Question

A colorblind carrier woman marries a normal man . Explain the different possibilities of

offsprings produced by them with the help of punnet square.​

Answer 1

Answer:

Colourblindness is a X-linked recessive disorder. One copy of the affected gene in males in each cell is sufficient to cause the disorder (X

c

Y). Females with two copies of the affected gene show the disorder (X

c

X

c

). Females heterozygous (X

c

X) for this trait are normal but serve as a carrier of the disease. According to the question, the female is colourblind (X

c

X

c

) and father is normal (XY). The affected mother will inherit the disease to 100% sons (X

c

Y) while the 100% daughter will be normal but will serve as carrier. Correct answer is option B.

Parent generation: (X

c

X

c

) x XY

Gametes :

(X

c

X

c

) -->

XY X

c

X

c

X X

c

X

carrier girl

X

c

X

carrier girl

Y X

c

Y

colourblind boy

X

c

Y

colourblind boy

Answer 2

Given:

A colorblind carrier woman marries a normal man.

To find:

The different possibilities of offsprings produced by them with the help of punnet square.

Solution:

The 1st thing to know about COLOUR-BLINDNESS is that:

• It is an X linked recessive genetic disorder.

• So, a colourblind carrier women will have one X_(c) chromosome and one normal X chromosome.

• And a normal male will have one X chromosome and one Y chromosome.

Now refer to Punnett Square attached in the diagram:

The different possibilities are:

• One colourblind carrier girl [XX_(c)]

• One colourblind boy [X_(c)Y]

• One normal girl [XX] and one normal boy [XY].

________________________________

Some additional informations:

• The same case would have been created in case of hemophilia too, because it's also a X linked recessive disorder.

• However , cases would have been different if the disorder were Alkaptonuria, Spherocytosis, etc. That's because they are autosomal disorders.

$\star$ Hope It Helps.