m07-part1 - UNIVERSITY OF TORONTO Faculty of Arts and...

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Unformatted text preview: UNIVERSITY OF TORONTO Faculty of Arts and Science APRIL/MAY 2007 EXAMINATIONS HM3265H1S Duration - 3 hours Examination Aids: Non-Programmable Calculators Instructions: Please indicate on the scantron sheet which form (A, B, C, or D) of the HMBZGSHIS exam you are writing. Bubble this under the Form section of the scantron sheet. The Form type is indicated in title on page 1 of the exam. Please correctly bubble in your student number, and surname. The exam consists of 70 multiple choice questions. Tables and figures are found on pages 17 and 18. _ There are 18 pages in total. Please refer to the "Marking Instructions” on the scantron sheet provided for information on filling in the scantron sheets. Please use HB pencil only. HMB265HIS 06-07 Form A Final Examination Name Student Number MULTIPLE CHOICE. Choose the one response that best completes the statement or answers the question. 1. Which of the following concerning the method to create a knockout mouse via homologous recombination is correct? A) DNA containing the targeting construct is injected into a zygote from a mouse with brown fur. B) The targeting construct containing the gene of interest and a selectable marker randomly integrates into the genome allowing for expression of a selectable marker. C) To remove the gene of interest, the gene is labeled by fluorescence in situ hybridization and then a laser is directed towards the fluorescence causing the gene to be mutated. D) The embryonic stem cells that are used in gene targeting have limited developmental potential. They can develop into certain cells in a mouse, but not others. E) To know if you have successfully created a germ-line mutation in a gene, you mate the chimeric mouse with a wild-type female and examine the DNA from the offspring. . The chemical colchicine may be used for which of the following? A) To induce mutant polyploidy individuals. B) To induce individual chromosomal duplications in experimental lines. C) To prevent non—disjunction in cell cultures. D) To induce chromosomal deletions in experimental cell lines. E) To induce mutant aneuploid individuals. . Many scientists believe that the evolution of multigene families, such as the genes for hemoglobin, is a result of which type of genetic rearrangement? A) Reciprocal translocation. B) N on—reciprocal translocation. C) Deletion. D) Inversion. E) Duplication. . The synonymous susbstitution rates across a given genome rarely differ by more than a factor of two, a difference that cannot account for the roughly 1,000 fold difference in nonsynonymous substitution rates observed between some genes. What is an explanation for this observation? A) Nonsynonymous substitutions are more common than synonymous substitutions. B) Natural selection has been acting on such genes. C) Mutation rates are far more variable than previously believed. D) Synonymous substitution rates are underestimated. E) Genetic drift is a stronger evolutionary force than previously believed. lof18 5. How many genes are estimated to be in the human genome? A) 20,000-30,000 B) 4,000-6,000 C) 100,000-200,000 D) 3—4 million E) LOCO—3,000 6. What is found in a packaging construct for a viral vector? A) A promoter to drive expression of the transgene. B) The transgene that you would like to transfer to the individual. C) The gene coding for hydroxyurea. D) Genes coding for viral structural proteins. E) None of the above. 7. Mutations that result in Iowa than the normal amount of a protein or a protein with limited function are known as A) null mutations. B) hypomorphic mutations. C) hypermorphic mutations. D) conditional mutations. E) neomorphic mutations. 8. What is the most commonly used method to introduce DNA into cells for gene therapy in humans? A) Electroporation. B) Viral vectors. C) Liposomes. D) Direct injection of naked DNA into tissues or blood. E) Particle bombardment. 9. The heterozygote genotype frequency term for a gene with two alternate alleles A (frequency p) and a (frequency q) in the Hardy-Weinberg equation is A) p2 B) (12 C)2Pq D) (p+q)2 E)p+q 10. Which of the follow statements concerning sequence-tagged sites (STS) is NOT correct? A) STS can be easily amplified by PCR. B) STS are unique markers along a chromosome. C) Single nucleotide polymorphisms (SNP) can be used as STS. D) STS can be used in linkage and physical mapping. E) STS represent regions of the genome that code for proteins. 