Problem SetsFall 1999mutants of mating type o_-.._ 1 2 3 4 51 + + -- + --2 + -- + --mutants ofmatingtypea 3 + + --4. + +5 +Give as much information as you can about your new canavanine resistant mutations.Indicate which mutations are dominant and which are recessive also state how many genesare representedandwhich mutationsliethe samegene. Again assumeeachstrain carriesonly a single mutation.2. A hypotheticalinsecthas blue eyes, but mutant insectsthat can not form blue pigmenthave white eyes. Productionof blue pigment involvesthe activities of enzymeA andenzyme B, both encoded by autosomal genes. First, suppose that the pathway forproduction of blue pigment involves enzyme A and enzyme B operating in series:Enzyme A Enzyme B_ Bluepigment(a) A true breeding strain with a recessive mutation in the gene for enzyme A is crossed toa true breedingstrainwith a recessivemutationin the gene forenzyme B. Will the resultingF1 progeny have blue or white eyes? When these F1 insects are then crossed amongthemselves, what will the phenotypic ratio of blue to white eyed insects be among the F2?I(b) A true breeding strain with a dominant mutation in the gene for enzyme A is crossed toa true breedingstrainwith a dominantmutationinthe genefor enzymeB. Willthe resultingF1 progeny have blue or white eyes? When these F1 insects are then crossed amongthemselves, what will the phenotypicratio of blue to white eyedinsects be among the F2?(c) A true breeding strain with a recessive mutation in the gene for enzyme A is crossed to"J a true breeding strain with a dominant mutation in the gene for enzyme B. Will the resultingF1 progeny have blue or white eyes? When these F1 insects are then crossed amongthemselves,what will the phenotypic ratioof blue to white eyed insects be among the F2?Nowsupposethat enzyme A and enzyme B act in parallel. That is,there are two differentways to make blue pigment and white-eyedinsects only result when the steps carried outby both enzyme A and enzyme B are inactive.Enzyme A>Blue pigmentEnzyme B>(d) A true breeding strain with a recessive mutation in the gene for enzyme A is crossed toa true breedingstrainwith a dominantmutationinthe gene for enzyme B. Giventhe parallelpathway model, will the resulting F1 progeny have btueor white eyes? When these F1insects are then crossed among themselves, what will the phenotypic ratio of blue to whiteeyed insects be among the F2?(e) Let's say that you are trying to distinguish the series model in part (c) from the parallelpathway model inpart (d),but you decide to lookat only eight flies from the F2 generation.Apply the chi-square test to all nine possible phenotypic ratios for eight flies to determinewhich observed ratios are consistent with the expected ratios for each of the two models(Usethe criteriathatthe hypothesiscan be rejectedonly if the p value is < 0.05). Howmany of the possible outcomes will not definitively distinguish the two models?(f) Finally,suppose that the gene for enzymeA is on the X-chromosome and that the genefor enzymeB is autosomal. Both males and femalesfrom a true-breedingstrainwith arecessive mutation in the gene for enzyme A are crossed to females and males from a truebreedingstrain witha dominantmutationin the gene for enzymeB. Given the parallelpathway model and the sex linkage of gene A, will the F1 progeny have blue or whiteeyes? (Specifymales or females). When the F1insectsare then crossed amongthemselves,what will the phenotypic ratio of blue to whiteeyed insectsbe among males inthe F2? What will the phenotypic ratio be for femalesin the F2?3, You have just been hired as a genetic counselor for a royaJ family that still engages in a.._. significantamountofinbreeding. As your firstassignmentyou are presentedwiththefollowing pedigree where the filled symbol represents a male in the royal family who has arare recessive disease.?Your job is to calculate the probability that the child indicated by ? will have the disease. Todo this, assume that no new mutations arise within the pedigree and that no unrelatedindividualis a carrier(becausethis is a very raredisease).(a) tf the disease is caused by an autosomal recessive allele, what is the probability thatthe child indicated by ? will have the disease?(b) If the disease is caused by an X-linkedrecessiveallele,what isthe probabilitythat ason will havethe disease? What is the probabilitythat a daughter will have the disease?(c) If the disease is caused by an autosomat recessive allele, and the first child has thedisease, what is the probability that the second child will have the disease?7.03 Problem Set #1 SolutionsFA1999la.Mutant 3 fails to complement with any other strain, but carries only a single mutation. It is likelythat this strain carries a dominant mutation,Mutants 1 and 5 are recessive mutations in the same complementation group. They are probablyalleles of the same gene.Mutant 2 is a recessive mutation in its own complementation group, as it complements the rest ofthe mutants. It is in a separate gene from the rest.Mutant 4 is a recessive mutation in its own complementation group, in a separate gene from therest.Our set of strains represent mutations in 3 or 4 different genes, with the ambiguity due to mutant3 being dominant. By this test, we can not determine in which gene mutant 3 is in.lb.Mutant 4 fails to complement with any other strain, but carries only a single mutation. It is likelythat this strain carries a dominant mutation.Mutants 1 and 2 are recessive mutations in the same complementation group. They are probablyalleles of the same gene.Mutant 3 is a recessive mutation in its own complementation group, as it complements the rest ofthe mutants. It is in a separate gene from the rest.Mutant 5 is a recessive mutation in its own complementation group, in a separate gene from therest.---- Our set of strains represent mutations in 3 or 4 different genes, with the ambiguity due to mutant4 being dominant. By this test, we can not determine in which gene mutant 4 is in.2a.All the F I progeny will have blue eyes. Both mutations are recessive to the wild-type allele, andthus, the two strains will complement each other in the F I generation. Using the punnet squaregives a ratio of 9:7 blue to white eyed progeny for the F: generationAB Ab aB abAB AABB AABb AaBB AaBbAb AABb AAbb* AaBb Aabb*aB AaBB AaBb aaBB* aaBb*ab AaBb Aabb* aaBb* aabb*Only progeny with either 2 "a" alleles or 2 "b" alleles (or both) will produce white eyed flies.2b.All the F1
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