10/21/2010Biochem: Enzymes II Enzymes II Andy Howard Introductory Biochemistry 21 October 2010

10/21/2010Biochem: Enzymes II

Enzymes II

Andy HowardIntroductory Biochemistry

21 October 2010

10/21/2010 Biochem: Enzymes II P. 2 of 55

What we’ll discuss

Enzymes Enzyme kinetics

Michaelis-Menten kinetics: overview

Kinetic Constants

Kinetic Mechanisms

Induced Fit

Enzymes (concluded) Bisubstrate reactions

Measurements and calculational tools

Inhibition Why study it? The concept Types of inhibitors


Kinetics, continued

In most situations more product will be produced per unit time if A0 is large than if it is small, and in fact the rate will be linear with the concentration at any given time:

d[B]/dt = v = k[A] where v is the velocity of the reaction and k is a constant known as the forward rate constant.

Here, since [A] has dimensions of concentration and d[B]/dt has dimensions of concentration / time, the dimensions of k will be those of inverse time, e.g. sec-1.

[B]

t


More complex cases More complicated than this if >1 reactant involved or if a catalyst whose concentration influences the production of species B is present.

If >1 reactant required for making B, then usually the reaction will be linear in the concentration of the scarcest reactant and nearly independent of the concentration of the more plentiful reactants.

In fact, many enzymes operate by converting a second-order reaction into a pair of first-order reactions!


Bimolecular reaction

If in the reactionA + D Bthe initial concentrations of [A] and [D] are comparable, then the reaction rate will be linear in both [A] and [D]:

d[B]/dt = v = k[A][D] = k[A]1[D]1

i.e. the reaction is first-order in both A and D, and it’s second-order overall


Forward and backward Rate of reverse reaction may not be the same as the rate at which the forward reaction occurs.

If the forward reaction rate of reaction 1 is designated as k1,the backward rate typically designated as k-1.


Multi-step reactions In complex reactions, we may need to keep track of rates in the forward and reverse directions of multiple reactions.

Thus in the conversion A B Cwe can write rate constantsk1, k-1, k2, and k-2

as the rate constants associated with converting A to B, converting B to A, converting B to C, and converting C to B.


Michaelis-Menten kinetics

A very common situation is one in which for some portion of the time in which a reaction is being monitored, the concentration of the enzyme-substrate complex is nearly constant. Thus in the general reaction

E + S ES E + P where E is the enzyme, S is the substrate, ES is the enzyme-substrate complex (or "enzyme-intermediate complex"), and P is the product

We find that [ES] is nearly constant for a considerable stretch of time.

[ES]

t


Michaelis-Menten rates

Rate at which new ES molecules are being produced in the first forward reaction is equal to the rate at which ES molecules are being converted to (E and P) and (E and S).

Formation of ES is first-order in both S and available [E]

Therefore: rate of formation of ES from left =vf = k1([E]tot - [ES])[S]because the enzyme that is already substrate-bound is unavailable!


Equating the rates We started with the statement that the rate of formation of ES and the rate of destruction of it are equal

Rate of disappearance of ES on right and left isvd = k-1[ES] + k2[ES] = (k-1+ k2)[ES]

This rate of disappearance should be equal to the rate of appearance

Under these conditions vf = vd.


Derivation, continued Thus since vf = vd by assumption, k1([E]tot - [ES])[S] = (k-1+ k2)[ES]

Km (k-1+ k2)/k1 = ([E]tot - [ES])[S] / [ES]

[ES] = [E]tot [S] / (Km + [S]) But the rate-limiting reaction is the formation of product: v0 = k2[ES]

Thus v0 = k2[E]tot [S] / (Km + [S])


Maximum velocity What conditions would produce the maximum velocity?

