Michaelis–Menten method of parameter estimation MCQs With Answer

Michaelis–Menten method of parameter estimation MCQs With Answer

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Michaelis–Menten method of parameter estimation MCQs With Answer is essential for B. Pharm students studying enzyme kinetics and drug metabolism. This introduction covers key concepts like Michaelis–Menten equation, Km, Vmax, kcat, catalytic efficiency (kcat/Km), steady-state and rapid-equilibrium assumptions, and common linear transforms (Lineweaver–Burk, Eadie–Hofstee, Hanes–Woolf). It also highlights practical aspects of parameter estimation: initial rate measurement, substrate range, experimental error, bias from double-reciprocal plots, and the advantages of non-linear regression. Mastery of these topics helps in enzyme characterization, inhibitor analysis, and pharmacokinetic modelling. Now let’s test your knowledge with 30 MCQs on this topic.

Q1. What is the Michaelis–Menten constant (Km) conceptually?

  • The substrate concentration at which reaction velocity is half of Vmax
  • The maximum reaction velocity achieved by the enzyme
  • The turnover number of an enzyme
  • The dissociation constant between enzyme and inhibitor

Correct Answer: The substrate concentration at which reaction velocity is half of Vmax

Q2. Which equation represents the Michaelis–Menten rate law?

  • v = (Vmax [S])/(Km + [S])
  • v = Vmax + Km[S]
  • v = kcat[E][S]
  • v = (Km [S])/(Vmax + [S])

Correct Answer: v = (Vmax [S])/(Km + [S])

Q3. In Lineweaver–Burk plot (double reciprocal), what does the y-intercept equal?

  • 1/Vmax
  • -1/Km
  • Km/Vmax
  • Vmax/Km

Correct Answer: 1/Vmax

Q4. Which transformation expresses the Lineweaver–Burk relationship?

  • 1/v = (Km/Vmax)(1/[S]) + 1/Vmax
  • v = Vmax – Km[1/[S]]
  • v/[S] = (Vmax/Km) – (1/Km)[S]
  • v = (1/Vmax) + (Km)[S]

Correct Answer: 1/v = (Km/Vmax)(1/[S]) + 1/Vmax

Q5. What is a major drawback of Lineweaver–Burk plots for parameter estimation?

  • They overweight low substrate concentration data and amplify error
  • They require knowledge of kcat to calculate Km
  • They cannot be used to estimate Vmax
  • They linearize cooperative enzyme kinetics accurately

Correct Answer: They overweight low substrate concentration data and amplify error

Q6. Which method is generally preferred today for accurate estimation of Km and Vmax?

  • Non-linear regression fitting of the Michaelis–Menten equation
  • Lineweaver–Burk linearization
  • Graphical eyeballing of Vmax
  • Using only initial and final data points

Correct Answer: Non-linear regression fitting of the Michaelis–Menten equation

Q7. The turnover number (kcat) is defined as:

  • Vmax divided by total enzyme concentration [E]t
  • The substrate concentration at half Vmax
  • The slope of the Lineweaver–Burk plot
  • The dissociation constant of enzyme-substrate complex

Correct Answer: Vmax divided by total enzyme concentration [E]t

Q8. Catalytic efficiency is best represented by which ratio?

  • kcat/Km
  • Km/kcat
  • Vmax/Km
  • [E]t/Vmax

Correct Answer: kcat/Km

Q9. The Eadie–Hofstee plot has which form?

  • v = -Km (v/[S]) + Vmax
  • 1/v = (Km/Vmax)(1/[S]) + 1/Vmax
  • v/[S] = (Vmax/Km) – (1/Km) [S]
  • v = Vmax [S]/(Km + [S])

Correct Answer: v = -Km (v/[S]) + Vmax

Q10. Why can Eadie–Hofstee plots be problematic despite less distortion of low [S] data?

  • Because the dependent variable v appears on both axes, creating correlated errors
  • Because they eliminate the need for measuring Vmax
  • Because they assume substrate inhibition always occurs
  • Because they require logarithmic transformation of data

Correct Answer: Because the dependent variable v appears on both axes, creating correlated errors

Q11. Hanes–Woolf plot uses which transformed variables?

  • [S]/v versus [S]
  • 1/v versus 1/[S]
  • v versus v/[S]
  • ln v versus ln [S]

Correct Answer: [S]/v versus [S]

Q12. Briggs–Haldane approach to Michaelis–Menten kinetics relies on which assumption?

  • Steady-state approximation for ES complex formation and breakdown
  • Rapid equilibrium between E and ES
  • Reaction is irreversible at all substrate concentrations
  • Only initial velocity is zero

Correct Answer: Steady-state approximation for ES complex formation and breakdown

Q13. When using non-linear regression, which error model is commonly assumed for enzyme initial rate data?

  • Normally distributed errors with constant variance (homoscedasticity) or weighted variance if heteroscedastic
  • Poisson-distributed errors independent of signal
  • Errors proportional only to Vmax
  • No errors; non-linear least squares is exact

Correct Answer: Normally distributed errors with constant variance (homoscedasticity) or weighted variance if heteroscedastic

Q14. Which plot often exaggerates experimental error at low substrate concentrations the most?

  • Lineweaver–Burk plot (double reciprocal)
  • Eadie–Hofstee plot
  • Hanes–Woolf plot
  • Non-linear Michaelis–Menten fit

Correct Answer: Lineweaver–Burk plot (double reciprocal)

Q15. What experimental design choice improves accuracy in estimating Km?

