Enzyme Kinetics PDF

ENZYME KINETICS Module 2 Dr. Ariel V. Melendres  ENZYMES Enzymes are protein that catalyzed biological products Enzymes are proteins that help speed up metabolism, or the chemical reactions in our bodies. They build some substance...

ENZYME KINETICS Module 2 Dr. Ariel V. Melendres  ENZYMES Enzymes are protein that catalyzed biological products Enzymes are proteins that help speed up metabolism, or the chemical reactions in our bodies. They build some substances and break others down. They are remarkable biomolecules with extraordinary specifity and catalytic power Examples of ENZYMES Dr. Ariel V. Melendres A typical Enzyme Enzymes are proteins which act as biological catalyst or in simple words, its a molecules which increases reaction rates within the body or elsewhere. Enzymes are highly specific, one enzyme will catalyze and act on one specific type of substance. Active-site Enzymes that breaks down molecules Lock-Key Complex Chymotrypsin is an enzyme that is used in the small intestine to break down proteins into individual amino acids. It specifically targets the aromatic amino acids, tyrosine, phenylalanine, and tryptophan  Action of enzyme on dipeptide substrate Chymotrypsin is an enzyme that is used in the small intestine to break down proteins into individual amino acids. It specifically targets the aromatic amino acids, tyrosine, phenylalanine, and tryptophan. Chymotrypsin has also seen some use in medicine, particularly in assisting cataract surgery. Simple Enzyme Kinetics with one Two Substrates S → P 𝑑𝑠 𝑑𝑝 𝑣=− = 𝑑𝑡 𝑑𝑡 𝑣: 𝑟𝑎𝑡𝑒 𝑖𝑛 𝑚𝑜𝑙𝑒 𝑝𝑒𝑟 𝑣𝑜𝑙𝑢𝑚𝑒 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑡𝑖𝑚𝑒 s and p: moles per unit volume of substrate and product respectively Let us develop a suitable mathematical expression for enzyme catalyzed reaction  Michaelis – Menten Kinetics 1. The rate of reaction is first order in concentration of substrate 2. As the substrate concentration is continually increased, the reaction order in substrate diminishes continuously from one to zero 3. The rate of reaction is proportional to the total amount of