Chance of passing a genetic disease &;y&; trait by the affected parents to children is 0.16. They plan to have two children. Prabability of both the children having &;y&; trait is –
**Core Concept**
The question is testing the understanding of the probability of inheritance of a genetic trait from affected parents to their children, specifically using the concept of genotype and phenotype probabilities in a Mendelian inheritance pattern.
**Why the Correct Answer is Right**
To solve this problem, we need to understand the probability of each parent passing on the recessive allele (let's denote it as 'a') that causes the genetic disease and trait. Let's assume the affected parents are both carriers, meaning they have one normal allele ('A') and one recessive allele ('a'). The genotype of each parent can be represented as 'Aa'. The probability of each parent passing on the recessive allele is 0.5 (50%). Since the parents are both carriers, the probability of both children inheriting the recessive allele (and thus the disease and trait) can be calculated using the Punnett square or by multiplying the probabilities. In this case, the probability of both children inheriting the recessive allele is 0.5 x 0.5 = 0.25. However, this is the probability of both children being affected homozygotes (aa). To find the probability of both children having the trait (which can also include being carriers, Aa), we need to consider the genotype probabilities of each child and multiply them together.
**Why Each Wrong Option is Incorrect**
**Option A:** This option is not provided, so we will not discuss it.
**Option B:** This option is not provided, so we will not discuss it.
**Option C:** This option is not provided, so we will not discuss it.
**Option D:** This option is not provided, so we will not discuss it.
**Clinical Pearl / High-Yield Fact**
When dealing with genetic inheritance, it's essential to understand the genotype and phenotype probabilities of each generation. This can be done using a Punnett square or by calculating the probabilities of each possible genotype combination.
**Correct Answer:**. 0.25