Question

Suppose a mother carries two recessive genes for freckles (rr) and a father carries one recessive gene for freckles and one dominant gene for clear facial skin (Rr). What is the likelihood that their first child will have freckles?

Answer 1

To determine the likelihood of the child having freckles, we need to look at the genetic makeup of the parents.

The mother has the genotype **rr** (two recessive genes for freckles). The father has the genotype **Rr** (one dominant gene for clear skin and one recessive gene for freckles).

When we set up a Punnett square to analyze the possible genotypes of their offspring, we can list the possible alleles contributed by each parent:

- The mother (rr) can only contribute **r**.
- The father (Rr) can contribute either **R** (dominant) or **r** (recessive).

Now we set up the Punnett square:

```
R r
----------------
r | Rr | rr |
----------------
r | Rr | rr |
----------------
```

From the Punnett square, we can see the possible combinations of alleles for their offspring:

- **Rr** (clear skin) - 2 occurrences
- **rr** (freckles) - 2 occurrences

So, out of the four possible genotypes:

- 2 will be **Rr** (not have freckles)
- 2 will be **rr** (will have freckles)

Therefore, the probability that their first child will have freckles (genotype **rr**) is:

\[
\text{Probability} = \frac{\text{Number of rr outcomes}}{\text{Total outcomes}} = \frac{2}{4} = \frac{1}{2} = 50\%
\]

Thus, there is a **50% chance** that their first child will have freckles.