enzyme present Enzymes follow zero order kinetics when substrate concentrations are high. Zero order means there is no increase in the rate of the reaction when more substrate is added. Given the following breakdown of sucrose to glucose and fructose Sucrase Sucrose + H20 Glucose + Fructose H HH O O H O H O H OH H H OH H H O Fructose HO HO H H OH H OH Glucose  Equation on Product form from Substrate catalyzed by Enzymes E S→P Detailed Elementary Reactions 𝑘1 E + S ⇌ ES (1) 𝑘−1 𝑘2 ES → P + E (2) Where: E = Enzyme S = Substrate P = Product ES = Enzyme-Substrate complex k1 rate constant for the forward reaction k-1 = rate constant for the breakdown of the ES to substrate k2 = rate constant for the formation of the products At start, enzyme E and substrate S combine to form complex, ES Then dissociate into product P and free enzyme E When the substrate concentration becomes large enough to force the equilibrium to form completely all ES, the second step in the reaction becomes rate limiting because no more ES can be made and the enzyme-substrate complex is at its maximum value. 𝑘1 E+S⇌ 𝑘 ES − R1 1 𝑘2 R2 ES → P + E d P  v= = k 2 ES (1) dt d ES = k1 E S − k −1 ES − k 2 ES (2) dt In the book Bailey and Ollis, small letters refers to mole concentrations such as (3) e = [E] = enzyme s = [S] = substrate es =[ES] =enzyme-substrate complex (4)  Assumption so 𝑑(𝑒𝑠) =0 𝑑𝑡 quasi-steady state, es is concentration constant s For practical purposes, Km is the p concentration of substrate which permits the enzyme to achieve half Vmax. An enzyme with a high Km has a eo = e + es low affinity for its substrate, and requires a greater concentration of substrate to achieve Vmax." eo es e Time 𝑘1 E + S ⇌ ES (1) eo=initial conc of enzyme= total enzyme 𝑘−1 𝑘2 es=enzyme-substrate complex ES → P + E (2) e=enzyme at any time  𝑠𝑒𝑜 Method 1 𝑒𝑠 = 𝐾𝑚 + 𝑠 The formation of product from enzyme- product complex is so slow, k-1>>k2, 𝑑𝑝 𝑘1 v= = 𝑘2 𝑒𝑠 E + S ⇌ ES 𝑑𝑡 𝑘−1 𝑘2 ES → P + E (so slow) v= 𝑘2 𝑒𝑠 -1 -1 Units: 𝑘1 (concentration time ) 𝑘2 𝑒𝑜 𝑠 𝑘−1 (time-1) v= 𝐾𝑚 +𝑠 𝑘2. (time-1) The velocity is max at total