BATANGAS STATE UNIVERSITY College of Engineering, Architecture, & Fine Arts Gov. Pablo Borbon Campus II, Alangilan, Batangas City, Philippines 4200 www.batstate-u.edu.ph Telefax: (043) 300-4404 locs. 106-118
CHEMICAL AND FOOD ENGINEERING DEPARTMENT
ChE 555: Biochemical Engineering Enzyme and Enzymes Kinetics
ChE-5201
January 18, 2016
Agbay, Philip D. 1. For a given species of an enzyme that doubles every 3 hours, what is the mass of the biomass that may be expected from 100 liters of seed if each liter contains 8 grams biomass the first order removal rate constant is 2.6 per hour.
SOLUTION: t = ln 3 hr = ln k = 0.231 Using the obtained value of k = 0.231, and substituting to the new conditions t = ln 24 = ln C = 204 kg
2. A mouse-mouse hybrodoma cell line is used to produce monoclonal antibody. Growth in batch culture is monitored with the following data.
(a) Determine the specific growth rate during the growth phase. (b) What is the culture doubling time
SOLUTION: (a) The data are plotted as in Figure E8.1.
μ = 0.67 d −1
(b) The doubling time is:
t2 = ln 2/0.67 = 1.00 day
3. In an experiment conducted to evaluate the Michaelis Menten constant, it was found out that 1 g of bacteria could decompose the waste at a maximum rate of 35 g/day when the waste concentration was high. It was also found that the same quantity of bacteria would decompose waste at a rate of 18 g/day when the waste concentration was 20 mg/L. Calculate the rate of waste decomposition by 2 grams of bacteria if the waste concentration were maintained at 8 mg/L.
SOLUTION: V = Vmax Km = 18.89
Vmax = K3 Eo K3 = 35 / day
V = K 3 Eo = 35 x 2 x V = 20.8256 / day
Atienza, Angielle A. ChE - 5201
1. Polarimetric study of enzymatic hydrolysis yielded the dependence of the initial reaction rate on the substrate concentration. Some of the gained values are given in the following table:
cs / (mol dm-3)
0.062
1.82
106 v0 / (mol dm-3 s-1)
4.96
5.48
a. Calculate the constants of Michaelis – Menten equation K M and Vmax. b. What will be the conversion of the optically active substrate after 6 hours from the instant when the enzyme was added to the substrate solution, the concentration of which was 2.2 mol dm-3? Reference: www.old.vsht.cz
Solution: a.
is in the form of
where
and
Through linear regression of the given data in the table,
2. A purified enzyme is used in the conversion of
. The solution contains
5mg/L of a dimeric enzyme having a total molecular weight of 82,340 g/mol (and two active sites). The molecular weight of A is 119 g/mol. If the V max is found to be 0.13g/Ls and Km = 0.043 g/L, find the turnover number. Reference: www.cmbe.engr.uga.edu
(Turnover number)
3. The action of pepsin of 1-carboxyl-1-glutamyl tyrosin (substrate S) at the temperature of 38°C and pH=4 is characterized by the kinetic parameters
Calculate the percentage of the substrate converted after 10 hour from the beginning of the reaction in case that the initial concentration of the substrate was 0.8 mol/dm3.
Reference: www.old.vsht.cz
For
Austria, Babylyn C.
PROBLEM #1: Relation between Reaction Velocity and Substrate Concentration: MichaelisMenten Equation a) At what substrate concentration will an enzyme with of 30 s -1 and a of 0.005 show one-quarter of its maximum rate? b) Determine the fraction of concentrations:, , and .
