What is the probability of obtaining pure homozygous individuals, if a cross is made between AaBbCcDd `×x` aaBBCCdd individuals ?

Options

A. `(2)/(16)`
B. `(1)/(16)`
C. `(16)/(248)`
D. `(1)/(64)`

Answer

To find the probability of obtaining pure homozygous individuals from the cross between AaBbCcDd and aaBBCCdd, we can follow these steps:

### Step 1: Identify the Genotypes
We have two parents:
- Parent 1: AaBbCcDd (heterozygous for all traits)
- Parent 2: aabbCCdd (homozygous for some traits)

### Step 2: Determine the Number of Gametes
To calculate the number of gametes produced by each parent, we use the formula \(2^n\), where \(n\) is the number of heterozygous traits.

- For Parent 1 (AaBbCcDd):
- A: Aa (heterozygous)
- B: Bb (heterozygous)
- C: Cc (heterozygous)
- D: Dd (heterozygous)

Here, \(n = 4\) (since all four traits are heterozygous).

Therefore, the number of gametes = \(2^4 = 16\).

- For Parent 2 (aaBBCCdd):
- A: aa (homozygous)
- B: BB (homozygous)
- C: CC (homozygous)
- D: dd (homozygous)

Here, \(n = 0\) (since there are no heterozygous traits).

Therefore, the number of gametes = \(2^0 = 1\).

### Step 3: Calculate Total Progeny
The total number of progeny produced from the cross is the product of the number of gametes from both parents:
- Total progeny = Number of gametes from Parent 1 × Number of gametes from Parent 2 = \(16 \times 1 = 16\).

### Step 4: Determine the Probability of Pure Homozygous Individuals
To find the probability of obtaining pure homozygous individuals, we need to consider each trait independently, as per the law of independent assortment.

- For trait A:
- Possible genotypes from the cross: Aa (from Parent 1) and aa (from Parent 2).
- Only one genotype is homozygous (aa).

- For trait B:
- Possible genotypes: Bb (from Parent 1) and BB (from Parent 2).
- Only one genotype is homozygous (BB).

- For trait C:
- Possible genotypes: Cc (from Parent 1) and CC (from Parent 2).
- Only one genotype is homozygous (CC).

- For trait D:
- Possible genotypes: Dd (from Parent 1) and dd (from Parent 2).
- Only one genotype is homozygous (dd).

Since each trait contributes independently to the overall genotype, the probability of obtaining a pure homozygous individual for each trait is:
- Probability for A = 1/2 (aa)
- Probability for B = 1/2 (BB)
- Probability for C = 1/2 (CC)
- Probability for D = 1/2 (dd)

### Step 5: Calculate Overall Probability
The overall probability of obtaining a pure homozygous individual is the product of the probabilities for each trait:
\[ P(\text{pure homozygous}) = P(A) \times P(B) \times P(C) \times P(D) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{16}. \]

### Final Answer
The probability of obtaining pure homozygous individuals from the cross is \( \frac{1}{16} \).

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