genetics06 - Student Name: Student Number: UNIVERSITY OF...

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Unformatted text preview: Student Name: Student Number: UNIVERSITY OF TORONTO Faculty of Arts and Science APRIL/MAY EXAMINATIONS 2006 HM8265H1 S Duration — 3 hours No aids allowed (Non-programmable calculators are permitted) Page 1 of 19 Student Name: Student Number: E— 1) PCR (polymerase chain reaction) was used to amplify a small region at the 5' end of the coding region of two mutant alleles in yeast. The following sequences were obtained when the PCR products were sequenced. Give the name of the mutations found in mutant 1 and mutant 2. Wild type: 5' GAA CTC GAA C'lT 3' Mutant 1: 5' AAA CTC GAA CW 3' Mutant 2: 5' GAA CTC AAC TT 3' A) Mutant 1: GO to AT transition; Mutant 2: insertion B) Mutant 1: GA to CT transversion; Mutant 2: deletion C) Mutant 1: GO to AT transition; Mutant 2: deletion D) Mutant 1: GC to AT transversion; Mutant 2: deletion E) Mutant 1: GA to CT transition; Mutant 2: insertion 2) The following piece of DNA inside a cell becomes completely deaminated. The cell undergoes two rounds of cell division—so that 4 cells are formed. it all DNA repair mechanisms are NOT functional, what would be the sequences of the DNA found in the 4 daughter cells? Give a complete list. Original DNA: 5’ACGTGCA3’ sTGCACGTs L SAUGTGUAS ETACACATS z EACGTGCA3 sTGCACGTs & SATGTGTAS sTACACATs 4 EACATACAs sTGTATGTs 5 SACAIACAs sTGUAUGTs A)L2;14 B)L3uL5 C)L2;i5 Dfl13mh4 E)L1 55 Page 2 of 19 Student Name: Student Number: 3) Which of the following statements describes the genetic defect observed in Huntington Disease? A) There is a CAG expansion of the promoter region of the gene that is defective in the disease resulting in lower amounts of gene expression. B) in families with individuals with alterations in the gene involved in the disease, earlier generations are more severely affected by the disease in comparison to later generations. C) There is a chromosomal inversion such that the protein-coding region of one gene is placed near the promoter of another gene and excess protein is made. D) There is CAG expansion in the coding region of the gene that is defective in the disease resulting in a long stretch of glutamines in the coded protein. E) Only males are affected with Huntington Disease, because there is CAG triplet expansion on the X chromosome, which makes the condition X-linked. 4) If there is G ' T base pairing as a result of an error during replication, which of the 5) 6) following systems would repair the error? A) Uracil DNA glycosylase. B) Photolyase. C) Alkyl transferase. D) General excision repair. E) Methyl-directed mismatch repair. Ultraviolet light usually causes mutations by a mechanism involving which of the following? A) Induction of thymine dimers. B) One-strand breakage in DNA. C) inversion of DNA segments. D) Deletion of DNA segments. E) Light-induced change of thymine to alkylated guanine. A scientist has mutated one allele of a specific gene in yeast. She then performs additional experiments to determine which type of mutation she has created. When she inserts an additional copy of the wild-type gene into the mutant yeast cells, the mutant phenotype is more severe. On the other hand, if she deletes one allele of the wild-type gene, the phenotype appears more like the wild-type phenotype. What type of mutation did she originally create? A) A gain-of-function neomorphic mutation. B) A Ioss—of-function neomorphic mutation. C) A gain-of-function hypomorphic mutation. D) A loss-of-function hypermorphic mutation. E) A gain-of—function hypermorphic mutation. Page 3 of 19 Student Name: Student Number: 7) A new mutant allele when homozygous gives a mutant phenotype. When combined with a wild-type allele, a wild-type phenotype is observed. When the mutant allele is paired with a deletion, the combination is lethal. What is the most likely explanation for this phenotype? A) The mutant allele is null. B) The mutant allele is hyperrnorphic. C) The mutant allele is hypomorphic. D) The mutant allele is neomorphic. E) None of the above. 