For the following crosses, indicate the probability of obtaining the indicated genotype in an offspring. remember, it is easiest to treat each gene separately as a monohybrid cross and then combine the probabilities.

1. cross: aabb × aabb
offspring: aabb
2. cross: aabb × aabb
offspring: aabb
3. cross: aabbcc × aabbcc
offspring: aabbcc
4. cross: aabbcc × aabbcc
offspring: aabbcc

1. Frequency of Aabb= 1/8

2. Frequency of aaBB= 1/16

3. Frequency of  AaBbCc = 1/2

4. Frequency of aabbcc = 1/32

Explanation:

1. Cross: AABb × AaBb

Offspring: Aabb

AA x Aa= 1/2 AA: 1/2 Aa

Bb x Bb = 1/4 BB: 1/2 Bb: 1/4 bb

Frequency of Aabb= 1/2 Aa x 1/4 bb= 1/8

2. Cross: AaBb × AaBb

Offspring: aaBB

Aa x Aa= 1/4 Aa: 1/2 Aa: 1/4 aa

Bb x Bb = 1/4 BB: 1/2 Bb: 1/4 bb

Frequency of aaBB= 1/4 aa x 1/4 BB= 1/16

3. Cross: AABbcc × aabbCC

Offspring: AaBbCc

AA x aa = All Aa

Bb x bb = 1/2 Bb : 1/2 bb

cc x CC = All Cc

Frequency of  AaBbCc  = 1 Aa x 1/2 Bb: 1 Cc = 1/2

4. Cross: AaBbCc × AaBbcc

Offspring: aabbcc

Aa x Aa= 1/4 Aa: 1/2 Aa: 1/4 aa

Bb x Bb = 1/4 BB: 1/2 Bb: 1/4 bb

Cc x cc = 1/2 Cc: 1/2 cc

Frequency of aabbcc = 1/4 aa x 1/4 bb x 1/2 cc= 1/32

•there must be variation for the particular trait within a population.

•the variation must be inheritable (that is, it must be capable of being passed on from the parents to their offspring).

•individuals with one version of the trait must produce more offspring than those with a different version of the trait.good

explanation:

i think the answer is b.