmented embryo–egg–pupa–larva–adult c. egg–segmented embryo–larva–pupa–adult d. egg–larva–segmented embryo–pupa–adult e. segmented embryo–larva–egg–pupa–adult ANSWER: c 5. There are ______ Hox genes in Drosophila melanogaster. a. 12 b. 8 c. 10 d. 25 e. 39 Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chapter 13: The Genetic Control of Development ANSWER: b 6. Homeosis is defined as the failure to a. form the correct number of segments. b. correctly polarize a developing embryo. c. develop antennae. d. correctly establish segmental identity. e. None of the answer options is correct.

ANSWER: d 7. The maternal-effect Bicoid gene codes for a a. signal protein. b. serine protease. c. DNA-binding transcription factor. d. cell-to-cell junction protein. e. transmembrane protein.

ANSWER: c 8. Establishment of the dorsoventral axis of the Drosophila embryo a. involves the production of localized DL mRNA on the ventral side of the developing embryo. b. occurs via at least two paracrine signals transmitted between the developing oocyte and the surrounding somatic follicle cells. c. involves the zygotically expressed DL morphogen. d. depends on the activity of anterior–posterior morphogens, such as BCD or HB-M. e. None of the answer options is correct.

ANSWER: b 9. The homeotic, gap, pair-rule, and segment polarity genes all control a. aspects of anterior–posterior pattern formation. b. the number of segments. c. the developmental fates (identities) of the individual segments. d. dorsal–ventral pattern formation. e. the production of the anterior–posterior morphogens.

ANSWER: b 10. There are ______ Hox genes in the laboratory mouse. a. 12 b. 8 c. 10 d. 25 e. 39 Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chapter 13: The Genetic Control of Development ANSWER: e 11. In Drosophila melanogaster, Hox genes are a. clustered into operons. b. clustered in one large gene complex. c. clustered together in two gene complexes. d. clustered together in three gene complexes. e. evenly distributed throughout the genome.

ANSWER: c 12. In the developing Drosophila embryo, the Hox genes are expressed a. in temporally restricted domains. b. uniformly throughout the embryo. c. only in homeotic mutants. d. in spatially restricted domains. e. None of the answer options is correct.

ANSWER: d 13. All eight Drosophila Hox genes encode proteins containing a highly conserved 60-amino-acid domain called the a. hox box. b. homeodomain. c. hox domain. d. heterodomain. e. chromodomain.

ANSWER: b 14. The homeodomain encodes a(n) a. helix-turn-helix motif. b. ATP-binding domain. c. leucine zipper. d. zinc-finger. e. protease.

ANSWER: a 15. Genes with products provided by the female to the egg are called a. paternal-effect genes. b. maternal-effect genes. c. sex-linked genes. d. embryo-specific genes. e. egg-specific genes.

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chapter 13: The Genetic Control of Development ANSWER: b 16. The genes required for proper organization of the anteroposterior body axis of the fly embryo are grouped into how many classes? a. 1 b. 2 c. 3 d. 4 e. 5 ANSWER: e 17. The class of genes required for establishing the anteroposterior axis is the a. maternal-effect genes. b. gap genes. c. pair-rule genes. d. segment polarity genes. e. Hox genes.

ANSWER: a 18. The class of genes required for establishing the formation of a contiguous block of segments is the a. maternal-effect genes. b. gap genes. c. pair-rule genes. d. segment polarity genes. e. Hox genes.

ANSWER: b 19. The class of genes required that act at a double-segment periodicity is the a. maternal-effect genes. b. gap genes. c. pair-rule genes. d. segment polarity genes. e. Hox genes.

ANSWER: c 20. The class of genes that affect patterning within each segment is the a. maternal-effect genes. b. gap genes. c. pair-rule genes. d. segment polarity genes. e. Hox genes.

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chapter 13: The Genetic Control of Development ANSWER: d 21. In Drosophila, the vast majority of A–P axis genes that contribute to pattern formation encode a. transcription factors. b. histones. c. sugar-utilizing enzymes. d. operons. e. ligands.

ANSWER: a 22. The early Drosophila embryo has multiple nuclei within a single cytoplasm and is considered a a. coenocyte. b. hyphae. c. syncytium. d. chorion. e. None of the answer options is correct.

ANSWER: c 23. The formation of stripe 2 in a Drosophila embryo is controlled by the binding of four transcription factors to the eve stripe 2 cis-acting regulatory element. This includes a. one maternal protein and three gap proteins. b. one paternal protein and three gap proteins. c. one pair-rule, one segment-polarity, and two gap proteins. d. two pair-rule and two gap proteins. e. all maternal proteins.

ANSWER: a 24. The doublesex (dsx) gene plays a central role in governing the sexual identity of somatic (non-germ-line) tissue. Null mutations in dsx cause a. males to develop as females. b. females to develop as males. c. females and males to develop as intermediate intersexes. d. only females to develop as intermediate intersexes. e. None of the answer options is correct.

ANSWER: c 25. The pre-mRNA for Sex-lethal, transformer, and doublesex are a. identical in both sexes. b. shorter in females. c. shorter in males. d. longer in females.

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chapter 13: The Genetic Control of Development e. longer in males.

ANSWER: a 26. The nematode worm Caenorhabditis elegans is comprised of approximately ______ somatic cells. a. 10 b. 100 c. 1000 d. 100,000 e. 100,000,000 ANSWER: c 27. In the nematode worm, C. elegans, embryonic development is controlled largely by controlling a. Bicoid. b. transcription. c. post-transcriptional regulation. d. protein processing. e. None of the answer options is correct.

ANSWER: c 28. The products of the gene let-7 is a well-known a. transcription factor. b. polarity protein. c. trans-acting activator of pair-rule genes. d. microRNA. e. homeotic protein.

ANSWER: d 29. Toolkit genes are a. not conserved among different animal phyla. b. typically never transcription factors. c. conserved genes that do not have the same function across different animal phyla. d. highly conserved among different animal phyla. e. generally never components of signaling pathways.

ANSWER: d 30. To determine the role of toolkit gene Sonic hedgehog (Shh) in chicken, researchers Cliff Tabin and colleagues did the following: a. caused the Shh protein to be expressed in the posterior region of the developing limb bud. b. deleted Shh expression in the posterior region of the developing limb bud. c. caused the Shh protein to be expressed in the anterior region of the developing limb bud. d. deleted Shh expression in the anterior region of the developing limb bud.

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chapter 13: The Genetic Control of Development e. None of the above.

ANSWER: c 31. The human epidermal growth factor receptor 2 (HER2 or ERBB2) is a. expressed only in humans. b. a homolog of the Drosophila torpedo gene. c. underexpressed in some breast cancers. d. a homolog of the chicken shh gene. e. a homolog of the Drosophila shh gene.

ANSWER: b Multiple Response 32. Shown below is a list of molecules that function in Drosophila development (A–G). Select all the molecules that are found primarily in the anterior end of the embryo. a. toll transmembrane receptor b. hunchback protein c. activated SPZ ligand d. inactive DL protein e. active DL protein f. NOS protein g. Bicoid protein ANSWER: b, g Essay 33. What characteristics allowed Drosophila melanogaster (an insect) to emerge as the leading genetic model of animal development?

ANSWER: Ease of rearing, rapid life cycle, amenable genetics, well-established cytogenetics, large collection of visible mutants, and decades of classical genetic analysis.

34. Shown below is a list of gene classes that play a role in Drosophila development. homeotic genes gap genes segment-polarity genes maternal-effect genes pair-rule genes (i) Describe the role of each class of genes and the impact that a mutation in the class has on Drosophila development. (ii) The different classes of genes function at different times during Drosophila development. Put the gene classes in order, from earliest to latest.

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chapter 13: The Genetic Control of Development ANSWER: (i) Homeotic genes establish the identity of segments. Mutations cause the development of body parts in the wrong location (such as legs in place of antennae).

Gap genes organize the formation of a spatially localized set of segments. Mutations result in flies that lack the proper sequential set of larval segments.

Segment-polarity genes define the anterior to posterior rows of cells within each of the 14 segments.

Mutations result in opposite orientation of the cells within a segment.

Maternal-effect genes establish the asymmetries and the chemical concentration gradients that the embryo uses for anterior–posterior differentiation. Mutations cause incorrect anterior–posterior differentiation.

Pair-rule genes ensure the development of the correct number of segments. Mutations result in the loss of certain segments. (ii) 1) maternal-effect genes 2) gap genes 3) pair-rule genes 4) segment-polarity genes 5) homeotic genes 35. Why would animals as diverse as mice, humans, and Drosophila have maintained nearly identical Hox genes?

ANSWER: Their deep common ancestry indicates that Hox genes play some fundamental role in the development of most animals. The extent of sequence similarity indicates very strong pressure to maintain the sequence of the homeodomain.

36. A Drosophila researcher recently isolated a number of mutant fruit flies. The phenotype of each mutant is described below (i–iv). For each mutant, describe the type of gene (maternal-effect gene, gap gene, segmentpolarity gene, or homeotic gene) that was most likely the cause of the altered phenotype. (i) Homozygous mutant females appear phenotypically normal but produce larvae that have two posterior ends. (ii) The flies contain four wings rather than two. (iii) The wings are oriented backward. (iv) The flies are missing two pairs of legs due to a shortened thoracic region.

ANSWER: (i) maternal effect gene (ii) homeotic gene (iii) segment-polarity gene (iv) gap gene 37. What is the single most important early feature that controls spatial expression of developmental genes in the Drosophila embryo?

ANSWER: Concentration-dependent response to graded inputs of Bicoid protein diffusion, which activates expression of other genes such as hunchback.

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chapter 13: The Genetic Control of Development 38. A scientist working in a Drosophila lab locates a mutation in an organism that caused antennae to be transformed into legs. He determines that this is a homeotic transformation. Do you agree?

ANSWER: Yes. This is a gain-of-function dominant mutation.

39. How are alternative forms of the Dsx protein in Drosophila generated?

ANSWER: Alternative forms of the Dsx protein are generated by alternative splicing of the primary dsx RNA transcript.

40. A new bird species was identified and its genome was fully sequenced. How would you determine whether the Sonic hedgehog (Shh) gene is expressed?

ANSWER: The expression of the Shh gene can be determined using bioinformatics to identify the homologous Shh gene based on sequence similarity.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chapter 14: Genomes and Genomics Multiple Choice 1. The genome of Hemophilus influenzae, the first free-living organism to have its genome sequenced, is how large? a. 0.5 megabase b. 1.0 megabase c. 1.8 megabase d. 12 megabase e. 18 megabase ANSWER: c 2. The human genome is comprised of how many base pairs (bp)? a. 1 billion b. 2 billion c. 3 billion d. 4 billion e. 5 billion ANSWER: c 3. The strategy for obtaining a genomic sequence can be divided into four steps: (1) overlap contigs for complete sequence, (2) sequence each fragment, (3) cut many genome copies into random fragments, and (4) overlap sequence reads. What is the order of the four steps? a. 1, 2, 3, and 4 b. 4, 3, 2, and 1 c. 3, 2, 4, and 1 d. 3, 2, 1, and 4 e. 2, 4, 3, and 1 ANSWER: c 4. The term coverage, when applied to genomics, means sequencing a. every base pair of the entire genome. b. every base pair of a single chromosome. c. the same gene in multiple individuals. d. the same gene in multiple species. e. the same base pair multiple times.

ANSWER: e 5. "Traditional" whole-genome shotgun (WGS) sequencing employs which of the following techniques? a. the construction of genomic libraries b. cloning of DNA in bacterial systems c. Sanger dideoxy DNA sequencing Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chapter 14: Genomes and Genomics d. All of the answer options are correct.

ANSWER: d 6. The best evidence to prove that a candidate gene is a disease gene is finding a. a start and stop codon. b. a CpG island upstream. c. that the gene is expressed. d. the homologous gene in many similar animals. e. a mutation in the homologous sequence from a disease sufferer.

ANSWER: e 7. Which of the following is/are FALSE concerning the comparisons between genetic, physical, and cytogenetic maps? a. The distance between two linked markers is the same in genetic and physical maps because crossing over occurs with equal frequency along the entire length of the chromosome. b. Restriction maps, contig maps, and STS maps are examples of physical maps. c. In physical maps, the distances between markers are given in megabases (Mb) where 1 Mb is approximately equal to 1 cM. d. The banding patterns of chromosomes created by different staining techniques are used in constructing cytogenetic maps. e. In genetic maps, the distances between various markers (for example, genes or RFLPs) are given in map units or centiMorgans.

ANSWER: a 8. Deducing the protein-encoding genes from genomic sequences involves a. ORF detection. b. direct evidence from cDNA sequences. c. predictions of binding site. d. predictions based on codon bias. e. All of the answer options are correct.

ANSWER: e 9. What is meant by "codon bias"? a. Codons are specific for the amino acids they specify. b. Certain viruses like to infect particular DNA sequences in our genome. c. Some organisms will use specific codons for amino acids they specify. d. Certain viruses like to infect particular DNA sequences in our genome, and some organisms will use specific codons for amino acids they specify. e. None of the answer options is correct.

ANSWER: c 10. The most closely related genes either within a single organism or between two different organisms are Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chapter 14: Genomes and Genomics called a. homologs. b. orthologs. c. paralogs. d. heterologs. e. alleles.

ANSWER: a 11. Genes that have been inherited from a common ancestor are called a. homologs. b. orthologs. c. paralogs. d. heterologs. e. alleles.

ANSWER: b 12. Genes that are related by gene-duplication events within the same genome are called a. homologs. b. orthologs. c. paralogs. d. heterologs. e. alleles.

ANSWER: c 13. The principle of parsimony refers to a. the most complicated explanation involving the smallest number of evolutionary changes. b. the most complicated explanation involving the largest number of evolutionary changes. c. the simplest explanation involving the smallest number of evolutionary changes. d. the simplest explanation involving the largest number of evolutionary changes. e. explanations that support your hypothesis.

ANSWER: c 14. When comparing different genomes, "synteny" is defined as a. the same genes in the same order. b. the same genes in essentially the same order. c. the same genes in reverse order. d. different genes found on similar chromosomes. e. duplication of entire genomes.

ANSWER: a 15. The "transcriptome" is defined as the Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chapter 14: Genomes and Genomics a. sequence and expression patterns of all transcripts. b. sequence and expression of all proteins. c. complete set of all physical interactions (i.e., protein/DNA or protein/protein). d. complete set of all metabolites. e. None of the answer options is correct.

ANSWER: a 16. The "proteome" is defined as the a. sequence and expression patterns of all transcripts. b. sequence and expression of all proteins. c. complete set of all physical interactions (i.e. protein/DNA or protein/protein). d. complete set of all metabolites. e. None of the answer options is correct.

ANSWER: b 17. The "interactome" is defined as the a. sequence and expression patterns of all transcripts. b. sequence and expression of all proteins. c. complete set of all physical interactions (i.e., protein/DNA or protein/protein). d. complete set of all metabolites. e. None of the answer options is correct.

ANSWER: c 18. Which of the following statements is/are TRUE about reverse genetics? a. Reverse genetics analysis starts with an altered phenotype in an organism. b. Reverse genetics analysis starts with a known DNA sequence, mRNA, or protein. c. Reverse genetics can be performed by random mutagenesis, targeted mutagenesis, or by phenocopying. d. Reverse genetics analysis starts with a known DNA sequence, mRNA, or protein, and it can be performed by random mutagenesis, targeted mutagenesis, or by phenocopying. e. All of the answer options are correct.

ANSWER: e 19. Which of the following methods can be used to inactivate a gene without changing its DNA sequence? a. targeted mutagenesis b. random mutagenesis c. gene knockout d. RNAi e. All of the answer options are correct.

ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chapter 14: Genomes and Genomics 20. The process of identifying all of the functional elements of the genome and determining what those genes do is known as a. bioinformatics. b. annotation. c. comparative genomics. d. functional genomics. e. forward genetics.

ANSWER: b 21. The study and analysis of the information content of genomes is called a. bioinformatics. b. annotation. c. comparative genomics. d. functional genomics. e. forward genetics.

ANSWER: a 22. The use of an expanding variety of methods, including reverse genetics, to understand gene and protein function in biological processes is known as a. bioinformatics. b. annotation. c. comparative genomics. d. functional genomics. e. forward genetics.

ANSWER: d Essay 23. Describe the information content of a genome.

ANSWER: A gene can be viewed as an open-reading frame (ORF) and a series of binding sites for RNAs and proteins. The binding sites would include regulatory elements, promoters, and tRNA and ribosome binding sites.

24. List three approaches to reverse genetics.

ANSWER: (1) Generate random mutations in the genome; (2) generate targeted mutations directly in the gene of interest; and (3) generate phenocopies by treating organisms with agents that interfere with transcription.

25. Describe the difference between draft quality, finished quality, and complete as it relates to genome sequencing.

ANSWER: A draft quality genome sequence is one in which the general outline is identified but significant typographical errors and grammatical errors exist. A draft quality genome also has gaps and sections that need rearranging. A finished quality draft has a very low rate of typographical errors and Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chapter 14: Genomes and Genomics limited missing sections, but everything that is currently possible has been done to fill in these sections. A complete sequence has no typographical errors and every base pair is correct.

26. What is the difference between forward genetics and reverse genetics?

ANSWER: Forward genetics is phenotype driven and seeks to understand what genes underlie a phenotype.

Reverse genetics is genotype driven and seeks to understand phenotypes associated with specific genes.

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chapter 15: DNA Damage, Repair, and Recombination Multiple Choice 1. A mutation that changes the codon UAA to the codon UAG in a mammalian cell line is called a(n) a. synonymous mutation. b. suppressor mutation. c. nonsense mutation. d. missense mutation. e. antisense mutation.

ANSWER: a 2. A mutation changes a codon from AAA (encoding lysine) to AGA (encoding arginine) in yeast, but no mutant phenotype is detected when the mutant strain is plated and grown on minimal or complete medium. This type of mutation is called a. synonymous. b. suppressor. c. nonsense. d. missense. e. frameshift.

ANSWER: d 3. A mutation does not affect the length of a gene but results in an abnormally short protein. The mutation is most likely of a type called a. silent. b. nonsense. c. missense. d. frameshift. e. deletion.

ANSWER: b 4. A small (one-base-pair) insertion in the middle of the coding region of a gene will cause a a. synonymous mutation. b. silent mutation. c. nonsense mutation. d. missense mutation. e. frameshift mutation.

ANSWER: e 5. In a haploid fungus, a transversion arises in the coding region of a gene necessary for the production of the amino acid leucine. The transversion substitutes the last base in the codon UAC (encoding cysteine), resulting in the stop codon UAG. This substitution will most likely cause a. the production of a truncated (shortened) protein. b. the production of a truncated (shortened) mRNA. Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chapter 15: DNA Damage, Repair, and Recombination c. the production of a shortened mRNA and of a shortened protein. d. no detectable phenotype. e. inability to grow on medium that lacks cysteine.

ANSWER: a 6. In a haploid fungus, a small in-frame deletion arises in the coding region of a gene necessary for the production of the amino acid leucine. The deletion removes the three base pairs corresponding to a UAC codon (encoding cysteine). This small deletion will most likely cause a. the production of a truncated (shortened) protein. b. the production of a truncated (shortened) mRNA. c. the production of a shorter mRNA and of a shorter protein. d. no detectable phenotype. e. inability to grow on medium that lacks cysteine.

ANSWER: c 7. A plant is homozygous for a point mutation in gene B. This plant produces wild-type B protein in wild-type amounts, but a more detailed analysis reveals that the B mRNA produced by this plant is two nucleotides shorter than wild type. The mutation is most likely a two-base-pair deletion a. downstream of the stop codon, in the last exon of gene B. b. in an intron of gene B away from the splice sites. c. in the open reading frame of gene B. d. that removed a splice site of gene B. e. within the promoter of gene B. ANSWER: a 8. A point mutation in a gene's promoter will most likely cause the production of a. a shortened mRNA and a truncated protein. b. a shortened mRNA and a wild-type protein. c. reduced amounts of mRNA and protein. d. reduced amounts of mRNA and wild-type amounts of protein. e. wild-type amounts of mRNA and reduced amounts of protein.

