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Vinil7 [7]
3 years ago
14

Human ABO blood type is controlled by the I locus and has three possible alleles. The alleles which code for type A blood (IA) a

nd type B blood (IB) are codominant, while the allele for type O blood (i) is recessive. In a cross between a person with a genotype of IAi and a person with genotype IAIB, what is the probability of seeing three offspring with type A blood?
Biology
1 answer:
Oxana [17]3 years ago
6 0

Answer:

1/8 (12.5 %)

Explanation:

Cross: IAi X  IAIB

F1: 1/4 IAIA 1/4 IAIB 1/4 IAi 1/4 IBi

Phenotypes:

- IAIA and IAi >> type A blood

- IBi >> type B blood

- IAIB >> type AB blood

Phenotypic frequency type A blood: 1/2  (i.e., 1/4 IAIA +  1/4 IAi) >>

Probability to have 3 offspring with type A blood: 1/2 x 1/2 x 1/2 = 1/8

The human ABO blood group is a system consisting of there different alleles: A, B, and O. This system is used in genetics as an example of codominance because both A and B alleles are expressed in heterozygous individuals (i.e., IAIB), whereas only the allele A is expressed in individuals that have A and O alleles (since O allele is recessive to the A and B alleles).

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A barred hen would have a copy of B on its Z chromosome, a non-barred rooster would have both copies of b on both Z chromosomes. Using Punnet squares to represent the crosses we get the following cases:

\begin{center}\begin{tabular}{ |c|c|c|c| }\ & Z^{B} & W \\ \ Z^{b} & Z^{B}Z^{b} & Z^{b}W \\ \ Z^{b} & Z^{B}Z^{b} & Z^{b}W \\ \end{tabular}\end{center}

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\begin{center}\begin{tabular}{ |c|c|c|c| }\ & Z^{b} & W \\ \ Z^{B} & Z^{B}Z^{b} & Z^{B}W \\ \ Z^{b} & Z^{b}Z^{b} & Z^{b}W \\ \end{tabular}\end{center}

That is a ratio of one barred heterozygote rooster to one barred hen to one non-barred rooster to one non barred hen.

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