Question

In a family of 10 children, where both parents are heterozygous for albinism, what mathematical equation predicts the probability that 8 are normal and 2 are albino?

Answer 1

The probability of a family with both parents heterozygous for albinism having 8 children with normal pigmentation and 2 with albinism is calculated using the binomial probability formula: P(2; 10, 1/4).

Step-by-step explanation:

The subject of this question is Biology, specifically genetics and probability in predicting inheritance patterns. The question asks for the probability that in a family with 10 children, where both parents are heterozygous for albinism, 8 children are normal and 2 are albino. Albinism is a recessive trait, so the parental genotype is Aa for both parents. Using a Punnett square for a monohybrid cross, we know that each child has a 1 in 4 chance of being albino (aa) and a 3 in 4 chance of having normal pigmentation (AA or Aa).

To calculate the probability of having exactly 8 children with normal pigmentation and 2 albino, we use the binomial probability formula: P(x; n, p) = (n choose x) * p^x * (1-p)^(n-x), where 'n' is the total number of trials (or children, 10 in this case), 'x' is the number of successes (2 albino children), and 'p' is the probability of a success on any given trial (1/4 for albino). Plugging in the values, we get: P(2; 10, 1/4) = (10 choose 2) * (1/4)^2 * (3/4)^8.

Answer 2

In a cross between two heterozygous parents, the probability that a child is homozygous for albinism is 1/4. So out of 10 children, the probability that 8 are normal and 2 are albino is

`\dbinom{10}2\left(\frac14\right)^2\left(1-\frac14\right)^(10-2)\approx0.2816`