Molecular Genetics Exam: Answering Questions about DNA, PCR, and Genetic Traits - Prof. St, Exams of Molecular biology

Download Molecular Genetics Exam: Answering Questions about DNA, PCR, and Genetic Traits - Prof. St and more Exams Molecular biology in PDF only on Docsity! Molecular Genetics FINAL, Thursday, Dec. 15, 2005 page 1 of 8 This is an actual exam given in 2005. There are no answers provided. I encourage you to use this as a guide to study by consulting the book and lecture notes. ------------------------------------------------------------------------------------------------------------- 1 _____________________ You have a yeast strain in the lab and you are characterizing it. You find that it is haploid, of the alpha mating type, and is phenotypically leu- met- (i.e. it requires added leucine and methionine for growth). You mate your strain with a wild-type strain of the opposite, a, mating type, allow sporulation, and characterize 100 tetrads. You find 47 tetrads with two leu- met- spores and two leu+ met+ spores. 52 tetrads with two leu- met+ spores and two leu+ met- spores. 1 tetrad with 1 leu- met- spore, 1 leu+ met+ spore, 1 leu+ met- spore and 1 leu- met+ spore. What can you conclude about the leu gene in your strain and the met gene in your strain? In particular, 1) (4 points) Are these two genes linked? 2) (4 points) Can you say that one, neither or both of these genes is centromere-linked (i.e. that they map within a few centimorgans of the centromere?) 3) (4 points) You want to carry out complementation tests with a series of test strains of the alpha mating type. The original strain was also of the alpha mating type, so you decide to use the progeny from your tetrad analysis. You have no trouble finding a met- spore of the a mating type, but observe that about 82% of the leu- spores are of the alpha mating type. Explain these results. What mapping information do they provide? 4. (4 points) Here is the sequence of the template strand of a DNA fragment: 5' GTACGACGAGTTCGACCTTCTCGCGAGCGCAGAA 3' Which of the following would be the complementary, nontemplate, strand: a) 5' TTCTGCGCTCGCGAGAAGGTCGAACTCGTCGTAC 3' b) 5' AAGACGCGAGCGCTCTTCCAGCTTGAGCAGCATG 3' c) 5' GTACGACGAGTTCGACCTTCTCGCGAGCGCAGAA 3' d) 5' CATGCTGCTCAAGCTGGAAGAGCGCTCGCGTCTT 3' 5. (5 points) You have a piece of DNA that includes the sequence: 5' - GATGAGGATGAGGAGAAGTACCGGCCGCCGCCTGCGCATCACAATATGTTCAGT - 3' Molecular Genetics FINAL, Thursday, Dec. 15, 2005 page 2 of 8 This is an actual exam given in 2005. There are no answers provided. I encourage you to use this as a guide to study by consulting the book and lecture notes. ------------------------------------------------------------------------------------------------------------- 2 _____________________ To amplify this DNA by PCR you would use a pair of primers containing which two of the following eight primers (you must get both primers right to get any credit -- circle two letters): a) 5' - GATGAGGATGAGGAGAAG- 3' b) 5' - CTACTCCTACTCCTCTTC - 3' c) 5' - CATCACAATATGTTCAGT- 3' d) 5' - GTAGTGTTATACAAGTCA- 3' e) 5' - GAAGAGGAGTAGGAGTAG - 3' f) 5' - CTTCTCCTCATCCTCATC - 3' g) 5' - TGACTTGTATAACACTAC- 3’ h) 5' - ACTGAACATATTGTGATG- 3' 6. (4 points) Four wild-type alleles of the marker locus STR7; 1, 2, 3 and 4; are present in a human population in Hardy-Weinberg equilibrium. The frequency of allele 1 is 0.5, the frequency of allele 2 is 0.4, the frequency of allele 3 is 0.08 and the frequency of allele 4 is 0.02. What fraction of the population is homozygous for allele 4? 7. (4 points) A very poorly preserved DNA specimen is found at a crime scene. It is typed for the STR7marker locus and found to have a signal for allele 4. Because the sample is so poorly preserved, it is possible that another allele was present but not detected. What fraction of people from this population would be expected to have allele 4. 8. (4 points) Ann’s father had Mount’s dementia, a fully penetrant autosomal recessive learning disorder. Ann does not have the disease. The overall frequency of this disease in the population is 1/40,000. Assuming that Ann’s husband has not been genetically tested, what is the risk to her children (i.e. what is the probability of their being affected)? 