Name:__________KEY______________ 1 2007 7.013 Problem Set 2 KEY Due before 5 PM on FRIDAY, March 2, 2007. Turn answers in to the box outside of 68-120. PLEASE WRITE YOUR ANSWERS ON THIS PRINTOUT. 1. Wild-type flies are brown in color. You have discovered a gene that controls body color in flies called gene A. You have two true-breeding mutant strains, both of which have black bodies. Strain One (a1/a1) is homozygous for a mutation in gene A that causes a dominant mutant phenotype. Strain Two (a2/a2) is homozygous for a mutation in gene A that causes a recessive mutant phenotype. Group Three are flies that result from mating Strain One to Strain Two. Group Four are flies that result from mating Strain One to wild-type (A/A). Group Five are flies that result from mating Strain Two to wild-type (A/A). Predict the genotypic ratio and the phenotypic ratio of the offspring resulting from the following crosses. Make sure to label each class in each of your ratios. Group Three are a1/a2 Group Four are A/ a1 Group Five are A/ a2 (a) Group Three and Group Five Genotypic ratio: Phenotypic ratio: 1 A/a1 to 1 a1/a2 to 1 a2/a2 to 1 A/a2 3 black to 1 brown -----black-------------------- brown This is the result of a cross of a1/a2 to A/ a2. (b) Strain Two and Group Three Genotypic ratio: Phenotypic ratio: 1 a1/a2 to 1 a2/a2 1 black to zero brown -----black---------- This is the result of a cross of a1/a2 to a2/ a2.Name:__________KEY______________ 2 (c) Group Four and Strain Two Genotypic ratio: Phenotypic ratio: 1 A/a2 to 1 a1/a2 1 black to 1 brown brown black This is the result of a cross of a1/A to a2/ a2. (d) Group Four and Group Five Genotypic ratio: Phenotypic ratio: 1 A/A to 1 A/a2 to 1 a1/a2 to 1 A/a1 1 black to 1 brown -----brown------- ---brown---------- This is the result of a cross of a1/A to A/ a2. 2. Consider a fictitious pathway that controls coat color in the mouse. The mouse genome contains gene A, which encodes enzyme A, an enzyme that converts the precursor (a white compound) to the intermediate (a tan compound). The mouse genome contains gene B, which encodes enzyme B, an enzyme that converts the intermediate to Compound #2 (a brown compound). The mouse genome contains gene C, which encodes enzyme C, an enzyme that converts the intermediate to Compound #1 (a gray compound). Mice that produce both compound #1 and compound #2 are brownish-gray. Genes A, B, and C are all autosomal genes that lie on different mouse chromosomes. Loss-of-function mutations in both homologous versions (maternal and paternal) of any one gene (A, B, or C) causes a recessive phenotype. antigen 1antigen 2intermediateenzyme Aenzyme Cenzyme B precursor Compound #2 (brown) Compound #1 (gray) (tan) (white)Name:__________KEY______________ 3 (a) Which color coat would each following mouse have? Some are filled in for you. AABBCC = brownish gray The introduction tells you that mice that produce both compound #1 and compound #2 are brownish-gray. aaBBCC = white Any mouse that is “aa” will be unable to convert the white precursor into the tan intermediate, and thus will never produce any color at all. This same explanation holds for the rest of the answers you had to fill in. AaBbcc = brownish-tan AabbCc = grayish-tan Aabbcc = tan aaBBcc = white aabbCc = white aabbcc = white (b) Two different true-breeding strains of mice have been isolated; neither strain can produce compound #1 or compound #2. When an individual from one strain is crossed with an individual from the other strain, all of the F1 mice produce both compounds. Write out the genotypes for both P generation strains, and the genotype for the F1 generation. (Use “A,” “B,” and “C” to designate the wild-type alleles and “a,” “b,” and “c” to designate the defective alleles of the three genes.) Also write out the coat color of each group of animals. Genotype Coat Color P generation, one strain AAbbcc tan P generation, other strain aaBBCC white F1 generation AaBbCc Brownish grayName:__________KEY______________ 4 If neither of the original strains produce compound #1 and neither produce compound #2, then the original strains must be one of the following three genotypes (remember, the two original strains are true-breeding so they are homozygous): AAbbcc aaBBCC aabbcc aabbCC aaBBcc If you investigate every pairwise mating, the only one that gives you all brownish gray in the F1 is AAbbcc mated to aaBBCC. To be brownish gray, a mouse must have at least one “A” and at least one “B” and at least one “C” allele. (c) Two of the F1 mice are crossed to one another. The possible phenotypes for the F2 progeny are shown below. What fraction of the F2 generation will be represented by each phenotype on average? (Note: some fractions in the table may be zero.) Coat Color Fraction of F2 that is this coat color white 16/64 tan 3/64 brown zero brownish-gray 27/64 brownish-tan 9/64 grayish-tan 9/64 AaBbCc crossed to AaBbCc gives the following proportion of genotypes (where “A_” for instance means at least one “A” allele): A_B_C_ = 3/4 X 3/4 X 3/4 = 27/64 browish-gray A_bbC_ = 3/4 X 1/4 X 3/4 = 9/64 grayish-tan A_B_cc = 3/4 X 3/4 X 1/4 = 9/64 brownish-tan A_bbcc = 3/4 X 1/4 X 1/4 = 3/64 tan aaB_C_ = 1/4 X 3/4 X 3/4 = 9/64 white aabbC_ = 1/4 X 1/4 X 3/4 = 3/64 white aaB_cc = 1/4 X 3/4 X 1/4 = 3/64 white aabbcc = 1/4 X 1/4 X 1/4 = 1/64 whiteName:__________KEY______________ 5 3. You are studying two recessive traits in the fruit fly Drosophila melanogaster. The “h” allele causes flies to have the recessive phenotype of hairy backs (wild-type flies have hairless backs). The “t” allele causes flies to have the recessive phenotype of thick legs (wild-type flies have thin legs). You mate females from a true-breeding strain with hairy backs and normal legs to males from a true-breeding strain with normal backs and thick legs. F1 females are then mated to males that have hairy backs and thick legs to produce F2 progeny. If you analyzed 1000 MALE progeny in the F2 generation, how many flies of each possible phenotypic class would you expect, given that: (a) The two traits are determined by two unlinked autosomal genes hairy thick: 250 hairy thin: 250 hairless thick: 250 hairless thin: 250 P
View Full Document
Unlocking...