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)

Question

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%.