ANSWER: c 9. A transversion somewhere within gene B in a laboratory plant results in the production of a mutant mRNA that is much longer than the wild type. This transversion is most likely located a. downstream of the stop codon, in the last exon of gene B. b. in an intron of gene B away from the splice sites. c. in the open reading frame of gene B. d. within a splice site of gene B. e. within the promoter of gene B. ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chapter 15: DNA Damage, Repair, and Recombination 10. A mutant plant with white flowers exists that lacks red anthocyanin pigment, normally made by enzyme P. Indeed, the petal tissue lacks all detectable activity for enzyme P. Despite the lack of enzyme activity, a study of homozygous mutant cells using antibodies against the wild-type enzyme demonstrated that the cells homozygous for the mutation still had the enzyme (i.e., the antibody showed the presence of the enzyme).

Which statement could explain these results? a. The mutant had another gene the researcher was not aware of which produced an enzyme that could function like the mutated enzyme. b. The mutant cells likely had large-scale chromosomal mutations that resulted in the expression of some gene similar to the gene for enzyme P. c. The mutant allele must have had a nonsense mutation that resulted in complete formation of a nonfunctioning enzyme. d. The mutant allele must have only missense mutations that simply knocked out enzyme function, yet the enzyme would still bind to the antibody. e. None of the answer options is correct.

ANSWER: d 11. The fluctuation test of Luria and Delbruck showed that a. a selecting agent can affect mutation rate in E. coli. b. mutations can arise spontaneously before exposure to the selecting agent. c. mutations in E. coli occur at a relatively high frequency. d. the mutation rate in E. coli fluctuates greatly from one generation to the next. e. the T1 phage can act as a mutagen as well as a selecting agent in E. coli.

ANSWER: b 12. In E. coli, a region of a gene with repeats of the sequence CTGG will be prone to a. deletions. b. frameshift mutations. c. missense mutations. d. reversion. e. triplet expansion.

ANSWER: b 13. In E. coli, a region flanked by two repeats of a sequence such as GTGGTGAA is prone to a. deletions. b. duplications. c. frameshift mutations. d. missense mutations. e. reversion.

ANSWER: a 14. Fragile X syndrome is caused by a. exposure to high doses of 5-bromouracil.

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chapter 15: DNA Damage, Repair, and Recombination b. exposure to high doses of free radicals. c. an inherited microdeletion. d. spontaneous depurination. e. trinucleotide expansion.

ANSWER: e 15. Imagine that exposure to the antibiotic streptomycin promotes mutations causing streptomycin resistance in E. coli. If you repeated Luria and Delbruck's test using streptomycin instead of the T1 phage, what results would you expect? a. None of the 20 individual cultures would show any streptomycin-resistant colonies. b. The 20 individual cultures would show a high variation in the number of streptomycin-resistant colonies. c. The 20 individual cultures would show a high variation in the size of streptomycin-resistant colonies. d. The 20 individual cultures would show comparable numbers of streptomycin-resistant colonies. e. The 20 individual cultures would show streptomycin-resistant colonies of unusually large size.

ANSWER: d 16. In a classical genetic experiment, Newcombe spread E. coli cells on complete medium. After several generations of growth, he replica-plated the colonies onto two plates with complete medium plus a selective agent. Colonies on the first replica were left untouched, while on the second replica, they were respread, allowing bacteria from each colony to be "moved" to a new location on the plate. Significantly more mutants after spreading (resistant to the selective agent) were observed than if they had not been respread. The experiment was used to a. demonstrate how to screen for environmental mutagens. b. demonstrate that in some cases mutations are caused by the selective agent itself. c. provide evidence that mutations occur in the absence of the selective agent. d. provide evidence that mutation occurs in prokaryotes as well as in eukaryotes. e. show a direct correlation between the amount of the selective agent used and the number of resistant mutants.

ANSWER: c 17. The rare enol form of thymine pairs with guanine. If a thymine enolization occurs during replication, what would be the mutational event? a. AT to TA b. CG to AT c. CG to GC d. GC to TA e. TA to CG ANSWER: e 18. After mutagen treatment, a molecule of 2-aminopurine (an adenine analogue) incorporates into DNA. During replication, the 2-AP protonates. The mutational event caused by this will be Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chapter 15: DNA Damage, Repair, and Recombination a. AT to CG. b. GC to AT. c. AT to TA. d. AT to GC. e. GC to CG. ANSWER: d 19. During mutagenic treatment with nitrous acid, an adenine deaminates to form hypoxanthine, which bonds like guanine. The mutational event would be a. AT to CG. b. AT to GC. c. AT to TA. d. GC to AT. e. GC to TA. ANSWER: b 20. The spontaneous reversion rate for a chemically induced mutation is 1 × 10–8. For EMS, the rate is 0.9 × 10– 8 , and for acridine, it is 2 × 10–5. What change was involved in the original mutation? a. an AT-to-CG transversion b. an AT-to-GC transition c. a frameshift mutation d. a GC-to-AT transition e. a GC-to-TA transversion ANSWER: c 21. A missense mutation in Neurospora will revert by treatment with nitrous acid, but not by hydroxylamine.

Hydroxylamine (HA) causes only G · C → A · T transitions, while nitrous acid causes both G · C → A · T and A · T → G · C transitions. The original mutation (not the reversion) must have been a. AT to CG. b. AT to GC. c. AT to TA. d. GC to AT. e. GC to TA. ANSWER: d 22. A Neurospora nonsense mutation known to be UAG is treated with two compounds, hydroxylamine and 5bromouracil. Hydroxylamine (HA) causes G · C → A · T transitions, while 5-bromouracil causes T · A → C · G transitions. Which of these agents do you expect to induce revertants? a. Both agents are expected to induce revertants. b. Both agents are expected to induce revertants, but hydroxylamine will be more effective. c. 5-bromouracil only is expected to induce revertants.

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chapter 15: DNA Damage, Repair, and Recombination d. Hydroxylamine only is expected to induce revertants. e. Neither hydroxylamine nor 5-bromouracil agents are expected to induce revertants.

ANSWER: c 23. The Ames test is used for determining whether a particular chemical acts as a mutagen. It does this by selecting for a. new His– mutations in Escherichia coli. b. revertants of His– mutations in Escherichia coli. c. new His– mutations in Salmonella typhimurium. d. revertants of His– mutations in Salmonella typhimurium. e. both revertants and new His– mutations in Escherichia coli.

ANSWER: d 24. A mutation that occurred in a plant petal would be best termed a. dominant. b. germinal. c. recessive. d. somatic. e. suppressor.

ANSWER: d 25. If an incorrect base is incorporated during DNA synthesis and is not corrected by DNA polymerase, it can be corrected by post-replication repair. Post-replication repair does NOT involve which of the following? a. detection of the mismatch b. photoreactivation repair c. a process similar to excision repair d. recognition of the methylation status of the DNA strands e. recombinational repair ANSWER: b 26. In E. coli, mutations arising during repair are mostly caused by a. base excision repair. b. mismatch repair. c. recombinational repair. d. SOS repair. e. thymine dimer splitting.

ANSWER: d 27. Which of the following diseases is LEAST likely to be caused by excessive exposure to UV light? a. Cockayne syndrome Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chapter 15: DNA Damage, Repair, and Recombination b. Huntington disease c. lung cancer d. skin cancer e. xeroderma pigmentosum ANSWER: b 28. Which of the following diseases is MOST likely to be caused by excessive exposure to UV light? a. Cockayne syndrome b. Huntington disease c. lung cancer d. skin cancer e. xeroderma pigmentosum ANSWER: d 29. E. coli cells that have null mutations in the gene encoding adenine methylase will have a. a higher spontaneous mutation rate than wild-type E. coli. b. the same spontaneous mutation rate as wild-type E. coli. c. a lower spontaneous mutation rate than wild-type E. coli. d. more double-strand breaks in their DNA than wild-type E. coli. e. fewer double-strand breaks in their DNA than wild-type E. coli.

ANSWER: a 30. E. coli cells that have null mutations in the gene encoding mutH will be defective in a. base excision repair. b. mismatch repair. c. recombinational repair. d. SOS repair. e. thymine dimer splitting.

ANSWER: b 31. Which mutagen causes single-nucleotide insertions or deletions resulting in frameshift mutations if the mutation occurred in the coding region? Hint: The mutagen is a flat planar molecule that slips between stacked nitrogen bases. a. 2 aminopurine b. UV light c. acridine orange d. aflatoxine e. ethylmethanesulfonate (EMS) ANSWER: c 32. Which mutagen causes transitions (usually GC to AT) by adding an alkyl group to a base (usually G), thus Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chapter 15: DNA Damage, Repair, and Recombination altering its pairing properties? a. 2 aminopurine b. UV light c. acridine orange d. aflatoxin e. ethylmethanesulfonate (EMS) ANSWER: e 33. Which mutagen causes AT-to-GC transitions by acting as a base analog for adenine? Hint: It is unstable and readily undergoes tautomeric shifts. Its imino form pairs with C rather than G. a. 2 aminopurine b. UV light c. acridine orange d. aflatoxin e. ethylmethanesulfonate (EMS) ANSWER: a 34. Which mutagen causes GC-to-TA transversions? Hint: This mutagen creates an apurinic site by breaking the base–sugar bonds. In order to correct the mutation, the SOS repair system preferentially adds an A opposite to G. a. 2 aminopurine b. UV light c. acridine orange d. aflatoxin e. ethylmethanesulfonate (EMS) ANSWER: d 35. Which mutagen makes pyrimidine dimers by making a bond between adjacent pyrimidines, thus interfering with the normal base pairing between the complementary DNA strands? a. 2 aminopurine b. UV light c. acridine orange d. aflatoxin e. ethylmethanesulfonate (EMS) ANSWER: b 36. Planar molecules that mimic base pairs and are able to slip in between the bases inside the DNA double helix are called a. intercalating agents. b. alkylating agents. c. mutagenic compounds.

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chapter 15: DNA Damage, Repair, and Recombination d. auxotrophs. e. methylating agents.

ANSWER: a 37. Which of the following is NOT true of the Ames Test? a. It is used to evaluate the safety of chemical compounds. b. It is used to determine the mutagenic activity of chemicals. c. It tests whether chemicals increase the frequency of mutations in bacteria. d. It can be used to detect malignant cells in tissue. e. It can be used to screen thousands of compounds to weed out potential carcinogens.

ANSWER: d 38. Which of the following repair mechanisms will remove damaged bases? a. mismatch repair b. base excision repair c. nucleoside excision repair d. damage repair e. recombinational repair ANSWER: b Multiple Response 39. Shown below is a list of statements (A–K). Check all the statements that describe a missense mutation. a. a mutation that changes UUA to UUG b. a mutation that gives methionine instead of leucine c. created by the addition of a nucleotide to a coding region d. a stop codon is read as an amino acid e. a chemically similar amino acid is replaced by the mutation f. a mutation that changes CCU to ACU g. deletion of a nucleotide in a coding region h. mutation does not alter the peptide i. a mutation changing UAU to UAG j. premature termination codon is responsible for this mutation k. a chemically different amino acid is replaced by the mutation ANSWER: b, e, f, k 40. Shown below is a list of statements (A–K). Check all the statements that describe a silent mutation. a. a mutation that changes UUA to UUG b. a mutation that gives methionine instead of leucine c. created by the addition of a nucleotide to a coding region d. a stop codon is read as an amino acid Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chapter 15: DNA Damage, Repair, and Recombination e. a chemically similar amino acid is replaced by the mutation f. a mutation that changes CCU to ACU g. deletion of a nucleotide in a coding region h. mutation does not alter the peptide i. a mutation changing UAU to UAG j. premature termination codon is responsible for this mutation k. a chemically different amino acid is replaced by the mutation ANSWER: a, h 41. Shown below is a list of statements (A–K). Check all the statements that describe a frameshift mutation. a. a mutation that changes UUA to UUG b. a mutation that gives methionine instead of leucine c. created by the addition of a nucleotide to a coding region d. a stop codon is read as an amino acid e. a chemically similar amino acid is replaced by the mutation f. a mutation that changes CCU to ACU g. deletion of a nucleotide in a coding region h. mutation does not alter the peptide i. a mutation changing UAU to UAG j. premature termination codon is responsible for this mutation k. a chemically different amino acid is replaced by the mutation ANSWER: c, g 42. Shown below is a list of statements (A–K). Check all the statements that describe a nonsense mutation. a. a mutation that changes UUA to UUG b. a mutation that gives methionine instead of leucine c. created by the addition of a nucleotide to a coding region d. a stop codon is read as an amino acid e. a chemically similar amino acid is replaced by the mutation f. a mutation that changes CCU to ACU g. deletion of a nucleotide in a coding region h. mutation does not alter the peptide i. a mutation changing UAU to UAG j. premature termination codon is responsible for this mutation k. a chemically different amino acid is replaced by the mutation ANSWER: i, j 43. Shown below is a list of statements (A–K). Check all the statements describe a synonymous mutation. a. a mutation that changes UUA to UUG b. a mutation that gives methionine instead of leucine Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chapter 15: DNA Damage, Repair, and Recombination c. created by the addition of a nucleotide to a coding region d. a stop codon is read as an amino acid e. a chemically similar amino acid is replaced by the mutation f. a mutation that changes CCU to ACU g. deletion of a nucleotide in a coding region h. mutation does not alter the peptide i. a mutation changing UAU to UAG j. premature termination codon is responsible for this mutation k. a chemically different amino acid is replaced by the mutation ANSWER: a, h 44. Shown below is a list of statements (A–K). Check all the statements that describe a nonsense suppressor mutation. a. a mutation that changes UUA to UUG b. a mutation that gives methionine instead of leucine c. created by the addition of a nucleotide to a coding region d. a stop codon is read as an amino acid e. a chemically similar amino acid is replaced by the mutation f. a mutation that changes CCU to ACU g. deletion of a nucleotide in a coding region h. mutation does not alter the peptide i. a mutation changing UAU to UAG j. premature termination codon is responsible for this mutation k. a chemically different amino acid is replaced by the mutation ANSWER: d 45. Below is a list of types of repair systems or repair molecules (A–G). Select all the systems and molecules that can repair (or play a role in the repair) of alkylation of G residues. a. photolyase b. base-excision repair c. SOS repair d. alkyltransferases e. mismatch repair f. homologous recombination g. nucleotide-excision repair ANSWER: b, d, g 46. Below is a list of types of repair systems or repair molecules (A–G). Select all the systems and molecules that can repair (or play a role in the repair) of depurination. a. photolyase Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chapter 15: DNA Damage, Repair, and Recombination b. base-excision repair c. SOS repair d. alkyltransferases e. mismatch repair f. homologous recombination g. nucleotide-excision repair ANSWER: b, g 47. Below is a list of types of repair systems or repair molecules (A–G). Select all the systems and molecules that can repair (or play a role in the repair) of pyrimidine dimers. a. photolyase b. base-excision repair c. SOS repair d. alkyltransferases e. mismatch repair f. homologous recombination g. nucleotide-excision repair ANSWER: a, c, g 48. Below is a list of types of repair systems or repair molecules (A–G). Select all the systems and molecules that can repair (or play a role in the repair) of deamination. a. photolyase b. base-excision repair c. SOS repair d. alkyltransferases e. mismatch repair f. homologous recombination g. nucleotide-excision repair ANSWER: b, g 49. Below is a list of types of repair systems or repair molecules (A–G). Select all the systems and molecules that can repair (or play a role in the repair) of double-stranded breaks. a. photolyase b. base-excision repair c. SOS repair d. alkyltransferases e. mismatch repair f. homologous recombination g. nucleotide-excision repair ANSWER: f Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chapter 15: DNA Damage, Repair, and Recombination Essay 50. A mouse is homozygous for a transversion that eliminated the splice site between the first exon and the first intron of gene G. The resulting mutant mRNA is longer than the wild type, but the protein that it encodes is much shorter than its wild-type counterpart. Provide a logical and plausible explanation for these observations.

ANSWER: The mutation causes the removal of a splice site, so an intron will remain part of the (mutant) mRNA. If this intron contains a stop codon that is in the same reading frame as the mRNA's ORF (i.e., a premature stop codon), the resulting protein will be shorter than the wild type (truncated).

51. A partial peptide sequence for the wild type and three mutant alleles of a gene, PET1, are shown below.

Each mutant was caused by a single point mutation.

a) Using the amino acid sequence of the wild type and the three mutants, deduce the exact DNA sequence of the coding strand of the wild-type allele. b) What type of mutation (transition vs. transversion vs. indel; missense vs. nonsense vs. frameshift) has occurred in each mutant? c) What consequence will each of the above-mentioned DNA mutations have on protein function?

ANSWER: a) Each mutation helps determine the correct codon possible for the amino acids. Students therefore cannot get full credit for indicating alternate codons for an amino acid (except for the lys codon, which could be AAA or AAG).

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chapter 15: DNA Damage, Repair, and Recombination

b) Mutant 1: Insertion of either a C in the first base of the third codon or a T, C, or A in the third base of the second codon; frameshift mutation.

Mutant 2: A to G transition in the second base of the fifth codon; missense mutation Mutant 3: A to C or T transversion in the third base of the third codon; missense mutation. c) Mutant 1: Truncated peptide. The protein is probably nonfunctional.

Mutants 2 and 3: Altered peptide sequence. Unknown effect on the protein's function.

52. Three mutations were obtained in a bacterial gene. An antibody is available for the protein product of this gene. Both Northern analysis (RNA separated by electrophoresis, blotted, and probed with DNA) and Western analysis (proteins separated by electrophoresis, blotted, and probed with antibodies) were performed on the mutants. The results are summarized below.

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chapter 15: DNA Damage, Repair, and Recombination For each mutation, what kind of mutation occurred and how do you know it? a) Mutant 1 b) Mutant 2 c) Mutant 3 ANSWER: a) Mutant 1 is a missense or frameshift mutation (no change in RNA or protein size). b) Mutant 2 is a nonsense mutation (RNA same size but protein shorter). c) Mutant 3 is possibly a promoter mutation as no RNA is transcribed, or possibly a deletion of the gene.

53. Some indel mutations can be created spontaneously as a result of errors during DNA replication. a) Use diagrams to show how (1) insertions and (2) deletions are spontaneously created as a result of the replication of tandem repeats. b) Give the term used to describe this process. c) Name two human diseases that arise as a consequence of expanded trinucleotide repeats in DNA. d) Describe two ways in which expanded trinucleotide repeats can cause disease.

ANSWER: a) Slippage of the template strand during replication results in deletion. Shown below is a deletion of two repeats (total of 6 bp).

Slippage of the daughter strand during replication results in insertions. Shown below is an insertion of two repeats (total of 6 bp).

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chapter 15: DNA Damage, Repair, and Recombination

b) replication slippage c) fragile X syndrome, Huntington disease, Kennedy disease, myotonic dystrophy d) (1) If the trinucleotide repeats are in the coding region, the protein could have additional amino acids (such as glutamine repeats), which could disrupt the function of the protein.

(2) Regulation of gene expression could be affected if the repeats are in the promoter region or the 5′ UTR or 3UTR.

54. Both cytosines and 5′ methyl-cytosines are susceptible to spontaneous deamination, which leads to GC-toAT transitions. Do you expect cytosine nucleotides to be a "hot spot" for spontaneous mutations? Briefly explain your answer.

ANSWER: No, (unmethylated) cytosines are not expected to be a hot spot for mutations. A deaminated cytosine is a uracil, which does not normally occur in DNA. The excision-repair mechanism will promptly identify it and remove it.

55. 5-Bromouracil is an analog of thymine that normally pairs with adenine. Its rare tautomeric form pairs with adenine. Show the steps (i.e., several replications) by which 5-bromouracil causes the GC base pair shown below to change to an AT base pair. Show the point at which the rare tautomeric form of 5-bromouracil occurs.

Show both products of all divisions in which 5-bromouracil is involved. Do not show chemical structures.

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chapter 15: DNA Damage, Repair, and Recombination ANSWER:

56. Five different mutations were derived from base-pair substitutions at a single codon. In this codon, the mutant alleles had arginine, leucine, glycine, serine, and cysteine. What was the wild-type codon?

Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chapter 15: DNA Damage, Repair, and Recombination ANSWER: Tryptophan (UGG) can give all five amino acids by single base-pair transitions or transversions at positions 1, 2, or 3.

