hemophilia is a sex linked recessive disorder. A man who has a hemophilia has a daughter who has a normal phenotype. She marries a man who also does not have hemophilia. what is the chance that they will have a daughter who has hemophilia? what chance will their son have inheriting hemophilia. if the couple has 4 sons . What is the probability that 4 boys will have hemophilia? show work by using punett squarehemophilia is a sex linked recessive disorder. A man who has a hemophilia has a daughter who has a normal phenotype. She marries a man who also does not have hemophilia. what is the chance that they will have a daughter who has hemophilia? what chance will their son have inheriting hemophilia. if the couple has 4 sons . What is the probability that 4 boys will have hemophilia? show work by using punett square

A sex-linked gene is on the X chromosome, and XX makes a girl. The daughter, getting the hemophilia gene from dad, is a carrier.

Since her husband can only give a normal X chromosome, the most her daughters could be is carriers, but by chance, half of the boys will have hemophilia.

Probability of all events occurring is found by multiplying the probability of the individual events.

To determine the probability of their children inheriting hemophilia, we need to understand the inheritance pattern for sex-linked recessive disorders:

1. Chance of having a hemophiliac daughter:
Since hemophilia is a sex-linked recessive disorder, the woman can only pass the defective gene to her children if she is a carrier (i.e., she inherited one normal copy of the gene from her mother and one defective copy from her hemophiliac father).

The Punnett square for this scenario would be as follows:
```
X X
Y XX XY
Y XX XY
```

In this case, all the possible offspring will be carriers of the hemophilia gene, but they will not display symptoms of the disorder.

2. Chance of having a hemophiliac son:
If the father does not have hemophilia, he must have inherited one normal copy of the gene from his mother and one normal copy from his father. Therefore, any son he has will not have hemophilia.

The Punnett square for this scenario would be as follows:
```
X Y
X XX XY
X XX XY
```

In this case, all possible offspring will not have hemophilia.

3. Probability of 4 boys with hemophilia:
If the couple has 4 sons, and we assume independent assortment (meaning the outcome of one birth does not affect the outcome of another), we can calculate the probability of having 4 boys with hemophilia by multiplying the individual probabilities.

Since the probability of having a hemophiliac daughter is 0, we don't need to consider it in this calculation. The probability of having a hemophiliac son is also 0 for the reasons explained earlier.

Therefore, the probability of having 4 boys with hemophilia is 0.

Note: Please consult a genetic counselor or healthcare professional for a comprehensive understanding of genetic disorders and inheritance patterns.

To determine the probability of certain traits being inherited, we can use Punnett squares. In the case of hemophilia, which is a sex-linked disorder, we need to consider the difference in inheritance patterns between males and females.

First, let's determine the chance that the couple will have a daughter who has hemophilia:

1. The man who has hemophilia (XhY) will contribute his X chromosome with the hemophilia gene to his daughter.

2. The woman who has a normal phenotype (XX) will contribute one of her X chromosomes without the hemophilia gene.

Using a Punnett square, we can combine the parental chromosomes:

Xh X
-----------------
Xh | XhXh XhX
X | XhX XX

As a result, there are two possible combinations: XhXh and XhX. This means that there is a 50% (1 out of 2) chance that their daughter will inherit hemophilia.

Next, let's determine the chance that their son will inherit hemophilia:

1. The father who has hemophilia (XhY) will contribute his X chromosome with the hemophilia gene to his son.

2. The mother who does not have hemophilia (XX) will contribute her X chromosome without the hemophilia gene.

Using a Punnett square again:

Xh X
-----------------
Xh | XhXh XhX
Y | XhY XY

As a result, there are two possible combinations: XhY and XY. This means that there is a 50% (1 out of 2) chance that their son will inherit hemophilia.

Now, let's determine the probability that all four sons will have hemophilia:

Since each son inherits one X chromosome from the mother and one Y chromosome from the father, we can calculate the probability using the multiplication rule:

Probability = (Probability of hemophilia in a son) ^ (Number of sons)

Using the information we determined earlier, the probability of a son having hemophilia is 50%, so:

Probability = (0.5) ^ 4
= 0.5 * 0.5 * 0.5 * 0.5
= 0.0625

Therefore, the probability that all four sons will have hemophilia is 6.25% or 0.0625.

Note: This calculation assumes that the parents are both carriers of the hemophilia gene.