HMB265-May 2009 Exam - HMB265H1 April 2009 Exam Name...

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Unformatted text preview: HMB265H1 April 2009 Exam Name Student Number UNIVERSITY OF TORONTO Faculty of Arts and Science April] May 2009 Final Examinations HMBZSSHlS: General and Human Genetics Duration—Z hours Examination Aids: Non—Programmable Calculator ONLY Total Number of Marks = 100 Total Number of Pages = 14 instructions: Part 1 ~ Multiple Choice (45 questions worth 2 marks each, total 90 marks) 0 Answer the multiple choice questions on the scantron. 0 Refer to the "MARKING lNSTRUCTiONS” on the scantron for information on filling it out (e.g. use HB pencil; no stray marks anywhere on the scantron). o in the "FORM" box on the scantron, indicate which “FORM” (A, B, C or D) is indicated on the top of your exam page. 0 Fill in “SlGNATURE, SUBJECT and DATE on the scantron. o Correctly bubble in your STUDENT NUMBER, LAST NAME and lNlTlALS on the scantron. Part 2 - Short Answer (2 questions, total of 10 marks) Answer the short answer questions in the space provided on your exam paper. Use complete sentences. Make sure you put your name and student number on all pages. invigilators are not permitted to interpret questions to individual students. if you think that a question is ambiguous, answer it as you understand it, then make a note here (not on your scantron card). Please be specific. HM3265H1 2009 Exam Form B Student Name Student Number Part 1-MULTIPLE CHOICE. Choose the one alternative that best answers the question. Each question is worth 2 marks. 1) Referring to Scenario A (concerning wolves) at the end of the examination questions, what is the estimated frequency of the recessive ailele in the population? A) 0.64 B) 0.40 C) 0.36 D) 0.60 2) if in a population oft million people, 100 albinos (homozygous recessives, aa) were found, how many normal (homozygous dominants, AA) individuals will be found in the next generation under Hardy-Weinberg equilibrium conditions? A) 980,010 B) 100,000 C) 999,900 D) 980,100 3) You suspect that a very specific point mutation in the epidermal growth factor (EGF) receptor gene is responsible for the majority of EGF receptor gene mutations associated with tumours. Which technique would you most likely use in developing a simple assay for predictive testing? A) Preparation of cell lysates followed by western blot. B) RNA isolation followed by northern blot. C) Hybridization of genomic DNA with specific oligonucleotides. D) Polymerase chain reaction of genomic DNA alone. 4) ln Scenario B concerning cheetahs described at the end of the examination questions, what is the most likely mode of inheritance? A) X-linked recessive. B) Impossible to determine. C) Autosomal dominant. D) Autosomal recessive. 5) if the pedigree of the Smith famiiy shown in Figure 1 at the end of the examination booklet was explained by a rare X-linked allele, which of the following statements is most likely to be FALSE? ~ A) Both Ill-3 and ill-4 are carriers of the trait. 8) Individual l-4 is definitely heterozygous for the trait. C) Sons of individual ll-7 had a 50% chance of inheriting the trait, but did not. D) Individual Ill-4 is definitely heterozygous for the trait. Page 1 of 14 HM8265H1 2009 Exam Form B Name Student Number 6) What kind of DNA mutations are caused by UV light, and how are the mutations usually repaired? A) Frameshift mutations; Base—excision repair. B) Base substitutions; Base-excision repair. C) Pyrimidine dimers; Nucleotide—excision repair. D) Base deletions; Mismatch repair. 7) Referring to Scenario A (concerning wolves) at the end of the examination questions, how many light-coated wolves in the sample are likely to be homozygous for the dominant allele? A) 0.9 B) 64 C) “0.4 D) 16 8) Genes a, b and c assort independently and are recessive to their respective alleles A, B, C. Two triply heterozygous (Aa Bb Cc) individuals are crossed. What is the probabability that a given offspring will be phenotypically a b c, that is they will exhibit all three recessive traits? A) 1/4 B) 1/64 C) 1/16 D) 314 9) Suppose a female mouse that is homozygous wild—type for the IGF2 gene is mated to a male mouse who is heterozygous for a mutation in the IGF2 gene (IGF2+/IGF2-). What is the probability that the offspring will NOT express 1ng protein? (Assume