A plant with genotype WWXxYyZZ is self pollinated. What proportion of the progeny will show the genotype WWxxyyZZ (if all four genes are independently assorting) :

Question

Accepted Answer

To determine the proportion of progeny with the genotype WWxxyyZZ from a self-pollinated plant with genotype WWXxYyZZ, we need to consider the independent assortment of each gene. The plant has four genes, each with two alleles, and they assort independently according to Mendel's laws.

1. **Gene W (WW):** 
   - The parent genotype is WW, which means all progeny will inherit the W allele from both parents. Therefore, the probability of progeny being WW is 1.

2. **Gene X (Xx):**
   - The parent genotype is Xx, which means the progeny can be either X or x. The probability of getting xx is the probability of getting x from both parents, which is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

3. **Gene Y (Yy):**
   - The parent genotype is Yy, which means the progeny can be either Y or y. The probability of getting yy is the probability of getting y from both parents, which is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

4. **Gene Z (ZZ):**
   - The parent genotype is ZZ, which means all progeny will inherit the Z allele from both parents. Therefore, the probability of progeny being ZZ is 1.

To find the overall probability of the progeny having the genotype WWxxyyZZ, we multiply the probabilities for each gene:

$$ 
1 \times \frac{1}{4} \times \frac{1}{4} \times 1 = \frac{1}{16} 
$$

Thus, the proportion of progeny with the genotype WWxxyyZZ is $\frac{1}{16}$, which corresponds to Option 2.

**Explanation of Incorrect Options:**

- **Option 1 ($\frac{1}{9}$):** This option is incorrect because it does not account for the correct probability calculation of independent assortment. The probability for each gene must be multiplied, not added or averaged, which results in $\frac{1}{16}$, not $\frac{1}{9}$.

- **Option 3 ($\frac{1}{64}$):** This option is incorrect because it suggests a lower probability than the actual calculation. The probability of $\frac{1}{64}$ would imply more restrictive conditions or additional genes, which is not the case here.

- **Option 4 ($\frac{1}{128}$):** This option is incorrect because it suggests an even lower probability than $\frac{1}{64}$. The calculation of $\frac{1}{16}$ is based on the independent assortment of four genes, two of which are homozygous, leading to a higher probability than $\frac{1}{128}$.

Therefore, the correct answer is Option 2, $\frac{1}{16}$.