57. a) A met+ strain of Neurospora was treated with a mutagen to create met– mutants. The mutants were reverted to met+ with HA (hydroxylamine), which causes GC-to-AT transitions. What was the original mutation on the molecular level? b) Two of the revertants showed odd results when crossed with a wild-type met+ strain.

Cross met+revertant A × met+ wild type

Progeny 88% met+ 12% met–

met+revertant B × met+ wild type

93% met+ 7% met–

Explain how these results occurred and why the above numbers of offspring were obtained.

ANSWER: a) HA causes GC-to-AT transitions, so the original mutation must have been AT to GC. b) This reversion is not true reversion but suppression. Furthermore, in case A the suppressor must be 24 m.u. away from the met locus, although in B it is 14 m.u. away.

58. How do mutation and DNA damage differ?

ANSWER: Mutation is a change in the information in DNA, which is nevertheless chemically normal. DNA damage is chemically abnormal DNA, which in many cases makes it recognizable by repair enzymes and susceptible to repair.

59. Photoreactivation repair is effective only in the presence of visible light. What other repair mechanism is available to repair pyrimidine dimers if visible light was not available? Describe the steps of this mechanism.

ANSWER: Nucleotide excision repair can be used to repair pyrimidine dimers in the dark. This system detects distortions in the double helix created by abnormal bases. UvrABC excinucleases cleave flanking bases at the damaged area to remove the abnormal bases. DNA polymerase I then repairs the missing segment and DNA ligase seals the newly synthesized bases to the original DNA strand.

60. Crossing over is a very precise process. Why is it important that the process of crossing over is so precise?

What impact would there be if it were not precise? Draw a picture depicting an imprecise crossover and show the result of this imprecise event.

ANSWER: Imprecise crossing over would lead to either the gain or loss of DNA (and possibly genes) as depicted in the following diagram.

Copyright Macmillan Learning. Powered by Cognero.

Page 18

Name:

Class:

Date:

Chapter 15: DNA Damage, Repair, and Recombination

1 61. The partial DNA sequences of a wild-type and mutant alleles of a gene and are listed below: Wild-type: TCCATTGGCAGACCAGAATC Mutant: TCCATTGGCTGACCAGAATC How has the substitution changed the gene? What are the consequences of such a mutation?

ANSWER: The nucleotide substitution seems to have created a stop codon (TGA). This can result in a truncated nonfunctional peptide.

62. Describe two types of spontaneous mutations induced by the cellular environment ANSWER: Depurination, deamination, and oxidative damage are the three spontaneous mutations induced by the cellular environment.

Depurination is the loss of a purine base, guanine or adenine caused by hydrolysis of the glycosidic bond between the base and deoxyribose sugar. Deamination is the hydrolytic removal of an amine group that alters the three DNA bases that contain an amino group (cytosine, adenine, and guanine).

Deamination converts cytosine to uracil, adenine to hypoxanthine, and guanine to xanthine. Oxidative DNA damage is caused by reactive oxygen species such as superoxide radicals, hydrogen peroxide, and hydroxyl radicals that are byproducts of normal aerobic metabolism of molecular oxygen by mitochondria resulting in lesions on DNA, breaks in DNA strands, and faulty links and base gaps in sequences. (Students can describe any two.)

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chapter 16: The Dynamic Genome: Transposable Elements Multiple Choice 1. Which of the following scientists discovered the Ac and Ds transposable elements in maize? a. Marcus Rhoades b. Rollins Emerson c. Barbara McClintock d. George Beadle e. All of these scientists made the discovery.

ANSWER: c 2. A corn plant is homozygous for a mutant allele that results in no pigment in the seed (i.e., white). The mutant is caused by Ds insertion that often exits late in seed development, when there is an active Ac element in the genome. The seeds of this plant will be a. without pigment (i.e., white). b. pigmented all over. c. white with small spots of pigment. d. white with large spots of pigment. e. weakly pigmented.

ANSWER: c 3. A corn plant is homozygous for a mutant allele that results in no pigment in the seed (i.e., white). The mutant is caused by Ds insertion that often exits late in seed development, when there is an active Ac element in the genome. If there is NO active Ac element, the seeds of this plant will be a. without pigment (i.e., white). b. pigmented all over. c. white with small spots of pigment. d. white with large spots of pigment. e. weakly pigmented.

ANSWER: a 4. Which of the following features do bacterial and corn transposons not have in common? a. Both may cause unstable mutations. b. Both may carry drug resistance genes in natural populations. c. Both may have inverted repeats. d. Both may move to new loci. e. Both may cause rearrangements.

ANSWER: b 5. An autonomous element a. requires no other elements for its mobility. b. requires one or more additional elements for its mobility. c. is found only in corn (maize).

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chapter 16: The Dynamic Genome: Transposable Elements d. is found only in plants. e. None of the answer options is correct.

ANSWER: a 6. A nonautonomous element a. requires no other elements for its mobility. b. requires one or more additional elements for its mobility. c. is found only in corn (maize). d. is found only in plants. e. None of the answer options is correct.

ANSWER: b 7. Bacterial transposon structure can be thought of as a. IS sequences flanked by inverted drug-resistance genes. b. drug-resistance gene(s) flanked by IS elements. c. drug-resistance gene(s) flanked by a pair of mu (µ) phages. d. a mu (µ) phage flanked by two IS elements. e. resistance gene(s) flanked by inverted resistance transfer factors (RTFs).

ANSWER: b 8. Hybridization of single-stranded wild-type DNA with DNA from mutations caused by IS elements characteristically shows (through electron microscopy) a. chi structures. b. unpaired tails. c. a single-stranded loop representing IS DNA. d. a single-stranded loop representing wild-type DNA. e. theta structures.

ANSWER: c 9. IS-induced mutations are different from missense mutations in that they are a. nonrevertible. b. less severe. c. more severe. d. nonpolar. e. not reverted by mutagens.

ANSWER: e 10. Retroviruses replicate using a. DNA polymerase. b. RNA polymerase. c. restriction endonuclease.

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chapter 16: The Dynamic Genome: Transposable Elements d. reverse transcriptase. e. topoisomerase.

ANSWER: d 11. Retrotransposons move via an intermediate that is a. a double-stranded lollipop. b. a retrovirus. c. double-stranded RNA. d. single-stranded DNA. e. single-stranded RNA. ANSWER: e 12. The first eukaryotic transposable elements to be characterized at the molecular level were identified within the genes of which organism? a. corn b. yeast c. E. coli d. mouse e. human ANSWER: b 13. Eukaryotic retrotransposons such as Ty1 and Copia are flanked by a. direct repeats (DRs). b. inverted repeats (IRs). c. long terminal repeats (LTRs). d. long inverted repeats (LIRs). e. None of the answer options is correct.

ANSWER: c 14. Transposable elements that transpose via an RNA intermediate are known as a. class 1 elements. b. class 2 elements. c. class 3 elements. d. alpha elements. e. D elements.

ANSWER: a 15. Transposable elements that are comprised of DNA transposons are known as a. class 1 elements. b. class 2 elements. c. class 3 elements.

Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chapter 16: The Dynamic Genome: Transposable Elements d. alpha elements. e. D elements.

ANSWER: b 16. In D. melanogaster, when M cytotype (lab stock) females are crossed to P cytotype (wild) males, the resulting F1 progeny are a. normal. b. defective. c. normal:defective in a 1:1 ratio. d. normal:defective in a 1:2:1 ratio. e. normal:defective in a 3:1 ratio.

ANSWER: b 17. In D. melanogaster, when P cytotype (lab stock) females are crossed to M cytotype (wild) males, the resulting F1 progeny are a. normal. b. defective. c. normal:defective in a 1:1 ratio. d. normal:defective in a 1:2:1 ratio. e. normal:defective in a 3:1 ratio.

ANSWER: a 18. Which of the following statements is/are TRUE about the P elements of Drosophila? a. P elements are examples of eukaryotic DNA transposons. b. P elements are mobilized when females with these elements are crossed to males that lack them. c. P elements are silenced in M cytotype. d. P elements are examples of eukaryotic DNA transposons that are mobilized when females with these elements are crossed to males that lack them. e. P elements are examples of eukaryotic DNA transposons that are silenced in M cytotype.

ANSWER: a 19. LINEs differ from retrotransposons in that LINEs do NOT a. encode transposase. b. encode reverse transcriptase. c. contain LTRs. d. transpose in a replicative manner. e. contain the transposase gene.

ANSWER: c 20. What percentage of the human genome is derived from transposable elements? a. less than 5% Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chapter 16: The Dynamic Genome: Transposable Elements b. 25% c. 50% d. 75% e. nearly 100% ANSWER: c 21. The grasses (such as barley, rice, sorghum, and corn) vary widely in the size of their genomes, yet they are descended from a common ancestor. What accounts for the variation in DNA content (i.e., the total amount of DNA)? a. large-scale duplication of genes b. large-scale loss of genes c. accumulation of LTR retrotransposons d. loss of LTR retrotransposons e. None of the answer options is correct.

ANSWER: c 22. Which of the following statements is/are FALSE about Tc1 elements found in C. elegans? a. Tc1 elements are retrotransposons. b. Tc1 elements can transpose only in somatic cells. c. Tc1 elements were used to study the RNAi based silencing mechanisms. d. Single Tc1 element can trigger the silencing of all copies of Tc1 elements in the C. elegans genome. e. None of the answer options is correct.

ANSWER: a 23. Match the following transposable elements with the species where it is found.

a. Ds element in maize, Ty1 element in yeast, L1 element in humans, P element in Drosophila. b. Ds element in humans, Ty1 element in Drosophila, L1 element in yeast, P element in maize. c. Ds element in maize, Ty1 element in Drosophila, L1 element in humans, P element in yeast. d. Ds element in yeast, Ty1 element in maize, L1 element in humans, P element in Drosophila. e. Ds element in humans, Ty1 element in yeast, L1 element in maize, P element in Drosophila.

ANSWER: a 24. The most abundant SINE in humans is called Alu. Which of the following is/are TRUE of Alu sequences? a. The human genome contains exactly 1 million whole Alu sequences. b. Alu sequences make up more than 20% of the human genome. c. Alu sequences are found within exons. d. The full Alu sequence is about 300 nucleotides long. e. All of the above.

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chapter 16: The Dynamic Genome: Transposable Elements ANSWER: d 25. What is one of the functions of RNAi in eukaryotes? a. To enhance the expression of active transposable elements in their genomes. b. To express active transposable elements in their genomes. c. To repress the expression of active transposable elements in their genomes. d. To repress the expression of inactive transposable elements in their genomes. e. To enhance the expression of inactive transposable elements in their genomes.

ANSWER: c 26. What are piRNAs? a. Long single-stranded RNA that is mainly expressed in germ line cells. b. Short single-stranded RNA that is mainly expressed in germ line cells and are approximately 23–30 nucleotides long. c. Short double-stranded RNA that is mainly expressed in germ line cells and are approximately 23–30 nucleotides long. d. RNA transcribed from the pi-clusters and then processed into double-stranded DNA of > 30 nucleotides in length. e. Long RNAs transcribed from the pi-clusters.

ANSWER: b Essay 27. Transposable elements can cause corn kernels to be spotted if they jump in or out of the color genes during the development of the kernel. However, the size of the spots can vary greatly. In words and diagrams, describe the difference between the two kernels shown below at the molecular and developmental levels.

ANSWER: Kernel development consists of many rounds of mitosis. When a transposon jumps out of a disrupted "color gene" in a cell, that cell, and all cells descending from it, will contain a functional copy of the color gene, creating a sector of purple cells surrounded by yellow cells. The earlier the transposition event occurs in development, the more descendants will arise. Thus, a bigger purple spot will be observed (see the diagram below). Therefore, transposition probably occurred earlier in kernel A. Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chapter 16: The Dynamic Genome: Transposable Elements

28. Compare and contrast autonomous and nonautonomous transposable elements.

ANSWER: Autonomous and nonautonomous elements contain inverted terminal repeats. Autonomous elements encode functional transposase enzymes. Nonautonomous elements do not encode functional enzymes. The gene may contain point mutations or, more likely, deletions that eliminate its function.

29. A maize plant is homozygous for an allele Cm, caused by the insertion of a Ds element into the coding region of a gene (C) that encodes an enzyme necessary for the synthesis of anthocyanin in maize kernels. a) If this plant also contains an Ac element in its genome, what will be the phenotype of its kernels? Justify your answers. b) If this plant does not contain an Ac element, what will be the kernel phenotype? Justify your answers. c) If the plant does not contain an Ac element, but it does contain an autonomous element for another family of transposon, what will be the kernel phenotype? Justify your answers.

ANSWER: a) spotted (The Ac element provides a transposase that can move the Ds element during kernel development.) b) colorless (no transposase = no movement = no restoration of color) c) colorless (The transposase enzyme is family specific; similar to part (b), there is no transposase that can recognize and move elements of the Ac/Ds family.)

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chapter 16: The Dynamic Genome: Transposable Elements 30. Compare and contrast IS elements, simple transposons, and composite transposons. Use diagrams to help in the comparison.

ANSWER: IS element: IS elements are short mobile sequences that contain a transposase gene flanked by two inverted repeats (IR). They do not contain any genes other than transposase.

Simple transposon: Simple transposons are similar to IS elements in that they are composed of a transposase gene flanked by IRs. However, they differ from IS elements in that they contain additional bacterial genes between the two IRs. Composite element: Composite transposons contain one or more genes (but no transposase) between two IS elements that are oriented in opposite directions. The movement of the element is dependent on the transposase encoded by one of the flanking IS elements. They are different from simple transposons in that their ability to move depends on a transposase made by either of the IS elements because no transposase gene resides between the two IS elements.

31. Approximately 45 percent of the human genome is derived from transposable elements, such as LINEs and SINEs. a) What are LINEs and SINEs? b) How do LINEs differ from SINEs? c) How is survival possible with this high a percentage of transposable elements in the human genome?

ANSWER: a) LINEs: long interspersed nuclear elements SINEs: short interspersed nuclear elements b) SINEs are the nonautonomous version of the LINEs. Therefore, SINEs can transpose only by using a reverse transcriptase encoded by LINEs. c) Several reasons: (1) Transposable elements inserted in introns are positively selected and do not harm gene expression. (2) The majority of elements are inactive due to the accumulation of mutations. Therefore, they can no longer jump into vital genes. (3) The active elements that are capable of increasing in copy number are rendered inactive by host regulatory mechanisms.

32. Why doesn't hybrid dysgenesis occur in Drosophila when a P cytotype female is mated with an M cytotype male, but does occur in the reciprocal mating?

ANSWER: The P elements of Drosophila are repressed in the germ line by the piRNA-Piwi complex early in embryonic development. When a P cytotype female is mated with an M cytotype male, these offspring are protected by piRNAs homologous to the P-element, which are sequestered in the embryo of P-type females and deposited in cells of the primordial germ line. As these protective piRNAs target the P-element mRNA for degradation, their presence reduces P-element activity and the resulting DNA damage that lead to hybrid dysgenesis, allowing healthy P (female) × M (male) Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chapter 16: The Dynamic Genome: Transposable Elements F1s.

In the reciprocal cross, an M female has no P elements in her pi-clusters; therefore, no maternal piRNAs complementary to Pelements are loaded into her embryos. The F1 receives P elements from the male parent but no maternally loaded piRNAs to activate the piRNA-Piwi complex to repress their movement. As development proceeds, the P elements cause genomic havoc in the germ line, and the M (female) × P (male) F1s are dysgenic.

33. A new insertion of a transposon into the genome of C. elegans generated antisense RNA. Postulate the response of the host cells.

ANSWER: The host responds by producing siRNAs that target the transposase mRNA, silencing the gene and preventing the movement of all transposable elements.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chapter 17: Large-Scale Chromosomal Changes Multiple Choice 1. A particular hybrid species is defined as 3n = 12. It is discovered that at meiosis, pairing is always univalent + bivalent (never trivalent) in this species. How many bivalents will be present in prophase of meiosis I? a. 2 b. 3 c. 4 d. 8 e. 12 ANSWER: c 2. A hybrid allotetraploid species (2n = 60) was backcrossed to one of the suspected parents (2n = 30). When the F1 underwent meiosis, the prophase chromosome configuration was examined. If the guess about the suspected parent was correct, what would the chromosome configuration look like? a. 30 pairs b. 45 singles c. 60 singles d. 15 pairs and 15 singles e. 30 pairs and 15 singles ANSWER: d 3. In a triploid of genotype B/b/b, what proportion of gametes will be B? a. 1/6 b. 1/4 c. 1/3 d. 1/2 e. 2/3 ANSWER: a 4. A triploid with five chromosomes in each set (3n = 15) is discovered. What is the probability of a meiosis in which all univalents pass to the same pole? a. 1/5 b. 1/10 c. 1/15 d. 1/16 e. 1/25 ANSWER: d 5. A man is found to be karyotypically 47, XYY. The presence of an extra Y chromosome most likely results from a. nondisjunction in a maternal meiocyte at meiosis I. Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chapter 17: Large-Scale Chromosomal Changes b. nondisjunction in a maternal meiocyte at meiosis II. c. nondisjunction in a paternal meiocyte at meiosis I. d. nondisjunction in a paternal meiocyte at meiosis II. e. unequal crossing over.

ANSWER: d 6. In Neurospora, the genes his2 and leu2 are closely linked. From a cross, his2 × leu2, which is a rare octad of the following type, arose: his2, + his2, + his2, + his2, + abort abort +, leu2 +, leu2 This octad most likely arose by a. crossover between his2 and leu2, followed by first-division nondisjunction. b. first-division nondisjunction. c. post-meiotic mitotic nondisjunction. d. rare triploidy. e. second-division nondisjunction.

ANSWER: e 7. In corn, R is a gene for red aleurone; its recessive allele r determines colorless aleurone. A cross is made between a diploid r/r female and a trisomic R/r/r male. If pollen grains with extra chromosomes are inviable, the ratio of red to colorless kernels expected in the progeny would be a. 1 red : 5 colorless. b. 1 red : 2 colorless. c. 1 red : 1 colorless. d. 2 red : 1 colorless. e. 5 red : 1 colorless.

ANSWER: b 8. In a tetraploid species, an individual with genotype B/b/b/b was crossed with an individual with genotype b/b/b/b. What is the expected genotypic ratio in the offspring? a. 1:1 b. 1:2 c. 1:4 d. 1:8 e. 1:16 Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chapter 17: Large-Scale Chromosomal Changes ANSWER: a 9. A triploid of genotype a b / a B / a b is crossed to a triploid of genotype A b / a b / a b. If genes A (a) and B (b) are 20 map units apart, what proportion of the progeny will be phenotypically A B? a. 1/3 b. 1/4 c. 1/6 d. 1/9 e. 1/36 ANSWER: c 10. Suppose that in guinea pigs ABO blood types are determined like in humans, where IA and IB are expressed codominantly and i (blood type O) is recessive to both. During a set of routine crosses you notice something unusual: repeated matings between a certain female with blood type AB and a certain male with blood type A produces F1 guinea pigs of all four blood types in the following ratios: 5A : 2B : 4AB : 1O. The genotypes of the parents are most likely: a. IA/IB for the female and IA/IA for the male. b. IA/IB for the female and IA/i for the male. c. IA/IB/i for the female and IA/IA for the male. d. IA/IB/i for the female and IA/i for the male. e. None of the answer options is correct.

ANSWER: d 11. In humans, males have only one X chromosome and females have two, yet housekeeping X chromosome genes are expressed in roughly equal amounts in males and females. How can this be explained? a. The X chromosome is hyperactivated in males. b. The X chromosomes are hypoactivated in females. c. One X chromosome in females is inactivated. d. The Y chromosome in males actually contains similar genes as the X chromosome.

ANSWER: c 12. A plant trisomic for a chromosome carrying the A gene has the genotype A/a/a. Which of the following represents a gamete this plant could NOT produce? a. A/a b. A c. a d. A/A ANSWER: d 13. The red fox has 17 pairs of large, long chromosomes. The arctic fox has 26 pairs of smaller shorter chromosomes. What do you expect to be the chromosome number in somatic tissues of a red fox/arctic fox Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chapter 17: Large-Scale Chromosomal Changes hybrid? a. 13 b. 17 c. 26 d. 43 ANSWER: d 14. In the plant genus Triticum there are many different polyploid species as well as diploid species. Crosses were made between three different species, and hybrids were obtained. The meiotic pairing was observed in each hybrid and is recorded in the following table. (A bivalent is two homologous chromosomes paired at meiosis, and a univalent is an unpaired chromosome at meiosis.)

