Lecture Handout: Biochemistry III and IV

MIT Department of Biology 7.014 Introductory Biology, Spring 2005 7.014 Handout Biochemistry III & IV G and Keq For the chemical equilibrium: products reactants [products] at equilibrium. [reactants] The equilibrium constant, Keq is defined as: K eq = Where [] symbolizes concentration, for example, [products] indicates concentration of products. The equation that relates the free energy of the reaction, ΔG, to the standard free energy, ΔGo' and the concentrations of reactants and products is: o ΔG = ΔG ' + RT ln where: if T = 25 ˚C ⎛ [products] ⎞ Equation (1) ⎝ [reactants]⎠ R= the universal gas constant T= the temperature in oK RT = 0.59 kcal mol if T = 37 ˚C RT = 0.61 kcal mol Under Standard Conditions where: [products] = 1 M and [reactants] = 1 M (M = moles/liter) (Equation 1) becomes: ΔG = ΔGo' therefore, the ΔGo' of the reaction is the free energy change (ΔG) under standard conditions. At equilibrium, ΔG = 0, and equation (1) becomes: o ΔG ' = −RT ln ( ) ⎛ [products] ⎞ = −RT ln Keq ⎝ [reactants]⎠ or: K eq = e ΔG o ' ⎞ ⎛ ⎝ RT ⎠ Significance of Keq: Keq > 1 ΔGo' has negative value, therefore reaction ⇒ is possible. Keq < 1 ΔGo' has positive value, therefore reaction ⇒ is possible if [react] > [prod]. Keq >> 1 ΔGo' has large negative value, therefore reaction ⇒ is irreversible. Keq << 1 ΔGo' has large positive value, therefore reaction ⇒ cannot occur. Enzymes (This handout is designed to supplement the coverage of enzymes in Purves et al.) A catalyst is a molecule that increases the rate of a reaction but is not the substrate or product of that reaction. Enzymes are usually proteins that catalyze chemical reactions in cells (the exception is catalytic RNA). A substrate is a molecule upon which an enzyme acts to yield a product. The part of the enzyme that binds substrate is called the active site. Consider the reaction: A (substrate) ------> B (product) A graph of the free energy against the reaction progress is shown below: Ea Energy of activation substrate (A) ΔG product (B) Reaction Progress ΔG is negative overall for forward reaction. The free energy of this reaction is not changed by the presence of the enzyme, BUT the enzyme can speed up the reaction by lower the Ea. Enzyme Mechanisms In a reaction like that above, Ea could represent an energetically unfavorable transitional state that the substrate must obtain before forming product. Enzymes can lower the Ea by stabilizing the substrate in that transitional state and thus speed up the reaction. For a reaction involving two substrates: A + B ---> C, the enzyme may increase the local concentration of A and B and drive the reaction forward. 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. Initial reaction velocity Example: Equilibrium concentration of product one reaction run with a particular [S], [Enz], pH, etc. 0.0 0.0 time elapsed 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. Vmax maximum reaction velocity Vo initial reaction velocity 0 0 ∞ [S] initial substrate concentration A few things about the V vs. [S] graph: Velocity is dependent on [S], but remember this graph was generated at one enzyme concentration. If more enzyme was added, the reaction velocity would increase. The velocity asymptotically approaches a maximum as [S] --> ∞ . Why isn't the graph linear? As [S] gets large, the enzyme becomes saturated. Every enzyme molecule is complexed with substrate and since the amount of enzyme used in this series of experiments is fixed, the rate asymptotically approaches a maximum. NOTE: the two preceding graphs look very similar, but represent different things. It is important to understand the difference between an individual reaction: [P] vs. t (from which you get Vo) and a kinetic graph of many such reactions: Vo vs. [S]. Enzymes that display this kinetic behavior can be modeled mathematically as was first shown by Michaelis and Menten in 1913: Michaelis-Menten Enzyme Kinetics For an enzymatic reaction: k1 k3 E+S ES E+P k2 The symbols represent (in our example): E enzyme S substrate ES enzyme-substrate complex P product k1 the rate at which E combines with S to form ES k2 the rate at which ES can dissociate into E and S k3 the rate at which ES can proceed to form E and P Michaelis and Menten modeled the catalytic rate of enzymes under steady-state conditions where the concentration of ES remains unchanged over time while the concentrations of starting materials and products are changing. At steady state the rate of ES formation equals the rate of ES breakdown. The following conditions are assumed in this model. 1) The uncatalyzed rate of S ---> P is negligible (the reaction does not go without enzyme). 2) Because the [S] is far greater than the [E], the amount of substrate bound by enzyme is negligible compare to the total [S]. Also assumed is that [S] does not vary with time (this reaction is under examination for a brief moment), and the reaction rate remains the same. 3) The reverse reaction (P ---> S) does not occur. 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) ⎛ [S] ⎞ ⎟ V o = k 3 [E]tot ⎜ (1) ⎝ [S] + K M ⎠ Where: V o = initial velocity of the reaction [E]tot = total concentration of enzyme in reaction [S] = substrate concentration k + k3 KM = the Michaelis constant, K M = 2 k1 Three cases will illustrate how this equation behaves. CASE 1: [S] very large ([S]>>KM => Enzyme is saturated with substrate) ⎛ [S] ⎞ V o = k 3 [E]tot ⎝ [S] ⎠ or: This is the rate at a large substrate concentration - (the maximal rate) which is called V max V max =k3[E]tot in this case [S] + KM ≈ [S], so (1) becomes: V 0 = k 3 [E]tot Substituting into (1) gives the Michaelis-Menten Equation as it is normally shown: Vmax [S] (2) KM + [S] Remember, this equation is derived for Vo, when very little product has formed and the back-reaction can be ignored. V0 = CASE 2: [S] small ([S]< linear range) in this case [S] + KM ≈ KM so (2) becomes: V0 = 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 ) V0 = V max [S] [S] + [S] or Vo = V max 2 KM is defined as the [S] that results in half-maximal reaction rate. SIGNIFICANCE OF KM and Vmax V max and KM are the two parameters which define the kinetic behavior of an enzyme as a function of [S]. moles moles moles or or etc. min sec min V max depends on the structure the enzyme itself and the concentration of enzyme present. V max is a rate of reaction. It will have units of: KM is a the concentration substrate required to approach the maximum reaction velocity if [S]>>Km then V o will be close to Vmax . moles moles KM is a concentration. It will have units of: (M),or ( M),etc. liter liter 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 V0. 2) Graph V0 vs. [S]. 3) Estimate Vmax from asymptote. 4) Calculate Vmax /2 5) read K M from graph. V0 Vmax Vmax /2 0 0 KM [S] ∞