Lab 6: Reversible Inhibition

Question 1

What type of inhibition was studied using what?

Answer 1

Reversible inhibition was studied using partially purified B-galactosidase

Question 2

How was the degree of inhibition measured?

Answer 2

By the amount of synthetic substrate ONPG that produces ONP (a yellow colour) when cleaved

Question 3

What determines first order?

Answer 3

[So] is much smaller than Km

Question 4

What determines if a reaction is zero order?

Answer 4

If [So] is much larger than Km, v approaches Vmax

Question 5

What is Km?

Answer 5

The substrate concentration when the reaction velocity is half its maximum velocity

Question 6

Why are linear transformations often used instead of the hyperbolic function?

Answer 6

Because of errors associated with Vmax and Km with a hyperbolic plot

Question 7

What do the X and Y intercepts, as well as the slope, show on a Lineweaver-Burk graph?

Answer 7

The slope shows Km/Vmax
The X-axis is for -1/Km
The Y-axis is for 1/Vmax

Question 8

What type of inhibitor was calcium?

Answer 8

Mixed-type inhibitor

Question 9

What type of inhibitor was sucrose?

Answer 9

Uncompetitive

Question 10

What type of inhibitor was IPTG?

Answer 10

Competitive

Question 11

What is Ki?

Answer 11

Dissociation constant for inhibitor binding to free enzyme

Question 12

What is Kies?

Answer 12

Dissociation constant for inhibitor binding to enzyme-substrate complex, Kies

Question 13

What are Kmapp and Vmaxapp?

Answer 13

the “apparent” values of Km and Vmax that are acutally observed in the presence of inhibitor for the Michaelis-Menten equation for inhibited enzymes

Question 14

What makes a inhibitor competitive?

Answer 14

Binds at the active site and competes with substrate for enzyme

Question 15

How do Vmax and Km change for competitive inhibition?

Answer 15

Vmax is unchanged, Km increases

Question 16

What makes a inhibitor mixed-type?

Answer 16

Inhibit enzyme activity by combining both enzyme and enzyme-substrate complex

Question 17

How do Vmax and Km change for mixed-type?

Answer 17

Vmax decreases, Km can increase or decrease

Question 18

What makes a uncompetitive inhibitor?

Answer 18

Binds only with the enzyme-substrate complex. Limiting case of mixed-type inhibition where Ki approaches infinity

Question 19

How do Vmax and Km change for uncompetitive?

Answer 19

Vmax decreases, Km decreases

Question 20

What makes a non-competitive inhibitor?

Answer 20

Special case of mixed type inhibition where Ki=Kies and Vmax decreases by the inhibition factor while Km is unchanged

Question 21

What is the equation for Km? What is it called?

Answer 21

The Michaelis- Menten Constant

Km= (K-1 + K2)/K1

Question 22

What did we change for the reaction? What was kept constant?

Answer 22

Inhibitor and Substrate concentrations were changed and the concentration of the enzyme was kept constant

Question 23

What were the reaction velocities measured for?

Answer 23

• for varying [S] with fixed [I]

- for varying [I] with fixed [S]

Question 24

What was the enzyme used?

Answer 24

B-galactosidase

Question 25

What is the substrate that is used?

Answer 25

ONPG

Question 26

What are the inhibitors used?

Answer 26

2 mg/mLIPTG, 200 mg/mL calcium, 1 M sucrose

Question 27

For the initial dilution of B-gal what is the rate of change needed for ONPG?

Answer 27

0.2 to 0.3 A405nm/min

Question 28

Why do you need to keep enzyme on ice?

Answer 28

to minimize degradation

Question 29

What do we use to blank the sample to read the velocities?

Answer 29

• B-gal assay buffer
• substrate
• diluted B-galactosidase enzyme

Question 30

What are the units for velocity used in the lab?

Answer 30

delta A405nm/minute read at 15 second intervals for 2 minutes

Question 31

What type of graph is Vo vs. [S]? Why?

Answer 31

• hyperbolic
• the substrate is able to interact maximally with the free enzyme. As the substrate-enzyme complex reaches equilibrium the reaction will slow down. At saturation point then it plateaus

Question 32

What happens at low [S] vs high [S] for the hyperbolic curve?

Answer 32

• low [S]: most of enzyme is in free E form. Rate of the reaction is proportional to [S] because the equilibrium of the reaction is pushed towards the formation of [ES] as [S] increases
• high [S]: most of the enzyme is in the ES form. The enzyme essentially becomes saturated. Vmax is reached because increasing [S] has little effect on the rate of the reaction. The reaction rate plateaus

Question 33

What can be found using the initial velocity vs [S] graph?

Answer 33

Vmax and Km (just an estimate)

Question 34

What can be found on the Dixon Plot? What are the axis?

Answer 34

1/Vo (Y) vs [I] (X axis) for each fixed substrate

Find the -Ki value (the x-axis where the lines intersect) which converts to a +Ki value

Question 35

Which types of inhibitors have Ki values?

Answer 35

Competitive and mixed-type (Ki is for binding to the free enzyme)

Question 36

Which type of inhibitors have Kies values?

Answer 36

Uncompetitive and mixed type (Kies is binding to the enzyme-substrate complex)

Question 37

What can be found on the [S]/Vo versus [I] graph?

Answer 37

• made for the fixed [S]
• Kies
• -Kies is the x-axis where the lines intersect (give + value)

Question 38

What makes competitive inhibition inherently different from uncompetitive and mixed-type inhibition?

Answer 38

• inhibitor binds at the active site

- means Vmax is unchanged because increasing the [S] can overcome the inhibition

Question 39

What does the inhibitor bind for competitive inhibition?

Answer 39

Free enzyme at active site

Question 40

What does the inhibitor bind for uncompetitive inhibition?

Answer 40

Enzyme-substrate complex NOT at active site

Question 41

What does the inhibitor bind for mixed-type inhibition?

Answer 41

Free enzyme NOT at active site AND Enzyme-substrate complex NOT at active site

Question 42

What does the Michaelis-Menten equation show?

Answer 42

The behavior of most enzymes but NOT the mechanism