If a plant with genotype AaBb is self-fertilized, the probability of g...
To determine the probability of obtaining a specific genotype from a self-fertilization cross, we need to consider the principles of Mendelian genetics and the laws of segregation and independent assortment.
In this case, the plant has the genotype AaBb, which means it carries one copy of the A allele and one copy of the B allele. During gamete formation, these alleles will segregate independently, meaning that each allele has an equal chance of being passed on to the offspring.
To calculate the probability of obtaining the AABB genotype, we need to determine the probability of each allele combination coming together. Since the alleles segregate independently, the probability of obtaining the AABB genotype would be the product of the probabilities of getting the A allele twice and the B allele twice.
The probability of getting the A allele twice is 1/2 * 1/2 = 1/4 (assuming A is dominant and a is recessive). Similarly, the probability of getting the B allele twice is also 1/4 (assuming B is dominant and b is recessive).
To find the probability of both events occurring together (AABB genotype), we multiply the probabilities:
Probability = (1/4) * (1/4) = 1/16
Therefore, the correct answer is D: 1/16.