Question

hemophilia is an x - linked recessive disorder.
what is the probability of the offspring having hemophilia for the cross that is shown in the punnett square?
0 percent
25 percent
50 percent

Explanation:

Step1: Analyze the Punnett Square

The Punnett square has four possible offspring genotypes: \(X^H X^h\), \(X^H Y\), \(X^h X^h\), and \(X^h Y\).

Step2: Determine Hemophilia Genotypes

Hemophilia is X - linked recessive, so males with \(X^h Y\) and females with \(X^h X^h\) have hemophilia. From the square, \(X^h X^h\) (female with hemophilia) and \(X^h Y\) (male with hemophilia) are two of the four genotypes.

Step3: Calculate Probability

The number of offspring with hemophilia is 2 out of 4. To find the percentage, we use the formula \(\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\times100=\frac{2}{4}\times100 = 50\%\)? Wait, no, wait. Wait, \(X^H X^h\) is a carrier female (no hemophilia), \(X^H Y\) is a normal male, \(X^h X^h\) is a hemophilic female, \(X^h Y\) is a hemophilic male. So two out of four have hemophilia? Wait, no, let's re - check. Wait, the mother's gametes are \(X^H\) and \(X^h\), father's gametes are \(X^h\) and \(Y\). So the four combinations:

1. \(X^H\) (mother) and \(X^h\) (father): \(X^H X^h\) (carrier female, no hemophilia)
2. \(X^H\) (mother) and \(Y\) (father): \(X^H Y\) (normal male)
3. \(X^h\) (mother) and \(X^h\) (father): \(X^h X^h\) (hemophilic female)
4. \(X^h\) (mother) and \(Y\) (father): \(X^h Y\) (hemophilic male)

So the number of offspring with hemophilia is 2 ( \(X^h X^h\) and \(X^h Y\)) out of 4. So the probability is \(\frac{2}{4}\times100 = 50\%\)? Wait, but wait, maybe I made a mistake. Wait, no, let's count again. Wait, the four squares: top left \(X^H X^h\) (no), top right \(X^H Y\) (no), bottom left \(X^h X^h\) (yes), bottom right \(X^h Y\) (yes). So 2 out of 4, which is 50%? Wait, but the options have 25, 50, 0. Wait, no, wait, maybe I misread the gametes. Wait, the father's gametes: the top of the Punnett square is \(X^h\) and \(Y\), mother's gametes are \(X^H\) and \(X^h\) (left side). So the four cells:

• \(X^H\) (mother) and \(X^h\) (father): \(X^H X^h\) (carrier, female, no hemophilia)
• \(X^H\) (mother) and \(Y\) (father): \(X^H Y\) (normal male)
• \(X^h\) (mother) and \(X^h\) (father): \(X^h X^h\) (hemophilic female)
• \(X^h\) (mother) and \(Y\) (father): \(X^h Y\) (hemophilic male)

So two of the four offspring have hemophilia. So the probability is \(\frac{2}{4}\times100 = 50\%\).

Answer:

50 percent