Question

problems:
how many type a individuals
how many type ab individuals
how many type b individuals
carriers of o type
what are the blood types of the
above parents? &

Explanation:

Step1: Analyze blood - type inheritance

In the ABO blood - type system, the alleles are \(I^{A}\), \(I^{B}\), and \(i\). \(I^{A}\) and \(I^{B}\) are codominant, and both are dominant over \(i\).
If both parents have the genotype \(I^{A}i\) (carriers of the \(i\) allele which is related to type O in a recessive sense), when they have children, we use the Punnett square method.
The cross is \(I^{A}i\times I^{A}i\).

Step2: Construct the Punnett square

The possible gametes from each parent are \(I^{A}\) and \(i\).
The Punnett square is:

	\(I^{A}\)	\(i\)
\(i\)	\(I^{A}i\)	\(ii\)

The genotype \(I^{A}I^{A}\) and \(I^{A}i\) result in type A blood. There are 3 out of 4 genotypes (\(I^{A}I^{A}\), \(I^{A}i\), \(I^{A}i\)) that give type A blood. But if we assume a simple 2 - by - 2 cross (where each parent contributes one allele), the number of type A individuals (genotypes \(I^{A}I^{A}\) and \(I^{A}i\)) is 4 (if we consider all 4 offspring positions in the Punnett square as individual cases). Type AB (\(I^{A}I^{B}\)) requires one \(I^{A}\) and one \(I^{B}\) allele, so there are 0. Type B (\(I^{B}I^{B}\) or \(I^{B}i\)) also requires \(I^{B}\) allele, so 0. Carriers of O type (genotype \(ii\)) need two \(i\) alleles. Here, if we assume the cross is \(I^{A}I^{A}\times I^{A}I^{A}\) (since all offspring are type A), there are 0 carriers of O type.
For the parents' blood types, if all offspring are type A, and considering the alleles in the cross (assuming no \(I^{B}\) or \(i\) in a recessive - showing way for non - A types in offspring), the parents' genotypes are \(I^{A}I^{A}\) (homozygous type A)

Answer:

How many type A individuals: \(4\)
How many type AB individuals: \(0\)
How many type B individuals: \(0\)
Carriers of O type: \(0\)
What are the blood types of the above parents? \(I^{A}I^{A}\)&\(I^{A}I^{A}\)