Answer:
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
