I think the answer to this would be A
I'm guessing Liver.
If it was in the stomach, we wouldn't get belly aches. If it were in the large intestine, we wouldn't poop. So that's my guess.
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:
C.
Explanation:
I have done this question before.
