BIOL 121 (2009) Fall Exam 3+ Key

3 pts mutant m2 b a plant has the genotype m1m1m2m2

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Unformatted text preview: part of the promoter, which is upstream of the TH coding sequence. Therefore, the additional base should have no effect on the amino acid sequence encoded by TH. c) Consider a mutation in which the GRE adjacent to the TH gene is deleted. Mice heterozygous for this mutation are mated with each other. What percentage of the offspring will transcribe TH in response to glucocorticoid? (5 pts) Answer: since the GRE acts in cis, the GRE mutation is recessive. 25% of the offspring will be homozygous for the mutation and will be unable to transcribe TH. The remaining 75% will show glucocorticoid- responsive TH transcription. 6 BIOL 121 Exam 2 11/04/09 NAME: 13) A cross between female Drosophila melanogaster heterozygous for three loci (x, y, z) and wild- type males yielded the following numbers of progeny phenotypes: Females: +++ 800 Males: +++ 48 ++z 6 +yz 290 +y+ 68 xyz 40 xy+ 2 x++ 266 x+z 80 i. On which chromosome(s) do these loci reside? (3 pts) sex chromosome ii. If any of the loci are on the same chromosome, indicate the map distance(s) between them. (7 pts) +yz and x++ are the parental types xy or ++ = (2+40+48+6)/800 =96 = 12 xz or ++ = 40+80+48+68/...
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This note was uploaded on 01/16/2014 for the course BIOL 121 taught by Professor Dustinbrisson during the Fall '11 term at UPenn.

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