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Homework Questions 2

Homework Questions 2 - epistatic to cinnabar so cn cn w w...

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Genetics 2153, Homework #2 (Due Feb. 13/15) Required Problems to hand in from book (4 points): 4.3, 4.5, 4.7, 4.8, 4.10, 4.12, 4.13, 4.18, 4.20, 4.21, 4.26, 4.27, 4.28. 25.1, 5.1, 5.4, 5.6, 5.7, 5.10,  5.12, 5.13, 5.17, 5.22 Recommended additional problems: 3.17, 3.18, 3.23, 3.25, 4.1, 4.2, 4.9, 4.11, 4.14, 4.17, 4.25,  5.8, 5.9, 5.11, 5.15. Additional required problems (6 points): A.  Using the information in problem 4.21 on the  brown  ( br ) and  scarlet  ( sc ) genes, How many  phenotypes will be produced in the progeny from a cross of a  br + br   sc + sc  fly with a white-eyed  fly?  What are the different phenotypes that will be produced?  What proportion of the progeny  will have each phenotype? B.   cn +  = wild type eyes (non-cinnabar)   cn  = cinnabar (bright red) eyes w +  = wild type eyes (non-white)    w  = white eyes ( an X-linked trait)  Cinnabar and white are each recessive to their respective wild type allele.  The white eye trait is 
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Unformatted text preview: epistatic to cinnabar, so cn cn; w w flies have white eyes. A wild-type female of genotype cn + cn ; w + w is crossed to a cn + cn ; w Y white eyed male. What progeny ratios for eye color will be produced among the male and female offspring? Note: The wild-type allele of each gene is dominant only to its own mutant allele. Epistasis is NOT a form of dominance. For recessive alleles, the “covering up” or “masking” of the phenotype that is called epistasis only occurs when both genes are in the homozygous recessive condition. Be careful to clarify your thinking about the difference between epistasis and dominance. Epistasis will not be defined on the exam....
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