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D
D
A dimpled dad and a non-dimpled mom are expecting a
child. If the presence of dimples is dominant over
absence of dimples, what is the probability that the
child will not have dimples?
○ 0%
○ 25%
O 75%
○ 100%
d
Dd
Dd
d
Dd
Dd


Sagot :

Certainly! Let's solve the problem step-by-step to determine the probability that the child will not have dimples.

First, let's denote the alleles:
- "D" represents the dominant allele for dimples.
- "d" represents the recessive allele for no dimples.

Given:
- The dad is heterozygous for dimples, so his genotype is "Dd".
- The mom is homozygous recessive, so her genotype is "dd".

We can use a Punnett square to determine the possible genotypes of their children. The Punnett square looks like this:

```
d d
+-----+-----+
D | Dd | Dd |
+-----+-----+
d | dd | dd |
+-----+-----+
```

Each box in the Punnett square represents a possible genotype of their child. There are a total of 4 possible outcomes:

1. Dd
2. Dd
3. dd
4. dd

Now, let's interpret these results:
- The genotypes "Dd" indicate the presence of dimples because the dominant allele "D" is present.
- The genotypes "dd" indicate the absence of dimples because only the recessive alleles "d" are present.

Count the outcomes:
- We have 2 outcomes with "Dd" (dimples)
- We have 2 outcomes with "dd" (no dimples)

So, out of 4 possible outcomes, 2 result in no dimples.

The probability that the child will not have dimples (genotype "dd") is calculated as:
[tex]\[ \text{Probability of no dimples} = \frac{\text{Number of "dd" outcomes}}{\text{Total number of outcomes}} = \frac{2}{4} = 0.5 \][/tex]

To express this probability as a percentage:
[tex]\[ 0.5 \times 100 = 50\% \][/tex]

Therefore, the probability that the child will not have dimples is 50%.