You have purified an enzyme using a series of chromatographic methods. It was observed that a $10 \mu \text{ g mL}^{-1}$ of this purified enzyme converted $10 \text{ mM}$ substrate per hour at 25$^{\circ}$C and pH 7. Its specific activity is _______ $\text{IU } \mu\text{g}^{-1}$. (rounded off to three decimal places)
Enzyme Specific Activity Calculation
Specific activity is defined as enzyme activity per unit mass of protein, typically expressed in International Units (IU) per microgram ($\mu$g).
Activity Rate Conversion
The enzyme's conversion rate of $10 \text{ mM}$ substrate per hour needs to be converted into standard units of $\mu$mol per minute, which define 1 IU.
- Given conversion rate: $10 \text{ mM substrate hour}^{-1}$
- Convert mM to $\mu$mol/mL: $1 \text{ mM} = 1 \text{ } \mu\text{mol mL}^{-1}$
- The rate is thus: $10 \text{ } \mu\text{mol mL}^{-1} \text{ hour}^{-1}$
- Convert hours to minutes: $1 \text{ hour} = 60 \text{ minutes}$
- Activity rate per mL per minute: $\frac{10 \text{ } \mu\text{mol}}{1 \text{ mL} \cdot 60 \text{ min}} = \frac{10}{60} \text{ } \mu\text{mol mL}^{-1} \text{ min}^{-1}$
Specific Activity Calculation
Specific activity is calculated by dividing the total activity rate (per mL) by the enzyme mass concentration (per mL).
- Total activity rate (per mL): $\frac{10}{60} \text{ } \mu\text{mol mL}^{-1} \text{ min}^{-1}$
- Enzyme concentration (per mL): $10 \text{ } \mu\text{g mL}^{-1}$
- Specific Activity = $\frac{\text{Activity rate (per mL)}}{\text{Enzyme concentration (per mL)}}$
- Specific Activity = $\frac{\frac{10}{60} \text{ } \mu\text{mol mL}^{-1} \text{ min}^{-1}}{10 \text{ } \mu\text{g mL}^{-1}}$
- Specific Activity = $\frac{10}{60 \times 10} \text{ } \frac{\mu\text{mol min}^{-1}}{\mu\text{g}}$
- Specific Activity = $\frac{1}{60} \text{ IU } \mu\text{g}^{-1}$
Final Value
Calculate the numerical value and round to the specified decimal places.
- The specific activity is $\frac{1}{60} \text{ IU } \mu\text{g}^{-1}$.
- As a decimal: $\frac{1}{60} \approx 0.016666...$
- Rounded to three decimal places: $0.017 \text{ IU } \mu\text{g}^{-1}$.
Important Questions from Enzyme Kinetics and Michaelis Menten Equation
- An enzyme (E) catalyzes the biochemical reaction $A \rightarrow B$ with $k_{cat}$ equal to $500 s^{-1}$. If the initial reaction velocity ($V_0$) is $10 \mu M.s^{-1}$ at the total enzyme concentration $[E_t]$ of 30 nM and substrate concentration $[A]$ of $40 \mu M$, the value of $K_m$ (in $\mu M$) is ________
Within the Michaelis-Menten framework, the ratio of $v_0/V_{max}$
when $[S] = 20 \times K_m$ is _________.
(Round off to two decimal places)
The enzyme $\alpha$-amylase used in starch hydrolysis has an affinity constant ($K_m$) value of $0.005$ M. To achieve one-fourth of the maximum rate of hydrolysis, the required starch concentration in mM (rounded off to two decimal places) is____.
An enzymatic reaction exhibits Michaelis-Menten kinetics. For this reaction, on doubling the concentration of enzyme while maintaining [S] >> [$E_o$],
- In an assay of the type II dehydroquinase of molecular mass 18 kDa, it is found that the $V_{max}$ of the enzyme is $0.0134 \ \mu mol.min^{-1}$ when $1.8 \ \mu g$ enzyme is added to the assay mixture. If the $K_m$ for the substrate is $25 \ \mu M$, the $k_{cat}/K_m$ ratio will be ____________________ $\times 10^4 \ M^{-1}.s^{-1}$.
