7.014 Handout, Study notes of Biochemistry

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7.014 Handout

Biochemistry III & IV

G and Keq

For the chemical equilibrium:

reactants products

The equilibrium constant, Keq is defined as: K (^) eq =

[products] [reactants]

at equilibrium.

Where [] symbolizes concentration, for example, [products] indicates concentration of products.

The equation that relates the free energy of the reaction, ∆G, to the standard free energy,

∆Go' and the concentrations of reactants and products is:

∆G = ∆Go^ ' + RT ln

[products] [reactants]

Equation (1)

where: R= the universal gas constant T= the temperature in oK

if T = 25 ˚C RT = 0.

kcal mol

if T = 37 ˚C RT = 0.

kcal mol

Under Standard Conditions where: [products] = 1 M and [reactants] = 1 M (M = moles/liter)

(Equation 1) becomes: ∆G = ∆Go'

therefore, the ∆Go' of the reaction is the free energy change (∆G) under standard conditions.

At equilibrium , ∆G = 0, and equation (1) becomes:

∆Go^ '= −RT ln

[products] [reactants]

= −RT ln K( (^) eq) or: Keq = e

• ∆G^

o (^) ' RT

 

 

Significance of Keq:

Keq > 1 ∆Go'^ has negative value, therefore reaction ⇒ is possible.

Keq < 1 ∆Go'^ has positive value, therefore reaction ⇒ is possible if [react] > [prod].

Keq >> 1 ∆Go'^ has large negative value, therefore reaction ⇒ is irreversible.

Keq << 1 ∆Go'^ has large positive value, therefore reaction ⇒ cannot occur.

Kinetics of enzyme reactions

One way to study an enzyme is to measure the formation of the product. If you were to perform an experiment under defined conditions at a given concentration of substrate and enzyme you could plot a time course of the enzyme catalyzed reaction.

Example:

velocity

Initial reaction Equilibrium concentration of product

time elapsed

one reaction run with a particular [S], [Enz], pH, etc.

Enzymes catalyze both forward and reverse reactions. In the reaction S -->P, the enzyme converts substrate (S) to product (P). Initially, the concentration of P is small and the net reaction is S -->P. As [P] increases, the rate of the reaction S -->P decreases until the rate of the forward reaction equals the rate of the reverse reaction. At this point the reaction is in equilibrium.

To measure the kinetic properties of a given enzyme, you must perform many experiments like the one above, holding the enzyme concentration constant and varying the substrate concentrations. The initial reaction velocity at each substrate concentration is measured, and the data from all the experiments is used to plot the initial reaction velocity, Vo, as a function of substrate concentration [S]. An example is found below.

V

initial

reaction

velocity

[S]

maximum reaction velocity

initial substrate concentration

0 ∞

0

V

max

o

Given the above assumptions and approximations, the dependence of Vo on [S] can be derived mathematically. (For further information, see "Biochemistry" 4th edition by Stryer; pp 192-193)

V (^) o = k 3 [E]tot

[S]

[S] + K M

Where: V (^) o = initial velocity of the reaction [E]tot = total concentration of enzyme in reaction [S] = substrate concentration

KM = the Michaelis constant, K (^) M =

k 2 + k (^3) k 1

Three cases will illustrate how this equation behaves.

CASE 1: [S] very large ([S]>>KM => Enzyme is saturated with substrate)

in this case [S] + KM ≈ [S], so (1) becomes: V 0 = k 3 [E]tot

[S]

[S]

 or:^ V (^) o = k 3 [E]tot

This is the rate at a large substrate concentration - (the maximal rate) which is called Vmax

Vmax =k3[E]tot

Substituting into (1) gives the Michaelis-Menten Equation as it is normally shown:

V 0 =

Vmax [S] KM + [S]

Remember, this equation is derived for Vo, when very little product has formed and the

back-reaction can be ignored.

CASE 2: [S] small ([S]<<KM => linear range)

in this case [S] + KM ≈ KM so (2) becomes: V 0 =

Vmax [S] KM

or: V 0 [S]

So, at low [S], Vo is linearly proportional to [S].

CASE 3: [S] = KM (The definition of KM )

V 0 =

V (^) max[S] [S] + [S] or

Vo = Vmax 2

KM is defined as the [S] that results in half-maximal reaction rate.

SIGNIFICANCE OF KM and Vmax

Vmax and KM are the two parameters which define the kinetic behavior of an enzyme as a

function of [S].

Vmax is a rate of reaction. It will have units of:

moles min

or

moles sec

or

moles min

etc.

Vmax depends on the structure the enzyme itself and the concentration of enzyme present.

KM is a the concentration substrate required to approach the maximum reaction velocity - if [S]>>Km then Vo will be close to Vmax.

KM is a concentration. It will have units of:

moles liter

(M),or

moles liter

( M),etc.

KM depends only on the structure of the enzyme and is independent of enzyme concentration.

Measuring KM and Vmax

The quantities KM and Vmax are experimentally determined and different for each

enzyme. Once you have an assay for enzyme activity, you can determine these parameters. You can estimate KM and Vmax from the graph of initial velocity versus [S].

1. Run a series of reactions each with a single [E]tot, varying [S], and measure V 0.

2. Graph V 0 vs. [S].

3. Estimate Vmax from asymptote.

4. Calculate Vmax/

5. read KM from graph.

0 [S] ∞

V 0 Vmax

Vmax /

KM