Enz Kinetics 2425 - Enzyme Kinetics PDF

4BICH001W Biochemistry Introduction to Enzyme Kinetics and Calculations 1. Chemical Kinetics: the law of mass action The rate of a chemical reaction is often directly proportional to the concentrations of the reactants. A → B reaction rate, v0 α [A] ...

4BICH001W Biochemistry Introduction to Enzyme Kinetics and Calculations 1. Chemical Kinetics: the law of mass action The rate of a chemical reaction is often directly proportional to the concentrations of the reactants. A → B reaction rate, v0 α [A]  Effect of [A] on initial rate of production of B for a first order reaction v0 (nmol B min-1) 250 200 150 P (nmol/min) 100 50 0 0 20 40 60 80 100 120 [A] (mM) There is often a simple linear relationship between v0 and [A]... 2. Enzyme Kinetics a) Why are enzymes important? Essentially all life processes are controlled by proteins. A typical cell will have thousands of different enzymes, carrying out a myriad of different reactions. Of the ~ 20,000 protein genes in the human genome, a substantial proportion correspond to enzymes… b) Enzymes as biological catalysts Highly catalytic Enzymes greatly enhance the rate of chemical reactions relative to the un-catalysed (no enzyme). Typically enzymes speed up reactions by a factor or 103 – 106. Enzymes are very efficient catalysts. Rate enhancement by some enzymes: Enzyme k uncatalysed rate k catalysed rate Rate (s-1) (s-1) enhancement OMP Decarboxylase 2.8 x 10-16 39 1.4 x 1017 AMP nucleosidase 1.0 x 10-11 60 6.0 x 1012 Triose phosphate isomerase 4.3 x 10-6 4,300 1.0 x 109 Chlorimate mutase 2.6 x 10-5 50 1.9 x 106 Carbonic anhydrase 1.3 x 10-1 1 x 106 7.7 x 106  Fig. 8-17 1 Substrates enter active site; enzyme changes shape such that its active site 2 Substrates held in enfolds the substrates (induced fit). active site by weak interactions, such as hydrogen bonds and ionic bonds. Substrates Enzyme-substrate complex 3 Active site can lower EA and speed up a reaction. 6 Active site is available for two new substrate molecules. Enzyme 5 Products are 4 Substrates are released. converted to products. Products  S + E ⇋ E:S → E + P Note the relative size of enzyme (grey) and substrate (blue) – this is typical for most enzyme:substrate complexes.  Mechanism of action? Consider a reaction that converts A → B: As A is converted to B, it goes through an intermediate state, called the transition state, X‡, that has a higher free energy (G) than either A or B. The difference in free energy between A and X‡ is known as the Activation Energy DG‡. *We will talk about DG later in the module. Note that in biochemistry we tend to use DG and DG‡ instead of Ea (or DH) as in chemical kinetics. Enzymes effectively reduce the DG‡ for a reaction – because the enzyme allows the reaction to proceed via a different reaction pathway to the un-catalysed reaction pathway. http://en.wikipedia.org/wiki/File:Enzyme_catalysis_delta_delta_G.png c) Enzyme kinetics Simple, one-substrate enzyme-catalysed reaction: S→P What will a plot of v0 versus [S] look like? Enzyme-catalysed reactions shows saturation at high [S]. At low [S] v0 is (approximately) proportional to [S] At high [S] v0 is independent of [S] Michaelis-Menten Kinetics Michaelis & Menten (1913) suggested a simple mathematical model to explain the substrate saturation observed in enzyme-catalysed reactions. k1 k2 Leonor Michaelis Maud Menten S + E ⇋ E:S → E + P k-1 S = Substrate E = Enzyme E:S = Enzyme:Substrate complex P = Product Effect of substrate concentration on rate  A simple model of enzyme action E + S [ES] E + P enzyme-substrate complex  Rate constants for formation and decomposition of ES k1 k2 E + S [ES] E + P k-1 A steady state exists when [ES] is constant  Rate constants for formation and decomposition of ES k1 k2 E + S [ES] E + P k-1 A steady state exists when [ES] is constant We assume that there is a velocity, Vmax at which ALL of the enzyme active sites are occupied by