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
AnswerExplanationnnnnnnnnnnnnnnn:
Answer:
A) histone coding genes
Explanation:
Histones are the proteins involved in the packing of DNA in eukaryotes. They are positively charged proteins and are not found in bacteria. To analyze the evolutionary relationship in a gene between eukaryotes and bacteria, one must choose the gene present in both of them. The tRNA, rRNA and hexokinase encoding genes are common to both bacteria and eukaryotes. Therefore, one of these genes may be chosen for the study. However, histone encoding genes are not present in the genomes of bacteria. Therefore, the study of the histone encoding gene would not be useful for the mentioned purpose.
Answer: Homologous structures
Explanation: They are structures which have similarities in various organisms but carry out opposite functions. This is the opposite of analogous structures