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In an isolated population of 1000 individuals, 444 are genotype MN, 478 are MM, and 78 are NN. What are the frequencies of the M and N alleles in this population?


    f(M) = 0.7, f(N) = 0.3
    f(M) = 0.5, f(N) = 0.5
    f(M) = 0.478, f(N) = 0.078
    f(M) = 0.4, f(N) = 0.6

Hello,

f(M) = 0.7, f(N) = 0.3

thank you ! could you explain a little how you got that?

Have you used the Hardy-Weinberg equation?

Okay it's a long explanation so here goes...

There are 478 MM, 444 MN, and 77 NN, so the M allelic frequency is:

number of M alleles/total number of M and N alleles

There are two alleles for each individual so the number M alleles in MM individuals is:
2 x 478 = 956

There is only one M allele in each heterozygote:
1 x 444 = 444

The total number of M alleles in this population is:
444 + 956 = 1400

The total number of both alleles (M and N) is
2(478 + 444 + 78) = 2000
*note there are two of each allele so must multiply be 2

The allelic frequency of the M allele is
f(M) = 1400/2000 = 0.7

And f(M) + f(N) = 1 so
0.7 + x = 1.0 ; f(N) = 0.3

Hope this helped! ;)

Wow, thanks.