Question

People with freckles have a dominant trait. Having no freckles is a recessive trait. If two parents are heterozygous (Ff) for freckles, what is the likelihood that their children will have freckles?

a
75%
b
25%
c
100%
d
50%

Answer 1

To determine the likelihood that the children of two heterozygous parents (Ff) will have freckles, we can use a Punnett square.

The alleles are:
- F = allele for freckles (dominant)
- f = allele for no freckles (recessive)

The possible combinations from two heterozygous parents (Ff x Ff) are:

```
F f
______________
F | FF | Ff |
-------------
f | Ff | ff |
```

From the Punnett square, the possible genotypes for the children are:
- 1 FF (homozygous dominant, will have freckles)
- 2 Ff (heterozygous, will have freckles)
- 1 ff (homozygous recessive, will not have freckles)

This means:
- 3 out of 4 combinations (FF and Ff) will have freckles.

Thus, the likelihood that their children will have freckles is:
- \( \frac{3}{4} \times 100 = 75\% \)

The correct answer is **a) 75%**.