enzyme eo 𝑘−1 𝑒𝑠 = 𝑘1 𝑠𝑒 At v= 𝑣𝑚𝑎𝑥, 𝑒𝑠 = 𝑒𝑜 𝑘−1 𝑠𝑒 Km = = (reactant/product) vmax= 𝑘2 𝑒𝑜 𝑘1 𝑒𝑠 Eq 3.8 Shuler eo = e + es 𝑣𝑚𝑎𝑥 𝑠 (page 62) 𝑣= Eq 3.3 Bailey &Ollis 𝑠(𝑒𝑜−𝑒𝑠) 𝐾𝑚 + 𝑠 (page 100) Km = 𝑒𝑠 Michaelis-Menten Equation  Method 2 Breakdown of the enzyme‐substrate complex to 𝑠𝑒𝑜 products is not slow in comparison with its 𝑒𝑠 = dissociation 𝐾𝑚 + 𝑠 𝑘1 𝑑𝑝 E + S ⇌ ES v= = 𝑘2 𝑒𝑠 𝑘−1 𝑑𝑡 𝑘2 ES → P + E v= 𝑘2 𝑒𝑠 𝑘2 𝑒𝑜 𝑠 𝑑(𝑒𝑠) = 𝑘1 𝑠𝑒 − (𝑘−1 +𝑘2)(𝑒𝑠) v= 𝑑𝑡 𝐾𝑚 +𝑠 Quasi steady state vmax= 𝑘2 𝑒𝑜 𝑑(𝑒𝑠) = 0= 𝑘1 𝑠𝑒 − (𝑘−1 +𝑘2)(𝑒𝑠) 𝑣𝑚𝑎𝑥 𝑠 𝑑𝑡 𝑣= 𝐾𝑚 + 𝑠 eo = e + es 𝑘−1 +𝑘2 𝑠𝑒 Km = = 𝑘1 (𝑒𝑠) 𝑠(𝑒𝑜−𝑒𝑠) Km = Km is a measure of affinity, Brigs and Haldane 𝑒𝑠 the higher the lower is the affinity Method  Plot of v vs s 1.2 Low Substrate Concentrations: vmax 0.8 As a result of this, we can essentially say v(M/min) 𝑣𝑚𝑎𝑥𝑠 that ≈. Thus, at low v=0.5xvmax v= concentrations of , the equation can 𝐾𝑚 +𝑠 0.4 change as follows. Km 0.0 High Substrate Concentrations: 0 2 4 6 8 10 s [mol/L] s ≈ If Km = s 𝑚𝑎𝑥 𝑣 𝑠 v= 𝑠+𝑠 Example: Given: vmax=4(mol/L)/min, Km=6 mol/L 𝑣𝑚𝑎𝑥 a) Find v at s=10 mol/L. 4 (𝑠) v= 2 b) Find s at v=3(mo/L)/min b) 3= 6+𝑠 𝑣𝑚𝑎𝑥 𝑠 𝑣= 𝐾𝑚 + 𝑠 s= 18 𝑚𝑜𝑙/𝐿 4 (10) 𝑚𝑜 a) 𝑣 = 6+10 = 2.5 ( 𝐿 )/𝑚𝑖𝑛  Integration of Michaelis-Menten Equation 𝐺𝑒𝑛𝑒𝑟𝑎𝑙 𝐶𝑎𝑠𝑒 𝐶𝑎𝑠𝑒 1: 𝐾𝑚 ≫ 𝑠 𝑣𝑚𝑎𝑥 𝑠 −𝑑𝑠 𝑣𝑚𝑎𝑥 𝑠 𝑣= 𝐾𝑚 + 𝑠 = 𝑑𝑡 𝐾𝑚 +𝑠 −𝑑𝑠 𝑣𝑚𝑎𝑥 𝑠 = −𝑑𝑠 = 𝑣𝑚𝑎𝑥 𝑠 𝑑𝑡 𝐾𝑚 +𝑠 𝑑𝑡 𝐾𝑚 −𝑑𝑠 𝑑𝑡 = 𝑣 𝐾𝑚 𝑑𝑠 𝑚𝑎𝑥 𝑠 𝑑𝑡 = −( ) 𝐾𝑚 + 𝑠 𝑣𝑚𝑎𝑥 𝑠 𝑑𝑡 = −( 𝐾𝑚 𝑑𝑠 + 𝑑𝑠 ) 𝐾𝑚 𝑠𝑜 𝑣𝑚𝑎𝑥 𝑠 𝑣𝑚𝑎𝑥 𝑡= ln 𝑣𝑚𝑎𝑥 𝑠 𝑡 𝑠 𝐾𝑚 𝑑𝑠 𝑑𝑠 𝐶𝑎𝑠𝑒 2: 𝐾𝑚 >s 𝑡= ln 𝑑𝑡 𝐾𝑚 𝑣𝑚𝑎𝑥 𝑠 (First order) −𝑑𝑠 Case 2 = 𝑣𝑚𝑎𝑥 Km1 𝑣𝑚𝑎𝑥 = ea k(T) 𝑘𝐵 𝑇 ∆𝑆 ∗ /𝑅 )(𝑒 −𝐸/𝑅𝑇 ) k(T)=( ℎ )(𝑒 Low Temp kB is Boltzmann's constant, h is Planck’s constant  is proportionality constant High Temp 𝑒 𝑘𝐵 𝑇 ∆𝑆∗ /𝑅 )(𝑒 −𝐸/𝑅𝑇 )] 𝑜 vmax=1+𝐾 [( ℎ )(𝑒 𝑑 Kd=(𝑒 ∆𝑆𝑑 /𝑅 )(𝑒 −∆𝐻𝑑 /𝑅𝑇 ) 𝛽𝑇𝑒 −𝐸/𝑅𝑇 )] vmax= 1+ (𝑒 ∆𝑆𝑑 /𝑅 )(𝑒 −∆𝐻𝑑 /𝑅𝑇 ) ∗ Where  includes , kB, h, eo and 𝑒 ∆𝑆 /𝑅 as over-all constant  Sample Problem: Given the following data obtained from the test of enzyme activity at different temperatures: Evaluate E and Hd. Experimental Data Calculated k/min T(oC) lnK T (K) 1/T(K) 21 3 3.045 276.15 0.00362122 35 21 3.555 294.15 0.003399626 41 