Chance of passing a genetic disease ‘y’ trait by the affected parents to children is 0.16. They plan to have two children. Probability of both the children having ‘y’ trait is ?

**Core Concept**
The question is based on the probability of passing a genetic trait from affected parents to their children, specifically the concept of Mendelian inheritance and the Punnett square. The genetic disease 'y' trait is likely an autosomal recessive disorder.

**Why the Correct Answer is Right**
To solve this problem, we need to calculate the probability of both children having the 'y' trait. Since the parents are affected, they must be homozygous recessive for the 'y' trait (yy). The probability of passing the 'y' trait to each child is 1 (certainty). The probability of both children having the 'y' trait can be calculated using the formula for the probability of two independent events: P(yy) = P(y) × P(y) = 0.16 × 0.16 = 0.0256.

**Why Each Wrong Option is Incorrect**
**Option A:** This option is not provided.
**Option B:** This option is not provided.
**Option C:** This option is not provided.
**Option D:** This option is not provided.

**Clinical Pearl / High-Yield Fact**
When dealing with genetic disorders, it's essential to understand the mode of inheritance (autosomal dominant, autosomal recessive, X-linked, etc.) to accurately calculate the probability of passing the disease to offspring.

**Correct Answer: Correct Answer: C. 0.0256.**

✓ Correct Answer: D. 0.0256