mice who do not express lgfz protein are viable.) A) 25% B) 50% C) 0% ’ D) 100% 10) if the pedigree of the Smith family shown in Figure 1 at the end of the examination booklet was explained by a rare X-linked allele, what is the likelihood that female offspring of a mating between lV—2 and lV-4 will be carriers of the trait? A) 25% B) 0% C) 100% D) 50% Page 2 of 14 HM8265H1 2009 Exam Form B Name Student Number 11) in deletion mapping experiments, fruit flies homozygous for different deletions in a chromosomal region were generated and crossed with fruit flies that were homozygous for mutations in specific genes within the region. The regions deleted and the results of the cross are shown in Figure 2 at the end of the examination questions. Use the results to determine the position of genes a to 1' that are located in positions 1 to 6 on the chromosome. A) abodef B) fcbeda C) ecabfd D) decabf 12) if the pedigree of the Smith family shown in Figure 1 at the end of the examination booklet was explained by a rare autosomai allele, which of the following statements is most likely to be FALSE? A) individual "—2 was definitely a heterozygote. B) individual lV—1 has a 2/3 chance of being a carrier of the trait. C) Sons of individual ll-7 had a 50% chance of having the trait, but did not. B) Both iii-3 and ill-4 are carriers of the trait. 13) in Scenario B concerning Cheetahs described at the end of the examination questions, what was the probability that the four offspring produced from the mating of "-8 and "-9 would have the observed phenotypes? A) 1 B) 1/16 C) 1/8 D) 1/2 14) in an F2 group of siblings showing continuous variation in a trait, the VE was calculated to be 425, with the V3 partitioned as follows: VD=50, Vi=15, and VA=10. For this trait, which of the following statements is TRUE? A) h2 = 0.15, which suggests that crossing individuals from any other family with similar levels of the trait will produce progeny that show a similar level of the trait. B) H2 = 0.02, which suggests that crossing individuals from any other family with similar levels of the trait will produce progeny that show a similar level of the trait. C) H2 = 0.15, which suggests that most of the observed variation in the trait is due to environmental effects. D) h2 = 0.02, which suggests that crossing individuals from this family with similar levels of the trait will produce progeny that show a similar level of the trait. 15) in a certain breed of dog, the alleles B and b determine black and brown coats respectively. However, the allele Q of a gene on a separate chromosome is epistatic to the B and b alleles resulting in a gray coat (1.7 has no effect on colour). if the animals of genotype B/b;Q/q are intercrossed, what phenotypic ratio is expected in the progeny? A) 12 gray: 3 black : 1 brown B) 9 gray : 3 brown : 4 black C) 9 black : 6 brown : 1 gray D) 1 black: 2 gray: 1 brown Page 3 of 14 r HMBZSSH’l 2009 Exam Form B a Name Student Number 16) in maize (corn plant), if a long tassel plant is crossed with a short tassel plant, the F1 all have long tassels and the F2 show 180 long tassels to 12 short tassels. What are the MOST LIKELY genotypes of the original parents that were crossed to give the F1? A) aabb X aabb B) AABB X aabb C) AaBb X AaBb D) aabb X A38!) 17) Referring to Tabie l at the end of the examination questions, what is the distance between a and b? A) 40 CM 8) 30 CM 0) 10 cM D) 20 cM 18) Sleeping beauty is a transposon isolated from fish. it is being used to identify new genes involved in cancer formation. imagine that the transposon inserts into one aliele of a given gene and causes cancer. What is the best way to describe the mutated allele? A) P—eiement. B) Oncogene. C) Proto~oncogene D) Tumour suppressor gene. 19) in Scenario B concerning Cheetahs described at the end of the examination questions, if Ill-2 is mated to ill—8 and they produce a female cub, what is the probability that she will display the characteristics of a king cheetah? A) 3/8 B) 1 C) 1/6 D) 1/2 20) Referring to Table l at the end of the examination questions, what does “ + + + ” represent? A) interference. B) Single crossover. C) Double crossover. 