Enzyme-substrate interactions

Name 6 factors that affect enzyme activity:

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• pH

• Denaturing reagents

• Temperature

• Enzyme concentration

• Substrate concentration

• Inhibitors

How does pH affect enzyme activity?

pH affects ionisation of side chains and active site

Extreme pH causes denaturation

What do denaturing reagents do?

Disrupt protein structure

Name some denaturing reagents:

• Detergents

• Urea

• Guanidinium hydrochloride

How does temperature affect enzyme activity?

Increases kinetic energy and collision rate

High temperatures cause protein unfolding.

How does enzyme concentration affect enzyme activity?

More enzymes means more active sites, which means higher maximum rate.

How does substrate concentration affect enzyme activity?

The rate increases with substrate concentration until enzymes are saturated.

How do inhibitors affect enzyme activity?

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 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 is the Michaelis-Menten curve?

It is a graph that shows how the rate of an enzyme-catalysed reaction changes as substrate concentration increases.

What are the axes on the Michaelis-Menten curve?

X-axis = Substrate concentration = [S]

Y-axis = Reaction rate = v

What does the Michaelis-Menten curve tell us?

• At low [S], the rate of reaction increases rapidly as more [ES] complexes form

• At high [S], the rate plateaus because all active sites are saturated

• The maximum rate reached is called the Vmax

What’s the Michaelis-Menten equation?

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).

What’s the downside of using the Michaelis-Menten plot?

• It can be difficult to decide when V_max is reached

What is the Lineweaver-Burk plot?

• The Michealis-Menten curve is curved, which makes it hard to accurately determine the Vmax and Km

• Therefore, the Michealis-Menten equation was rearranged into a reciprocal (inverse) form, giving a straight line graph

What are the axes of the Lineweaver-Burk plot?

X-axis = 1/[S]

Y-axis = 1/v

What does the y-intercept and x-intercept represent in the Lineweaver-Burk Plot?

• Y-intercept = 1/V_max

• X-intercept = -1/K_m

What’s the advantage of using the Lineweaver-Burk plot?

Higher precision

What are the disadvantages of using the Lineweaver-Burk plot?

• Lower accuracy

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

What is the Direct-Linear plot?

• This plot uses the original Michaelis-Menten equation

• Plots multiple straight lines from experimental data points

• Finds where those lines intersect to estimate Vmax and K_m

What are the axes in the Direct Linear Plot?

• x axis = K_m

• y axis = Vmax

• Each data point gives a straight line; all lines intersect near true Km and Vmax

• Gives median value

What’s the advantages of the Direct Linear plot?

It makes no assumptions about the errors

It’s accurate

What is k_cat?

The number of substrate molecules converted to product per enzyme per second when the enzyme is fully saturated with substrate (turnover number).

How do you calculate k_cat?

k_cat = V_max ÷ Amount of enzyme.

What’s the unit for k_cat?

s^-1

What does k_cat/K_m represent?

A measure of enzyme efficiency at low substrate concentration, combining substrate binding (k_m) and catalytic rate (k_cat), indicating how effectively an enzyme converts substrate to product.

What are 6 assumptions of the steady state model (Michealis-Menten)?

• Substrate is in large excess over enzyme ([E] « [S])

• [S] remains constant

• Measured rate is initial rate (linear start of curve)

• [ES] remains constant

• The product binds weakly to enzyme

• The backwards reaction is neglible

If V_max = 0.5 what does that mean?

The maximum rate of the enzyme catalysed reaction is 0.5 units per unit time. 

What does K_m tell us?

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

What does a low K_m 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 K_cat  show?

How fast the enzyme can convert substrate to product once the substrate is bound.

Unit conversions

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].

V_max

• 1/V_max = 0.102 min/nmol

• 1/V_max = 0.102 min nmol^-1

• V_max = 9.8 nmol min^-1

K_m

• -1/K_m = -3.45 mM^-1

• 1/K_m = 3.45 mM^-1

• K_m = 0.29 mM

K_cat

• 0.14 mg of enzyme with M_W of 47,146 Da

• 47,146 Da = 47,146 gmol^-1

• 0.14mg = 0.00014g

• mol = 0.00014g / 47146 gmol^-1

• mol = 2.97 × 10^-9

• K_cat = Vmax / Amount of enzyme

• 9.8nmol min-1 = 9.8 × 10^-9 mol min^-1

• K_cat = 9.8 × 10^-9 mol min^-1 / 2.97 × 10^-9 mol

• K_cat = 3.30 min^-1

• K_cat = 0.0550 s^-1

K_cat/K_m

• 0.0550 s^-1 / 0.29mM = 0.190 mM^-1 s^-1

• 0.190 mM^-1 s^-1

• K_cat/K_m = 190 M^-1 s^-1