1

SDS PAGE = SDS polyacrylamide gel electrophoresis

• sodium dodecyl sulfate, SDS (or SLS): CH3-(CH2)11- SO4--

• CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-SO4--

SDS

All the polypeptides are denatured and behave as random coilsAll the polypeptides have the same charge per unit lengthAll are subject to the same electromotive force in the electric fieldSeparation based on the sieving effect of the polyacrylamide gelSeparation is by molecular weight onlySDS does not break covalent bonds (i.e., disulfides) (but can treat with mercaptoethanol for that) (and perhaps boil for a bit for good measure)

2

Disulfides between 2 cysteines can be cleaved in the laboratory by reduction, i.e., adding 2 Hs (with their electrons) back across the disulfide bond.

One adds a reducing agent: mercaptoethanol (HO-CH2-CH2-SH). In the presence of this reagent, one gets exchange among the disulfides and the

sulfhydryls:

Protein-CH2-S-S-CH2-Protein  + 2 HO-CH2CH2-SH  --->

Protein-CH2-SH + HS-CH2-Protein   +  HO-CH2CH2-S-S-CH2CH2-OH

The protein's disulfide gets reduced (and the S-S bond cleaved), while the mercaptoethanol gets oxidized, losing electrons and protons and itself forming a disulfide bond. 

3

Molecular weight markers

(proteins of known molecular weight)

P.A.G.E.

e.g., “p53”

12 18

48

80 110 130 160

140

4

Sephadex bead

Molecular sieve chromatography(= gel filtration, Sephadex chromatography)

5

Sephadex bead

Molecular sieve chromatography

6

Sephadex bead

Molecular sieve chromatography

7

Sephadex bead

Molecular sieve chromatography

8

Sephadex bead

Molecular sieve chromatography

9

PlainFancy4oC (cold room)

10

Non-spherical molecules get to the bottom faster

Larger molecules get to the bottom faster, and ….Non-spherical molecules get to the bottom faster

~infrequent orientation

11Handout 4-3: protein separations

12

Most chargedand smallest

Largest and most spherical

Lowest MW

Largest and least spherical

Similar to handout 4-3, but Winners &

native PAGE added

Winners:

Winners:

13

Enzymes = protein catalysts

14

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 nEach arrow = an ENZYME

Each arrow = an ENZYME

15

H2 + I2 2 HI

H2 + I2 2 HI + energy

“Spontaneous” reaction:

Energy releasedGoes to the rightH-I is more stable than H-H or I-I herei.e., the H-I bond is stronger, takes more energy to break itThat’s why it “goes” to the right, i.e., it will end up with more products than reactantsi.e., less tendency to go to the left, since the products are more stable

Chemical reaction between 2 reactants

16

Ch

ang

e in

En

erg

y (F

ree

En

erg

y)

H2 + I22 HI

{

-3 kcal/mole

2H + 2I

say, 100 kcal/mole

say, 103kcal/mole

Atom pulled completely apart(a “thought” experiment)

Reaction goes spontaneously to the right

If energy change is negative: spontaneously to the right = exergonic: energy-releasingIf energy change is positive: spontaneously to the left = endergonic: energy-requiring

17

H2 + I2 2 HI

2 HIH2 + I2

2 HIH2 + I2

Different ways of writing chemical reactions

H2 + I2 2 HI

2 HIH2 + I2

18

Ch

ang

e in

En

erg

y (F

ree

En

erg

y)

H2 + I22 HI

2H + 2I

{

-3 kcal/mole

say, 100 kcal/mole

say, 103kcal/mole

But: it is not necessary to break molecule down to its atoms in order to rearrange them

19

H H

+I I

H H

IIII

H H

Transition state (TS) +

H

I

H

I

(2 HI)

H H+

I I

(H2 + I2)

Products

Reactions proceed through a transition state

20

Ch

ang

e in

En

erg

y

H2 + I2

2 HI

2H + 2I

{

-3 kcal/mole

~100 kcal/mole

H-H| |I-I(TS)

