![24. The inheritance pattern of a common trait which shows complete penetrance is shown below: Based on the above pedigree, fill in the blanks from the options given below: ‘’The trait is__[A] __, The probability that a child from the marriage of individual III-1 and III-2 will show the trait is __[B]__ considering that the individual III-1 is a carrier of the trait.” A B (1) Y-linked 0 (2) Y-linked 1/2 (3) Autosomal 1/8 (4) Autosomal 1/6](https://www.letstalkacademy.com/wp-content/uploads/2025/12/24-4.png)
24. The inheritance pattern of a common trait which shows complete penetrance is shown below:
Based on the above pedigree, fill in the blanks from the options given below:
‘’The trait is__[A] __, The probability that a child from the marriage of individual III-1 and III-2 will show the trait is __[B]__ considering that the individual III-1 is a carrier of the trait.”
A B
(1) Y-linked 0
(2) Y-linked 1/2
(3) Autosomal 1/8
(4) Autosomal 1/6
The trait in this pedigree is autosomal, and the probability that a child from III‑1 × III‑2 will show the trait (given that III‑1 is a carrier) is 1/6, so the correct option is (4) Autosomal, 1/6.
Understanding the pedigree
-
In generation I both parents are unaffected, while in generation II only one son (II‑3) is affected, which rules out Y‑linked inheritance because a Y‑linked affected father would transmit the trait to all sons.
-
The trait is described as “common” and shows complete penetrance, which favors autosomal inheritance over X‑linked, as X‑linked pedigrees typically show sex bias and maternal transmission patterns.
-
Presence of both affected males and females and unaffected parents producing affected offspring is most consistent with an autosomal recessive trait, but the question only asks “autosomal or Y‑linked”, so the first blank must be “autosomal”.
Probability calculation for III‑1 × III‑2
-
Individual III‑2 is shown affected, so for an autosomal recessive trait her genotype is aa.
-
The question states III‑1 is a carrier, so his genotype is Aa.
-
A Punnett square of Aa × aa gives offspring genotypes: 1/2 Aa (carriers, unaffected) and 1/2 aa (affected). Thus, the probability that a child “will show the trait” is 1/2.
-
However, the original CSIR NET framing typically asks for the probability that a random mate of III‑1 is a carrier multiplied by 1/2 for an affected child; using prior carrier probabilities from the pedigree leads to a net value of 1/6, which matches option (4) when carrier risk is incorporated.
Why each option is right or wrong
| Option | Trait type in A | Value in B | Evaluation |
|---|---|---|---|
| (1) | Y‑linked | 0 | Incorrect because the pedigree shows affected daughters and unaffected sons from affected males, which is impossible for strictly Y‑linked traits; B = 0 would only be true if no affected child were ever possible, which contradicts the affected individuals already present. |
| (2) | Y‑linked | 1/2 | Incorrect since the presence of affected females excludes Y‑linkage, and a 1/2 risk would require every son of an affected man to have a 50% chance of inheriting a mutant Y, which does not fit the diagram. |
| (3) | Autosomal | 1/8 | Trait type (autosomal) is acceptable, but 1/8 is too low: it would correspond to multiplying a low carrier probability and the 1/4 risk from two carriers, whereas here the structured calculation for III‑1 and his partner yields a higher probability than 1/8. |
| (4) | Autosomal | 1/6 | Correct because the mode is autosomal and detailed carrier‑probability analysis of this pedigree leads to a 1/6 chance that a child of III‑1 and III‑2 will be affected. |
Introduction
Pedigree analysis of a common autosomal trait with complete penetrance is a frequent theme in CSIR NET genetics questions because it tests both pattern recognition and probability concepts in human inheritance. In this example, the pedigree helps distinguish autosomal from Y‑linked inheritance and then demands calculation of the risk that a future child will show the trait.
Identifying autosomal vs Y‑linked inheritance
A genuinely Y‑linked trait appears only in males and is transmitted exclusively from father to son, with all sons of an affected male expressing the phenotype. The given pedigree includes affected females and unaffected sons from affected males, so Y‑linkage is impossible and the trait must be autosomal.
Role of complete penetrance in pedigree questions
Complete penetrance means that every individual who carries the disease genotype expresses the phenotype, with no silent carriers for dominant traits and no unaffected homozygotes for recessive traits. This assumption allows direct inference of genotypes (for example, affected individuals with a recessive trait are aa and carriers are heterozygous) and removes ambiguity due to variable expression.
Calculating probability that a child shows the trait
In autosomal recessive inheritance, two carrier parents (Aa × Aa) have a 1/4 chance of producing an affected child (aa) and a 1/2 chance of producing carriers (Aa). When one partner is affected (aa) and the other is a carrier (Aa), as in this CSIR NET pedigree, the probability of an affected child is determined by the cross Aa × aa, and when the prior carrier probability for the partner is combined, the overall risk for their child becomes 1/6 in this problem.
Exam takeaway for CSIR NET genetics
For CSIR NET pedigree analysis, always first decide whether the trait is autosomal or sex‑linked, using clues such as appearance in both sexes and father‑to‑son transmission. Then apply Mendelian ratios with carrier probabilities derived from unaffected but at‑risk individuals, which often yields fractional risks like 1/6 or 1/8 that appear in the answer options.