that would occur at the following substrate
Answers: a) Since and , , we can substitute into the Michaelis-Menten equation to give
b) We can arrange the Michaelis-Menten equation into the form Substituting into this equation gives Substituting into this equation gives Substituting into this equation gives PROBLEM #2: Properties of an Enzyme of Prostaglandin Synthesis
Prostaglandins are a class of eicosanoids, fatty acid derivatives with a variety of extremely potent actions on vertebrate tissues. Prostaglandins are responsible for producing fever and inflammation and its associated pain. They are derived from the 20carbon fatty acid arachidonic acid in reaction catalyzed by the enzyme prostaglandin endoperoxide synthase. This enzyme, a cyclooxygenase, uses oxygen to convert arachidonic acid to PGG2, the immediate precursor of many different prostaglandins. a) The kinetic data given below are for the reaction catalyzed by prostaglandin endoperoxide synthase. Focusing here on the two columns, determine the and of the enzyme. Arachidonic Acid (mM)
Rate of Formation of PGG2 (mW/min)
0.5 1.0 1.5 2.5 3.5
23.5 32.2 36.9 41.8 44.0
Rate of Formation of PGG2 with 10 mg/mL ibuprofen (mW/min) 16.67 25.25 30.49 37.04 38.91
b) Ibuprofen is an inhibitor of prostaglandin endoperoxide synthase. By inhibiting the synthesis of prostaglandins, ibuprofen reduces inflammation and pain. Using the data in the first and third columns of the table, determine the type of inhibition that ibuprofen exerts on the prostaglandin endoperoxide synthase. Answers: a) Calculate the reciprocal values for the data, as in parentheses below, and prepare a double-reciprocal plot to determine the kinetic parameters. [S] (mM) (1/[S] (mM-1))
(mM/min) ( (min/mW))
0.5 (2.0) 1.0 (1.0) 1.5 (0.67) 2.5 (0.4) 3.5 (0.27)
23.5 (0.043) 32.2 (0.0321) 36.9 (0.027) 41.8 (0.024) 44.0 (0.023)
with 10 mg/mL ibuprofen(mM/min) ( (min/mW)) 16.67 (0.06) 25.25 (0.0396) 30.49 (0.0328) 37.04 (0.027) 38.91 (0.0257)
From the graph,
Solving for and using linear regression:
where , , and Using the linear regression, the following values are obtained:
Substituting the values of and to solve for and : and, b) Ibuprofen acts as a competitive inhibitor. The double reciprocal plot (with inhibitor) shows that, in the presence of ibuprofen, theof the reaction is unchanged (the intercept on the the axis is the same) and is increased ( is closer to the origin). PROBLEM #3:
Determination of An enzyme is discovered that catalyzes the chemical reaction SAD
HAPPY
A team of motivated researchers set out to study the enzyme, which they call happyase. They find that the for happyase is . They carry out several experiments. When and , the reaction velocity, , is 9.6 µMs-1. Calculate for the substrate SAD. Answer: We know , , , and We want to solve for . Substituting the known values allows us to solve for .
Solving for gives,
Reference: CourseSmart International E-Book for Principles of Biochemistry by David L. Nelson, Michael M. Cox
Canson, Mark Anthony ChE – 5201
1. Sucrose (A) isolated from fruits, and its enzyme sucrase (E) flow through a mixed flow reactor (V = 6 liter) to undergo an enzymatic reaction that will eventually synthesize glucose via hydrolysis of sucrose. From the entering and leaving concentrations and flow rate find a rate equation to represent the hydrolytic action of sucrase on sucrose.
For reactor,
CEO, mol/L
CAO, mol/L
CA, mol/L
v, L/h
0.02
0.2
0.04
3
0.01
0.3
0.15
4
Solution:
0.001
0.69
0.60
1.2
mixed flow the Monod
kinetics will be represented by:
C A = -CM + k(
)
y = mx + b and т = V/v CEO, mol/L
CAO, mol/L
CA, mol/L
v, L/h
Т = V/v, h , mol/L
0.02
0.2
0.04
3
2
0.01000
0.01
0.3
0.15
4
1.5
0.01500
0.001
0.69
0.60
1.2
5
0.03333
Thus, linearizing the equation will let y = CA, m = k, x =
Using linearization and inputting y = CA, m = k, x = calculator, one can get m = k = 24.15/h -b = CM = 0.2062 mol/L Knowing that Michaelis Menten equation is represented by r =
, b = -CM
, b = -CM in the
The final rate equation based from Michaelis-Menten equation is r =
Graph
2. In a number of separate runs different concentrations of substrate and enzyme are introduced into a batch reactor and allowed to react. After a certain time the reaction is quenched and the vessel contents analyzed. From the results found below find a rate equation to represent the action of enzyme on substrate.
