Respuesta :
Answer:
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