Enzyme Concentration Calculation: Determining Substrate Molarity
This solution calculates the substrate concentration based on the given enzyme concentration, molecular weight, and molar ratio.
Enzyme Concentration Conversion
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Convert the enzyme molecular weight (MW) from kDa to g/mol:
MWenzyme = 40 kDa = 40,000 g/mol
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Convert the enzyme concentration from mg/ml to g/L:
Enzyme Concentration = 12 mg/ml = 12,000 mg/L = 12 g/L
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Calculate the molar concentration (molarity) of the enzyme:
Molarity = Concentration (g/L) / MW (g/mol)
Enzyme Molarity = $ \frac{12 \text{ g/L}}{40,000 \text{ g/mol}} $ = 0.0003 mol/L
Enzyme Molarity = 0.3 mmol/L = 0.3 mM
Substrate Concentration Calculation
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Use the given molar ratio of enzyme to substrate (1:3):
$ \frac{\text{Molarity}_\text{enzyme}}{\text{Molarity}_\text{substrate}} = \frac{1}{3} $
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Calculate the substrate molarity:
Molaritysubstrate = Molarityenzyme $ \times $ 3
Molaritysubstrate = 0.3 mM $ \times $ 3 = 0.9 mM
The corresponding substrate concentration is 0.9 mM.
Important Questions from Enzyme Kinetics Michaelis Menten K_m V_{max}
- An enzyme following Michaelis-Menten kinetics, catalyses a reaction with an initial velocity ($V_0$) of $2\ \mu\text{M s}^{-1}$ at the substrate concentration of $10\ \mu\text{M}$. If the turnover number ($k_{\text{cat}}$) of the enzyme for the given substrate is $500\ \text{s}^{-1}$ and the enzyme concentration in the reaction is $0.01\ \mu\text{M}$, then the value of the Michaelis-Menten constant ($K_m$) would be__________ $\times\ 10^{-6}\ \text{M}$ (in integer).
The graph below shows the activity of enzyme pepsin in the presence of inhibitors aliphatic alcohols (P) or N-acetyl-1-phenylalanine (Q). Which ONE of the following represents the nature of inhibition by P and Q, respectively?
The following plot represents the Lineweaver-Burk equation of an enzymatic reaction both in the presence and the absence of inhibitor. Here, V is the velocity of reaction and S is the substrate concentration.
The nature of inhibition shown in the plot is
- For an enzyme catalyzed reaction, the plot that correctly represents the relationship between the rate and temperature is
- In an enzyme catalyzed reaction, the initial reaction velocity is only one fourth of its maximum velocity. If the substrate concentration is $3.0 \times 10^{-3}$ mM, the value of $K_m$ in micro molar ($\mu$M) will be ....
