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MIT 7 03 - Problem Set #1

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2005 7.03 Problem Set 1Due before 5 PM on WEDNESDAY, September 21, 2005.Turn answers in to the box outside of 68-120.PLEASE WRITE YOUR ANSWERS ON THIS PRINTOUT.1. Wild-type flies are brown in color. You have discovered two genes that control bodycolor in flies -- gene A and gene B, which are on different autosomal chromosomes. Youhave three true-breeding mutant strains, all of which have black bodies.Strain One (A*/A*) is homozygous for a dominant mutation in gene A.Strain Two (A–/A–) is homozygous for a recessive mutation in gene A.Strain Three (B–/B–) is homozygous for a recessive mutation in gene B.Group Four are flies that result from mating Strain One to Strain Two.Group Five are flies that result from mating Strain One to Strain Three.Group Six are flies that result from mating Strain Two to Strain Three.Group Seven are flies that result from mating Strain One to wild-type.Group Eight are flies that result from mating Strain Two to wild-type.Group Nine are flies that result from mating Strain Three to wild-type.Predict the phenotypic ratio (the numerical ratio and the phenotype of each phenotypic class)of the offspring resulting from a cross between:(a) Group Five and Group Nine(b) Group Six and Group Eight(c) Group Four and Group Seven7.03 Fall 2005 Problem Sets & Answers1 of 1159/5/072(d) Strain Three and Group Four(e) Strain Two and Strain Three(f) Strain One and Strain Two(g) List all of the above six crosses (parts (a) – (f)) that are proper complementation testswhich clearly reveal whether two mutations are in the same gene or different genes.2. You are studying a type of yeast that has two different chromosomes in its genome. Youhave isolated three mutations, “a,” “b” and “d,” each of which causes the same phenotype.When you mate a strain containing any one of these three mutations to wild-type, theresulting diploid exhibits the wild-type phenotype. You are in the process of doingcomplementation tests with these mutants. You discover that “a” and “b” do complementeach other, but “a” and “d” do not. The corresponding wild-type alleles are “A,” “B” and “D.”Draw in the correct alleles that exist at each of these loci (A, B, and D) in each of the nineyeast cells drawn below. Make sure to put the alleles in their correct locations, asdetermined by those already drawn in for you. Also make sure to draw in the chromosomesto any cell whose chromosome(s) is/are missing.First cross: You mate haploid yeast of genotype “a” to haploid yeast of genotype “b.”X=ba7.03 Fall 2005 Problem Sets & Answers2 of 1159/5/073Second cross: You mate haploid yeast of genotype “a” to haploid yeast of genotype “d.”Third cross: You mate haploid yeast of genotype “b” to haploid yeast of genotype “d.”3. You are studying the inheritance of feather color in a new species of bird. You cross atrue-breeding dark green bird to a true-breeding pale green bird. All of the resulting F1 birdsare medium green. You then cross two medium green F1 birds, and analyze the resultingF2 generation. You obtain 50 birds: 13 are dark green, 23 are medium green, and 14 arepale green. We have not covered linkage and sex-linkage yet, so do not take thoseconsiderations into account during this problem.(a) Propose a one-gene genetic model that explains the inheritance of feather color in thisbird that is consistent with these results. By “propose a genetic model,” we mean define allpossible genotypes and their associated phenotypes. Then give the genotypes of the birdsin each generation of each cross described.X=X=7.03 Fall 2005 Problem Sets & Answers3 of 1159/5/074(b) You cross a true-breeding dark green bird to a true-breeding blue bird. All of the resultingF1 birds are blue. You then cross two blue F1 birds, and analyze the resulting F2generation. You obtain 50 birds: 37 are blue, 4 are dark green, 7 are medium green, and 2are pale green.Propose a two-gene genetic model that explains the inheritance of color in this bird that isconsistent with all of the data in this problem.(c) Use chi-square analysis to test whether the numbers of F2 progeny you saw in part (b)correlate with the expected numbers based on your genetic model. For the chi square test youdo, give the numbers of observed and expected organisms in each phenotypic class, thedegrees of freedom, and your calculated value for χ2. Also state what your conclusion is based onthe results of this chi-square test.p value:.995.9750.90.50.10.050.0250.010.005df = 1.000.000.016.462.73.85.06.67.9df = 2.01.05.211.44.66.07.49.210.6df = 3.07.22.582.46.37.89.311.312.87.03 Fall 2005 Problem Sets & Answers4 of 1159/5/075(d) Based on your model, how many different results might you expect to get if you crossedone randomly selected light green bird from the F2 to one randomly selected blue bird fromthe F2? For each possible result, state the phenotypic ratio present in the progeny obtained.4. This problem deals with the following pedigree, which shows the inheritance of a veryrare trait.?= female= male(a) Assume that the inherited disorder is expressed with complete penetrance and thatthere are no new mutations. What mode(s) of inheritance is/are consistent with thispedigree? (Your choices are: X-linked dominant, X-linked recessive, autosomalrecessive, autosomal dominant.)127.03 Fall 2005 Problem Sets & Answers5 of 1159/5/076(b) For each consistent mode of inheritance, what are the probabilities that Individuals 1and 2 will have:… an affected son?… an affected daughter?… an unaffected son?… an unaffected daughter?(c) Use Bayes' theorem to calculate the probability that the next child of Individuals 1 and 2will be affected with the disorder, given the new knowledge that the couple already has twohealthy sons. Do this calculation for each mode of inheritance consistent with the pedigree.7.03 Fall 2005 Problem Sets & Answers6 of 1159/5/072005 7.03 Problem Set 1 ANSWER KEY1. Wild-type flies are brown in color. You have discovered two genes that control bodycolor in flies -- gene A and gene B, which are on different autosomal chromosomes. Youhave three true-breeding mutant strains, all of which have black bodies.Strain One (A*/A*) is homozygous for a dominant mutation in gene A. A*A* B+B+ blackStrain Two (A–/A–) is homozygous for a recessive mutation in gene A. A–A– B+B+ blackStrain Three (B–/B–) is homozygous for a recessive mutation in gene B.


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MIT 7 03 - Problem Set #1

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