9. (4 points) Mount’s dementia has one known disease allele (G56X) that accounts for 50% of identified mutations and a test for this allele is available. Neither Ann nor her husband has this allele. Now what is the risk to her children? Molecular Genetics FINAL, Thursday, Dec. 15, 2005 page 5 of 8 This is an actual exam given in 2005. There are no answers provided. I encourage you to use this as a guide to study by consulting the book and lecture notes. ------------------------------------------------------------------------------------------------------------- 5 _____________________ b) A hypomorphic mutation produces much less of a protein while an antimorphic mutation produces a protein with reduced function. 27. a) Yeast (S. cerevisiae) can grow vegetatively as either haploid or diploid cells. b) Yeast (S. cerevisiae) grow vegetatively as diploid cells; haploid cells mate soon after sporulation. 28. a) Small regulatory RNAs like lin-4 RNA may play a much larger role in development than originally suspected, and not only in nematodes. b) Small regulatory RNAs like lin-4 RNA are a novel feature of gene regulation in nematodes. 29. a) Recombinant inbred strains are obtained by transformation of an inbred strain with recombinant DNA. b) Recombinant inbred strains are obtained by systematically inbreeding the progeny of a cross. 30. a) Congenic mouse strains are identical except for a single locus and are generated by repeated backcrossing. b) Congenic mouse strains are obtained by transformation of a standard (non-inbred) strain with recombinant DNA. Molecular Genetics FINAL, Thursday, Dec. 15, 2005 page 6 of 8 This is an actual exam given in 2005. There are no answers provided. I encourage you to use this as a guide to study by consulting the book and lecture notes. ------------------------------------------------------------------------------------------------------------- 6 _____________________ For each of the following statements, indicate for which of the five model organisms (Saccharomyces cerevisiae, Drosophila melanogaster, Caenorhabditis elegans, Arabidopsis thaliana, Mus musculus) it is true. These are 45 true-false questions, worth 1 point each. Statement. S. c. D. m. C. e. A. t. M. m. Undergoes mitotic divisions both as a diploid and as a haploid. Gene disruption using homologous recombination is routinely used for reverse genetics. This organism does not have sex. XX individualss are females and XY individuals are males. Self-fertilization allows a heterozygous allele in a single individual to be homozygous in the next generation. There are fewer than 10,000 genes. The genome size is under 300 Mb.. Many genes undergo trans-splicing. The majority of genes have no introns. (5 points) Define bulk segregant analysis. Molecular Genetics FINAL, Thursday, Dec. 15, 2005 page 7 of 8 This is an actual exam given in 2005. There are no answers provided. I encourage you to use this as a guide to study by consulting the book and lecture notes. ------------------------------------------------------------------------------------------------------------- 7 _____________________ In diagramming developmental signaling pathways, the symbol ---| is used to indicate repression; the activity of one gene negatively regulates the activity of the next. For the pathway A ---| B ---> C if A is on, then B will be off. If B is on, then C will also be on. You are studying mutations that affect vulval development in C. elegans and you have defined two genes, vuv-1 and vuv-2. Loss-of-function mutations in vuv-1 result in hermaphrodites with the multivulva phenotype while loss-of-function mutations in vuv-2 result in a vulvaless phenotype. (5 points) Which of the following regulatory pathways would be consistent with these results? vuv-1 ---> vuv-2 ---> vulva formation. vuv-1 ----| vuv-2 ---> vulva formation. vuv-1 ---> vuv-2 ----| vulva formation. vuv-1 ----| vuv-2 ----| vulva formation. vuv-2 ---> vuv-1 ---> vulva formation. vuv-2 ----| vuv-1 ---> vulva formation. vuv-2 ---> vuv-1 ---| vulva formation. vuv-2 ----| vuv-1 ---| vulva formation. (5 points) In further studies you find that a vuv-1; vuv-2 double mutant looks identical to a vuv-2 single mutant (i.e. vulvaless). Which of the pathways is most consistent with this result? vuv-1 ---> vuv-2 ---> vulva formation. vuv-1 ----| vuv-2 ---> vulva formation. vuv-1 ---> vuv-2 ----| vulva formation. vuv-1 ----| vuv-2 ----| vulva formation. vuv-2 ---> vuv-1 ---> vulva formation. vuv-2 ----| vuv-1 ---> vulva formation. vuv-2 ---> vuv-1 ---| vulva formation. vuv-2 ----| vuv-1 ---| vulva formation. ----------------------------------------------- (5 points) Define hybrid dysgenesis