Question

multiple attempts not allowed. this test can only be taken once. force completion once started, this test must be completed in one sitting. do not leave the test before clicking save and submit. this test does not allow backtracking. changes to the answer after submission are prohibited. your answers are saved automatically. remaining time: 42 minutes, 52 seconds. question completion status: question 38 2 points a woman who is a carrier for hemophilia marries a man who is a normal. which of the following statements is true? 1. since the father is normal, no children will have hemophilia 2. the sons will have a 50% chance of being hemophiliacs 3. the daughters will all be carriers for hemophilia 4. the gene for hemophilia is found on the y chromosome

Explanation:

Step1: Analyze hemophilia inheritance

Hemophilia is an X - linked recessive disorder. Let \(X^h\) represent the hemophilia - carrying X chromosome and \(X\) represent the normal X chromosome. The woman (carrier) has the genotype \(X^hX\), and the man (normal) has the genotype \(XY\).

Step2: Use Punnett square for cross

The possible gametes from the woman are \(X^h\) and \(X\), and from the man are \(X\) and \(Y\). The Punnett square for the cross \(X^hX\times XY\) is:

	\(X\)	\(Y\)
\(X\)	\(XX\) (normal daughter)	\(XY\) (normal son)

Step3: Evaluate each option

• Option 1: Sons can have hemophilia (\(X^hY\)), so this is false.
• Option 2: Out of the two possible sons (\(X^hY\) and \(XY\)), there is a \(50\%\) chance of a son having hemophilia. This is true.
• Option 3: Daughters can be either \(X^hX\) (carrier) or \(XX\) (normal), not all carriers. So this is false.
• Option 4: Hemophilia gene is on the X chromosome (X - linked), not Y. So this is false.

Answer:

1. the sons will have a 50% chance of being hemophiliacs