20f18 11. Gene therapy trials that were performed in France to treat boys with X-linked severe combined immunodeficiency illustrated one of the difficulties involved in gene therapy. What was the difficulty? A) Exaggerated immune responses to viral vectors were observed in one of the boys and resulted in his death. B) It was very difficult to get enough DNA into cells to correct the genetic defect, and therefore, the boys had to be treated by other methods. C) It was very difficult to get DNA into all of the cells in the body, and therefore, gene therapy was not effective. D) The transgene inserted into a protooncogene in some of the boys resulting in abnormally high expression of the gene, which led to leukemia. E) The in viva gene therapy resulted in the incorporation of the transgene into the germline, and therefore, the transgene was transmitted to some of the offspring of the boys when the boys grew up and had children. 12. Scientists performed ENU mutatgenesis on male mice and mated the mice with wild—type females. They then determined the blood glucose levels in all of the offspring. One of the offspring, a male, was found to have abnormally high glucose levels. The male was mated to several wild—type female mice. Half of the offspring had high blood glucose levels, and there were an equal number of males and females with the high glucose phenotype. What type of mutation was induced by ENU treatment? A) X—linked dominant. B) Y—linked. C) X—linked recessive. D) Autosomal dominant. E) Autosomal recessive. 13. A paracentric inversion heterozygote contains a normal chromosome with a gene order GENE in the long arm and an inverted chromosome with the gene order GNEE in the long arm. They form an inversion loop to pair during prophase I of meiosis. If a single crossover occurs between the G and the E segments of the loop, what will be the gene orders in the resulting recombinant dicentric chromosome and acentric fragment? Assume crossing over occurs at the four-strand stage involving two nonsister chromatids. A) Dicentric chromosome: GENE; Acentric fragment: GNEE B) Dicentric chromosome: GEN G; Acentric fragment: EENE C) Dicentric chromosome: GE; Acentric fragment: GNEENE D) Dicentric chromosome: GNEE; Acentric fragment: GENE E) Dicentric chromosome: EENE; Acentric fragment: GENG 14. Which is the only human autosomal trisomy in which the individuals may survive to adulthood? A) Edward syndrome B) Patau syndrome C) Down syndrome D) Turner syndrome E) Klinefelter syndrome 30f18 15. Common red clover, Trifolium pretense, is a diploid with 14 chromosomes per somatic cell. What would be the somatic chromosome number of a trisomic variant and an autotetraploid variant of this species? A) trisomic: 46; autotetraploid: 56 B) trisomic: 16; autotetraploid: 28 C) trisomic: 21; autotetraploid: 17 D) trisomic: 15; autotetraploid: 56 E) trisomic: 15; autotetraploid: 28 16. Which of the following is NOT one of the assumptions of the Hardy—Weinberg law? A) The population is very large. B) There is non-random mating within the population. C) Mutations in the alleles do not occur. D) N o migration occurs into or out of the population. E) The ability of all genotypes for survival and reproduction is the same. 17. Which of the following individuals would appear phenotypically male? A) An individual with an XY genotype with a loss-of—function mutation in the SRY gene. B) An individual with an XY genotype who is homozygous for a loss—of—function mutation in the Sox 9 gene. C) An individual with an XX genotype who is carrying a copy of the SRY gene on an X chromosome. D) An individual with an XY genotype who is homozygous for a loss—of—function mutation in the androgen receptor. E) An individual with an XY genotype who is homozygous for a loss-of-function mutation in the gene coding for the enzyme that converts testosterone to DHT. 18. What is the explanation for how retinoblastoma appears to be dominant in pedigrees, however, the underlying inherited defect is a recessive loss-of—function mutation? A) Individuals in these families can inherit mutations in both Rb alleles, but the mutation is in a different part of the Rb gene. B) Individuals in these families have one loss-of—function Rb allele in only some of their somatic cells and a subsequent somatic mutation occurs in one of these cells. C) Individuals in these families can inherit one loss-of—function Rb allele in all of their cells, and there is an increased chance that a second somatic mutation will occur in one of the cells. D) Individuals in these pedigrees are known to also have loss-of-function mutations in the gene coding for p53 which increases the chances that retinoblastoma will form. E) None of the above. 