In `F_(2)` generation, 960 garden pea pods are produced during dihybrid cross due to self pollination of heterozygous parents. How many pods would be green and inflated

Options

A. 240
B. 180
C. 60
D. 540

Answer

To solve the question regarding the number of green and inflated garden pea pods produced in the F2 generation during a dihybrid cross, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Traits**:
In a dihybrid cross involving garden pea pods, we are dealing with two traits:
- Pod color: Green (dominant) and Yellow (recessive)
- Pod shape: Inflated (dominant) and Constricted (recessive)

2. **Identify the Genotypes**:
The heterozygous parents for both traits can be represented as:
- Green and Inflated: GgIi (where G = green, g = yellow, I = inflated, i = constricted)

3. **Determine the Phenotypic Ratio**:
From Mendel's dihybrid cross, the phenotypic ratio for the offspring is:
- 9:3:3:1
This means:
- 9 parts will be Green and Inflated (dominant for both traits)
- 3 parts will be Green and Constricted
- 3 parts will be Yellow and Inflated
- 1 part will be Yellow and Constricted

4. **Calculate Total Pods in F2 Generation**:
We are given that a total of 960 pods are produced in the F2 generation.

5. **Calculate the Number of Green and Inflated Pods**:
According to the 9:3:3:1 ratio, the fraction of pods that are Green and Inflated is:
\[ \text{Fraction of Green and Inflated} = \frac{9}{16} \]
To find the actual number of Green and Inflated pods, we calculate:
\[ \text{Number of Green and Inflated Pods} = \frac{9}{16} \times 960 \]

6. **Perform the Calculation**:
\[ \text{Number of Green and Inflated Pods} = \frac{9 \times 960}{16} = \frac{8640}{16} = 540 \]

### Final Answer:
Thus, the number of green and inflated pods in the F2 generation is **540**.