Question

In a family, father has blood group A and mother has a blood group B. Their children show $50\%$ probability for a blood group AB indicating that
A. Father is heterozygous
B. Mother is heterozygous
C. Either of parent is heterozygous
D. Mother is homozygous

Answer 1

In the punnett square, use the genotypes in the form ${I^X}{I^Y}$ where $X$ and $Y$ blood groups can be A, B, O. Heterozygous genotypes $X$ and $Y$ are different and for homozygous genotypes, $X$ and $Y$ are the same.

Step by step answer: Heterozygous genotype has two different alleles unlike homozygous genotype, which has similar alleles.
-The blood group of a person can be expressed in the form ${I^X}{I^Y}$ where $X$ and $Y$ blood groups can be A, B, O. The heterozygous genotypes $X$ and $Y$ are different and for homozygous genotypes, $X$ and $Y$ are the same.
-For a person having blood group A, the genotypes will be ${I^A}{I^A}$ (for homozygous genotype) and ${I^A}{I^O}$ (for heterozygous genotype).
-For a person having blood group B, the genotypes will be ${I^B}{I^B}$ (for homozygous genotype) and ${I^B}{I^O}$ (for heterozygous genotype).
-For a person having blood group AB, the genotype will be ${I^A}{I^B}$.
-In option A, the father has heterozygous genotype of blood group A. The mother, on the other hand, can have either heterozygous or homozygous genotype of blood group B. So, the punnett square for heterozygous father and homozygous mother.

| ${I^A}$| ${I^O}$
---|---|---
${I^B}$| ${I^A}{I^B}$| ${I^O}{I^B}$
${I^B}$| ${I^A}{I^B}$| ${I^O}{I^B}$

The punnett square for heterozygous father and heterozygous mother.

| ${I^A}$| ${I^O}$
---|---|---
${I^B}$| ${I^A}{I^B}$| ${I^O}{I^B}$
${I^O}$| ${I^A}{I^O}$| ${I^O}{I^O}$

-For $50\%$ progeny to have blood group AB, only 2 out of 4 alleles should have the genotype of the form ${I^A}{I^B}$. As in the first punnett square, there are only two progenies with blood group AB, the other having blood group B. The other punnett squares do not satisfy the question needs because only 1 progeny out of 4 has blood group AB.
-In option B, the mother is having a heterozygous genotype for blood group B. But the father can have either heterozygous or homozygous genotype of blood group A. The punnett square for heterozygous mother and homozygous father.

| ${I^A}$| ${I^A}$
---|---|---
${I^B}$| ${I^A}{I^B}$| ${I^A}{I^B}$
${I^O}$| ${I^A}{I^O}$| ${I^A}{I^O}$

The punnett square for heterozygous mother and heterozygous father.

| ${I^A}$| ${I^O}$
---|---|---
${I^B}$| ${I^A}{I^B}$| ${I^O}{I^B}$
${I^O}$| ${I^A}{I^O}$| ${I^O}{I^O}$

For $50\%$ progeny to have blood group AB, we know that only 2 out of 4 alleles should have the genotype of the form ${I^A}{I^B}$. As we can see in the first punnett square, there are only two progenies with blood group AB, the other having blood group A. The other punnett squares do not satisfy the question needs because only 1 progeny out of 4 has blood group AB.
In option C, it is mentioned that one of the parents needs to be heterozygous for $50\%$ progeny to have blood group AB. In the punnett squares made above, only one of the parents to be heterozygous will fulfill the need of the question. So this is the correct option.
Hence option C is correct.

Note: The punnett squares are widely used in the field of genetics. Punnett squares are used to determine all the possible genotypes of next-generation based on genotypes of the parent generation. It has been widely used by Mendel to explain the concept of inheritance using pea plants and their simple characteristics.