Problem Set 4 Key

Problem Set 4 Key - BICD100 FA2011 Problem Set #4 KEY 2 1...

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BICD100 FA2011 Problem Set #4 KEY 1.) This pedigree shows the pattern of inheritance for a rare trait that is not fully penetrant. a. Which is a more likely mode of inheritance for this trait, autosomal dominant or X-linked dominant, and why? In your explanation, refer as needed to the numbered individuals shown in the pedigree. Autosomal dominant. #10 must have the trait in order to pass it on to his daughter (since the trait is rare, we assume that #13 did not inherit the trait from her mother). Incomplete penetrance explains why #10 does not express the trait. However, an X-linked trait cannot be passed from a father (#5) to his son (#10 - we assume that #10 did not inherit the trait from his mother since the trait is rare). Therefore the trait cannot be X-linked dominant. b. Assuming the mode of inheritance you indicated in part a, what percent penetrance for this trait is indicated by the data in this pedigree? 6/7 (this would be the correct answer for either autosomal or sex linked dominant) c. If the affected female labeled 13 marries an unaffected man, what is the probability that they will have an affected child? Factor incomplete penetrance into your calculation. 13 is heterozygous so each child has a ½ chance of inheriting the trait from her (since the trait is rare we will assume the unaffected male does not have it). If the child inherits the trait, the data in this pedigree indicates the probability of being affected is 6/7. Overall probability of affected child = ½ X 6/7 = 6/14 or 3/7. 2.) Wild-type eye color in Drosophila is red; many mutations have been recovered that affect eye color. A geneticist performs a mutagenesis and isolates a new, recessive eye color mutation showing a variable phenotype; homozygous “variable” mutants have eye colors ranging from white to wild type red. To determine whether this mutation defines a previously unknown gene involved in eye color development, the geneticist crosses flies with the new mutation (“variable”) to flies homozygous for other recessive eye color mutations: white, buff, vermillion, and brown. The results are summarized in the following table, with the male parent for each cross indicated in the left-most column, and the female parent at the top of each column. white buff variable vermillion brown white buff variable red red buff variable red red variable red red vermillion red brown 1 1 3 2 6 10 7 8 4 13 11 5 12 9
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a. Organize the mutations shown in this table into “complementation groups” (groups of mutations that fail to complement one another). How many different genes altogether are affected by these 5 mutations? Group 1: white, buff, variable Group 2: vermillion Group 3: brown Three groups so three different genes affected b. Which of the following contribute to the variable nature of the new mutant’s phenotype? Circle all that apply: 1. incomplete dominance >2. variable expressivity 3. epistasis >4. incomplete penetrance*
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This note was uploaded on 01/16/2012 for the course BICD 100 taught by Professor Nehring during the Fall '08 term at UCSD.

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Problem Set 4 Key - BICD100 FA2011 Problem Set #4 KEY 2 1...

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