Activationenergy

Say,~20 kcal/mole

21

Ch

ang

e in

En

erg

y (n

ew s

cale

)

H2 + I2

2 HI

{

3 kcal/mole

Activation energy

HHII(TS)

Allows it to happen

determines speed = VELOCITY = rate of a reaction

Energy neededto bring molecules together to forma TS complex

Net energy change:Which way it will end up. the DIRECTIONof the reaction, independent of the rate

2 separate concepts

22

Concerns about the cell’s chemical reactions

• Direction– We need it to go in the direction we want

• Speed– We need it to go fast enough to have the

cell double in one generation

23

3 glucose’s 18-carbon fatty acid

Free energy change: ~ 300 kcal per mole of glucose used is REQUIRED

Biosynthesis of a fatty acid

So: 3 glucose 18-carbon fatty acid

So getting a reaction to go in the direction you want is a major problem(to be discussed next time)

Example

24

Concerns about the cell’s chemical reactions

• Direction– We need it to go in the direction we want

• Speed– We need it to go fast enough to have the

cell double in one generation

– Catalysts deal with this second problem, which we will now consider

25

The catalyzed reaction

The velocity problem is solved by catalysts

The catalyst takes part in the reaction, but it itself emerges unchanged

26

Ch

ang

e in

En

erg

y

H2 + I2

2 HI

Activation energywithoutcatalyst

HHII(TS)

TS complexwith catalyst

Activation energyWITH thecatalyst

27

Reactants in an enzyme-catalyzed reaction = “substrates”

28

Reactants (substrates)

Not a substrate

Active site or

substrate binding site(not exactly synonymous,

could be just part of the active site)

29

Substrate Binding

Unlike inorganic catalysts, enzymes are specific

30Small molecules bind with great specificity to pockets on ENZYME surfaces

Too far

31Unlike inorganic catalysts,

enzymes are specific

                                      succinic dehydrogenase

HOOC-HC=CH-COOH <-------------------------------> HOOC-CH2-CH2-COOH +2H

fumaric acid                                                     succinic acid

NOT a substrate for the enzyme: 1-hydroxy-butenoate:    HO-CH=CH-COOH (simple OH instead of one of the carboxyls)

Maleic acid

Platinum will work with all of these, indiscriminantly

32

Enzymes work as catalysts for two reasons:

1. They bind the substrates putting them in close proximity.

2. They participate in the reaction, weakening the covalent bonds of a substrate by its interaction with their amino acid residue side

groups (e.g., by stretching).

+

33Dihydrofolate reductase, the movie: FH2 + NADPH2 FH4 + NADP

or: DHF + NADPH + H+ THF + NADP+

Enzyme-substrate interaction is oftendynamic.

The enzyme protein changes its 3-D structureupon binding thesubstrate.

http://chem-faculty.ucsd.edu/kraut/dhfr.mpg

34

Chemical kinetics

Substrate Product (reactants in enzyme catalyzed reactions are called substrates)S PVelocity = V = ΔP/ Δ tSo V also = -ΔS/ Δt (disappearance)From the laws of mass action:ΔP/ Δt = - ΔS/ Δt = k1[S] – k2[P]

For the INITIAL reaction, [P] is small and can be neglected:ΔP/ Δt = - ΔS/ Δt = k1[S]

So the INITIAL velocity Vo = k1[S]

back reaction

O signifies INITIAL velocity

35

P vs. tSlope = Vo

Vo = ΔP/ Δ t

36

P

t

[S1]

[S2]

[S3]

[S4]

Effect of different initial substrate concentrations on P vs. t

0.0

0.2

0.4

0.6

37

P

Vo = the slope in each case

t

[S1]

[S2]

[S3]

[S4]Effect of different initial substrate concentrations

0.0

0.2

0.4

0.6

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

Slope = k1Vo = k1[S]

Dependence of Vo on substrate concentraion

38

We can ignore the rate of the non-catalyzed reaction (exaggerated here to make it visible)

Now, with an enzyme:

39

Vo proportional to [S]

Vo independent of [S]

Enzyme kinetics (as opposed to simple chemical kinetics)

Can we understand this curve?

