1 p t [s1] [s2] [s3] [s4] vo = the slope in each case effect of different initial substrate...

1

P

t

[S1]

[S2]

[S3]

[S4]

Vo = the slope in each caseVo = the slope in each case

Effect of different initial substrate concentrations

0.0

0.2

0.4

0.6

2

Considering Vo as a function of [S](which wil be our usual useful consideration)

No enzyme

3

We can ignore the rate of the non-catalyzed reaction

Now, with an enzyme:

4

Vo proportional to [S]

Vo independent of [S]

Enzyme catalyzed kinetics (as opposed to simple chemical kinetics)

5Michaelis and Menten mechanism for the action of enzymes (1913)

6Michaelis-Menten mechanism

• Assumption 1. E + S <--> ES: this is how enzymes work, via a complex

• Assumption 2. Reaction 4 is negligible, when considering INITIAL velocities (Vo, not V).

• Assumption 3. The ES complex is in a STEADY-STATE, with its concentration unchanged with time during this period of initial rates. 

(Steady state is not an equilibrium condition, it means that a compound is

being added at the same rate as it is being lost, so that its concentration remains constant.)

X

7

System is at equilibriumConstant levelNo net flow

Steady state is not the same as equilibrium

System is at “steady state”Constant levelPlenty of flow

8

E + S

ES

E + P

System is at “steady state”Constant levelPlenty of flow

9

Michaelis-Menten Equation(s)

[(k2+k3)/k1] +[S]

k3[Eo][S]Vo =

If we let Km = (k2+k3)/k1, just gathering 3 constants into one, then:

k3 [Eo] [S]Vo =

Km + [S]

See handout at your leisure for the derivation (algebra)

10

k3 [Eo] [S]Vo =

Km + [S]

Rate is proportional to the amount of enzyme

Otherwise, the rate is dependent only on S

At low S (compared to Km),rate is proportional to S:

Vo ~ k3Eo[S]/Km

At high S (compared to Km),Rate is constant

Vo = k3Eo

All the k‘s are constants for a particular enzyme

11

At high S, Vo here = Vmax = k3Eo

So the Michaelis-Menten equation can be written:

Vmax [S]Vo =

Km + [S]

k3 [Eo] [S]Vo =

Km + [S]Simplest form

12Now, Vmax = k3Eo

So: k3 = Vmax/Eo

= the maximum (dP/dt)/Eo, = the maximum (-dS/dt)/Eo

k3 = the TURNOVER NUMBER

• the maximum number of moles of substrate converted to product per mole of enzyme per second;

• max. molecules of substrate converted to product per molecule of enzyme per second

• Turnover number then is: a measure of  the enzyme's catalytic power.

13Some turnover numbers

• Succinic dehydrogenase: 19 (below average)

• Most enzymes: 100 -1000

• The winner:

Carbonic anhydrase (CO2 +H20 H2CO3)

600,000

That’s 600,000 molecules of substrate per molecule of enzyme per second.

Picture it!

You can’t.

14Km ?

Consider a Vo that is 50% of Vmax

So Km is numerically equal to the concentration of substrate required to drive the reaction at ½ the maximal velocityTry it: Set Vo = ½ Vmax and solve for S.

Vmax/2 is achieved at a [S] that is numerically equal to Km

15

The equilibrium constant for this dissociation reaction is:

Consider the reverse of this reaction(the DISsociation of the ES complex):

ESk2

k1

E + S

Kd = [E][S] / [ES] = k2/k1

(It’s the forward rate constant divided by the backward rate constant. See the Web lecture if you want to see this relationship derived)

Another view of Km:

(Note: This equation has been corrected from the original )

16

ESk2

k1

E + S Kd = k2/k1

Km = (k2+k3)/k1

IF k3 << k2, then: Km ~ k2/k1

But Kd = k2/k1 (from last graphic)

so Km ~ Kd for the dissociation reaction

(and 1/Km = ~ the association constant)

So: the lower the Km, the more poorly it dissociates.That is, the more TIGHTLY it is held by the enzyme

And the greater the Km, the more readily the substrate dissociates,so the enzyme is binding it poorly

{Consider in reverse

17

Km ranges

• 10-6M is good• 10-4M is mediocre• 10-3M is fairly poor

So Km and k3 quantitatively characterize how an enzyme does the job as a catalyst

