Gene a has two alleles (a1 and a2). a1a1 homozygotes are twice as likely to survive from birth to reproductive age as heterozygotes or a2a2 homozygotes (w11=1, w12=0.5, w22=0.5). a population starts with 40 individuals of each genotype allowed to mate at random. their progeny then survive according to the fitness values given above. what will be the allele frequencies in those progeny when they reach reproductive age

a. 0.40 a1, 0.60 a2 b. 0.50 a1, 0.50 a2 c. 0.60 a1, 0.40 a2 d. 0.80 a1, 0.20 a2 e. all a1

The correct answer is C: 0.60 A1 and 0.40 A2 will be the allele frequencies in those progeny when they reach reproductive age

Explanation:

Genotype                                A1A1  A1A2 A2A2

Relative aptitude, w                  1           0.5    0.5

Number of individuals                 40            40     40

Initial allelic frequency      p0= (40+20)/120=0.5    q0= (40+20)/120=0.5

Zygote frequency                  p2= 0.252pq=0.25q2=0.625

Relative contribution         0.25x1=0.25     0.5x0.5=0.250.25x0.5=0.125

of each genotype              

Average aptitude W              W= 0.025 + 0.25 + 0.125 = 0.625

Population                    AA= 0.25/0.625    AB=0.25/0.625  BB=0.125/0.625

Genotype frequency             AA= 0.4             AB=0.4          BB=0.2  

New Allelic frequency       p1=0.4+(0.4/2)=0.6  q1=0.2+(0.4/2)=0.4

                  (number of homozygote individuals + half number of  

                   heterozygote individuals) / Total number of individuals

                           Zygote frequency x Relative aptitude

                    wA1A1 x p2 + WA1A2 x 2 x p x q + WA2A2 x q2

                Relative contribution of each genotype / Average aptitude

                      Homozygote population genotype frequency +  half  

                           heterozygote population genotype frequency

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