7.03 Genetics: Problem Set 1 Solutions (2000)

ProblemSet1 Solutions. ir la. Those mutants that make tan colonies when crossed to another mutant can be assumed to carry recessive mutations, because the white phenotype is not present in a heterozygote. Those mutants that never make tan colonies as a diploid likely carry dominant mutations, although we can not completely rule out the possibility that these mutants 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, 4, 5, 9, and 10 form one complementation gene. Likewise, 2 and 7 fail to complement complementation complements we can conclude that 1, group (Group A) and are mutations in the same and are members of a second _oup (Group B) representing mutations in a second gene. Mutant 6 members of both Group A and Group B and therefore represents a third gene. Mutants 8 and 12 both complement Group A and Group B mutants as well as mutant 6 and therefore represent at least one more gene and possibly two genes. We have no data regarding the phenotype of the 8 x 12 diploid and can't determine whether they are mutations in the same or different genes. Taken together, we can say that the recessive mutations represent at least 4 genes. :_i We are unable to make any conclusions regarding the number of new genes represented by the dominant mutants, 3 and 8. It is possible that 3 and 8 are mutations in the same gene. It is also possible that one or both of these mutations is in one of the genes that we recovered a recessive mutation in above. It is also possible that the white phenotype in one or more of the mutants is caused by mutations in more the one gene. Thus, the best answer for the number of mutants represented 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 would have to sporulate, for example the 8 x wild type diploid, and select for a haploid spore of the appropriate mating type that shows the white phenotype. A second ambiguity is whether 3 and 11 represent unique genes. One way to approach this question would be to generate the-3 x 11 diploid, sporulate, and look at the segregation pattern of the white phenotype in the resulting haploids. If any of the haploids 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 a mutation in a single gene. To determine this, you would cross each mutant to wild type, . I_. sporulate the resulting diploids, and look for 2 tan : 2 white segregation in the haploids. Any other segregation pattern is inconsistent with the white phenotype being caused by mutation iJf a single gene" An extreme example of this possibility would be if a single strain carried recessive mutations in genes representing all of the complementation •_oups identified. Such a mutant would appear "dominant" by the type of crosses performed in the table, because it would fail to complement every other mutant. To avoid this possibility, you would cross each mutant to a wild type haploid and look at the diploid phenotype. 2a. Both traits are X-linked recessive. They must be recessive because all F1 offspring appear normal. We can deduce that both traits are X-linked because, of the 32 progeny shown, 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 are unlinked and segegate randomly. Now we must choose our classes for the Chi Square test. First, note that since all F2 females are normal, we can tell nothing about their genotype and must disregard them. The best way to test for linkage is to group X chromosomes of the F2 males as either parental or recombinant based on the an'angement of alleles for traits 1 and 2. For example, the F1 female is a carrier for both traits and both traits are present on the same X chromosome, so her genotype can be represented as Xt'aX. Thus, there are two parental chromosome types: X and X 1'2. The other two possible chromosome types (X 1 and X 2) must be due to a crossover between the genes for traits 1 and 2 and are recombinant. The Chi Square table would look as follows: Chromosome Observed Expected Type Parental 13 8 Recombinant 3 8 The observed number of parental chromosomes- is the sum of the number of normal males and the number of males expressing both traits. The observed number of recombinant chromosomes is the sum of the number of males expressing only trait 1 and the number of males expressing only trait 2. Based on our hypothesis, if both traits are segregating randomly, then parental and recombinant equal frequency. chromosomes would be expected to arise with .° _ m Thus, Z_ =_, (O-E) z__ E 2 - 52 +-50-=6.25 8 8 df = l Using the Chi Square table with df=l, 0.025> p > 0.01. Therefore, an expectation the two traits are unlinked differs significantly from the observed data. that An alternative method for solving this problem would be to consider each chromosome or genotype type as a different class. While this method is acceptable, it is less robust than grouping by parental and recombinant. This is because as the number of classes increases, the class sizes decrease and the degrees of freedom increase. With the Chi Square test it is important to have as large a data set for each group as possible. As shown below, this method results in not being able to reject the hypothesis. Male X Genotype X Q_ K' Z a = za Observed 6 Expected 4 X1 2 4 Xz 1 4 X 12 7 4 (O - E) a - 2a +--+--+--=6.5 22 32 3 2 E 4 4 4 4 dr=3 In this case, using the Chi Square table with df=3, 0.1> p > 0.05. Therefore, expectation that the two traits are unlinked does not differ significantly observed data. 2c. Number of recombinants = number of non-parental chromosomes an from the = 3. m.u.: 100(_61 = 18.75 The map distance between the genes for trait I and 2 is approximately _I__ " 18.75 cM. Recessive. Person 1 A=wt a=mutant. Genotype aa Cross • 2x5 P(aa) 1/2 2 Aa 6x4 1/4 3 aa 7x4 1/4 4 Aa 3x8 5 aa 6 Aa 7 As 8 aa 1 Dominant. A = wt A a°m = dominant mutant Person J_ Genotype Ad°mA 2 Cross 2x5 AA " P(Aa°mA or Ad°mAd°m) 1/2 6x4 0 7x4 0 3x8 3/4 I 3 4 5 A_°mA AA Aa°mA 6 AA 7 AA 8 A_°mA X-linked. X A "-" wild type X a = mutant. Note that the X-linked trait must be recessive, tf the trmt were X-linked dominant then progeny of individual #5 would show the trait. Person 1 Genotype XW Cross Male p(xay) Female 2 xax A 3 XaX_ 2x5 1/2 1/2 4 xAy 6X4 1/2 1/2 5 XaY 7x4 1/2 0 6 xax A 3x8 7 XaX A 8 X_Y i P(X"X a) 1