What are the genotypic and phenotypic ratios for a cross between a heterozygous type A mother ($I^A i$) and a homozygous type B father ($I^B I^B$) in the ABO blood group system?

Answer

Cross $I^A i \times I^B I^B$: the mother makes gametes $I^A$ and $i$, and the father makes only $I^B$ gametes. Offspring are 50% $I^A I^B$ (type AB) and 50% $I^B i$ (type B). So the genotypic ratio is $1\, I^A I^B: 1\, I^B i$ and the phenotypic ratio is $1\, AB: 1\, B$.

Explanation

What you are crossing in ABO blood types

ABO blood type is controlled by three alleles: $I^A$, $I^B$, and $i$. The alleles $I^A$ and $I^B$ are codominant (both show if together), and $i$ is recessive.

List the parental genotypes and gametes

Mother (heterozygous type A): $I^A i$ Gametes: $I^A$ or $i$
Father (homozygous type B): $I^B I^B$ Gametes: only $I^B$

Combine gametes to get offspring genotypes

Possible children:

$I^A$ (mom) + $I^B$ (dad) $\rightarrow I^A I^B$ (phenotype AB)
$i$ (mom) + $I^B$ (dad) $\rightarrow I^B i$ (phenotype B)

Ratios (genotype and phenotype)

Because the mother produces $I^A$ and $i$ gametes equally often, the outcomes are split evenly:

Genotypic ratio: $1\, I^A I^B: 1\, I^B i$
Phenotypic ratio: $1\, AB: 1\, B$

Quick check

No child can be type A or type O here because the father always contributes an $I^B$ allele.