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 = 0.25 2pq=0.25 q2=0.625

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

Total number of individuals: 120 Initial allelic frequency:

(number of homozygote individuals + half number of

heterozygote individuals) / Total number of individuals

Relative contribution of each genotype:

Zygote frequency x Relative aptitude

Average aptitude W: It is the sum of relative contribution of each genotype to the next generation.

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

Population Genotype frequency:

Relative contribution of each genotype / Average aptitude

Allelic frequency:

Homozygote population genotype frequency + half

heterozygote population genotype frequency