27 3.714 300.15 0.003331667 52 42 3.951 315.15 0.003173092 45 49 3.807 322.15 0.003104144 33 55 3.497 328.15 0.003047387 20 60 2.996 333.15 0.003001651 slope1 -2052.27 R2 0.99740 15 63 2.708 336.15 0.002974862 slope2 9355.72 R2 0.99346 12 65 2.485 338.15 0.002957267 -2052.27=-E/R lnk = lnA-E/R (1/T) (Low Temperature curve) E=17,062.58 Joules/mol where R=8.314 Joules/mol K 9355.72=(Hd-E)/R lnk = lnA+(Hd-E)/R (1/T) (High Temperature curve) Hd=R(9355.72)+17,062.58=94846.03 Joules/mol d (Tmax)= (E+RT max)/(Hd-E-RTmax) =. 17,062.58+8.314x(49+273.15)/(94846.03-17062.58-8.314x(49+273.15)) =0.263 Knowing d (Tmax) and Hd, we can calculate Gd from Eq. 3.68 Gd=3575.55Jouls 𝑘 ∗ /𝑅 Where: =( ℎ𝐵 )(𝑒 ∆𝑆 ) Substrate Inhibition -substrate itself inhibits the enzyme reaction or cell growth Types of Substrate Inhibitions (Ref: Enzymatic Reaction with Inhibition) Competitive 𝑣𝑚𝑎𝑥𝑠 v= 𝑖 Enzymatic reaction E + S ⇌ ES 𝑠 +𝐾𝑠(1+𝐾 ) 𝑖 E + I ⇌ EI 𝜇𝑚𝑎𝑥𝑠 k ES → P + E  = 𝑠 Cell growth substrate inhibition 𝑠+𝐾1(1+𝐾 ) 2 Noncompetitive 𝑣𝑚𝑎𝑥𝑠 E + S ⇌ ES v= 𝑖 Enzymatic reaction ( 𝐾𝑠 +𝑠)( 1+𝐾 ) E + I ⇌ EI 𝑖 ES + I ⇌ ESI 𝜇𝑚𝑎𝑥𝑠 = 𝑠 Cell growth substrate inhibition EI + S ⇌ ESI (𝐾1 +𝑠)( 1+𝐾 ) 2 ES → k P+ E 𝑣𝑚𝑎𝑥𝑠 Uncompetitive v= 𝑖 Enzymatic reaction 𝐾𝑠 +𝑠(1+𝐾 ) E + S ⇌ ES 𝑖 E S + I ⇌ ESI = 𝜇𝑚𝑎𝑥𝑠 Cell growth substrate inhibition 𝑠 k P+ E 𝐾1+𝑠(1+𝐾 ) ES → 2 For substrate inhibition, i is replaced with s. The symbol for Ks and Ki is replaced with K1 and K2 with the same definition.  Example 1 Given: max = 50/h K1=2 K2=10 a) Plot the curve of u vs s for competitive, non-competitive, uncompetitive substrate inhibition and compare with no inhibition curve. b) Determine from the plot the sopt and the value of  at sopt Competitive Answer for (a) 𝜇𝑚𝑎𝑥𝑠 = 𝑠 70 𝑠+𝐾𝑠(1+𝐾 ) 𝑖 60 Noncompetitive 50 𝜇𝑚𝑎𝑥𝑠 = u(/h) 𝐾𝑠+𝑠(1+𝐾 ) 𝑠 40 𝑖 30 No Inhibition Uncompetitive 20 Competitive 𝜇𝑚𝑎𝑥𝑠 = 𝑠2 10 Noncompetitive 𝐾1 +𝑠+𝐾 Uncompeti tive 2 0 Answer for (b) 0 5 10 15 20 smax = 6.00 g/L u (uncompetitive) = 35.00 /h s (g/L) u (noncompetitive) = 31.10 /h Optimum Substrate Concentration, sopt Uncompetitive Inhibition 𝜇𝑚𝑎𝑥𝑠 𝐾2 𝜇𝑚𝑎𝑥𝑠 = 𝑠2 = 𝐾1 