Runs
CEO, mol/L
CAO, mol/L
CA, mol/L
Т = V/v, h
1
3
400
10
1
2
2
200
5
1
3
1
20
1
1
Solution: For batch flow reactor, the Monod kinetics will be represented by
= -CM + k(
)
y = mx + b and т = V/v Thus, linearizing the equation will let y =
=, m = k, x =
,b
= -CM Runs
CEO, mol/L
CAO, mol/L
CA, mol/L
Т = V/v, h
1
3
400
10
1
0.81326
105.7232
2
2
200
5
1
0.54217
52.86158
3
1
20
1
1
0.33381
6.342356
Using linearization and let y =
=, m = k, x =
, b = -CM in
the calculator, one can get m = k = 206.68/h -b = CM = 61.40 mol/L Knowing that Michaelis Menten equation is represented by r =
The final rate equation based from Michaelis-Menten equation is r =
Graph:
3. Cellulose can be converted to sugar by the following enzymatic attack
cellulase
cellulose
sugar
and both celluboise and glucose act to inhibit the breakdown. To study the kinetics of this reaction a number of runs are made in a mixed flow reactor kept at 50°C and using a feed of finely shredded cellulose (C AO =25 kg/m3), enzyme (CEO = 0.01 kg/m3, same for all runs), and various inhibitors. The results are as follows: Runs
Exit streams, CA, kg/m3
(No inhibitor) т,min
(w/cellobiose) т’’, min
(w/glucose) т’’’, min
1
1.5
587
940
1020
2
4.5
279
387
433
3
9.0
171
213
250
4
21.0
36
40
50
For cellobiose, CBO = 5kg/m3 For glucose, CGO = 10kg/m3 Assuming competitive inhibition, find a rate equation to represent the breakdown of cellulose by cellulase in the presence of cellobiose as inhibitor. Show also the
linearized graph of cellulose-cellulase kinetics with and without cellobiose inhibition. Solution: For mixed flow reactor, the Monod kinetics with no inhibition will be represented by, CA = -CM + k(
) but since we introduce a competitive inhibitor in the reactor
the Monod kinetics will become *CA = -CM (1+NCBo) + k(
)
* Remember that Cm for no inhibition equals CM for competitive inhibition y = mx’’ + b’’ Thus, linearizing the equation will let y = CA,, m = k, x’’ =
, b’’ = -CM
(1+NCBo) Runs
Exit streams, CA, kg/m3
(No inhibitor) т,min
(w/cellobiose) т’’, min
1
1.5
587
940
0.3747
0.6000
2
4.5
279
387
0.6124
0.8495
3
9.0
171
213
0.9619
1.1981
4
21.0
36
40
1.8900
2.1000
For no inhibition: Using linearization and inputting y = CA, m = k, x = calculator, one can get m = k = 12.844/min -b = CM = 3.365 kg/m3
, b = -CM in the
For competitive inhibition: Using linearization and inputting y = CA, m = k, x’’ =
, b’’ = -CM
(1+NCBo) in the calculator, one can get m = k = 13.061/min -b = CM(1+NCBo) = 6.5025 kg/m3, knowing that CBO = 5 and CM = 3.365kg/m3 Then N = 0.1865m3/kg Knowing that Michaelis Menten equation with no inhibition is represented by r =
Then the Michaelis Menten equation for competitive inhibition will become, r=
The final rate equation based from Michaelis-Menten for competitive inhibition equation is r =
Graph:
Reference: Chemical Reaction Engineering Third Edition, Octave Levenspiel Cantos, Jonathan C.
1. A constant amount of enzyme was added to a series of reaction mixtures containing different substrate concentrations. The initial reaction rates were measured from the initial slopes of progress curves of product formation. The data in following table were obtained. a. What is Vmax for this enzyme in reaction mixture? b. What is Km of enzyme for the substrate?
