In Garden Pea, round shape is dominant over wrinkled shape. A pea plant heterozygous for round shape of seed is selfed and 1600 sedds produced during the cross are subsequently germinated. How many offspring will have parental phenotype

Options

A. 1600
B. 800
C. 400
D. 1200

Answer

To solve the problem, we need to determine how many offspring will exhibit the parental phenotype when a heterozygous pea plant is self-fertilized. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Genotypes
In the garden pea, round shape (R) is dominant over wrinkled shape (r). A heterozygous plant for round shape has the genotype Rr.

### Step 2: Set Up the Punnett Square
When a heterozygous plant (Rr) is self-fertilized, we can set up a Punnett square to determine the genotypes of the offspring:

```
R r
---------------
R | RR | Rr |
---------------
r | Rr | rr |
---------------
```

### Step 3: Determine the Genotypic Ratio
From the Punnett square, we can see the following genotypes:
- 1 RR (homozygous dominant)
- 2 Rr (heterozygous)
- 1 rr (homozygous recessive)

This gives us a genotypic ratio of:
- 1 RR : 2 Rr : 1 rr

### Step 4: Determine the Phenotypic Ratio
The phenotypes corresponding to these genotypes are:
- RR and Rr both show the round shape (dominant phenotype)
- rr shows the wrinkled shape (recessive phenotype)

Thus, the phenotypic ratio is:
- 3 round (RR and Rr) : 1 wrinkled (rr)

### Step 5: Calculate the Number of Offspring
Given that a total of 1600 seeds are produced, we can calculate the number of offspring for each phenotype based on the phenotypic ratio.

- Total offspring = 1600
- Number of round shape offspring = (3/4) * 1600 = 1200
- Number of wrinkled shape offspring = (1/4) * 1600 = 400

### Conclusion
The number of offspring that will have the parental phenotype (round shape) is **1200**.

---