Suppose that in a population the frequency of a particular recessive disease is 1/400. assume the presence of only two alleles: dominant allele (a) and a recessive allele (a) in the population and that the population is at hardy-weinberg equilibrium. what is the frequency of the recessive allele that causes the condition?

Question

Laurel Hansen

Biology

19 April, 05:07

Suppose that in a population the frequency of a particular recessive disease is 1/400. assume the presence of only two alleles: dominant allele (a) and a recessive allele (a) in the population and that the population is at hardy-weinberg equilibrium. what is the frequency of the recessive allele that causes the condition?

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Meyer · Answer 1

Given:

Frequency of recessive disease = 1/400

Under Hardy-Weinbert equilibrium, we assume

p = frequency of dominant allele

q = frequency of recessive allele,

Then frequency of recessive disease=qq, or

q=sqrt (1/400) = 1/20=0.05

=>

p=1-q=1-0.05=0.095

Answer: frequency of the recessive allele, q, is 1/20, or 0.05, or 5%