2006problemset3key

2006problemset3key - Problem Set #3 KEY BICD 100 W' 06 (L....

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Unformatted text preview: Problem Set #3 KEY BICD 100 W' 06 (L. Smith) 1.) The following pedigrees show the pattern of inheritance for two different, rare traits in human families. Indicate for each the likely mode(s) of inheritance (autosomal recessive, X-linked recessive, autosomal dominant, X-linked dominant, or maternal inheritance... if two or more are both likely, so state). First one: X-linked recessive. Second one: maternal inheritance and autosomal dominant are both likely (if autosomal dominant, we expect that 1/2 the progeny of all affected individuals will be affected here we see that of 11 progeny of affected individuals, 6 are affected). 1 2.) In Drosophila (fruit flies), miniature wings results from an X-linked recessive mutation,w. Sepia eyes results from an autosomal recessive mutation, s (wild-type eye color for fruit flies is red). a. A female fly with miniature wings and sepia eyes is crossed to a true-breeding wild-type male (normal wings, red eyes). The F1 are intercrossed to produce an F2 generation. What wing and eye color phenotypes do you expect to find among the F2 females, and in what proportions? What phenotypes do you expect to find among the F2 males, and in what proportions? XwXw ss x X+Y SS -> all female progeny XwX+ Ss, all male progeny XwY Ss F1 male gametes: Xw S, Xw s, Y S, Y s F1 female gametes: Xw S, Xw s, X+ S, X+ s F2 progeny males and females same: 3/8 miniature wings + red eyes, 1/8 miniature wings + sepia eyes, 3/8 normal wings + red eyes, 1/8 miniature wings + sepia eyes b. A female from a true-breeding line of flies with sepia eyes and normal wings is crossed to a male from a true breeding line of flies with miniature wings and red eyes. The F1 are intercrossed to produce an F2 generation. What wing and eye color phenotypes do you expect to find among the F2 females, and in what proportions? What phenotypes do you expect to find among the F2 males, and in what proportions? X+X+ ss x XwY SS -> all female progeny X+Xw Ss, all male progeny X+Y Ss F1 male gametes: X+ S, X+ s, Y S, Y s F1 female gametes: X+ S, X+ s, Xw S, Xw s F2 progeny females: 3/4 normal wings + red eyes, 1/4 normal wings + sepia eyes F2 progeny males: 3/8 normal wings + red eyes, 1/8 normal wings + sepia eyes, 3/8 miniature wings, red eyes, 1/8 miniature wings + sepia eyes 3.) A certain type of deafness in humans is inherited as an X-linked recessive trait. A man who suffers from this type of deafness marries a woman of normal hearing, and they are expecting their first child. They find out that they are distantly related as shown in the following pedigree (see original problem set for pedigree, not shown on key). What would you tell these parents is the probability that their child will be affected if it is a girl? 1/16 If it's a boy? 1/16 2 4.) Consider the following pedigree: a. Could this trait be recessive with incomplete penetrance? Yes If yes, estimate the maximum percent penetrance based on the information given. 8 people show the trait + 4 more (all unaffected people in 3rd generation) must also be homozygous mutant, but don't show the trait = 12 homozygotes, 8/12 show trait 2/3 b. Could it be dominant with incomplete penetrance? Yes If yes, estimate the maximum percent penetrance based on the information given. 8 people show the trait + 2 more (one of the grandparents on each side of the family) must have the mutation but don't show the trait = 10 have mutation, 8/10 show trait 4/5 c. If this deafness trait shows variable expressivity in addition to incomplete penetrance, describe in one or two sentences how you would expect this to manifest itself. Variable expressivity could manifest itself as a variable degree of deafness among affected individuals, or variability in the age of onset, or variability in whether deafness is unilateral (one ear) or bilateral (two ears). These are not necessarily the only possible correct answers... 5.) 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 white buff ivory buff variable variable variable variable 3 vermillion red red red vermillion brown red red red red brown white buff variable vermillion brown 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, and variable Group 2: vermillion A total of 3 genes are affected. Group 3: brown c. 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 5. pleiotropy 4. incomplete penetrance 6. co-dominance 5). Wild-type snapdragons have long, personate (asymmetric) flowers and are true breeding for this shape. An gardening enthusiast who likes to cultivate new, interesting variants brings you seeds from a true breeding line she has established that forms short, peloric (symmetric) flowers. Describe the series of crosses you would do to determine whether short, peloric flowers are due to a single, pleiotropic mutation in a gene controlling both flower length and asymmetry, or two different mutations, one of which is responsible for shortness and the other for peloria. Clearly indicate the expected outcome for each possibility (phenotypic progeny classes and their proportions), assuming that the mutation(s) responsible for shortness and peloria are recessive. Then answer the question again with the assumption that the mutation(s) responsible for shortness and peloria are dominant. shortness and peloria RECESSIVE: One pleiotropic gene: long personate x short peloric -> In F1 expect all long personate self-> In F2 expect 3/4 long personate, 1/4 short peloric Alternatively, F1 long personate x short peloric -> In F2 expect 1/2 long personate, 1/2 short peloric Two genes: long personate x short peloric -> In F1 expect all long personate self-> In F2 expect 9/16 long personate, 3/16 short personate, 3/16 long peloric, 1/16 short peloric Alternatively, F1 long personate x short peloric -> In F2 expect 1/4 long personate, 1/4 long peloric, 1/4 short personate, 1/4 short peloric shortness and peloria DOMINANT: One pleiotropic gene: Long personate x short peloric -> In F1 expect all short peloric self-> In F2 expect 3/4 short peloric, 1/4 long personate Alternatively, F1 short peloric x long personate -> In F2 expect 1/2 short peloric, 1/2 long personate Two genes: Long personate x short peloric -> In F1 expect all short peloric self-> In F2 expect 9/16 short peloric, 3/16 long peloric, 3/16 short personate, 1/16 long personate Alternatively, F1 short peloric x long personate -> In F2 expect 1/4 long personate, 1/4 long peloric, 1/4 short personate, 1/4 short peloric 4 ...
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This note was uploaded on 04/17/2008 for the course BICD 100 taught by Professor Nehring during the Spring '08 term at UCSD.

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