2008_06_05_13_07_54

2008_06_05_13_07_54 - BICD 100, Summer 2007 Final...

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Unformatted text preview: BICD 100, Summer 2007 Final Examination Enter each answer into the appropriate box on the answer sheet. Turn in only the answer sheet. You may use scratch paper, but you may not turn it in. Genetic studies in a particular population have discovered four unrelated individuals with a recessive condition involving skin which is very la): and easily torn. An investigator isolated a complex molecule called collagen 2A front skin biopsies from these people. lt proved to contain two protein chains called 2Al and 2A2. Analvsis of these )roteins ram normal eole and from the our mutant individuals revealed the following (table). Individual 2A1aa16t r 1. Which individuals have mutants which are homoalleles ofeach W14 -_ 2A1 33 16t . other? Enter all correct choices into box 1. a) A and B . 2A13316 he andC c)AandD d)BandC e)BandD f)Can D “ 2A2 33 16 he “ g) none are homoalleles 2. Which individuals have mutants which are hetetoalleles ofeach other? Enter all correct choices into box 2 a A and B b) A and C c) A and D (EBB and C e) B and D t) C and D g) none are heteroalleles 3. Which individuals have mutants which are not alleles of each other? Enter all correct choices in box 3 fi/lh and B b) A and C ® and D d) B and C (ei B and D @C and D g) all are alleles H 4. The progeny ofa cross between individuals A and B would have normal skin. T or (9 2 5. The progeny ofa cross between individuals A and C would have normal skin. T or z 6. The progeny ofa cross between individuals A and D would have normal skin. <9 or F A human disease called xeroderma pigntentosum shows a tal recessive inheritance. Aflected individuals cannot repair DNA that has been exposed to U V radiatiotmmmt cancers and other problems. In culture, cells from aflected people fail to show “unscheduled DNA synthesis " (low levels ofDNA synthesis outside the S period) afier UV exposure, because repair requires Cutting and patching damaged chromosomal DNA. Cells front normal people do show unscheduled DNA synthesis. An investigator collected and cultured cells from several unrelated aflected individualsfi-om around the world. One of the things he did with these cells was to make lntltrids. meaning he caused, for example, the cells of individual A to jitse with cells of individual B to make All hybrids. /'l (2 pts). AB hybrid cells did demonstrate unscheduled DNA synthesis after UV exposure. This indicates that the mutations in individuals A and B definitely were in different genes in) on different chromosomes 0] not linked d) linked e) in the same gene affecting ferent amino acids I) in the same gene affecting different base pairs g) in the same gene affecting the same base pair h) some other answer i) can't tell -‘ 8 (2 pts). AC hybrid cells did not demonstrate unscheduled DNA synthesis after UV exposure. This indicates that the mutations in individuals A and C definitely were a) in different genes b] on different chromosomes c) not linked @inked e) in the same gene affecting different amino acids l) in the same gene affecting different base pairs g) in the same gene affecting the same base pair h) some other answer i) can't tell 9 (2 pts). The mutation in person D affects the same base pair as the mutation in person A. This means that the affected protein in person D will have the Same amino acid error as that in person A. T or © I0 In person A the uvrA protein has an abnormal amino acid it 16. which is tyr but is supposed to be cys. In person E the same protein is abnormal. Therefore the mutat n in A and the mutation in E are definitely @lleles b) homoalleles c} heteroaileles d) not alleles @can’t tell from this information 11 (2 pts). The uvrA protein in individual E is 15 amino acids long, whereas the urvA protein in individual A is 227 amino acids long. With this additional information it is possible to say that the A and E mutations definitely are a) not alleles b) homoalleles @wteroalleles (1) this information does not add anything useful rim/4L em W ji"e/ 12 The mutation in individual B and that ' dividual F fail to complement. This means BF hybrids would show unscheduled DNA synthesis. T or Q Eric has a recigrocal translocation between chromosomes 4 and 7. 