Final answer:
The number of offspring with the aaBb genotype from a cross between a female lizard with genotype AaBB and a male with genotype AaBb is 2/16, or equivalently, 1/8. The correct answer is A.2/16.
Step-by-step explanation:
To predict the number of offspring with the aaBb genotype from a cross between a female lizard with genotype AaBB and a male with genotype AaBb, we can use a Punnett square to determine the likelihood of each genotype.
Since we are dealing with a dihybrid cross, there are four possible types of gametes for each parent: AB, Ab, aB, and ab. The cross can be represented as:
- Female gametes: AB, Ab, aB, ab
- Male gametes: AB, Ab, aB, ab
Upon fertilization, the aa genotype will arise from an a gamete from the female and an a gamete from the male. Similarly, the Bb genotype will result from either a B gamete from the female and a b gamete from the male, or a b gamete from the female and a B gamete from the male.
The probability of the offspring being aa is 1/4 (as there is a 50% chance from each parent to pass on the a allele), and the probability of the offspring being Bb is 1/2 (since there are two ways to get a Bb genotype: B from the female and b from the male, or b from the female and B from the male).
We multiply these probabilities to find the overall likelihood of an offspring having the aaBb genotype:
(1/4 for aa) * (1/2 for Bb) = 1/8
Therefore, the number of offspring with the aaBb genotype is 2/16, which is the same as 1/8.