Question

An alternative to using the expanded binomial equation and Pascal's triangle in determining probabilities of phenotypes in a subsequent generation when the parents' genotypes are known is to use the following equation: \(\frac{n !}{s ! t !} a^{s} b^{t}\) where \(n\) is the total number of offspring, \(s\) is the number of offspring in one phenotypic category, \(t\) is the number of offspring in the other phenotypic category, \(a\) is the probability of occurrence of the first phenotype, and \(b\) is the probability of the second phenotype. Using this equation, determine the probability of a family of 5 offspring having exactly 2 children afflicted with sickle-cell anemia (an autosomal recessive disease \()\) when both parents are heterozygous for the sickle-cell allele.

Short answer

Answer: The probability is \(\frac{135}{512}\).

Step-by-step solution

Calculate the probabilities of each phenotype

To find the probability of children being afflicted or not afflicted with sickle-cell anemia, we need to consider the possible genotypes of the parents. Both parents are heterozygous, which means they carry one normal allele (A) and one sickle-cell allele (S). Therefore, their genotypes are AS. When two heterozygous parents (AS) reproduce, the possible outcomes for their offspring are: 1. AA (normal): Probability = 1/4 2. AS (carrier): Probability = 1/2 3. SA (carrier - same as AS): Probability = 1/4 4. SS (sickle-cell anemia): Probability = 1/4 We will consider the SS genotype as the affected offspring, and we'll group AA, AS, and SA genotypes as non-affected offspring. So, the probabilities are: - Probability of having sickle-cell anemia (a): 1/4 - Probability of not having sickle-cell anemia (b): 3/4

Apply the given equation to determine the probability of having 2 children with sickle-cell anemia

Now, we'll use the equation provided: \(\frac{n !}{s ! t !} a^{s} b^{t}\) Where: - n = total number of offspring (5) - s = number of offspring with sickle-cell anemia (2) - t = number of offspring not having sickle-cell anemia (3) - a = probability of having sickle-cell anemia (1/4) - b = probability of not having sickle-cell anemia (3/4) Plugging in the values in the equation, we have: \(\frac{5 !}{2 ! 3 !} (1/4)^{2} (3/4)^{3}\)

Solve the equation

Let's evaluate the equation step by step: 1. Find the factorials: 5! = 120, 2! = 2, 3! = 6 2. Substitute the factorials into the equation: \(\frac{120}{2 * 6} (1/4)^{2} (3/4)^{3}\) 3. Simplify the factorial fraction: \(\frac{120}{12} (1/4)^{2} (3/4)^{3}\) = 10(1/4)^{2}(3/4)^{3}$ 4. Calculate the probabilities raised to their respective powers: \((1/16)(27/64)\) 5. Multiply the fractions together: \(10(27/1024)\) 6. Simplify the final fraction: \(10(27/1024) = \frac{270}{1024} = \frac{135}{512}\) The probability of a family of 5 having exactly 2 children with sickle-cell anemia when both parents are heterozygous is \(\frac{135}{512}\).