Enzyme Activity Calculation in Katal
Enzyme activity quantifies the rate at which an enzyme converts substrate into product. The standard unit for enzyme activity is the Katal (kat), defined as one mole of substrate transformed per second.
Calculating Enzyme Activity Rate
The problem states that the enzyme converts 420 µmoles of substrate in one minute. We need to calculate this rate in Katals.
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Convert substrate amount to moles:
Amount = 420 µmoles
Since 1 µmole = $ 1 \times 10^{-6} $ moles, the amount is:
Amount = $ 420 \times 10^{-6} $ moles
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Convert time to seconds:
Time = 1 minute
Since 1 minute = 60 seconds, the time is:
Time = 60 seconds
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Calculate the reaction rate:
Rate = Amount converted / Time taken
Rate = $ \frac{420 \times 10^{-6} \text{ moles}}{60 \text{ seconds}} $
Rate = $ 7 \times 10^{-6} $ moles/second
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Express rate in Katals:
The rate calculated is $ 7 \times 10^{-6} $ moles/second, which is equal to $ 7 \times 10^{-6} $ Katals.
The question asks for the activity in the format '__________ $ \times 10^{-6} $ Katal'.
Therefore, the missing value is 7.
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 ____________________ $\times 10^4 \ M^{-1}.s^{-1}$.
