(Test Bank) An Introduction To Genetic Analysis

Griffiths,Wessler,Carroll,Doebley, 11

Medicine

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11th Edition

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Test Bank for An Introduction To Genetic Analysis, The downloadable file consists of 314 pages, and the total size of the file is 4.45 MB. Purchase now only for highly discounted price.

Author(s): Griffiths, Wessler, Carroll, Doebley

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MULTIPLE-CHOICE QUESTIONS Section 1.1 (The birth of genetics ) 1. The early 1900s was an important period for genetics due to which of the following major events? A) the rediscovery of Grego r Mendel’s scientific findings B) Watson and Crick solv ing the structure of DNA C) Walter Sutton and Theodore Boveri hypo thesizing that chromosomes are the hereditary elements D) All of the answer options are correct . E) the rediscovery of Gregor Mendel ’s scientific findings and Walter Sutton and Theodore Boveri hypothesiz ing that chromosomes are the hereditary elements 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 stran d must be: A) 3'-ATGC-5'. B) 3'-TACG-5'. C) 5'-ATCG-3'. D) 5'-CGTA-3'. Test Bank for An Introduction to Genetic Analysis 11e Anthony Griffiths Susan Wessler Sean Carroll John Doebley (All Chapters) Chapter 1 The Genetics Revolution 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 FALSE about genes ? A) Genes are located on chromosomes . B) Genes come in variants known as allele s. C) Genes always encode protein products . D) All of the answer options are correct . E) None of the answer options are correct. Answer: C 6. Wild cats ( Felis silvestris ) and common mice ( Mus musculus ) are diploid. In wild cats 2n = 38, while in common mice 2 n = 40. Based on this information, we can conclude that wild cat cells have : A) less DNA than common mouse cells. B) smaller genome s 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) guanidine E) tyrosine Answer: D 8. Which of the following is/are TRUE of the DNA structure solved by Wats on and Crick? A) It is a double -helical structure . B) Sugar –phosphate backbone is always toward the outside of the DNA . C) There are three hydrogen bonds between A and T and two hydrogen bonds between C and G . D) All of the answer options are correct . E) It is a doub le-helical structure with the sugar –phosphate backbone always toward the outside of the DNA . Answer: E 9. Which of the followi ng 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 are correct . Answer: C Section 1.2 ( After cracking the code ) 10. A gene is transcribed into an mRNA , and this mRNA is 110 nucleo tides long. Which of the following proteins could it encode? A) an enzyme that is 330 amino acids long B) a ribosomal protein that is 360 amino acids long C) a regulatory protein that is 36 amino aci ds long D) a signaling protein that is 10 amino acids lo ng E) a structural protein that is 110 amino acids long Answer: D 11. The nuclear genome of a mouse nerve cell is compared to that of a mouse skin cell. What differences can we expect to see be tween these two cells’ nuclear genomes? A) The two cells have almost identical genomes, but the nerve cell has more nerve - specific genes. B) The two cells have almost identical genomes, but the skin cell has more skin -specific genes. C) The two cells have different genomes: the nerve cell has nerve -specific gen es but not skin-specific genes, and the skin cell has skin -specific genes but no nerve -specific genes. D) The two cells have similar genomes, but the nerve cell has more nerve -specific genes and the s kin cell has more skin -specific genes. E) The two cells have the same genome; there will be no differences. Answer: E 12. 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 muta tion, where has this mutation most likely taken place? A) in Cindy’s gametes B) in the cells that should normally have given rise to Cindy’s tail C) in the cells that should normally have given rise t o Cindy’s and her puppies’ tails D) in all of Cindy’s ce lls (including her gametes) E) in a gamete of one of Cindy’s parents Answer: A 13. Which of the following features make 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 14. Using molecular techniques, researchers have knocked out both copies of gene “ G” in a series of genetically identical mouse embryos. These mice develop nor mally, except for their forelimbs, which are missing sev eral 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 prot ein 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 dev eloping mice. Answer: C MATCHING QUESTIONS Sections 1.1 and 1.2 15. Choose the best answer from the list below. Not all of the answers will be used. a. alleles b. de novo mutations c. point mutations d. nucleases e. DNA polymerase f. Tatum and Beadle g. Gregor Mendel 1. ___f__one-gene -one–enzyme hypothesis 2. ___a__gene variants 3. ___d__cut DNA at specific location 4. ___b__unique DNA variants that exist in a child but neither of its parents. 5. ___e__DNA replication OPEN -ENDED QUESTIONS Section 1.1 ( The birth of genetics ) 16. 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 tempera ture 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 orde r for it to denature. 17. Arabidopsis thaliana is a di ploid plant model organism with 2 n = 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 chromosome s does the nucleus of an Arabidopsis thaliana leaf cell contain? Answer: a) 2 b) 2 c) 5 Section 1.2 ( After cracking the code ) 18. 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 . 