Question

A woman has a rare abnormality of the eyelids called ptosis. This condition is caused by a dominant allele P. The woman's father had ptosis, but mother had normal eyelids. Her father's mother had normal eyelids. What proportion of the woman's children will have ptosis if she marries a man with normal eyelids?

• A. 1/2
• B. 1/3
• C. 1/4
• D. 2/3

Hint:

Crossing a heterozygote (\( Pp \)) with a homozygous recessive individual (\( pp \)) is a standard genetic test cross.
The resulting phenotypic ratio is always \( 1:1 \), meaning exactly \( 50\% \) or \( 1/2 \) of the progeny will express the dominant phenotype.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Ptosis is inherited as an autosomal dominant trait, denoted by the dominant allele \( P \).
Individuals exhibiting the normal eyelid phenotype are homozygous recessive, represented by \( pp \).
An individual displaying the ptosis phenotype must possess at least one dominant allele, meaning their genotype is either \( PP \) or \( Pp \).
By reviewing the parental phenotypes, we can deduce the exact heterozygous status of the parents and predict offspring frequencies.

Step 2: Detailed Explanation:
Let us establish the genotypes of the family members systematically.
1. The woman's mother has normal eyelids, which means her genotype is homozygous recessive, or \( pp \).
2. The woman has ptosis, indicating she has at least one dominant allele \( P \).
3. Since her mother is \( pp \), the mother can only pass a recessive \( p \) allele to her daughter.
4. Therefore, the woman must have inherited the \( P \) allele from her father and the \( p \) allele from her mother, establishing her genotype as heterozygous \( Pp \).
5. The woman marries a man with normal eyelids, who must have the homozygous recessive genotype \( pp \).
Now, let us cross the heterozygous woman (\( Pp \)) with the homozygous recessive man (\( pp \)):
The Punnett square of the cross \( Pp \times pp \) yields:
- \( 50\% \) heterozygous \( Pp \) (exhibiting the ptosis phenotype).
- \( 50\% \) homozygous recessive \( pp \) (exhibiting the normal eyelid phenotype).
Thus, the expected proportion of her children with ptosis is \( 1/2 \) or \( 50\% \).

Step 3: Final Answer:
The proportion of the woman's children expected to have ptosis is \( 1/2 \).