Total 1120 seeds were obtained from a cross AaBb `xx` AaBb. Find the number of plants produced in F2 having both the resseive traits

Options

A. 210
B. 420
C. 70
D. 480

Answer

To solve the problem of finding the number of plants produced in the F2 generation that have both recessive traits from the cross AaBb x AaBb, we can follow these steps:

### Step 1: Understand the Cross
We are dealing with a dihybrid cross (AaBb x AaBb). Here, A and a represent one trait, and B and b represent another trait. The recessive traits we are interested in are represented by the genotype 'aabb'.

### Step 2: Determine the Total Number of Seeds
According to the question, a total of 1120 seeds were obtained from this cross.

### Step 3: Calculate the Frequency of Recessive Traits
In a dihybrid cross, the phenotypic ratio for the traits is 9:3:3:1. The recessive phenotype (aabb) corresponds to the 1 in the 16 total parts of the ratio. Therefore, the frequency of the recessive phenotype (both traits recessive) in the F2 generation is:

\[ \text{Frequency of aabb} = \frac{1}{16} \]

### Step 4: Calculate the Number of Plants with Recessive Traits in F2
To find the number of plants with the recessive traits, we multiply the total number of seeds by the frequency of the recessive phenotype:

\[ \text{Number of aabb plants} = \frac{1}{16} \times 1120 \]

Calculating this gives:

\[ \text{Number of aabb plants} = 70 \]

### Conclusion
Thus, the number of plants produced in the F2 generation that have both recessive traits (genotype aabb) is **70**.