substrate. At other substrate concentrations, the velocity (V0) is assumed to be proportional to the percentage of active sites occupied. e.g. If 40% of the active sites are occupied then V0 = 0.4 Vmax. Low [S] ( Km)  v0 = initial velocity of reaction Vmax [S] v0 = [S] = substrate concentration [S] + Km Vmax = maximum velocity of the reaction Km = Michaelis constant - [S] required to give an initial velocity equal to 1/2 Vmax Michaelis-Menten equation VMax S VMaxS V0  V0  Km  S Km  S V0 is initial velocity [S] is substrate concentration VMax is maximum velocity Km is the Michaelis constant But what do the values of Km and Vmax “mean” (i.e. why are they useful to know)? Km tells us the dependency of v0 on [S] (usually 10-1 - 10-6 M). Vmax tells us the maximum rate of reaction.  Vmax The maximum velocity attainable by a known quantity of the enzyme Vmax is measured in units of velocity, e.g. mg/min mmol/sec  The Michaelis Constant, Km  [S]  VMaxS V0 Vmax  V0   K d  [S] Km  S The simplest interpretation of Km is that it is the dissociation constant of the enzyme-substrate complex. Unfortunately, it isn’t quite that simple.  Rate constants for formation and decomposition of ES k1 k2 E + S [ES] E + P k-1 A steady state exists when [ES] is constant  rate of decomposit ion of ES Km  rate of formationof ES K 1  K 2 K 1 Km  Kd  K1 K1 Km only equals Kd if K-1 >> K2 Not necessarily true, but Km is generally taken as indicating the affinity of the enzyme for its Km the Michaelis Constant A more rigorous definition is “the concentration of substrate required to give V0 that is half of VMax”. or [S] required Km is to half-saturate measured in units of the substrate concentration, enzyme. e.g. mmol/L or mg/mL  Vmax S  V0  K m  S  If we insert 0.5Vmax for V0. Vmax S  0.5Vmax  K m  S   0. 5  S K m  S  0.5K m  0.5S  S  K m  S  2S  K m 2S   S  [ S ] The LOWER the Km the FASTER the reaction, unless [S] is high enough for V0 to approach Vmax. V0 Low Km High Km [S]  What is the effect of [E] on Vmax & Velocity Km? High [E] Vmax Vmax Vmax Low [E] Km [S] Turnover Number (kcat) is a measure of Maximum catalytic activity (molar or molecular activity) and is the number of reactions per second catalysed by each molecule of enzyme It is the same as k2 under initial velocity conditions and can be obtained from Vmax/([E]+[ES]) Units are sec-1 Values range from 40 million sec-1 for catalase to 0.5 sec-1 for lysozyme and >1.5 sec-1 for RuBisCO. Km & Vmax vary independently Low Km, High VMax High Km, High VMax Low Km, Low VMax High Km, Low VMax Enzyme Reaction Progress Curve Velocity mg/min Time After the initial period, a number of different factors may reduce the reaction rate: 1.Enough substrate may be used up for [S] to be substantially decreased. 2.As product accumulates, the reverse reaction may become important, or in some enzymes product inhibition may occur. 3.The enzyme may break down under the reaction conditions. Enzyme Reaction Progress Curve Velocity mg/min Time Enzyme Reaction Progress Curve Velocity Slope of the line mg/min is V0 initial velocity i.e. velocity at time zero when [S] is still that at T0 (approximately). Time  Generally we refer to an “activity” of an enzyme rather than a ‘concentration’. This is the rate of conversion of substrate into product; S P ‘activity’ = velocity v0 (or rate) of substrate consumption or product formation Units of activity International Unit (IU) (μmol min-1) Most common katal (mol sec-1) A true SI unit, but rare because it is ridiculously big – One katal of invertase would hydrolyse 3/4 of a one pound bag of sugar in a second. How do we determine Km and Vmax?  lmax = 405 nm https://www.researchgate.net/figure/ Absorption-spectral-change-from-PNPP-100- mM-to-PNP-catalyzed-by-ALP-0-10-and-100- U-L_fig2_319866514 Can you see any problems with using this M-M plot to accurately determine values for Vmax and Km? And the solution? The Lineweaver - Burk equation (a simple