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If the recessive allele for an x-linked recessive disease in humans has a frequency of 0.02 in the population, what proportion of individuals in the population will have the disease? assume that the population is 50 : 50 male:female.

Respuesta :

To have the disease, males need only one copy of the affected X.

So, the frequency of affected males will be the frequency of the disease allele = 0.02

Females must inherit two copies and be homozygotes, so the frequency of affected females will be the double of frequency of the disease allele = (0.02)

Since the ratio between female and male is 50:50 that is half of the population is male and half female

Then the overall frequency of the disease will be,

(0.5) (0.02) + (0.5) (0.0004) = 0.01 + 0.0002

Thus, the frequency of affected individuals in this population is 0.0102.

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