Okay what shall i help you with
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%
Passive transport:
For simple diffusion = fatty acids, amino acids, steroid hormones (because they’re non-polar)
Facilitated diffusion = N2 and Cl-
