In garden pea, yellow colour of cotyledons is dominant over green and round shape of seed is dominant over wrinkled. When a plant with yellow and round seeds is crossed with a plant having yellow and wrinkled seeds, the progeny showed segregation for all the four characters. The probability of obtaining green round seeds in the progeny of this cross is

Options

A. `1//4 `
B. `1//8`
C. `1//16`
D. `3//16`

Answer

To solve the problem of determining the probability of obtaining green round seeds from the cross of two garden pea plants, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Dominant and Recessive Traits**:
- Yellow cotyledon color (Y) is dominant over green (y).
- Round seed shape (R) is dominant over wrinkled (r).

2. **Determine the Genotypes of the Parent Plants**:
- The first plant has yellow and round seeds. Since yellow is dominant, it can be either homozygous (YY) or heterozygous (Yy) for color. Since it has round seeds, it can also be either homozygous (RR) or heterozygous (Rr) for shape. For simplicity, we will assume it is heterozygous for both traits: YyRr.
- The second plant has yellow and wrinkled seeds. It is homozygous for yellow (YY) and homozygous for wrinkled (rr): YYrr.

3. **Cross the Parent Plants**:
- The cross will be YyRr (yellow round) x YYrr (yellow wrinkled).
- The gametes produced by YyRr will be: YR, Yr, yR, yr.
- The gametes produced by YYrr will be: Yr, Yr.

4. **Determine the F1 Generation**:
- The F1 generation will be obtained by combining the gametes:
- From YyRr (yellow round):
- YR with Yr → YYRr (yellow round)
- Yr with Yr → YYrr (yellow wrinkled)
- yR with Yr → YyRr (yellow round)
- yr with Yr → Yyrr (yellow wrinkled)

Thus, the F1 generation will consist of:
- 2 yellow round (YYRr and YyRr)
- 2 yellow wrinkled (YYrr and Yyrr)

5. **Selfing the F1 Generation**:
- To find the F2 generation, we can self-cross the F1 generation (YyRr x YyRr).
- The gametes will be: YR, Yr, yR, yr.

6. **Construct a Punnett Square**:
- Create a 4x4 Punnett square to determine the phenotypes of the F2 generation:
```
YR | Yr | yR | yr
-------------------------------
YR | YYRR | YYRr | YyRR | YyRr
-------------------------------
Yr | YYRr | YYrr | YyRr | Yyrr
-------------------------------
yR | YyRR | YyRr | yyRR | yyRr
-------------------------------
yr | YyRr | Yyrr | yyRr | yyrr
```

7. **Count the Phenotypes**:
- From the Punnett square, we can count the phenotypes:
- Yellow Round (YYRR, YYRr, YyRR, YyRr): 9
- Yellow Wrinkled (YYrr, Yyrr): 3
- Green Round (yyRR, yyRr): 3
- Green Wrinkled (yyrr): 1

8. **Calculate the Probability of Green Round Seeds**:
- The total number of offspring is 16.
- The number of green round seeds is 3.
- Therefore, the probability of obtaining green round seeds is:
\[ \text{Probability} = \frac{\text{Number of Green Round Seeds}}{\text{Total Offspring}} = \frac{3}{16} \]

### Final Answer:
The probability of obtaining green round seeds in the progeny of this cross is **3/16**.

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