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
<u>Answer:</u>
<u>Answer:Patterns of Inheritance. </u><em>The phenotype of an individual is determined by his or her genotype. The genotype is determined by alleles that are received from the individual's parents (one from Mom and one from Dad). These alleles control if a trait is “dominant” or “recessive”.</em>
Answer:
Glucose is quickly broken down by liver and muscle cells to provide energy, which demonstrates that its type of chemical bonds is related to its function.
Answer:
RNA Analysis
Explanation:
the test shows that in archaea, the RNA polymerase is complex and contains more subunits, than the RNA polymerase in bacteria.
B)Bacteria
Cause viruses are not prokaryotes
