Can you help me with Question 2 please Conservationists arc

Can you help me with ^{Question 2} please

Conservationists arc concerned for the fate of the endangered Amur leopard. Recent surveys of their native habitat show that there are 110 leopards left in the wild. However, 75 of these arc males and 35 of these arc females. What is the effective population size for Amur leopards? Why is this a concern? Before World War II the frequency of individuals with attached carlobes (an autosomal recessive, selectively neutral trait) was estimated to be 64%. Radiation following the bombing of Japan was thought to raise the mutation rate by 10 fold. Suppose the mutation rates for car lobe attachment was raised to: A rightarrow a = 10^-4 and a rightarrow A = 10^-7. a. What would the frequency of the A and a alleles be in the generation born after World war II? b. Assuming these mutation rates remained constant, and individuals continued not to select their mates based on earlobe appearance, what will be the new equilibrium frequency of the two alleles in the population? Ninety rats live on a island, and the frequency of a recessive coat color allele in this population is 0.8. Ten rats from the mainland arc inadvertently brought to the island by a ferry boat. The frequency of the coat color allele in the mainland rat population is .20. After migration, what is the frequency of the coat color allele in the new rat population on the island?

Solution

2A.

According the Hardy Weinberg law p+q =1.

Assume that allele ‘A’ has frequency p and allele ‘a’ has frequency q. There were 64% chances of recessive trait. Therefore, before world war the recessive allele ‘a’ has the frequency 0.64 and allele ‘A’ has frequency p 0.36 (1-0.64). After war there is mutation change in A to ‘a’ is 10⁴ and ‘a’ to ‘A’ is 10⁷.

To calculate the new frequency of alleles we use following formula:

Frequency affected through mutation = f+ u(1-f) where u is the mutation rate and f is frequency of particular allele.

Now new frequency q’= f+ 10^-4 (1-0.64) = 0.64 + 0.36 x 10^-4

In addition, for allele ‘A’ frequency ‘p’= 0.36 + 0.64 x 10⁷.

2B.

The frequency p is much lesser than frequency q. Therefore, in the new equilibrium phase, t the new frequency of allele ‘a’ will remain 0.64 + 0.36 x 10^-4 and ‘A’ will at 0.36.