Answer: very high substrate concentration ([S] >> [E]tot),for which all the enzyme would be bound up with substrate. Thus under those conditions we getVmax = v0 = k2[ES] = k2[E]tot


Using Vmax inM-M kinetics

Thus sinceVmax = k2[E]tot,

v0 = Vmax [S] / (Km+[S]) That’s the famous Michaelis-Mentenequation


Graphical interpretation

Michaelis-Menten kinetics

0

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

Substrate conc, M

Initial velocity v0, Ms-1

Vmax = 0.01 Ms -1

Km = 0.03M

[E] tot = 10-7M

kcat = 105 s-1


Physical meaning of Km As we can see from the plot, the

velocity is half-maximal when [S] = Km

Trivially derivable: if [S] = Km, thenv0 = Vmax[S] / ([S]+[S]) = Vmax /2

We can turn that around and say that the Km is defined as the concentration resulting in half-maximal velocity

Km is a property associated with binding of S to E, not a property of turnover

Michaelis Constant:British Christian hiphop band


kcat

We’ve already discussed what Vmax is; but it will be larger for high [E]tot than otherwise.

A quantity we often want is the maximum velocity independent of how much enzyme we originally dumped in

That would be kcat = Vmax / [E]tot

Oh wait: that’s just the rate of our rate-limiting step, i.e. kcat = k2


Physical meaning of kcat

Describes turnover of substrate to product:Number of product molecules produced per sec per molecule of enzyme

More complex reactions may not have kcat = k2, but we can often approximate them that way anyway

Some enzymes very efficient:kcat > 106 s-1


Specificity constant, kcat/Km kcat/Km measures affinity of enzyme for a specific substrate: we call it the specificity constant or the molecular activity for the enzyme for that particular substrate

Useful in comparing primary substrate to other substrates (e.g. ethanol vs. propanol in alcohol dehydrogenase)


Dimensions Km must have dimensions of concentration (remember it corresponds to the concentration of substrate that produces half-maximal velocity)

Vmax must have dimensions of concentration over time (d[A]/dt)

kcat must have dimensions of inverse time

kcat / Km must have dimensions of inverse time divided by concentration, i.e.inverse time * inverse concentration


Typical units for kinetic parameters Remember the distinction between dimensions and units!

Km typically measured in mM or µM

Vmax typically measured in mMs-1 or µMs-1

kcat typically measured in s-1

kcat / Km typically measured in s-1M-1


Kinetic Mechanisms If a reaction involves >1 reactant or >1 product, there may be variations in kinetics that occur as a result of the order in which substrates are bound or products are released.

Examine eqns. 13.48, 13.49, 13.50, and the unnumbered eqn. on p. 430 in G&G, which depict bisubstrate reactions of various sorts. As you can see, the possibilities enumerated include sequential, random, and ping-pong mechanisms.


Historical thought Biochemists, 1935 - 1970 examined effect on reaction rates of changing [reactants] and [enzymes], and deducing the mechanistic realities from kinetic data.

In recent years other tools have become available for deriving the same information, including static and dynamic structural studies that provide us with slide-shows or even movies of reaction sequences.

But diagrams like these still help!


Sequential, ordered reactions Substrates, products must bind in specific order for reaction to complete A B P Q_____________________________E EA (EAB) (EPQ) EQ E

W.W.Cleland


Sequential, random reactions Substrates can come in in either order, and products can be released in either order A B P Q EA EQ__E (EAB)(EPQ) E EB EP B A Q P


Ping-pong mechanism First substrate enters, is altered, is released, with change in enzyme

Then second substrate reacts with altered enzyme, is altered, is released

Enzyme restored to original state A P B Q

E EA FA F FB FQ E


Induced fit

Conformations of enzymes don't change enormously when they bind substrates, but they do change to some extent. An instance where the changes are fairly substantial is the binding of substrates to kinases.

Daniel Koshland

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

Cartoon from textbookofbacteriology.net


Kinase reactions

unwanted reactionATP + H-O-H ⇒ ADP + Pi

will compete with the desired reaction

ATP + R-O-H ⇒ ADP + R-O-P Kinases minimize the likelihood of this unproductive activity by changing conformation upon binding substrate so that hydrolysis of ATP cannot occur until the binding happens.

Illustrates the importance of the order in which things happen in enzyme function


iClicker quiz, question 1

The Michaelis constant Km has dimensions of (a) concentration per unit time (b) inverse concentration per unit time

(c) concentration (d) inverse concentration (e) none of the above


iClicker quiz question 2 kcat is a measure of

(a) substrate binding (b) turnover (c) inhibition potential (d) none of the above


Hexokinase conformational changes

G&G Fig. 13.28


Measurements and calculations

The standard Michaelis-Menten formulation is v0=f([S]), but it’s not linear in [S]. We seek linearizations of the equation so that we can find Km and kcat, and so that we can understand how various changes affect the reaction.