transformation on the M-M equation)! y= m. x + c This is the equation of a straight line! So we can plot 1 / v0 (y-axis) versus 1 / [S] (x-axis)! Lineweaver- Burk (L-B) plot Lineweaver- Burk (L-B) plot 1/Vmax  1 Km 1 1   V0 VMax [S] VMax For 1/V0 = 0 (i.e. the x-axis intercept) Km 1 1 0  VMax [S] VMax 1 Km 1 -  VMax VMax [S] 1 1 Therefore Km can -  [S] Km be determined from the x-axis Lineweaver- Burk (L-B) plot 1/Vmax - 1/Km An Example… [S] (mM) v0 (Product) (mM min-1) 0.00 0.00 3.00 52.0 5.00 72.5 10.0 112.5 30.0 169.0 90 202.5  Michaelis - Menten plot of 250 v0 versus [S] 200 v0 (Product) 150 (mM min-1) 100 50 0 0 10 20 30 40 50 60 70 80 90 100 [S] (mM) Transform the data into a form suitable for a L-B plot… [S] (mM) v0 (Product) (mM min-1) 0.00 0.00 3.00 52.0 5.00 72.5 10.0 112.5 30.0 169.0 90.0 202.5 Transform the data into a form suitable for a L-B plot… [S] (mM) v0 (Product) (mM min-1) 1/[S] mM-1 1/v0 (mM-1 min) 0.00 0.00 3.00 52.0 0.333 0.0192 5.00 72.5 0.200 0.0138 10.0 112.5 0.100 0.00889 30.0 169.0 0.0333 0.00592 90.0 202.5 0.0111 0.00494  Lineweaver - Burk plot of 1/v0 versus 1/[S] 0.025 1 / v0 (mM-1 min) 0.02 f(x) = 0.0447496314978725 x + 0.00448731992466823 R² = 0.998739733790827 0.015 0.01 - 1/Km = - 0.101 mM-1 0.005 1/Vmax = 0.0045 mM-1 min 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 1/[S] (mM-1) 1/Vmax = 0.0045 mM-1 min Vmax = 1 / 0.0045 mM-1 min = 222 mM min-1 (3 sf) - 1/Km = - 0.101 mM-1 1/Km = 0.101 mM-1 Km = 1/0.101 mM-1 Km = 9.90 mM (3 sf) quiz for Lectures 7, 8 and 10 (Dr Paul Cu eins and enzyme kinetics Answers will be given in your Seminar sessions – with further discussion. You must attempt before your seminar session. These quizzes are part of your learning for the Biochemistry module They will aid your on-going studies at the University of Westminster 1) Which of the following statements about proteins is correct? a)Proteins are made from a single polypeptide chain b)Proteins have up to 5 levels of structural organization c)Denaturing a protein disrupts the native 3-D structure by breaking all the bonds in the protein d)Location of di-sulphide bonds is part of the primary structure e)Amino acid side chains are only relevant for the proteins secondary structure.  2) The 3-D structure of proteins is essential for their function. Bonds and interactions which determine protein 3-D structure include? a)Hydrogen bonds, di-sulphide bonds, peptide bonds, hydrophobic interactions b)Peptide bonds, hydroxide bonds, electrostatic interactions, di- sulphide bonds c)Electrostatic interactions, beta-sheet bonds, peptide bonds, phosphodiester bonds d)Di-sulphide bonds, glycosidic bonds, peptide bonds, phosphate bonds e)Peptide bonds, hydrophobic interactions, alpha-helix bonds, di- sulphide bonds 3) Proteins which have a specific binding site for a molecule (small ligand or other protein) are always enzymes. a)This statement is true. b)This statement is false. c) All proteins are enzymes. 4) Which of the following statements is incorrect? a)The central dogma of modern biology is DNA to RNA to protein. b)Proteins can be structural, enzymes, hormones, channels, receptors or transporter molecules c)Proteins often have non-polypeptide components. d)Proteins with similar functions must have similar primary sequences. e)Regulation of protein activity can be by adding or removing phosphate groups. 5) Enzymes are vital for life processes. Which of the following statements is incorrect? a) Specific enzymes increase the rate of specific chemical reactions. b) The specificity of the substrate binding pocket comes from the amino acids. These can be far apart in the primary structure. c) The rate of reaction will always increase with increasing substrate concentration. d) Plotting the inverse of substrate concentration against the inverse of reaction velocity will give a straight line plot. e) The Michaelis-Menten constant for an enzyme relates substrate concentration to rate of reaction.