18

-

Apparent Km increases

Inhibitor looks like the substrate and like the substrate binds to the substrate binding site

19Competitive inhibitor resembles the substrate

20

Competitive inhibitor can be swamped out at high substrate concentrations

Handout 5-3b

21

-

Apparent Km increases

Inhibitor looks like the substrate and like the substrate binds to the substrate binding site

22

-

Inhibition eventually swamped out as S is increased

23Biosynthetic pathway to cholesterol

24

Zocor(simvastin)

25

½ Vmax w/o inhibitor

½ Vmax withyet more inhibitor

Km remains unchanged. Vmax decreases.

26

Substrate Non-competitive inhibitor

Example: Hg ions (mercury) binding to –SH groups in the active site

27

Non-competitive inhibitor exampleSubstrate still bonds OK

But an essential participant in the reaction is blocked(here, by mercury binding a cysteine sulfhydryl)

--CH2-SH

Hg++

--CH2-SH

28

29

= allosteric inhibitor = substrate

+

Active Inactive

Allosteric inhibitor binds to a different site than the substrate,So it need bear no resemblance to the substrate

Active

The apparent Km OR the apparent Vmax or both may be affected.

Inhibitor binding site

Allosteric inhibition

30

Feedback inhibition of enzyme activity, orEnd product inhibition

First committed step is usually inhibited

Allosteric inhibitors are used by the cell for feedback inhibition of metabolic pathways

End product

End product

End product

31              Thr deaminase

glucose  ......  --> --> threonine -----------------> alpha-ketobutyric acid 

Substrate

Allosteric inhibitorAlso here: Feedback inhibitor(is dissimilar from substrate)

protein

protein

isoleucine  (and no other aa)

A

B

C

32

60 minutes, in a minimal medium 20 minutes !, in a rich medium

33

g l u c o s e

monomers

MacromoleculesPolysaccharides LipidsNucleic AcidsProteins

biosy

nthet

ic p

athw

ay

intermediates

F l o w o f g l u c o s e i n E . c o l i

E ac h a rro w = a sp e c ific c h em ica l re ac tio n

Direction of reactions in metabolism

34

} Energy difference

determines the direction of a chemical reagion

Free

35For the model reaction A + B C + D,

written in the left-to-right direction indicated:

Consider the quantity called the change in free energy associated with a chemical reaction, or: Δ G

Such that:• IF Δ G IS <0:

THEN A AND B WILL TEND TO PRODUCE C AND D(i.e., tends to go to the right).

• IF Δ G IS >0:THEN C AND D WILL TEND TO PRODUCE A AND B.(i.e., tends to go to the left)

• IF Δ G IS = 0:THEN THE REACTION WILL BE AT EQUILIBRIUM: NOT TENDING TO GO IN EITHER DIRECTION IN A NET WAY.

36ΔG = Δ Go+ RTln([C][D]/[A][B])

• where A, B, C and D are the concentrations of the reactants and the products AT THE MOMENT BEING CONSIDERED.(i.e., these A, B, C, D’s are not the equilibrium concentrations)

• R = UNIVERSAL GAS CONSTANT = 1.98 CAL / DEG K MOLE

• T = ABSOLUTE TEMP. (0oC = 273oK; Room temp = 25o C = 298o K)

• ln = NATURAL LOG

• Δ Go = a CONSTANT: a quantity related to the INTRINSIC properties of A, B, C, and D

37Also abbreviated form:Δ G = Δ Go+ RTlnQ

Where Q = ([C][D]/[A][B])

Qualitative term Quantitative term

Josiah Willard Gibbs(1839 - 1903)

38Δ Go

STANDARD FREE ENERGY CHANGE of a reaction.If all the reactants and all the products are present at 1 unit

concentration, then:

Δ G = Δ Go + RTln(Q) = Δ Go + RTln([1][1] / [1][1])Δ G = Δ Go + RTln(Q) = Δ Go + RTln(1)or Δ G = Δ Go +RT x 0,or Δ G = Δ Go,

when all components are at 1M….. a special case(when all components are at 1M)

39So Δ G and Δ Go are quite different,

and not to be confused with each other.

Δ Go allows us to compare all reactions under the same standard reaction conditions that we all agree to, independent of concentrations.

