Enzyme Kinetics

Question 1

Catalyst:

Answer 1

substance that increases the rate of a chemical reaction without being consumed in the reaction

provides an alternative pathway for the reaction in which the rate-determining step has a lower Gibbs activation energy than that of the non-catalysed reaction

Question 2

Catalysis

Answer 2

1. Does not appear in the overall equation for the reaction although it may appear in the rate equation
2. Some substances slow down chemical reactions and these are often called negative catalysts or inhibitors
3. Does not affect the amount of product formed—only the rate of its formation
4. Small amount is usually needed
5. Not changed chemically in the reaction and can be recovered at the end
6. Sometimes may be changed physically

Question 3

At a given temperature, the rate constant is greater for the catalysed reaction, so the reaction is ______

Answer 3

faster

Question 4

Non-catalysed reaction proceeds through a _____ _______ state, whereas the mechanism for the catalysed reaction involves the formation of an intermediate

Answer 4

single transition

Question 5

Enzyme-Catalysed reactions

Answer 5

Enzymes catalyse a wide range of reactions in biochemical systems
(human body dilute solution at 37 °C). Without the presence of enzymes, the reactions would take place much too slowly to sustain life.

Question 6

Define enzymes;

Answer 6

protein molecules, catalyse a specific reaction, unchanged by
the reactions they catalyse

Question 7

Whats the name of the model - enzyme and substrate bind?

Answer 7

Lock and Key Model

Question 8

name of site on enzyme where substrate binds to?

Answer 8

Active site

Question 9

Lock and key model - step by step

Answer 9

1) Substrate binds to enzyme (in active site) reversibly
2) Non-covalent interactions (enzyme-substrate complex) > reaction takes place
3) enzyme-product complex
4) product leaves to make free enzyme and free substrate

Question 10

What INDUCED fit mean vs lock and key model

Answer 10

the active site of an enzyme will undergo a conformational change when binding a substrate, to improve the fit.

Chemical and shape compatibility of a substrate with an enzyme active site (substrate fits to active site precisely)

Question 11

in the active sites what are two things that are present?

Answer 11

> hydrophobic region
charges +/-

Question 12

Enzymatic reaction Profile -
Enzyme works as catalyst providing an _______ pathway for the reaction in which the rate-determining step has a lower ______ activation energy than that of the non-catalysed reaction

Answer 12

alternative

Gibbs

Question 13

LOWER the activation energy…

Answer 13

more sufficient the reaction is = the FASTER the rate of reaction

Question 14

Two distinct steps in enzyme catalysed reaction:

Answer 14

1) first step - reversible
2) second step - irreversible

Question 15

Michaelis Menten model was developed to understand the kinetics of enzymic reactions.

Answer 15

E + S <> (k1/k-1)
ES > E + R (k2)

Question 16

A number of assumptions are applied to the Michaelis Menten model:

Answer 16

The enzyme binds a single substrate (the model assumes cofactors are already bound)
There are two distinct steps. The substrate binds reversibly but product formation is irreversible
A steady state approximation applies (quantitative analysis of the kinetics of enzyme-catalysed reactions)

Question 17

Whats the name of the state at the strart of an enzymatic reaction?

Answer 17

The steady state

Question 18

Rate of formation of ES = Rate of consumption of ES (conversion to E + P)

Answer 18

( 𝒅[𝑬𝑺])/𝒅𝒕=(Rate of formation of ES - Rate of consumption of ES)=0

Question 19

=k1 [E] [S] -k2 [ES] - k-1 [ES]

Answer 19

( 𝒅[𝑬𝑺])/𝒅𝒕=k1 [E] [S] –(k2 - k-1 )[ES]=0

Question 20

Assumption is made that: [E]0 = [ES] + [E] where [E]0=

Answer 20

total conc of enzyme

Question 21

Concentration of the substrate is usually much larger than that of the enzyme, so that [S] is effectively ______

Answer 21

constant

Question 22

In the initial phase of the reaction: [E]^ and [ES]^

Answer 22

[ES] and [E] remain steady over most of the reaction time.
Only at the end of the reaction, when [S] is exhausted, does [ES] decline

Question 23

The Michaelis Menten Equation

Answer 23

𝑽 (𝒓𝒂𝒕𝒆 𝒐𝒇 𝒓𝒆𝒂𝒄𝒕𝒊𝒐𝒏)= ( 𝑽𝒎𝒂𝒙 [𝑺])/(𝑲𝑴+[𝑺])

Question 24

What kind of linear when the velocity of an enzyme catalysed reaction increases

Answer 24

a non-linear fashion

Question 25

KM=the Michaelis constant.

Answer 25

It is (k2 +k-1)/k1

Question 26

Maximal Velocity Vmax, when is Vmas reached on graph?

Answer 26

At the highest point before the gradient goes flat

When independent of the [S] = the rate of reaction remains constant at the maximum value, Vmax, and the reaction is zero order with respect to [S].

Question 27

When [S] is large, [S]&raquo_space; KM…

Answer 27

a) 𝑽 (𝒓𝒂𝒕𝒆 𝒐𝒇 𝒓𝒆𝒂𝒄𝒕𝒊𝒐𝒏)= ( 𝑽𝒎𝒂𝒙 [𝑺])/([𝑺]) = Vmax =K2 [E]0
(S cancelles each other)

b) Rate of reaction is now independent of [S]

c) All of the enzyme molecules have S attached during the reaction and [ES] = [E]0

d) The enzyme is said to be saturated. If the [S] increases, no more enzyme–substrate complexes can be formed, and the rate of the reaction is independent of [S]

Question 28

Vmax;

Answer 28

the maximum rate that a given enzyme can achieve, indicates the velocity of the reaction when the enzyme active site is saturated and indicates the maximum number of moles of S that can be processed in a unit of time (mol s-1)

Question 29

How do we know it is first order?

Answer 29

The rate of reaction increases linearly with [S]
It is first order with respect to [S].

Question 29

How do we know it is first order?

Answer 29

The rate of reaction increases linearly with [S]
It is first order with respect to [S].

Question 30

Define the turnover number;

Answer 30

number of substrate molecules converted into product by an enzyme molecule in a unit time when the enzyme is fully saturated with substrate

Question 31

What is the turnover number related to?

Answer 31

It is related to Vmax and is equivalent to the rate constant k2

Question 32

Turnover number equation

Answer 32

Turnover Number = K2 = Vmax/[E]0 (s-1)

Range from 1 to 104 s-1 in MOST enzymes

Question 33

Km, whats it used for?

Answer 33

The Michaelis constant = (k2 +k-1)/k1

gives indication of how tightly the enzyme binds its substrate

Question 34

KM= substrate concentration ([S]) at which the reaction velocity is ____ of V maximum (Vmax/2)

Answer 34

half

Question 35

Weak substrate; large or small Km?

Answer 35

large Km

a high [S] is needed to achieve Vmax/2 (and even more to reach Vmax!)

Question 36

A good substrate has a ____ Km;

Answer 36

low

Only a low [S] needed to reach Vmax.

Question 37

KM= The [S] at which the enzyme active site is ____ saturated

Answer 37

half

Question 38

Lineweaver-Burk Equation

Answer 38

The y and x axis intercepts give us the important kinetic terms Vmax and KM from easily measured properties (rate and substrate concentration)

Question 39

Why use a straight line graph than a curve?

Answer 39

easier to plot