1 7.03 Problem Set 1Due before 5 PM on Thursday, September 20, 2001 Hand in answers in recitation section or in the box outside the class 1. You have isolated 10 new mutant yeast strains that are defective in synthesis of leucine, an amino acid. These Leu- mutants do not grow on minimal medium, but they do grow on minimal medium supplemented with leucine. Your Leu-mutants (numbered 1 through 10) were all isolated in a strain of mating type a (MAT a). As it turns out, your high school classmate, now at Harvard, has independently isolated 10 new Leu-mutants (numbered 11 through 20), all in a strain of mating type a (MAT a). You and your high school classmate decide to combine your resources and determine how many different genes are represented by your 20 mutant strains. You cross each of the MAT a strains to each of the MAT a 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. strains of mating type a wild-type 1 2 3 4 5 6 7 8 9 10 wild-type 11 + + + + --+ + + + + -+ -+ + + + + + + + 12 + + - + + + + + + + + 13 + + - + - + + + + + + strains of mating type a 14 15 16 + + -+ + ----+ + -+ + -+ --+ --+ + -+ + -+ + --+ -17 + + - + + + + + + + + 18 + + - - + + + + - + + 19 + - - + + + + + + + + 20 + + - + + + + + + + + (a) What property do mutants 2 and 16 share? (b) Which mutations do you know to be in the same gene?2 (c) Could mutations 3 and 16 be in the same gene? (d) Based on these experiments, what is the minimum number of genes required for leucine synthesis? (e) What is the maximum number of genes that these 20 mutants could represent? 2. Being a well-rounded geneticist, you also maintain a colony of mice. (a) One day you spot a mouse in your colony with a novel and interesting phenotype: a kinked tail. You breed the kinked-tail mouse (a male) with several wild-type females and observe that about half of the offspring (both males and females) have kinked tails and half have normal tails. Is the kinked-tail mutation dominant or recessive to wild type? (b) When two of the kinked-tail offspring from part (a) are crossed, what fraction of the resulting mice would you expect to have kinked tails? (c) When you cross kinked-tail offspring from part (a), you find that one third of the resulting kinked-tail males produce no sperm and thus are sterile. The other two thirds of the resulting kinked-tail males (and all of the normal-tail males and all of the females) are fertile. Propose a model to account for these findings. (d) Not to be outdone, your high school classmate informs you that he has isolated a pure-breeding mouse strain in which males produce no sperm but have normal tails, and in which females are phenotypically normal (fertile; normal tails). You explain to your high school classmate that this is impossible. Why? (e) Eventually your classmate convinces you that the male sterile mutant that he has discovered displays autosomal recessive inheritance. You decide to test whether your male-sterile mutant and your classmate’s male-sterile mutant are in the same gene. Diagram the series of crosses required to conduct an appropriate complementation experiment, including the expected ratios (in the final generation) of infertile to fertile males for two competing models: 1) two mutations in the same gene and 2) mutations in two different genes on different chromosomes.3 (f) Breeding experiments involving the two mutant strains are simplified when your classmate discovers that his mutant, with recessive male sterility, also displays a dominant phenotype in both sexes: short tail. Through extensive breeding, your classmate identifies a series of short-tail females that, when crossed with wild type, produce exclusively short-tail progeny. You cross these short-tail females with fertile, kinked-tail males and observe the following offspring: 35 short-kinked-tail females, 32 short-nonkinked-tail females, 28 short-kinked-tail males, 36 short-nonkinked-tail males (all offspring fertile). Propose a model to account for these ratios. (g) To test your model from part (f), you select, from among the offspring from part (f), short-kinked-tail females and short-kinked tail males, and you cross them. If your model is correct, what phenotypic classes do you expect to observe, and in what ratios? 3. Each of the families below exhibits a different, extremely rare genetic disorder. Individuals expressing the trait (the disorder) are indicated by solid symbols. Assume that no new mutations have arisen in any of the individuals shown. Consider the following possible modes of inheritance: (i) X-linked recessive with complete penetrance, (ii) autosomal recessive with complete penetrance, (iii) autosomal recessive with 70% penetrance, (iv) autosomal dominant with complete penetrance, (v) autosomal dominant with 70% penetrance. For each pedigree state which, if any, of these five modes of inheritance are not possible. For the modes of inheritance that are possible, calculate the probability that the individual indicated by a “?” is affected.4 (a) (b) (c) (d) ? ? ? ?7.03 Problem Set 1 Answer Key Fall 2001 1a. Mutants 2 and 16 have dominant Leu- alleles. Remember dominant or recessive alleles are tested by crossing a mutant to wildtype and observing the phenotype of the heterozygote. 1b. The following mutants are in the same genes: (1,19), (3,8,18), (4,13), (5,6,11,15), and (10,14). Mutant alleles in these strains fail to complement each other in the heterozygote. Nothing can be said definitively about strain 2 and 16 because complementation tests cannot be done with dominant alleles. 1c. Yes. Different alleles of the same gene can be both dominant and recessive. Again, complementation tests cannot be done with dominant alleles. 1d. There is a minimum of seven genes. Five groups defined by recessive alleles that noncomplement each other. One group minimally defined by mutants 7 and 9, and a second group defined by mutants 12, 17, and 20. 7 and 9 complement 11-20, excluding 16, and thus are not a part of any previously defined complementation group. They may however be in the same gene because we do not have data about complementation between mutants 7 and 9. Similar logic follows that 12,17, and 20
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