Question

2. Suppose that you and 19 of your classmates (giving a final population of 10 males and 10 females) are on a cruise, and your ship sinks near a deserted island. You and all of your friends make it to shore and start a new population isolated from the rest of the world. Two of your friends carry the recessive allele (i.e., are heterozygous) for phenylketonuria. If the frequency of this allele does not change as the population on your island increases, what will be the incidence of phenylketonuria on your island?

Answer 1

Answer:

0.25%

Explanation:

20 people start the new population. So there are 20 genes or 40 alleles for the recessive disorder phenylketonuria. 2 out of 40 alleles are recessive for the condition hence frequency of the allele = 2/40 = 0.05

Frequency of the allele does not change when the population increases so it is in Hardy-Weinberg equilibrium. According to it, if q is the frequency of recessive allele, q² = frequency of the recessive condition

Here, q = 0.05 So,

q² = (0.05)² = 0.0025

In percentage, it is 100 * 0.0025 = 0.25%

Hence, incidence of phenylketonuria in the new population is 0.25%

Answer 2

Hence, ​0.25% of the population will be born with Phenylketonuria.

Hardy-Weinberg principle:

The principle states that the allelic frequency remains constant through generations and the gene pool remains constant. This phenomenon is called genetic equilibrium.

The population of 20 people comprises 40 alleles. Out of these 40 alleles, 2 are for phenylketonuria.

Assigning this allele to $q$,

We get,

$q=\frac{2}{40}\\ q=5\%$

Now as per the Hardy-Weinberg principle, if the frequency of the recessive allele in a population is $q$ then the incidence/frequency of  recessive allele individuals will be,

$q^2=q\times q\\=0.05\times 0.05\\q^2=0.0025\\=0.25\%$

Learn more about the topic Hardy-Weinberg principle:

https://brainly.com/question/906952