MidtermReview
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MidtermReview

Course Number: BIOL 1201, Fall 2012

College/University: LSU

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BICD 100: Midterm Review Kevin Wu Mastering Genetics Problem Set 2, Part 7A You decide to conduct a genetic analysis of mutant lines by crossing each with a pure-breeding wild-type line. The numbers in the F2 indicate the number of progeny in each phenotypic class. Mastering Genetics Problem Set 2, Part 7A From these results, determine the relationship between the mutant allele and its corresponding wild-type...

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100: Midterm BICD Review Kevin Wu Mastering Genetics Problem Set 2, Part 7A You decide to conduct a genetic analysis of mutant lines by crossing each with a pure-breeding wild-type line. The numbers in the F2 indicate the number of progeny in each phenotypic class. Mastering Genetics Problem Set 2, Part 7A From these results, determine the relationship between the mutant allele and its corresponding wild-type allele in each line. Mastering Genetics Problem Set 2, Part 7A T T t Tt Tt t Tt Tt T t T TT Tt t Tt tt Mastering Genetics Problem Set 2, Part 7A ~54:18 ~48:16 34:17:~17 3:1 3:1 2:1:1 Mutant is dominant to wild type. Mutant is neither dominant or recessive to wild type. Mastering Genetics Problem Set 2, Part 7B You designate the forked mutant allele as F (wild type = f+) and the pale mutant allele as p (wild type = P). 3 phenotypes for color. 2 phenotypes for leaf. Mastering Genetics Problem Set 2, Part 7B Mastering Genetics Problem Set 2, Part 7B Mastering Genetics Problem Set 2, Part 7B Mastering Genetics Problem Set 2, Part 7B Mastering Genetics Problem Set 2, Part 7B Mastering Genetics Problem Set 2, Part 7C a. b. c. d. e. The forked mutation is epistatic to the twist mutation. The twist mutation is incompletely dominant to the forked mutation. The forked mutation has recombined with the twist mutation. The forked mutation is incompletely dominant to the twist mutation. The forked mutation and the twist mutation are codominant alleles of the same locus. f. The twist mutation is epistatic to the forked mutation. Mastering Genetics Problem Set 2, Part 7C a. b. c. d. e. The forked mutation is epistatic to the twist mutation. The twist mutation is incompletely dominant to the forked mutation. The forked mutation has recombined with the twist mutation. The forked mutation is incompletely dominant to the twist mutation. The forked mutation and the twist mutation are codominant alleles of the same locus. f. The twist mutation is epistatic to the forked mutation. Mastering Set Genetics Problem 2, Part 7D You decide to designate the twist allele as FT to distinguish it from the forked allele F. Identify the genotypes of the three F2 classes. Mastering Genetics Problem Set 2, Part 7D You decide to designate the twist allele as FT to distinguish it from the forked allele F. Identify the genotypes of the three F2 classes. Mastering Genetics Problem Set 4, Part 1A You are investigating the inheritance of two rare conditions (A and B) in an extended family of thoroughbred racehorses. Mastering Genetics Problem Set 4, Part 1A Mastering Genetics Problem Set 4, Part 1A Mastering Genetics Problem Set 4, Part 1A Aa Aa aa Condition A is autosomal recessive. Mastering Genetics Problem Set 4, Part 1A Mastering Genetics Problem Set 4, Part 1A XBXB XbY XBXb XbY XbY XBXb XbXb Condition B is x-linked recessive. Mastering Genetics Problem Set 4, Part 1B Mastering Genetics Problem Set 4, Part 1B Aa Aa aa AA A_ A_ A_ AA A_ Mastering Genetics Problem Set 4, Part 1B XbY XBXb XBY XBX_ XBY XBXB XBY Mastering Genetics Problem Set 4, Part 1B Mastering Genetics Problem Set 4, Part 1C Mastering Genetics Problem Set 4, Part 1C 1/2 2/3 2/3 1/2 1/2 1/2 1. IV-3 will have condition A? 2/3 x 1/2 x 1/2 x 2/3 x 1/2 x 1/2 = 1/36 Mastering Genetics Problem Set 4, Part 1C 1/2 1/2 1/2 2. IV-3 will have condition B? 1/2 x 1/2 x 1/2 = 1/8 Mastering Genetics Problem Set 4, Part 1C 3. Probability that IV-3 will have both conditions? 1/36 x 1/8 = 1/228 Mastering Genetics Problem Set 4, Part 1C 4. Probability that IV-3 will have neither conditions? (1-1/36) x (1-1/8) = 35/36 x 7/8 = 245/228 Mastering Genetics Problem Set 4, Part 1C 5. Probability that IV-3 will have at least one of the two conditions? 1-245/288 = 43/288 Mastering Genetics Problem Set 4, Part 1D Individual IV-3 is born as a male affected with condition B but not condition A. His parents are bred again, and an ultrasound shows that the fetus is another male. A_ XBXb A_ XBY 1/36 1/2

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