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.

Pedigree chart F0-F3 with symbols: unaffected male, affected male, carrier female, unaffected female, affected female; F3 individuals marked with ?

1. 1/4

2. 1/2

3. 1/8

4. Zero

Doubtly Asked 30 Oct 2026 NEET UG 2025 Practice test neet Save WhatsApp Report

1 answer

Best answer

Answer: 1. 1/4

Key idea: The legend shows carriers only among females, so the trait is X-linked recessive.

The F2 mother is a carrier: \(X^AX^a\). The F2 father is affected: \(X^aY\).
Offspring: \(X^AX^a\) (carrier daughter), \(X^aX^a\) (affected daughter), \(X^AY\) (normal son), \(X^aY\) (affected son), each 1/4.
A child with no disease who carries the mutation must be a carrier daughter: probability 1/4.

Option 1.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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