1n the region of chromosome 4 between the ’L telomere and the breakpoint lies a variable EcoRI site called El . Eric is heterozygous at this site, with the I” /~/ translocated 4 having the GAA TTC (cut) sequence, and the untranslacated 4 having GGATT C, .2 D54 is homologous to the region including this EcoRI site. When this cloned sequence was labeled and used as a I probe to hybridize to blots of limit! restricted DNA from Eric and his family. the following patterns were seen. Eric Wire SonA SonB SonC DanA DauB DanC l3 (2 pts). Son A most likely arose from a alternate A segregation (adjacent I segregation @mjacent ll 5 segregation d) nondisjunction e) crossmg over in the {1" loop 0 crossing over in the quadrivalent g) some other 4 mechanism h) can’t tell :2 14 (2 pts). Son B most likely arose fl-om lternate segregation b) adjacent! segregation c) adjacent II segregation d) nondisjunction e) crossmg over in the loop t) crossing over in the quadrivalent g) some other mechanism h) can’t tell “=5 15 (2 pts). Daughter A most likely arose from a) alternate segregation b) adjacent I segregation @djacent H segregation d) nondisjunction e) crossing over in the loop f) crossing over in the quadrivalent g) some other mechanism h) can’t tel] 416 (2 pts). Daughter B most likely arose from a) alternate segregation b) adjacent I segregation c) adjacent II segregation ' ondisjunction e) crossing over in the loop 0 crossing over in the quadrivalent g) some other mechani h) can’t tell X 17. Eric’s wife’s band pattern indicates she is homozygous at the El site. ('9 or F l8 (2 pts). The fact that the band seen in the wife is not the same as either of Eric’s bands indicates that a) she has a different variant at the 1 site b) her copies of chromosome 4 differ from Eric’s at another nearby EcoRI site 0) she has a deletion he does not have the translocation e) something else . l9. Kleinfelter’s Syndrome is an example of triploidy. @ or F 20. In normal oogenesis, meiosis II starts upon ovulation. T or @ 21. In meiosis I and mitosis the two sister chromatids separate, while in meiosis II the two homologous chromosomes separate from each other. T or When a particular prototrophic Hfr strain was crossed to an F - strain with multiple nutritional mutations, Hfi' genes were transferred at the following times: arg+ at 15 min, bio+ at 20 min, lac+ at 5 min, lys+ at J 5 min and thr+ at 10 min 22. Which of the Hfr alleles appeared least often in the progeny of this cross? a) arg+ bio+ c) 1ac+ d) lys+ e) thr+ i) all appeared to the same extent g) not enough information to tell 23 (2 pts). When an arg+ bio+ lys- Hfr was crossed to an arg- bio- lys+ F- and prototrophic progeny selected for, 992 were recovered per 5000 progeny. When an arg- bio+ lys+ Hfr was crossed to an arg+ bio- lys- F-, 56 were recovered per 5000. What is the order of loci? a) arg - bio — lys @io — arg —— lys c) bio - lys — arg d) some other order e) can’t tell When DNA from a wildtype strain of E. coli was applied to cells with the asp- bio- leu- pro- thr- genotype, the following transformants were obtained (table). mmmm mm _——_ cloned sequence fl 24. The asp locus is on one end, when these loci are put in order. @ or F 25. The bio locus is on the opposite end from asp. T or @ 26.'I‘heorderoflociis a)bio-leu—thr—pro~asp b)bio-pro-leu-thr-asp c)bio-thr-leu-pro— asp ro—bio—thr—leu-asp d)thr-pro-bio-asp-leu e)leu—pro-bio-asp—thr Osomeother A weed heterozygous at five loci was crossed to a homozygous recessive plant, with the resulting 4,979 progeny shown below. A_ causes spring flowering, aa flowers in autumn. B_ causes narrow leaves, bb leaves are broad. C_ causes white flowers, cc are colored. D_ causes white roots, dd are dark. E_ causes short stems, ee are elongated. Onl recessive traits are mentioned in the table. An trait not mentioned is resent in the dominant arm. broad colored dark elonated = 