19. The Aniridia gene in humans is involved in eye formation. Although eye development is very different in flies, this gene is highly conserved. What is its homolog in Drosophila? A) ey B) Fax-6 C) noeye D) ommatidia E) eyeless 40f 18 20. Which of the following phenotypes is not usually observed in Turner syndrome (XO)? A) unusually short stature B) infertility C) skeletal abnormalities D) unusually long limbs E) widely spaced nipples 21. Which of the following BEST describes the order of steps that is often taken in positional cloning? 1) Compare protein-coding sequences from individuals with the disease and those without the disease. 2) Obtain DNA from several families with multiple members that are afflicted with an inherited disease. 3) Obtain a molecular marker that fails to recombine with the disease gene. 4) Examine sequences for possible exon regions. 5) Perform linkage analysis with known markers to determine linkage to a chromosome arm. A) Order 1, 2, 3, 4, 5 B) Order 3, 2, 1, 4, 5 C) Order 5, 3, 4, 1, 2 D) Order 2, 5, 3, 1, 4 E) Order 2, 5, 3, 4, 1 22. The human papilloma virus (HPV) causes cancer in part because it carries a gene that codes for a protein that inactivates the p53 protein. The fact that the loss of p53 function is oncogenic suggests that A) p53 normally functions to prevent uncontrolled cell division. B) The HPV protein is encoded by a tumor-suppressor gene. C) p53 gene expression is upregulated by the HPV protein. D) The HPV protein functions at origins of replication on DNA. E) p53 is a proto-oncogene. 23. Which of the following statements concerning transposons is correct? A) Transposons are found in the same chromosomal location in all members of a species. B) Transposons make up only a small percentage of the total DNA found in humans. C) Movement of transposons from one location to another requires reverse transcriptase. D) Transposase is coded by transposons and is needed for transposition of transposons. E) Transposons always insert into non-coding DNA and therefore cannot cause mutations. 24. A monosomic cell would produce gametes with how many chromosomes? A) N -1 B) N C) N and N + 1 D) N + 1 E) N and N —1 50f18 25. Igf2 is a gene that is maternally imprinted. Inheritance of one wild-type allele is required for normal growth of the embryo, and without it the mice are small. Scientists have generated mice in which the Igf2 gene is mutated so that it no longer codes for a functional protein. If mice heterozygous for the mutation are mated with pure breeding wild—type mice, which of the following observations would be observed? A) The sex of the heterozygous mutant mice used for the cross does not have an affect on the phenotype of the progeny. B) Mice born from a cross of heterozygous male and wild-type female will all be of normal size. C) Half of the mice born from a cross of heterozygous female and wild-type male will be small and half will be of normal size. D) Half of the mice born from a cross of heterozygous male and wild—type female will be small and half will be of normal size. E) All of the male mice born from a cross of heterozygous females and wild—type males will be of normal size, while all of the female mice will be small. 26. What is the frequency of the a allele in the beetle population described in the previous question? A) 0.86 B) 0.91 C) 0.09 D) 0.45 E) 0.55 27. Which of the following is NOT a mechanism employed by repressor proteins to decrease transcription of a specific gene? A) The repressor associates with a promoter element, blocking RNA polymerase from binding promoter element. B) The repressor binds to the activation domain of an activator, eliminating its ability to increase transcription. C) The repressor binds to the DNA—binding domain of an activator, eliminating its ability to associate with an enhancer. D) The repressor binds to a DNA sequence in an enhancer, eliminating access to sequence by an activator. E) The repressor binds to RNA polymerase II, blocking its ability to associate with the promoter element. 28. In some insect larvae, polytene chromosomes are produced when A) a cell divides several times before the chromosomes are duplicated. B) a segment of a chromosome is duplicated during an unequal crossing-over event. C) characteristic banding patterns form on the chromosomes. D) a chromosome breaks and rejoins with another section of the same chromosome. E) chromosomes are duplicated repeatedly without a corresponding nuclear division. 29. What is the formula for LOD score calculation? A) LOD = In [P(1inl<age) / P(no linkage)] B) LOD = lnz [P(no linkage) / P(linkage)] C) LOD = log10 [P(no linkage) / P(linl<age)] D) LOD = logm [P(linl<age) / P(no linkage)] E) LOD = log10P(linkage) + log10P(no linkage) 60f18 30. Which of the following concerning RNA interference is correct? A) Short RNAs must be introduced from exogenous sources for RNA interference to take place in cells. B) MicroRNAs are considered to be cis-acting elements--they must be coded by the gene that they silence. C) MicroRNAs are processed from larger primary RNA transcripts that are produced upon transcription by RNA polymerase II. D) RISC is required to create short, double-stranded RNAs from larger precursors. E) A single type of microRNA is capable of silencing transcription of several genes. 