Question

Baldness is the loss of human hair on the head and is a recessive sex-linked trait. A male with the recessive genotype and a female heterozygous for the genotype have children. What is the probability that a female child will inherit the baldness trait?
Responses

0%
0%

25%
25%

50%
50%

100%
100%

Answer 1

To determine the probability that a female child will inherit the baldness trait given the provided genetic information, we first need to define the genotypes involved.

Let:
- \( B \) = dominant allele (no baldness)
- \( b \) = recessive allele (baldness)

Since baldness is a recessive sex-linked trait, for males (who have one X chromosome and one Y chromosome), the genotype that expresses baldness is \( bY \) (because they can only inherit one allele from the X chromosome, they can only have either the baldness trait or not). Females have two X chromosomes, so their possible genotypes could be \( BB \) (homozygous dominant), \( Bb \) (heterozygous), or \( bb \) (homozygous recessive).

Given:
- The male genotype is \( bY \) (bald)
- The female genotype is \( Bb \) (heterozygous, not bald)

When these two have children, we can set up a Punnett square for their X chromosomes to see the possible genotypes of their daughters:

| | X (b) | X (B) |
|---------|-----------|-----------|
| **Y** | X(b)Y | X(B)Y |

From this Punnett square, we can see that the potential offspring X chromosomes are:
- \( X(b) \) -> offspring with the genotype \( X(b)Y \) (this is a son with baldness)
- \( X(B) \) -> offspring with the genotype \( X(B)Y \) (this is a son without baldness)
- Female offspring can inherit the X chromosome from their mother, which could either be \( b \) or \( B \).

The combinations for daughters specifically would be:
1. \( X(b)X(b) \) = This would be a daughter with baldness.
2. \( X(B)X(b) \) = This would be a daughter who is a carrier but does not exhibit baldness.

Since we are interested in the daughters specifically, the possibilities based on their genotypes are:
- Daughters can either inherit the \( b \) or \( B \) from their mother.

For daughter probabilities from \( Bb \) (heterozygous, mother) and \( bY \) (bald, father):
- \( 50\% \) will be \( Bb \) (not bald)
- \( 50\% \) will be \( bb \) (bald)

Therefore, the probability that a female child will inherit the baldness trait (\( bb \)) is **50%**.

So, the correct response is:

**50%**