A plant that is heterozygous for two genes was selfed. A total of 1000 seeds were collected. Find out the total number of seeds heterozygous for the first gene and homozygous for the second gene.

Options

A. 125
B. 250
C. 300
D. 500

Answer

To solve the problem, we need to determine the total number of seeds that are heterozygous for the first gene (A) and homozygous for the second gene (B) from a selfed plant that is heterozygous for both genes.

### Step-by-Step Solution:

1. **Identify the Genotypes of the Parents**:
- The plant is heterozygous for two genes. We can denote the alleles for the first gene as A (dominant) and a (recessive), and for the second gene as B (dominant) and b (recessive).
- Therefore, the genotype of the parent plant is AaBb.

2. **Perform a Dihybrid Cross**:
- Since the plant is selfed, we will cross AaBb with AaBb.
- The possible gametes from each parent are: AB, Ab, aB, ab.
- The Punnett square for this dihybrid cross will give us the following combinations:

```
AB Ab aB ab
---------------------------------
AB | AABB AABb AaBB AaBb
Ab | AABb AAbb AaBb Aabb
aB | AaBB AaBb aaBB aaBb
ab | AaBb Aabb aaBb aabb
```

3. **Count the Genotypes**:
- From the Punnett square, we can summarize the genotypes:
- 1 AABB
- 2 AABb
- 2 AaBB
- 4 AaBb
- 1 AAbb
- 2 Aabb
- 1 aaBB
- 2 aaBb
- 1 aabb
- Total combinations = 16 (4x4 Punnett square).

4. **Identify the Required Genotype**:
- We need to find seeds that are heterozygous for the first gene (Aa) and homozygous for the second gene (BB or bb).
- The relevant genotypes are:
- AaBB (2 occurrences)
- Aabb (2 occurrences)
- Total occurrences = 2 (AaBB) + 2 (Aabb) = 4.

5. **Calculate the Ratio**:
- The ratio of seeds that are heterozygous for the first gene and homozygous for the second gene is:
- 4 (favorable outcomes) out of 16 (total outcomes) = 4/16 = 1/4.

6. **Calculate the Number of Seeds**:
- Given that there are 1000 seeds in total, we can calculate:
- (1/4) * 1000 = 250.

### Final Answer:
The total number of seeds that are heterozygous for the first gene and homozygous for the second gene is **250**.

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