Respuesta :
Answer:
Frequencies of allele [tex]A_{1}[/tex] and [tex]A_{2}\\[/tex] are [tex]0.74[/tex] and [tex]0.26[/tex] respectively.
Frequencies of Individuals with genotype [tex]A_{1}A_{1}, [tex]A_{2}A_{2} and [tex]A_{1}A_{2} are [tex]0.55, 0.128, 0.32[/tex] respectively.
Explanation:
As per Hardy Weinberg's equation -
[tex]p^{2} +q^{2} +2pq= 1[/tex] ----------Equation (A)
[tex]p+q= 1[/tex]-----------Equation (B)
Where "p" represents the frequency of "[tex]A_{1}[/tex]
"q" represents the frequency of "[tex]A_{2}[/tex]
[tex]p^{2}[/tex] represents frequency of individual [tex]A_{1}A_{1}[/tex]
[tex]q^{2}[/tex] represents frequency of individual [tex]A_{2}A_{2}[/tex]
[tex]pq[/tex] represents frequency of individual [tex]A_{1}A_{2}[/tex]
Here genotype frequencies are -
[tex]A_{1}A_{1} = 77\\p^{2} = \frac{77}{140} \\= 0.55\\[/tex]
[tex]A_{2}A_{2} = 18\\q^{2} = \frac{18}{140} \\= 0.128\\[/tex]
Substituting this values in equation A, we get
[tex]0.55 + 0.128 + 2pq = 1\\2pq = 1-(0.128+ 0.55)pq = 0.321[/tex]
Frequencies of allele [tex]A_{1}[/tex] and [tex]A_{2}\\[/tex] are -
For [tex]A_{1}[/tex][tex]= \sqrt{p^{2} } \\= \sqrt{0.55} \\= 0.74[/tex]
Substituting this value in equation B, we get
[tex]p+q=1\\0.74 + q = 1\\q = 1-0.74\\q = 0.26[/tex]
Frequencies of allele [tex]A_{1}[/tex] and [tex]A_{2}\\[/tex] are [tex]0.74[/tex] and [tex]0.26[/tex] respectively.
Frequencies of Individuals with genotype [tex]A_{1}A_{1}, A_{2}A_{2}, A_{1}A_{2}[/tex]are [tex]0.55, 0.128, 0.32[/tex] respectively.
