Learning Outcomes. k 1

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1 Module 1DHS - Data Handling Skills Unit: Applied Maths Lecturer: Dr. Simon Hubbard (H13), Simon.Hubbard@umist.ac.uk Title: Equilibria & Michaelis-Menten This lecture and problem class will introduce you to the basic maths that lays behind simple uilibria, which we will extend to consider the basic uation of all enzyme kinetics, the Michaelis-Menten uation. You will meet this uation again and again in Biochemistry, and you should be familiar with it, and what the constants in the uation represent. Learning Outcomes After this lecture you should: Be able to calculate uilibrium constants from uilibrium uations. Understand the terms and be able to use the Michaelis-Menten uation Be able to use graphical analysis to examine the kinetic properties of an enzyme-controlled reaction using the Michaelis-Menten uation In the last class, we examined some simple first order reactions and their kinetic properties. Although, these types of essentially irreversible process occur often in biochemistry, many processes do not fit this model. This is because they are not one-way processes. To describe them we need to consider uilibria. The simplest possible uilibrium is shown below: [ A] [ B ] In this case, the reaction is reversible and the product B can revert to the substrate A. The processes are dynamic, with [A] [B] and [B] [A] occurring simultaneously. To describe this process we need two rate constants, since as products accumulate the reverse reaction becomes important. k1 [ A] [ B] k 1 The forward reaction has a rate constant k1 and the reverse reaction k-1. Such a reaction proceeds until it reaches an uilibrium state at which the forward reaction is going as fast as the backward reaction. At this point, the following is true: or V = k [ A] + k [ B] = [ B] [ A] k1 = = K k 1 learn this one! Where K is an uilibrium constant. If both the forward and reverse reactions are first-order then K is ual to both the ratio of products to reactants and to the ratio of rate constants (forward to backward). Hence, if we know the uilibrium constant and one of the concentrations (at uilibrium) we can calculate the other one.

2 Similarly, K tells us about the relative amounts of products and reactants at uilibrium. if K > 1.0, then [B] > [A] if K = 1.0, then [B] = [A] if K < 1.0, then [B] < [A] Many reactions are more complex. The next simplest uilibrium we will consider is shown below. k [ A] [ B] + [ C] k 1 1 Here, there are 2 products produced from 1 reactant. In this case, the rate constant is expressed as: K B C = [ ] [ ] [ A] Let s consider the units of these uations, as they will inform us of the units of the uilibrium rate constants. The [ ] symbol refers to a concentration. This is usually expressed as moles per litre (= mol l -1 = mol dm -3 = M). Moles = mass in g / molecular weight. In the simplest uilibrium: B M K = [ ] [ A] = M no units and for the second case B C MM K = [ ][ ] A =. [ ] M units = M Many enzyme-catalysed reactions follow a slightly different path. The substrate is bound into the active site of the enzyme (reversibly), it passes through some transition state, and then products are released (irreversibly). This can be described in the uation below. k1 E + S ES E + P k 1 This is the minimal uation needed to described an enzyme catalysed reaction properly, which considers a one-substrate, one-product system. You will notice some notational differences. The rate constant describing the product release step is called kcat. The uilibrium constant for the substrate binding is usually referred to as KS. It is the dissociation constant of the enzyme-substrate complex. This constant is also sometimes uivalent to the Michaelis constant, KM. k cat

3 Using similar mathematical approaches as described in the class on rates, it is possible to derive an uation which quantifies the rate of such a reaction. One form of this uation which you will meet a lot in Biochemistry is the famous Michaelis-Menten uation, shown below. [ E] S k v = 0[ ] K + [ S] M (Additional info to see how this uation is derived is in Matthews & Van Holde). cat learn this one! This contains the Michaelis constant, KM and kcat (sometimes referred to as the turnover number) which are the most quoted values that describe the catalytic properties of an enzyme reaction. You will see them, and the ratio of the two, kcat/km repeatedly in research papers and textbooks. We can investigate the mathematical nature of the relationship between the velocity of the reaction v, and the substrate concentration [S]. Such a plot is shown right. The velocity of the reaction increases quickly as the substrate concentration is increased from 0 and slows down as [S] becomes large. We will examine the properties of this plot and some particularly important constants. VNM/s [S] in M

4 1DHS - PROBLEM SHEET. Equilibria & Michaelis-Menten WRITE YOUR NAME BELOW: Note: it is not a group exercise this week. You should EACH hand in one answer sheet. For the following simple uilibria, calculate the concentration of reactant or product using the other information you are given. Remember to state your units. 1. [A] [B] [A] = 10 mm, Kc = 0.03 What is [B] at uilibrium? 2. [A] [B] + [C] [B] = 3.2 mm, [C] = 3.2 mm, Kc = 0.25 M What is [A] at uilibrium? 3. [A] + [B] [C] [A] = 0.08 M, [B] = 0.04 M, Kc= 2 mm -1 What is [C] at uilibrium?

5 4. Calculate the uilibrium constant for the following reaction 2[A] + [B] 2[C] [A] = 55 µm, [B] = 170 nm, [C] = 24 nm 5. The following reaction is part of the glycolytic pathway. What is the uilibrium constant for the reaction, given the concentrations of the reactants and products below: 2 2 [lactate] [ATP] 3- glucose + 2 PO ADP 2 Lactate + 2 ATP K c = 2 [glucose][phosphate] [ ADP ] 2 At a temperature of 310K, the following concentrations are measured: [glucose] = 25 mm [phosphate] = 3 mm [ATP] = 90 mm [lactate] = 100 µm [ADP] = 10 mm 5. Use the Michaelis-Menten uation to examine the effect on the reaction rate v, when the substrate concentration [S] is increased from 0. Plot your values in the axes drawn over the page. The axes have been deliberately left without annotation, so you can choose appropriate scales and use the units you consider to be most appropriate. [ E] S k v = 0[ ] K + [ S] M Use the following data: K M = 9.34 mm, k cat = s -1 and [E] 0 = 5 mm. You will probably want to calculate and plot values of [S] from 0 to 40 mm to model the enzymatic behaviour. cat

6 Now answer the following questions: a) Describe briefly what happens to v as [S] increases? b) What is the value of v when [S] = Km (estimate this from your graph)? c) Estimate a value for Vmax from your graph, where Vmax is the maximum velocity of the reaction (the highest value of v). d) Actually, when [S] = Km, v = Vmax / 2. Using this information and the Michaelis-Menten uation, derive an expression for Vmax and calculate a more precise value for it than in c).