Question

A parent with heterozygous B blood has a child with a parent who is type O. What is the probability of that child having each of these blood types (USE FRACTIONS)

Type A

Type B

Type AB

Type O

Answer 1

To determine the probability of the child having each blood type, we first need to understand the genotypes involved:

- A parent with heterozygous B blood has the genotype \(I^B I\) (where \(I^B\) represents the B allele and \(I\) represents the O allele).
- A parent with type O blood has the genotype \(ii\).

Next, we can set up a Punnett square to determine the possible genotypes of the children:

\[
\begin{array}{c|c|c}
& I^B & i \\
\hline
i & I^B i & ii \\
\hline
\end{array}
\]

From this cross, we have the following potential genotypes for the child:
1. \(I^B i\) (which corresponds to type B blood)
2. \(ii\) (which corresponds to type O blood)

Now, let's determine the probabilities:

- **Type A**: A child would need to have the genotype \(I^A i\), but since neither parent has the A allele, the probability is \(0\).

- **Type B**: The probability of getting the genotype \(I^B i\) is \(1/2\) (1 out of the 2 outcomes from the Punnett square).

- **Type AB**: A child would need to have the genotype \(I^A I^B\), but since there is no A allele in either parent's genotype, the probability is \(0\).

- **Type O**: The probability of getting the genotype \(ii\) is \(1/2\) (1 out of the 2 outcomes from the Punnett square).

In summary, the probabilities are:
- Type A: \(0\)
- Type B: \(\frac{1}{2}\)
- Type AB: \(0\)
- Type O: \(\frac{1}{2}\)

Thus, the final probabilities are:
- **Type A:** \(0\)
- **Type B:** \(\frac{1}{2}\)
- **Type AB:** \(0\)
- **Type O:** \(\frac{1}{2}\)