In a cross between AABB `xx`aabb, the ratio of `F_(2)` genotype between AABB, AaBB,Aabb and aabb would be

Options

A. `9:3:3:1`
B. `2:1:1:2`
C. `1:2:2:1`
D. `7:5:3:1`

Answer

To solve the problem, we need to determine the genotype ratios in the F2 generation resulting from a cross between AABB and aabb. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Parent Genotypes
The parent genotypes are:
- Parent 1: AABB
- Parent 2: aabb

### Step 2: Determine the Gametes Produced by Each Parent
- Parent 1 (AABB) can produce only one type of gamete: AB.
- Parent 2 (aabb) can produce only one type of gamete: ab.

### Step 3: Perform the F1 Cross
When we cross AABB with aabb, all offspring (F1 generation) will be heterozygous:
- F1 genotype: AaBb

### Step 4: Self-Cross the F1 Generation
Now, we will perform a self-cross of the F1 generation (AaBb x AaBb) to produce the F2 generation.

### Step 5: Determine the Gametes from F1 Generation
The F1 generation (AaBb) can produce the following gametes:
- AB
- Ab
- aB
- ab

### Step 6: Create a Punnett Square for the F2 Generation
We can set up a Punnett square to find the combinations of these gametes:

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

### Step 7: Count the Genotypes
Now, we will count the occurrences of each genotype from the Punnett square:
1. AABB: 1
2. AABb: 2
3. AaBB: 2
4. AaBb: 4
5. AAbb: 1
6. Aabb: 2
7. aaBB: 1
8. aaBb: 2
9. aabb: 1

### Step 8: Group the Genotypes
Now, we will group the genotypes as per the question:
- AABB: 1
- AaBB: 2
- Aabb: 2
- aabb: 1

### Step 9: Write the Final Ratio
The final ratio of the genotypes AABB : AaBB : Aabb : aabb is:
- 1 : 2 : 2 : 1

### Conclusion
Thus, the ratio of the genotypes in the F2 generation is **1 : 2 : 2 : 1**.

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