Question

9. In another population the frequency of individuals with each genotype is MM = 0.25, MN=0.45, NN=0.30. a. Given the following relative fitness values for each genotype, calculate the total population fitness: WMM-1, WMN=0.9, WNN=0.8.

Answer 1

A and c part in image.

B. Based on fitness value we can state that directional selection has been enacted which is favoring MM genotype as the fitness value of MM individuals is the largest I.e., 1.

h as frequency of genetype MM - 0. 25 MN - 0.45 NN -0.30 MM = p² MN = 2pq NN = 92 So allele frequency M & 0.25 = 62 = 0.5 we know, prq = 1 q= 1-0.5 =) 0.5 WMM 1 WMN - 0.9 WWN - 0.8 Mean population fitness = W W = p²wmm + 2pq WmN + q ² WNN = 0.25 x 1 + 0.45 ×0.9 + 0.30X0.8 = 0.25 + 0.405 + 0.24 = 0.895

- 1-0.5 =) 0.5 wmm - 1 WMN = 0.9 WWN & Ons Mean population fitness = T W = p²w mm + 2pq Wmn + q ² WNN = 0.25 x 1 + 0.45 X0.9 + 0.30 80x8 = 0.25 + 0.405 + 0.24 = 0.895 0.25x1 -0.219 After 1 generation , MM = $wmm = frequency of ho w It has Increased from 0.25. 0.895 1028

MN = 2bq WMN = 0.45X 0.9 0.452 0.895 10.45 frea of MN 269 w It is similar to 0.895 nous previous frequency trea of NN = q ² WNN = 0.30x0.8 = 0.2687 It has decreased from 0.30 So, allele M is getting fixed, as its It can also frequency is increasing after I generation.