2009_Final

2009_Final - Biol 280 - Spring 2009 Final Exam 5/09/2008...

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Biol 280 - Spring 2009 Final Exam 5/09/2008 Total: 150 points 60 questions (2.5 points each) Use the answer sheet (bubble sheet) because the question booklet will not be graded. Be sure the answer sheet is filled in with #2 pencil. Be sure to fill in your identifying information (name, student ID number) completely and accurately. Good luck!
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1. In laboratory class, you were assigned to study an unknown mutation in Drosophila that had a whitish eye. You crossed females from your true breeding mutant stocks to wild-type (brick red-eyed) males, recovering all wild-type F 1 flies. In the F 2 generation, the following offspring were recovered in the following proportions: Wild type 5/8 Bright red 1/8 Brown eye 1/8 White eye 1/8 Given the Drosophila has no recombination in males, which of the followings can best explain the arrangement of these genes? a) The white-eyed phenotype is caused by two autosomal genes and they are on different chromosomes. b) The white-eyed phenotype is caused by two autosomal genes and they are closely linked on the same chromosome. c) The white-eyed phenotype is caused by two autosomal genes and they are far apart (unlinked) on the same chromosome. d) The white-eyed phenotype is caused by two X-linked genes. e) The white-eyed phenotype is caused by one X-linked gene and one autosomal gene. 2. A mouse that carries two wild-type alleles of Igf2 is normal in size, whereas a mouse that carries two mutant Igf2 alleles lacks a growth factor and is a dwarf. It is known that the Igf2 gene undergoes female-specific imprinting. Thus, if a dwarf-sized heterozygous ( lgf2/Igf2) male mouse is mated with normal-sized homozygous ( +/+) female, what is the phenotypic ratio? a) All normal. b) 3 normal : 1 dwarf. c) 1 normal : 1 dwarf. d) All dwarf. e) only sons are dwarf. 3. What is the molecular basis for the Bar D mutation in Drosophila? a) The methylation of cytosines at specific loci. b) The expansion of tri-nucleotide repeats in certain genes. c) The alteration of gene functions by specific point mutations. d) The increase of gene copy number by duplication. e) The phosphorylation of histone proteins at specific loci. 4. Wild type and mutant forms of glucose-6-phosphate dehydrogenase (G- 2
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6-PD), an X-linked gene, can be separated by electrophoretic analysis. A cell line was generated from one single cell isolated from a female heterozygous for G-6-PD. When the cellular proteins from this cell line were separated by electrophoresis, only the wild type variant was detected. This is because: a) The wild type variant is dominant to the mutant. b) Gene conversion of wild allele to the mutant variant. c) Random inactivation of X chromosome in human dosage compensation. d) The G-6PD locus is female-specifically imprinted
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2009_Final - Biol 280 - Spring 2009 Final Exam 5/09/2008...

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