The question is incomplete and the complete question is
Suppose that ear length in rabbits is controlled by two additive genes, each of which has two alleles. A true-breeding female (aabb) with 6-cm ears is mated to a true-breeding male (AABB) with 16-cm ears.
Answer:
AABb or AaBB
Explanation:
We know that,
aabb genotype - 6 cm
AABB genotype- 16-cm
To calculate the length of earlobe contributed by each allele in a genotype is :
1. length of aabb/4 or 6/4= 1.5 cm (a and b contribute for 1.5 cm each)
2. Length of AABB/4 or 16/4= 4 cm (A and B contribute for 4 cm each)
Now to have the earlobe to be 13.5 cm long then the genotype must be
13.5 = 4+4+4+1.5 or A+A+B+b or A+a+B+B
Therefore, the genotype will be-either AABb or AaBB
Explanation:
B
............ I need 20 lettera
It might be clone? As all the genetic material is the same
