Question

in humans, hemophilia is a sex linked trait. females can be normal, carriers, or have the disease. males will either have the disease or not (but they wont ever be carriers)
$x^h x^h$ = female, normal
$x^h x^h$ = female, carrier
$x^h x^h$ = female, hemophiliac
$x^h y$ = male, normal
$x^h y$ = male, hemophiliac
show the cross of a man who has hemophilia with a woman who is a carrier

1. what is the probability that their children will have the disease? ______

Explanation:

Part 1: Show the cross (Punnett Square)

Step 1: Determine the genotypes of parents

• Man with hemophilia: Genotype is $X^hY$ (since males with hemophilia have $X^h$ and $Y$).
• Woman who is a carrier: Genotype is $X^HX^h$ (carrier females have one normal $X^H$ and one hemophilia - causing $X^h$).

Step 2: Determine the gametes

• Gametes from the man ($X^hY$):
• He can produce two types of gametes: $X^h$ and $Y$.
• Gametes from the woman ($X^HX^h$):
• She can produce two types of gametes: $X^H$ and $X^h$.

Step 3: Construct the Punnett Square

	$X^h$ (from man)	$Y$ (from man)
$X^h$ (from woman)	$X^hX^h$ (female, hemophiliac)	$X^hY$ (male, hemophiliac)

Part 2: Probability that their children will have the disease

Step 1: Count the total number of possible offspring genotypes

From the Punnett Square, we have 4 possible genotypes of offspring: $X^HX^h$, $X^HY$, $X^hX^h$, $X^hY$.

Step 2: Count the number of offspring with the disease

• Offspring with hemophilia (disease) have genotypes $X^hX^h$ (female) and $X^hY$ (male). So there are 2 out of 4 offspring with the disease.

Step 3: Calculate the probability

The probability is calculated as the number of favorable outcomes (offspring with disease) divided by the total number of outcomes (total offspring). So the probability $P=\frac{2}{4}=\frac{1}{2}$ or 50%.

Answer:

for Punnett Square:

	$X^h$	$Y$
$X^h$	$X^hX^h$	$X^hY$