The correct answer is option (3) 2/3. Individual 6 has a 2/3 probability of being a heterozygous carrier for the autosomal recessive trait represented in this pedigree.​

Understanding the pedigree

This pedigree shows an autosomal recessive trait, meaning affected individuals are homozygous recessive (aa), while unaffected individuals are either homozygous dominant (AA) or heterozygous carriers (Aa). Individual 7 is affected (aa), but both parents 4 and 5 are unaffected, so they must each be obligate carriers (Aa × Aa).​

From an Aa × Aa cross, the genotypic ratio among children is:

• 1/4 AA (unaffected, non‑carrier)

• 1/2 Aa (unaffected carrier)

• 1/4 aa (affected)​

Individual 6 is unaffected, so aa is excluded. Among unaffected offspring (AA, Aa, Aa), 2 out of 3 genotypes are carriers (Aa). Therefore, the conditional probability that individual 6 is heterozygous is 2/3.​

Option‑wise explanation

Option (1) 1/4

A value of 1/4 arises if one directly takes the Aa frequency from a simple Punnett square (1/4 AA, 1/2 Aa, 1/4 aa) and mistakenly reads “1/4” from this distribution. However, this ignores the given information that individual 6 is unaffected, which eliminates the aa class and requires a conditional probability based only on unaffected genotypes.​

Option (2) 1/2

The value 1/2 corresponds to the probability that an arbitrary child from an Aa × Aa mating is a carrier, without any condition on phenotype. Since the pedigree explicitly shows that individual 6 is unaffected, the calculation must again be restricted to unaffected offspring only, making 1/2 an underestimate.​

Option (3) 2/3 ✅

Among unaffected children of Aa × Aa parents, the possible genotypes are AA, Aa, Aa, so carriers constitute 2 out of 3 unaffected genotypes. Because individual 6 is clearly unaffected yet comes from definite carrier parents, the correct probability that 6 is a heterozygote is therefore 2/3.​

Option (4) 1/3

A probability of 1/3 would imply 1 carrier among 3 unaffected genotypes, which contradicts the Punnett square for Aa × Aa (where two of the three unaffected outcomes are carriers). There is no reasonable genetic or conditional argument from this pedigree that leads to 1/3, so this option is incorrect.​

SEO‑friendly introduction

Pedigree analysis questions on autosomal recessive inheritance often test conditional probability, especially when the genotype of an unaffected individual must be inferred from an affected sibling. In this CSIR NET problem, the task is to calculate the probability that individual 6 is a heterozygote in an autosomal recessive pedigree where both parents are carriers and one offspring (individual 7) is affected. By combining a Punnett square for an Aa × Aa cross with the information that individual 6 is unaffected, the correct answer is obtained as 2/3, and each multiple‑choice option can be logically evaluated.​