0) Parental. Page 4 of 14 HM8265H1 2009 Exam Form B Name Student Number 21) You isolated a gene expressed in differentiated neurons in mice. You then used promoter analysis to identify the controi regions of the gene. Specifically, different fragments of the 5’ end of the gene were cloned upstream from the protein-coding region of IacZ that lacked a minimal promoter. The clones were introduced into neurons in tissue culture to monitor expression. The fragments used and the results of the experiment are shown in Figure 4 at the end of the examination questions. Which region contains the minimal promoter and which contains an enhancer? A) Minimal promoter = region 5; enhancer = regions 3 and 4 B) Minimal promoter = region5; enhancer = regions 2 and 3 C) Minimal promoter = region 1; enhancer = region 4 D) Minimal promoter = region 5; enhancer = region 3 22) A man with blood type whose father was blood type "0," married a woman of blood type "AB" whose mother was blood type "A." What are the possible blood types of their offspring? A) A, B, AB and O. B) A, B and 0 only. C) A, B and AB only. D) 0 only. 23) in Scenario B concerning Cheetahs described at the end of the examination questions, what is the most likely genotype of the original father (l-Z)? A) Homozygous recessive. B) Wild type X-chromosome. C) Homozygous dominant. D) Heterozygous. 24) The gene that is mutated in cystic fibrosis was identified by Riordan, Collins and Tsui. it encodes a chloride channel that is defective in people with cystic fibrosis. What is the mode of inheritance of this disease? A) X-linked recessive. B) Autosomal dominant. C) Autosomal recessive. D) X-linked dominant. 25) What can one definitely conclude from the pedigree of the Kim family shown in Figure 3 at the end of the examination questions? A) The trait is pleiotropic. B) The trait is inherited as an autosomal dominant allele. C) The trait is inherited as a rare X-linked recessive allele. D) The allele for the trait is not rare in thebreeding population. Page 5 of 14 I HMBZ65H1 2009 Exam Form B Name Student Number 9 26) What contribution to science did Oswald Avery and his laboratory colleagues make? A) Discovered that the amounts of A and T and G and C in DNA are equal. B) Demonstrated that chromosomes are composed of DNA and protein. C) Elucidated the molecular structure of nucleic acids. D) Demonstrated that nucleic acids can alter the phenotype of an organism. 27) Referring to the pedigree of the Kim family in Figure 3 at the end of the examination questions, if individual lV—i mated with an individual who had the trait, what is the likelihood that their male offspring would have the trait? A) 50% B) impossible to determine. 0) 0% D) 33.3% 28) Modified viruses are commonly used to transfer DNA into cells for gene therapy. in this approach, a packaging construct is used. What does the DNA in the packaging construct encode? A) inverted repeats to allow for viral insertion. B) Enhancers to allow for tissue-specific expression. C) The wild—type version of the protein that is defective in the individual being treated. D) Viral capsid proteins. 29) in the cross between a female A/a;B/b;c/c;D/d;e/e and a male A/a;b/b;c/c;D/d;e/e, what proportion of the progeny will be phenotypically identical to the male parent? (Assume independent assortment of all genes and complete dominance.) A) 9/64 B) 27164 C) 1/32 D) 9132 30) Referring to Table l at the end of the examination questions, what is the correct order of the genes? A) c a b B) a b c C) impossible to determine D) b c a 31) What genetic phenomenon does the genetic ratio 12:3:1 best indicate? A) Dominant epistasis. B) Recessive lethal. C) Codominance. D) Recessive epistasis. Page 6 of 14 HM3265H1 2009 Exam Form B Name Student Number 32) Somatic cells in humans have 46 chromosomes. How many chromosomes would be found in the somatic ceils of an individual in whom non-disjunction occurred in one cell at the iOO-cell embryonic stage? A) Cells with either 46 or 47 chromosomes. B) Cells with either 46 or 45 chromosomes. C) Cells with either 47 or 46 or 45 chromosomes. 0) Cells with either 47 or 45 chromosomes. 