Lecture 12: Enzymes as Biological Catalysts

Question 1

What is the function of an enzyme?

Answer 1

Catalyze conversion of substrate to product

Question 2

What is the mechanism of catalysis by an enzyme?

Answer 2

Lowering the activation energy

Question 3

What does free energy tell you about a reaction?

Can the free energy be changed by enzymes?

Answer 3

It tell you whether the reaction is favorable or not, and whether it will occur spontaneously or not.

But it does not tell you about the rate of the reaction.

It CANNOT be changed by enzymes

Question 4

How does enzymes interact with substrates?

What type of bonding interactions are involved?

Answer 4

Induced fit – enzyme changes conformation.

There is some complimentarity between the enzyme and the substrate, which allows them to interact and form bonds. The binding interactions results in the conformational change of the enzyme.

Bind substrate via weak non-covelent interaction( electrostatic, H bonds, Van der waals, hydrophobic effects)

Question 5

Describe the active site of an Enzyme?

Answer 5

Highly specific for substrate

In the cleft of the protein- water excluded

Can include prosthetic groups of and cofactor

In accordance with protein 3D structure

Question 6

How can you monitor enzymatic reaction?

Answer 6

Rate of substrate disappearance

Rate of product appearance

Question 7

How can you determine the Initial Velocity Experimentally?

Answer 7

• Examine initial Velocity at different substrate concentration
• Constant enzyme concentration

PIC: Top graph- substrate depleted– substrate dependent (line curved)

bottom graphs: high [substrate]– substrate independent (line initially linear and then curves)

Question 8

Describe the Velocity Vs [substrate] curve

Answer 8

At low substrate concentration, the velocity is substrate dependent and the graph is curved, at high substrate concentration, the enzymes are saturated and the velocity reaches a maximum and is substrate independent [Vmax] and is [enzyme] dependent.

Question 9

Michaelis Menten Equations

Question 10

What does the Km mean?

Answer 10

• Km is the substrate concentration at which the velocity is 1/2 maximal
• Km=[S] at 1/2Vmax
• it also tell you the affinity of the substrate ( high Km- low affinity, low Km- high affinity)
• it would allow you to measure how much enzyme is bound to substrate to form Enzyme:Substrate [ES] complex ( example when k3<<<>>>K3, you are forming less product and the dissociation of the ES complex is larger than product formation)

k3-rate of product formation

k2- rate of ES dissociation.

Question 11

Explain the Km/affinities of Hexokinase and Glucokinase

Answer 11

Hexokinase- wide distributions throught the body and low Km (0.1) higher affinity for Glucose (working at it Vmax at blood glucose 3mM)

Glucokinase- in the liver, high Km(5mM) lower affinity for glucose and working at 37% Vmax at glucose blood level of 3mM, after big meal worked at 58% Vmax

Question 12

When are enzymatic reaction most sensitive?

Answer 12

Enzymatic reactions are most sensitive when the substrate concentration is low

[S] <<<< Km

Question 13

What does V max mean?

Answer 13

• It is the maximum velocity possible for a specific concentration of enzyme
• Velocity unit
• Can be changed by changing the enzyme concentration
• increasing [enzyme]= increase Vmax ( increase enzyme→ increase binding sites for substrates→ increase Vmax)

Question 14

Describe Kcat (K3)

Answer 14

• It is a constant independent of enzyme concentration
• It is propotional to Vmax without affects of enzym
• It is a rate constant
• tell you how much substrate that can be made into product.

Question 15

Describe the relation of Kcat/Km

Answer 15

• High Kcat/Km tells you that there is more products being formed from a given about of ES complex
• Lower Kcat/Km indicated lower product formation

Kcat- rate of ES use/turnover.

Km- rate of ES formation

Question 16

How can you estimate Km and Vmax more accurately graphically?

Answer 16

Lineweaver-Burk Transform ( double reciprocal plot)

1/v=1/Vmax + (Km/Vmax)/(1/[S])

Y intercept = 1/Vmax

Slope = Km/Vmax

X intercept = -1/Km

Question 17

Describe Competitive Inhibitors?

Answer 17

• Compete with substrate for the active site
• Inhibition can be overcomed by the addition of more substrate
• Changes Km, increases the Km because you need more substrate to compete away the inhibitor.
• Vmax does not change

Question 18

Describe Non-competitive inhibitors

Answer 18

• compete with subtrate by binding to another site other than the active site.
• you cannot overcome inhibition by increasing substrate concentration
• the Km is not changed
• the Vmax is lowered because the enzyme concentration is lowered.

Question 19

What is Ki?

Answer 19

• It is the dissociation constant of the enzyme inhibitor complex
• you can determine the strength in which a inhibitor binds the enzyme
  
  • Higher Ki = weak interaction between enzyme and inhibitor.
  • Lower Ki = tighter interaction between enzyme and inhibitor
  • EI → E + I