Species crossed to make hybrid 1. T. turdigum × T. monococcum 2. T. aestivum × T. monococcum 3. T. aestivum × T. turdigum

Pairing in hybrid 7 bivalents + 7 univalents 7 bivalents + 14 univalents 14 bivalents + 7 univalents

The first cross suggests a diploid and tetraploid individual were involved with one set of seven chromosomes in common. Based on the other crosses, which species is the likely diploid individual? a. T. turdigum b. T. monococcum c. T. aestivum d. T. turdigum or T. monococcum ANSWER: b 15. The n + 1 female gametophytes (embryo sacs) produced by trisomic plants are usually more viable than the n + 1 male gametophytes (pollen grains). If 50% of the functional embryo sacs of a selfed trisomic plant are n + 1 but only 10% of the functional pollen grains are n + 1, what percentage of the offspring will be diploid? a. 0 b. 5 c. 45 d. 90 ANSWER: c 16. In the following pedigree, the shaded symbols represent a rare X-linked disease of the blood. The propositus has 45 chromosomes and is sterile. Propose a single mechanism to account for both the sterility and the blood disease in this female.

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chapter 17: Large-Scale Chromosomal Changes a. The woman's father was a carrier for the disease on his X chromosome. b. The woman is likely demonstrating a case of Turner syndrome. c. The woman is likely demonstrating a case of Down syndrome. d. The woman's mother's eggs likely experienced a nondisjunction event.

ANSWER: b 17. In the flowering plants (the angiosperms), the process of double fertilization occurs. This is where one sperm nucleus from the pollen grain fertilizes the egg, and the other fuses with the two polar nuclei in the female gametophyte to produce the triploid (3n) endosperm tissue. The aleurone layer (the outer layer of cells of the endosperm) can be colored or colorless. The allele C causes color (purple), and the allele c gives colorless aleurone. A plant is of genotype Cc and self-pollinates. What ratio of colored to colorless aleurone layered endosperms are expected in the progeny? a. 1:1 b. 2:1 c. 3:1 d. 10:1 ANSWER: c 18. A wild-type chromosome can be represented as ABC[*]DEFGH, and from this a chromosomal aberration arises that can be represented as ABC[*]DEGFH, where [*] represents the centromere. This aberration is known as a a. deletion. b. duplication. c. paracentric inversion. d. pericentric inversion. e. translocation.

ANSWER: c 19. In an animal bearing the heterozygous inversion ABCDE[*]FGHI / ABGF[*]EDCHI, in one meiocyte, a crossover occurred between the D and E loci and another crossover occurred between the F and G loci. The two crossovers involved the same two chromatids. What will be the proportion of abnormal meiotic products from that meiosis? (Note: [*] = centromere.) a. 0% b. 25% c. 50% d. 100% e. It depends on the distance between D and E and between F and G. ANSWER: a 20. In an animal bearing the heterozygous inversion ABCDE[*]FGHI / ABGF[*]EDCHI, in one meiocyte a crossover occurred between the D and E loci and another crossover occurred between the F and G loci. The two crossovers involved all four chromatids. What will be the proportion of abnormal meiotic products from that meiosis? (Note: [*] = centromere.)

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chapter 17: Large-Scale Chromosomal Changes a. 0% b. 25% c. 50% d. 100% e. It depends on the distance between D and E and between F and G. ANSWER: d 21. Assume you are studying an organism in which duplications or deficiencies in the gametes are not viable.

An individual is heterozygous for a pericentric inversion. A four-chromatid double crossover, with one of the two crossovers within the inversion and the other outside of it, would lead to a. four viable gametes, two of them parental and two recombinant. b. four viable gametes, two of them with the inversion and two without. c. two viable gametes, both recombinant. d. two viable gametes, both parental. e. zero viable gametes.

ANSWER: c 22. Assume you are studying an organism in which duplications or deficiencies in the gametes are not viable.

An individual is heterozygous for a pericentric inversion. A four-chromatid double crossover, with both crossovers occurring within the inversion, would lead to a. four viable gametes, two of them parental and two recombinant. b. four viable gametes, two of them with the inversion and two without. c. two viable gametes, both recombinant. d. two viable gametes, both parental. e. zero viable gametes.

ANSWER: e 23. Assume you are studying an organism in which duplications or deficiencies in the gametes are not viable.

An individual is heterozygous for a paracentric inversion. A four-chromatid double crossover, with one of the two crossovers within the inversion and the other outside of it, would lead to a. four viable gametes, two of them parental and two recombinant. b. four viable gametes, two of them with the inversion and two without. c. two viable gametes, both recombinant. d. two viable gametes, both parental. e. zero viable gametes.

ANSWER: e 24. The two loci Y (y) and R (r) are normally 20 map units apart. A particular dihybrid individual is heterozygous for an inversion that spans 98% of the region between Y and R but does not include Y or R. If this dihybrid is testcrossed to an individual with normal chromosomes, the recombinant frequency in progeny will be a. more than 20%.

Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chapter 17: Large-Scale Chromosomal Changes b. 20%. c. less than 20%, but more than 10%. d. 10%. e. less than 10%.

ANSWER: e 25. In rabbits, genes T and R are located on the same chromosome, 20 map units apart. A colleague sends you one of his pure-breeding t r/t r males, which you cross with a T R / T R individual. Each of the T R / t r F1 heterozygotes is then crossed back with your colleague's double homozygous recessive. Altogether, the backcrosses produce 50 rabbits, of which 24 are T R / t r, 24 are t r / t r, 1 is T r / t r, and 1 is t R / t r. Which of the following explanations BEST justifies the results? a. The colleague's rabbit is heterozygous for an inversion that spans at least 16 map units of the region between genes t and r. b. The colleague's rabbit is heterozygous for an inversion that spans at least 18 map units of the region between genes t and r. c. The colleague's rabbit is homozygous for an inversion that spans at least 16 map units of the region between genes t and r. d. The colleague's rabbit is homozygous for an inversion that spans at least 18 map units of the region between genes t and r. e. The colleague's rabbit carries an inversion that spans the whole region between genes t and r.

ANSWER: c 26. A pure line of plants of genotype a/a ; b/b ; c/c ; d/d ; e/e (all recessive to wild type) was crossed with a wild type. One F1 individual expressed the recessive alleles d and e. This individual arose most likely from a. a deletion in the quadruple homozygous recessive parent. b. a deletion in the wild-type parent. c. a gene mutation in the wild-type parent. d. position effect variegation in the F1. e. a reversion in the quadruple homozygous recessive parent.

ANSWER: b 27. In the diagram below, each of the five lines indicates the extent of a Drosophila deletion. A through G represent adjacent chromosomal regions (for example, in line 3 regions D and E are deleted).

When a radioactive DNA fragment corresponding to a Drosophila gene of interest was incubated with chromosomes carrying each of the five deletions, it bound only to chromosomes carrying deletion 1 or deletion Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chapter 17: Large-Scale Chromosomal Changes 5. Based on this information, the gene of interest is located in region a. A or B. b. C. c. C or D. d. D. e. E, F, or G. ANSWER: d 28. A meiocyte of an organism heterozygous for a reciprocal translocation goes through meiosis and results in four viable meiotic products. The most likely explanation is that a. there was adjacent segregation. b. there was alternate segregation. c. a suppressor mutation occurred. d. the translocation breakpoints were very close to the centromeres. e. the translocation reverted.

ANSWER: b 29. The fungal cross ad, pan ∞ +, + (where ad and pan are two linked loci) gave only the following two types of tetrads: Spore 1 Spore 2 Spore 3 Spore 4

Tetrad 1 ad, pan aborted aborted +, +

Tetrad 2 ad, pan ad, + +, pan +, +

These results suggest a(n) a. deletion in the wild-type parent. b. duplication in one or both parents. c. inversion in one of the parents. d. translocation in one of the parents, with adjacent segregation. e. translocation in one of the parents, with alternate segregation.

ANSWER: c 30. In a haploid fungus, genes a and b are located on chromosomes 2 and 7, respectively. A cross is performed between a strain that is genotypically a ; b and has normal chromosomes, and a strain that is genotypically wild type but has a reciprocal translocation between chromosomes 2 and 7. If crossing over never occurs in this fungus, what is the expected proportion of tetrads resulting from the cross?

Spore 1 Spore 2 Spore 3 Spore 4

Tetrad 1 aborted aborted aborted aborted

Tetrad 2 a;b a;b aborted aborted

Copyright Macmillan Learning. Powered by Cognero.

Tetrad 3 A;B A;B aborted aborted

Tetrad 4 A;B A;B a;b a;b

Tetrad 5 A;B aborted a;b aborted Page 8

Name:

Class:

Date:

Chapter 17: Large-Scale Chromosomal Changes a. All five tetrads will be present, in equal proportions. b. Only tetrads 1 and 4 will be present, in a 1:1 ratio. c. Only tetrads 2 and 3 will be present, in a 1:1 ratio. d. Only tetrads 2, 3, and 5 will be present, in a 1:1:2 ratio. e. Only tetrad 5 will be present.

ANSWER: b 31. In yeast, the gene phe1 is on chromosome 3 and lys4 is on chromosome 7. The progeny (spores) from the cross + ; + × phe1 ; lys4 were 47% 44% 4% 5%

+;+ phe1 ; lys4 phe1 ; + + ; lys4

This result suggests that: a. a nondisjunction occurred. b. one parent has a deletion. c. one parent has an inversion. d. one parent has a translocation. e. unequal crossovers occurred.

ANSWER: d 32. Suppose that in a certain diploid fish genes A and T are on chromosomes 4 and 7, respectively. A female fish is heterozygous at both loci and is also heterozygous for a reciprocal translocation between chromosomes 1 and 4. The translocation breakpoint on chromosome 4 is 10 map units away from the a allele (A is located on the normal chromosome 4). If this female is crossed to a male of genotype a/a ; t/t and with normal chromosomes, what percentage of the progeny is expected to be phenotypically A ; T and have only normal chromosomes? a. 0% b. 2.5% c. 22.5% d. 25% e. 45% ANSWER: c 33. A reciprocal translocation happened between chromosomes 1 and 2 in a wild species of strawberries. Below is a diagram of (unreplicated) chromosomes 1 and 2 for the translocated strain of strawberries. N1 and N2 indicate the normal chromosomes 1 and 2, respectively; T1 and T2 are their translocated counterparts.

Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chapter 17: Large-Scale Chromosomal Changes

Which chromosomes must segregate from one another at anaphase I to form viable meiotic products? a. N1 and N2 separate from T1 and T2. b. N1 separates from N2, T1, and T2. c. N1 and T1 separate from N2 and T2. d. N2 and T1 separate from N1 and T2. e. T1 separates from T2, N1, and N2.

ANSWER: a 34. The diagram below shows the chromosomes of a plant that is heterozygous at six loci as well as for a reciprocal translocation involving the chromosomes that carry these loci.

This plant is crossed to a plant with normal chromosomes that is homozygous recessive at all loci, and the progeny was comprised of about 250 individuals, half of them phenotypically ABDEFG and half abdefg. This result suggests that a. multiple crossovers occurred. b. no crossovers occurred. c. segregation was always adjacent. d. segregation was always alternate. e. the six loci are unlinked.

ANSWER: b 35. Which of the following diagnostic features is indicative of an inversion having occurred? a. formation of chromosomal "loop" structures in meiosis I Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chapter 17: Large-Scale Chromosomal Changes b. reduced fertility c. pseudodominance d. formation of chromosomal loops during meiosis and reduced fertility e. reduced fertility and pseudodominance ANSWER: d 36. Which of the following chromosomal aberrations can result in unmasking recessive alleles resulting in their expression? a. deletions b. duplications c. inversions d. translocations ANSWER: a 37. For autosomes in diploid organisms, the aneuploid 2n + 1 is a. monosomic. b. trisomic. c. nullisomic. d. haploid. e. amphidiploid.

ANSWER: b 38. For autosomes in diploid organisms, the aneuploid 2n – 1 is a. monosomic. b. trisomic. c. nullisomic. d. disomic. e. amphidiploid.

ANSWER: a 39. For autosomes in diploid organisms, the aneuploid 2n – 2 is a. monosomic. b. trisomic. c. nullisomic. d. disomic. e. amphidiploid.

ANSWER: c 40. For autosomes in haploid organisms, the aneuploid n + 1 is a. monosomic. b. trisomic.

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chapter 17: Large-Scale Chromosomal Changes c. nullisomic. d. disomic. e. amphidiploid.

ANSWER: d 41. An individual member of a normally diploid species that has only one chromosome set (n) is called a a. polyploid. b. triploid. c. haploid. d. monoploid. e. amphidiploid.

ANSWER: d 42. A deletion in one homolog that allows the phenotypic expression of recessive alleles in the other homolog is an example of a. epistasis. b. pseudodominance. c. unmasking. d. pleiotropy. e. gene dosage.

ANSWER: b 43. The human conditions of Down syndrome, Klinefelter syndrome, and Turner syndrome are welldocumented examples of a. triploidy. b. monoploidy. c. nulliploidy. d. aneuploidy. e. tetraploidy.

ANSWER: d Multiple Response 44. Below is a list of potential diagnostic features (A–G). Select all the diagnostic features that may apply to this chromosomal aberration: DELETION. a. formation of chromosomal "cruciform" structures in meiosis I b. formation of chromosomal "loop" structures in meiosis I c. Recombinant frequencies between genes on the same chromosome are (much) lower than expected. d. reduced fertility (other than semisterility) e. semisterility f. pseudodominance Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chapter 17: Large-Scale Chromosomal Changes g. pseudolinkage (two genes on separate chromosomes do not segregate independently).

ANSWER: b, c, f 45. Below is a list of potential diagnostic features (A–G). Select of all the diagnostic features that may apply to this chromosomal aberration: DUPLICATION. a. formation of chromosomal "cruciform" structures in meiosis I b. formation of chromosomal "loop" structures in meiosis I c. Recombinant frequencies between genes on the same chromosome are (much) lower than expected. d. reduced fertility (other than semisterility) e. semisterility f. pseudodominance g. pseudolinkage (two genes on separate chromosomes do not segregate independently).

ANSWER: b 46. Below is a list of potential diagnostic features (A–G). Select of all the diagnostic features that may apply to this chromosomal aberration: INVERSION. a. formation of chromosomal "cruciform" structures in meiosis I b. formation of chromosomal "loop" structures in meiosis I c. Recombinant frequencies between genes on the same chromosome are (much) lower than expected. d. reduced fertility (other than semisterility) e. semisterility f. pseudodominance g. pseudolinkage (two genes on separate chromosomes do not segregate independently).

ANSWER: b, c, d 47. Below is a list of potential diagnostic features (A–G). Select of all the diagnostic features that may apply to this chromosomal aberration: TRANSLOCATION. a. formation of chromosomal "cruciform" structures in meiosis I b. formation of chromosomal "loop" structures in meiosis I c. Recombinant frequencies between genes on the same chromosome are (much) lower than expected. d. reduced fertility (other than semisterility) e. semisterility f. pseudodominance g. pseudolinkage (two genes on separate chromosomes do not segregate independently).

ANSWER: a, c, e, g Essay 48. What are the pairing possibilities for tetraploids at meiosis I and what type of gametes do the pairings produce?

ANSWER: There are three different pairing possibilities at meiosis I. The four homologous chromosomes may pair as two bivalents or as a quadrivalent, and each can produce functional gametes. The third Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chapter 17: Large-Scale Chromosomal Changes possible pairing is a trivalent plus a univalent which produces nonfunctional gametes.

49. Assume that x is a new mutant allele in corn. An x/x plant is crossed with a triple-10 individual (trisomic for chromosome 10) carrying only normal dominant alleles at the x locus. Trisomic progeny for chromosome 10 are recovered and crossed back to x/x. a) What ratio of dominant to recessive phenotypes is expected if the x locus is not in chromosome 10? b) What ratio of dominant to recessive phenotypes is expected if the x locus is in chromosome 10?

ANSWER: a) 1:1 ratio of X/x and x/x b) Gametes Progeny X/X X/X/x x x/x X/x X/x/x X/x X/x/x X X/x X X/x 5:1 if n + 1 pollen are viable (ratio of gametic types is 2X/x : 2X : 1x : 1X/X). 2:1 if n + 1 pollen are not viable (ratio of gametic types is 2X/x : 1x/x).

50. In Zinnia plants (n = 16), the chromosomes of autotetraploids pair in twos. The locus for flower color (R = red, r = white) is close to the centromere on chromosome 5, and the locus for plant height (D = tall, d = dwarf) is also closely linked to the centromere on chromosome 11. Both dominant alleles show full dominance over any number of recessive alleles. a) If you are observing meiosis in tetraploid Zinnias, what is the total number of bivalents (chromosome pairs) that will be visible during prophase I? Explain. b) In a pollen cell from a tetraploid plant, how many chromosomes will be present? c) If a tetraploid plant of genotype R/R/r/r ; D/D/D/D is selfed, what proportion of progeny will be white flowered and tall? d) If a tetraploid plant of genotype R/r/r/r ; D/D/D/D is selfed, what proportion of progeny will be white flowered and tall? e) If a tetraploid plant of genotype R/R/r/r ; D/D/d/d is selfed, what proportion of progeny will be white flowered and dwarf? What proportion will be red flowered and tall?

ANSWER: a) Total chromosomes = 64 paired as 32 bivalents b) 32 c) r/r/r/r ; D/D/D/D = 1/36 × 1 = 1/36 d) r/r gametes = 1/2, so r/r/r/r = 1/4 and r/r/r/r ; D/D/D/D also = 1/4 e) r/r/r/r ; d/d/d/d = 1/36 × 1/36 = 1/1296 R- ; D- = 35/36 × 35/36 = 1225/1296 = 0.95

51. An allotetraploid plant with 2n = 30 is backcrossed to one of its progenitor species that has an n = 8. A Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chapter 17: Large-Scale Chromosomal Changes sterile progeny is produced. How many chromosomes will this sterile individual possess, and what are the chromosome numbers of the two progenitor species?

ANSWER: The allotetraploid can be represented as 2n = 2n1 + 2n2. Let n2 = 8, so 2n1 + 16 = 30; therefore, n1 = 7. The chromosome complement of the sterile progeny is 2n1 + n2 = 14 + 8 = 22.

52. You make meiotic chromosome preparations of two species of tetraploid plants. In one, there is normal pairing of chromosomes and tetrads (quadivalents) are observed, while in the other, there is abnormal pairing in some preparations and trivalents and univalents are seen. How would you account for the differences between these two species?

ANSWER: The first is an allotetraploid, and the chromosomes from the two progenitor species pair independently and normally. The other is an autotetraploid, and so there are four homologous chromosomes that have to pair in prophase I. This often leads to unusual pairing, and trivalents and univalents will be seen.

53. a) Diagram the pairing structure of the inversion heterozygote abcdefg / abfedcg. Assume the centromere is to the left of a. Is this a paracentric or pericentric inversion? b) Diagram the early anaphase configuration of the bivalent in part a if a chiasma occurred between d and e.

ANSWER: a) paracentric

b)

54. Two strains of Australian wombat were crossed. Meiotic metaphase configurations of the F1 hybrids are diagrammed below.

Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chapter 17: Large-Scale Chromosomal Changes

In both strains of wombats, n = 4. The black dots are the centromeres, a and b are genetic markers located on chromosome 2, c is located on chromosome 3, and d is located on chromosome 4. a) How do the chromosomes of the two parental strains differ? Use terminology that most accurately describes each chromosomal rearrangement.

Chromosome 1: Chromosome 2: Chromosome 3: Chromosome 4: b) In the two parental strains, the map distance between the a and b loci is not the same. Explain why. c) Assume that during meiosis in the F1 above, a crossover always occurs between the a locus and the centromere of chromosome 2. (i) What proportion of gametes produced by the F1 hybrid will be viable? (Note: You must consider the behavior of all four chromosomes in answering this question.) (ii) What will be the genotypes of the viable gametes at the a and b loci? (iii) What will be the genotypes of the viable gametes at the c and d loci?

