Answer:
1. Allele frequency of b = 0.09 (or 9%)
2. Allele frequency of B = 0.91 (0.91%)
3. Genotype frequency of BB = 0.8281 (or 82.81%)
4. Genotype frequency of Bb = 0.1638 (or 16.38%)
Explanation:
Given that:
p = the frequency of the dominant allele (represented here by B) = 0.91
q = the frequency of the recessive allele (represented here by b) = 0.09
For a population in genetic equilibrium:
p + q = 1.0 (The sum of the frequencies of both alleles is 100%.)
(p + q)^2 = 1
Therefore:
p^2 + 2pq + q^2 = 1
in which:
p^2 = frequency of BB (homozygous dominant)
2pq = frequency of Bb (heterozygous)
q^2 = frequency of bb (homozygous recessive)
p^2 = 0.91^2 = 0.8281
2pq = 2(0.91)(0.9) = 0.1638
The primary function of the active site of an enzyme is to catalyze the reaction associated with the enzyme (Option c). It is a fundamental structure in the enzyme.
<h3>What is the active site of an enzyme?</h3>
The active site of the enzyme is It is a fundamental structure in the enzyme that has catalytic activity.
The active site of the enzyme is a site that binds to the substrate to form the enzyme-substrate complex.
The formation of this complex leads to the generation of one or more products of a given chemical reaction.
Learn more about enzymes here:
brainly.com/question/1596855
Sex is a reproduction system involving both male and female to partake in sexual activity intercourse to create an offspring between the two beings creating a living organism
D. Lipids would be your answer
Answer:
25%
Explanation:
Let's assume that the recessive allele "p" imparts diseased conditions in the homozygous genotypes. The genotype of each of the carrier parents would be "Pp". A cross between Pp and Pp would produce progeny in the following phenotype ratio=
Pp x Pp= 3/4 Normal : 1/4 Affected.
Therefore, there are 1/4 or 25% chances for this couple to have a child with PKU.