19. Using a concept map, indicate all the relationships that exist among the following: chromosomes, DNA, genes, genomes, proteins, histones, chromatin, nucleus, and mitoc hondria. Answer: Exam ple: - chromosomes: [are comprised of] DNA, histone (p roteins) [carry] genes [are located in] the nucleus, in mitochondria (each “term” should be related to at least three other terms) Section 1.3. ( Genetics today ) 20. Mutations are often vi ewed 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. 1 Chapter 2 Single -Gene Inheritance MULTIPLE -CHOICE QUESTIONS Sections 2.1. and 2.2. (Single gene inheritance, The chromosomal basis of single -gene inheritance patterns ) 1. If a plant of genotype A/a is selfed, and numerous offspring are scored, what p roportion 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 2 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) C) 50% wild type and 50% mutant ( d) D) 25% wild type and 75% mutant ( d) E) all mutant ( d) Answer: C Section 2.3. ( The m olecular basis of Mendelian inheritance patterns ) 6. Mice ( Mus musculus ) have 40 chromosomes per diploid cell (2 n = 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 D NA molecules and 20 chromosomes B) 40 D NA molecules and 40 chromosomes C) 40 DNA molecules and 80 c hromosomes D) 80 D NA molecules and 40 chromosomes E) 80 D NA molecules and 80 chromosomes Answer: D 7. A mutation occurs in a ge rm 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 c hiasmata B) formation of the kinetochore C) pairing of homologous chromosomes D) segregation of homologo us chromoso mes during meiosis I 3 E) segregation of sist er 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 th is 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 10. The diagram below shows a part of the biochemical pat hway responsible for fruit colo r in peppers ( Caspicum annuum ). Enzyme 1 is responsible for catalyzing t he reaction that turns the colorless precursor into yellow pigment, wh ereas 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? Enzyme 1 Enzyme 2 Colo rless precursor Yellow pigment Red pigment 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 colo r in the fruit fly Drosophila melanogaster is dark red, as a result of a mixture o f 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 mutati on that completely inactivates the “a” gene. What proportion of their offspring will have a recessive eye colo r phenotype? A) all of the offspring B) 3/4 C) 1/2 D) 1/4 E) none of the offspring 4 Answer: D Section 2.4. ( Some genes discovered by o bservi ng segregation ratios) 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 c ommonly 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 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 F 1 is comprised of five rabbits with a c′ phenotype, two with cch phenotype, and three with c phenotype. Of the phenotypically c rabbits , two ar e females and are back crossed 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 bot h c′ and cch. Answer: E 15. A plant with small red flowers is crossed to a plant with large white flowers. The resulting F 1 is comprised of 75 plants with small red flowers and 72 plants with small white flowers. If flower color and flower size are cont rolled by a single gene each, what can be concluded from these results? A) Flower color is controlled by a sex -linked gene. 5 B) Red color and small size are dominant to white color and large size, respectively. C) Small size is dominant to large size, but w e can’t determine which color is dominant. D) We can’t 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 phenot ypic 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% wil l be black . Answer: E Section 2.5. (Sex -linked single -gene inheritance patterns ) 17. A very common type of red –green colorblindness in humans is caused by a mutation in a gene located on the X chromosome. Knowing that the mutant allele is recessive to t he wild type, what is the probability that the son of a woman whose father is colorblind is going to also be colorblind? A) 0% B) 25% C) 50% D) 75% E) 100% Answer: C 18. A phenotypically normal woman is heterozygous for the recessive Mendelian allele ca using 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 nor mal 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 6 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 yield ed 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 f or 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 F 1 generation all females are wil d-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 do minant for b. D) recessive for both a and b. E) X -linked for a and autosomal for b. 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 aff ected 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 Section 2.6. ( Human p edigree analysis) 7 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 r andomly 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 gumba lls. 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 % 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 8 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 o f inheritance of this disease? A) autosomal dominant B) autosomal recessive 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 earlob es (white symbol). Both alternative phenotypes are quite common in human populations. 9 If the phenotypes are determined by alleles of one gene, then attached earlobes are inherited as : A) an autosomal dominant trait. B) an autosomal recessive trait . C) a dominant trait that could be either autosomal or X -linked. D) a recessive trait that could be either autosomal or X -linked. E) an X -linked dominant trait. Answer: A 28. In the human pedigree shown below, black symbols indicate individuals suffer ing 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 domin ant D) X-linked recessive E) Y-linked Answer: C 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