Answer:
If the genotype of the parents are <em>Iᵃi </em>and <em>I</em>ᵇ<em>i, </em>then four type of offspring will be produced.The genotype of the offspring are, <em>IᵃIᵇ, Iᵃi,Iᵇi, </em>and <em>ii.</em>
Explanation:
<em>IᵃIᵇ = </em>As the alleles are co-dominant in nature, so both type of alleles are expressed. The blood group will be AB. So, both A and B type of antigen will be found in plasma membrane of RBC.
<em>Iᵃi= </em>In this type of genotype only A type of antigen will be expressed in the membrane of RBC. The blood group will be A type.
<em>Iᵇi= </em>In this type of genotype only B type of antigen will be expressed in the membrane of RBC. The blood group will be B type.
<em>ii= </em>This is a recessive type of genotype. So, no antigen will be found on the membrane of the RBC. And the blood group will be O type.
The offspring's ratio will be 1:1:1:1.
We solve the problem using the Hardy-Wineberg equation.
We have the frequency for gray trait as p = = 0.8
∴ p² = 0.64
64 % of the horses are homozygous for the gray trait.
Solving for the frequency of white chestnut trait by using the fact that,
p + q = 1
∴ q = 1 - p = 1 - 0.8 = 0.2
Now we can calculate 2pq in p² + 2pq + q², which is the frequency for heterozygous gray trait,
∴ 2pq = 2 x 0.8 x 0.2 = 0.32
Therefore, in a population of 100 horses, 32% are heterozygous for gray trait.
Hence the total number of gray horses in a population of 100 will be,
32 (heterozygous gray horses) + 64 (homozygous gray horses) = 96 (gray horses)
Hence, 96% of the horses are gray in color.
B, Magnitude of past earthquakes in the area
Choice A is false. RNA uses the 5 carbon sugar Ribose