So it allows a comparison of the stabilities of the bonds in the reactants vs. the products.

It is useful.

AND,

It is easily measured.

40Because,

• at equilibrium, Δ G = Δ Go + RTln(Q) = 0 and at equilibrium Q = Keq =(a second special case).

• So: at equilibrium, Δ G = Δ Go + RTln(Keq) = 0

• And so: Δ Go = - RTln(Keq) • So just measure the Keq,

• Plug in R and T• Get: ΔGo, the standard free energy change

[C]eq [D]eq

[A]eq [B]eq

41E.g., let’s say for the reaction A + B C + D, Keq happens to be:

[C]eq[D]eq

[A]eq[B]eq

Then Δ Go = -RTlnKeq = - 2 x 300 x ln(2.5 x 10-3)

= -600 x -6 = +3600 3600 cal/mole (If we use R=2 we are dealing with calories)

Or: 3.6 kcal/mole3.6 kcal/mole ABSORBED (positive number)So energy is required for the reaction in the left-

to-right directionAnd indeed, very little product accumulates at

equilibrium (Keq = 0.0025)

= 2.5 x 10-3

42Note:

If ΔGo = +3.6 for the reaction A + B < --- >C + D

Then ΔGo = -3.6 for the reaction C + D <--- > A + B

(Reverse the reaction: switch the sign)

And:For reactions of more than simple 1 to 1 stoichiometries:

aA + bB <--> cC + dD, ΔG = ΔGo + RT ln [C]c[D]d

                           [A]a[B]b

43Some exceptions to the 1M standard condition:

• 1) Water: 55 M (pure water) is considered “unit” concentration instead of 1M

The concentration of water rarely changes during the course of an aqueous reaction, since water is at such a high concentration.

• So when calulating Go, instead of writing in “55” when water participates in a reaction (e.g., a hydrolysis) we write “1.”

• This is not cheating; we are in charge of what is a “standard” condition, and we all agree to this: 55 M H20 is unit (“1”) concentration for the purpose of defining Go.

Exception #1:

44Exception #2:In the same way,

Hydrogen ion concentration, [H+]: 10-7 M is taken as unit concentration, by biochemists.

since pH7 is maintained in most parts of the cell despite a reaction that may produce acid or base.

This definition of the standard free energy change requires the designation ΔGo’

However, I will not bother.

But it should be understood we are always talking about ΔGo’ in this course.

45Summary

ΔG = Go + RTln(Q)

This combination of one qualitative and one quantitative (driving) term tell the direction of a chemical reaction in any particular circumstance

ΔGo = - RTln(Keq)

The ΔGo for any reaction is a constant that can be looked up in a book.

46

e eA-R

ATP, a small molecule in the cell that helps in the transfer of energy from a place where it is generated to a place where it is needed.

47

The hydrolysis of ATP:

ATP + HOH ADP + Pi

A= adenine (2 rings with N’s)

R = ribose (5-carbon ring sugar)

ADP

The Go of this reaction is about -7 kcal/mole.

Energy is released in this reaction.

This is an exergonic reaction

Strongly to the right, towards hydrolysis, towards ADP

AMP

48

O O O || || ||

A-R-O-P-O-P-O-P-O- + HOH

| | | O- O- O-

O O O || || ||

A-R-O-P-O-P-O- + -O-P-O-

| | | O- O- O-

ATPAdenosine triphosphate

ADPAdenosine diphosphate

Pi

Inorganic

phosphateThe Go of this reaction is about -7 kcal/mole.

Energy is released in this reaction.

This is an exergonic reaction

Strongly to the right, towards hydrolysis, towards ADP

49“High energy” bonds:

Go more negative than ~ -7 kcal/mole is released upon hydrolysis

• Designated with a squiggle (~):• ATP = A-P-P~P• Rationalized by the relief of electrical repulsion upon

hydrolysis:

ΔGo = -7 kcal/mole

50

A-R-P~P~P

ADP

AMP

ΔGo = -7 kcal/mole

Not a high energy bond

ATP + HOH + ATPase ADP + Pi + heat

Keq ~ 100,000 Prob set 4

51

• The cell often uses the hydrolysis of ATP to release energy.

• The released energy is used to drive reactions that require energy.

• How does this work ??