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. (a)`1.568xx10^(3)`
B. (b)`15.68`
C. (c )`1.568xx10^(4)`
D. (d)`2.136xx10^(4)`

Explanation

To find the minimum molecular weight of the peroxidase anhydrase enzyme given that the percentage of selenium (Se) in it is 0.5% by weight and the atomic weight of selenium is 78.4, we can follow these steps:

### Step-by-Step Solution:

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

2. **Set Up the Equation**:
The percentage by weight of an element in a compound can be expressed using the formula:
\[ \text{Percentage of Se} = \left( \frac{\text{Atomic weight of Se}}{\text{Molecular weight of the enzyme}} \right) \times 100 \]
Here, we denote the molecular weight of the enzyme as \( M \).

3. **Substitute the Known Values**:
Substitute the known values into the equation:
\[ 0.5 = \left( \frac{78.4}{M} \right) \times 100 \]

4. **Rearrange the Equation**:
To isolate \( M \), rearrange the equation:
\[ M = \frac{78.4 \times 100}{0.5} \]

5. **Calculate \( M \)**:
Now perform the calculation:
\[ M = \frac{7840}{0.5} = 15680 \text{ g/mol} \]

6. **Express in Scientific Notation**:
Convert the result into scientific notation:
\[ M = 1.568 \times 10^4 \text{ g/mol} \]

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

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