Question

With the help of given pedigree, find out the probability for the birth of a child having no disease and being a carrier (has the disease mutation in one allele of the gene) in F₃ generation.

Options

• 1/4

• 1/2

• 1/8

• Zero

Answer 1

1/2

Explanation:

From the pedigree:

Step 1: Determine Mode of Inheritance From the pedigree:

• Affected individuals are present in both sexes.
• Disease skips generations.
• Affected offspring (black squares and circles) appear from carrier parents.

This suggests it is an X-linked recessive disorder.

Genotype key:

• XX or XY: Normal (non-carrier, non-affected)
• X^aX: Carrier female (one mutant allele, no disease)
• X^aX^a and X^aY affected female and male

Step 2: Look at F₂ Generation Couple

We are examining the F₂ couple who produce the F₂ children.

From the pedigree:

• Mother (F₂): A carrier female → Genotype = X^aX
• Father (F₂): An affected male → Genotype = X^aY

Their cross is:

X^aX (carrier) × X^aY (affected)

Step 3: Use a Punnett Square

Cross: X^aX × X^aY

	Xa	Y
Xa	XaXa	XaY
X	XaX	XY

Results:

• X^aX: 25% (carrier female, not affected).
• X^aX^a: 25% (affected female).
• XY: 25% (unaffected male).
• X^aY: 25% (affected male).

Step 4: What the Question Asks 

We want the probability of a child who is Unaffected and Carrier.

25% males are unaffected, and 25% females are carriers but not diseased. So, the probability for the birth of a child having no disease and being a carrier is 50% or 1/2.