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Introduction Catalysis is a central concept in chemistry and biology, essential in life and industriPhysical Chemistry — Thermodynamics Chemistry Question

Enzyme kinetics by catalase

Introduction
Catalysis is a central concept in chemistry and biology, essential in life and industrial processes. Enzymes are catalysts for biochemical reactions. In this experiment, the Michaelis–Menten kinetics of hydrogen peroxide decomposition (2 H2O2 → 2 H2O + O2) by catalase in potato juice will be investigated. Catalase is well known for its extremely high reaction rate. One catalase molecule can decompose 40 million hydrogen peroxide molecules in one second. Such a high rate is needed to scavenge reactive oxygen species and protect cellular components in the oxidative environment. The figure below shows a 3–dimensional structure of catalase from E. coli determined by X–ray crystallography.

[VISUAL]

The number of moles of the evolved oxygen gas can be determined from its volume measured using a buret or from the pressure change in an enclosed reaction vessel. Reaction rate can be expressed as the number of moles of oxygen per unit time.

An enzyme (E) combines with a substrate (S) and produces an enzyme–substrate complex (ES) with a rate constant k1. ES could be decomposed back to E and S with a rate constant k2 or concerted to a product (P) with a rate constant k3. The steady state condition for ES can be determined by solving the following rate equations:

d[ES] / dt = k1([E]tot – [ES]) [S] , where [E]tot = [E] + [ES]
– d[ES]/dt = k2 [ES] + k3 [ES]
[S]([E]tot – [ES]) / [ES] = (k2 + k3) / k1

(k2 + k3) / k1 is defined as the Michaelis–Menten constant, KM.

Solving the last equation for [ES] one gives [ES] = [E] [S] / (KM + [S]).

Let v is the initial rate for the evolution of oxygen: v = k3 [ES]. If the enzyme is present only as ES, v will approach to a maximum value, vmax = k3 [E]tot . From these relations, one gets the Michaelis–Menten equation.

v = vmax [S] / (KM + [S])

Obviously, KM is the value of [S] when v = vmax / 2. Taking the inverse of the Michaelis–Menten equation one gets the celebrated Lineweaver–Burk equation (see Figure), which is one of the most frequently used equations in chemistry.

[VISUAL]

1/v = (KM / vmax) (1 / [S]) + 1 / vmax

Chemicals and materials
* hydrogen peroxide,
* fresh potato,
* catalase

Apparatus
* blender,
* ice bath,
* boiling water bath

Procedure
(1) Prepare 0.5, 1, 2, 3, 4, 6% solution of hydrogen peroxide by diluting the given 30% hydrogen peroxide with deionized water.
(2) Make potato juice by blending pieces of potato with approximately equal mass of water. Squeeze the juice with cheesecloth. Keep the juice in an ice bath.
(3) Add 2 cm3 of the juice to 30 cm3 of the diluted hydrogen peroxide solutions and shake. As a control, use 30 cm3 of deionized water.
(4) Measure the volume of oxygen produced using a set–up shown below. Make a soap bubble with the rubber bulb and measure time needed to produce a certain volume (20 cm3 for example) of oxygen gas at room temperature.

[VISUAL]

(5) Repeat with 6% hydrogen peroxide using the juice heated in boiling water bath for 10 min to denature the enzyme.
(6) If pure catalase is available, repeat the whole experiment using catalase at a known concentration (1 micromolar, for example).

Treatment of data and questions

36.1.

Calculate the molar concentration of hydrogen peroxide, [S].

36.2.

Calculate the amount of substance of oxygen produced in a given time for each [S].

36.3.

Calculate v for each [S].

36.4.

Plot v against [S] and see if it approaches a maximum value.

36.5.

Develop a Lineweaver–Burk plot to determine KM and Vmax.

36.6.

If [E]tot is known, calculate k3 from Vmax = k3 [E]tot. What is the turn–over number of catalase per second?

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