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}$.
To calculate the specific activity of lactate dehydrogenase, we need to first determine the reaction rate in terms of $ \mu mol.min^{-1} $ and then normalize it to the enzyme concentration in terms of mg of protein.
Step 1: Calculate the reaction rate in $ \mu mol.min^{-1} $.
Given the decrease in absorbance per minute is $0.14 \ min^{-1}$, we use Beer-Lambert Law:
$ A = \varepsilon \cdot c \cdot l $
Where $A$ is absorbance, $\varepsilon$ is molar extinction coefficient, $c$ is concentration, and $l$ is path length. Given $\varepsilon = 6220 \ M^{-1}.cm^{-1}$ and $l = 1 \ cm$, we find the change in concentration per minute (by setting $c$ equal to the change in concentration, which is $-\Delta [NADH]$ as absorbance decreases):
$ \Delta A = \varepsilon \cdot (-\Delta [NADH]) \cdot 1 $
Thus, $-\Delta [NADH] = \frac{0.14}{6220} = 2.25 \times 10^{-5} \ M.min^{-1}$
Convert this to $ \mu mol.L^{-1}.min^{-1} $:
$2.25 \times 10^{-5} \ mol.L^{-1}.min^{-1} = 22.5 \ \mu mol.L^{-1}.min^{-1} $
Since the total volume is $3 \ mL = 0.003 \ L$, the total reaction rate is:
$ 22.5 \ \mu mol.L^{-1}.min^{-1} \times 0.003 \ L = 0.0675 \ \mu mol.min^{-1} $
Step 2: Calculate the specific activity.
The enzyme solution added was $25 \ \mu L$ of $1 \ mL = 0.025 \ mL$, and the concentration of protein in that solution is $5 \ \mu g.mL^{-1}$:
$ \text{Total protein} = 5 \ \mu g.mL^{-1} \times 0.025 \ mL = 0.125 \ \mu g $
Convert $0.125 \ \mu g$ to mg:
$0.125 \ \mu g = 0.000125 \ mg $
The specific activity is given by the reaction rate divided by the enzyme mass:
$ \text{Specific Activity} = \frac{0.0675 \ \mu mol.min^{-1}}{0.000125 \ mg} = 540 \ \mu mol.min^{-1}.mg^{-1} $
This value falls within the expected range of 525,555, confirming its accuracy.
Conclusion: The specific activity of the enzyme is $540 \ \mu mol.min^{-1}.mg^{-1}$.
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 ____________________ $\times 10^4 \ M^{-1}.s^{-1}$.
