Chemical and Enzyme Kinetics

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Chemical and Enzyme Kinetics Chemical and enzyme kinetics D. Gonze & M. Kaufman November 24, 2016 Master en Bioinformatique et Mod´elisation Contents 1Definitions 4 1.1 Reactionrate .................................. 4 1.2 Examples .................................... 6 1.3 Systemsofchemicalreactions . 10 1.4 Chemical equilibrium . 11 1.5 Effectoftemperature-Arrheniusequation . ... 12 2Enzymekinetics 13 2.1 Enzymes..................................... 13 2.2 Mechanismofenzymereactions . 14 2.3 Equilibrium approximation: Michaelis-Menten equation . .16 2.4 Quasi-steady state assumption: Briggs-Haldane equation . ...... 17 2.5 Reversible Michaelis-Menten kinetics . 19 2.6 Inhibition .................................... 20 2.7 Activation.................................... 23 2.8 Two-substrateenzymekinetics. .24 2.9 Substratecompetition ............................. 27 2.10 Cooperativity: Hill function . 29 2.11Allostericmodel................................. 34 2.12 Zero-order ultrasensitivity . .38 3Generegulation 42 3.1 Transcription, regulation, and transcription factors . ...... 42 3.2 Transcriptional activation . 44 3.3 Transcriptional activation with auto-regulation . .... 47 3.4 Transcriptional activation with multiple binding sites . .49 3.5 Transcriptional activation by a dimeric complex . .52 3.6 Transcriptional inhibition with an inducer . 54 3.7 Combining transcriptional activation and inhibition . .56 4Appendix 58 4.1 Reaction in series - derivation . 58 4.2 Quasi-steadystateapproximation . .60 2 4.3 Validity of the quasi-steady state approximation . ... 63 4.4 Comparison of developed vs compact Michaelis-Menten kinetics . .... 64 4.5 Examples of kinetic values . 65 4.6 Competitive inhibition . 66 4.7 A rapid (but approximative) derivation of the Hill function . ... 68 5References 69 5.1 Textbooks.................................... 69 5.2 Papers...................................... 70 3 1Definitions 1.1 Reaction rate Consider the chemical reaction that transforms the substrates A and B into the products CandD: A+B C+D (1) → The variation in time of the concentration of the substrates (A and B)andtheproducts (C and D), dA dB dC dD , , ,and (2) dt dt dt dt is determined by the rate at which the reaction proceeds. 1 Product 0.8 0.6 0.4 Concentration 0.2 Substrate 0 0 1 2 3 4 5 Time Figure 1: Time evolution of the concentration of the substrate and product. For a chemical reaction to occur, the reacting species must collide, have sufficient energy and be well oriented. The number of collisions is proportional to the concentration of the reacting species. For the reaction (1), the rate law is given by the mass action law: v = kAB (3) Not all collisions are reactive. The rate constant k accounts for the probability that the molecules are well oriented and have sufficient energy to react. The variation in time of the concentration of the substrates and the products is given by dA dB dC dD = = kAB and = = kAB (4) dt dt − dt dt The sign in the right-hand side of these equations stands for the fact that, each time the reaction proceeds, one molecule (mole) of A (and B) disappears while one molecule (mole) of C (and D) appears. More generally, for an (elementary) reaction in which m molecules of A react with p molecules of B and in which the products (C and D) do not affect the reaction rate: mA + pB qC + rD (5) → 4 the rate law is: v = kAmBp (6) Note that the sum m + p is called the order of a reaction. Consider now for example the following reaction: 3A + B A+C (7) → According to Eq. (6), the rate of this reaction is: v = kA3B (8) When we write the evolution of the concentration of A, we must take into consideration the fact that each time this reaction occurs, only two molecules of A are transformed (one is conserved). So, the variation of A is given by: dA = 2v = 2kA3B (9) dt − − The coefficient “2” is the balance for the species A in reaction (7) and the sign “-” stands because, globally, A is consumed. Since v must have the unit [concentration]/[time], the units of k depend on the order of the reaction. In the general case, for a reaction in which for each n molecules (moles) of X transformed p molecules (moles) are recovered at the end: n X + ... p X + ... (10) → the evolution