270 broad elonated = 157 autumn colored dark = 156 27 All expected types of progeny are present. T or @ 28. The number of independently asserting loci or groups of loci is 2) none b) l 6 2 d) 3 e) 4 i) 5 29. a is linked to b. T or @ 30. b is linked to c. © or 3l.cislinkedto d. or 32. d is linked to e. ‘ or 33. The A and E alleles were in coupling in the heterozygous parent. T or CE) 34. The B and C alleles were in coupling in the heterozygous parent. T or ® 35 (9 pts). Map the five loci, and draw your map in box 35 on the answer sheet 36. If two loci are linked, then they are also syntenic. ® or F 37. If two loci are syntenic, then they are also linked. T ot© 38 (2 pts). If two loci assort independently, the recombination fraction will be a) 0 b) l @5 d) .25 e) .75 0 some other answer 39. Two syntenic loci in male Drosophila may assort independently. or @ 40. Two syntenic loci in female Drosophila may assort independently. (if) or F The following represent the beginning sequences of an mRNA derived from a normal version ofa gene, and 5 mRNAs derived from mutant versions of the same gene: (Dots at the end mean the sequence continues) wt: 7M-GAAUUCGGGGAG'AUQAAApGCIUCC‘AAQGGQAUGfiGUUAACCGAAA... l ‘3tytflf? {ha ~ g «r 7 f f In ll m1 : 7M-GAAUUCGGGGAGiAUGiviAiUGCinChUGiSGXUpdGUUhIidCGKAA...l m2: 7M—GAAUUCGGGGAGAUGMA‘UCECUcchéuhbchucb’cuifficccm... m3: 7M-GAAUUCGGGGAGAUGAAPIUGCUQC‘CAAUAGGGAUCGG'UUA‘ACCCAAZA... m4: 7M—GAAUUCGGGGAGhUGhAAUEiKUcCAAUGGGAUGGGUU CCGAAA... m5: 7M—GAAUUCGGGGAG‘AUGAAA‘UGCUccAAUprhUGbGUb ’ECGAAA... ‘ t 1 L! r 6 F" a? 41 (2 pts). How many amino acids are there in the wildtype protein? Write the answer in box 4l, or if the number cannot be determined, write X in the box 42 (2 pts). How many amino acids are there in the mutant 1 protein? Write the answer in box 42, or if the number cannot be determined, write X in the box 43 (2 pts) How many amino acids are there in the mutant 2 protein? Write the answer in box 43, or if the number cannot be determined, write X in the box 44 (2 pts) How many amino acids are there in the mutant 3 protein? Write the answer in box 44, or if the number cannot be determined, write X in the box 45 (2 pts) How many amino acids are there in the mutant 4 protein? Write the answer in box 45, or if the number cannot be determined, write X in the box 46 (2 pts) How many amino acids are there in the mutant 5 protein? Write the answer in box 46, or if the number cannot be determined, write X in the box 47. The type of mutation in mutant 1 is a) nonsense b) missense c) silent d) insertion frameshifi @eletion frameshifi t) impossible to tell 48. The type of mutation in mutant 2 is a) nonsense .@missense c) silent d) insertion frameshift e) deletion frameshift t) impossible to tell 49. The type of mutation in mutant 3 is a) nonsense b) missense c) silent sertion frameshifi e) deletion frameshifi t) impossible to tell 50. The type of mutation in mutant 4 is @onsense b) missense ‘c) silent d) insertion frameshifl e) deletion frameshifi f) impossible to te 51. The type of mutation in mutant 5 is a) nonsense b) missense @silent d) insertion frameshifl e) deletion frameshifi t) impossible to tell Stacy was adopted as an infant. and never learned that she was the ofispring of a brother/sister mating. Because of the incest Stacy '3 parents were put into foster care. Stacy '3 mother eventually married, and she and her husband had a son, Edward, who while in college made a donation to a sperm bank. Stacy conceived with Edward '3 sperm, and produced a son, Louis. Later, Edward married and had a daughter, Lena. Lena married Louis, and they are expecting. 52 (2 pts). What is the inbreeding coefficient of the child of Lena and Louis? @ .172 b) .008 c) .0156 d) .125 e) .0313 t) .45 g) some other number 00/53. What is the probability the child will have the recessive disorder xeroderma pigmenlosum if the frequency of that disorder in the population is .00000025? a) .0000117 b) .00443 0) .000765. @.0000862 e) some other number 54. Calculate the ratio of the risk of having xeroderma pigmentosum for the child in #52 to the risk for a random bred person in this population and write the number in box 54. 