31. Deduce the phenotypic proportions in the progeny of the following cross: a+/a/a/a X a/a/a/a of autotetraploids in which the a+/a locus is very close to the centromere. Assume that the four homologous chromosomes of any one type pair randomly two—by-two and that only one copy of the a+ allele is necessary for the wild—type phenotype. A) 5 wild-type: 1 mutant B) 1 wild-type: 1 mutant C) 3 wild—type: 1 mutant D) 11 wild-type: 1 mutant E) 2 wild—type: 1 mutant 32. Which type of DNA repair is mostly likely to result in changes to the DNA sequence in comparison with the original sequence? A) Base—excision repair. B) Non-homologous end joining. C) Nucleotide excision repair. D) Mismatch repair. E) Repair by photolyase. 33. What is the name of a substitution in which a mutation in particular protein results in the replacement of a serine amino acid with a glycine residue (both are hydrophobic amino acids)? A) synonymous B) conservative C) non—conservative D) favorable E) neutralized 34. Assume that for a given gene a mutation creates an allele that functions as a dominant negative. The gene codes for a protein that forms a trimer within the cell. If at least one of the subunits has the mutant structure the entire protein is inactivated. For a heterozygous individual, what percent of the trimers present in the cell will be inactive? A) 100% B) 25% C) 50% D) 6.25% E) 12.5% 7of18 35. The highest resolution map of a genome can best be generated from A) haplotype mapping. B) linkage mapping. C) sequence analysis. D) FISH. E) deletion mapping. 36. DNA regions that are transcriptionally silenced by hypercondensation of chromatin have often undergone a modification of their cytosine residues in a process called: A) hyperoxidation. B) methylation. C) dephosphorylation. D) phosphorylation. E) ubiquitination. 37. A mutation that is characterized by a change in the DNA sequence, but no change in the resulting protein sequence, is called a A) frameshift mutation. B) missense mutation. C) silent mutation. D) nonsense mutation. E) deletion mutation. 38. Which of the following concerning gene regulation is correct? A) Repressors bind to DNA via a DNA-binding domain and recruit histone acetylases to the region via a repressor domain. B) Mutation of a cis-element that results in loss of transcription of a specific gene can be corrected by the introduction of wild—type cis-element on any chromosome. C) A locus control region can influence the transcription of a cluster of genes in a sequential fashion during development. D) Mutations of an enhancer site for a gene generally have no effect on the expression of a gene in a given cell type- E) Chromatin modifications that influence transcription are always erased during cell division of somatic cells. 39. Hardy-Weinberg equilibrium in populations is defined as conditions that produce: A) only heterozygotes. B) many lethal alleles. C) genetic drift. D) constant allele frequencies that do not change from generation to generation. E) All of the choices are correct. 80f18 40. Assume that a new low—calorie sweetener is developed. The structure is novel and is tested with the Ames test for mutagenicity. The results are shown in Table 1 at the end of this examination booklet. What conclusion is most consistent with this data? A) The sweetener is not mutagenic. B) Rat liver enzymes are highly mutagenic. C) The sweetener is not mutagenic, but can be converted into a mutagen. D) The sweetener is mutagenic and is inactivated within the body. E) The sweetener and its conversion products are equally mutagenic. 41. Which of the following is NOT a type of epigenetic modification? A) Hypermethylation of DNA. B) Hyperacetylation of histones. C) Incorporation of histone variants into chromatin. D) Chromatin remodeling. E) Alteration in the nucleotide sequence of a gene. 42. UV light is a mutagen that can cause A) depurination. B) deamination. C) alkylation. D) thymine dimers. E) oxidation. 43. The retinoblastoma (Rb) protein regulates progression into S phase by regulating the activity of which of the following proteins? A) cyclin D B) p53 C) CDK4 D) EZF E) CDC28 90f 18 ...
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m07-part1 - UNIVERSITY OF TORONTO Faculty of Arts and...

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