A hidden population has been discovered in Turkey. Researchers have calculated that 22% of the population have Uner Tan Syndrome. Using the Hardy-Weinberg formulas, calculate what percentage of the population are heterozygous carriers for this condition.

Thus, 50% of the population are are heterozygous carriers for this condition

Explanation:

Researchers have calculated that 22% of the population have Uner Tan Syndrome. This condition follows the recessive mode of inheritance.

22% have the condition = q² (genotypic frequency uu) = 22/100 = 0.22

using the formular p + q = 1, since q² = 0.22, q = √0.22 = 0.4690

Thus p = 1-q where  is 0.4690 = 1-0.4690

  p = 0.5310

Then p² = 0.5310² = 0.2820 = 28% (genotypic frequency of homozygous normal)

using the formula p² + 2pq + q² where 2pq is the genotypic frequency of the heterozygous carriers. Thus, we have

2 x 0.5310 x 0.4690 = 0.498 = 50%

Thus, 50% of the population are are heterozygous carriers for this condition

b

explanation:

true is the final answer!