8) Which of the following DNA mutations could result in the expression of a mutant 9) protein that was smaller than normal and was missing its carboxy—terminal end? Give a complete list. 1. Frameshift 2. Nonsense 3. Conservative 4. Semiconservative 5. Splice-site 6. lntragenic suppressor A) 1, 2, 5 B) 1,2,4, 5,6 C) 2, 5, 6 D) 3, 4 E) 2, 5 Which procedure is an example of phenocopy? A) Use a mutagen and select for a specific trait. B) Incubate cells with a chemical that is known to bind to a specific protein and examine the resulting phenotype. C) Create a mouse by targeted gene replacement and examine the resulting phenotype. D) Use RNAi to examine the effects of knocking down the levels of a specific mRNA. E)BandD 10. Plant species A (n = 7) produces a fertile hybrid when crossed to species B (n = 10). What is the likely number of chromosomes in the fertile hybrid? A)? B) 10 C) 14 D) 17 E)34 Page 4 of 19 Student Name: Student Number: 11. Which of the following sex chromosome complements in a gamete could _n_ot be the result of nondisjunction of the sex chromosomes during meiosis ll (only) in human males during gametogenesis? A) no sex chromosomes. B) XX. C) W. D) XY. E) X. 12. A human somatic cell with 45 chromosomes is typical for which of the following? A) Monosomy B) Trisomy C) Monoploidy D) Triploidy E) Euploidy 13. For the parental cross A/A/A/A x a/a/a/a, what proportion of the F2 will be A/A/a/a? A)va B)5p C)3K3 D)w9 E)4E) 14. How many chromosomes would a Turner Syndrome girl with non-familial Down Syndrome have? A)44 B)45 C)46 D)47 E)49 15. Ebony body (e) in flies is an autosomal recessive trait. A true-breeding ebony female is mated with a true-breeding wild-type male that has been irradiated. Among the progeny, which is mostly wild-type, there is a single ebony male. What kind of chromosomal mutation likely occurred in the male parent? A) paracentric inversion B) pericentric inversion C) reciprocal translocation D) Robertsonian translocation E) microdeletion Page 5 of 19 Student Name: Student Number: 16. In a newly isolated strain of fruit flies with normal viability of offspring, two linked loci map much closer than in other strains. This could be due to the fact that the new strain is: A) homozygous for an inversion. B) heterozygous for an inversion. C) homozygous for a reciprocal translocation. D) heterozygous for a reciprocal translocation. E) homozygous for a duplication. 17. in studying a new sample of fruit flies, a geneticist noted phenotypic Variegation, semisterility and the nonlinkage of previously linked genes. These flies were probably: A) homozygous for a reciprocal translocation. B) heterozygous for a reciprocal translocation. C) homozygous for an inversion. D) heterozygous for an inversion. E) homozygous for a duplication. 18. A plant species that had been subjected to radiation for a long time in order to produce chromosomal mutations was then inbred until it was homozygous for all of - those mutations. it was then crossed to the original unirradiated plant, and meiosis of the F1 hybrids was examined. it was noticed that a cell with a dicentric chromosome and a fragment occurred at a low frequency in anaphase l of the hybrid. What kind of chromosomal mutation occurred in the irradiated plant? A) paracentric inversion B) pericentric inversion C) reciprocal translocation D) Robertsonian translocation E) microdeletion 19. The best evidence to prove that a candidate gene is a disease gene is A) finding a start and stop codon B) finding a promoter region upstream C) finding that the gene is expressed D) finding the homologous gene in many similar animals E) finding a mutation in the sequence from a disease sufferer 20. Which of the following gene cloning techniques would require a genomic library? A) Functional complementation cloning B) Positional cloning C) Degenerative oligonucleotide based cloning D) Expression cloning E) None of the above Page 6 of 19 Student Name: Student Number: 21. Twelve uridine-requiring Neurospora mutants (a - I) were isolated after ultraviolet irradiation. Complementation testing was done with the following results, where “0” denotes no growth and “+” denotes growth of the heterokaryon (diploid) on minimal medium: ‘w“:Q*@QoUm oo++oo+oo+oom oo++oo+oo+oc +++++++++oo oo++oo+ooq oo++oo+om ++++++ow oo++oom oo++o: +++o~ ++oN co» . How many different genes area represented among the twelve mutants? A)2 B)4 C)5 D)8 E) 12 22. The Mouse Mammary Tumour Virus (MMTV) causes cancer in mice by insertional mutagenesis; integration of the MMTV proviral DNA can activate