Enzyme-substrate interactions

Name the factors that affect enzyme activity:

• pH

• Denaturing reagents

• Temperature

• Enzyme concentration

• Substrate concentration

• Inhibitors

pH

Affects ionisation of side chains and active site; extreme pH causes denaturation.

Denaturing reagents

Detergents, urea, guanidinium hydrochloride disrupt protein structure.

Temperature

Increases kinetic energy and collision rate; high temperatures cause protein unfolding.

Enzyme concentration

More enzyme → more active sites → higher maximum rate.

Substrate concentration

Rate increases with substrate until enzymes are saturated.

Inhibitors

Reduce enzyme activity by blocking or altering the active site.

What 4 things do simple models assume?

• Only one molecule of one substrate binds to the enzyme

• Enzyme and substrate form a [ES] complex

• Enzyme converts substrate to product and product release is fast

• Products bind weakly to enzymes

What does the Michaelis–Menten curve show?

The relationship between reaction rate (v) and substrate concentration [S], showing saturation at high [S].

What’s the Michaelis-Menten equation?

What is v?

v = rate

What is [S]?

[S] = substrate concentration

What is Vmax?

Vmax = rate when all enzyme active sites are occupied

What is Km?

Km = [S] at which v = ½ Vmax.

The substrate concentration when half of all enzyme active sites are occupied.

What does it mean for “enzyme reactions to saturate”?

All enzyme active sites are occupied, the rate has reached Vmax, and increasing substrate concentration no longer increases the reaction rate.

Why do enzyme reactions saturate at high substrate concentrations?

Because all enzyme active sites become occupied, so adding more substrate cannot increase the rate further.

How can Km and Vmax be measured directly?

By plotting v against [S] (Michaelis–Menten plot).

Michaelis-Menten Plot

• Plots v against [S]

• Can be difficult to decide when V_max is reached

• Usually requires a computer programme

Lineweaver-Burk Plot

• Plots 1/v against 1/[S]

• Y-intercept = 1/V_max

• X-intercept = -1/K_m

• Higher precision;

• Lower accuracy;

• Errors are not equal at all points (least squares regression is not appropriate).

Direct Linear Plot

• Plots v against [S]

• Each data point is used to draw a line, not just a point

• Connect points;

• Intersections are estimates of K_m and V_max;

• Gives median value;

• Makes no assumptions about the errors;

• Difficult to deviations from ideal behaviour.

What is kcat?

kcat = Vmax ÷ amount of enzyme.

It is the first-order rate constant for conversion of substrate to product.

What does kcat/Km represent?

A measure of enzyme efficiency (specificity constant) that allows comparison between different enzymes.

What are the assumptions of the steady state model?

• Enzyme concentration is much smaller than substrate concentration ([E] « [S]) (typically nM vs. mM);

• [S] remains constant; Measured rate of reaction is linear at beginning of curve (‘initial rate’);

• [ES] remains the same throughout course of reaction (‘The steady-state approximation’);

• Product binds weakly to enzyme;

• Rate of conversion of product to substrate is negligible (no backward reaction)-

If V_max = 0.5 what does that mean?

The maximum rate of the enzyme catalysed reaction is 0.5 units per second. When the enzyme is working as fast as it possibly can (all enzyme active sites are full of substrate), it can make 0.5 units of product per unit time.

What does Km tell us?

It tells us how strongly an enzyme binds to its substrate.

What does a low Km mean?

A low Km means the enzyme binds tightly (it works well even at low substrate levels).

What does a high Km mean?

A high Km means the enzyme binds weakly (it needs more substrate to work efficiently).

What does Kcat  show?

Kcat shows how fast the enzyme can convert substrate to product once the substrate is bound.

Exam Question:

In an enzyme kinetic experiment, the value for 1/V_max was 0.102 min/nmol and -1/K_m was -3.45 mM^-1 from a Lineweaver-Burk (double reciprocal) plot. Each assay contained 0.14 mg of enzyme with a molecular weight (M_W) of 47 146 Da. Showing your workings, calculate values for V_max, K_m, k_cat, k_cat/K_m, assuming one active site per monomer (hint: k_cat = V_max divided by amount of enzyme) [10 marks].

• 1/V_max = 0.102 min/nmol

• -1/K_m = -3.45 mM^-1

• Lineweaver-Burk Plot

• Each assay contained 0.14 mg of enzyme with a molecular weight (M_W) of 47 146 Da.

Calculate values for V_max, K_m, k_cat, k_cat/K_m