Answer:
Frequencies of allele
and
are
and
respectively.
Frequencies of Individuals with genotype
respectively.
Explanation:
As per Hardy Weinberg's equation -
----------Equation (A)
-----------Equation (B)
Where "p" represents the frequency of "![A_{1}](https://tex.z-dn.net/?f=A_%7B1%7D)
"q" represents the frequency of "![A_{2}](https://tex.z-dn.net/?f=A_%7B2%7D)
represents frequency of individual ![A_{1}A_{1}](https://tex.z-dn.net/?f=A_%7B1%7DA_%7B1%7D)
represents frequency of individual ![A_{2}A_{2}](https://tex.z-dn.net/?f=A_%7B2%7DA_%7B2%7D)
represents frequency of individual ![A_{1}A_{2}](https://tex.z-dn.net/?f=A_%7B1%7DA_%7B2%7D)
Here genotype frequencies are -
![A_{1}A_{1} = 77\\p^{2} = \frac{77}{140} \\= 0.55\\](https://tex.z-dn.net/?f=A_%7B1%7DA_%7B1%7D%20%3D%2077%5C%5Cp%5E%7B2%7D%20%3D%20%5Cfrac%7B77%7D%7B140%7D%20%5C%5C%3D%200.55%5C%5C)
![A_{2}A_{2} = 18\\q^{2} = \frac{18}{140} \\= 0.128\\](https://tex.z-dn.net/?f=A_%7B2%7DA_%7B2%7D%20%3D%2018%5C%5Cq%5E%7B2%7D%20%3D%20%5Cfrac%7B18%7D%7B140%7D%20%5C%5C%3D%200.128%5C%5C)
Substituting this values in equation A, we get
![0.55 + 0.128 + 2pq = 1\\2pq = 1-(0.128+ 0.55)pq = 0.321](https://tex.z-dn.net/?f=0.55%20%2B%200.128%20%2B%202pq%20%3D%201%5C%5C2pq%20%3D%201-%280.128%2B%200.55%29pq%20%3D%200.321)
Frequencies of allele
and
are -
For ![A_{1}](https://tex.z-dn.net/?f=A_%7B1%7D)
![= \sqrt{p^{2} } \\= \sqrt{0.55} \\= 0.74](https://tex.z-dn.net/?f=%3D%20%5Csqrt%7Bp%5E%7B2%7D%20%7D%20%5C%5C%3D%20%5Csqrt%7B0.55%7D%20%5C%5C%3D%200.74)
Substituting this value in equation B, we get
![p+q=1\\0.74 + q = 1\\q = 1-0.74\\q = 0.26](https://tex.z-dn.net/?f=p%2Bq%3D1%5C%5C0.74%20%2B%20q%20%3D%201%5C%5Cq%20%3D%201-0.74%5C%5Cq%20%3D%200.26)
Frequencies of allele
and
are
and
respectively.
Frequencies of Individuals with genotype
are
respectively.