Question

In humans, dimples are a dominant trait. If two parents are both heterozygous for dimples, what percent of their children will also be heterozygous?(1 point)
Responses

0%
0%

25%
25%

50%
50%

75%
75%

100%

Answer 1

To determine the percentage of children who will be heterozygous for dimples when both parents are heterozygous, we can use a Punnett square.

Let's denote:
- D = dominant allele for dimples
- d = recessive allele for no dimples

If both parents are heterozygous, their genotypes are:
- Parent 1: Dd
- Parent 2: Dd

We can set up a Punnett square as follows:

```
D d
----------------
D | DD | Dd |
----------------
d | Dd | dd |
----------------
```

From the Punnett square, we see the genotypes of the potential offspring:
- DD (homozygous dominant): 1
- Dd (heterozygous): 2
- dd (homozygous recessive): 1

Now, we can count the genotypes:
- 1 DD
- 2 Dd (heterozygous)
- 1 dd

Thus, the percentage of children who will be heterozygous (Dd) is:
\[
\frac{2 \text{ (Dd)}}{4 \text{ (total offspring)}} \times 100 = 50\%
\]

So, the answer is **50%**.