Lineweaver-Burk Simple linearization of Michaelis-Menten:

v0 = Vmax[S]/(Km+[S]). Take reciprocals:

1/v0 = (Km +[S])/(Vmax[S])= Km /(Vmax[S]) + [S]/(Vmax[S]) 1/v0 = (Km/Vmax)*1/[S] + 1/Vmax

Thus a plot of 1/[S] as the independent variable vs. 1/v0 as the dependent variable will be linear with Y-intercept = 1/Vmax and slope Km/Vmax

Dean Burk

Hans Lineweaver


How to use this Y-intercept is useful directly:computeVmax = 1/(Y-intercept)

We can get Km/Vmax from slope and then use our knowledge of Vmax to get Km; or

X intercept = -1/ Km

… that gets it for us directly!


Demonstration that the X-intercept is at -1/Km X-intercept means Y = 0 In Lineweaver-Burk plot, 0 = (Km/Vmax)*1/[S] + 1/Vmax

For nonzero 1/Vmax we divide through: 0 = Km /[S] + 1, -1 = Km/[S], [S] = -Km.

But the axis is for 1/[S], so the intercept is at 1/[S] = -1/ Km.


Graphical form of L-B

1/[S], M-1

1/v0, s L mol-1

1/Vmax,s L mol-1

-1/Km, L mol-1

Slope=Km/Vmax


Are those values to the left of 1/[S] = 0 physical? No. It doesn’t make sense to talk about negative substrate concentrations or infinite substrate concentrations.

But if we can curve-fit, we can still use these extrapolations to derive the kinetic parameters.


Advantages and disadvantages of L-B plots

Easy conceptual reading of Km and Vmax

(but remember to take the reciprocals!)

Suboptimal error analysis [S] and v0 values have errors Error propagation can lead to significant uncertainty in Km (and Vmax)

Other linearizations available(see homework)

Better ways of getting Km and Vmax

available


Don’t fall into the trap!

When you’re calculating Km and Vmax from Lineweaver-Burk plots, remember that you need the reciprocal of the values at the intercepts

If the X-intercept is -5000 M-1, thenKm = -1/(X-intercept) =(-)(-1/5000 M-1) = 2*10-4M

Remember that the X intercept is negative, but Km is positive!


Sanity checks Sanity check #1:typically 10-7M < Km < 10-2M (table 13.3)

Typically kcat ~ 0.5 to 107 s-1 (table 13.4),so for typical [E]tot =10-7M,Vmax = [E]totkcat = 10-6 Ms-1 to 1 Ms-1

If you get Vmax or Km values outside of these ranges, you’ve probably done something wrong


iClicker quiz: question 3 The hexokinase reaction just described probably operates according to a (a) sequential, random mechanism (b) sequential, ordered mechanism (c) ping-pong mechanism (d) none of the above.


iClicker quiz #4 If we alter the kinetics of a reaction by increasing Km but leaving Vmax alone, how will the L-B plot change?

Answer X-intercept Y-intercept

a Moves toward origin

Unchanged

b Moves away from origin

Unchanged

c Unchanged Moves away from origin

d Unchanged Moves toward origin


iClicker question 5

Enzyme E has a tenfold stronger affinity for substrate A than for substrate B. Which of the following is true? (a) Km(A) = 10 * Km(B)

(b) Km(A) = 0.1 * Km(B)

(c) Vmax(A) = 10 * Vmax(B)

(d) Vmax(A) = 0.1 * Vmax(B) (e) None of the above.


Another physical significance of Km Years of experience have led biochemists to a general conclusion:

For its preferred substrate, the Km value of an enzyme is usually within a factor of 50 of the steady-state concentration of that substrate.

So if we find that Km = 0.2 mM for the primary substrate of an enzyme, then we expect that the steady-state concentration of that substrate is between 4 µM and 10 mM.