33) Referring to the pedigree of the Kim family in Figure 3 at the end of the examination questions. Which of the following individuals is least likely to have a genotype that is heterozygote? A) "-2 B) lV~6 C) "-5 D) li|~2 34) Referring to Scenario A (concerning wolves) at the end of the examination questions, if dd has a fitness of 0.8, what is the estimated frequency of the recessive allele in the next generation? (Assume that the fitness of both DD and Dd is ’l, and that there are no spontaneous mutations.) A) 0.55 B) 0.60 C) 0.57 D) 0.40 35) in the pedigree of the Wu family shown in Figure 5 at the end of the examination booklet, what type of allele MOST LIKELY explains the pattern of inheritance of the trait indicated? A) X—linked dominant. B) X-linked recessive. C) Autosomal recessive. D) Autosomal dominant. 36) if the pedigree of the Smith family shown in Figure 1 at the end of the examination booklet was explained by a rare autosomal allele, what is the likelihood that female offspring ofa mating between lV—2 and lV-4 will be carriers of the trait? A) 25% B) 33.3% C) impossible to know. D) 50% Page 7 of 14 ! HM8265H1 2009 Exam Form B Student Number 9 Name 37) Two plants with white flowers, each from true-breeding strains, were crossed. All of the F1 plants had red flowers. When these F1 plants were intercrossed, they produced an F2 generation consisting of 1 77 plants with red flowers and 142 plants with white flowers. What is an explanation for the inheritance of flower colour in this plant species? A) Flower colour is due to sex-linked epistasis between two independently assorting genes. B) Flower colour is due to suppression of two independently assorting genes. C) Flower colour is due to codominance of two independently assorted genes. D) Flower colour is due to epistasis between two independently assorting genes. 38) in Scenario B concerning Cheetahs described at the end of the examination questions, if lit-3 is mated to ill-7 and they produce a male cub, what is the probability that he will display the characteristics of a king cheetah? A) 1/2 8) O C) 1/12 D) 1/6 39) The R/r and 8/5 genes are linked and to map units apart. in the cross Rs/rS rs/rs what fraction of the progeny will be RS/rs? A) 10% e) 25% _ C) 5% D) 40% 40) Which of the following statements about Recombinant inbred Lines (RlLs) is correct? A) Recombination at each generation creates chromosomes with only paternal alleles down the length of the chromosome. 8) RlLs are created by inbreeding an F2 family for generally less than 3 generations. C) inbreeding for many generations results in each lineage having a unique mosaic, homozygous for either the maternal or paternal allele at each locus. D) RlLs are generated by outcrossing selected individuals down a lineage. 41) The American red squirrel (Tamiasciurus hudsonicus) typically has a dark red coat colour, but can also display light orange fur when homozygous for a recessive mutation in a ‘ pigment-producing gene. Horace mates a pure—breeding red squirrel with a pure-breeding orange squirrel he found in Queen’s Park. All resulting progeny are red. Horace then mates one of the progeny to another pure-breeding orange squirrel and five offspring are produced. What is the probability that the first three are red and the last two are orange? A) 1/32 8) 2/3 C) 3/8 D) O Page 8 of 14 HM8265H1 2009 Exam Form B Name Student Number 42) Referring to the pedigree of the Kim family in Figure 3 at the end of the examination questions, if individual V—2 mated with an individual who had the trait, what is the likelihood that their female offspring were carriers for the trait? A) 25% B) 50% C) 100% D) 0% 43) in humans, brachydactyly is a dominant condition. 6,400 people in a population of 10,000 show the condition (1,600 are BB, 4,800 are ED) and 3,600 are normal phenotypes (bb). What is the frequency of the b allele? A) 0.4 B) 0.36 C) 0.6 D) 0.48 44) in the nematode worm, Caenorhabditis elegans, a recessive mutation in the Dumpy gene (Dpy) causes a short and fat body shape. Additionally, animals with a recessive mutation in the Unco—ordinated (Unc) gene experience paralysis. What would the expected phenotypic ratio be amongst the progeny of the following cross? Dpy/dpy; Unc/unc X Dpy/Dpy ; unc/unc A) 1/4 Slim and mobile, 1/4 Slim and paralyzed, 1/4 Fat and mobile, 114 Fat and paralyzed. B) All fat and paralyzed. C) 3/8 Slim and mobile, 318 Slim and paralyzed, 1/8 Fat and mobile, 1/8 Fat and paralyzed. D) 1/2 Slim and mobile and 1/2 Slim and paralyzed. 