Question

You have purified an enzyme using a series of chromatographic methods. It was observed that a 10 \(\mu\)g mL^-1 of this purified enzyme converted 10 mM substrate per hour at 25\(^{\circ}\)C and pH 7. Its specific activity is IU \(\mu\)g^-1. (rounded off to three decimal places)

Hint:

1 IU equals 1 micromol of substrate converted per minute; convert the given mM per hour rate to micromol per minute in 1 mL, then divide by the micrograms of enzyme in that same mL.

Answer 1

Correct Answer: 0.017

Step 1: Recall what an International Unit (IU) means.
One International Unit (IU) of enzyme activity is the amount of enzyme that converts 1 \(\mu\)mol of substrate per minute under the stated assay conditions. Specific activity is the enzyme activity expressed per unit mass of enzyme protein, here in \(\text{IU}\ \mu\text{g}^{-1}\).

Step 2: Convert the substrate conversion rate to \(\mu\)mol per minute.
The enzyme converts 10 mM substrate per hour. Taking a 1 mL reaction as the basis, matching the given 10 \(\mu\)g mL\(^{-1}\) enzyme concentration, 10 mM in 1 mL is:
\[ 10\ \text{mmol L}^{-1} \times 1\ \text{mL} = 10\ \text{mmol L}^{-1} \times 0.001\ \text{L} = 0.01\ \text{mmol} = 10\ \mu\text{mol} \]
converted per hour, in 1 mL of reaction.

Step 3: Convert per hour to per minute.
\[ \frac{10\ \mu\text{mol}}{60\ \text{min}} = 0.16667\ \mu\text{mol min}^{-1} \]
This is the enzyme activity in IU, since 1 IU = 1 \(\mu\)mol min\(^{-1}\), so 0.16667 IU are present in that 1 mL.

Step 4: Divide by the mass of enzyme protein to get specific activity.
The 1 mL reaction contains \(10\ \mu\text{g}\) of enzyme protein, since the enzyme is at 10 \(\mu\)g mL\(^{-1}\). So:
\[ \text{Specific activity} = \frac{0.16667\ \text{IU}}{10\ \mu\text{g}} = 0.016667\ \text{IU}\ \mu\text{g}^{-1} \]
Rounded to three decimal places, this is 0.017 IU \(\mu\)g\(^{-1}\).

Final Answer:
The specific activity of the purified enzyme is \[ \boxed{0.017\ \text{IU}\ \mu\text{g}^{-1}} \]