𝐾2 +𝐾2 𝑠+𝑠 2 𝐾1 +𝑠+ ) 𝐾2 𝑑𝜇 𝑑 𝐾2 𝜇𝑚𝑎𝑥𝑠 = ( 𝑑𝑠 𝑑𝑠 𝐾 2 ) 1 𝐾2 +𝐾2 𝑠+𝑠 (𝐾1 𝐾2 + 𝐾2𝑠 + 𝑠 2 )(𝐾2 𝜇𝑚𝑎𝑥)− 𝐾2 𝜇𝑚𝑎𝑥𝑠(𝐾2 +2s) =0 𝐾1𝐾2 + 𝐾2 𝑠 + 𝑠 2 2 Optimum s (𝐾1 𝐾2 + 𝐾2 𝑠 + 𝑠 2 )− 𝑠(𝐾2 +2s)=0 2.500 sopt 2.000 1.500 2 u(/h) 𝐾1 𝐾2 − 𝑠 =0 1.000 0.500 0.000 𝑠𝑜𝑝𝑡 = 𝐾1 𝐾2 0 5 10 15 20 25 30 s (g/L) Optimum substrate concentration Optimum Substrate Concentration, sopt Noncompetitive Inhibition 𝜇𝑚𝑎𝑥𝑠 = 𝑠 (𝐾1 +𝑠)( 1+ ) 𝐾2 𝜇𝑚𝑎𝑥𝑠 𝐾2 𝜇𝑚𝑎𝑥𝑠 = 𝐾 𝑠 𝑠2 = 𝐾2 𝐾1 +𝐾1 𝑠+𝐾2 𝑠+𝑠 2 𝐾1 + 1 +𝑠+ ) 𝐾2 𝐾2 𝑑𝜇 𝑑 𝐾2 𝜇𝑚𝑎𝑥𝑠 = 𝑑𝑠 𝑑𝑠 (𝐾 2) 2 𝐾1 +𝐾1 𝑠+𝐾2 𝑠+𝑠 (𝐾2 𝐾1 +𝐾1 𝑠+𝐾2 𝑠+𝑠2 )𝐾2 𝜇𝑚𝑎𝑥 −𝐾2 𝜇𝑚𝑎𝑥 𝑠(𝐾1 +𝐾2+2𝑠) (𝐾2 𝐾1 +𝐾1 𝑠+𝑠2 ) 2 =0 Optimum s or Maximum s 𝑠opt 𝐾1 𝐾2 − 𝑠 2 =0 2.500 2.000 1.500 u(/h) 𝑆𝑜𝑝𝑡 = 𝐾1 𝐾2 1.000 Optimum substrate concentration 0.500 0.000 0 5 10 15 20 25 30 s (g/L) Evaluation of and max , K1 and K2 Uncompetitive Substrate Inhibition 𝜇𝑚𝑎𝑥𝑠 𝜇𝑚𝑎𝑥 = 𝑠2 = 𝐾1 𝑠 = (𝐾1 +𝑠+ ) +1+ 𝐾2 𝑠 𝐾2 𝐾1 𝑠 1 𝑠 +1+𝐾 2 = 𝜇𝑚𝑎𝑥 At high substrate concentration K1/s negligible 𝑠 1 1+𝐾 1 𝑠 2 = 𝜇𝑚𝑎𝑥 = 𝜇𝑚𝑎𝑥 +𝐾 2 𝜇𝑚𝑎𝑥 1 1 𝑠  = 𝜇𝑚𝑎𝑥+ 𝐾2 𝜇𝑚𝑎𝑥 Slope= 1 𝐾2 𝜇𝑚𝑎𝑥 1 1 y-intercept= Plot of vs s  𝜇𝑚𝑎𝑥 −1 x-intercept= 𝐾2  Evaluation of and max , K1 and K2 Noncompetitive Substrate Inhibition 𝜇𝑚𝑎𝑥𝑠 = 𝐾 𝑠 𝑠2 = 𝐾1 + 1 +𝑠+ 𝐾2 𝐾2 𝐾1 𝑠 𝑠2 𝐾 1 𝐾1 𝑠 1 𝐾1 + +𝑠+ + +1+ 𝐾2 𝐾2 𝑠 𝐾2 𝐾2 = 𝜇𝑚𝑎𝑥𝑠 = 𝜇𝑚𝑎𝑥 At high substrate concentration K1/s is negligible 𝐾1 𝑠 1 +1+ 1 𝐾1 𝑠 𝐾2 𝐾2 = 𝜇𝑚𝑎𝑥 = 𝜇𝑚𝑎𝑥 ( 𝐾2 +1)+ 𝐾2 𝜇𝑚𝑎𝑥 1 1 𝐾1 +𝐾2 𝑠 = 𝜇𝑚𝑎𝑥 ( 𝐾2 )+ 𝐾2 𝜇𝑚𝑎𝑥 1 Slope= 1 𝐾2 𝜇𝑚𝑎𝑥 Plot of  vs s y-intercept=( 𝐾1 +𝐾2 ) 𝐾2 𝜇𝑚𝑎𝑥 x-intercept=−(𝐾1 + 𝐾2)  Example 2 Given the following data, a) Evaluate, sopt, opt from the given data b) Evaluate K1, K2 and max from uncompetitive and noncompetitive inhibition models c) Find the ratio of opt with max s (g/L)  (1/h) 0.01 8.32 0.02 14.20 0.04 21.83 0.06 26.41 0.08 29.33 0.10 31.25 0.20 34.48 0.30 34.09 0.40 32.79 0.50 31.25 0.60 29.70 0.70 28.23 0.80 26.85 0.90 25.57 1.00 24.39 1.10 23.31 Evaluation of and K1 and K2, max , a) From the give data, sopt and opt is evaluated from the plot of  and s shown below 40 s (g/L)  (1/h) 0.01 8.32 35 0.02 14.20 30 0.04 21.83 0.06 26.41 25 0.08 29.33 u(/h) 20 0.10 31.25 0.20 34.48 15 0.30 34.09 10 