Question

Consider three independently assorting gene pairs, \(A / a, B / b,\) and \(C / c,\) where each demonstrates typical dominance \((A-, B-, C-)\) and recessiveness \((a a, b b, c c) .\) What is the probability of obtaining an offspring that is \(A A B b C c\) from parents that are \(A a B b C C\) and \(A A B b C c ?\)

Short answer

Answer: The probability of obtaining an offspring with the genotype \(AABbCc\) from these parents is 0.

Step-by-step solution

Find the probability of obtaining an \(AA\) allele from the parents

To get an \(AA\) allele, the offspring has to inherit the dominant allele \(A\) from both parents. The first parent has a genotype of \(Aa\), which means the probability of passing on the \(A\) allele is 1/2 (since there's 1 dominant allele and 2 alleles in total for this gene pair). The second parent has a genotype of \(AA\), which means the probability of passing on the \(A\) allele is 1 (since there are 2 dominant alleles and 2 alleles in total for this gene pair). Multiply the probabilities from each parent to get the probability of obtaining an \(AA\) offspring: \(P(AA) = \frac{1}{2} * 1 = \frac{1}{2}\)

Find the probability of obtaining a \(Bb\) allele from the parents

To get a \(Bb\) allele, the offspring has to inherit one dominant allele \(B\) and one recessive allele \(b\) from the parents. Both parents have genotypes of \(Bb\) for this gene pair, which means the probability of passing on the \(B\) allele is 1/2, and the probability of passing on the \(b\) allele is also 1/2. To obtain a \(Bb\) offspring, it can either inherit \(B\) from the first parent and \(b\) from the second, or inherit \(b\) from the first parent and \(B\) from the second. Therefore, we have to add the probabilities of these two cases: \(P(Bb) = (\frac{1}{2} * \frac{1}{2}) + (\frac{1}{2} * \frac{1}{2}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\)

Find the probability of obtaining a \(Cc\) allele from the parents

To get a \(Cc\) allele, the offspring has to inherit the recessive allele \(c\) from both parents. The first parent has a genotype of \(CC\) and the second parent has a genotype of \(Cc\). Since the first parent does not have a \(c\) allele, the probability of obtaining a \(Cc\) offspring from these parents is 0. \(P(Cc) = 0\)

Calculate the probability of obtaining an \(AABbCc\) offspring from the parents

To find the probability of obtaining an \(AABbCc\) offspring, we need to multiply the probabilities we found in steps 1-3: \(P(AABbCc) = P(AA) * P(Bb) * P(Cc) = \frac{1}{2} * \frac{1}{2} * 0 = 0\) The probability of obtaining an offspring with the genotype \(AABbCc\) from these parents is 0.