a nearby proto- oncogene. (This is due to the strong promoter/enhancer in the Long Terminal Repeats (LTR) at each end.) MMTV integration was used to identify the int (integration) genes that are responsible for tumorigenesis. Which technique that we discussed is this most analogous to? A) Gene tagging B) Gene (enhancer) trapping C) Functional complementation D) cDNA cloning E) Low stringency hybridization using viral genes 23. Which of the following methods would be described as cytogenetic mapping? A) Fluorescence in situ hybridization (FISH) B) Conducting dihybrid crosses to compute map distances C) Chromosome walking D) Use of RFLPs in crosses E) Using YACs and cosmids to construct a contig Page 7 of 19 Student Name: Student Number: a 24. A geneticist used primers to nine different STSs to test their presence (+) or absence (—) along five different YAC (.yeast artificial chromosome) clones. The results are show below: STSs 1 2 3 4 5 6 7 8 9 YACs 1 - - - - - + + — + 2 + — — - + - - + - 3 - - - + - + - - - 4 - + + - - - - + _ 5 - - + - - - - - + What is the order of the SSTs along the contig? A) 9 7 6 4 2 3 8 5 1 B)6 7 9 1 5 8 4 2 3 C)1 8 5 2 3 9 6 7 4 m4 6 7 9 3 2 8 5 1 E)1 2 4 5 7 3 6 8 9 25. The study of polypeptide motifs allows researchers: A) to determine which amino acids are found together. B) to discern possible biochemical properties of a protein. C) to detect expression patterns. D) to detect the cellular location of the polypeptide. E) C and D 26. Which of the following is _n_o§ a reason why the proteome is larger than the genome of a given species? A) Alternative splicing of pre-mRNAs B) Covalent modification of proteins C) DNA rearrangement of antibody and T-cell receptor genes D) Gene families such as the globins encoding proteins with different sequences E) Proteolytic processing of pro-proteins 27. Which of the following is associated with functional genomics? A) Characterizing the transcriptome present in a cell at a specific developmental stage B) Comparing the arrangement of genes in a different organisms C) Developing a physical map D) Mapping a gene using LOD scores E) Sequencing BAC clones aligned in a contig using a shotgun approach Page 8 of 19 Student Name: Student Number: 28. The mutation causing the recessive disease allele for sickle-cell anemia removes one Mstll restriction site from the globin gene so that a probe, instead of hybridizing to two fragments of 1.1 kb and 0.2 kb, hybridizes to one 1.3-kb fragment. For two parents to have a 25 percent chance of a child with sickle-cell anemia A) They must both show only fragments of 1.1 and 0.2 kb. B) They must both show fragments of 1.3 and 0.2 kb. C) One must show only a 1.1- and the other only a 0.2-kb fragment. D) One must show a 1.3- and the other a 1.1- and a 0.2-kb fragment. E) They must both show all three types of fragments. . 29. Over six thousand STS markers have been placed on the human genomic map and they are on average about 500 kb apart. If 30,000 markers were available, what would be the average distance between markers? A) 10 kb B) 50 kb C) 100 kb D) 200 kb E) 500 kb 30. The fru gene has been cloned, and both genomic and cDNA clones for fru are available. One way to more fully understand its function is to identify genes for proteins that interact with the protein product of the fru gene. Which of the following techniques would you use? A) Gene trapping B) Expression microarray analysis C) Yeast two-hybrid analysis D) Protein microarray analysis E) Chromatin immunoprecipitation 31. An RFLP marker is located 1 million base pairs away from a gene of interest. Your goal is to start at this RFLP marker and “walk” to this gene. The average insert size in the library is 55,000 bp and the average overlap at each end is 5,000 bp. Approximately how many steps will it take to get there? A) 18 B) 20 C) 25 D) 36 E) 200 Page 9 of 19 Student Name: Student Number: R 32. Which of the following is n_ot a common consequence of mutations that eliminate cell-cycle checkpoints? A) increased DNA damage B) Decreased frequency of cell division C) Polyploidy D) Aneuploidy E) increased chromosome loss 33. Which of the following would cause constitutive cell cycling? A) Mutation of the cyclin-binding domain of CDK2. B) Mutation of pRB phosphorylation sites. C) Deletion of the E2F binding domain of