ANSWER: a) Chromosome 1 appears to be similar in both species. Chromosome 2 has a pericentric inversion in one of the parents. Chromosomes 3 and 4 have reciprocal translocations in one of the parents. b) In the species where a and b are on opposite sides of the centromere, they are farther apart. It would have a higher rate of recombination than the species where a and b are closer together on the same side of the centromere. c) (i) (1/2 to 1/3 for chromosomes 3, 4) × (1/2 for chromosome 2) = 1/4 to 1/6 (ii) +/+ and a/b (iii) +/+ and c/d

55. In a lab strain of Drosophila, cinnabar (cn) and brown (bw) are recessive eye color mutations known to be 41 map units apart on chromosome 2. When similar mutant alleles were induced in a strain from nature, the same linkage of cn and bw was observed. However, when a wild-type strain from nature was crossed with a cn bw / cn bw lab strain to create the genotype + + / cn bw, and females of this type were testcrossed to cn bw / cn bw males from the lab strain, the following phenotypic proportions were obtained in the progeny: + cn cn +

+ bw + bw

25,200 21,009 11 36

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chapter 17: Large-Scale Chromosomal Changes a) What is unexpected about these results? b) What is the most likely explanation? (Summarize your model with a chromosome diagram with labeled genes.) c) On your diagram show the precise origin of the cn + and the + bw classes.

ANSWER: a) The RF is much less than the expected 41%. b) There must be a large inversion in the cn to bw region.

Strain from nature cn ---[inversion spanning most of region]--- bw Strain from lab strain + ---[noninverted sequence ]--- + These two chromosomes are paired in the F1, with the inverted region looped. c) These genotypes must arise from crossovers in the small homologous noninverted regions shown with dashes in the diagram.

56. In the fungus Neurospora, a standard lab strain carrying the auxotrophic mutation m was crossed to a prototrophic strain isolate from a burned sugarcane field. There were three types of asci as shown below, where the symbol ab means an aborted ascospore whose m locus constitution could not be tested.

Type 1 m m m m + + + +

Type 2 ab ab ab ab ab ab ab ab

Type 3 m m ab ab ab ab + +

Explain the origin of each ascus type.

ANSWER: This type of pattern suggests heterozygosity for an inversion. First, it is not the all dead/all alive pattern expected for a translocation. Second, it seems that all MII patterns have been converted into 50% lethality within those asci, suggesting that crossing over is the cause of death. So, type I is from a meiosis with no crossover, type 2 is from a four-strand double crossover, and type 3 is from a single crossover.

57. In Drosophila, h determines hairy body (h+ = smooth) and se determines sepia eyes (se+ = red). The loci are approximately 40 m.u. apart on an autosome. A cross was made: se h+ / se h+ × se+ h / se+ h, and the F1 were all wild type. F1 females were testcrossed to se h / se h males, and the offspring were 39 sepia eyes 42 hairy body a) What progeny and in what proportions were expected? b) Provide an explanation of the observed results if they differ from expectations.

ANSWER: a) Expected: 30% sepia, 30% hairy, 20% wild type, 20% sepia/hairy b) The recombinant classes are missing. The most likely explanation is that one of the parental lines Copyright Macmillan Learning. Powered by Cognero.

Page 17

Name:

Class:

Date:

Chapter 17: Large-Scale Chromosomal Changes was homozygous for an inversion spanning most of the se-h region.

58. In Drosophila, the phenotype bar-eye is due to a tandem duplication of several bands on the X chromosome.

Occasionally (1 out of 1600), all true-breeding stocks of bar-eyed flies yield an even more extreme phenotype called ultrabar, which is in fact due to a triplication of this chromosomal segment. The hypothesis used to explain ultrabar's origin is that a slight mispairing of chromosomes can occur whenever there is a duplication. If crossing over occurs (in females), then recombinants would form as follows:

What other evidence, not involving cytological observation, would you seek to substantiate this hypothesis? Be as exact as you can. There are at least two possible answers.

ANSWER: There should be as many wild-type males as there are ultrabar males in each generation.

If females that are homozygous for bar eyes are made heterozygous for markers on either side of the bar locus, the ultrabar males they produce should be recombinant for these markers.

59. You discover a Drosophila male that is heterozygous for a reciprocal translocation between the second and third chromosomes, each break having occurred near the centromere (which for these chromosomes is near the center). a) Draw a diagram showing how these chromosomes would synapse at meiosis. b) You find that this fly has the recessive genes bw (brown eyes) and eb (ebony body) on the second and third chromosomes, respectively, and wild-type alleles on the translocated ones. The fly is crossed with a female having normal chromosomes and homozygous for bw and eb. What phenotypic classes and in what ratio would be expected in the progeny of this cross? (Hint: Zygotes with an extra chromosome arm or deficient for one do not survive. Also, remember that there is no crossing over in Drosophila males.)

ANSWER: a)

b) Centromeres 1, 4 / 2, 3 Centromeres 1, 3 / 2, 4 Copyright Macmillan Learning. Powered by Cognero.

Inviable gametes Viable gametes Page 18

Name:

Class:

Date:

Chapter 17: Large-Scale Chromosomal Changes 50% brown, ebony (normal chromosomes) 50% + + (translocation heterozygote) 60. A pure-breeding flowering plant with a long stem and small leaves was crossed to a pure-breeding individual of the same species that has a short stem and large leaves. The F1 is comprised entirely of individuals with long stems and large leaves. When these F1 individuals are testcrossed, they produced only about half as many individuals as expected. The phenotypic ratios were: 462 short stems and large leaves, 450 long stems and small leaves, 52 long stems and large leaves, and 50 short stems and small leaves. These 1022 plants were further crossed to a tester, and many of them displayed semisterility; in particular: Of the 462 plants with short stems and large leaves, 362 were semisterile.

Of the 450 plants with long stems and small leaves, 90 were semisterile.

Of the 52 plants with long stems and large leaves, 40 were semisterile.

Of the 50 plants with short stems and small leaves, 10 were semisterile.

Provide a plausible, comprehensive genetic explanation for these results.

ANSWER: From the F1 phenotype, we can deduce that large leaves and long stems are dominant phenotypes, and the semisterility (when testcrossed, the F1 individuals produce only half as many individuals as expected) is suggestive of a translocation. Most likely, the F1 individuals are translocation heterozygous. This is confirmed by the fact that about half of the individuals issued from this testcross were revealed to be semisterile when testcrossed.

The numerical data allow us to deduce that the loci controlling stem length and leaf size are linked and, moreover, they are also linked to the translocation breakpoint. Map distances can be calculated from the testcross data.

Phenotype long stem, small leaves, normal fertility: short stem, large leaves, semisterility: long stem, large leaves, normal fertility: short stem, small leaves, semisterility: long stem, small leaves, semisterile: short stem, large leaves, normal fertility: long stem, large leaves, semisterile: short stem, small leaves, normal fertility:

Number of individuals 450 − 90 = 360 362 52 − 40 = 12 10 90 462 − 362 = 100 40 50 − 10 = 40

Total = 1022 The data can be analyzed as for a regular three-factor testcross. Data inspection suggests that the "leaves size" is located between the "leaves size gene" and the translocation breakpoint.

RF between "leaves size gene" and "stems length gene" = (12 + 10 + 40 + 40)/1022 = 0.099 → 9.9 map units RF between "leaves size" and breakpoint = (12 + 10 + 90 + 100)/1022 = 0.207 → 20.7 map units Map distance between "stem length" and translocation breakpoint = 9.9 m.u. + 20.7 m.u. = 30.6 m.u.

Copyright Macmillan Learning. Powered by Cognero.

Page 19

Name:

Class:

Date:

Chapter 18: Population Genetics Multiple Choice 1. Which of the following is a question that population genetics would address? a. How does cancer spread from one tissue to another? b. How does the cell copy DNA? c. How does a single nucleotide substitution cause sickle cell anemia? d. How many people have color blindness in Utah? e. How does phenylketonuria affect brain development?

ANSWER: d 2. A gene or a trait is said to be polymorphic if a. the allele frequency is about 50:50 for each of its alleles present in the population. b. the allele frequency of one allele is higher than that of all others. c. more than one form (morph or allele) exists in the population. d. one of its alleles occurs at a frequency lower than 1 %. e. only one form exists in the population.

ANSWER: c 3. After a thorough analysis of the DNA from 150 university students, a small region of the human X chromosome is found to have three SNP sites. Two of those sites have three different variants, while the third site has four. How many different haplotypes were present in each student's cells? a. 1 or 2 b. 3 or 6 c. 10 or 20 d. 32 or 64 e. It is impossible to determine based on this information alone.

ANSWER: a 4. Sequencing of a population of 110 dogs has revealed nine polymorphic sites within a small region of chromosome 11. All four possible variants were observed for each site. Based on these data, one can say that a. all the dogs in the sample population have different haplotypes. b. different haplotypes have different frequencies in the sample population. c. not all possible haplotypes are present in the sample population. d. the sample population is not representative of the entire population. e. some of the dogs in the sample population have the same haplotype.

ANSWER: c 5. Which of the following best describes the frequency of alleles in a Hardy–Weinberg population? a. p2 b. q2 c. p + q Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chapter 18: Population Genetics d. 2pq e. None of the answer options is correct.

ANSWER: c 6. In a human population, the genotype frequencies at one locus are 0.5 AA, 0.4 Aa, and 0.1 aa. The frequency of the A allele is a. 0.20. b. 0.32. c. 0.50. d. 0.70. e. 0.90.

ANSWER: d 7. Human albinism is an autosomal recessive trait. Suppose that you find a village in the Andes where onefourth of the population is albino. If the population size is 1000 and the population is in Hardy–Weinberg equilibrium with respect to this trait, how many individuals are expected to be heterozygotes? a. 50 b. 250 c. 300 d. 500 e. 750 ANSWER: d 8. You are studying an X-linked trait. There are two alleles, one showing complete dominance over the other. In females, 84% show the dominant phenotype. What percentage of the males will show the dominant phenotype, assuming that individuals in this large population mate at random with respect to this trait? a. 84% b. 60% c. 40% d. 36% e. 16% ANSWER: b 9. Some forms of cleft palate are caused by recessive alleles of a gene on the X chromosome. In an Icelandic population, 3.6 × 10–3 of the females have this form of cleft palate. What proportion of the males would you expect to be affected, assuming Hardy–Weinberg equilibrium? a. 1.3 × 10–5 b. 3.6 × 10–3 c. 1.6 × 10–3 d. 6.0 × 10–2 Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chapter 18: Population Genetics e. 0 ANSWER: c 10. In a tropical human population in Hardy–Weinberg equilibrium for an autosomal locus determining presence/absence of pigment in the skin, the frequency of albinism (aa) is 1 in 10,000. The frequency of heterozygotes is approximately a. 1 in 25. b. 1 in 50. c. 1 in 75. d. 1 in 100. e. 1 in 1000.

ANSWER: b 11. Suppose that you are a genetic counselor, and a couple seeks your advice about BHT tasting (people who can taste BHT in processed food are recessive homozygotes for the "taster" allele and nontasters are homozygous dominant or heterozygous for the "nontaster" allele). Both prospective parents are nontasters, but a careful analysis of the husband's pedigree reveals that he is a carrier of the taster allele. The two alleles are in Hardy–Weinberg equilibrium, and the proportion of BHT tasters in the population is 16%. What is the probability that the couple's first child will be a nontaster of BHT? a. 0.0 b. 0.32 c. 0.81 d. 0.86 e. 1.0 ANSWER: d 12. In a population in Hardy–Weinberg equilibrium, what will be the proportion of matings between homozygotes? a. p2 + 2pq b. p4 c. p4 + q4 d. p4 + q4 + 2p2q2 e. 4p2q2 ANSWER: d 13. Given below are the genotypic frequencies for a single gene with two alleles for three different populations.

Population 1 Population 2 Population 3

AA 0.25 0.35 0.49

Aa 0.50 0.56 0.42

Copyright Macmillan Learning. Powered by Cognero.

aa 0.25 0.09 0.09 Page 3

Name:

Class:

Date:

Chapter 18: Population Genetics Which of the following is/are NOT correct about these three populations? a. One must assume Hardy–Weinberg equilibrium to calculate the allele frequencies. b. Only two of the populations are in Hardy–Weinberg equilibrium. c. Population 1 is in Hardy–Weinberg equilibrium; the frequency of allele A is 0.5. d. Population 2 is not in Hardy–Weinberg equilibrium; the frequency of allele a is 0.37. e. Population 3 is in Hardy–Weinberg equilibrium; the frequency of allele a is 0.3.

ANSWER: a 14. In a population of rats, 35% of the individuals are genotypically B/B, 10% are B/b, and 55% are b/b.

Assuming there are no mutations, no selection, and no migrations, the predicted frequency of the b/b genotype after one generation of random mating is a. 25%. b. 42.25%. c. 47.5%. d. 55%. e. 57.5%.

ANSWER: b 15. If the frequency of the recessive allele that causes phenylketonuria (PKU) when homozygote is about 0.01 in European and American populations, what is the estimated heterozygote frequency in these populations?

(Assume that people mate randomly with respect to PKU, and that the mutant allele does not have any effects on fitness.) a. about 0.0001 b. about 0.01 c. about 0.0198 d. about 0.99 e. It is impossible to make an estimate.

ANSWER: c 16. The genetic frequencies of two separate populations are Population 1 Population 2 Population 3

AA .36 .33 .55

Aa .48 .33 .10

aa .16 .33 .35

Which of these populations is/are in Hardy–Weinberg equilibrium? a. Population 1 is in Hardy–Weinberg equilibrium, but populations 2 and 3 are not. b. Population 2 is in Hardy–Weinberg equilibrium, but populations 1 and 3 are not. c. Population 3 is in Hardy–Weinberg equilibrium, but populations 1 and 2 are not. d. Populations 1 and 2 are in Hardy–Weinberg equilibrium, but population 3 is not. e. None of the populations is in Hardy–Weinberg equilibrium.

ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chapter 18: Population Genetics 17. If 64% of the people in a population are blue-eyed and the population is in Hardy–Weinberg equilibrium, what is the percentage of heterozygotes in the population? a. 16% b. 20% c. 32% d. 36% e. It is impossible to make an estimate from these data.

ANSWER: c 18. Inbreeding in populations that are normally outbreeding leads to which of the following? a. a higher rate of genetic drift b. a higher rate of mutation c. an increase in the frequency of heterozygotes d. more individuals affected by rare diseases e. a smaller population ANSWER: d 19. A population is currently in Hardy–Weinberg equilibrium with respect to the A locus. If (positive) assortative mating occurs for one generation, the expected outcome is an increase in the frequency of the a. A allele. b. a allele. c. A– phenotype. d. heterozygous genotype (A/a). e. homozygous genotypes (A/A and a/a).

ANSWER: e 20. Suppose that a population of 100 black-footed ferrets has an initial heterozygosity (H) of 0.5. How many generations will it take for this population to decline to a heterozygosity of 0.25? a. 1 b. 34 c. 68 d. 100 e. 138 ANSWER: e 21. Suppose that a population of 25 black-footed ferrets has an initial heterozygosity (H) of 0.5. How many generations will it take for this population to decline to a heterozygosity of 0.25? a. 1 b. 34 c. 68 Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chapter 18: Population Genetics d. 100 e. 138 ANSWER: b 22. A horticulturist plans to self a plant, then (individually) self each of the F1 plants, and repeat the same procedure until he obtains a plant that is heterozygous at no more than one gene. If the parental plant has 15,000 genes and is heterozygous for all of them, how many generations will this process take? (For the purposes of this question, count the parental plant as "1 generation.") a. 8 generations b. 15 generations c. 30 generations d. 15,000 generations e. 30,000 generations ANSWER: b 23. What is the probability that individual A in the pedigree below is homozygous by descent?

a. 1/8 b. 1/16 c. 1/32 d. 1/64 e. 1/256 ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chapter 18: Population Genetics 24. What is the inbreeding coefficient for individual B in the following pedigree?

a. 1/8 b. 1/16 c. 1/32 d. 1/64 e. 1/256 ANSWER: c 25. Sequencing of a pine tree population reveals four variants of a particular SNP. Eighty percent (80%) of the trees have the G, 15% the A, 3% the T, and 2% the C variant. What is the genetic diversity of the pine tree population with respect to this SNP? a. 0.007% b. 2.0% c. 33.6% d. 66.4% e. 80% ANSWER: c 26. What group of organisms has, overall, the highest genetic diversity? a. humans b. invertebrates c. plants d. vertebrates other than humans e. unicellular eukaryotes ANSWER: e Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chapter 18: Population Genetics 27. Two chromosomes having the same combination of alleles at multiple loci are said to be of similar a. genotype. b. haplotype. c. phenotype. d. sporotype.

ANSWER: b 28. If two populations of birds with frequencies of an allele a of 0.2 and 0.3 are forced together by unusually high winds, the new frequency of the a allele in the combined population will be a. 0.2. b. 0.25. c. 0.3. d. 0.5. e. It is impossible to determine unless we know the population sizes.

ANSWER: e 29. A boatload of Swedish tourists, all of whom bear the MM blood group, is marooned on Haldane Island, where they are met by an equally sized population of Islanders, all bearing blood group NN. In time, the castaways become integrated into Island society. Assuming random mating, no mutation, no selection (based on blood group), and no genetic drift, what would you expect the blood group distribution to be among 4000 progeny of the new Haldane Island population? a. 4000 MN b. 2000 MN + 2000 MM c. 2000 MM + 2000 NN d. 2000 MN + 2000 NN e. 1000 MM + 2000 MN + 1000 NN ANSWER: e 30. In a population of rats, the relative fitness and frequencies of genotypes at present are HH Hh hh

Frequency 0.4 0.5 0.1

Fitness 1 0.8 0.6

What will be the frequency of the h allele in the next generation? a. 0.3 b. 0.4 c. 0.5 d. 0.6 e. 0.7 ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chapter 18: Population Genetics 31. For many generations, the following genotypic frequencies were observed in a large population of dinosaurs: 4% AA, 32% Aa, and 64% aa. The climate changed abruptly and resulted in the death of all homozygous recessive dinosaurs (s = 1). In the next generation, the recessive homozygotes died shortly after birth. What percentage of these newborn dinosaurs died? a. 4% b. 8.5% c. 19.7% d. 36.8% e. 44% ANSWER: c 32. You identify a population of mice (Peromyscus maniculatus) on an island. Their coat color is controlled by a single gene: BB mice are black, Bb mice are gray, and bb mice are white. You take a census of the population and record the following numbers of mice: Black 200 Gray 160 White 40 As a consequence of human activity, the climate on the island is changed permanently. All 200 mice with black fur die from heatstroke, but the other mice survive. If the climate change causes mice with black fur to die before reproducing, which of the following statements is CORRECT? a. The B allele will disappear from the population in one generation. b. The B allele will disappear from the population in two generations. c. The fitness of mice with gray fur (WBb) must be equal to 0.5. d. The fitness of mice with black fur (WBB) is 0. e. At Hardy–Weinberg equilibrium, f(B) will equal 0.135.

ANSWER: d 33. Which of the following conditions results in the faster evolution of a population? a. mutation b. selection c. nonrandom mating d. natural disaster ANSWER: b 34. In a population, the D → d mutation rate is 4 × 10–6. If p = 0.8 today, what will p be after 50,000 generations? a. 0.1 b. 0.25 c. 0.65 d. 0.8 e. None of the answer options is correct.

ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chapter 18: Population Genetics 35. Linkage disequilibrium will decay over time because of a. population size. b. genetic drift. c. mutation rates. d. recombination. e. migration.

ANSWER: d 36. A period of one or several consecutive generations of contraction in population size is known as a. genetic drift. b. natural selection. c. bottleneck. d. adaptation. e. directional selection.

ANSWER: c 37. The type of bias in mate choice that occurs when similar types mate is a. positive assortative mating. b. negative assortative mating. c. disassortative mating. d. isolation by distance. e. inbreeding.

ANSWER: a Objective Short Answer 38. What are the mechanisms that allow new genetic variation to enter a population?

ANSWER: New variation is introduced by mutation, migration, or gene flow.

39. What types of mating biases cause genotype frequencies to deviate from Hardy–Weinberg expectations?

ANSWER: The types of bias in mate choice are assortative mating, isolation by distance, and inbreeding.