Example:hexokinase isozymes

Hexokinase catalyzeshexose + ATP hexose-6-P + ADP

Most isozymes of hexokinase prefer glucose; some also work okay mannose and fructose

Muscle hexokinases have Km ~ 0.1mM so they work efficiently in blood, where [glucose] ~ 4 mM

Liver glucokinase has Km = 10 mM, which is around the liver [glucose] and can respond to fluctuations in liver [glucose]

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

Mutant human type I hexokinasePDB 1DGK, 2.8Å110 kDa monomer


Using kinetics to determine mechanisms

In a reaction involving substrates A and B, we hold [B] constant and vary [A].

Then we move to a different [B] and again vary [A].

Continue through several values of [B] That gives us a family of Lineweaver-Burk plots of 1/v0 vs 1/[A]

How those curves appear on a single plot tells us which kind of mechanism we have.


L-B plots for ordered sequential reactions

http://www-biol.paisley.ac.uk/kinetics/Chapter_4/chapter4_3.html

Plot 1/v0 vs. 1/[A] for various [B] values;flatter slopes correspond to larger [B]

Lines intersect @ a pointin between X intercept and Y intercept


L-B plots for ping-pong reactions Again we plot 1/v vs 1/[A] for various [B]

Parallel lines (same kcat/Km);lower lines correspond to larger [B]

http://www-biol.paisley.ac.uk/kinetics/Chapter_4/chapter4_3_2.html


Using exchange reactions to discern mechanisms Example: sucrose phosphorylase and maltose phosphorylase both cleave disaccharides and add Pi to one product:

Sucrose + Pi glucose-1-P + fructose Maltose + Pi glucose-1-P + glucose Try 32P tracers with G-1-P:G-1-P + 32Pi Pi + G-1-32Pi

… so what happens with these two enzymes?


Sucrose & maltose phosphorylase

Sucrose phosphorylase doescatalyze the exchange;not maltose phosphorylase

This suggests that SucPase usesdouble-displacement reaction;MalPase uses a single-displacement

Sucrose + E E-glucose + fructoseE-glucose + Pi E + glucose-1-P

Maltose + E + Pi Maltose:E:Pi

Maltose:E:Pi glucose-1P + glucose

Sucrose phosphorylaseBifidobacteri

um113 kDa dimer

PDB 1R7A, 1.77Å

EC 2.4.1.7


Why study inhibition?

• Let’s look at how enzymes get inhibited.• At least two reasons to do this:

• We can use inhibition as a probe for understanding the kinetics and properties of enzymes in their uninhibited state;

• Many—perhaps most—drugs are inhibitors of specific enzymes.

• We'll see these two reasons for understanding inhibition as we work our way through this topic.


The concept of inhibition An enzyme is a biological catalyst, i.e. a substance that alters the rate of a reaction without itself becoming permanently altered by its participation in the reaction.

The ability of an enzyme (particularly a proteinaceous enzyme) to catalyze a reaction can be altered by binding small molecules to it


Inhibitors and accelerators Usually these alterations involve a reduction in the enzyme's ability to accelerate the reaction; less commonly, they give rise to an increase in the enzyme's ability to accelerate a reaction.


Why more inhibitors than accelerators? Natural selection: if there were small

molecules that can facilitate the enzyme's propensity to speed up a reaction, nature probably would have found a way to incorporate those facilitators into the enzyme over the billions of years that the enzyme has been available.

Most enzymes are already fairly close to optimal in their properties; we can readily mess them up with effectors, but it's more of a challenge to find ways to make enzymes better at their jobs.


Distinctions we can make

Inhibitors can be reversible or irreversible

If they’re reversible, where do they bind? At the enzyme’s active site At a site distant from the active site.

If they’re reversible, to what do they bind? To the unliganded enzyme E To the enzyme-intermediate complex or the enzyme-substrate complex (ES)

To both (E or ES)


Types of inhibitors Irreversible Inhibitor binds without possibility of release

Usually covalent Each inhibition event effectively removes a molecule of enzyme from availability

Reversible Usually noncovalent (ionic or van der Waals)

Several kinds Classifications somewhat superseded by detailed structure-based knowledge of mechanisms, but not entirely

Documents

10/21/2010Biochem: Enzymes II Enzymes II Andy Howard Introductory Biochemistry 21 October 2010