An animal has the genotype AaBbCcDD. Gene A is on chromosome 1, gene B is on chromosome 2, gene C is on chromosome 3, and gene D is on chromosome 4. What is the probability that this animal will produce ABCD gametes?
To calculate the probability of producing ABCD gametes, we need to determine the probability of each gene being included in the gametes individually and then multiply them together.
Given that gene A is on chromosome 1, gene B is on chromosome 2, gene C is on chromosome 3, and gene D is on chromosome 4, we can assume that they will segregate independently during meiosis.
Let's break down the probability calculation step-by-step:
1. Probability of gene A being included: Since the animal has genotype AaBbCcDD, it has two copies of gene A (Aa). During meiosis, one copy of gene A can be passed into the gametes with equal likelihood. Therefore, the probability of including gene A is 1/2 (50%).
2. Probability of gene B being included: Similar to gene A, the animal has two copies of gene B (Bb). The probability of including gene B is also 1/2 (50%).
3. Probability of gene C being included: The animal also has two copies of gene C (Cc). Therefore, the probability of including gene C is 1/2 (50%).
4. Probability of gene D being included: Gene D is present in a homozygous dominant state (DD) in the animal. Since there is no gene d allele present, there is a 100% chance that gene D will be included in the gametes.
To calculate the overall probability, we multiply the individual probabilities together:
Probability of ABCD gametes = Probability of gene A * Probability of gene B * Probability of gene C * Probability of gene D
= (1/2) * (1/2) * (1/2) * 1
= 1/8 or 0.125 (12.5%)
Therefore, the probability that this animal will produce ABCD gametes is 1/8 or 0.125, which is equivalent to 12.5%.