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A normal woman whose father was colour b...

A normal woman whose father was colour blind marries a normal man .the progeny would be

A

Sons normal duaghters colour blind

B

Sons colour blind, daughter normal

C

50% sons colour blind, remaining 50 % sons and all daughters phenotypically normal

D

Both sons and daughter are colour blind.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of determining the progeny of a normal woman whose father was color blind and a normal man, we can follow these steps: ### Step 1: Determine the Genotypes - The woman is normal but has a color-blind father. Color blindness is an X-linked recessive trait. Therefore, her genotype must be X^C X (where X^C represents the normal vision allele and X represents the color-blind allele). Since her father was color blind, she must have inherited the color-blind allele from him. - The normal man has the genotype XY, where X represents the normal vision allele. ### Step 2: Set Up the Punnett Square - We will set up a Punnett square to determine the possible genotypes of the offspring. The alleles from the mother (X^C X) and the father (XY) will be placed along the top and side of the square. ``` X Y ---------------- X^C | X^C X | X^C Y | ---------------- X | X X | X Y | ---------------- ``` ### Step 3: Analyze the Offspring - From the Punnett square, we can see the following combinations: - X^C X (Daughter, carrier for color blindness) - X^C Y (Son, color blind) - X X (Daughter, normal vision) - X Y (Son, normal vision) ### Step 4: Determine the Progeny Ratios - The offspring can be categorized as follows: - Daughters: 1 carrier (X^C X) and 1 normal (X X) → 50% normal, 50% carrier - Sons: 1 color blind (X^C Y) and 1 normal (X Y) → 50% color blind, 50% normal ### Conclusion - Therefore, the progeny would be: - 50% sons color blind, 50% sons normal - All daughters would be phenotypically normal but 50% would be carriers. ### Final Answer The correct option is: **50% sons color blind, remaining 50% sons normal, and all daughters phenotypically normal (Option 3).** ---
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