Assume that genes A and B are not linked. If the probability of allele A in a gamete is 1/2 and the probability of allele B in a gamete is 1/2, then the probability that both A and B are in the same gamete is

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The question asks about the probability that both allele A and allele B appear in the same gamete, given that A and B are not linked and each allele has a 1/2 chance to be present in a gamete.
First, recognize that the events “A is in the gamete” and “B is in the gamete” are independent when genes are unlinked. Independence means we multiply their probabilities to get the joint probability.
Next, compute the probability for A: P(A) = 1/2. Then compute the probability for B: P(B) = 1/2. Since they're independent, P(A and B) = P(A) × P(B) = (1/2) × (1/2) = 1/4.
Therefore, the probability that both A and B are in the same gamete is 1/4.
Note: The provided data set lists the answer as "1/2 x 1/2" but does not supply individual answer options to evaluate; the underlying calculation simplifies to 1/4, as shown by multiplying the two independent probabilities. In terms of unit fractions, 1/4 is equivalent to 0.25 in decimal form.
If you had been given multiple-choice options such as (a) 1/2, (b) 1/4, (c) 1/2 × 1/2, or (d) 1, the correct choice would be the one showing the product (1/2) × (1/2) or its simplified form 1/4. The key misconception to avoid would be treating the occurrence of A and B as dependent events or assuming any linkage that would alter the independence.

Assume that genes A and B are not linked. If the probability of allele A in a gamete is 1/2 and the probability of allele B in a gamete is 1/2, then the probability that both A and B are in the same gamete is

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