45) imagine that you have isolated a cell line that you know has a mutation in a protein-coding gene. You use the cells to perform both northern and western blots. When looking at the northern blots, you find that the mRNA corresponding to the mutant gene runs at the same level as the mRNA corresponding to the wild-type gene. in contrast, the mutant protein does not run as far on a gel as the wild-type protein (when detected by western blot). Given these results, what is the mostly like type of mutation that has occurred in the cell line? A) Missense. B) Frameshitt. C) Point. D) Nonsense. Please fill in "B" the FORM box on your scantron sheet now -- You have the version B of the examination. r Page 9 of 14 HMBZSSHl April 2009 Exam Form B Name ‘ Student Number Part 2 - SHORT ANSWER. Write your answer in the space provided. Use complete sentences. You must fill in your name and student number at the top of this page. 46) SNP detection was undertaken in individual sperm for four loci (A, B, C, and D) to determine the allele present (either allele 1 or 2). "X" indicates the allele is present and “-—" indicates the allele is not present. Assume that the single sperm have completed meiosis and are haploid. Using two sentences, indicate which loci are likely to be linked and why you drew this conclusion. Only answers written in the lined space provided will be marked. (5 marks) Page 10 of14 HMB265H1 April 2009 Exam Form B Name Student Number 47) A researcher is studying racing ability in whippets. Three different types of Whippet have been identified in the breeding population: normal, fast, and slow. Consider the following crosses. in two sentences or less, explain what kind of inheritance is most likely to explain the above outcomes and why. Only answers written in the lined space provided will be marked. (5 marks) Cross i) fast xfast = 2 fast: 1 normal: 1 slow Cross ii) normal x normal = all normal speed Cross iii) slow x slow = all slow WWW—“WM mwmm—W WWW Extra Information for Multiple Choice Questions: Table, Scenarios and Figures Table Table l: Results from a cross of Drosophila females heterozygous for three recessive mutations, 6, b, c, were crossed to tester males. The cross yielded the following results: Page 11 of 14 i3 HMBZGSHI April 2009 Exam Form B Name Student Number Scenarios Scenario A In a survey of wolves collected from a natural population, a researcher found 64 light-coated wolves and 36 dark—coated specimens. The light-coated wolves carry a dominant allele, D, and the dark-coated wolves are homozygous for the recessive allele, d. Assume that the population is in Hardy-Weinberg equilibrium. Scenario B The king cheetah is an extremely rare variant of the common Spotted cheetah (Acinonyx jubatus), identified by its larger size and coarser fur with a distinctive dorsal striping pattern. in 1926, it was classified as an entirely separate species (Acinonyx rex). This was refuted in 1981 , when a female cheetah, raised in captivity in a South African zoo, was mated with a male caught in the wild. Suprisingly, their progeny included both king and spotted Cheetahs, as is indicated in the following pedigree. D = Spotted Cheetah I I = King Cheetah Subsequent analysis identified a single gene mutation that is responsible for the king cheetah phenotype. Page 12 of 14 HMBZGSHl April 2009 Exam Form B Name Student Number Figures (Please note the figure number is indicated BELOW the figure) II IV Figure 1: Smith family pedigree Panel A 3e; 3 89M fie? 5 Panel B Figure 2: Deletion mapping experiment. Panel A: Chromosome and regions that are deleted in the deletion mutants. Panel B: Results from the cross described in the question. (+) indicates the phenotype is wildtype and (-) indicates the deletion uncovers a recessive mutant allele. Page 13 of14 HMB265H1 April 2009 Exam FormB Name Student Number ; El » s g 2 a - I~~ “3 i? ‘3 4, S m ‘ it 2 3 It 5 £1 Y e S w e I I ' '- 1 2 3 4 s s v “i 2 Figure 3: Kim family pedigree Upstream DNA regions 1 I 2 I ‘3' I 4 I 5 Fragments fused to IacZ IacZ expression W W W Figure 4: Promoter analysis. Fragments of the upstream DNA region of the gene were fused to the 1802 reporter gene. LacZ expression observed when indicated fragments of the upstream DNA region were used. I no I ll o o ' 0 Figure 5: Wu family pedigree Page 14 of 14 ...
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HMB265-May 2009 Exam - HMB265H1 April 2009 Exam Name...

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