Answers to 7.03 Exam #2

Answers to 7.03 Exam #2 1 a) No. You would expect that most offspring come from non-recombinant gametes. Since the children are small thumbed, normal eyed or normal thumbed, cross eyed these are most likely the non-recombinant genotypes. Therefore the mother's genotype is xye +/+ smt, so one grandparent was only cross eyed and the other only small thumbed. 1 b) 5%. From the definition of map units we know there is a 10% chance of being a recombinant gamete for xye and stm. Remeber that half of those recombinants will be wt and half will be stm and xye. So there is only a 5% chance the child will be double mutant. I c) 20%. From the definition of map units we know that 10% of the gametes are recombinant. There must be crossovers in 20% of the oocytes to yield 10% recombinant gametes. 20/100 oocytes with a cross over --> 40/400 -->10% recombinant gametes. 2 a) Type I = NPD Type 2 = TT 2 b) 35 m.u. Just apply the tetrad formula 35 m.u. = 100134 + 6(6)/2(100)] recombinant chromosomes--> Type 3 = PD for linked genes. 2 c) 34 TT expected. The fact that the trpl- and red1- cross yielded no tetratypes means that the two genes are unlinked, and both are completely linked to their respective centromeres. For the trpl-, red2- cross we know the markers are unlinked (because red1 is linked to red 2 and red1 and trpl are unlinked). Any crossover event that gives a tetratype tetrad for markers red1 and red2 (the situation in problem 2b) will lead to a tetratype tetrad between markers red2 and t-rpl. So the answer is you'd expect the same number of tetratypes as in 2b, 34 tetratypes. 2 d) expect 17 or 16.6. The data in the problem suggests the two marker s are unlinked. expect a 1:4:1 ratio of PD:TT:NPD. Thus 1/6 of the tetrads should be of PD (2 red, trp+ trp-). 1/6(100)=16.6 or 17. 3 a) 4%. 8 m.u.= 100(# recombinant phage)/total The number of recombinant phage is 8%. Remember only half of the recombinants other half are tsl-,ts2-. So the fraction of wt will be .08(1/2)=.04 So you 2 white are wt the 3b) 4 m.u. Of the phage growing at 40°C (tsl+) 4% of them are recombinant (mot-). mu = # recombinant phage out of the total. We are missing half the recombinant phage because they are tsl-,mot+ and can not grow at 40°C. However we are also missing half the total because one of the parentals tsl-,mot- also can not grow at 40°C. So map units is simply .04(100). 3c ) The rare class is tsl+, ts2+, mot+, they represent only 4% of the plaques growing on the plate. must be the result of a double crossover event. Therefore tsl is between mot and ts2. This