Question

Suppose you are investigating an autosomal recessive disease known as "studius toxicosis" which occurs at a rate in the American college student population of 1 in 16 individuals. Students who inherited two recessive alleles (tt) have the disease. If we assume Hardy-Weinberg equilibrium in a large population of college students, what is the percentage of heterozygous carriers of the studius toxicosis allele in the population? That is, individuals who have the genotype Tt. Refer to the Hardy-Weinberg equation. Be sure to answer in a percentage, not decimal, but do not include the % sign. Example: answer 45, not 0.45.

Answer 1

Answer:

37.5

Explanation:

Acording to the the Hardy-Weinberg equation:

$p^2 +pq+q^2=1$

$p+q = 1$

Where p is the frequency of the dominant allele "T", and q is the frequency of the recessive allele "t".

Therefore, p^2 is the frequency of the dominant homozygous genotype "TT", q^2 is the frequency of the recessive homozygous genotype "tt" and

2pq is the frequency of the heterozygous genotypes "Tt" and "tT".

Since 1 in 16 individuals have inherited two recessive alleles (tt) and have the disease:

$q^2=\frac{1}{16} \\q=\sqrt{\frac{1}{16}} \\q= 0.25$

Thus, the fraction of heterozygous carriers of the studius toxicosis allele in the population is given by:

$p=1-q\\p = 1 - 0.25 = 0.75\\2pq=2*(0.75)*(0.25)\\2pq= 0.375$

37.5 percent of the population are heterozygous carriers