  • Include multiple substrate concentrations around Km (both below and above)
  • Use only very high substrate concentrations ([S] >> Km)
  • Use a single substrate concentration equal to Km
  • Measure only at initial and saturating substrate concentrations

Correct Answer: Include multiple substrate concentrations around Km (both below and above)

Q16. In competitive inhibition, how are Km and Vmax affected (apparent values)?

  • Apparent Km increases, Vmax unchanged
  • Apparent Km decreases, Vmax unchanged
  • Vmax decreases, Km unchanged
  • Both Km and Vmax decrease

Correct Answer: Apparent Km increases, Vmax unchanged

Q17. In pure noncompetitive inhibition, what is the effect on apparent Km and Vmax?

  • Km unchanged, apparent Vmax decreased
  • Km increased, Vmax unchanged
  • Km decreased, Vmax decreased
  • Both Km and Vmax increased

Correct Answer: Km unchanged, apparent Vmax decreased

Q18. Substrate inhibition deviates from Michaelis–Menten kinetics by showing what behavior at high [S]?

  • Rate decreases as substrate concentration becomes excessive
  • Rate continues to increase linearly indefinitely
  • Rate becomes independent of enzyme concentration
  • Rate reaches zero immediately above Km

Correct Answer: Rate decreases as substrate concentration becomes excessive

Q19. Which parameter can directly inform about enzyme specificity toward a substrate?

  • kcat/Km (catalytic efficiency)
  • Km alone
  • Only Vmax
  • 1/Vmax

Correct Answer: kcat/Km (catalytic efficiency)

Q20. When using Lineweaver–Burk plots to analyze inhibition, what feature indicates competitive inhibition?

  • Lines intersect on the y-axis (same 1/Vmax intercept)
  • Lines intersect left of the y-axis at the same x-intercept
  • Lines are parallel with different y-intercepts
  • Lines do not intersect

Correct Answer: Lines intersect on the y-axis (same 1/Vmax intercept)

Q21. Which of the following is true about Km under Michaelis–Menten steady-state conditions?

  • Km approximates the substrate concentration at which the enzyme is half-saturated; relates to rate constants
  • Km is always equal to the dissociation constant Kd
  • Km is independent of catalytic step and only depends on inhibitor binding
  • Km equals Vmax at physiological temperature

Correct Answer: Km approximates the substrate concentration at which the enzyme is half-saturated; relates to rate constants

Q22. Which technique reduces bias from reciprocal transformation when estimating Km and Vmax?

  • Non-linear regression of the original Michaelis–Menten equation with appropriate weighting
  • Using only the Lineweaver–Burk plot but excluding low [S] points
  • Taking logarithms of both sides and using linear regression
  • Estimating parameters visually from raw data

Correct Answer: Non-linear regression of the original Michaelis–Menten equation with appropriate weighting

Q23. In units, what is the typical unit for kcat if Vmax is in μmol·min⁻¹·mg⁻¹ and enzyme concentration in mg?

  • min⁻¹
  • μM
  • μmol·min⁻¹
  • mg·min⁻¹

Correct Answer: min⁻¹

Q24. Which statement about initial velocity (v0) measurements is correct?

  • Initial rates should be measured before significant substrate depletion or product inhibition occurs
  • Initial rates are taken after the reaction reaches equilibrium
  • Initial rates depend only on product concentration
  • Initial rates require complete conversion of substrate to product

Correct Answer: Initial rates should be measured before significant substrate depletion or product inhibition occurs

Q25. When the assumption of rapid equilibrium holds, which simplification applies?

  • Formation and dissociation of ES is fast relative to catalysis, so Km approximates Kd
  • Catalysis is infinitely fast and [S] does not matter
  • ES complex is never formed
  • Vmax becomes zero

Correct Answer: Formation and dissociation of ES is fast relative to catalysis, so Km approximates Kd

Q26. What does a Hanes–Woolf plot reduce compared to Lineweaver–Burk?

  • It reduces the disproportionate weighting of low [S] values and gives more even error distribution
  • It completely eliminates experimental error
  • It makes v appear on both axes causing correlated errors
  • It always yields higher Km than true value

Correct Answer: It reduces the disproportionate weighting of low [S] values and gives more even error distribution

Q27. In mixed (noncompetitive) inhibition where inhibitor binds both E and ES with different affinities, what is observed?

  • Apparent Km changes and apparent Vmax decreases
  • Km unchanged and Vmax increases
  • Both Km and Vmax increase equally
  • No change in kinetics

Correct Answer: Apparent Km changes and apparent Vmax decreases

Q28. Which parameter estimation method allows inclusion of measurement error estimates for each data point?

  • Weighted non-linear regression
  • Simple linear regression on Lineweaver–Burk plot
  • Graphical interpolation
  • Unweighted Hanes–Woolf linear fit

Correct Answer: Weighted non-linear regression

Q29. What is a practical sign that your enzyme assay data may not follow Michaelis–Menten kinetics?

  • Rate versus [S] curve is sigmoidal, suggesting cooperativity
  • Rate increases hyperbolically and plateaus
  • Rate is directly proportional to enzyme concentration
  • Initial rate is zero at all substrate concentrations

Correct Answer: Rate versus [S] curve is sigmoidal, suggesting cooperativity

Q30. When fitting data to the Michaelis–Menten model, what is a recommended practice to ensure reliability of estimated parameters?

  • Measure multiple replicates across a wide substrate range, use non-linear regression, and check residuals and confidence intervals
  • Rely on a single high-concentration data point to define Vmax
  • Use only linear transforms and ignore residual patterns
  • Exclude replicates to reduce scatter

Correct Answer: Measure multiple replicates across a wide substrate range, use non-linear regression, and check residuals and confidence intervals

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