0.40 32.79 0.50 31.25 5 0.60 29.70 0 0.70 28.23 0.80 26.85 0 0.2 0.4 0.6 0.8 1 1.2 0.90 25.57 s (g/L) 1.00 24.39 1.10 23.31 sopt = 0.22 g/l  opt = 34.55 /h  Solution for (b) 1/ s (g/L)  (1/h) 1/ (All s) (High s) 0.01 8.32 0.120 - Make a plot of 1/ vs s and locate the region of 0.02 14.20 0.070 - high concentration of s. At high concentration 0.04 21.83 0.046 - of s, calculate the slope, y-intercept or x- 0.06 26.41 0.038 - 0.08 29.33 0.034 - intercept using a) Uncompetitive, b) 0.10 31.25 0.032 - Noncompetitive 0.20 34.48 0.029 - 0.30 34.09 0.029 0.029 0.40 32.79 0.031 0.031 0.50 31.25 0.032 0.032 0.60 29.70 0.034 0.034 0.70 28.23 0.035 0.035 0.80 26.85 0.037 0.037 0.90 25.57 0.039 0.039 1.00 24.39 0.041 0.041 1.10 23.31 0.043 0.043  Evaluation of and K1 and K2, max b) Solution using uncompetitive inhibition 1 1 𝑠  = 𝜇𝑚𝑎𝑥 + 𝐾2 𝜇𝑚𝑎𝑥 0.14 0.08 1  vs s 0.12 high and low s 0.06 1 0.10 Slope= high s 𝐾2𝜇𝑚𝑎𝑥 0.08 y = 0.0173x + 0.0236 u(1/h) u(/h) R² = 0.9963 0.04 1 y-intercept= 0.06 𝜇𝑚𝑎𝑥 −1 0.04 0.02 x-intercept= 𝐾2 0.02 0.00 0.00 0 0.2 0.4 0.6 0.8 1 1.2 s (g/L) slope 0.0173 y-intercept 0.0236 x-intercept -1.367 𝑠𝑜𝑝𝑡 = 𝐾1 𝐾2 max 42.371 /h 0.22 = 𝐾1 1.367 K2 1.367 g/l 𝐾1 = 0.0354 Ratio uopt / umax opt/max 34.55/42.37=0.82 uopt is 82% of umax  b) Solution using noncompetitive inhibition 1 1 𝐾 𝑠 =𝜇 (𝐾1 +1)+ 𝐾 𝑚𝑎𝑥 2 2 𝜇𝑚𝑎𝑥 0.14 0.08 0.12 1 high and low s 0.06 Plot of vs s 0.10  high s 0.08 y = 0.0173x + 0.0236 u(1/h) u(/h) R² = 0.9963 0.04 0.06 1 Slope= 0.04 𝐾2 𝜇𝑚𝑎𝑥 0.02 0.02 𝐾1 +𝐾2 0.00 0.00 y-intercept=( ) 0 0.2 0.4 0.6 0.8 1 1.2 𝐾2 𝜇𝑚𝑎𝑥 s (g/L) 1 0.0173= 𝐾2 𝜇𝑚𝑎𝑥 𝐾1 +𝐾2 𝐾1 = 0.03655 g/L 0.0236 =( ) Solve Eq. 1 and Eq. (2) 𝐾2 𝜇𝑚𝑎𝑥 𝜇𝑚𝑎𝑥= 1/(slope K2)=43.54/h 0.0236/0.0173 = 𝐾1 + 𝐾2 0.0484 + 𝐾2 = 1.3642 Ratio opt / max 𝐾1 + 𝐾2 = 1.3642 Eq (1) 𝐾2 𝐾22 − 1.3642𝐾2 + 0.0484 opt / max = 34.55/43.54=0.7935 0.22 = 𝐾1 𝐾2 From sopt Solving by quadratic equation uopt is 79.32% of umax 0.0484 = 𝐾1 𝐾2 𝐾1 = 0.0484 Eq. (2) 𝐾2 = 1.32764 g/L 𝐾2 Example 3. The data exhibiting substrate inhibition gave a slope of 0.01 and y-intercept of 0.15 for plot of 1/ vs s at high substrate concentration. If sopt=2 