7.03 Genetics: Answer Key to Problem Set #2 (2002)

2002 Problem Set #2 Answer Key 1) a) This three factor cross only gives us four different phenotypic classes because two of the mutations of interest give the same phenotype. b) The distance between pr1 and pr2 can be calculated by recognizing that half of the crossovers between pr1 and pr2 will give rise to progeny with normal eyes and the other half will carry both the pr1 and pr2 alleles and thus be purple. Since a crossover event between pr1 and pr2 is the only way to get flies with normal eyes, we can state that the number of progeny that resulted from a crossover between pr1 & pr2 is twice the number of normal eyed progeny. 30 (total normal * 2 (we only observe half of the crossover = 60 (# of recombinants eyed progeny) classes as normal eyed progeny) between pr1 & pr2) Map distance = from pr1 to pr2 60 (# of recombinants between pr1 & pr2) * 100 = 6cM 1000 (total # of progeny) c) Pr1 and pr2 are mutant alleles that are dominant to the normal eyes allele and result in a purple-eyed phenotype. The pr1, pr2 double mutant could have a separate phenotype than the individual pr mutants, but in this case the only two phenotypes observed are normal and purple eyes. We know that the normal allele is recessive to either pr allele and that there is no normal allele present in a pr1, pr2 double mutant, so the only possible phenotype of a pr1, pr2 double mutant is purple eyes. d) We know the parental classes are: st + pr2 (stubby legs & purple eyes) + (normal legs & purple eyes) AND pr + because these are what the heterozygous mother’s chromosomes looked like before any crossing over. The least frequent class is stubby legs & normal eyes (only 3 progeny), which we assume is one of the double crossover classes. These individuals received neither mutant pr allele from their mother, but did receive her mutant st allele. Therefore the other double crossover event resulted in progeny with both mutant pr alleles and the wild-type st allele so they have normal legs & purple eyes. From these observations we know that the double crossover classes are: st + + (stubby legs & normal eyes) pr2 (normal legs & purple eyes) AND + pr1 When we compare the parental classes to the double crossover classes, we see that the same st and pr1 alleles are together in both classes but with different pr2 alleles. Therefore, pr2 must be in between pr1 and st and the gene order is: pr1_______________pr2________________st e) The distance between any one pr gene and the st gene can not be calculated directly since the two mutant pr genes give the same phenotype and one is each copy of the parental chromosome. Therefore, we must indirectly calculate this distance by taking advantage of the known distance between the two pr genes, the known number of double crossovers, and the relationship between single and double crossovers in this interval. The probability that a double crossover with respect to the three genes of interest takes place in between pr1 and st is the product of the probability that a crossover occurs in between pr1 & pr2 and between pr & st. p(double x-over) = p(x-over between pr1 & pr2) * p(x-over between pr2 & st) - We know the probability of a crossover between pr1 & pr2 is 60/1000 since we observe 30 progeny with normal eyes which is one of the two classes produced as a result of a pr1, pr2 crossover. (see 3b) - We know that the probability of a double crossover is 6/1000 since we observe 3 progeny of one of the double crossover classes and assume that the other indistinguishable double crossover class has about the same number of progeny. Therefore: (6/1000) = (60/1000) * p(x-over between pr2 & st) p(x-over between pr2 & st) = 0.1 distance between pr2 & st = .01 * 100 = 10cM 2.) a.) For each pair of markers, you can “cover” the column corresponding to the marker in which you are not interested with a pen and just classify each tetrad as P, N, T on the basis of the other two. Tetrad A B C D E F G H I J K L Number 17 1 47 3 5 2 3 4 12 9 16 1 a/m P N T T P N T T T T N P a/h P N P N T T T T T T P N m/h P P T T T T P N T T N N b.) P >> NPD for two markers to be linked, if unlinked the genes would assort independently and result in a much closer number of apparent P and N type tetrads: a/m: P = 17+5+1 = 23 N = 1+2+16 = 19 T = 47+3+3+4+12+9 = 78 P=N (a, m unlinked) a/h: P = 17+47+16 = 80 N = 1+3+1 = 5 T = 5+2+3+4+12+9 = 35 P>N (a, h linked) m/h: P = 17+1+3 = 21 N = 4+16+1 = 21 T = 47+3+5+2+12+9 = 78 P=N (m, h unlinked) a-h map distance = 100 cM x (T + 6N)/2Σ = 100 cM x (35 + 6(5)) / 2(120) = 27 cM c.) Tetrad Type Non-Parental Ditype (N) Tetratype (T) Parental Ditype (P) Phenotype 2His+:2His- 1 His+:3 His­ 0 His+:4 His- Genotypes His2+ His4+ His2+ His4+ His2- His4­ His2- His4­ His2+ His4+ His2+ His4­ His2- His4+ His2- His4­ His2+ His4­ His2+ His4­ His2- His4+ His2- His4+ d.) For every 1 N tetrad counted, we can assume there are 4 total double crossover tetrads (1 P, 1 N, 2 T), which means double crossover probability among these 100 tetrads is 4/100 = 0.04. Single crossover tetrad probability must therefore be v0.04 = 0.2, which means that there are 0.2(100) = 20 tetrads with single crossovers. Now here is where the solution encounters ambiguity and two possible solutions: i. Assuming that single crossovers and double crossovers are exclusive events, this means we have 20 (single crossover) T tetrads + 2 (double crossover = 2NPD) T tetrads = 22 T. We also know from the problem there is 1 N tetrad so by process of elimination there are 100 – 22 – 1 = 77 P tetrads. ii. Assuming that double crossover tetrads are a subclass of single crossover tetrads that have just experienced one extra crossover, we have to subtract 4 (=4NPD) from the 20 single crossover tetrads but add back 2 (=2 NPD) double crossover (T) tetrads to get the total number of 20 – 4 +2 = 18 T tetrads. We also know from the problem there is 1 N tetrad so there are 100 – 18 – 1 = 81 P tetrads. iii. On a completely tangential note, this problem could have been definitively solved rather complicatedly with a Poisson distribution, but such was by no means expected. All three answers would have been accepted. e.) i: his2-his4 map distance = 100cM x (T+6N)/2Σ = 100cM x (22+6(1)/2(100)) = 14 cM OR ii: distance = 100 cM x (18 + 6(1)/2(100)) = 12 cM 3.) a.) Although there are only 80 observable plaque-forming units, you have to account for the equal number of reciprocal recombinants (ts-1 ts-2 double mutants) that cannot be seen because they do not form plaques at the non-permissive temperature: ts1-ts2 distance = 100 mu x #crossovers/#total = 100 m.u. x (2 x 80)/10000 = 1.6 mu b.) 1.6 mu = x bp 100 mu 5x104 bp x = 800 bp c.) small, temperature-resistant plaques are the least frequent phenotypic class, hence it must be the result of double crossovers, and only the sm-ts2-ts1 order (or its reverse, ts1ts2-sm) can accommodate this: ----+----------------+-----------------ts1-------sm---------------ts2-----------------+---For the same reasons employed in problem 1e, looking at a singly crossed over class of known distance can give us the recombination frequency between the other two linked loci. Looking among temperature-resistant plaque forming units (wildtype at both the ts2 and ts1 loci), we count the frequency of double crossovers (to yield small plaques at 45ºC). Sm-tm2 map distance = 100 mu x #small tempr/#total = 100 mu x 25/100 = 25 mu.