7.03 Genetics: Problem Set 6 Solutions (2000)

' Problem 7.03 Problem Set 6 Solutions Fall 2000 1i • a) f(NT/NT) = q2 = 0.3 f(NT)= q = 0.55 = 55% b)p = 1 -q p = 1-0.55 = 0.45 f(NT/T) = 2pq = 2(0.55)(0.45) c) (0.3)/(0.3+1/2(0.5)) = 0.50 = 50% = 0.545 = 55% : d) p(a taster is heterozygous) = (2pq)/(p2+2pq) = (0.5)/(0.2+0.5) = (0.5/0.7) p(taster offspring) = 1 - p(non-taster offspring) p(both parents were heterozygous and both give NT allele) = (0.5/0.7)(0.5/0.7)(1/4) = 0.1276 p(PTC taster) = 1 - 0.1276 = 0.87 = 87% e) as above, only this time one parent is certain to Nve the NT allele, as he or she is a non-taster (NT/NT): 1 - (0.5/0.7)(1/2)(1) = 0.64 = 64% ,< : _.i: Alternatively, this could be figured out by looking at the ways &making a taster. p(taster parent is T/T and gives T alIele)+ p(taster parent is NT/T and gives T allele) = (0.2/0.7)(1) + (0.5/0.7)(t/2) = 0.64 = 64% f) f(NT/NT)= q2 = 0.01 q= 0.1 g)p= l-q=0.9 f(NT/T) = 2pq = 2(0.1)(0.9) = 0.18 = 18% h) (0.01)/(0.01+1/2(0.18))= 0.1 = 10% i) p(each parent was a heterozygote and gave the NT allele): (5/7)(18/99)(1/4) = 0.032 = 3% Problem 2: a) S=1; q= 0.0004 = Sq2 ,a = 0.00000016 b) = 1.6 x 10-7 ZXqmut+ kqsel = 0 t.t - ½(2Shetpc0 - Sq 2 = 0 =Sh_tpq +Sq 2 Sq2 + Shetpq - !_ = 0 q2 + O.Ol(q)(l-q) - 0.00000016 = 0 0.99q 2 + O.Olq - 0.00000016 = 0 (a quadratic equation) q= 1.57 x 104 = 1.6 x 10-5 What ffwe make the approximation, 1-q (=p) is about equal to 1? 9 -Then we get: q- + 0.01q -p. = 0 Solving forq, we get: q = 1.597 x 10 -s = 1.6 x 10: s Thus, in this case, the approximation is valid. c) Problem Again, Aqmut + Aqs_I= 0 Using the above approximation (that p = I): Sq 2- hq - ._ = 0 (see lecture notes for derivation of Aqs_l) q2 _0.01q_ 0.00000016 = 0 q = 1.0x 10 -_ 3: a) for M: q2 = 0.0004; q = 0.02 forN: (Fist cousins)(q)(f(1 st col.lsirlmarriages)) + q2 = f(aa) (1/16)q(0.1) + (9/10)q 2 = 0.0004 (0.9)q- + (i/160)q- 0.0004 = 0 q=1.8x10 -z b) Cross BL6 mouse carrying the mutation to a DBA mouse. The reatlting progeny carry genetic material that is 50% BL6 and 50°/; DBA in origin. These progeny can be screened for the presence of the mutation of interesx by looking for the SSR which was inseparable from that mutation. Such mice, which carry the mutation, can be again bred to DBA mice, resulting in progenywhich carry, genetic material that is 75% DBA and 25% BL6. As with their parent before them,, mice which carry the original mutation can be identified due to the presence of the linked SSR_ Continuing this lo_c, one can determine how many generations of such matings are n_essary to obtain a mouse which is homo_gous for DBA alleles at geater than 99°'6 of a!l genes in the genome: number of breedings (n) required: 1- (tA)n = proportion of genes from DBA thus, for n=7, 1- (V2)7 = 0.9922. So 7 breedings wiI1 be required to get a mouse with 99% of its loci DBA in origin. This mouse mtlst r_henbe bred with its siblings (the "infamous" brother - sister matings) to produce a home_,'gous mouse. (This entire process Hill likely take more than 2.5 years of ?'our life). "-.