Answer:
F(M) = 0.7 F(N)=0.3
Explanation:
We need to take into account that individuals NN and MM only N or M aleles respectively while MN gives just half of the allelic contribution for each. With this in mind we have the following probabilities for getting each of the alleles: MM => p(M) = 1.0 and p(N)=0.0
MN => p(M ) = 0.5 and p(N) = 0.5
NN => P(M) = 0.0 and p(N) = 1.0
With this we can calculate the frequencies of the M and N alleles in the population with 100 individuals using the following formulas
For f(M):
[tex]f(M)=\frac{1.0 (MM) + 0.5 (MN)}{Total individuals}[/tex]
[tex]f(M)=\frac{1.0(50)+0.5(40)}{100}[/tex]
[tex]f(M) = \frac{50+20}{100}[/tex]
[tex]f(M)=\frac{70}{100}=\frac{7}{10}=0.7[/tex]
The frequency of the M allele is 0.7
For f(N):
[tex]f(N)=\frac{1.0 (NN) + 0.5 (MN)}{Total individuals}[/tex]
[tex]f(N)=\frac{1.0(10)+0.5(40)}{100}[/tex]
[tex]f(N) = \frac{10+20}{100}[/tex]
[tex]f(M)=\frac{30}{100}=\frac{3}{10}=0.3[/tex]
The frequency of the N allele is 0.3
