Lecture 9: Measuring enzyme activity

Question 1

What can we learn from a progress curve of an enzyme catalysed reaction?

Answer 1

A progress curve measures the appearance of product or disappearance of substrate with time.

This can help determine initial reaction rate (Vo)

Question 2

What is the relationship between V vs [S] on a V vs [S] curve for an enzyme catalysed reaction at low substrate concentrations?

Answer 2

At low substrate concentrations, the relationship is approximately linear.

This linear relationship reflects first-order kinetics, where the reaction rate 𝑉 is directly proportional to the substrate concentration [𝑆]. In this range, the enzyme is not saturated with substrate, and the rate of product formation increases as more substrate is available to bind to the enzyme.

Question 3

Consider a fixed amount of enzyme when substrate is in excess for a V vs [S] curve:

Answer 3

• The initial phase often reflects a first-order reaction (linear) (dependent on substrate concentration)
• while the plateau may indicate zero-order kinetics as all the active sites get occupied (independent of substrate concentration).

Question 4

What is Vmax?

Answer 4

maximum reaction velocity possible, when [S] = infinity

Question 5

What is Km and how do we find it?

Answer 5

Km is the Michaelis constant

Km = the substrate concentration at which V = Vmax/2

Question 6

What is the Michaelis Menten equation?

Answer 6

V = Vmax x [S] / Km + [S]

Can be used to describe V vs [S] curve to determine their kinetic parameters (Vmax and Km)

Question 7

When there is an excess of substrate, reaction velocity is _________ to enzyme concentration.

Answer 7

proportional to enzyme concentration

Question 8

Why does an enzyme catalysed reaction reach a maximum velocity when there is excess substrate but a fixed concentration of enzyme?

Answer 8

V max is reached when the enzyme is saturated with substrate, meaning that every enzyme molecule is constantly occupied and working at full capacity. Adding more substrate beyond this point does not increase the reaction rate because the limiting factor is the enzyme concentration, not the substrate availability.

Question 9

What equation can be used to describe a V vs [S] curve for a monomeric enzyme?

Answer 9

Michaelis Menten equation baby

Question 10

What is plotted on a Lineweaver-Burk plot (both axis, slope? Y-intercept? X-intercept?)?

Answer 10

X axis = 1/[S]

Y axis =1/V

Slope = Km / Vmax

Y-intercept = 1 / Vmax

X intercept = -1/ Km

Question 11

What is the significance of Km?

Answer 11

Low Km = high affinity between enzyme and substrate

High Km = low affinity between enzyme and substrate

(is the substrate concentration needed to reach half Vmax)

approx Km = k-1 / k1 (see slide 14 to understand)

Question 12

If an enzyme has multiple substrates, how can we use KM to determine substrate preference?

Answer 12

The substance with the lower Km indicates a higher affinity for that substance and therefore will prefer that substance

Question 13

What does Kcat for an enzyme catalysed reaction represent?

Answer 13

Kcat is the turnover number

It is the number of substrate molecules converted to product, per enzyme, per unit time, when the enzyme is saturated with substrate

k2 = Kcat = Vmax / [E]t (see slide 17 to understand)

Question 14

How do we define catalytic efficiency? What would a ‘peak’ enzyme have?

Answer 14

We can define catalytic efficiency by Kcat / Km

Thus a ‘peak’ enzyme will have:
High Kcat (high turnover)
Low Km (high affinity for substrate)

Question 15

What assumptions can be made to simplify a Michaelis-Menten model reaction? (there are 3)

Answer 15

1. Product is not converted back to substrate
2. Haldane’s steady state assumption: the rate of enzyme-substrate formation equals the rate of its breakdown
3. Measuring initial rate means [S] does not change significantly (and [S] > [E])