Question

If a male with hypophosphatemia (X‐linked dominant trait) marries a normal female, which of the following prediction about their potential progeny would be true?

• A. All of their sons would inherit the disease
• B. All of their daughters would inherit the disease
• C. About 50% of their sons would inherit the disease
• D. About 50% of their daughters would inherit the disease

Hint:

For any X-linked dominant condition, an affected father will pass the trait to all of his daughters but none of his sons, as sons only inherit his Y chromosome.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
X-linked dominant traits are expressed in both hemizygous males (\(X^D Y\)) and heterozygous females (\(X^D X^d\)).
The inheritance pattern of sex-linked traits is defined by which sex chromosomes are passed from parents to offspring.

Step 2: Detailed Explanation:
Let us establish the genotypes of the parents:
- The father has hypophosphatemia, an X-linked dominant trait.
His genotype must be: \[ \text{Father's genotype} = X^D Y \]
- The mother is phenotypically normal, so she must carry only the recessive normal alleles.
Her genotype must be: \[ \text{Mother's genotype} = X^d X^d \]
During reproduction, the parents pass sex chromosomes to their offspring as follows:
Sons: Receive the Y chromosome from their father and one of the \(X^d\) chromosomes from their mother.
Their genotype will be \(X^d Y\). All sons will have a normal phenotype.
Daughters: Must receive the \(X^D\) chromosome from their father and an \(X^d\) chromosome from their mother.
Their genotype will be \(X^D X^d\). Because the trait is dominant, all daughters will express the disease.
Thus, all daughters will inherit the disease, and all sons will be normal.

Step 3: Final Answer:
All of their daughters would inherit the disease.