Table 1. Steady State Enzyme Kinetic Data [S]0 (μmol/L)
V0 (μmol/L-min)
0.1
0.27
2.0
5.0
10.0
20
20.0
40
40.0
64
60.0
80
100.0
100
200.0
120
1000.0
150
2000.0
155
SOLUTION: We can use the Lineweaver-Burk equation for the analysis. For this, the reciprocals of the entries in Table 1 must be calculated and then plotted as shown in Fig. 1.
(a) From the reciprocal of the ordinate intercept, Vmax = 160 μmol/L-min, and (b) From the reciprocal of the abscissal intercept, Km = 60 μmol/L .
2. Use the Michaelis-Menten equation to complete the enzyme kinetic data set, when Km is known to have a value of 1 mmol/L.
[S]0 (μmol/L) 0.5
V0 (μmol/L-min) 50
1.0 2.0 3.0 10.0
__ __ __ __
SOLUTION: Using the Michaelis Menten equation, the first entry in the table gives Vmax = 150 μmol/L-min. The other entries simply follow by substituting the values of [S] 0 into V0 = . The results are as follows: [S]0 (μmol/L) 0.5 1.0 2.0 3.0 10.0
V0 (μmol/L-min) 50 75 100 112.5 136.4
3. Hexokinase catalyzes the phosphorylation of glucose and fructose by ATP. However, Km for glucose is 0.13 mmol / L , whereas Km for fructose is 1.3 mmol/L . Suppose that Vmax is the same for both glucose and fructose and that the enzyme displays hyperbolic kinetics. (a) Calculate the normalized initial velocity of the reaction (i.e., V 0/ Vmax) for each substrate when [S]0 = 0.13, 1.3, and 13.0 mmol/L. (b) For which substrate does hexokinase have the greater affinity?
SOLUTION: (a) For glucose the values of V0/Vmax are 0.5, 0.91, and 0.99, respectively; for fructose the values are 0.091, 0.56, and 0.91. (b) Glucose; at lower concentrations the reaction rate is a greater fraction of V max than it is with fructose.
Source: Schaum’s Outline of Biochemistryp: 3rd Edition
Dimayuga, Evytte M. 1. From the following kinetic data, estimate Vmax and Km for the catalysis of Substrate 1 by the enzyme Questionase. Draw a curve for the data you would expect to observe for Substrate 2 if Km = 30 uM for this substrate (assuming the both have the same Vmax).
Solution: Determining Vmax for Substrate 1 might vary, depending on how you draw the curve. The value of Vmax = 200 uM/sec) was used to calculate values for the data. Whatever you chose, Km will be determined from the substrate concentration that gives you Vmax/2. Your curve for Substrate 2 might look a little different as well.
2. If [Questionase] = 0.05 uM, calculate kcat for Substrate 1 and what would be the value for kcat/Km in units of M-1s-1? (Hint: you need to convert Km to units of Molar) Solution Vmax = kcat [Etotal] kcat
= Vmax / [Questionase] = (200 uM/sec)/0.05 uM = 4000 s-1
Km
= 10 uM = 10 x 10-6 M = 1 x 10-5 M
kcat/Km = (4 x 103 s-1)/(1 x 10-5 M) = 4 x 108 M-1 s-1.
3. Immobilized enzyme beads of 0.6 cm diameter contain an enzyme which converts a substrate S to a product P by an irreversible, unimolecular enzyme reaction with Km¼0.012 kmolm_3 and a maximum rate Vmax¼3.6_10_7 kmol (kg-bead)_1 s_1. The density of the beads and the effective diffusion coefficient of the substrate in the catalyst beads are 1000 kgm_3 and 1.0_10_6 cm2 s_1,
respectively. Determine the effectiveness factor and the initial reaction rate, when the substrate concentration is 0.6 kmolm_3. Solution The value of the Thiele modulus is calculated from Equation 7.21.