55. In this same population 16% of people cannot taste PTC (inability to taste PTC is an autosomal recessive trait, so they are tt). What is the fiequency of the t allele? a) .l b) .16 c) .2 d) .25 e) .3 f) .5 g) .6 @some other number 56. Calculate the ratio of the risk of being unable to taste PTC for the child in #52 compared to the risk for a random bred person in this population and write the number is box 56. 57. If a random bred person in this population can taste PTC, what is the probability that person is Tt? a) .2 b) .24 c) .36 d) .4 e) .5 t) .6 .57 h) some other number —-—’ 58. What is the probability that the child in #52 will be Tt? a) .022 b) .723 c) .498 d) .472 e) .828 f) 0 @some other number Individuals represented by filled symbols in the following pedigree sufi'er from a rare inherited disease. A variant RFLP locus. cd100, is known to map close to the region where the gene causing the disease is found. It has 4 haplorypes A, B. C, and D. After testing the members of the family, the RF LP genotyges are shown below: AA: 11-4, 111-1, 111-14, IV-1, IV-2 ¢ 14 5 BB: 11-8, 11-10. III-9 I CC: [-1, 111-10 Lg DD: 11-1, 11-6, [11-6, 111-17. IV-4, IV-5, IV-7 IL:— oo 4 C 6 n no 44 515 AB: 1.2, 111.7, 111-11, 111-12. IV-8 I O I 0 AC: 11-2, 11.5, 11-7, III-8, 1V-11, IV-IZ, IV-l4 ‘ " ‘4 f ( - AD.-111-2,111-3,111.4,1V-3 m: A 0, Mg, ,, M C (a, w,’ *6 , A 80113. 11.9.111-13,111-15. III-I6,IV-9,1V-10, o I o I I o I o I ' 1r )‘4 < 6 ‘3' V 4 M H 1 BD:IV-l5, IV-16, IV-l8 —m A -,.. o a. ., c a -. . . . 4. .. CD: 111-5, IV-6, IV-l7, IV-19 o I I o O I I I o I o 1t 5 n! j 6 9/ ,V( [a mu] {‘1 (Hg/749,7 59. The disease segregating in this family could be inherited as an autosomal recessive. T or 60. The disease segregating in this family could be inherited as an X-linked recessive. T or ® 61. The disease segregating in this family could be inherited as an autosomal dominant. @ or F 62. The disease segregating in this family could be inherited as an X-linked dominant. T or (a regard to the disease gene and cleO locus? all of them b) 11-2 and 11-3 c) 11-5, 11-7, and 11-9 d) only 11-5 M 63 (2 pts). Which of the descendants of [-1 @2 in generation II are impossible to identify as either N or R with e) all except 11-5 0 some other combination g) none of them 64 (2 pts). How many of the descendants of [-1 and l-2 in generation 111 can be definitely identified as N with regard to the disease gene and cleO locus? a) all of them b) 7 c) 8 d) 9 e) 10 0 some other number g) none of them 65. How many of the descendants of H and 1-2 in generation IV can be definitely identified as N with regard to the disease gene and cleO locus? a) all of them b) 8 c) 9 d) 10 e) 11 0 some other number g) none of them 66. Which of the descendants of H and I-2 in generation 11 can be identified as R with regard to the disease gene and cleO locus? a) all of them b) 11-3 c) 11-5 (1) 11-7 e) 11-2 and 11-9 f) some other combination g) none of them 67 (2 pts). Which of the descendants of H and 1-2 in generation 111 can be identified as R with regard to the disease gene and cleO locus? a) all of them b) 111-2, 111-3, and III-ll c) 111-2 and 11141 (1) only III-16 e) III-3 and Ill-16 i) some other combination g) none of them 68. Which of the descendants of 1-1 and 1-2 in generation IV can be identified as R with regard to the disease gene and cleO locus? a) all ofthem b) IV-1, IV-1 l, and IV-lS c) IV-2 and IV-12 d) only IV-14 e) IV-16, IV-18, and IV -19 i) lV-4, IV-6, and IV-lO g) IV-8, lV-9, IV-lO, and IV-15 h) some other combination i) none of them 69. How many definite N offspring have you identified altogether in this family? (Write number in box 69) 70. How many definite R offspring have you identified altogether in this family? (Write number in box 70) 71. What is the most likely map distance between the disease locus and cleO locus, based on the data from this family? ...
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2008_06_05_13_07_54 - BICD 100, Summer 2007 Final...

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