7.03 Genetics: Answers to Problem Set 1 (2002)

1 Answers: 7.03 Problem Set 1 1. You and your lab partner have isolated 20 new mutant yeast strains that are defective in synthesis of threonine, an amino acid. These Thr- mutants do not grow on minimal medium, but they do grow on minimal medium supplemented with threonine. Ten of your Thr- mutants (numbered 1 through 10) were isolated in a strain of mating type a (MAT a). The other 10 Thr- mutants (numbered 11 through 20) were isolated in a strain of mating type α (MAT α). You and your lab partner cross each of the MAT a strains to each of the MAT α strains, and you include crosses to the appropriate wild-type strains. Your experimental observations are shown in the table below, where (-) indicates diploids that did not grow on minimal medium and (+) indicates diploids that did grow on minimal medium. (a) Unfortunately, when all the data were collected and the plates discarded, your dog munched on your notebook, swallowing some data. From the data that remains, see if you can reconstruct the full table. strains of mating type α wild-type strains of mating type a 1 2 3 4 5 6 7 8 9 10 wild-type 11 + + + + + + + + + + + - + + - + + + 12 + - + + + + - - + - + 13 + - + + + + - - + - + 14 - - - - - - - - - - - 15 + + - + + - + + + - + 16 + + + + + + + + - - 17 + + + + - - - + 18 + - - + - + 19 + + + + + + + - - + 20 + + + + + - + - + + + + + + + - - + + - - - + + Note: The table was filled in assuming that there are four complementation groups (see part d). (b) Which mutations are recessive? All except 9 and 14. 2 (c) Which mutations are dominant? 9 and 14. (d) Which mutations do you know to be in the same gene? [1, 6, 7, 12, 13, 18] [2, 5, 11, 15, 20] [3, 10, 16] [4, 8, 17, 19] However, there are no data in the table that exclude the possibility that the third and fourth groups listed above actually form one large complementation group, i.e. [3, 10, 16, 4, 8, 17, 19]. Thus there are either three or four complementation groups. (e) Based on these experiments, what is the minimum number of genes required for threonine synthesis? Four (or three; see d above) (if the dominant mutations 9 and 14 are in genes represented by the complementation groups). This is the number of complementation groups. (f) What is the maximum number of genes that these 20 mutants could represent? Six (or five; see d above) (if mutations 9 and 14 are in genes not represented by the complementation groups). This is the number of complementation groups plus the number of dominant mutations. 2. Being a well-rounded geneticist, you also maintain a colony of chickens. In a cross between a female chicken from a true-breeding strain with three toes per foot and a male from a true-breeding strain with seven toes per foot, all of the F1 progeny have five toes per foot. (a) Based on the information that you have at this stage, is it possible that a single gene determines the differences in toe count among the two parental strains and progeny? If so, what toe counts should appear in the F2 generation and at what frequencies? One gene exhibiting incomplete dominance could be responsible for the observed data. For example: AA – seven toes Aa – five toes aa – three toes The progeny of a mating between five-toed chickens (Aa) would yield a variant on the typical 1:2:1 phenotypic ratio: 25% seven-toed (AA), 50% five-toed (Aa), and 25% three-toed (aa). (b) In fact, when F1 chickens are crossed among themselves, the following F2 progeny are produced: 46 chickens with five toes per foot, 30 chickens with four toes per foot, 28 3 chickens with six toes per foot, 8 chickens with seven toes per foot, and 7 chickens with three toes per foot. Propose a genetic model to account for the existence of the 5 phenotypic classes and their observed frequencies. In your answer, provide the genotypes of the parental chickens, the F1 chickens, and each class of F2 chickens. The spectrum of phenotypes suggests that more than one gene is involved, and the appearance of phenotypes not seen in either the P or F1 generations suggests incomplete dominance. A model consistent with the data invokes two genes (e.g., A and B), both exhibiting incomplete dominance, which act additively to determine toe number. The basal number of toes, given by phenotype aabb, is three toes, and each “big” allele contributes one additional toe. For example: P F1 F2 AABB (seven toes) × aabb (three toes) ↓ AaBb (five toes) ↓ 1/16 AABB – seven toes 1/4 AaBB, AABb – six toes 3/8 AaBb, AAbb, aaBB – five toes 1/4 Aabb, aaBb – four toes 1/16 aabb – three toes (c) Use the Chi-square test to show that the observed frequencies fit with the expected frequencies based on your model. For your answer, give the observed and expected phenotypic ratios, the degrees of freedom, your calculated value for χ2, and a rough estimate of the p value. Observed Expected (Obs-Exp) (Obs-Exp)2/Exp Three toes 7 7.44 0.44 0.026 Four toes 30 29.75 0.25 0.002 Five toes 46 44.6 1.375 0.042 Six toes 28 29.75 1.75 0.103 Seven toes 8 7.44 0.56 0.042 0.215 Total 119 119 2 χ = 0.215, and degrees of freedom = [(5 classes) –1] = 4. From the chi-square table, p ˜ 0.9. Thus we are unable to reject the hypothesised model. 