pRB. D) Overexpression of normal pRB protein. E) Deletion of the DNA-binding domain of E2F 34. Which of the following mutations is frequently seen in many different types of cancer? A) Chromosomal translocation involving a transcriptional enhancer and BCL2 gene B) Chromosomal translocation of the BCR and ABL genes (the Philadelphia chromosome) C) Missense mutations of the p53 gene D) Missense mutations of the breast cancer BRCA1 gene E) Amplification and overexpression of the NEU gene 35. Which of the following would M lead to formation of an oncogene? A) A mutation in the promoter for Platelet-Derived Growth Factor that leads to an increase in the efficiency of transcription initiation. B) A mutation affecting a regulatory domain of a nonreceptor kinase that causes it to always be active. 0) Amplification of the gene encoding the early response transcription factor MYC. D) Amplification of the gene encoding the RAS—GTPase activating protein (GAP). E) Translocation of the coding region of BCL2 next to a constitutively transcribed gene. 36. Cancer is a disease that develops in stages. Which of the following applies? A) Younger people get cancer more than older people. B) Benign tumors become malignant. C) Proto-oncogenes become tumor-suppressor genes. D) Small numbers of individual mutational events occur, separated by long periods of time. E) None of the above. Page 100f19 Student Name: Student Number: 37. Which of the following mutations might result in an oncogene? A) Deletion of the entire coding region of a proto-oncogene. B) Deletion of an enhancer that lies 3’ to the coding region. C) A'translocation that places the gene near constitutive heterochromatin. D) Deletion of a silencer that lies 5' to the coding region. E) The introduction of a premature stop codon in the first coding exon. 38. Signals that control apoptosis include all of the following except A) cellular damage that triggers leakage of mitochondria. B) action of Bel proteins that block release of cytochrome c. C) activation through binding of cytochrome c to Apaf. D) secreted survival factors. E) binding of steroid hormones to cytoplasmic receptors. 39. Cyclins act to control the progression of the cell cycle by A) phosphorylating E2F. B) interacting with microtubule proteins in building the spindle apparatus. C) complexing with and activating cyclin—dependent kinases. D) dephosphorylating Rb protein. E) none of the above 40. Which of the following would suggest that a disease does n_ot have a genetic basis? A) Karyotypes from people with the disease have a characteristic translocation chromosome. B) The frequency of the disease is less likely in relatives that live apart compared to those that live together. C) The frequency of the disease is unusually high in a small group of related individuals in southern Spain. D) The disease symptoms usually begin around 40. E) It is more likely that both monozygotic twins will be affected by the disease compared to dizygotic twins. 41. Cytochalasin B is a drug that disrupts actin microfilaments. If it is administered to a C. elegans zygote (Po cell), where would you find the P granules? A) At the anterior side of the cytoplasm of the Po cell B) At the posterior side of the cytoplasm of the Po cell C) Distributed evenly throughout the cytoplasm of the Po cell D) In the nucleus of the Po cell E) Totally absent Page 11 of 19 Student Name: Student Number: 42. Which of the following would lead to phenotypic normal male adult flies? A) XX homozygous mutant for SisB. B) XX homozygous mutant for le. C) XX homozygous mutant for Tra. D) XX homozygous mutant for sz. E) XX homozygous mutant for M312. 43. Which of the following would result in a phenotypic male? A) XY with an inactivating mutation in the Testes-determining factor (TDF) gene. B) XX homozygous mutant for the Testicular-feminization syndrome (TFS) gene. C) XY with a block in testosterone synthesis. D) XX with a single copy of the SRY due to non-reciprocal translocation. E) XY with Testicular feminization syndrome. 44. In early Drosophila development, which of the following statements about the pattern of gap gene expression is Q93 true? A) It is dependent upon previous expression of the segment polarity genes. B) It is determined by the activity of the anterior-posterior morphogens BCD and HB-M. C) It determines the pattern of expression of the pair-rule genes. D) It determines the pattern of expression of the homeotic genes. E) lt divides the anterior-posterior axis into several domains. 