Question

If a woman carrier for color blindness marries to a color blind man and produce several children, the expected phenotype (color blind or normal) of the offspring would be -

• A. All the offspring will be color blind
• B. All male offspring will be color blind and all females will be normal
• C. All male offspring will be normal and female will be color blind
• D. All female offspring will be carriers of color blindness or color blind

Hint:

For X-linked recessive crosses, remember that a daughter can only be color blind if her father is color blind and her mother is at least a carrier.
Sons always inherit their X chromosome from their mother, so a color-blind father cannot pass the trait directly to his sons.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question involves understanding the inheritance pattern of color blindness, which is a classic example of an X-linked recessive trait in humans.
In such traits, the gene responsible for the condition is located on the X chromosome.
Women have two X chromosomes (\(XX\)), while men have one X and one Y chromosome (\(XY\)).
A carrier female has one normal allele and one defective allele on her two X chromosomes, usually not expressing the trait herself.
A color-blind male has the defective allele on his single X chromosome, thus expressing the trait.

Step 2: Detailed Explanation:
1. Let us denote the normal vision allele as \(X\) and the color blindness allele as \(X^c\).
2. The woman is a carrier, so her genotype is \(X X^c\).
3. The man is color blind, so his genotype is \(X^c Y\).
4. During gamete formation, the woman produces two types of eggs: those carrying the normal \(X\) chromosome and those carrying the recessive \(X^c\) chromosome.
5. The man produces two types of sperm: those carrying the \(X^c\) chromosome and those carrying the \(Y\) chromosome.
6. By creating a Punnett square to predict the offspring genotypes:
- Egg \(X\) + Sperm \(X^c\) \(\rightarrow\) \(X X^c\) (Carrier Daughter with normal vision)
- Egg \(X\) + Sperm \(Y\) \(\rightarrow\) \(XY\) (Normal Vision Son)
- Egg \(X^c\) + Sperm \(X^c\) \(\rightarrow\) \(X^c X^c\) (Color Blind Daughter)
- Egg \(X^c\) + Sperm \(Y\) \(\rightarrow\) \(X^c Y\) (Color Blind Son)
7. Analyzing the results, we find that 50% of the sons are normal and 50% are color blind.
8. For the daughters, 50% are carriers (having normal vision but carrying the gene) and 50% are color blind.
9. Therefore, all female offspring will either be carriers or manifest color blindness.
Final Answer:
Based on the genetic cross, the only accurate statement among the options regarding the phenotypes is that all female offspring are either carriers or color blind.
This corresponds to option (D).