This question involves basic Mendelian genetics, specifically the ABO blood group system, which has three alleles: $I^A$, $I^B$, and $i$ (often represented as $I^O$). Alleles $I^A$ and $I^B$ are codominant, and both are dominant over $i$. Blood group 'O' has the genotype $ii$ ($I^O I^O$).
The mother is heterozygous for 'A' blood group. This means her genotype is $I^A i$ ($I^A I^O$). She has the 'A' antigen but also carries the recessive 'O' allele.
The father is heterozygous for 'B' blood group. This means his genotype is $I^B i$ ($I^B I^O$). He has the 'B' antigen but also carries the recessive 'O' allele.
We can use a Punnett square to determine the possible genotypes of the offspring. Each parent can produce two types of gametes:
| $I^B$ | $i$ ($I^O$) | |
|---|---|---|
| $I^A$ | $I^A I^B$ | $I^A i$ ($I^A I^O$) |
| $i$ ($I^O$) | $I^B i$ ($I^B I^O$) | $i i$ ($I^O I^O$) |
The Punnett square shows four possible genotypes for the offspring:
Each genotype has an equal probability of occurring, which is 1 out of 4, or 25%.
The genotype required for blood group 'O' is $ii$ ($I^O I^O$). This occurs in 1 out of the 4 possible combinations.
Therefore, the probability of having a child with blood group 'O' is 25%.
| List I | List II |
| A. Incomplete dominance | I. Human skin colour |
| B. Co-dominance | II. Inheritance of flower colour in Antirrhinum sp. |
| C. Pleiotropy | III. Phenylketonuria disease in humans |
| D. Polygenic inheritance | IV. ABO blood groups |
| List I | List II |
| A. Incomplete dominance | I. Human skin colour |
| B. Co-dominance | II. Inheritance of flower colour in Antirrhinum sp. |
| C. Pleiotropy | III. Phenylketonuria disease in humans |
| D. Polygenic inheritance | IV. ABO blood groups |