1 lec. 6 intro bio 2011 protein structure cont. prosthetics groups membrane proteins protein...

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Lec. 6 Intro Bio 2011 Protein structure cont. Prosthetics groups Membrane proteins Protein purification methods Ultracentrifugation Native gel electrophoresis SDS gel electrophoresis Gel filtration Chemical kinetics Enzyme kinetics Michaelis-Menton equation turnover number, Km

Last updated: 9/22/11 12:44 PM

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Riboflavin~ vitamin B2 Heme

Tetrahydrofolic acid~ vitamin B9

Most prosthetic groups are bound tightly via ONLY weak bonds.

Hemoglobin

α2β2 tetramer

heme

3

Metal ions can be prosthetic groups

Zinc (Zn++)

A “zinc finger” protein domain

with two histidines and two cysteines shown

(Voet and Voet, Biochemistry)

4

Membrane proteins

Hydrophobic side chains on the protein exterior for the portion in contact with the interior of the phospholipid bilayer.

Anions are negatively charged.

Cations are positively charged

5

Small molecules bind with great specificity to pockets on protein surfaces

Too far

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A ligand nestled in a tight-binding pocket

7

Estrogen receptor binding estrogen, a steroid hormone

estrogen estrogen

detail

Estrogen receptor is specific, does not bind testosterone

8

Protein binding can be very specific

Testosterone Estrogen

Estrogen (green) binding to amino acid side chains in the estrogen receptor

9

Protein separation methods

Ultracentrifugation

Mixture of proteins

10Ultracentrifuge

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centrifugal force = mω2r

m = massω = angular velocity r = distance from the center of rotation

Causing sedimentation:

Opposing sedimentation = friction = foV.

Constant velocity is soon reached; then, no net forceSo now: centrifugal force = frictional force (balanced each other out)

And so: mω2r =  foV   V = mω2r/fo,

Or:  V = [ω2r] x [m / fo]

fo = frictional coefficient (shape)V = velocity

V proportional to mass (MW). So higher MW moves faster

V inversely proportional to fo;V inversely proportional to “non-sphericity”

So most spherical shape moves fastest

1212

Intermediate:Large, +++high positive charge

Loser:Large, +low positive charge

Winner: Small, +++High positive charge

Intermediate:Small, +Low positive charge

Molecules shown after several hours of electrophoresis

Glass plates

Sample loaded here

+ + +

+++ +++ ++++

++

+++++++++

poly-acrylamidefibers

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Upper resevoir

Cut out for contactof buffer with gel

Upper resevoir

Cut out for contactof buffer with gel

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Reservoir for bufferElectrode connection

(~ 150 V)

Clamped glass sandwich

Cut out of glass platefor contact of buffer with gel

Platinum wire electrode

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Tracking dyes

Happy post-doc

Power supply

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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

Polypeptides denatured, act as random coilsAll have same charge per unit lengthAll 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 treat with mercaptoethanol for that, boil for a bit for good measure)

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Reduction of protein disulfide bonds :

Protein-CH2-S-S-CH2-Protein  + 2 HO-CH2CH2-SH) (mercaptoethanol)[ a reducing agent ]

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

+ HS-CH2-Protein  

Protein gets reduced; mercaptoethanol gets oxidized

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Molecular weight markers

(proteins of known molecular weight)

P.A.G.E.

e.g., “p53”

12 18

48 80 110 130 160

140 Kd

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Sephadex bead

Molecular sieve chromatography(= gel filtration, Sephadex chromatography)

Mild conditions (pH, temperature), proteins in their native state

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Sephadex bead

Molecular sieve chromatography

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Sephadex bead

Molecular sieve chromatography

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Sephadex bead

Molecular sieve chromatography

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Sephadex bead

Molecular sieve chromatography

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PlainFancy4oC (cold room)

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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

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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:

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Enzymes = protein catalysts

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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

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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

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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

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H2 + I2 2 HI

2 HIH2 + I2

2 HIH2 + I2

Different ways of writing chemical reactions

H2 + I2 2 HI

2 HIH2 + I2

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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

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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

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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

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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

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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 time

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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

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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 time

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

Got this far

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The catalyzed reaction

The velocity problem is solved by catalysts

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

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Ch

ang

e in

En

erg

y

H2 + I2

2 HI

Activation energywithoutcatalyst

HHII(TS)

TS complexwith catalyst

Activation energyWITH thecatalyst

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Reactants in an enzyme-catalyzed reaction = “substrates”

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Reactants (substrates)

Not a substrate

Active site or

substrate binding site(not exactly synonymous,

could be just part of the active site)

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Substrate Binding

Unlike inorganic catalysts, enzymes are specific

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Small molecules bind with great specificity to pockets on ENZYME surfaces

Too far

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Unlike 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

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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 bonds).

+

49Dihydrofolate 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

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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:Initial ΔP/ Δt = - ΔS/ Δt = k1[S]

So the INITIAL velocity Vo = k1[S]

back reaction

O signifies INITIAL velocity

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P vs. tSlope = Vo

Vo = ΔP/ Δ t

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P

t

[S1]

[S2]

[S3]

[S4]

Effect of different initial substrate concentrations on P vs. t

0.0

0.2

0.4

0.6

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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

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

Now, with an enzyme:

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Vo proportional to [S]

Vo independent of [S]

Enzyme kinetics (as opposed to simple chemical kinetics)

Can we understand this curve?

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

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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

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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

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E + S

ES

E + P

System is at equilibriumConstant levelNo net flow

System is at “steady state”Constant levelPlenty of flow

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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)

=

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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

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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

=

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Understanding 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.

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Some turnover numbers (per second)

• Succinic dehydrogenase: 19 per second (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.

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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

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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:

==

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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

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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

.

.

.

And any enzyme can be characterized by its turnover number and its Km.