Enzyme kinetics & The Michaelis-Menten Equation

Question 1

Quantitative description of enzyme catalysis

What is rate of reaction?

What is enzyme activity?

Answer 1

Rate of reaction = Change in [substrate] or [product] per unit time (Mmin^-1 or molL^-1min^-1)

= conc of substrate used ÷ time or

= conc of product formed ÷ time (in M or mM)

Enzyme activity = moles substrate converted per unit time

= rate x volume

Enzyme activity represents the quantity of enzyme present

SI unit: 1 katal = 1 mol s^-1 (very large unit, impractical)

1 enzyme unit = 1 μmol min^-1 = 10^-6 mol min^-1

(widely used)

Question 2

What is specific activity?

Answer 2

Specific activity = activity per unit mass of enzyme (Activity per unit mass of total protein)

Moles substrate converted per unit time per unit mass of enzyme (protein) = Enzyme activity/mass of enzyme

• SI units: katal kg^-1
• Practical units: μmol min^-1 mg^-1 or μmol min^-1 μg^-1
• When two different pure enzymes are compared, specific activity is a measure of enzyme efficiency
  
  • e.g. enzyme-1 has SA of 50 μmol min^-1 μg-1, enzyme-2 has an SA of 25 μmol min^-1 μg^-1; enzyme-1 is twice as efficient
• When pure and impure samples of same enzyme are compared,
• specific activity is a measure of enzyme purity
  
  • e.g. pure enzyme has a specific activity of 10 μmol min^-1 mg^-1
  • sample with a specific activity of 2 μmol min^-1 mg^-1 is 20% pure
  • less pure sample includes impurities making up 80% of its mass

Question 3

What is molar activity? What is it equal to?

Answer 3

Molar activity = activity per mole of enzyme

Moles substrate converted per unit time per mole of enzyme (protein) mol min^-1 mol^-1 = min^-1

• Molar activity = specific activity x molar mass of enzyme (with necessary unit conversions!!!)
• If n moles of substrate are converted per second per mole of enzyme, then n molecules of substrate are converted per molecule of enzyme
• Molar activity is equal to the turnover number, the number of catalytic reaction cycles per molecule of enzyme per second

Question 4

What is starting sample specific activity? Its purity?

After 1st purification? Second?

Answer 4

Question 5

Sample calculations: reaction of chymotrypsin

Start: 25.0x10^-4 M peptide
After 10 min: 18.6x10^-4 M peptide

Reaction contains 1.50 μg chymotrypsin in 2.5 mL total volume Molar mass of chymotrypsin is 25,000 g mol^-1 (25 kDa)

Answer 5

Peptide substrate consumed = (25.0-18.6) x 10^-4 M in 10 minutes

Rate of reaction = 6.4 x 10^-5 mol L^-1 min^-1

Enzyme activity = 6.4 x 10^-5 x 2.5 x 10^-3 mol min^-1

(rate x volume) = 1.6 x 10^-7 mol min^-1

Specific activity = 1.6 x 10^-7 mol min^-1 ÷ 1.5 μg (activity/mass enzyme)= 1.1 x 10-7 mol μg-1 min-1

Turnover number = 1.1 x 10-7 mol μg-1 min-1 x 25000 x 106 μg mol-1 (spec act x mol mass) = 2.7 x 103 min-1

= 45 s-1

Question 6

Kinetics: mathematical analysis of how reaction rate varies as a function of …

Answer 6

Kinetics: mathematical analysis of how reaction rate varies as a function of reactant concentration

• Plot substrate or product concentration over period of time
  
  • this is a progress curve
• Initial rate vo is taken from the slope of the curve at time zero
• Measure several rates at different initial [S]

Question 7

What is zero order, first order, second order?

What does the plot as faction of [S] look like?

Answer 7

• Zero order rate = k[S]⁰
• First order rate = k[S]¹
• Second order rate = k[S]²
• Normal chemical reactions follow simple rate laws
• Enzyme reaction does not follow a simple rate law

Question 8

Analysis of enzyme reaction as two steps

How does an enzyme catalytic reaction work? Who first explored this?

Answer 8

• First explored by Henri in France 1905, with systematic experimental verification by Michaelis and Menten in 1913
• The derivation that follows is due to Briggs and Haldane, 1926
• Since enzyme is recycled, it is not consumed, [E]_total is constant
• Work at time = 0, so [P] = 0, and reverse reaction at step 2 can be ignored
• Individual steps have rate constants k₁, k_-1, k₂

Question 9

Explain rate of reaction

Answer 9

Rate of reaction = rate of appearance of P

• Initial rate v_o = k₂ [ES]
• Although we know [E]_total, we don’t know how much enzyme is empty [E] and how much is occupied [ES]

Question 10

How can we determine [ES]?

Answer 10

Question 11

What is the Michaelis-Menten equation?

Answer 11

Question 12

Michaelis and Menten verified this equation experimentally in 1913

What did they also demonstrate?

What does the equation show?

How can the equation also be written?

Answer 12

• They also demonstrated the importance of using initial rates measured at time = 0
• The equation shows how v_o varies as a function of [S]
• Every enzyme has characteristic values of the two constants, V_max and K_M (the Michaelis constant)
• The equation may also be written in the fractional form, useful for solving enzyme kinetics problems

Question 13

What does V_max tell you about an enzyme?

Answer 13

• V_max is the upper limit for rate
  At high [S], all enzyme molecules have bound substrate and are engaged in catalytic activity, so reaction can’t go any faster
• V_max is a pseudo-constant, constant only if amount of enzyme is fixed

V_max = k₂[E]_total
k₂ is the true constant and is the turnover number of the enzyme

Question 14

What does K_M tell you about an enzyme?

Answer 14

• K_M is the concentration of substrate [S] at which rate vo is equal to 50% of the maximum rate V_max

typical K_M values are between 10^-6 M (1 μM) and 10^-2 M (10 mM)

• If [S] = K_M, then v₀/V_max = K_M/K_{M +} K_M = 0.5
• A low K_M indicates that the enzyme binds and utilizes substrate well; less [S] is needed to occupy the enzyme
• A high K_M indicates that the enzyme binds and utilizes substrate poorly; more [S] is needed to occupy the enzyme