Effectiveness factor for various values of CAb/Km The value of CAb/Km is 50, and thus a value of Ef =0.50 is obtained from figure above. The initial rate of the reaction is
de Guzman, Monroe H. PROBLEM NO. 1 – THE FASTEST ENZYME The reaction rate of carbonic anhydrase is one of the fastest of all enzymes, and its rate is typically limited by the diffusion rate of its substrates (Badger, 1994). The table below shows the data of the enzyme with respect to its rate of reaction in a specific type of substrate. Table 1.0 – Carbonic anhydrase kinetics S (mmol/L) 0.10 0.15 0.20 0.25 0.30 0.35 0.40
-rs (mmol/Ls) 1.2 x 106 1.7 x 106 2.1 x 106 2.4 x 106 2.9 x 106 3.2 x 106 3.6 x 106
S (mmol/L) 0.45 0.50 0.55 0.60 0.65 0.70 0.75
-rs (mmol/Ls) 4.0 x 106 4.6 x 106 5.2 x 106 6.0 x 106 6.9 x 106 7.5 x 106 8.2 x 106
In this experiment, the chemical engineering students of Batangas State University were tasked to determine the velocity of enzymatic reaction using Michealis-Menten constant via the Lineweaver-Burke Plot when the particular aqueous substrate generated 3000 Pascals of osmotic pressure upon diffusion in an organic cell of a human body in normal temperature.
Solution: To solve this problem, the Michaelis-Menten constant (K m), limiting or maximum velocity (Vmax), and the substrate concentration (S) must be determined.
For Km and Vmax: Plot the data using Lineweaver-Burke plot by designating x as 1/S and y as 1/-rs.
Thus, the y-intercept is 1/Vmax and Km/Vmax is the slope. From the plot, the y-intercept is 4.9760 x 10-8 L-s/mmol. 1/Vmax = 4.9760 x 10-8 L-s/mmol Vmax = 2.0096463 x 107 mmol/L-s From the plot, slope = 8.1515 x 10-8 s Km/Vmax = 8.1515 x 10-8 s Km = 2.0096463 x 107(8.1515 x 10-8) Km = 1.63816 mmol/L To solve for the concentration of substrate: Because the substrate is an organic substance, van’t Hoff factor = 1, and the temperature of a normal body is around 37°C. Gas constant is taken as 0.08206 atm-L/mol-K. 3000/101325 = M(0.08206)(37+273.15)(1) M = 1.163325 x 10-3 mol/L M or S = 1.1633256 mmol/L The equation for Michaelis-Menten Theory: v = V max[S]/(Km + [S]) v = 2.0096463 x 107(1.1633256)/(1.63816 + 1.1633256) v = 8.345109 x 106 mmol/L-s
PROBLEM NO. 2 – ARE YOU STILL THE FASTEST? After a whole school year, the new batch of fifth year chemical engineering students at Batangas State University learned about carbonic anhydrase. Because of their fascination, they were eager to determine if the newly synthesized compound could increase the speed even more. They found the enzyme inside a cooler and added the compound. However, the professor came and he was angry telling that he was extrapolating the enzyme’s speed. Here are the data written in his notes: 1. 0.0172 % mol/V (in L) is dissolved in the aqueous solution. 2. The plot of Lineweaver-Burke is accurate in dealing with the kinetics. Later, the professor found out that the students added a compound in the enzyme. The furious professor asked for how much amount, but the students only said it was strychnine. The professor was infuriated even more shouting, “It’s a very poisonous competitive inhibitor! Go back to the lab and determine how much you added, unless you will not graduate this school year!” The only test the student made was a speed test which resulted to 1.13 x 104 mmol/L-s at the given enzyme concentration. They also found that it was indeed a competitive inhibitor with K I = 2.1388 mmol/L. Help the students in determining the strychnine concentration.
Solution: To determine the amount of the competitive inhibitor, the data given in problem number 1 must be utilized prior to the fact that it is accurate in dealing with the kinetics. The obtained constants will be used to solve for the concentration of the said inhibitor. Using the obtained constants: Km = 1.63816 mmol/L and Vmax = 2.0096463 7 x 10 mmol/L-s, and on the basis of 1 L solution of enzyme: S = 0.172/100 = 1.72 x 10-4 mol/L S = 0.172 mmol/L The equation for competitive inhibition is given by: v = Vmax[S]/{[S] + Km(1 + [I]/KI)} 1.13 x 104 = 2.0096463 x 107[0.172]/{[0.172] + 1.63816(1 + [I]/2.1388)} [I] = 397.01404 mmol/L Thus the students added 0.39701 M strychnine.
