Question

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

Answer 1

The probability of obtaining specific genotypes in offspring depends on the alleles and their assortment during the crosses. Calculations reveal probabilities of 1/8, 1/16, 1, and 1/64 for the respective offspring genotypes given in the question.

Step-by-step explanation:

To determine the probability of obtaining a specific genotype in offspring for each cross, we can use the probability method and the concept of independent assortment. Below is the calculation for each cross:

1. Cross: AABb × AaBb
   
   For genotype Aabb:
   
   A from AABb has a probability of 1, a from Aa has a probability of 1/2.
   
   B from AABb has a probability of 1/2 for b, B from AaBb also has a 1/2.
   
   The combined probability: 1 * (1/2) * (1/2) * (1/2) = 1/8.
2. Cross: AaBb × AaBb
   
   For genotype aaBB:
   
   a from Aa has a probability of 1/2, a from the other Aa also has 1/2.
   
   B from Bb has a probability of 1/2, and B from the other Bb also has 1/2.
   
   The combined probability: (1/2) * (1/2) * (1/2) * (1/2) = 1/16.
3. Cross: AABbcc × aabbCC
   
   For genotype AaBbCc:
   
   A from AABb has a probability of 1, a from aabb also has 1.
   
   B from AABb has a probability of 1, b from aabb also has 1.
   
   Cc is heterozygous, so it's 100% or a probability of 1.
   
   The combined probability: 1 * 1 * 1 * 1 * 1 * 1 = 1/1 or 100% certainty.
4. Cross: AaBbCc × AaBbcc
   
   For genotype aabbcc:
   
   a from Aa has a probability of 1/2; a from the other Aa also has 1/2.
   
   b from Bb has a probability of 1/2; b from the other Bb also has 1/2.
   
   c from Cc has a probability of 1/2, c from the other cc is 1.
   
   The combined probability: (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * 1 = 1/64.

Answer 2

1. Frequency of Aabb= 1/8

2. Frequency of aaBB= 1/16

3. Frequency of AaBbCc = 1/2

4. Frequency of aabbcc = 1/32

Step-by-step 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