Question

Baldness is an x-linked recessive trait. A bald man marries a woman whose father was bald but she is not.

What are the chances that they have a bald daughter? %

Answer 1

To determine the chances of the bald couple having a bald daughter, let's analyze the genetic information involved:

1. The gene for baldness is X-linked recessive, which means that:
- Males have one X chromosome (XY), and if they carry the allele for baldness on their X chromosome, they will express the trait.
- Females have two X chromosomes (XX). They need two copies of the baldness allele (one on each X chromosome) to express the trait, but if they have one copy, they will be a carrier without expressing the baldness.

2. The bald man (who we can denote as being affected by the baldness trait) has the genotype \( X^bY \), where \( X^b \) represents the X chromosome with the baldness allele.

3. The woman is not bald and her father was bald, which means she must be a carrier for the baldness allele. Therefore, her genotype is \( X^bX \).

To determine their children's genotypes, we can set up a Punnett square:

- The father (bald man): \( X^bY \)
- The mother (carrier): \( X^bX \)

The possible offspring genotypes from these combinations are:

- From \( X^b \) (father) and \( X^b \) (mother): \( X^bX^b \) (bald daughter)
- From \( X^b \) (father) and \( X \) (mother): \( X^bX \) (carrier daughter)
- From \( Y \) (father) and \( X^b \) (mother): \( X^bY \) (bald son)
- From \( Y \) (father) and \( X \) (mother): \( XY \) (normal son)

Now let's summarize the potential offspring:

- Daughters:
- \( X^bX^b \) (bald)
- \( X^bX \) (carrier)

- Sons:
- \( X^bY \) (bald)
- \( XY \) (normal)

To find the chances of having a bald daughter:
- There are 2 daughters in total: 1 bald and 1 carrier.
- Therefore, the probability of having a bald daughter is \( \frac{1}{2} \), or 50%.

So, the chances that they have a bald daughter are **50%**.