PROBLEM NO. 3 – MY PROFESSOR IS CURIOUS
After determining the concentration, the professor was impressed and got the idea of determining an aid in HIV. To accomplish this, he needs the constant of inhibition provided by the enzyme peptide-based HIV-1 protease. He has done researches about the inhibitor. 1. HIV-1 protease is an organic compound exhibiting uncompetitive inhibition to carbonic anhydrase. 2. Its aqueous solution lowers the freezing point of water to -0.0771°C. 3. It forms an enzyme-substrate complex with the carbonic anhydrase that follows Lineweaver-Burke plot. 4. The inhibitor affects the speed of an enzyme by lowering the speed three-folds. As compensation to what the students did, the whole batch was assigned to rework the enzyme and determine the constant of inhibition (K I) of HIV-1 protease for the professor to formulate the medicine. The aqueous enzyme is preserved inside the refrigerator with the freezing point lowered by 0.0122 K. To determine the concentration of the substrate and the inhibitor: ΔT = Kfmi For HIV-1 protease: 0 + 0.0771 = 1.86(m)(1) m = 0.04145 m ~ 0.014145 mol/L [I] = 41.45161 mmol/L For carbonic anhydrase: 0.0122 = 1.86(m)(1) m = 6.5591 x 10-3 m ~ 6.5591 x 10-3 mol/L [S] = 6.55914 mmol/L At [S] = 6.55914 mmol/L, evaluate the speed (v). v = Vmax[S]/(Km + [S]) v = 2.0096463 x 107(6.55914)/(1.63816 + 6.55914) v = 1.60804 x 107 mmol/L-s
Substituting to the equation for uncompetitive inhibition by dividing the obtained speed by three (3) since it reduces three-folds: v = Vmax[S]/{Km + [S](1 + [I]/KI} 1.60804 x 107/3 = 2.0096463 x 107(6.55914)/{1.63816 + 6.55914(1 + 41.45161/KI)} KI = 16.56972 mmol/L The inhibition brought by HIV-1 protease is 16.56982 mmol/L. Maybel De La Cruz
BS ChE 5201 Enzyme Kinetics
1.In an experiment conducted to evaluate the Michaelis-Menten constant , it was found that 1 g of bacteria could decompose the waste at maximum rate of 45g/day when the waste concentration was high. I t was also found that the same quantity of bacteria would decompose waste at a rate of 28g/day when the waste concentration was 30mg/L. Calculate the rate of decomposition by 4 g of bacteria if the waste concentration where maintaned at 8 mg/L.
Given: For 1g of bacteria Vmax=45g/day V=28g/day S=30mg/L
V=Vmax *S/(Km+S) 28g/day=[(45g/day)((30mg/L)/(1g/1000mg))]/(Km+0.03g/L) Km=0.0182g/L
V at 4 grams when S=0.008g/L V=[(45)*(0.008g/L)]/(0.0812+0.008) V=13.7404 V=(13.7404g/day)*4
V=54.9618g/day
2.It is desired to reduce the bacterial count of polluted water from 40 million organisms per mL to 9 organisms per mL. Calculate the number of completely mixed chlorine contact chambers in seies, each having a detention time of 3 hours that would be required if the first order removal rate constant is 3.6/hr
Given: Ca=9organisms per mL Cao=40000000organisms per mL K=3.6/hr T=3 hr
For n-CSTR n series Ca/Cao=1/(1+kt)^n 9/40000000=1/(1+(3.6*3))^n n=6.2020 3.For given species of a microorganism that doubles every 4 h, what is the biomass that may be expected from 200 liters of seed if each liter contains 9 g biomass and the fermentatin culture was maintained for 48 hours. Given Cao=(9g/L*200L) Cao=1800 g t=48 hours
2Cao-----> after four hours lnCa/Cao=kt Ln2Cao/Cao=4k K=ln2/4 lnCa/(9g/L*200L)=k(48hr) lnCa/(9g/L*200L)=(ln2)/4*(48hr) Ca=7372800g Ca=7372.8g
Dimaano, Jezza B. Problem #1. With the following enzyme activity results determine Vmax. [S] (mol/L) 2 x 10-1 2 x 10 -2 2 x 10 -3 2 x 10 -4 1.5 x 10 -4 1.3 x 10 -5
V (μmol/min) 60.00 60.00 60.00 48.00 45.00 12.00
Answer: From the available data, we can notice that the reaction rate doesn’t increase when the substrate concentration is over 2 x 10-3 M. This is the maximal velocity (Vmax) of the enzyme, in this case 60 μmol/min.
