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I) In a population of 100 individuals 33% are of the BB for eye colour. What percentage is expected to be BG assuming Hardy-Weinberg equilibrium? A. 9% B. 21% C. 49% D. 51% E. insufficient information to answer question Answer: C q^2 = 33% or .33, so q = .574 and p=.426. The genotype BG is heterozygous so the frequency = 2pq = 2(.426)(.574) = 48.9% II) The lactase gene is involved in the breakdown of lactose in the human, allele + (p=0.8) functions typically, allele – (in a recessive pattern) causes the development of lactose intolerance at young adulthood. Assume Hardy-Weinberg equilibrium has been reached. 1. What is the frequency of all possible genotypes? In the year 2115, the dairy industry has infiltrated every government in the world (all 2 of them) and has bribed enough politicians to manipulate the food industry into including some dairy derivative in all food production. This all of a sudden reduced the relative fitness of lactose intolerant people to half that of their pizza-eating peers (not because of any real deleterious effects but rather because it is nearly impossible to find a date when afflicted by unchecked flatulence). 2. Within 2 generations, what is the allele frequency of lactase – ? Answers: #1 p = 0.8 q = 1-p = 0.2 G0 f(+/+) = p^2 = 0.64 G0 f(+/-) = 2pq = 0.32 G0 f(-/-) = q^2 = 0.04 G0 = generation 0 (parental) #2 w+/+ = 1 w+/- = 1 w-/- = 0.5 G0 Waverage = f(+/+)*w+/+ + f(+/-)*w+/- + f(-/-)*w-/- = 0.98 After 1 generation: G1 f(+/+) = w+/+ * G0 f(+/+) / Waverage = 0.65 G1 f(+/-) = w+/- * G0 f(+/-) / Waverage = 0.33 G1 f(-/-) = w-/- * G0 f(-/-) / Waverage = 0.02 At G1, q = 0.02 + 0.33/2 = 0.185 G1 Waverage = G1 f(+/+)*w+/+ + G1 f(+/-)*w+/- + G1 f(-/-)*w-/- = 0.99 After 2 generations: G2 f(+/+) = w+/+ * G1 f(+/+) / Waverage = 0.66 G2 f(+/-) = w+/- * G1 f(+/-) / Waverage = 0.33 G2 f(-/-) = w-/- * G1 f(-/-) / Waverage = 0.01 At G2, q = 0.01 + 0.33/2 = 0.175 Since q is getting smaller with each generation, there seems to be an erosion of genetic diversity regarding the lactase gene. III) Individuals who are A/A are considered wild type and their red blood cells remain in circulation for 120 days. Individuals that are A/a are heterozygous for the sickle cell trait and their red blood cells function normally, however have a reduced lifespan providing them resistance against the parasite Plasmodium falciparum whose development requires an incubation period longer than that provided by the heterozygotes. Individuals with a/a are affected with sickle cell anemia and if left untreated will die before reaching reproductive age. An outbred population within a large village in Botswana is quite frisky and mating randomly. 1234 individuals genotyped and the following results were obtained 334 individuals were A/A, 764 were A/a, 136 were a/a. Medical treatment there is weak and only 1/20 a/a individuals will survive to reproductive age, 10/20 that are A/A will die of malaria before reproducing and all heterozygotes are healthy. a) What will be the A and a allele frequencies after the following generation? b) What will be the allele frequencies at equilibrium? Answers: a) In the present generation (G0): f(A)=p={334 +764(0.5)}/1234=0.58 f(a)=q=1-0.58=0.42 f(A/A) = 334/1234= 0.27 f(A/a) = 764/1234= 0.62 f(a/a) = 136/1234= 0.11 W A/A=0.5 W A/a= 1 W a/a= 0.05 G0 Waverage = 0.5*0.27 + 1*0.62 + 0.05*0.11 = 0.76 After one round of selection: f(A/A) = 0.27*0.5/0.76 = 0.178 f(A/a) = 0.62*1/0.76 = 0.816 f(a/a) = 0.11*0.05/0.76 = 0.007 p= 0.178 + 0.816/2 = 0.585 q= 1-0.585= 0.415 b) At equilibrium p= s/(s+t) s = selection co-efficient against the a/a genotype (w= 1 - s) t = selection co-efficient against the A/A genotype (w= 1 - t) p =0.95/(0.95+0.50) = 0.65 q=1 - 0.65 = 0.35 or (t/s+t)= 0.50/(0.95+0.50)= 0.35 Reference p.629 in textbook IV) You have two markers linked to QTL on separate chromosomes that contribute to plant height in cat-tail (Typha latifolia). The RF value for marker 1 and its locus is 0.2; that for marker 2 and its locus is 0.05. Each marker has two alleles characterized by the following phenotypic effects (a unit refers to the allele’s contribution to the phenotypic differences between the pure-breeding lines mentioned below) M1 = 8 units m1 = 1 unit M2 = 4 units m2 = 0 units You have two pure-breeding lines, one short, one tall and cross them to generate an F1 which is then allowed to self to produce an F2 segregating for the marker alleles. The genotype of the tall line is M1/M1;m2/m2 and that of the short is m1/m1;M2/M2 If your aim is to increase plant height which markers should you select for in the F2? How many units of progress can you make by selecting these markers? Answers: You would select progeny that are M1/M1;M2/M2. The average phenotypic effects of the allele classes are as follows M1 – 8(0.8)+1(0.2) = 6.6 m1 – 8(0.2)+1(0.8) = 2.4 M2 – 4(0.95)+0(0.05) = 3.8 m2 – 4(0.05)+0(0.95) = 0.2 The phenotypic effect of gametes carrying: [M1 ; M2] = 6.6+3.8 = 10.4 [m1, m2] = 2.4+0.2 = 2.6 M1/M2 homozygotes = 10.4 + 10.4 = 20.8 height units m1/m2 homozygotes = 2.6 + 2.6 = 5.2 height units The difference (20.8-5.2 = 15.6 units) reflects maximum height units of improvement toward by selectively breeding plants containing M1 and M2 markers. V) Consider the following pedigree: I II III IV ? The effected individuals have a rare (fictional) disorder that involves amnesia, dyslexia, color blindness, and an extremely acute sense of hearing, called Heinsbergen syndrome. The wild type allele (H+) is present in the population at a frequency of 0.995. Assuming that the gene responsible for Heinsbergen is in Hardy-Weinberg equilibrium, what is the probability that child IV-1 will be affected? Answer: If H+ is 0.995 frequent, we can write p=0.995, and q=1-p or 0.005 is the frequency of the H- allele. From this we calculate the probability that individual III-1 is a carrier (P(III-1)) as P(H+/H-)=2pq. I H+/H- H+/H- II P (H+/H-) = 2pq =2*(0.995)(0.005) =0.00995 III P (H+/H-) = 2/3 IV ? P (H-/H-) = P (H-/H- | III-1, III-2)P (III-1)P (III-2) =(1/4)(0.0095)(2/3) = 0.00166 Probability that the father is a carrier: Probability that the mother is a carrier: 2/3 (using pedigree information) 0.00995 (using population estimates) Probability that each will pass on H-: ½*2/3*1/2*0.00995 = 0.00166 (probability of affected child)