Home
Class 12
BIOLOGY
A woman with normal vision but with colu...

A woman with normal vision but with colurblind father marries a coloublind man The fouth child of the couple is a boy .This boy

A

May or may not be colourblind

B

Must be colourblind

C

Must have nrormal vision

D

Will be partially coloublind due to being heterozygous

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the genetic inheritance of colorblindness, which is a recessive sex-linked trait. Here’s a step-by-step breakdown of the solution: ### Step 1: Understand the Genetics of Colorblindness Colorblindness is a recessive trait linked to the X chromosome. This means that males (XY) are more likely to express the trait because they have only one X chromosome. Females (XX) can be carriers if they have one normal vision allele and one colorblind allele. ### Step 2: Determine the Genotypes of the Parents - The woman has normal vision but has a colorblind father. Therefore, she must be a carrier for colorblindness. Her genotype is X^N X^c (where X^N is the normal vision allele and X^c is the colorblind allele). - The man is colorblind, so his genotype is X^c Y. ### Step 3: Set Up the Punnett Square We can set up a Punnett square to determine the possible genotypes of their children. The woman (X^N X^c) can pass on either X^N or X^c, while the man (X^c Y) can pass on either X^c or Y. | | X^c (from father) | Y (from father) | |---------|--------------------|------------------| | X^N | X^N X^c (carrier girl) | X^N Y (normal boy) | | X^c | X^c X^c (colorblind girl) | X^c Y (colorblind boy) | ### Step 4: Analyze the Possible Outcomes From the Punnett square, we can see the possible genotypes of the children: 1. X^N X^c - Carrier girl (normal vision) 2. X^c X^c - Colorblind girl 3. X^N Y - Normal vision boy 4. X^c Y - Colorblind boy ### Step 5: Determine the Outcome for the Fourth Child The question states that the fourth child is a boy. The possible outcomes for a boy are: - X^N Y (normal vision) - X^c Y (colorblind) This means that the boy may or may not be colorblind. There is a 50% chance he will be colorblind and a 50% chance he will have normal vision. ### Final Answer The correct option is: **may or may not be colorblind.** ---
Promotional Banner

Topper's Solved these Questions

  • CHROMOSOMAL BASIS OF INHERITANCE

    DINESH PUBLICATION ENGLISH|Exercise Check Your Grasp|57 Videos

  • CHORDATA- PROTOCHORDATA

    DINESH PUBLICATION ENGLISH|Exercise CYG|11 Videos

  • COELENTERATA AND CTENOPHORA (OR CNIDARIA AND ACNIDARIA)

    DINESH PUBLICATION ENGLISH|Exercise CYG|10 Videos

Similar Questions

Explore conceptually related problems

A woman with normal vision, but whose father was colour blind, marries a colourblind man. Suppose that the fourth child of this couple was a boy. This boy :

A woman with albinic father marries an albinic man. The proportion of her progeny is

If a colourblind woman marries a normal visioned man. Their sons will be

If a colourblind woman marries a normal visioned man. Their sons will be

A boy with normal brother and colourblind sister has his parents

If a colourblind woman marries a normal visioned man, their sons will be -

A woman with normal vision whose mother was colourblind married a man with normal vision. The woman is pregnant with her first child. What is the probability that the cild will be a colour blind boy?

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

A women with normal vision marries a man with normal vision and gives birth to a colourblind son. Her husband dies and she marries a colourblind man. Wht is the probability of heer children having the abnormality :-

Colour blindness is a recessive trait. A couple with normal vision has two sons - one colourblind and one with normal vision. If the couple has two daughters, what proportion of them will have normal vision?