Subjective Short Answer 40. In a Pygmy group in central Africa, the frequencies of alleles determining the ABO blood groups were estimated as 0.2 for IA, 0.1 for IB, and 0.7 for i. Assuming random mating with respect to blood group, determine the expected frequencies of the blood types A, B, AB, and O. ANSWER: f(A) = 0.32 = 0.22 + 2(0.2)(0.7) f(B) = 0.15 = 0.12 + 2(0.1)(0.7) f(AB) = 2(0.2)(0.1) = 0.04 2 f(O) = 0.49 = 0.7 Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chapter 18: Population Genetics 41. What processes increase genetic variation in a given population?

ANSWER: mutation, recombination and gene flow 42. You identify a population of mice (Peromyscus maniculatus) on an island. Their coat color is controlled by a single gene: BB mice are black, Bb mice are gray, and bb mice are white. You take a census of the population and record the following numbers of mice.

Black 200 Gray 160 White 40 a) What are the frequencies of the two alleles? b) What are the Hardy–Weinberg equilibrium frequencies for these three phenotypes? c) A heat wave hits the island. All 200 mice with black fur die from heatstroke, but the other mice survive.

What are the new allele frequencies for the population? d) If the population suffers no further cataclysms after the heat wave, and the surviving animals mate randomly, what will be the frequency of mice with black fur in the next generation?

ANSWER: a) f(B) = 0.7 f(b) = 0.3 b) f(BB) = 0.49 f(Bb) = 0.42 f(bb) = 0.09 c) f(B) = 0.4 f(b) = 0.6 d) f(black mice) = 0.16 Essay 43. How is the genetic composition of a population defined?

ANSWER: The genetic composition of the population can be defined as the collection of frequencies of different genotypes in the population. These frequencies are the consequence of processes that act at the level of individual organisms to increase or decrease the number of organisms of each genotype.

44. The table below shows the haplotypes relative to a small region on chromosome 2 for 10 strains of laboratory mice. All strains are homozygous with respect to the region in question.

Strain SNP1 +1 A 2 A 3 A 4 A 5 T 6 A 7 A 8 T

SNP2 G T G G G T G G

SNP3 G C C C C C A C

SNP4 T T T T G T T G

Copyright Macmillan Learning. Powered by Cognero.

SNP5 A G A A A A A G

SNP6 C C C C C T T C Page 11

Name:

Class:

Date:

Chapter 18: Population Genetics 9 10

A A

C G

C T

T A

A A

G G

a) Based on the data in the table, what strains are genetically most closely related? b) How many different haplotypes are represented in the table? c) What SNP has the highest number of variants/alleles?

ANSWER: a) strains 3 and 4 (they have identical haplotypes) b) nine different haplotypes c) SNP3 45. A recessive X-linked character appears in 40 percent of males and 16 percent of females in a randomly interbreeding population. Assume only two alleles are present. What are the allele frequencies? How many females are heterozygotes? How many males are heterozygotes?

ANSWER: The frequency of the recessive allele must be 0.4 because the frequency of males with the trait will directly reflect the allele frequency. The frequency of females equals (0.4)2, which is consistent with a Hardy–Weinberg equilibrium. Therefore, the number of heterozygotes in females should be 2pq or 2 (0.4)(0.6) = 0.48. There are no heterozygous males (X-linked gene).

46. Red hair (autosomal recessive) is found in approximately 4 percent of the people in Norway. If we assume that the Norwegian population is in Hardy–Weinberg equilibrium with respect to hair color: a) what are the frequencies of the red hair (r) and non-red hair (R) alleles? b) what is the frequency of heterozygotes? c) what proportion of all marriages could potentially have a child with red hair?

ANSWER: a) b) 2 × 0.2 × 0.8 = 0.32 c)

R/r × R/r R/r × r/r r/r × r/r

0.32 × 0.32 0.32 × 0.04 × 2 0.04 × 0.04

= 0.1024 = 0.0256 = 0.0016 Total 0.1296

12.96 47. Tay-Sachs disease is inherited as an autosomal recessive. In a certain large eastern European population, the frequency of Tay-Sachs disease is 1%. a) If the population is assumed to be in Hardy-Weinberg equilibrium with respect to Tay-Sachs, what is the frequency of the allele that causes Tay-Sachs? b) What would be the frequency of heterozygotes? c) What is the probability of two heterozygotes marrying?

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chapter 18: Population Genetics d) In children of such marriages, what would be the frequency of Tay-Sachs disease? e) What proportion of all Tay-Sachs births are produced by such marriages?

ANSWER: a) b) 2 × 0.9 × 0.1 = 0.18 c) (0.18)2 = 0.032 d) 0.25 e) 0.032 × 0.25 = 0.0081, which is 81% of the total Tay-Sachs cases (0.01) 48. The MN blood group in humans is under the control of a pair of codominant alleles, LM and LN. In a group of 556 individuals, the following genotypic frequencies are found: 167 MM 280 MN 109 NN a) Calculate the frequency of the LM and LN alleles. b) Test (using the chi-squared test) whether the genotypic frequencies conform to the Hardy–Weinberg distribution.

ANSWER: a) If p = f(LM); q = f(LN) p = (2 × 167 + 280)/(2 × 556) = 0.552 q = (2 × 109 + 280)/(2 × 556) = 0448 b) MM MN NN

Observed 167 280 109

Expected 169 275 112

χ2 = 0.195 d.f. = 1 0.5 < p < 0.7

49. Apert syndrome (acrocephalosyndactyly) results from a dominant mutant allele. Among 322,182 births to normal parents, two infants were found with this syndrome. What is the mutation rate per gamete for this genetic disease?

ANSWER: 2/(322,182 × 2) = 3.1 × 10–6 There are 2 × 322,182 gametes that produced these children, and only two of those gametes had the mutation.

50. The table below shows the haplotypes relative to a small region on chromosome 2 for 10 strains of laboratory mice. All strains are homozygous with respect to the region in question.

What SNP shows the highest genetic diversity?

Strain SNP1 1 A 2 A 3 A 4 A 5 T

SNP2 G T G G G

SNP3 G C C C C

SNP4 T T T T G

Copyright Macmillan Learning. Powered by Cognero.

SNP5 A G A A A

SNP6 C C C C C Page 13

Name:

Class:

Date:

Chapter 18: Population Genetics 6 A T C T A T 7 A G A A T 8 T G C G G C 9 A C C T A G 10 A G T A A G ANSWER: SNP1 and SNP5: 1 – (0.64 + 0.04) = 0.32 SNP2 and SNP4: 1 – (0.49 + 0.04 + 0.01) = 0.46 SNP3: 1 – (0.49 + 0.01 + 0.01 + 0.01) = 0.48 SNP6: 1 – (0.36 + 0.04 + 0.04) = 0.56 The SNP with the highest genetic diversity is SNP6.

51. Samples of brook trout from two mountain streams yielded the following data for a polymorphism for restriction fragment length (3.3, 3.9, 4.1 kilobases) at the gene encoding the enzyme alcohol dehydrogenase.

Population 1 3.3-kb homozygotes 4.1-kb homozygotes 3.3-/4.1-kb homozygotes

Population 2 49 9 42

3.3-kb homozygotes 3.9-kb homozygotes 3.3-/3.9-heterozygotes

4 64 32

a) What are the allele frequencies in each population? b) Are the populations at Hardy–Weinberg equilibrium? c) What effect would migration of salmon from population 2 into the river where population 1 lives have on the allele frequencies in the progeny of population 1?

ANSWER: a) Population 1 f(3.3-kb allele) = 0.7 Population 2 f(3.3-kb allele) = 0.2 b) Both populations appear to be in Hardy–Weinberg equilibrium. c) Migration would introduce the 3.9-kb allele into population 1 and would change the relative frequencies of all alleles.

52. A population of lab rats recently escaped into the sewer system of Washington, D.C. A brave researcher sampled this population and estimated the following phenotypic frequencies, which were caused by a single gene with two alleles: E and e.

61% normal eyes (EE) 26% cross-eyed (Ee) 13% blind (ee) a) If this population were to reach a Hardy–Weinberg equilibrium (HWE) in one generation, what would the phenotypic frequencies be in that new HWE generation? b) Suppose that the phenotypes of the original adult population were not equally likely to survive and reproduce (selection). The fitness coefficients are as follows: WEE = 1 WEe = 0.6 Wee = 0.2 What would be the expected phenotypic ratio of babies in the next generation (if selection acts only on adults)?

Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chapter 18: Population Genetics ANSWER: a) We can calculate p and q from the genotypic frequencies (which are the same as the phenotypic frequencies in this case). f(E) = p = 0.74 f(e) = q = 0.26 The HWE should then be EE 0.55 Ee 0.38 Ee 0.07 b) If selection acts before breeding, you cannot use the frequencies calculated in part (a) to answer part (b). The original frequencies must be used.

EE Ee ee 0.61 0.26 0.13 1 0.6 0.2 = W(fitness) - the proportion surviving 0.61 0.16 0.03 = the proportions of the original adults in each class after selection 0.76 0.20 0.04 = the normalized frequencies after selection, calculated by taking the previous values and dividing each by the frequency of all the surviving rats: (0.61 + 0.16 + 0.03 = 0.8) If these rats that have survived now mate at random, their offspring will be in HWE, but with new allele frequencies: p = 0.86; q = 0.14.

EE 0.74 Ee 0.24 Ee 0.02 53. A large population of hawkmoths lives in a valley in Canada. Genetic tests reveal the following frequencies of alleles for wing color.

Frequency 0.5 0.4 0.1

Allele C ct c–

Phenotype Black wing (dominant to ct and c–) Tan wing (dominant to c–) White wing (recessive to both other alleles)

a) If moths continue to mate randomly, what would be the frequencies of moths with black, tan, and white wings in the next generation? b) A small group of the moths flies to neighboring Montana and starts a new population. After several generations, there is a large randomly mating population of hawkmoths. Observations of the Montana moths reveal the following: Phenotype Frequency Black wings 0.75 Tan wings 0 White wings 0.25 What are the allele frequencies for C, ct, and c–? c) The change in allele frequency in the Montana population as compared with the original Canadian population Copyright Macmillan Learning. Powered by Cognero.

Page 15

Name:

Class:

Date:

Chapter 18: Population Genetics is an example of _______. d) A group of Western Kingbirds migrates into the area occupied by the Montana hawkmoths. The birds find it easier to spot and catch the white-winged moths, compared with the black moths. The relative fitness of each moth is as follows: Phenotype Fitness Black wings 1 White wings 0.2 What will be the allele frequencies for C, ct, and c– after one generation of selection? e) Imagine that instead of birds, hawkmoths from Idaho migrate into the Montana population. The Montana population contains 10,000 moths, and the migrants from Idaho number 2500. The Idaho moths are all whitewinged. What is the allele frequency for c– in the next generation?

ANSWER: a) f(black) = 0.52 + 2(0.5)(0.4) + 2(0.5)(0.1) = 0.75 f(tan) = 0.24 = 0.42 + 2(0.4)(0.1) 2 f(white) = 0.1 = 0.01 b)

f(C) = 0.5 f(ct) = 0 f(c–) = 0.5 You can calculate f(c–) by taking the square root of 0.25, a procedure you can perform only if you assume Hardy–Weinberg equilibrium. c) either genetic drift or founder effect d)

f(C) = 0.625 f(ct) = 0 f(c–) = 0.375 e) Since the Montana population is in HWE, you can assume that there are 2500 homozygous black moths and 5000 heterozygous black moths. When you add 2500 white moths, you then can calculate: f(c–) = (5000 + 5000/2)/12,500 = 0.6.

Copyright Macmillan Learning. Powered by Cognero.

Page 16

Name:

Class:

Date:

Chapter 19: The Inheritance of Complex Traits Multiple Choice 1. A high variance indicates that a. the mean value is very high. b. most values are very close to the mean. c. most values are higher than the mean. d. most values are lower than the mean. e. the variation among the values is high.

ANSWER: e 2. Which of the following statements is/are a central assumption of the multifactorial inheritance hypothesis? a. Only one locus is associated with the trait. b. Several loci are associated with the trait. c. Environment plays no role in the final phenotype. d. Only one locus is associated with the trait, and environment interacts with this locus to produce the final phenotype. e. None of the answer options is correct.

ANSWER: b 3. The amount of milk produced per day by a cow is an example of a a. continuous trait. b. discrete trait. c. dominant trait. d. meristic trait. e. threshold trait.

ANSWER: a 4. The mean weight in a population of fruit flies (Drosophila melanogaster) grown in identical conditions is 1.10 mg, and the standard deviation is 0.05. The heaviest fly in this population weighs 1.30 mg. The genetic deviation of this individual is a. 0.05 mg. b. 0.15 mg. c. 0.20 mg. d. 0.84 mg. e. It is impossible to determine based on these data.

ANSWER: c 5. A large number of genetically identical tomato plants are grown in a greenhouse. The mean height in this plant population is 84 cm, and the standard deviation is 2.5 cm. The variation in height within this population is most likely due to a. a correlation between genotype and environment. b. the difference between the greenhouse population mean and the whole population mean.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chapter 19: The Inheritance of Complex Traits c. differences in the plants' microenvironment. d. random genetic differences within the population. e. the use of a nonrepresentative sample to compute the mean.

ANSWER: c 6. Three populations of true-breeding C. elegans larvae are each split into four subpopulations of equal sizes, and each subpopulation is grown in a different environment. The mean adult body lengths in each environment are as follows: Environment 1: Environment 2: Environment 3: Environment 4:

Population A 1.07 mm 1.02 mm 0.99 mm 0.97 mm

Population B 1.03 mm 0.98 mm 0.95 mm 0.93 mm

Population C 1.08 mm 1.03 mm 1.00 mm 0.98 mm

Based on these data, what environment has the highest environmental deviation? a. both Environments 3 and 4 b. Environment 1 c. Environment 2 d. Environment 3 e. Environment 4 ANSWER: b 7. When all of the variation in a population is due to environmental sources and there is no genetic variation, broad-sense heritability (H2) is a. zero and all of the phenotypic variability is due to loci at other locations. b. zero and all of the phenotypic variability is due to environment. c. some negative value indicating the phenotypic variability is due to some recessive loci. d. some positive value indicating the phenotypic variability is due to some dominant loci.

ANSWER: b 8. A quantitative geneticist measures the broad-sense heritability for bill length in an isolated population of ducks to be 0.75. This result suggests that a. the population tested has a high variance in bill length. b. in the population tested, 75% of the variance in bill length is due to genetic differences among individuals. c. in the population tested, 75% of the variance in bill length is due to environmental differences among the individuals' microhabitats. d. if all the ducks in the population tested were raised in identical environments, 75% of them would have identical bill length. e. in the population tested, 75% of the ducks have a bill length that is within one standard deviation of the mean.

ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chapter 19: The Inheritance of Complex Traits 9. Two different pure-breeding lines of tobacco were crossed and their F1 showed a variance in flower length of 8.76. The total variance in flower length in the F2 was 40.96. The F1 and the F2 plants were bred, grown, and kept in identical environments.

What is the broad-sense heritability of flower length in tobacco in this experiment? a. 8.76% b. 21.38% c. 32.20% d. 78.61% e. The broad-sense heritability cannot be calculated based on these data.

ANSWER: d 10. Two different pure-breeding lines of tobacco were crossed, and their F1 showed a variance in flower length of 8.76. The total variance in flower length in the F2 was 40.96. The F1 and the F2 plants were bred, grown, and kept in identical environments.

The variance in flower length observed in the F1 generation is due to a. changes in the environment between the parental and the F1 generation. b. differences in heritability for flower length between the parental and the F1 plants. c. differences in the individual F1 tobacco plants' environments. d. genetic differences among the individual F1 tobacco plants. e. genetic differences among the individual F1 tobacco plants as well as differences in their environments.

ANSWER: c 11. Two different pure-breeding lines of tobacco were crossed, and their F1 showed a variance in flower length of 8.76. The total variance in flower length in the F2 was 40.96. The F1 and the F2 plants were bred, grown, and kept in identical environments, yet the variance in flower length was 8.76 for the F1 and 40.96 for the F2 plants.

This result indicates that for this experiment, the environmental variance (Ve) a. does not play a role in determining flower length. b. is constant and equal to zero. c. is constant between the F1 and F2 plants, but not between the F1 and the parental plants. d. is constant and not equal to zero. e. is not constant.

ANSWER: d 12. You are studying a highly inbred strain of barley. After many generations of inbreeding, you now assume that the genetic variation is zero. Suppose that you raise many individuals of your barley strain in a field in southeastern Michigan and then measure their height after one month of growth. You find that the variance in Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chapter 19: The Inheritance of Complex Traits their height is 4.0. The broad-sense heritability of this trait is 0. From this particular study, you can conclude that a. genetics does not play a major role in determining the height of barley plants. b. if you were to repeat the same experiment in Iowa, you would obtain a different broad-sense heritability. c. if you were to repeat the same experiment in Iowa, you would obtain the same broad-sense heritability. d. in barley plants, height is not inheritable. e. the line of barley plants that you are studying is not inbred enough.

ANSWER: c 13. The variance in tail length in a large population of genetically diverse mice is found to be 21.30. The broadsense heritability for this trait in an identical population was found to be 0.8. Based on this information, what is the predicted genetic variance (Vg) for these mice? a. 4.26 b. 17.04 c. 20.50 d. 21.30 e. 26.62 ANSWER: b 14. Ten pairs of monozygotic twins were tested for their ability to solve geometry problems. The highest score obtained was 46 points, while the lowest was 6 points. The variance in geometry test score within the 20 subjects (measured in points2) was 208. The covariance for test score between twins was 125. In this study, the broad-sense heritability of geometry problem-solving ability is a. 26%. b. 32%. c. 40%. d. 60%. e. It is impossible to calculate.

ANSWER: d 15. It is difficult to measure any type of heritability without the ability to perform specific crosses. However, vast population studies enable estimation of heritability for some characters in human beings. Some estimated broad-sense heritabilities include the following: Character stature body weight systolic blood pressure diastolic blood pressure twinning

H2 0.85 0.62 0.57 0.44 0.50

Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chapter 19: The Inheritance of Complex Traits Which of these characteristics is most likely to "run in families"? a. body weight b. diastolic blood pressure c. stature d. systolic blood pressure e. twinning ANSWER: c 16. Two pure lines of plants had mean heights of 26 cm and 42 cm, respectively. The F1 had a mean of 34 cm and a variance of 20. The F2 also had a mean of 34 but a variance of 60. The broad-sense heritability (H2) is a. 0.13. b. 0.34. c. 0.50. d. 0.67. e. 0.80.

ANSWER: d 17. Narrow-sense heritability (h2) is a quantification of the proportion of total variance due to a. additive genetic variance. b. dominance variance. c. environmental variance. d. phenotypic variance. e. total genetic variance.

ANSWER: a 18. Suppose that gene A controls the synthesis of blue pigment in the petals of a flowering plant. Plants with genotype A/A produce blue pigment in their petals, while those with genotype A/a or a/a do not produce any blue pigment. The expected narrow-sense heritability (h2) for the production of blue pigment in this plant's flowers is a. exactly 0. b. exactly 0.5. c. higher than 0, but lower than 1. d. exactly 1. e. dependent on the environmental variance.

ANSWER: d 19. Consider a diploid fish in which tail length is solely controlled by the additive action of two unlinked loci, F (f) and G (g), without any effect of the environment. A female of genotype F/F ; G/G, with a 3-cm-long tail, is crossed to a male of genotype f/f ; g/g, with a 2-cm-long tail. All the F1 fish have 2.5-cm-long tails.

If the F1 fish are crossed to each other, how many different tail length phenotypes will be present in F2?

Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chapter 19: The Inheritance of Complex Traits a. 16 b. 5 c. 4 d. 3 e. 2 ANSWER: b 20. Consider a diploid fish in which tail length is solely controlled by the additive action of two unlinked loci, F (f) and G (g), without any effect of the environment. A female of genotype F/F ; G/G, with a 3-cm-long tail, is crossed to a male of genotype f/f ; g/g, with a 2-cm-long tail. All the F1 fish have 2.5-cm-long tails.

