Answer:
(attached Punnett square below)
100% of the offspring will have black feathers
0% of the offspring will have white feathers
Explanation:
All the possible offspring have the genotype Bb, and since B (black) is dominant it will be the phenotype regardless of the presence of a b allele.
Therefore, all the offspring will be black (100%) and since none of the genotypes are bb there's no chance for white feathered chickens (0%)
hope this helps!
I think you forgot to give all the options in the question. I have researched and found the correct answer. I hope that it helps you. As a result of insertion, the incorrect sequence of amino acids will be translated into a protein, resulting in a mutation. Amino acids contain both a carboxyl and an amino group to form a simple organic compound.
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