g/L, a) Calculate the values of K1, K2, max and opt for substrate noncompetitive inhibition model b) Plot the u vs s and locate sopt and opt Solve Eq. 1 and Eq. (2) Solution: (a) 1 4 Slope= + 𝐾2 = 15 𝐾2 𝜇𝑚𝑎𝑥 𝐾2 𝐾1 +𝐾2 y-intercept=( ) 𝐾22 − 15𝐾2 + 4=0 𝐾2 𝜇𝑚𝑎𝑥 1 𝐾2 = 14.73g/L = 0.01 𝐾2 𝜇𝑚𝑎𝑥 𝐾1 = 0.2716 g/L 𝐾1 +𝐾2 max = 6.79 /h ( )=0.15 𝜇𝑚𝑎𝑥𝑠𝑜𝑝𝑡 𝐾2 𝜇𝑚𝑎𝑥 opt = 𝐾 𝑠 𝑠𝑜𝑝𝑡2 0.15 𝐾1 + 1𝐾 𝑜𝑝𝑡+𝑠𝑜𝑝𝑡+ 𝐾 𝐾1 + 𝐾2 = = 15 2 2 0.01 opt = 5.26/h 2 = 𝐾1 𝐾2 opt / max =5.26/6.79=0.78 4 = 𝐾1 𝐾2  4 𝐾1 = 𝐾2  𝜇𝑚𝑎𝑥𝑠 = 𝐾1 𝑠 𝑠2 𝐾1 + +𝑠+ 𝐾2 𝐾2 s (g/L) m (/h) opt 6 0 0 0.400 3.937 5 0.800 4.808 1.200 5.119 4 u(h) 1.600 5.236 3 2.000 5.263 2.400 5.245 2 2.800 5.201 3.200 5.141 1 sopt 3.600 5.073 4.000 5.000 0 4.400 4.924 0 1 2 3 4 5 6 4.800 4.847 s (g/L)  Product Inhibitions Competitive 𝟏 𝟏 𝑲𝒔𝒂 𝟏 𝜇𝑚𝑎𝑥𝑠 = + = 𝑝 𝝁 𝝁𝒎𝒂𝒙 𝝁𝒎𝒂𝒙 𝒔 𝑠+𝐾𝑠(1+ ) 𝑝 𝐾𝑝 𝒂=1+ 𝐾𝑝 s (g/L) u (/h) normal) u (1/h) (high product) 1/s 1/u (normal) 1/u (high p) 0.18 1.02 1.34 5.64 0.980 0.745 11.90 1.637 1.426 0.08 0.61 0.70 16.91 2.079 1.754 0.06 0.48 0.57 28.64 2.827 2.455 0.03 0.35 0.41 45.09 3.386 3.043 0.02 0.30 0.33 68.82 3.980 3.633 0.01 0.25 0.28 112.73 4.377 4.136 0.01 0.23 0.24 Slope 4.74 6.50 Interecept 0.192 0.180 1.60 1.40 R2 0.998 0.999 1.20 1.00 umax 5.218 5.560 1/s 0.80 Ks 24.717 36.149 0.60 0.40 x intercept -0.040 -0.028 0.20 0.00 alpha 1.37 0.00 0.05 0.10 1/s 0.15 0.20 p 2.00 Kp 5.37  Product Inhibition High concentrations of product can be inhibitory for microbial growth. Alcohol fermentation provides an example of product inhibition Noncompetitive 𝟏 𝒂 𝑲𝒔 𝒂 𝟏 𝜇𝑚𝑎𝑥𝑠 = + 𝝁 𝝁𝒎𝒂𝒙 𝝁𝒎𝒂𝒙 𝒔 = 𝑝 𝑝 (𝐾𝑠 +𝑠)( 1+ ) 𝒂=1+ 𝐾𝑝 𝐾𝑝 u (/h) high 1/s 1/u (normal) 1/u (high p) s (g/L) u (1/h) normal product 0.22 1.30 1.69 4.641 0.768 0.591 0.10 0.76 0.98 9.700 1.309 1.021 0.07 0.58 0.76 13.923 1.711 1.316 0.04 0.44 0.56 23.166 2.269 1.777 0.03 0.36 0.47 37.127 2.778 2.137 0.02 0.32 0.41 55.645 3.174 2.450 0.01 0.28 0.36 92.818 3.582 2.756 1.80 1.60 Slope 5.03 6.53 1.40 Interecept 0.227 0.291 1.20 umax 4.410 3.431 1.00 1/u Ks 22.161 22.405 0.80 x intercept -0.045 -0.045 0.60 0.40 alpha 1.30 0.20 p 1.00 0.00 Kp 3.34 0.00 0.05 0.10 1/s 0.15 0.20 0.25