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A rare genetic disease which is due to a recessive allele (a) and that is lethal when homozygous, occurs within a specific population at a frequency of one in a hundred thousand. In other words, one out of a hundred thousand individuals are homozygous recessive with the disease. How many individuals in a town with a population of 20,000 can be expected to carry this allele?

Respuesta :

Answer:

40 individuals

Explanation:

The homozygous recessive percentage is equal to q2.

Recessive genotype frequency, q 2 is 1/1000000  = 0.000001  

therefore, q = [tex]\sqrt{q2}[/tex]  = [tex]\sqrt{0.000001 }[/tex]  = 0.001 =  allele "a" frequency  

and p + q = 1, thus, p = 1 – q = 1 – 0.001 = 0.999  

thus, the frequency of allele a = q = 0.001  and the frequency of allele A = p = 0.999

Carriers are heterozygous and are equal to 2pq.  

So, 2pq = 2  x 0.999 x 0.001 x  20,000 = 39.96 (round of to 40)

Hence, 40 individuals will be expected to carry the recessive allele.

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