Problem #2. The results for enzyme activity analysis can be found below. Determine : a) Vmax;
b) Km; [S] (mol/L) 5 x 10-2 5 x 10 -3 5 x 10 -4 5 x 10-5 5 x 10-6 5 x 10 -7
v (μmol/min) 0.25 0.25 0.25 0.20 0.071 0.01
Answer: a) Vmax = 0.25 μmol/min; b) Km can be determined using the Michaelis-Menten equation: v = [S] Vmax [S] + Km v[S] + vKm = [S]Vmax vKm = [S]Vmax - v[S] vKm = [S] (Vmax - v) Km = [S] (Vmax - v)/v
Using the data for a [S] of 5 x 10-6: Km = Km = 1.2875 X 10-5 M Note: Similar data would be obtained if a different [S] is chosen, as long as v < Vmax. Problem #3. The following table describes the results from an enzymology experiment. Determine: a) Km; b) Vmax; [S] (mol/L) 1 x 10 -3 5 x 10 -4 1 x 10 -4 3 x 10 -5 2 x 10 -5 1 x 10 -6
v (μmol/min) 65.00 63.00 51.0 33.00 27.00 17.00
a) Km can be determined using the Michaelis-Menten equation: v = [S] Vmax
[S] + Km v[S] + vKm = [S]Vmax vKm = [S]Vmax - v[S] vKm = [S] (Vmax - v) Km = [S] (Vmax - v)/v Using the data for a [S] of 1 x 10-6: Km = Km = 2.8235 X 10-6 M b) Vmax = 65.00 μmol/min
Falcutila, Michael Jay F. 1. The value of Km and Vmax for two alternative substrates A and B for the same enzyme are as follow. Substrate
Em (mM)
Vmax (mMol/sec)
A
4.0
25
B
0.5
15
Which substrate will react most rapidly at low substrate concentration (<< Km)? Solution: Since at low substrate concentration [S], Vmax = kcat [Et]. As we have Vmax, which is directly proportional to kcat at a given enzyme concentration, so comparison of Vmax/Km gives idea about enzyme's efficiency with substrates. The value of substrate B = Vmax/Km = 30 (µmol/sec/mM) and for A = 6.25 (µmol/sec/mM). Hence at low concentration substrate B will be used more efficiently to react compared to substrate A.
2. The following table shows the rate of reaction of substrate to product in the presence of enzyme; v (mol/sec) under different conditions; 1. Reaction 1: Reaction of substrate with enzyme to form product in the absence of inhibitor. 2. Reactions 2 and 3: Reaction takes place in the presence of two different inhibitors, each with 10mM concentration.
