HMB265-May 2007 Exam part 2 - 44 45 46 47 Shown below are...

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Unformatted text preview: 44. 45. 46. 47. Shown below are the deletions found in a series of heterozygous deletion mutants. The deleted region is indicated as ( .... ..) and the intact region as . The numbers indicate the different mutants. 1 ( ........ ..) 2 ( ........ ..) 3 ( ....... ..) 4 ( .......... .. ) 5 ( ......... ..) A series of point mutants having homozygous mutations in genes A-E were used in a deletion mapping experiment (i.e. the point mutants were crossed with the heterozygous deletion mutants). The results are Shown in Table 3 at the end of the examination booklet. Wild-type progeny are indicated by (+), (0) indicates no wild—type progeny. Indicate the order of genes that is most consistent With the data. A) CADBE B) ACBDE C) BADCE D) ABDEC E) CEADB A male has a mutation such that the enzyme that methylates the insulator and H19 promoter region does not function. Therefore, he is unable to imprint this region as usual. He has a child with a woman who has normal imprinting mechanisms. Assuming the child is viable. What would you expect to observe? A) The child would have no H19 protein and twice the level of Igf2 protein. B) The child would have normal levels of both H19 and Igf2 protein C) The child would have no H19 protein and no Igf2 protein. D) The child would have twice the level of H19 protein and twice the level of Igf2 protein. E) The child would have twice the level of H19 protein and no Igf2 protein. Three cloned fragments of DNA from a specific chromosomal region, and sequence—tagged sites (STS) that are detected on each fragment are shown below. What is the correct order of the STS on the chromosome? Clone A: STS content = 2, 4, 5, 6 Clone B: STS content = 1, 4, 5 Clone C: STS content = 2, 3, 6, 7 A) 1, 2, 3, 4, 5, 6, 7 B) 4, 5, 6, 2, 7, 3, 1 C) 2, 6, 5, 4, 1, 3, 7 D) 1, 4, 5, 6, 2, 7, 3 E) 2, 4, 5, 6, 7, 1, 3 Which of the following concerning p53 is correct? A) p53 is activated in response to growth factors. B) p53 is a transcription factor that directly activates E2F. C) p53 enhances entry into S phase. D) p53 is a transcription factor that enhances expression of p21. E) p53 expression is reduced following ionizing radiation. 10 of 18 48. The principle evolutionary mechanism that gave rise to the diverse immunoglobulin supergene family was A) gene conversion. B) exon shuffling. C) genetic drift. D) concerted evolution. E) transposition. 49. You are investigating the genetic causes of a dominantly inherited disease by performing linkage analysis in families having multiple members with the condition and using single nucleotide polymorphisms (SNP) as markers (designated A1 to A5). Assume that those marrying into the family are homozygous recessive and those with the dominant phenotype are heterozygotes. The pedigree for this study is shown in Figure 1 at the back of the examination booklet. What is the Odds ratio for the third generation? A) 1 B) 1.4 C) 4.3 D) —1.4 E) —4.3 50. Assume that a series of compounds has been discovered in Neurospora. Compounds A-F appear to be members of an enzyme pathway. Several mutations have been identified and strains 1—4 each contain a single mutation. Table 4 at the end of the examination booklet shows the results from experiments in which the mutants were grown in media supplemented with the indicated compound. Choose the pathway that best fits the data presented. [Growth in minimal media with supplements is shown by (+), no growth is shown by (0)]. A) A—->B——>C—->D——>E—->F B) A——>B——>C——>F-—>D——>E C) F——>B-—->C——>D——>A—->E D) A——>B——>C—->D—->F—->E E) A—->B—->F—->E-->C——>D 51. Mutations in the CF gene that cause cystic fibrosis are carried by 1 in 20 individuals of Caucasian ancestry. The disease is clearly detrimental, yet the allele is maintained at a relatively high level in the population. Which of the following is NOT a possible explanation to explain this paradox? A) Individuals who are homozygous for the mutation able to reproduce before they die from the disease. B) There may be some benefit to having the allele in the heterozygous state. C) The mutant allele is likely to be closely linked to another allele that provides some benefit to the individual. D) Having the mutation at a high level in the population keeps population numbers low—this benefits the entire population. E) There may be some benefit to having the allele in the homozygous state. 52. Which of the following concerning fragile X syndrome is correct? A) Because it is an X—linked condition, it is observed in males, but not in females. B) It is caused by an increase in the number of CGG repeats in the 3' untranslated region of the PMR—l gene. C) It is caused by a point mutation in the PMR—l gene on the X chromosome. D) The FMR—l gene is not expressed in people with the syndrome, because of gene silencing. B). When individuals with the premutation allele for fragile X syndrome have children, the children have fewer fragile X symptoms in comparison with their parents. 