45. Actinomycin D is an antibiotic that inhibits RNA synthesis. Suppose that you add Actinomycin D to newly fertilized Drosophila eggs (i.e. eady embryos); levels of which of the following proteins would be affected? A) Bicoid (BCD) B) Nanos (NOS) C) Gurkin (GRK) D) Spaetzle (SPZ) E) Knirps (KM) 46. Where would you find the DL protein if you injected SPZ protein into the perivitelline space (the space between the vitelline membrane and the plasma membrane) on the dorsal side of a wild type early embryo? A) in the cytoplasm of all cells B) In the nucleus of all cells C) in the nucleus of cells on the ventral side only D) in the nucleus of cells on the dorsal side only E) Throughout the cell (cytoplasm and nucleus) in all cells Page 12 of 19 Student Name: Student Number: 47. What phenotype would you expect if the Antennapedia (Antp) gene was expressed in the abdominal segments where abd-A is normally expressed? A) Antennae attached to the abdominal segments B) Legs attached to the head instead of antennae C) Legs attached to the abdominal segments D) Two anterior regions and no abdomen E) Antennae instead of legs in the thoracic region 48. Below is a diagram comparing a wild type (left) and a mutant (right) larva. In what type of gene is the mutation? A) Maternal effect gene B) Gap gene C) Pair-rule gene D) Segment polarity gene E) Homeotic gene 49. A researcher has a Drosophila strain that has a loss-of-function mutation in the bicoid gene (bicoid-A). A female fly homozygous for the mutant was obtained, and mRNA from the oocytes was analyzed using in situ hybridization. The results for the mutant compared to a wild type fly are shown below. Where would you predict the mutation to be? wild type bicoid-A A) Promoter region B) 3’-untranslated region C) 5’-untranslated region D) 3’-end of the coding region E) 5’-end of the coding region Page 13 of 19 Student Name: Student Number: 50. How many generations are required for Hardy-Weinberg proportions to be established in a diploid, sexually reproducing population in regard to one locus with 10 alleles, when random mating begins? A) 1 B) 2 C) 10 D) 20 E) varies 51. Tay—Sachs disease is inherited as an autosomal recessive. In a certain large eastern European population, the frequency of Tay—Sachs disease is 1%. If the population is assumed to be in Hardy-Weinberg equilibrium with respect to Tay—Sachs, what is the probability of two heterozygotes marrying? A) 0.18 B) 0.032 C) 0.081 D) 0.25 E) 0.0081 52. In a sample of 100 persons from Nepal where the 0 allele is missing, 54 were of blood type A, 32 were of blood type AB, and 14 were of blood type B. What is your best estimate of the frequency of the B allele? A) 0.14 B) 0.30 C) 0.37 D) 0.80 E) 0.70 53. This is one line from the table giving frequencies of gametic types for the MNS system in various human populations: Am“ Population MS Ms NS Ns Ainu 0.024 0.381 0.247 0.348 Calculate the haplotype heterozygosity for this subpopulation. A) 0.878 B) 0.840 C) 0.872 D) 0.438 E) 0.338 Page 14 of 19 Student Name: Student Number: 54. Rebecca breeds Maine coon cats. A pedigree for her cats is given below; litterrnates are analogous to dizygotic twins. Sam wants to buy a pair of Rebecca’s cats to breed, but wants to make sure they share as few genes as possible to minimize the risk that their kittens will inherit homozygous disorders. He wants Farfel, but which female cat should he buy to be Farfel’s mate? mammal: A) Marbles B) Juice C) Angie D) Fluffy E) Yang 55. An extra row of eyelashes is an autosomal recessive trait that occurs in 900 of the 10,000 residents of an island in the south Pacific. Greta knows that she is a heterozygote because her eyelashes are normal, but she has an affected parent. She wants to have children with a homozygous dominant man, so that the trait will not affect her offspring. What is the probability that a person with normal eyelashes in this population is a homozygote for this gene? A) 0.18 B) 0.49 C) 0.54 o) 0.70 E) 0.91 Page 15 of19 ...
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genetics06 - Student Name: Student Number: UNIVERSITY OF...

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