equation for the concentration of X is: dX = η v with η = p n (11) dt X X − ηX is called the stoechiometric coefficient of compound X. This coefficient is positive if, globally, the species is produced (p>n) and negative if the species is consumed (n>p). For example, for the following reaction: A+2B 3A + C (12) → the stoechiometric coefficients of the different species are: η =3 1=2,η =0 2= 2, η =1 0=1. (13) A − B − − C − and the evolution equations are: dA dB dC =2kAB2, = 2kAB2, = kAB2 (14) dt dt − dt 5 1.2 Examples 1st-order kinetics Consider the reaction of 1st-order: A A* (ex: conformational change of a molecule) (15) → or A B + C (dissociation of a molecule into two molecules) (16) → By definition (eq 3), the rate of this reaction is v = kA (17) and the time evolution of the concentration of the substrate A is: dA = kA (18) dt − After integration, we find: −kt A(t)=A0e (19) where A0 is the initial concentration of substrate A (A0 = A(0)). We observe an exponential decrease of the concentration of A with time: 3 A 2.5 0 2 1.5 1 Concentration of A 0.5 0 0 1 2 3 4 5 Time Figure 2: 1st-order kinetics: exponential decrease of the concentration of substrate A. Remark: On the long run, A will converge to 0, i.e. A will be fully degraded. The half-life τ1/2 of A is defined by the time it takes for its concentration A to decrease from its initial value A0 to half of this value, A0/2: A0 − = A e kτ1/2 2 0 ln 2 τ = (20) 1/2 k Note that τ1/2 does not depend on the initial value A0. 6 2nd-order kinetics Let’s take now the reaction: 2A B (2 molecules A fuse together to give one single molecule) (21) → or 2A B+C (2 molecules A react together to give two different molecules) (22) → Its rate is: v = kA2 (23) and the time evolution of the substrate A is dA = 2kA2 (24) dt − After integration, we find: A A(t)= 0 (25) 1+2A0kt where A0 is the initial concentration of substrate A. Here, we observe an hyperbolic decrease of the concentration of A with time: 3 A 2.5 0 2 1.5 1 Concentration of A 0.5 0 0 1 2 3 4 5 Time Figure 3: 2nd-order kinetics: hyperbolic decrease of the concentration of substrate A. 7 Reactions in series Let’s consider the following reactions in series: k k A 1 B 2 C(26) → → Evolution equations for the compounds A, B, and C write: dA = k A (27) dt − 1 dB = k A k B (28) dt 1 − 2 dC = k B (29) dt 2 Assuming that we start with A(0) = A0 and B(0) = C(0) = 0, we find −k1t A(t)=A0e (as before) (30) and, after integration k A B(t)= 1 0 e−k1t e−k2t (31) k k − 2 − 1 1 ! " C(t)=A 1 k e−k1t k e−k2t (32) 0 − k k 2 − 1 # 2 − 1 $ The derivation is detailed in the Appendix. ! " 10 9 C 8 7 A 6 k1=1 k2=0.5 5 A =10 0 B =C =0 4 0 0 Concentration 3 B 2 1 0 0 1 2 3 4 5 6 7 8 9 10 Time Figure 4: Kinetics of reactions in series. 8 Now we can consider the case where A is continuously supplied, so that A remains constant over time: k k k k 0 A 1 B 2 C 3 (33) → → → → The evolution equations for the compound B is, with A(t)=A0 = k0/k1: dB = k A k B (34) dt 1 0 − 2 and the evolution of B is then given by: k A (k A k B )e−k2t B(t)= 1 0 − 1 0 − 2 0 (35) k2 Now, on the long run and regardless of its initial concentration, B reaches a steady state: k A B( )= 1 0 (36) ∞ k2 The derivation of the equation for C(t) is left as an exercise. 22 20 18 B 16 14 12 10 A 8 Concentration 6 k1=1 4 k2=0.5 2 A=10 (const) 0 0 1 2 3 4 5 6 7 8 9 10 Time Figure 5: Kinetics of reactions in series with A = A0 =constant. 9 1.3 Systems of chemical reactions We are usually interested by systems of coupled chemical reactions. n X +n X + ... p X +p X ... 11 1 21 2 → 11 1 21 2 n X +n X + ... p X +p X ... 12 1 22 2 → 12 1 22 2 ... n X +n X + ... p X +p X ... (37) 1R 1 2R 2 → 1R 1 2R 2 The variation of a given compound Xi involved in R reactions is defined by: dX R i = η v = η v + η v + ... + η v (38) dt ir r i1 1 i2 2 iR R r=1 % where vr = rate of reaction r (with r =1, 2, ...R): nir vr = kr Xi i & η = p n = stoechiometric coefficient of compound X in reaction r. ir ir − ir i We illustrate this on the following example: r reaction rate ηXr ηYr 1Ak1 X v = k A η =1 η =0 −→ 1 1 X1 Y 1 2B+Xk2 Y+C v = k BX η = 1 η =1 −→ 2 2 X2 − Y 2 32X+Yk3 3X v = k X2Y η =1 η = 1 −→ 3 3 X3 Y 3 − 4Xk4 D v = k X η = 1 η =0 −→ 4 4 X4 − Y 4 The evolution equations for X and Y are given by (see eq.
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