Question

black and white are codomimant in a population of kittens. the population has 20 black, 40 gray, and 43 black. let B represent black and b represent white for the alleles. a field experiment was done on the kittens to learn about survival and bottlenecks. for each genotype, 100 kittens were placed in the experiment. the survival rates were: RR 120,Rr 20, rr 20.

1.calculate the fitness for the kitten genotypes

2. determine the next generation of kittens based on the fitness from the F1 generation

3. determine how allele frequencies have changed in the F2 generation.

*43 WHITE not black

Answer 1

1.) Fitness of Kihen's genotuber Black (BB) => (120/120) = 2 (Wh) 1 Grey (Bb) (20/120) = 0.17 (W12) | Afbroy. While (bb) 37 (20/120) = 0,17 (W) 22). less (W) = (0.19x19 +(0.39x0.17)+(0.42X0(7) = 0:33 (abb.). Frequency of Genoly be BB=( 20603) = 0.19 : Bb= (49103) = 0.39 bb 43/103) = 0.42 (Fitness & next generation (h) genoty be forequensayo BB =>}(6-1981)/0,331 - 0058 -7 58 blachy Abbas Bb = 40.398 0.17%.3322 0.20 20 Browny AP Арһәт. bbs ģ C6-42X017)/2.33420:22 -> 22 while ! 3 Next generatiou genoty be foréquena BB=>$(0.58X1)/0,3341 .761 Bb=> (8.20 x0.97) /01334= 0.10 Abbaox, Bb 3 500.22401179/61334= 0 Allere frequency, B = A: 76 +0.10/21 - 1.81 b => 0.1 +60110/23 -0.15 Population are not in HWE, so don't expect Btb will be equal to a an selection bressure is impose on population.