7.03 Problem Set 2 Solutions

J / 7.03 Problem Set 2 Solutions 7.03 Problem Set 2 Solutions Problem 1 Part (a) (5 points) Two true-breeding parents had offspring that included the F1 females that were test crossed with truebreeding recessive males to give the progeny data. By finding which markers are linked and are traveling on the same chromosome in the F1 females, you can get the phenotypes of the true-breeding parents. Begin by evaluating the combinations of 2-factor cross data: (i) analysis for markers a &b ab = 2+ t1+ 1 +9=23 (rec) ++=2 + 10 +3 + 12 =27 (rec) a + = 34 + 208 + 31 + 205 = 478 (par) + b = 41 + 200 + 36 + 195 = 472 (par) a & b are linked => distance = I00 x 50/1000 = 5 cM a+ parental => ...... +b (ii) analysis for markers a & c ac=2+ 11 +34+205 =252 ++=2 + l0 +41+ 195 =248 a+ = 1 +208 +9+31 =249 +c=3 +200+ 12+36=251 all classes are equally represented, so a &c are unlinked. (iii) analysis for markers a & d ad=2 + 1 + 34 + 31 =68 (rec) ++=2 +3 +41 +36= 82 (rec) a + = I I + 208 + 9 + 205 = 433 (par) +d = 10+200+ 12+ 195 =417 (par) a &d are linked => distance = 100 x 150/1000 = 15 cM a+ parental => ....... +d (iv) 2-factor cross between b &c confirms that b is unlinked to c (v) analysis for markers b & d bd=2+ 1 +200+ 195=398 ++=2+3 +208 +205 =418 b+=lt +41 +9+36=97 '/ +d=10+34+12+3I=87 '_ b & d are linked => distance = I00 x I84/I000 = 18.4 cM bd parental => ....... ++ (vi) 2-factor cross between c & d confirms that they are unlinked. genotype for F1 females (order not accurate at this point) a++ +bd Therefore, one parent exhibited trait "a", and the other parent exhibited the "b" & "d" traits Only one parent exhibited the "c" trait, but it is unclear which one, since "c" is unlinked and segregates independently. Part Co) (5 points) We have three linked markers, so we can to a 3-factor cross to determine order: abd=2+I=3 ab+=ll+9=20 +++=2+3 =5 ++d=lO+ 12=22 a + + = 208 + 205 = 413---the fact that these classes are the highest confirms + b d = 200 + 195 = 395---the chromosomal marker locations for the F1 females above a+d=34+31 =65 +b+=41 +36=77 Because the abd and +++ reciprocal class is the smallest, a crossover on both sides of the middle marker would have to generate those recombinants. The order that fits this condition is: b + I J X d I X I l l + a + so "a" is the middle marker. Map: b [._ 5cM a [ 15cM d I c (different chromosome) I. The distance calculated with a two-factor cross between the outer markers will tend to be less than the sum of the internal intervals, because the calculation is not accounting for double crossovers that restore the parental gametes. Also, preliminary i_._ appraisal of the data shows that there are four classes that have the same high frequency. This means with that not Because "c" segregates b &all d of the traits are linked, one of them is unlinked/independent. and with "a" at the same frequency, it must be in a different linkage group than a, b, & d. The 2-factor crosses confirm this. Problem 2 Part (a) (3 points) Since physical distance in males and females is the same, then the # of chiasma per meiosis is twice as high in females than in males (ie. 100 chiasma per meiosis), because genetic map distancerecombination rate = physical distance The I0 cM distance in this question was arrived at in the same way as the 8 cM distance in part c: by crosses in the random population, in which males are as equally likely to be the heterozygote as the females. This question is asking for the chiasma in the males only. F_st you should calculate the distance in just the males: y = male distance 2y = female distance (since the genetic distance in females is twice that in males) (y + 2y)/2 = 10 y = 6.666 2y = 13.333 Now we can use the male genetic distance of 6.666 cM to determine the number of chiasma: 6.666 cM = 100 x (# ofrecombinants)/total # ofrecombinants = 2 x # of chiasma 6.666 cM = 100 x (2 x # of chiasma)/total # of chiasmaJtotal = 1/30 Part (c) (3 points) One way to look at this problem is that the human equivalent of a test cross was done with homozygous males to check the passing of alMes from the mother, and also with homozygous females to check the recombination in the father. The data had a half-and-half distribution of the crosses, and the resulting distance was 8cM. The questions is asking for the distance calculated if only the crosses with heterozygous males and homozygous females are used. Ify = distance using only male recombination data then 2y = distance using only female recombination data (y + 2y)/2 = 8cM y = 5.333 cM Problem Set 2 Correction 2b. Method 1 If you thought the 10 cM distance given in the problem was from both males and females, proceed as in the original solutions to get the male-only distance of 6.666 cM. 6.666 cM = !00 X 6.666 recombinant I00 total qametes gametes Now we can use a few conversion factors to "change units" from recombinant gametes/total gametes to chiasma/meiotic 6.666 recombinant qametes I00 total gametes X ! chiasma 2 recombinant gametes X 4 qametes = 13.333 chiasma meiotic cell I00 meiotic cells 13.333% of the male meiotic cells you examined would have a chiasma between the two features. Method 2 If you thought the 10 cM distance given in the problem was from males only, use that distance in the same way as above. I0 cM = I00 X i0 recombinant i00 total qametes gametes I0 recombinant qametes X 2 recombinant 1 chiasma i00 total gametes gametes X meiotic 4 qametes 20 chiasma cell = i00 meiotic cells 20% of the male meiotic cells you examined would have a chiasma between the two features. P cells: