Color blindness is X linked trait and is carried on X chromosome so when a <span> A mother with normal color vision and a color blind father have a color blind daughter the offspring will be
</span><span>. Some of their sons can have normal color vision
</span>because male is XY and female XX
so i conclude option B is correct<span />
Answer:
200 approximate
Explanation:
depends how many in common
Phylum: Chordata
Class: Aves
Order: Strigiformes
Family: Strigidae
Genus: Bubo
Species: B. scandiacus
from species down we have:
species
population
organism
organ system
organ
tissue
cell
organelle
macromolecule
molecule
atom
elementary particle
I hope this helps you.
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
