L11: Measuring and Comparing the Activity of Enzymes

Question 1

This measures the appearance of a product (or disappearance of a product) over time.

Answer 1

Progress curve

Question 2

Initial velocity is proportional to enzyme concentration when?

Answer 2

Substrate concentration is in excess

• this basically means that enzyme concentration affects reaction velocity when substrate concentration outcompetes number of enzymes (zero order kinetics - rate does not depend on substrate because the amount of substrate doesn’t change the amount of enzyme)

Question 3

Why does Michaelis Menten enzymes have a hyperbolic shape?

Answer 3

Because the rate will increase in a linear way at first and as the active sites become activated, the rate of reaction stops increasing (because there’s no more enzyme to accomodate substrate)

Question 4

What does first rate kinetics depend on?

Answer 4

Substrate concentration

Question 5

What is Vmax?

Answer 5

Vmax is the maximum reaction velocity when substrate concentration is in excess.

Question 6

What is Km?

Answer 6

Km is the substrate concentration at Vmax/2. It is also the Michaelis constant.

Question 7

What are the 3 main assumptions of Michael Menten model?

Answer 7

1) product is not converted back to substrate.

2) the rate of enzyme-substrate formation is equal to the rate of its breakdown.

3) measuring initial rate means [S] does not change significantly ([S] > [E])

Question 8

What is on the y intercept, x intercept of the Lineweaver Burk plot?

Answer 8

y intercept (velocity) = 1/vmax

x intercept (substrate concentration) = 1/Km

Question 9

What is the significance of Km?

Answer 9

Km signifies one enzyme substrate pair. Means enzymes can act on different substrates and have different Km values for each => can be used for substrate preference indication

Question 10

What is the physiological significance of Km? What does this mean for rate control?

Answer 10

In physiological enzyme-substrate interaction, [S] is often below the Km.

> > means rate control is effective because the substrate molecules aren’t queuing.

Question 11

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

Answer 11

Kcat

Question 12

A peak enzyme should have what 2 things?

Answer 12

A high Kcat (turnover number)

A low Km (for higher affinity for substrate => requires a lower substrate concentration required to get up to speed)

Question 13

What is the equation for the overall measure of enzyme activity?

Answer 13

Kcat/Km