Enzyme kinetics 1 and 2

Question 1

Kinetics of uncatalyzed reactions

Answer 1

A+B-> Q+P (rate of reaction is determined by the law of mass action)
Law of mass action : observed rate Vo is proportional to [A][B]k with (k being the rate constant)

rxn velocity (vo) has a unit of concentration/ time and can be based on the conversion (neg) of substrate or the creation (pos) of product: Vo= d [P]/ d t = -d [S]/d t

Question 2

Reaction order

Answer 2

law of mass action states that Vo is proportional to the collisions between A and B
reaction order is the number of species that must collide for the reaction to occur
A+B-> Q+P (is a second order when both substrates are limiting)

if B were in excess, the rxn is in first order in A (only dependent on [A]) and in zero order for B

Question 3

Kinetics of enzyme catalyzed reactions

Answer 3

Simple Enzyme: S binds to E and ES complext undergoes the catalyzed reaction to from P

E +S -> ES -> E + P

K1, K2
 more complicated (E +S -> ES -> EP->  E + P

K1, K2, K3 (rate constants)

Assumptions:
Initial Rate (when rate is restricted to early stage reaction right after the pre steady state)
[S]= amount of S added doesnt change the conversition to P, minimal P is formed and no reverse reaction. ES-> E+P is treated as irreversible (k-2=0)

Question 4

Initial rate analysis

Answer 4

Exess of substrate, so [S] does not change significantly thru out the analysis

The amount of P made initially is not significant so we assume there is no K-2 (how are you going to make P->ES if theres almost no P)

Assumptions:
Initial Rate assumption
Steady State assumption (d [ES] = 0

The assumptions allow us to isolate the catalytic properties of the enzyme of interest (it tells us the affinity of enzyme for substrate and the rate of conversion from ES to P

Question 5

Michaelis menton equation

Answer 5

vo= ( Vmax[S] ) / ( Km + [S] ) Velocity

Vmax= k2 [Etot] max velocity under saturated conditions

Km= (k-1 + k2)/ k1 (affinity of ES, higher km, weaker affinity

Question 6

Km and Kd

Answer 6

Km= (k-1 + K2)/ K1
Km= [s] at 1/2 Vmax

Kd= K-1/K1 (the dissociation constant at equilibrium)

If an enzyme has a very very small K2, the enzyme does not turn ES-> EP before the substrate dissociates, and the Kd and Km are very close

if K2 is much larger than K-1 the enzyme will have a low affinity for the S (due to high km) but the Km is much larger than the Kd

Question 7

Catalytic efficiency

Answer 7

Kcat/ Km is a way to measure K2 in relation to an enzymes affinity for substrate, enzymes with a higher catalytic efficiency are able to make lots of P without having to bind tighlty to S

Enzymes with a low Km work well when there is a SMALL amount of substrate because theres a high chance of grabbing and holding on to substate

If Km is small the Kcat is likely to be small as well

Kcat (1/seconds) and Km (concentration)

Remeber a high Km means a weaker affinity to S (you only want to hold on tightly to S when there isnt a lot of S)

Question 8

Line weaver burk plots

Answer 8

MM plot, intial diagonal line due to [S]«> Km

Km is the [S] at 1/2 vmax

Lin weaver uses a double reciprical of MM to give a linear plot: 1/v on the y axis and 1/[s] on the x axis

The y intercept is 1/Vmax and the x intercept is -1/ Km and the slope is Km/Vmax

The y=mx + b LB equation is (derived fromMM equation)

1/vo= (Km/Vmax *[s]) + 1/Vmax)

Question 9

Reversible inhibitors

Answer 9

Competitive inhibitors:
I binds to E not EX, Km increases, overcome with large [S]
slope (km/Vmax) increases, Vmax stays the same, Km increases

Non competitive inhibitors:
I binds to E AND ES complex, Vmax decreased, Km stays the same, Affect the ES-> E+ P, slope (km/vmax) goes up, affinity for E and ES is exactly the same

UN competitive inhibitor:

I binds only to the ES, So KM is decreased and Vmax is decreased, slope (Km/vmax doesnt change)

Mixed inhibition:
I binds to both E and ES, but at different affinities so the Km could increase, decrease or remain unchanged, Vmax is always decreased. The main differnce between mixed and competitive is the fact that mixed inhib cant be overcome by increasing S

Question 10

irreversible inhibitors

Answer 10

Irr. I binds tightly (usually covalent to the enzyme) and inactivate it (clinical inhibitor are specific to only one class and do not destroy the enzyme but only the target activ site. They are like noncompetitive inhibition (but are nor reverirsible)

Ex penicillins

Question 11

Allosteric enzymes and inhibitors

Answer 11

Allosteric enzymes have a regulatory (R) site distinct from the active site that allow inhibitors or activators to bind and change catalytic site C activity

Asp. transcarbamoylase (CTP stabilises inactive state, ATP stabilizes activ state)

Cooperativity , multiple interacting subunits, lowers Km and increases Vmax at other subunits

Kinetics dont follow MM kinetics, they have sigmoidal shaped curves (as opposed to hyperbolic