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zaharov [31]
3 years ago
6

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 s

hip 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?
Biology
1 answer:
Trava [24]3 years ago
3 0

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%

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1) Given

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Since the trait is displayed when the people have two lowercase letters (aa), and people can be unaffected and carriers, (Aa), this trait has to be recessive.

If you make a Punnett Square, you will see that half of the offspring can result in an Aa, while the other half can result in an aa, meaning there's a 50/50 chance of the offspring of the F1 (first) generation will have the trait.

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