Answer:
The correct answer is - 9 out of 16.
Explanation:
It is common case of the dihybrid cross where independent assortment takes pace. In the case of the given cross the genotype and produced gametes from AaBb:
AB, Ab, aB, ab.
From these gametes cross with similar that are produced following genotypes are will form phenotypes in the 9:3:3:1 where the dominant gene is present in the 9 out of 16 offspring.
Thus, the correct answer is - 9 out of 16.
