FOM 2.2.3

Question 1

In an reaction why doesn’t the concentration of ES go to zero at long time points?

Answer 1

When the system reaches equilibrium, product is going to be turned into reactant and this will create a ES complex

Question 2

Draw a graph of time vs concentration for the reaction:

E+S <—> ES

ES <—> EP

EP <—> E+P

Draw the lines for S, E, ES and P Also note where the transient state, equilibrium state and steady state are on the graph.

Answer 2

Slide 8 on Quantification of Enzyme reactions Just remember it and dont be a pussy

Question 3

In a graph for time vs concentration for product formation, what does the slope of it equal?

Answer 3

Slope equals the rate of the reaction Vo

Question 4

What is the equation for Vo?

Answer 4

Vo = (Vmax)/2 or Vo= k[E][S]

Question 5

Why does substrate concentration dependence for determining Vo disappear at a high concentration?

Answer 5

The enzyme becomes saturated with substrate and it cannot work any faster

Question 6

In a graph of substrate concentration vs Vo, how can you determine the amount of substrate present?

Answer 6

By knowing how fast the reaction is occurring you can derive the substrate concentration

Question 7

What is Vmax?

Answer 7

The maximum rate by which a reaction can occur

Question 8

What does the Vmax depend on? How can it be altered?

Answer 8

Vmax depends on the concentration of Enzyme present. Increasing the amount of enzyme present can increase the Vmax and decreasing the amount of enzyme can decrease the Vmax

Question 9

What is Kcat?

Answer 9

The turnover number, the amount of reactions in time. It is independent of [E]

Question 10

What is the formula for the formation of [ES]?

Answer 10

[ES] = k1[E][S]

Question 11

What is the formula for the breakdown of [ES]?

Answer 11

[ES] = (k2+k3)[ES]

Question 12

What is the formula for Km?

Answer 12

(K2+K3)/K1

Question 13

What is the formula for Vmax?

Answer 13

(k3)([Etotal])

Question 14

What is the michaelis-menten equation for Vo?

Answer 14

Vo = (Vmax[S])/(Km + [S])

Question 15

What is Km?

Answer 15

Km is the substrate concentration when the reaction is going at 1/2 Vmax

Question 16

What happens to Km when the concentration of enzyme in a system is increased or decreased?

Answer 16

The Km is unchanged the only thing that is changed is the Vmax

Question 17

What it the effect of changing the [Total Enzyme] on Km, Kcat, Vo, and Vmax?

Answer 17

Km - No change Kcat - No change it is independent of the enzyme concentration (Turnover Number) Vo - No change Vmax - Increases or decreases depending on whether enzyme is added or lost

Question 18

What is the effect of changing the substrate concentration on the Vmax, Kcat, Vo, and Km?

Answer 18

Vmax - no change Kcat - no change Vo - is the observed rate this will change with substrate concentration Km - no change

Question 19

What will happen to enzyme activity is the concentration of co-enzyme changes?

Answer 19

Concentration of co-enzyme will alter the Vmax of a reaction similar to how the concentration of total enzyme will

Question 20

What factors influence the kinetic constants?

Answer 20

Temp, pH, salt conc, and mutation in enzyme

Question 21

What is a basic strategy for detecting reactions when signals are not available from substrate or product?

Answer 21

This happens in many lab tests. The reaction is coupled with another reaction that produces a color change that can be observed. The color will change at a rate that can determine things like rate of reaction or substrate concentration

Question 22

Explain this picture and why the test might not be able to function at very high levels of glucose

Answer 22

The chromagen is the limiting factor, because eventually at a high enough level of glucose, it will run out.

Question 23

Aspergillus niger glucose oxidase

Normal adult blood glucose: 65-120 mg/dL = 3.6 – 6.7 mM

Diabetic adults (uncontrolled): ≥ 200 mg/dL = 11.1 mM

550 mg/dL or more = 30 mM

Published Km range: 1.9 to 11.9 mM,

depending on expt’l conditions

pH, temperature, strip matrix

What other parameter could you adjust in order to choose conditions that would allow results to quickly be read by the patient?

A. [Glucose oxidase] B. [H2O2] C. [Glucose] D. [O2]

Answer 23

To make the test quicker it would be easiest to add more glucose oxidase to the reaction.

Question 24

Aspergillus niger glucose oxidase

Normal adult blood glucose: 65-120 mg/dL = 3.6 – 6.7 mM

Diabetic adults (uncontrolled): ≥ 200 mg/dL = 11.1 mM

550 mg/dL or more = 30 mM

Published Km range: 1.9 to 11.9 mM,

depending on expt’l conditions

pH, temperature, strip matrix

Changed reaction conditions can alter Km. Which of the following would be the most appropriate value?

A. 1.9 mM B. 6.7 mM C. 11.9 mM D. 30 mM

Answer 24

It would be most appropriate to have it at 11.9mM because that would allow for the detection of normal concentrations as well as acutely high conditions

Question 25

Do this

Answer 25

Question 26

When [S] equals Km what can be said about the velocity of the reaction?

Answer 26

It is operating at 1/2 Vmax

Question 27

An enzyme is suddenly introduced to a much higher amount of glucose than normally expected. What would happen to the Km of the reaction?

Answer 27

It would be unchanged

Question 28

In the stomach a bunch of new enzyme is created to handle the meal that was just eaten. What could is happening to the Km and Vmax?

Answer 28

The Km is still unchanged you dipshit. The new enzyme will increase the Vmax of the reaction though.

Question 29

What is going on in this picture?

Answer 29

The loss of NADH in reaction 2 is going to be associated with a loss of absorbance at 340 nm. This is going to give a clear picture of how much pyruvate is forming.