Percentage of Se in peroxidase anhydrase enzyme is `0.5%` by weight (at. Wt. `=78.4)`, then minimum molecular weight of peroxidase anhydrase enzyme is:

Options

A. `1.568xx10^(3)`
B. `15.68`
C. `2.168xx10^(4)`
D. `1.568xx10^(4)`

Explanation

To find the minimum molecular weight of the peroxidase anhydrase enzyme given the percentage of selenium (Se) in it, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Percentage of Se in peroxidase anhydrase enzyme = 0.5%
- Atomic weight of Se = 78.4 g/mol

2. **Assume a Total Weight:**
- For simplicity, assume the total weight of the peroxidase anhydrase enzyme is 100 grams.

3. **Calculate the Weight of Se:**
- Since Se is 0.5% of the total weight, the weight of Se in 100 grams of the enzyme is:
\[ \text{Weight of Se} = \frac{0.5}{100} \times 100 = 0.5 \text{ grams} \]

4. **Determine the Amount of Enzyme per Gram of Se:**
- If 0.5 grams of Se is present in 100 grams of the enzyme, then in 1 gram of Se, the weight of the enzyme can be calculated as:
\[ \text{Weight of enzyme} = \frac{100 \text{ grams}}{0.5 \text{ grams}} = 200 \text{ grams} \]

5. **Calculate the Minimum Molecular Weight of the Enzyme:**
- Now, we need to find out how much enzyme corresponds to 78.4 grams of Se:
\[ \text{Weight of enzyme for 78.4 grams of Se} = 200 \text{ grams} \times \frac{78.4 \text{ grams}}{1 \text{ gram}} = 15680 \text{ grams} \]

6. **Express the Minimum Molecular Weight:**
- The minimum molecular weight of the peroxidase anhydrase enzyme can be expressed as:
\[ \text{Minimum Molecular Weight} = 15680 \text{ g/mol} = 1.568 \times 10^4 \text{ g/mol} \]

### Final Answer:
The minimum molecular weight of peroxidase anhydrase enzyme is \( 1.568 \times 10^4 \) g/mol.

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