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Question

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 ________

This question requires calculating the Michaelis constant ($K_m$) using the Michaelis-Menten equation parameters provided.

Solution Steps

Step 1: Calculate Maximum Velocity ($V_{max}$)

The maximum velocity ($V_{max}$) is related to the turnover number ($k_{cat}$) and the total enzyme concentration ($[E_t]$) by the formula:

$V_{max} = k_{cat}[E_t]$

Given:

  • $k_{cat} = 500 s^{-1}$
  • $[E_t] = 30 \text{ nM} = 30 \times 10^{-3} \mu M$ (converting nM to $\mu M$ for unit consistency)

Calculation:

$V_{max} = (500 s^{-1}) \times (30 \times 10^{-3} \mu M)$

$V_{max} = 15 \mu M.s^{-1}$

Step 2: Calculate Michaelis Constant ($K_m$)

The Michaelis-Menten equation relates initial velocity ($V_0$), maximum velocity ($V_{max}$), substrate concentration ($[S]$), and $K_m$:

$V_0 = \frac{V_{max}[S]}{K_m + [S]}$

We need to rearrange this equation to solve for $K_m$. First, let's isolate the term containing $K_m$:

$K_m + [S] = \frac{V_{max}[S]}{V_0}$

Now, solve for $K_m$:

$K_m = \frac{V_{max}[S]}{V_0} - [S]$

Given values for this step:

  • $V_{max} = 15 \mu M.s^{-1}$ (from Step 1)
  • $[S] = [A] = 40 \mu M$
  • $V_0 = 10 \mu M.s^{-1}$

Calculation:

$K_m = \frac{(15 \mu M.s^{-1}) \times (40 \mu M)}{10 \mu M.s^{-1}} - 40 \mu M$

$K_m = \frac{600}{10} \mu M - 40 \mu M$

$K_m = 60 \mu M - 40 \mu M$

$K_m = 20 \mu M$

Conclusion

The calculated value of $K_m$ is $20 \mu M$. This value falls within the specified range.

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Important Questions from Enzyme Kinetics and Michaelis Menten Equation

  1. The catalytic efficiency of an enzyme following Michaelis-Menten kinetics is defined by
  2. You are characterizing a new enzyme isolated and purified in the laboratory. If the maximum velocity of the enzyme is $1800 \text{ } \mu moles \text{ L}^{-1}  \text{min}^{-1}$ and the total concentration of the enzyme in the reaction mixture is $1.5 \mu \text{M}$, then the turnover number of the enzyme is _______ $\text{s}^{-1}$. (answer in integer)

  3. 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)

  4. 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)

  5. The activity of lactate dehydrogenase can be measured by monitoring the following reaction: 

    Pyruvate + NADH $ \longrightarrow $ Lactate + $NAD^+$ 

    The molar extinction coefficient of NADH at 340 nm is $6220 \ M^{-1}.cm^{-1}$. $NAD^+$ does not absorb at this wavelength. In an assay, $25 \ \mu L$ of a sample of enzyme (containing $5 \ \mu g$ protein per mL) was added to a mixture of pyruvate and NADH to give a total volume of 3 mL in a cuvette of 1 cm pathlength. The rate of decrease in absorbance at 340 nm was $0.14 \ min^{-1}$. The specific activity of the enzyme will be ____________________ $ \mu mol.min^{-1}.mg^{-1}$.

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