Question

Suppose that 35 percent of a population of individuals as hitchhikers thumb, which is Inherited as a recessive. Suppose also that 10 individuals, 5 of which with hitchhikers thumb from that population go deep sea diving and are stranded on an island and mate with eachother. What are the genotype and allele frequencies for both new populations in the next generation?

Answer 1

Answer is given below-

. Hitchhikers thumb is recessive. Persons having hitchhikers thumb will have genotype aa . suppose, In population A total number of individuals =loo 35% are aa .. Frequency of aa (92) = 350 = 0.35 9 = 50.35 = 10.592 we know, onar P+4=1 p .P=1-4=1-592 = 10.408 = 1-9 In Population A Allele frequency of A=p=0.408 of a = 4-0.592 P+4 =1 p²+2pq+q2 = 1 (AA) (Aa) (aa) Genotype frequency of AA = P²-(0.408)² = 0.166 Aa=209=2x0.592X0.408 = 0.483 da = 9²= (0.592)2 = 0.35 10 individuals Total= 100 aa = 35 5=aa Population A Population B Total = 100-10 = 90 Aa=35-5 - 30 Current population A

In population B Number of total individuals = 10 Number of aa = 5 Frequency of aa = 5 92= 0.5 4= 50.5 = 10.707 p=1-4-1-0.707 =10.293 a p+4=1 > (P+4)²=1 - p²+Qpq+q² = 1 (AA) Ha) (aa) senotype frequency of AA=P2= (0.293)² = [0.086 Aa = 209 = 2X0.293XO. ZOZ=10.414 da = q² = (0.707) 2 = 10.5 Allele frequency of A = 10. 293] a = 10.707 If the population is in Hardy Weinberg Equilibrium, then the genotype and allele frequency will remain same in next generation. In Population A (new) Frequency of aa (42) = 30 = 0.333 & 2 0.333 = 10.577 P=1-0.577 = 10 423] In Hardy-Weinberg Equilibrium , allelic and genetype frequency will remain same in next generation Allele frequency of A =P = 10.423] a=4 = 10.577 Genotype frequency of AA = p2= (0.423)² - 10.179 Aa= 204 = 2x0. 423 X0.577= 10.488 da = 82=(0.577) 2 = 10.333