Enzyme Kinetics Practical

Question 1

Enzyme Kinetics

Answer 1

the study of the rates of an enzyme catalyzed reactions.

In enzyme kinetics,

Question 2

Michaelis Menten - Low substrate concentration

Answer 2

At low substrate concentration, there is a steep increase in the rate of reaction with increasing substrate concentration.
The catalytic site is empty, waiting for the substrate to bind for much of the time.
The rate at which the product is formed is limited by the concentration of substrate.

Question 3

The initial rate of an enzyme-catalyzed reaction

Answer 3

(V0)
Is directly proportional to the concentration of substrate [S]
In this region, the reaction is first order with respect to the substrate.

Question 4

What happens when the concentration of substrate increases

Answer 4

The enzyme becomes saturated with substrate.
At high concentrations of substrate, there is a decreasing response as substrate concentration increases, until the maximum velocity (Vmax) is achieved at substrate saturation.
The reaction is now zero order with respect to the substrate

The rate of formation now depends on the activity of the enzyme itself, and adding more substrate will not affect the rate of reaction to any significant effect.

Question 5

zero-order rate

Answer 5

a zero-order reaction has a rate that is independent of the concentration of the reactant(s). As such, increasing or decreasing the concentration of the reacting species will not speed up or slow down the reaction rate.

Question 6

first order rate of reaction

Answer 6

A first-order reaction can be defined as a chemical reaction in which the reaction rate is linearly dependent on the concentration of only one reactant.

Question 7

Draw the Michaelis Menten plot - the relationship between substrate concentration and initial velocity in an enzyme-catalyzed reaction

Answer 7

–

Question 8

What is Michaelis Menten

Answer 8

Relationship between Substrate

Concentration and Initial Velocity in an Enzyme-Catalysed Reaction

Question 9

What is the Michaelis constant

Answer 9

Km , Km is the Michaelis-Menten constant which shows the concentration of the substrate when the reaction velocity is equal to one half of the maximal velocity for the reaction. It can also be thought of as a measure of how well a substrate complexes with a given enzyme, otherwise known as its binding affinity.

Question 10

what is the Michaelis menten equation

Answer 10

V0 = Vmax[S]/ Km + [S]

Question 11

Ks

Answer 11

The dissociation constant of the enzyme-substrate complex

Question 12

1/Ks

Answer 12

The affinity constant of the enzyme for the substrate

Question 13

Km and affinity

Answer 13

Km provides a measure of the affinity of an enzyme for the
substrate
A large value of Km indicates low affinity for substrate, a small Km indicates a high affinity for substrate

Question 14

The Lineweaver Burk equation

Answer 14

When 1/ vo is plotted against 1/[S], a straight line is obtained, from which
the Michaelis constants Km and Vmax can be obtained as in Figure 2

created by plotting the inverse initial velocity (1/V0) as a function of the inverse of the substrate concentration (1/[S]). The Vmax can be accurately determined and thus KM can also be determined with accuracy because a straight line is formed.

Question 15

Enzyme inhibition

Answer 15

Enzyme inhibitors are molecules that interact with an enzyme in some way and keep it from working in its usual manner

Question 16

Competitive inhibitors

Answer 16

Competes with the substrate for binding at the enzyme active site

Question 17

Competitive inhibitors

Answer 17

Competes with the substrate for binding at the enzyme active site
The binding of an inhibitor prevents substrate binding
A competitive inhibitor often has a similar chemical structure to the substrate, however, it is not acted on by the enzyme.
Thus, every time an inhibitor binds to the active site, the enzyme will no longer be able to catalyze the reaction and the rate will decrease.

Question 18

enzyme-inhibitor complex

Answer 18

A competitive inhibitor reacts reversibly with the enzyme to form the enzyme-inhibitor complex. This is analogous to the enzyme-substrate complex.

Question 19

Ki

Answer 19

The inhibitor constant. The dissociation of the enzyme- inhibitor complex.

Question 20

What will a competitive inhibitor do to the Km of the enzyme.

Answer 20

Will increase the apparent Km of the enzyme for the substrate.
It causes it to require a higher substrate concentration to achieve Vmax.

Question 21

Why is Vmax unchanged by a competitive inhibitor

Answer 21

Vmax can still be achieved if enough substrate is added to out-compete the inhibitor

Question 22

A series of double-reciprocal plots , competitive inhibitor

Answer 22

A series of double reciprocal plots (Lineweaver Burke) plotted against each other at different concentrations, will all intersect at the same point.
1/Vmax on the Y-axis

Question 23

Uncompetitive inhibition

Answer 23

Occurs when an inhibitor binds to the enzyme-substrate complex and in doing so renders the enzyme catalytically inactive.

Question 24

Vmax and Km in uncompetitive inhibition

Answer 24

As some of the enzyme is rendered inactive, regardless of how much substrate is present Vmax is reduced, and hence in a double reciprocal plot 1/Vmax is increased.

Question 25

The plot of 1/Vo against 1/[S] diff conc. uncompetitive inhibitor

Answer 25

A series of parallel lines,
the x-intercept shifts to the left
because km decreases and hence 1/km becomes larger

Question 26

Noncompetitive inhibition

Answer 26

Occurs when the inhibitor binds at a site different from the active site.
The binding of the inhibitor at this site generally reduces the enzyme’s ability to convert substrates into products in the active site.

Question 27

Vmax and Km - noncompetitive inhibitor

Answer 27

Vmax reduces as the enzyme can no longer function as well as it could in the absence of the inhibitor.
Vmax reduces, 1/Vmax will increase. The Km for the reaction will not change, because while the enzyme isn’t catalyzing the reaction well, it can still bind the substrate just as well as it did before.

Question 28

Noncompetitive inhibition, Lineweaver burke

Answer 28

Plot of 1/Vo against 1/[S]