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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The answer is d. <span>animals with transmissible diseases separated from healthy animals.
An isolation ward is a separate ward used to isolate animals (or people) suffering from some transmissible diseases in order to reduce a</span><span> risk of passing a potential infection on to others.</span>