11 of 18 53. Scientists used positional cloning to identify the gene that is mutated in cystic fibrosis. They used fine mapping to narrow down the chromosomal region that contained the defective gene. In this region, four genes were predicted. How did the scientists determine which of the four genes is involved in cystic fibrosis? A) They sequenced the entire chromosomal region and compared the sequences found in patients with cystic fibrosis and those without the disease. B) They used fluorescence in situ hybridization with cDNA isolated from cystic fibrosis patients to identify the gene. C) They performed northern analysis to determine which of the four genes is expressed in the lungs and pancreas-—tissues that are affected in cystic fibrosis. D) They performed restriction digests on the chromosomal region isolated from patients with cystic fibrosis and those without the disease to look for differences. E) They created four lines of knockout mice each having a mutation in one of the four genes and looked to see which had a phenotype that was similar to a person with cystic fibrosis. 54. Assume that a wild-type sequence is 5'AGCCTAC3'. Indicate the sequence that might be produced by a transversion. A) 5'AGTCTAC3' B) 5'AGCCGCCGCCGCCTAC3' C) 5'AGCCCAC3' D) 5'ATCCTAC3' E) 5'AGCCTGC3' 55. What is the consequence of a homozygous translocation? A) Abnormal pairing during meiosis. B) An alteration in the linkage relationships of genes. C) Formation of abnormal chromatids following crossing over. D) Gene duplications and deletions. E) Semisterility. 56. For a gene A, a geneticist studying a population of beetles found the following genotypes: AA 45 beetles Au 347 beetles aa 8 beetles What is the observed genotypic frequency of All individuals? A) 0.86 B) 1.15 C) 0.02 D) 0.90 E) 0.13 12 of 18 57. What approach is used in hierarchical genome sequencing? A) Generate plasmid libraries containing DNA from single chromosomes, sequence the DNA and use computational methods to look for sequence overlap. B) Generate BAC and plasmid libraries containing DNA from all 46 chromosomes, sequence the DNA, use computational methods to look for sequence overlap. C) Perform fluorescence in situ hybridization to identify regions on chromosomes coding for proteins, isolate these regions and sequence them. D) Perform linkage analysis on large numbers of model organisms, use the linkage map to create a physical map. E) Generate BACs containing fragments of individual chromosomes, order the BACs by looking for overlapping molecular markers to create a physical map, subclone the BACs and sequence. 58. The majority of clinical trials in gene therapy in the past few years have been performed to test treatments for cancer. In one type of cancer therapy, the gene for thymidine kinase is introduced into cells. Which of the following statements concerning this therapy is correct? A) The gene for thymidine kinase is introduced into brain tumour cells by direct injection of DNA into the tumour. B) Fibroblasts carrying the thymidine kinase gene are injected into the site of a brain tumour and the thymidine kinase gene spreads to brain tumour cells. C) Both tumour cells and healthy brain tissue are destroyed when patients are subsequently treated with ganciclovir. D) The gene for thymidine kinase is transferred to healthy brain cells to protect them against cell death induced by ganciclovir. E) None of the above. 59. You compare the sequence of a gene coding for a protein with the mature mRNA produced upon transcription of the gene and processing of the primary transcript. Which of the following sequences would be found only in the mRNA and not in the gene? A) The 5'untranslated region. B) The 3' untranslated region. C) A poly A sequence. D) Intron sequences. E) Exon sequences. 60. Which of the following concerning X—inactivation is correct? A) XIST RNA coats one X chromosome ensuring that the X chromosome remains active. B) Females that are heterozygous for a mutation in an X—linked gene sometimes have a mild form of disease because of random inactivation of the wild-type X chromosome. C) X chromosomes are occasionally inactivated in adult human males resulting in X—linked diseases. D) X—inactivation remains during formation of gametes (germ cells) in females and is passed on to all offspring. E) In human females, X inactivation occurs in embryos at the two—cell stage 13 of 18 61. In order for the D5 mutations in corn to be stable, A) the DNA must not be replicated. B) Ac elements must be absent. C) an Ac element must be present. D) the Ds must contain the gene for transposition. E) None of the above. 62. Which of the following is NOT an aneuploidy? A) monosomy B) tetraploidy C) trisomy D) tetrasomy E) cells from an individual with Down syndrome. 63. What was a surprising result of the paper that was presented in class on the examination of mutations found in kinase genes isolated from different cancer types? A) It was surprising that several of the mutations that were identified were also found in the normal cells of the individuals from which the cancer was isolated. This implies that cancer is inherited at a higher level than previously expected. B) It was surprising that so many of the mutations in the kinase genes were identified as causing cancer (i.e. driver mutations). This implies that cancer can be caused by many more genes than originally expected. C) It was surprising that kinase genes were identified as being mutated in many cancers. Scientists had not previously known that mutations in kinase genes can cause cancer. D) It was surprising that there were so many kinase genes in humans. Scientists had not previously known the number of kinase genes in the human genome. E) It was surprising that so many of the mutations in the kinase genes had nothing to do with cancer development (i.e. passenger mutations). This implies that this type of study will not provide useful information about how cancer develops. 