3. Your friends Ben and Jane are contemplating having children and seek your genetic counsel. Ben and Jane are unrelated, but both have younger brothers with the same, extremely rare genetic trait. (In fact, Ben and Jane met at a support group meeting for 4 children and families of children with this rare disorder.) Ben’s parents and Jane’s parents are all unaffected. (a) Assume that the trait is autosomal recessive with complete penetrance. What is the probability that Ben and Jane’s first child will be affected with the trait? If not affected, what is the probability that the child is a carrier? Aa Aa Aa Aa ? Let’s call the recessive allele for the trait ‘a’ and its wild type counterpart ‘A.’ We know that the genotype of the grandparents must be Aa because both Ben and Jane have brothers who are affected. Therefore: p(child will be affected) = p(Jane is Aa) x p(Ben is Aa) x p (child is aa) The probability that either parent is Aa is 2/3 because we already know that they are not affected (conditional probability). Therefore: p(child will be affected) = 2 x 2 x 1 = 1 3 3 4 9 For the next question let’s calculate the probability that the child is a carrier first. p(child is a carrier) = p(Jane is AA)x p(Ben is Aa) x p(child is Aa) + p(Ben is AA)x p(Jane is Aa) x p(child is Aa) + p(Jane is Aa) x p(Ben is Aa) x p(child is Aa) = 1/3 x 2/3 x 1/2 + 1/3 x 2/3 x 1/2 + 2/3 x 2/3 x 1/2 = 4/9 Since we know that the child is not affected the new total is: 1 – p(child will be affected) = 1 - 1/9 = 8/9 Therefore, given that it is not affected, p(child is a carrier) is 4/9 = 1/2 8/9 5 (b) Assume that the trait is autosomal recessive with 80% penetrance. What is the probability that Ben and Jane’s first child will be affected? Since the trait is now autosomal recessive with 80% penetrance, we cannot longer assume that both Ben and Jane are Aa. Jane and Ben can be: - AA = 1/4 - Aa = 1/2 - aa not expressing the trait (represented as aanot) = 1/4 x 20% = 1/20 Since we know that both parents do not express the trait in question, the new total is the sum of probabilities of the different genotypes Jane and Ben can be: 1/4 + 1/2 + 1/20 = 16/20 Therefore the probabilities of the different genoytpes that Jane and Ben can be (given that they do not express the trait) are: p(Aa) = 1/2__ = 5/8 16/20 not p(aa ) = _1/20_ = 1/16 16/20 Therefore: p(child is aa) = p(Jane is Aa & Ben is Aa) x p(child is aa) + p(Jane is aanot & Ben is Aa) x p(child is aa) + p(Jane is Aa & Ben is aanot) x p(child is aa) + p(Jane & Ben are aanot) x p(childis aa) = (5/8 x 5/8 x 1/4 x 80%) + 2(1/16 x 5/8 x 1/2 x 80%) + (1/16 x 1/16 x 1 x 80%) = 9/80 (c) Assume that the trait is X-linked recessive with complete penetrance. What is the probability that Ben and Jane’s first child will be affected? What is the probability that the first child will be a carrier? XAXa XAY XAXa ? XAY 6 Since the trait is now X-linked recessive, males will be affected (XaY) and females will most likely be carriers (XAXa). Therefore: p(child is a carrier) = p(Jane is a carrier) x p(child is a carrier) = 1/2 x 1/4 = 1/8 p(child is affected) = p(Jane is a carrier) x p(child is XaY) = 1/2 x 1/4 = 1/8 (d) (extra credit) Assume that the trait is autosomal dominant with 80% penetrance in heterozygotes and 100% penetrance in homozygotes. What is the probability that Ben and Jane’s first child will be affected? If not affected, what is the probability that the child is a carrier? Aanot aa aa Aanot ? Since the trait is now autosomal dominant with 80% penetrance and is extremely rare, we assume that only one of the grandparents from each family is a carrier (heterozygote not expressing the trait) and that the other one is wild type. Therefore Ben and Jane could either be Aanot (heterozygous for the trait but not affected), or wild type, denoted by aa (note that since the trait is now dominant, ‘A’ is being used to denote the trait in question and ‘a’ its wild type counterpart). Jane and Ben can be: p(Aanot) = 1/2 x 20% = 1/10 p (aa) = 1/2 Since we know that both parents do not express the trait in question, the new total is the sum of probabilities of the different genotypes that Jane and Ben can be: 7 1/10 + 1/2 = 6/10 Therefore the probabilities of the different genotypes that Jane and Ben can be, given that they do not express the trait are: p(Aanot) = 1/10 = 1/6 6/10 p (aa) = 1/2 = 5/6 6/10 Therefore: p(child will be affected) = p(Jane is Aanot & Ben is Aanot) x p(child is AA) + p(Jane is Aanot & Ben is Aanot) x p(child is Aa) + p(Jane is Aanot & Ben is aa) x p(child is Aa) + p(Ben is Aanot & Jane is aa) x p(child is Aa) = 1/6 x 1/6 x 1/4 x 100% + 1/6 x 1/6 x 1/2 x 80% + 2(1/6 x 5/6 x 1/2 x 80%) = 31/240 For the next question let’s calculate the probability that the child is a carrier first. p(child is a carrier) = p(Jane is Aanot & Ben is Aanot) x p(child is Aanot) + p(Jane is Aanot & Ben is aa) x p(child is Aanot) + p(Jane is aa & Ben is Aanot) x p(child is Aanot) = 1/6 x 1/6 x 1/2 x 20% + 2(1/6 x 5/6 x 1/2 x 20%) = 11/360 Since we know that the child is not affected the new total is: 1- 31/240 = 209/240 Therefore, if not affected, p(child is a carrier) is 11/360 209/240 = 2/57