Answer:
The correct answer is C: <em>0.60 </em><em>A1</em> and <em>0.40 A2</em> will be the allele frequencies in those progeny when they reach reproductive age
Explanation:
<em>Genotype</em> <u><em>A1A1 </em></u><em> </em><u><em>A1A2</em></u><em> </em><u><em>A2A2
</em></u>
<em>Relative aptitude, w</em> 1 0.5 0.5
<em>Number of individuals</em> 40 40 40
<em>Initial allelic frequency</em> p0= (40+20)/120=0.5 q0= (40+20)/120=0.5
<em>Zygote frequency</em> p2= 0.25 2pq=0.25 q2=0.625
<em>Relative contribution</em> 0.25x1=0.25 0.5x0.5=0.25 0.25x0.5=0.125
<em>of each genotype</em>
<em>Average aptitude W</em> W= 0.025 + 0.25 + 0.125 = 0.625
<em>Population</em> AA= 0.25/0.625 AB=0.25/0.625 BB=0.125/0.625
<em>Genotype frequency</em> AA= 0.4 AB=0.4 BB=0.2
<em>New Allelic frequency</em> p1=0.4+(0.4/2)=0.6 q1=0.2+(0.4/2)=0.4
- <em>Total number of individuals</em>: 120
- <em>Initial allelic frequency</em>:
(number of homozygote individuals + half number of
heterozygote individuals) / Total number of individuals
- <em>Relative contribution of each genotype</em>:
Zygote frequency x Relative aptitude
- <em>Average aptitude W</em>: 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
- <em>Population Genotype frequency</em>:
Relative contribution of each genotype / Average aptitude
- <em>Allelic frequency</em>:
Homozygote population genotype frequency + half
heterozygote population genotype frequency
