A total of 512 seeds are collected from the cross between `AaBb xx AaBb`. Find the number of plants produced with the genotype Aabb

Options

A. 35
B. 350
C. 64
D. 285

Answer

To solve the problem of finding the number of plants produced with the genotype Aabb from the cross AaBb x AaBb, we can follow these steps:

### Step 1: Identify the parental genotypes
The parents are both heterozygous for two traits:
- Male parent: AaBb
- Female parent: AaBb

### Step 2: Determine the gametes produced by each parent
Since both parents are heterozygous for two traits, they can produce four types of gametes:
- From AaBb, the possible gametes are:
1. AB
2. Ab
3. aB
4. ab

### Step 3: Set up the Punnett square
To find the offspring genotypes, we can set up a Punnett square using the gametes from both parents. Each parent can produce four types of gametes, leading to a 4x4 Punnett square:

```
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 4: Identify the target genotype
We are interested in the genotype Aabb. From the Punnett square, we can see that Aabb appears once in the offspring.

### Step 5: Calculate the probability of obtaining Aabb
The total number of different genotypes produced in the Punnett square is 16 (4x4). The genotype Aabb appears once, so the probability of getting Aabb is:
\[ \text{Probability of Aabb} = \frac{1}{16} \]

### Step 6: Calculate the expected number of Aabb plants
Given that a total of 512 seeds are collected, we can calculate the expected number of Aabb plants:
\[ \text{Number of Aabb plants} = \text{Total seeds} \times \text{Probability of Aabb} \]
\[ \text{Number of Aabb plants} = 512 \times \frac{1}{16} = 32 \]

### Final Answer
The number of plants produced with the genotype Aabb is **32**.

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