64. You would like to determine the tissues and developmental stage at which a protein of interest (protein X) is expressed in mice. You do not have antibodies available against the protein, so you decide to make a transgenic mouse and use a reporter gene approach. What pieces of DNA should you ligate together to form the transgene that will be injected into a mouse zygote? A) The gene control region of the GFP gene ligated with the protein-coding region of the GFP gene. B) The gene control region of the mouse protein X gene ligated with the protein-coding region of the GPP gene. C) The gene control region of the human protein X gene ligated with the gene control region of the GPP gene. D) The protein-coding region of the mouse protein X gene ligated with the protein-coding region of the GFP gene. E) The gene control region of the GFP gene ligated with the protein—coding region of the protein X gene. 14 of 18 65. Six different strains of Drosophila have been isolated, each shows a recessive white eye trait. Crosses were performed between the mutant strains. The results from the crosses are shown in Table 2 at the end of the examination booklet. "w" indicates White—eyed progeny, "R" indicates wild—type red eyes. Based on these crosses, how many different genes are present and What strains have mutations in the same gene as strain "a"? A) 1 gene. Strains b, c, d, e, f have mutations in the same gene as strain a. B) 2 genes. Strains e and fhave mutations in the same gene as strain a. C) 3 genes. Strain e has mutations in the same gene as strain a. D) 4 genes. Strain e has mutations in the same gene as strain a. E) 6 genes. None of the strains have mutations in the same gene as strain a. 66. Scientists working on the cancer genome project sequenced the exons of kinase genes in several cancer types and found many mutations. Mutant kinase genes that cause cancer are likely to be what class of cancer gene and do the mutations have to be homozygous to induce cancer development? A) They are called oncogenes. No they do not have to be homozygous. B) They are called proto-oncogenes. No they do not have to be homozygous. C) They are called tumour—supressor genes. Yes they have to be homozygous. D) They are called tumour—surpressor genes. N 0 they do not have to be homozygous. E) They are called oncogenes. Yes they have to be homozygous. 67. DNA sequences that are often found adjacent to a gene and serve as binding sites for gene regulatory proteins are said to be: A) cross-reacting. B) origins of transcription. C) trans—acting. D) cis-acting. E) transcription factors. 68. You are investigating the genetic causes of a dominantly inherited disease by performing linkage analysis in families having multiple members with the condition and using single nucleotide polymorphisms (SNP) as markers (designated A1 to A5). Assume that those marrying into the family are homozygous recessive and those with the dominant phenotype are heterozygotes. The pedigree for this study is shown in Figure 1 at the back of the examination booklet. What is the recombination frequency for the third generation? A) 8.3% B) 16.6% C) 33.3% D) 50% E) 83% 15 of 18 69. Two girls with severe combined immunodeficiency were the first humans to receive gene therapy. Ten years after the end of their treatment, one girl (patient 1) had a significant proportion of lymphocytes that expressed the transgene, while the second girl (patient 2) had very few transgenic cells. What was the reason that was postulated for this difference? A) The transgene caused leukemia in patient 2. Therefore all of her lymphocytes had to be destroyed to fight the leukemia. B) The transgene did not enter the lymphocytes of patient 2. Therefore, the lymphocytes that were transferred initially never were transgenic. C) The transgene incorporated into the genome of patient 1, but did not incorporate into the genome of patient 2. D) Patient 2 mounted an immune response to various components of the gene transfer system. Therefore, the transgenic cells were selectively destroyed. E) Patient 1 was younger than patient 2. Therefore her body more readily accepted the treatment. 70. SNRPN is the gene that has been identified in Prader-Willi syndrome. It is maternally imprinted. A man is heterozygous for a loss-of—function mutation in the SNRPN gene. His wife is homozygous wild—type. What are the chances that they will have a child with Prader—Willi syndrome? A) 0% B) 25% C) 50% D) 75% E) 100% 16 of 18 Table 1: Ames test to determine the mutagenicity of a novel sweetner. Sam 1e I umber of colonies in medium 'thout histidine _ 5 Table 2: Results from complementation tests in which mutant flies were crossed to determine the number of uni ue -enes that influence e e colour. Table 3: Results from a deletion mapping experiment performed to determine the order of «enes A to E in a s n ecific chromosomal region. _ W IIIII I utant _E- __ _E- _E- _E- Table 4: Results of an experiment to determine a biochemical pathway in Neurospora. _ -UHI “Hun alumna _uu l7 of18 m Ma Ma A134 Aim we Figure l: Pedigree from an analysis of linkage of a dominantly inherited disease with markers A1 to A5. 18 ofl8 ...
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HMB265-May 2007 Exam part 2 - 44 45 46 47 Shown below are...

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