40

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

41

Michaelis-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

42

S ystem is a t equ ilib riumC onstant leve lN o net flow

S ystem is a t “steady state”C onstant leve lP lenty o f flow

Steady state is not the same as equilibrium

43

E + S

ES

E + P

System is at equilibriumConstant levelNo net flow

System is at “steady state”Constant levelPlenty of flow

44

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 5-1 at your leisure for the derivation (algebra, not complicated, neat)

=

45

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

46

At high S, Vo here = k3Eo, = Vmax

So the Michaelis-Menten equation can be written:

Vmax [S]Vo =

Km + [S]

k3 [Eo] [S]Vo =

Km + [S]Simplest form

=

47Understanding Vmax:( the maximum intital velocity achievable with a given amount of enzyme )

Now, 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;

• the maximum number of molecules of substrate converted to product per molecule of enzyme per second

• Turnover number (k3) then is: a measure of  the enzyme's catalytic power.

48

Some turnover numbers (per second)

• 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.

49

Km ?

Consider the 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 in the M.M. equation and solve for S.

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

50

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:

==

51

ESk2

k1

E + S Kd = k2/k1

Km = (k2+k3)/k1 (by definition)

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

But k2/k1 = Kd (from last graphic)so Km ~ Kd for the dissociation reaction (i.e. the equilibrium constant)

(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

52

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

k3, how good an enzyme is in facitiating the chemical change (given that the substrate is bound)

Km, how well the enzyme can bind the substrate in the first place

53

Got this far

Exam one material ends at this point.

54

A competitive inhibitor resembles the substrate

Enzyme inhibition: competitive, non-competitive, and allosteric

Competitive:

55

A competitive inhibitor can be swamped out at high substrate concentrations

Handout 5-3b

56

-

Apparent (measured) Km increases

Inhibitor looks like the substrate And, like the substrate, binds to the substrate binding site

Substrate concentration

Vo

+

57Biosynthetic pathway to cholesterol

58

Zocor(simvastatin

)

59

½ Vmax w/o inhibitor

½ Vmax withyet more inhibitor

Km remains unchanged. Vmax decreases.

60

Substrate Non-competitive inhibitor

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

61

Non-competitive inhibitor exampleSubstrate still binds OK

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

--CH2-SH

Hg++

--CH2-SH

62

63

= 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.The effects on the Vo vs. S curve are more complex and ignored here

Inhibitor binding site

Allosteric inhibition

64

End product

End product

End product

Feedback inhibition of enzyme activity, or “End product inhibition”

First committed step is usually inhibited

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

P Q R S T U V

65              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

66

60 minutes, in a minimal medium

20 minutes !, in a rich medium

Rich medium = provide glucose + all 20 amino acids and all vitamins, etc.

67

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

68

} Energy difference

determines the direction of a chemical reagion

Free

69

For 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.

70ΔG = 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 here are not the equilibrium concentrations)

• R = UNIVERSAL GAS CONSTANT = 1.98 CAL / DEG K MOLE (R =~2)

• T = ABSOLUTE TEMP ( oK ) 0oC = 273oK; Room temp = 25o C = 298o K (T =~ 300)

• ln = NATURAL LOG

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

71Also abbreviated form:

Δ G = Δ Go+ RTlnQ (Q for “quotient”)

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

Qualitative term Quantitative term

Josiah Willard Gibbs(1839 - 1903)

72

Δ 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 1….. a special case(when all components are at 1)“1” usually means 1 M

73

So 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.

74

Because,

• 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

75E.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

76

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

Note:

77Some exceptions to the 1M standard condition:

• 1) Water: 55 M (pure water) is considered the “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:

78

Exception #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.

79Summary

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.