Question

Problem 2: A) In an industrialized region of the UK, the relative fitness (w) of Biston betularia, the peppered moth, is 1.0 for the dark form (DD and Dd) and 0.20 for the light form (dd). The frequency of the Dallele (p) is 0.90 at a certain time. Place the appropriate values in the first and second rows of Table A and obtain the numerical values for all the expressions given in rows 3, 4, and 5 of the table (including 91 and Aq). B) Use the same information in part A, except substitute p=0.20. Calculate the values of qı and Ag Compare them with those obtained in A. Was the change smaller or larger in part B? Why?

Answer 1

12 13 JU6 ? TUESDAY 18 19 20 21 25 26 27 28 Problem2: la f(1) = p = Biston betularia 0.90 :9=1-6= 0.10 S = selection co-efficient DD Dd dd Total tale 0.81 0.18 0.01 I o.lo t 1-5 -0.2 Initial Zygote. - frequency Relative fitness: (W) Contribut of each: to next gen. s=0.8 0.81 0.18 0.01x02 1-sqz = 0.002 = 0.992 0.18 0.002 0.81 Normalized freg.: .992 (after selection) = 0.816 0.992 0992 0.181 0.003 AV. Change in allele a b тизику за - Spa 2 = 0.007258 -0.007258 qa qo - Sa T-sar 0.092

USO- JUN 12 13 14 19 20 21 26 27 28 WEDNESDAY B. I pain þa 0-20 i qo 0.80 1-sqz 0.80 -0.880.82 1 -0.880.82 = 0.590 Aqsa Spar 1-Squ - 50.8 x 0.2 x 0.82 7 t 1 - 0.88-82- 0 – 0.209 TES: NOTES: Aql > laalf Change was larger in Part B. because mihat or selection was working on a glleater allelic frequ in Part B. So