Answer:
Option A, [tex]32[/tex]%
Explanation:
Number of chickens having featherless legs [tex]= 128[/tex]
Number of chickens having feathered legs [tex]= 200-128[/tex][tex]= 72[/tex]
Frequency of chickens having featherless legs
[tex]= q^{2}[/tex]
[tex]q^{2} = \frac{128}{200}\\q^{2} = 0.64[/tex]
Frequency of alleles for featherless legs
[tex]= \sqrt{q^{2}} \\= \sqrt{0.64} \\= 0.8[/tex]
As per first equation of Hardy Weinberg-
[tex]p+q=1[/tex]
Substituting the given values we get -
[tex]p = 1-0.8\\p=0.2[/tex]
As per second equation of Hardy Weinberg-
[tex]p^{2} + q^{2} + 2pq=1[/tex]
Substituting the given values we get -
[tex]0.64+0.04+2pq=1\\2pq=1-(0.64+0.04)\\2pq= 0.32\\2pq=32[/tex]%