What will be the most frequent tail length among the F2 individuals? a. 2 cm b. 2.25 cm c. 2.5 cm d. 2.75 cm e. 3 cm ANSWER: c 21. Loppins (fictitious diploid invertebrates) have variable numbers of dots on their abdomen. This variability is entirely due to genetics, with no influence of the environment, so the broad-sense heritability for this trait is 1.

The mean number of dots per individual in a large population is 37. A breeder crosses the two animals with the highest number of dots, 45. In the progeny, the mean number of dots is 43.4. These results suggest that the phenotype of interest is controlled a. at least partly by genes that have an additive mode of action. b. at least partly by genes that have a dominant mode of action. c. by genes that have both an additive and a dominant mode of action. d. only by genes that have an additive mode of action. e. only by genes that have a dominant mode of action.

ANSWER: b 22. Presume that the number of leaves in clovers is controlled by the combined actions of two genes, A and B, with no contributions from the environment. A/A results in four leaves, A/a in three leaves, and a/a in two leaves. Moreover, the genotypes B/B and B/b result in one additional leaf (b/b has no effect on leaf number).

Based on the above information, a. A has an additive mode of action and B has a dominant mode of action. b. A has a dominant mode of action and B has an additive mode of action. c. both A and B have a dominant mode of action. d. both A and B have an additive mode of action. e. the respective modes of action of A and B cannot be determined.

ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chapter 19: The Inheritance of Complex Traits 23. Presume that the number of leaves in clovers is controlled by the combined actions of two genes, A and B, with no contributions from the environment. A/A results in four leaves, A/a in three leaves, and a/a in two leaves. Moreover, the genotypes B/B and B/b result in one additional leaf (b/b has no effect on leaf number).

In the model described, the additive effect is a. 0. b. 1/2. c. 1. d. 2. e. 4.

ANSWER: c 24. Presume that the number of leaves in clovers is controlled by the combined actions of two genes, A and B, with no contributions from the environment. A/A results in four leaves, A/a in three leaves, and a/a in two leaves. Moreover, the genotypes B/B and B/b result in one additional leaf (b/b has no effect on leaf number).

In the model described, the dominance effect is a. 0. b. 1/2. c. 1. d. 2. e. 4.

ANSWER: b 25. A behavioral biologist suspects the "N" locus to be involved in the learning ability of dogs. Learning is measured through a standardized test in which each dog can score from 0 (very poor) to 20 points (excellent). In an experiment, the mean score was 15.6 for dogs of genotype N/N, 12.1 for dogs of genotype N/n, and 7.2 for those of genotype n/n. What is the dominance/additivity ratio for this locus? a. 0 b. higher than 0, but lower than 0.5 c. 0.5 d. higher than 0.5, but lower than 1 e. 1 ANSWER: b 26. The phenotypic deviation that is transmitted from parents to their offspring is known as a. additive genetic variance. b. dominance variance. c. environmental variance. d. total genetic variance. e. transmissible variance.

Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chapter 19: The Inheritance of Complex Traits ANSWER: a 27. In an experiment carried out in a chicken population, measurements of fat content reveal that the genetic variance for this trait is 120 g2, the environmental variance is 80 g2, and the variance due to dominance is 38 g2.

The narrow-sense heritability (h2) of fat content in these chickens is a. 0.24. b. 0.41. c. 0.52. d. 0.74. e. 0.90.

ANSWER: b 28. A population of beetles has a mean weight of 30 g. In a controlled, constant environment, individuals of 12 g are selected and allowed to interbreed, and their progeny have a mean weight of 21 g with a variance of 4.5 g2. What is the additive variance in this experiment? a. 0.50 b. 0.57 c. 0.70 d. 1.75 e. 2.25 ANSWER: e 29. You are studying the inheritance of antennae length in loppins using strains from British Columbia (BC) and Washington State (WA). BC loppins have, on average, much longer antennae than their WA counterparts. You have identified three QTLs that control this trait, and you obtained the following data:

QTL1 QTL2 QTL3

Genotype and mean antennae length BC/BC BC/WA WA/WA 4 mm 4.8 mm 5.9 mm BC/BC BC/WA WA/WA 4.4 mm 4.2 mm 4 mm BC/BC BC/WA WA/WA 3.9 mm 4.5 mm 5.7 mm

Which QTL(s) has/have the highest dominance effect? a. QTL1 b. QTL2 c. QTL3 d. QTLs 1 and 2 e. QTLs 1 and 3 ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chapter 19: The Inheritance of Complex Traits 30. In the course of estimating the degree of linkage of a marker and a quantitative trait of interest, researchers announced that they obtained a LOD score of 4.5. This finding indicates that a. there is a 4.5 % chance that an individual with the marker will also show the trait of interest. b. the marker is 4.5 map units away from the QTL controlling the trait of interest. c. the marker is most likely not linked to the QTL controlling the trait of interest. d. the odds that the marker is linked to the QTL controlling the trait of interest are 4.5:1. e. the odds that the marker is linked to the QTL controlling the trait of interest are 4500:1.

ANSWER: e 31. Imagine that an association mapping study revealed a particular variant of a SNP to be strongly associated with susceptibility to insomnia. From these data, you could infer that the a. SNP is closely linked to a locus that plays a role in insomnia. b. SNP is located in a gene that determines insomnia. c. SNP provides evidence that insomnia is a genetic condition. d. variant SNP most likely causes insomnia. e. variant SNP plays a role in insomnia.

ANSWER: a 32. Which of the following has more important value to the plant or animal breeder? a. broad-sense heritability b. variance c. correlation d. additive deviation e. dominance deviation ANSWER: d 33. Which of the following is an example of a meristic trait? a. number of toes on the forefoot b. height of a flower c. weight of a cow d. shape of a trout e. size of a turtle shell ANSWER: a 34. A complex trait is a trait that a. arises from a single gene. b. does not show simple Mendelian inheritance. c. is not categorical or continuous. d. is not affected by the environment. e. behaves according to Mendelian inheritance laws.

ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chapter 19: The Inheritance of Complex Traits 35. The genes that control variation in complex traits are known as a. quantitative trait loci. b. gene variants. c. gene mutations. d. gene substitutions. e. additive genes.

ANSWER: a Essay 36. Three inbred lines of potatoes are cultured in three different environments. The mean tuber weight for each line in each environment is reported below: Environment 1: Environment 2: Environment 3:

Line A 131 g 180 g 97 g

Line B 93 g 142 g 59 g

Line C 160 g 209 g 126 g

a) What are the genetic deviations (g) for each of the three lines? b) What is the environmental effect (e) on tuber weight for each of the three environments?

ANSWER: a) Overall mean tuber weight: (131 + 93 + 160 + 180 + 142 + 209 + 97 + 59 + 126)/9 = 121.8 [g] mean for line A: 102.6 [g] g (line A) = 102.6 – 121.8 = –19.2 g mean for line B: 98 [g] g (line B) = 98 – 121.8 = –23.8 g mean for line C: 165 [g] g (line C) = 165 – 121.8 = 43.2 g b) mean for environment 1: 128 [g] e (env1) = 128 – 121.8 = 6.2 g mean for environment 2: 177 [g] e (env2) = 177 – 121.8 = 65.2 g mean for environment 3: 94 [g] e (env3) = 94 – 121.8 = –27.8 g 37. Two highly inbred populations of the annual weed Capsella bursa-pastoris grow in adjacent fields. One field is mowed regularly; the other is never mowed. Seeds from the plants are taken to a greenhouse where they can be grown under uniform conditions. The mean height of plants at the time of flowering is measured for each population, and for the F1 and F2 generations from crosses between representatives of each population. The data from these experiments are as follows:

a) How much of the observed variance in plant height at flowering is caused by the environment in which the plants are grown? b) How much of the variance is due to genotype?

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chapter 19: The Inheritance of Complex Traits c) What is the broad-sense heritability (H2) for this character?

ANSWER: a) environmental variance = (12.6 + 4.0 + 10.2)/3 = 8.9 total variance = 20.3 b) genotypic variance = total variance – environmental variance = 11.4 c) H2 = 11.4/20.3 = 0.56 38. In a natural population of outbreeding annual plants, the variance of the total number of seeds per plant is 16. From the natural population, 20 plants are taken into the laboratory and each selfed for 10 generations. The average variance in the 10th generation in each of the 20 sets is about equal, and averages 5.8 across all the sets.

Estimate the broad-sense heritability (H2) of seed number in this population.

ANSWER: Repeated selfing results in homozygosity, so plants within each of the 20 sets were very close to being genetically identical after 10 generations. Therefore, the variance that remains in the 10th generation must be environmental variance, which can be subtracted from the total variance to get the genetic variance.

H2 = (16 – 5.8)/16 = 0.68 39. Two pure lines of Nicotianum (tobacco) have significantly different corolla lengths. In line 1, the average length is 30 mm, ranging from 25 mm to 35 mm with a variance of 10. In line 2, the average corolla length is 60 mm, ranging from 45 mm to 65 mm with a variance of 12. The F1 has a corolla length of 45 mm, ranging from 40 to 50, and the variance is 9. Upon selfing, an F2 is produced that has an average flower length of 45 mm but ranges from 10 mm to 80 mm with a variance of 55. a) Provide a general explanation of these results. b) Calculate the broad-sense heritability (H2) for corolla length in this study.

ANSWER: a) Multiple genes with an additive mode of action can explain the general results. There is no segregation in the P or F1 generations, but in the F2, the genes with additive mode of action of the F1 have segregated to give the complete range of genotypes. For example: A/A ; B/B ; C/C ; d/d ; e/e ; f/f

a/a ; b/b ; c/c ; D/D ; E/E ; F/F

F1 A/a ; B/b ; C/c ; D/d ; E/e ; F/f F2 ranges from a/a ; b/b ; c/c ; d/d ; e/e ; f/f to A/A ; B/B ; C/C ; D/D ; E/E ; F/F b) The variance of the P and F1 must be all environmental, and these are all approximately equal, averaging 10.3. The F2 variance must be a combination of genetic and environmental variance.

Therefore: H2= (55 − 10.3)/55 = 81% 40. An individual plant is heterozygous for loci that affect height and have a strictly additive mode of action. If this plant is selfed, a) how many height classes will there be among the progeny? (Assume no environmental effects.) b) what will be the frequency of the tallest class of progeny? (Assume no environmental effects.)

Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chapter 19: The Inheritance of Complex Traits ANSWER: a) There can be a range of 0 to 12 genes acting with an additive mode of action for a total of 13 phenotypic classes. The general formula is 2N + 1, where N is the number of loci. b) (1/4)6 = 1/4096 41. In a population of beetles, the total variance of body weight is 130. It is estimated that the environmental variance is 35, and dominance genetic variance is 45. Calculate the narrow-sense heritability (h2) of body weight in these beetles.

ANSWER: Additive genetic variance must be 130 – 45 – 35 = 50 h2 = 50/130 = 0.39 42. In a large flock of turkeys, average body weight is 5.0 kg. In a test to determine if the average weight of the flock can be increased by selective breeding, 6.2-kg birds are removed and allowed to interbreed. Their progeny have an average weight of 5.2 kg. Calculate the narrow-sense heritability (h2).

ANSWER: The selection differential is 6.2 – 5.0 = 1.2 The selective gain is 5.2 – 5.0 = 0.2 h2 = 0.2/1.2 = 0.17 43. The narrow-sense heritability (h2) for several traits in domesticated cattle follows: Trait Percent protein in milk Feed efficiency Milk yield Calving interval

h2 0.54 0.34 0.30 0.01

a) Which of these traits would be most responsive to artificial selection? b) Which would be least responsive?

ANSWER: a) The traits with the higher heritability (such as milk protein) would have the greatest response to selection. b) Calving interval would have the least response to selection.

44. a) What is the role of marker genes for the identification of quantitative trait loci? b) How do they differ from candidate genes?

ANSWER: a) Marker genes are used to localize the position of quantitative trait loci on a chromosome or a portion of a chromosome. They should be picked to have phenotypes quite different from that of the quantitative trait. Molecular markers are commonly used. b) Candidate genes are genes that are suspected to make a significant contribution to the variation of the character under investigation.

45. What are the advantages of association mapping over QTL mapping?

ANSWER: Association mapping does not require controlled crosses or to work with human families with known parent–offspring relationships because it is performed with random-mating populations.

Association mapping tests many alleles at a locus at once, while QTL mapping compares only two alleles. Association mapping assays all the alleles in a population at the same time and can also lead to the direct identification of the genes at the QTL without the need for subsequent fine-mapping Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chapter 19: The Inheritance of Complex Traits studies.

46. What does variance measure, and what are the types and causes of variance?

ANSWER: Variance measures the extent to which individuals deviate from the population mean. Genetic variance is due to genetic factors, and environmental variance is due to environmental factors.

Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chapter 20: Evolution of Genes and Traits Multiple Choice 1. Which of the following is/are NOT a principle of Darwinian evolution? a. There is variation in morphology, physiology, and behavior among individuals in a population. b. Offspring resemble their parents more closely than unrelated individuals. c. In any given environment, some individuals are more successful at surviving and reproducing than others. d. Darwinian evolution fully describes where all living things come from. e. All of the answer options are correct.

ANSWER: d 2. Which of these principles of evolution, as described by Darwin's theory, is correctly matched with its role in evolution? a. Principle of variation: Offspring must resemble their parents more than they resemble unrelated individuals. b. Principle of selection: Variation in morphology, physiology, and behavior must be present in a population for selection to occur. c. Principle of variation: Variation in morphology, physiology, and behavior must be present in a population for selection to occur. d. Principle of heredity: Some genotypes of a population should be more successful at surviving and reproducing than other genotypes in a given environment. e. None of the answer options is correct.

ANSWER: c 3. Which of the following are requirements for evolution by natural selection?

I environmental change II differential survival and reproduction III heritability of phenotypic variation IV variation in phenotype V sexual reproduction a. II, III, V b. II, III, IV c. I, II, IV d. III, IV, V e. II, IV, V ANSWER: b 4. Why don't we find different human populations with different fixed alleles at a certain locus? a. Inbreeding is high. b. Migration between populations is high. c. The mutation rate is lower than the critical value. d. Population sizes are too large.

Copyright Macmillan Learning. Powered by Cognero.

Page 1

Name:

Class:

Date:

Chapter 20: Evolution of Genes and Traits e. Allelic variation is low.

ANSWER: b 5. Which of the following processes will increase variation between populations? a. earthquakes b. mutation c. migration d. balancing selection e. inbreeding ANSWER: e 6. Which of the following processes will increase genetic variation within populations? a. incompatible selection b. mutation c. directional selection d. asexual reproduction e. genetic drift ANSWER: b 7. Which of the following reduce heterozygosity? a. migration b. directional selection c. large population size d. balancing selection e. mutation ANSWER: b 8. Which researcher proposed that change was caused by the environment acting directly on an organism, and that those changes acquired in an organism's lifetime were passed on to its offspring? a. Gould b. Darwin c. Wallace d. Lamarck e. Allison ANSWER: d 9. The neutral theory of evolution proposed that most genetic mutations are functionally neutral or nearly neutral. Neutral mutations would be fixed in a population by a. natural selection. b. random genetic drift. c. balancing selection.

Copyright Macmillan Learning. Powered by Cognero.

Page 2

Name:

Class:

Date:

Chapter 20: Evolution of Genes and Traits d. asexual reproduction. e. the invariance of DNA. ANSWER: b 10. When considering synonymous and nonsynonymous mutations within a population, which of the following is TRUE? a. Nonsynonymous mutations will often be removed by purifying selection. b. Synonymous mutations will often be removed by purifying selection. c. The two types of mutations are not distinguished by purifying selection. d. Only nonsynonymous mutations are known to occur. e. Only synonymous mutations are known to occur.

ANSWER: a 11. When considering the rate of nucleotide divergence for a particular gene over time, which of the following has been observed to be TRUE? a. Nonsynonymous mutations will become established less frequently. b. Adenine nucleotides are more readily mutated than other nucleotides. c. Synonymous mutations will become established less frequently. d. Nucleotide changes over time are impossible to measure. e. Only noncoding sequences display change.

ANSWER: a 12. Tony Allison examined the allele frequency of hemoglobin (HbA or HbS) in populations of Kenyon tribes.

A surprising result of this study was that a. the HbS and HbA alleles were evenly distributed among Kenyan tribes. b. the HbS alleles appeared more frequently in men than in women. c. tribes living in arid or highland regions had high frequencies of HbS alleles. d. tribes living near Lake Victoria had a low frequency of HbS alleles. e. tribes living in arid or highland regions had low frequencies of HbS alleles.

ANSWER: e 13. Plasmodium falciparum is a parasitic protist that can be carried among host organisms by a mosquito.

Infection generates anemia via a process of a. infecting the bone marrow of its hosts. b. causing uncontrolled bleeding from tissues. c. invading red blood cells and dividing within them. d. mosquito-based feeding upon the blood during transmission. e. Anemia is not a symptom of Plasmodium infection.

ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

Page 3

Name:

Class:

Date:

Chapter 20: Evolution of Genes and Traits 14. When considering hemoglobin alleles, scientists concluded that the proportions of the two alleles within a population was strongly affected by the a. severity of malaria. b. atmospheric pressure of O2. c. rate of elimination of harmful HbS alleles. d. nutritional intake of each individual. e. severity of malaria and the rate of elimination of harmful HbS alleles.

ANSWER: b 15. Balanced polymorphism is illustrated by a. a situation where the selective pressure enhances the frequency of heterozygotes over homozygotes. b. a situation where the selective pressure enhances the frequency of homozygotes over heterozygotes. c. only situations where disease impacts allele frequency. d. situations where allelic homozygosity is impossible. e. All of the answer options are correct.

ANSWER: a 16. The fitness advantage of the AS allelic combination can be calculated by a. measuring O2-carrying capacity in the blood of AS individuals. b. comparing the actual frequency of AS to the frequency predicted by the Hardy–Weinberg equation. c. measuring the frequency of AS in the population. d. infecting the blood of AS individuals with the parasite. e. measuring the frequency of AA and subtracting that value from the frequency of AS. ANSWER: b 17. In human populations, the AS allele arose from a. the insertion of a mutagenic transposon into the Hb gene. b. one individual located in central Africa. c. five independent events, in separate populations. d. a frameshift mutation in the Hb gene. e. a plasmodium-induced change in DNA sequence.

ANSWER: c 18. The molecular basis of sickle cell hemoglobin is a. a frameshift mutation. b. a single nucleotide substitution. c. a translocation event. d. a virus-induced inversion. e. exposure to a chemical mutagen.

ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

Page 4

Name:

Class:

Date:

Chapter 20: Evolution of Genes and Traits 19. What term could be used to describe the type of selection occurring with the sickle-cell and normal hemoglobin alleles among areas where malaria is present? a. directional selection b. balancing selection c. stabilizing selection d. genetic drift e. inbreeding depression ANSWER: b 20. The gene for which of the following proteins is likely to have the higher neutral rate of mutation? a. cytochrome c b. alpha subunit of hemoglobin c. beta subunit of hemoglobin d. fibrinopeptides e. actin ANSWER: d 21. Organophosphate resistance in the sheep blowfly (Luciliacuprina) is an example of a. the toxicity of insecticides. b. the generation of neutral mutations over time. c. a lethal frameshift mutation. d. cumulative selection. e. evolutionary selection of a single amino acid change in a protein.

ANSWER: e 22. An excess of nonsynonymous fixed mutations between species is likely due to a. natural selection. b. random genetic drift. c. a catastrophic event. d. inbreeding. e. immigration.

ANSWER: a 23. Some of the most striking and best understood examples of morphological divergence are found in a. populations of antibiotic-resistant bacteria. b. virology. c. animal body-color patterns. d. mouse eye-color mutants. e. blood-type examples in mammals.

ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

Page 5

Name:

Class:

Date:

Chapter 20: Evolution of Genes and Traits 24. The melanocortin 1 receptor protein displays four amino acid substitutions between dark- and light-colored populations of pocket mice. In dark mice, these unique amino acids cause the receptor to a. fold inappropriately in the cell membrane. b. incorrectly identify ligand. c. be nonfunctional in regulating gene expression. d. be internalized and degraded. e. be constitutively active.

ANSWER: e 25. In southwestern Arizona, the melanocortin 1 receptor mutation-bearing mice (dark hair) are commonly found in a. rocky lava outcrops. b. stream beds. c. sandy desert regions. d. mixed rodent populations. e. high mountain regions.

