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BIOL201 (60)
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# b207p05a.pdf

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School
Department
Biology (Biological Sciences)
Course
BIOL201
Professor
Michael Harrington
Semester
Fall

Description
Biology 207 - Practice questions 5 - Mendelian genetics - Answers Question 1 Type I albinism is caused by the absence of the enzyme Tyrosinase which converts the amino acid tyrosine into the pigment melanin. This enzyme is encoded by the TYR gene at 11q14-q21. a) Describe the gene's cytologtical location. The TYR gene is on the long arm of chromosome 11 between region 1 band 4 and region 2 band 1. b) Why are mutant alleles of this gene recessive? People with one functioning copy of the gene have a WT phenotype because this allele can make a sufficient amount of Tyrosinase. Question 2 Both members of an Edmonton couple are heterozygous for recessive mutations in the TYR gene. They have two children. a) Write out the genotype and phenotype for each parent. Let the A-gene be responsible for albinism A/_ = non-albino, a/a = albino Mother = A/a (non-albino), Father = A/a (non-albino) b) What are the possible genotypes of the gametes, and their frequencies, produced by the father? The mother? Half the gametes will be ‘A’, half will be ‘a’ for both the mother and the father. c) What is the probability that the first child is albino? 1/4 d) What is the probability that the second child is albino? 1/4 e) What is the probability that both children are albino? p(both children albino)= p(1st child albino) x p(2nd child albino) = 1/4 x 1/4 = 1/16 f) What is the probability that both children are phenotypically identical with regard to skin color? p(phenotypically identical) = p(both albino) + p(both non-albino) = (1/4 x 1/4) + (3/4 x 3/4) = 5/8 g) What is the probability that both children will be genetically identical at that locus? p(genetically identical) = p(both A/A) + p(both a/a) + p(both A/a) = (1/4 x 1/4) + (1/4 x 1/4) + (1/2 x 1/2) = 3/8 Question 3 You have just done a screen for orange-eyed mutant Drosophila. You now have five strains, each in its own sto
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