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MIT 7 03 - PROBLEM SETS

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Problem SetsFall 2000%2. Consider the following mouse breeding experiment involving two different rare traits. A_ male mouse with both traits is crossed to a normal female and all of the offspring appearnormal. A female offspring from this cross is mated multiple times to a normal male to produceseveral litters of offspring. A total of 32 offspring are scored as having the followingcharacteristics:16 normal females6 normal males2 males with trait 11 male with trait 2•7 males with both traits(a) What is the mode of inheritance of each of the two traits? Explain your reasoning.(b) Use the chi-square test to determine whether the two traits appear to be linked. For thistest, you are trying to determine whether or not an expectation that the two traits are unlinkeddiffers significantly from the observed data. Note that there are a number of different ways toset up this test, but there is one best way to test for linkage. Show your work and use the tablebelow which gives p values as a function of chi square values and degrees of freedom. Foryour final answer use a p value < 0.05 as the cut-off for significant deviation from expectation.p value: .995 .975 0.9 0.5 0.1 0.05 0.025 0.01 0.005df = 1 .000 .000 .016 .46 2.7 3.8 5.0 6.6 7.9df = 2 .01 .05 .21 1.4 4.6 6.0 7.4 9.2 10.6df = 3 .07 .22 .58 2.4 6.3 7.8 9.3 11.3 12.8i(c) Based on the data give your best estimate of the distance between the genes for trait 1 andtrait 2.3. The producers of a soap opera have hired you as a consultant. The story line includes twofamilies, each with individuals that have a rare trait. The families are diagramed below--individuals are numbered and those expressing the trait are represented by the filled symbols.The scriptwriters are contemplating a number of different couplings between individuals in thetwo families. Because they are concerned with the genetic accuracy of the story they want youto figure out what the offspring from each possible mating might be like.1 2 6i__4 (_)7 8(a) Assume that the rare trait is recessive. Consider the possible matings given below. Foreach, calculate the probability that the child will have the rare trait.Female 2 and Male 5Female 6 and Male 4Female 7 and Male 4Female 3 and Mate 8(b) Now assume that the rare trait is dominant. Again for each of the possible matings givenbelow, calculate the probability that the child will have the rare trait.Female 2 and Male 5Female 6 and Male 4Female 7 and Male 4Female 3 and Male 8(c) Finally, assume that the rare trait is X-linked. For each of the possible matings givenbelow, calculate the probability that the child will have the rare trait. Explicitly give each prob-ability in the cases where the probabilities for a boy. or a girl having the trait differ.Female 2 and Male 5Female 6 and Male 1Female 7 and Male 4Female 3 and Male 8ProblemSet1Solutions.irla. Those mutants that make tan colonies when crossed to another mutant can beassumed to carry recessive mutations, because the white phenotype is not present in aheterozygote. Those mutants that never make tan colonies as a diploid likely carrydominant mutations, although we can not completely rule out the possibility that thesemutants carry mutations in multiple genes (discussed in-part C). Thus, mutations 1, 2, 4,5, 6, 7, 8, 9, 1O, and 12 are recessive. Mutations 3 and 11 are most likely dominant.lb. Based on non-complementation of the recessive mutations, we can conclude that 1,4, 5, 9, and 10 form one complementation group (Group A) and are mutations in the samegene. Likewise, 2 and 7 fail to complement and are members of a secondcomplementation _oup (Group B) representing mutations in a second gene. Mutant 6complements members of both Group A and Group B and therefore represents a thirdgene. Mutants 8 and 12 both complement Group A and Group B mutants as well asmutant 6 and therefore represent at least one more gene and possibly two genes. Wehave no data regarding the phenotype of the 8 x 12 diploid and can't determine whetherthey are mutations in the same or different genes. Taken together, we can say that therecessive mutations represent at least 4 genes.:_i We are unable to make any conclusions regarding the number of new genesrepresented by the dominant mutants, 3 and 8. It is possible that 3 and 8 are mutations inthe same gene. It is also possible that one or both of these mutations is in one of thegenes that we recovered a recessive mutation in above.It is also possible that the white phenotype in one or more of the mutants is causedby mutations in more the one gene. Thus, the best answer for the number of mutantsrepresented is at least 4.lc. The most obvious ambiguity is whether 8 and i2 are mutations in the same gene.This could be resolved by generating the 8 x 12 diploid. Note that to do this you wouldhave to sporulate, for example the 8 x wild type diploid, and select for a haploid spore ofthe appropriate mating type that shows the white phenotype.A second ambiguity is whether 3 and 11 represent unique genes. One way toapproach this question would be to generate the-3 x 11 diploid, sporulate, and look at thesegregation pattern of the white phenotype in the resulting haploids. If any of thehaploids form tan colonies then 3 and 11 are likely to be mutations in different genes.A third ambiguity is whether or not the white phenotype in each case is due to amutation in a single gene. To determine this, you would cross each mutant to wild type,. sporulate the resulting diploids, and look for 2 tan : 2 white segregation in the haploids.I_. Any other segregation pattern is inconsistent with the white phenotype being caused bymutation iJfa single gene" An extreme example of this possibility would be if a singlestrain carried recessive mutations in genes representing all of the complementation•_oups identified. Such a mutant would appear "dominant" by the type of crossesperformed in the table, because it would fail to complement every other mutant. Toavoid this possibility, you would cross each mutant to a wild type haploid and look at thediploid phenotype.2a. Both traits are X-linked recessive. They must be recessive because all F1 offspringappear normal. We can deduce that both traits are X-linked because, of the 32 progenyshown, all females are normal and only males are affected.2b. To use the Chi Square test, we first need to formulate a hypothesis that can be tested.In this case an appropriate hypothesis would be: The genes for trait 1 and trait 2 areunlinked and


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MIT 7 03 - PROBLEM SETS

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