ANSWER: a 26. Inherited blindness in some cave fish arose through a. an individual mutant fish that displayed reproductive success. b. atrophy of the eye when unused in darkness. c. viral infection of the optic nerve. d. a defect in gene splicing. e. separate incidences of inactivating mutation in the Oca2 gene.

ANSWER: e 27. In fruit flies, mutations in cis-acting regulatory sequences are associated with a. abnormal functioning of operator sequences. b. body form/color variations. c. changes in fruit fly behavior. d. changes in reproductive fitness. e. an alteration of DNA replication.

ANSWER: b 28. What process reduces genetic variation and preserves DNA and protein sequences over eons of time. a. purifying selection b. natural selection c. genetic drift d. positive selection e. evolution Copyright Macmillan Learning. Powered by Cognero.

Page 6

Name:

Class:

Date:

Chapter 20: Evolution of Genes and Traits ANSWER: a 29. What theory of molecular evolution offers a baseline expectation of how DNA should change over time when natural selection is absent? a. neutral evolution b. negative selection c. genetic drift d. positive selection e. purifying selection ANSWER: a Essay 30. Darwin's theory of evolution was developed around three central principles: the principle of variation, the principle of heredity, and the principle of selection. Define each of these principles, and explain their role in evolution.

ANSWER: Principle of variation: Variation in morphology, physiology, and behavior must be present in a population for selection to occur. The "best" individual will be selected and will have a reproductive advantage.

Principle of heredity: The difference in traits that are being selected must be heritable (controlled by genes). Thus, offspring should resemble their parents more than they resemble unrelated individuals.

Principle of selection: In order for natural selection to occur, some genotypes of a population should be more successful at surviving and reproducing than other genotypes in a given environment.

31. A strawberry breeder wants to increase the yield of strawberry plants. To do so, he selects the two highestyielding plants from a segregating population and uses them as parents that he hopes will produce higheryielding offspring. How would you use Darwin's evolutionary theory to explain the following results? a) The breeder was able to obtain a better-yielding offspring. b) The breeder was unable to obtain a better-yielding offspring.

ANSWER: a) The variation observed for yield in the initial segregating population must have had a genetic basis. Therefore, the individuals that were phenotypically superior in the environment were also genotypically superior. In other words, for the breeder to produce higher-yielding offspring, this variation must have been highly heritable. b) The variation observed for yield must not have had a genetic basis but was instead due to differences in the environmental or field conditions. Thus, the individuals that were phenotypically superior were not necessarily genotypically superior. In other words, the trait (high yield) was not heritable or shows very low heritability and hence does not appear in the next generation.

32. An orchid breeder is interested in increasing the number of stems and the length of stems that can be produced by an orchid plant. She finds a segregating population that contains a wide range of both stem number (ranging from 2 to 15 stems per plant) and stem length (ranging from 2 to 7 inches in length). She selected plants from this population that either had the most stems per plant or the longest stems per plant (shown as Copyright Macmillan Learning. Powered by Cognero.

Page 7

Name:

Class:

Date:

Chapter 20: Evolution of Genes and Traits shaded regions in the diagrams below). She then interbred the plants with the most stems (or the longest stems) and grew the resulting populations of orchids. Following an additional round of selection and breeding, the breeder drew graphs to look at the distribution of plants in each generation. Shown below are graphs representing the distribution of plants in each generation in terms of stem number and stem length (in inches).

For each trait, determine whether or not the trait is heritable. Justify your answers.

ANSWER: Stems per plant: This trait is heritable. Each generation, the population, on average, was improved in terms of number of stems per plant.

Stem length: This trait is not heritable. Although she selected plants with the longest stems and intermated them, the distribution of stem length in the offspring is identical to that of the initial generation. If the trait were heritable, then the resulting generation would be shifted in the direction of longer stems per plant.

33. Can behaviors be the product of selection, even when their heritability is either very low or zero? Justify your answer.

ANSWER: Yes. Selection can lead to reduced heritability, as favorable alleles become fixed, so finding a trait with low heritability may actually mean that it has been the product of intense selection. Also, behaviors can often be flexible, but the capacity for that flexibility can be heritable. In such cases, the capacity or form of the learning of the behavior is the more appropriate trait to analyze genetically.

34. The rate of change is not identical for all genes. The following graph shows the rate of change of four hypothetical human proteins (A–D). Use this graph to answer the following questions.

Copyright Macmillan Learning. Powered by Cognero.

Page 8

Name:

Class:

Date:

Chapter 20: Evolution of Genes and Traits

a) Which of the proteins is changing at the fastest rate? b) Which protein is most conserved? c) Which evolutionary force (drift, selection, or migration) plays the biggest role in causing the observed difference in the rate of change between proteins A–D? d) If you do Southern hybridization to determine if the genes encoding each of these human proteins (A–D) also exist in rice, which would give the strongest hybridization signal? Justify your answer. e) If you were going to use one of the proteins to build a phylogenetic tree of various mammalian species (which diverged during the last 50 million years), which protein sequence would be most useful? Justify your answer. f) If you were going to use one of the proteins to build a phylogenetic tree of various plant species (which diverged during the past 500 million years), which protein sequence would be most useful? Justify your answer. g) If you were going to use one of the proteins to build a phylogenetic tree of a mixture of plant and animal species (which diverged 1200 million years ago), which protein sequence would be most useful? Justify your answer.

ANSWER: a) A b) D c) Selection. There is a lot of selection against mutations in Protein D. There is less selection against mutations in the other proteins. Thus, Protein D is likely to play a vital role in the cell, and mutations would likely result in the death of the cell. d) D. The rate of change is the slowest, so the DNA sequences will be most similar between species.

Because the DNA sequences will be highly similar, the ability for a probe to hybridize will be Copyright Macmillan Learning. Powered by Cognero.

Page 9

Name:

Class:

Date:

Chapter 20: Evolution of Genes and Traits greater. The gene for Protein A will be very different between human and rice; therefore, the probe might not hybridize to the gene in rice. e) A. The other proteins change too slowly. Thus, all mammals would probably have identical sequences for these proteins. However, they would probably have different sequences for protein A. These differences could be used to create a tree. f) B. Protein A would be too divergent to be useful (because there would be a lot of differences between species). Proteins C and D would be pretty much identical between species. B would have the right amount of sequence divergence to be useful. g) C. A and B would be too divergent to be useful (because there would be a lot of differences between species). Protein D would be too similar between species. C would have the right amount of sequence divergence to be useful.

35. Assume that the following amino acid sequences were obtained for a highly conserved gene, gene X, in eukaryotes. Using this information, determine the most probable ancestral amino acid sequence.

Human G D V E K G K K I F I M K C S Q C H T Chimpanzee G D V E K G K K I F I M K C S Q C H T Horse G D V E K G K K I F V Q K C A Q C H T Dog G D V E K G K K I F V Q K C A Q C H T Chicken G D I E K G K K I F V Q K C S Q C H T Tuna G D V A K G K K T F V Q K C A Q C H T Drosophila G D V E K G K K L F V Q R C A Q C H T Neurospora G D S K K G A N L F K T R C A E C H G Wheat G N P D A G A K I F K T K C A Q C H T ANSWER: In order to find the ancestral amino acid sequence, you need to look at every amino acid and determine which is most common among the organisms.

Human Chimpanzee Horse Dog Chicken Tuna Drosophila Neurospora Wheat

G G G G G G G G G

D D D D D D D D N

V V V V I V V S P

E E E E E A E K D

K K K K K K K K A

G G G G G G G G G

K K K K K K K A A

K K K K K K K N K

I I I I I T L L I

F F F F F F F F F

I I V V V V V K K

M M Q Q Q Q Q T T

K K K K K K R R K

C C C C C C C C C

S S A A S A A A A

Q Q Q Q Q Q Q E Q

C C C C C C C C C

H H H H H H H H H

T T T T T T T G T

Therefore, the most ancestral amino acid sequence is GDVEKGKKIFVQKCAQCHT. 36. Define the molecular clock.

ANSWER: The molecular clock measures how much divergence has taken place between two species over evolutionary time. The constant rate of neutral nucleotide substitutions predicts that if the number of nucleotide differences between two species were plotted against time since they diverged from a common ancestor, the result should be a straight line with a slope equal to μ, the rate of neutral mutation replacement, and so we can say that evolution proceeds at a rate equal to this molecular clock.

Copyright Macmillan Learning. Powered by Cognero.

Page 10

Name:

Class:

Date:

Chapter 20: Evolution of Genes and Traits 37. Suppose that a population composed of 100 individuals existed on an island in the Caribbean Sea. Two different alleles of the A gene were present in this population (a1 and a2) at equal frequency. The following events occurred to this unlucky population. Determine what evolutionary force (mutation, migration, drift, or selection) is represented by each event. In addition, tell the effect of each event on the genetic composition of the population. a) A devastating fire caused the death of 10 individuals. b) A disease swept through the island and killed 13 individuals. The a1 confers resistance to the disease. c) A disease swept through the island, killing 13 individuals. Neither a1 nor a2 confers resistance to the disease. d) A ship transporting individuals from a different island containing a1, a2, and a3 alleles landed on the island. e) The president (whose genotype is a1a1) declared that all a1 individuals had more rights than a2 individuals.

All a2 individuals moved to another island. f) An individual who was working in a cave stumbled upon a new rock and decided to put it in his pants pockets to show his wife. Some of their babies had an a4.

ANSWER: a) Drift. It could either wipe out an allele or kill equal numbers of each allele. The effect on the genetic composition cannot be determined. b) Selection. It increases the frequency of the a1 allele and decreases the frequency of the a2 allele in the population. c) Drift. It could either wipe out an allele or kill equal numbers of each allele. The effect on the genetic composition cannot be determined. d) Migration. It increases the frequency of the a3 allele in the population and decreases the frequency of a1 and a2. e) Selection. It increases the frequency of the a1 allele in the population. f) Mutation. It increases the frequency of the a4 allele in the population and decreases the frequency of a1 and a2.

38. Following a gene duplication event, the additional copy of the gene is identical to the original copy. a) Does the duplication event, by itself, create a novel function? b) How does a gene duplication event promote the creation of a gene with a novel function?

ANSWER: a) No, because the gene is identical, it will not have a novel function. The duplication could allow more protein to be made, though, which could be either beneficial or harmful for the individuals. b) Before a duplication event occurs, the gene, especially if it performs a vital function for the cell, is fairly constrained in the amount of mutations it can accumulate and in the amount of "exploring" of new functions it can do. However, following a gene duplication event, the second copy is free to mutate without constraint because the original copy of the gene can continue to perform its function while the new copy mutates and explores new functions.

39. There are several programs that allow researchers to search DNA databases with a newly acquired DNA sequence. One program, BLAST (www.ncbi.nlm.nih.gov/BLAST/), allows researchers to compare their sequence with all publicly available sequences. The search engine returns a list of similar sequences, with an Copyright Macmillan Learning. Powered by Cognero.

Page 11

Name:

Class:

Date:

Chapter 20: Evolution of Genes and Traits estimate of similarity score (the greater the similarity between two sequences, the larger the score), and reveals the function of these similar sequences (if known). Suppose that you have just cloned and sequenced a gene from a gorilla, but you do not know what it does in a cell. You do a BLAST search using your gorilla sequence.

Some of the output from the search is shown below: Database ID NM_001799

Description Homo sapienscyclin-dependent kinase 7 (CDK7), mRNA XM_215467 Rattus norvegicuscyclin-dependent kinase 7 (Cdk7), mRNA NM_009874 Mus musculuscyclin-dependent kinase 7 (Cdk7), mRNA XLP40PK XenopusMO15 mRNA for 40 kDa protein kinase, cdc2-related CRAMO15CDK Goldfish mRNA for MO15(cdk7) kinase, complete cds AF188750 Caenorhabditis elegansputative tyrosine kinase receptor

Score (bits) 1814 878 767 90 78 42

a) What is the likely function of the gorilla gene? b) Why does the output include sequences from species other than gorilla? c) Why are some sequences more similar than others?

ANSWER: a) The gorilla gene is probably a CDK (helps regulate the cell cycle). b) All species were derived from a single ancestor. This means that most genes found in the various species also evolved from the same ancestral gene. For genes with an important function, the DNA sequence has been conserved to allow the function to be maintained. Therefore, this gene is likely important for the survival of the organism and cell, explaining why a similar sequence has been found in such a wide range of organisms (from humans to C. elegans). c) Gorillas are more closely related to humans than to rats, mice, or goldfish. DNA sequences from organisms that are more closely related have had less time to accumulate mutations, whereas the sequences from organisms that diverged a long time ago (such as gorilla vs. C. elegans) have acquired many more mutations.

40. Suppose that you have cloned and sequenced a gene that controls fruit size in strawberry. However, you do not know the function of the gene. a) Describe how you could determine if this gene is also present in Arabidopsis (which has been fully sequenced). b) Describe how you could determine if the gene is present in day lily (which has not been fully sequenced). c) How could you determine a putative function for this gene?

ANSWER: a) Search an Arabidopsis gene database for a similar sequence. (Or do a Southern hybridization as described in part b.) b) Do a Southern hybridization. Probe lily DNA with the cloned strawberry gene. If a similar gene is present in lily, the probe will hybridize. If not, the probe will not hybridize.

Copyright Macmillan Learning. Powered by Cognero.

Page 12

Name:

Class:

Date:

Chapter 20: Evolution of Genes and Traits c) Look at the Arabidopsis database for similar genes. It is hoped that the function of the Arabidopsis genes is known. You can hypothesize that the strawberry gene has a similar function to the Arabidopsis gene.

41. Describe the relationship between overall DNA change and functional change.

ANSWER: The relationship is highly variable. In general, however, there are more opportunities for overall DNA change that is not functional than there are for functional changes.

42. Why do you expect different rates of evolution for synonymous versus nonsynonymous nucleotide substitutions?

ANSWER: Because fewer nonsynonymous mutations will be preserved (many will be deleterious and therefore lost). In one sense, the rate of these two mutations should be identical, but not all will persist long enough for us to detect them. So, our detection of more synonymous changes is the result of two processes: mutation and persistence.

43. In what way can synonymous mutations still have a selective component?

ANSWER: They can change the probability for mistakes in splicing and copying, on the stability and lifetime of mRNA and on the use of the available pool of tRNA molecules, which will affect the speed of translation.

44. Based on your knowledge of the endogenous causes of mutation, what features of an organism's phenotype or ecology should affect the molecular clock?

ANSWER: Some possible answers: exposure to ultraviolet light (and the extent to which UV light can reach tissues of the organism); metabolic rate; body temperature (both homeostatic temperature, and ambient temperature); exposure to environmental pollutants.

45. What will determine the rate of a molecular clock for a gene that codes for a functional protein?

ANSWER: The rate will be determined by the degree to which changes in the protein will decrease the fitness of the organism. If the protein is very sensitive to change (cannot function optimally with even small changes), then any detected clock will be extremely slow. If the protein can function adequately even with many changes, perhaps to sites that are not immediately associated with an active site on the protein, then any detected clock will be relatively fast.

46. Most genetic variation for many human loci lies within local populations rather than between populations or races. What does this observation tell you about human genetic evolution?

ANSWER: It suggests that humans have had persistent migration between populations that has prevented localized differentiation for most genes.

47. In a reforestation project, a fast-growing pine tree (fast-growing due to a dominant A allele) was selfed, and the resulting seed were planted on two islands (X and Y). One year after planting, there was a severe termite attack on Island X. The termites, which killed all slow-growing trees, destroyed 25% of the trees on the island.

A severe cyclone struck Island Y and randomly killed 25% of its trees. Justify all answers. a) With the information provided above, can you suggest the genotype of the original fast-growing tree that was selfed to obtain seed? b) What evolutionary force is at work on Island X? Copyright Macmillan Learning. Powered by Cognero.

Page 13

Name:

Class:

Date:

Chapter 20: Evolution of Genes and Traits c) What evolutionary force is at work in Island Y? d) Following the termite and cyclone attacks, which of the pine populations (X or Y) will have the most genetic variation? e) Suppose that the cyclone that struck Island Y did so early in the year, before the emergence of the slowgrowing trees. In this case, what evolutionary force is at work on Island Y? ANSWER: a) Aa. This can be determined based on the information about the damage caused by the termites on Island X. Because 3/4 of the progeny are fast growing (A–) and 1/4 are slow growing (aa), the original plant must have been Aa. b) directional selection (selectively killing aa plants, thus decreasing the frequency of the a allele and increasing the frequency of A allele) c) random drift (randomly kills trees, regardless of their genotype) d) Island Y. Island X has only AA and Aa individuals, whereas island Y probably has AA, Aa, and aa individuals. e) directional selection (selectively killing A– plants, thus decreasing the frequency of the A allele and increasing the frequency of a allele) 48. Three types of selection are "directional selection," "balancing selection," and "selection against heterozygotes." a) Shown below are three graphs (A–C). Each graph represents a different type of selection. In each case, certain individuals are being selected against (represented by the shaded regions). For each graph, write the name of the type of selection that is depicted.

b) Described below are three examples of selection. For each example, write the name of the type of selection being described. i) A maize breeder selects for increased ear size in maize in order to increase yield. ii) Susceptibility to malaria and sickle-cell anemia is controlled by the same gene. The HbA allele provides resistance to sickle-cell anemia but not to malaria (thus, individuals who have the genotype HbAHbA die from malaria), and the HbS allele provides resistance to malaria but not to sickle-cell anemia (thus, individuals with the genotype HbSHbS die from sickle-cell anemia). HbAHbS individuals are resistant to both diseases. iii) Lower-yielding beans are resistant to anthracnose, whereas higher-yielding varieties are susceptible to the disease. Anthracnose is present only in certain areas of the region. So, breeders want to select higher- and lower-yielding plants to fulfill different breeding goals.

ANSWER: a) A = selection against heterozygotes Copyright Macmillan Learning. Powered by Cognero.

Page 14

Name:

Class:

Date:

Chapter 20: Evolution of Genes and Traits B = directional selection C = balancing selection b) i = directional selection ii = balancing selection iii = selection against heterozygotes 49. Mutations within the Oca2 gene are associated with albinism and blindness in cave fish, while mutations in MC1R have been associated with increased pigmentation in the hairs of some pocket mice. With regard to protein function, compare the mutations generating each alteration of phenotype.

ANSWER: Mutations in MC1R that generate a darkening of coat color generate a gain-of-function characteristic in the encoded protein. The receptor protein is active, even in the absence of ligand, generating hyper-stimulation of the MC1R-dependent signaling pathway, and increased pigment synthesis. Mutations in Oca2 that are associated with albinism and blindness generate a loss-offunction effect upon this protein.

50. A key step in the generation of new genes appears to be increases in the number of copies of existing genes.

Define the mechanisms that can increase the copy number of a particular gene.

ANSWER: Four known mechanisms for increasing gene number are (1) production of polyploids, (2) individual or regional gene duplication, (3) replicative transposition, and (4) retrotransposition.

51. Explain why the sickle-cell trait is found at higher frequency in humid, low-lying regions where malaria is prevalent.

ANSWER: The sickle-cell hemoglobin mutation is under balancing selection in malarial zones and conveys a large survival advantage in heterozygotes. Individuals who are homozygous with genotype AA are susceptible to malaria, and homozygous individuals with genotype SS would succumb to sickle-cell anemia. A heterozygous individual with genotype AS has an advantage over either homozygote and is favored by natural selection because the importance of combating malaria counterbalances the deleterious effect of the sickle-cell mutation.

52. What are three important facets of the evolutionary process?

ANSWER: (1) Evolution can and does repeat itself. The same mutations can arise and spread repeatedly. (2) Fitness is a relative, conditional status. Whether a mutation is advantageous, disadvantageous, or neither depends on environmental conditions. (3) Natural selection acts on whatever variation is available, and not necessarily by the best means imaginable.

53. What are the possible gene outcomes that can arise from the duplication of the entire coding and regulatory region of a gene?

ANSWER: The allele bearing the duplicate could be lost from the population before it rises to any significant frequency. The duplication could survive and develop new mutations within the duplicate gene pair and result in neofunctionalization, subfunctionalization, or the formation of pseudogenes.

Copyright Macmillan Learning. Powered by Cognero.

Page 15