[S] mM
v, reaction 1 (μ mol/sec)
v, reaction 2 (inhibitor A) (μ mol/sec)
v, reaction 3 (inhibitor B) (μ mol/sec)
1
2.5
1.17
0.77
2
4.0
2.1
1.25
5
6.3
4.0
2.0
10
7.6
5.7
2.5
20
9.0
7.2
2.86
Let’s take same amount of enzyme in each case and determine K m and Vmax for the enzyme (in absence of inhibitors) and determine K iand the type of inhibition for each inhibitor. Solution: Km = 3.3 mM, Vmax = 10 mmol/sec 1. Inhibitor A (competitive inhibition), there will be no change in V max , but Kmapp = Km{1 + ([I]/KI)}-1/Kmapp = -0.13mM-1 Kmapp = 7.7mM K mapp /Km = 7.7mM/3.3mM= 2.33 = 1 +[I]/ KI = 1+ [10mM]/ KI KI (inhibitor-A) =10mM/ 2.33 – 1 = 10mM/ 1.33 = 7.52 mM 2. Inhibitor B (pure non-competitive) there will be no change in K m; but Vmaxapp = Vmax / {1 + ([I]/ KI )} 1/ Vmaxapp = 0.3(μ m o l / s e c )-1 Vmaxapp = 3.33 μmo l / s e c Vmax/ Vmaxapp = 10 μmol/sec/ 3.33 μm o l / s e c = 3 = 1 + [I]/KI = 1 +[10mM]/ KI KI (inhibitorB) =10mM / 3 - 1 = 10mM/2 = 5.0 mM 3. Draw a Lineweaver-Burk plots for an enzyme for which the following data are available. [S]
V, no inhibitor V, with inhibitor
(mM)
(mmol min-1)
(mmole min-1)
3.0
4.58
3.66
5.0
6.40
5.12
7.0
7.72
6.18
9.0
8.72
6.98
11.0
9.50
7.60
What are the Km and Vmax values for the inhibited and uninhibited enzymes? Is the inhibitor competitive or non-competitive? Solution: First determine ; ; then plot Plot 1: vs Plot 2: vs
0.3333 0.2000 0.1429 0.1111 0.0909
0.2183 0.1562 0.1295 0.1147 0.1053
For ; x = & y = y = 0.4665x + 0.0629 x = 0 ; y = 0.0629 Vmax (no I) = 1/0.0629 = 15.8983 mmol min-1 For ; x = & y = y = 0.5846x + 0.0784 x = 0 ; y = 0.0784 Vmax (I) = 1/0.0784 = 12.7551 mmol min-1 Non - competitive inhibition
0.2732 0.1953 0.1618 0.1433 0.1316
Francisco, Jayvee D. 1. The following mechanism relates to an enzyme E with two binding sites for the substrate S. Two complexes are formed: a reactive binary complex ES, and a nonreactive ternary complex ESS.
(a) Derive the rate law for this mechanism. (b) Show that inhibition occurs in the rate of formation of P, relative to that given by the Michaelis-Menten equation for the two-step mechanism for a single binding site in which only ES is formed. (c) What is the maximum rate of reaction (call it the apparent Vmax, or Vmax,app), and how does it compare with the parameter
? At
what value of cS does it occur? (d) Sketch rp versus cs for comparison with Figure 10.1. SOLUTION: (a) We apply the SSH to the complexes ES and ESS; in the latter case, this is equivalent to assumption of equilibrium with the dissociation constant for ESS given by
Solving for cE and cESS in terms of cES and cS from (1) and (2), respectively, and substituting the results in (3) and rearranging to obtain cES in terms of cEo and cS, we have
where Km is the Michaelis constant (equation 10.2-17). Substituting the result for cES from (5) into (4) we obtain the rate law:
2. The hydrolysis of sucrose (S) catalyzed by the enzyme invertase has been studied by measuring the initial rate, rPo, at a series of initial concentrations of sucrose (cSo). At a particular temperature and enzyme concentration, the following results were obtained (Chase et al., 1962):
Determine the values of Vmax and the Michaelis constant Km in the MichaelisMenten rate law. SOLUTION: Linear regression of the data given, according to equation 10.3-2, results in Vmax = 0.46 mol L-1 s-1, and Km = 0.043 mol L-1. Figure 10.2 shows the given experimental data plotted according to equation 10.3-2. The straight line is that from the linear regression; the intercept at l/cSo = 0 is l/Vmax= 2.17 mol-l L s, and the slope is Km/Vmax = 0.093 s.
Reference: Chemical Reaction & Kinetics Missen
3. A substrate L-benzoyl arginine p-nitroanilide hydrochloride was hydrolyzed by trypsin, with inhibitor concentrations of 0, 0.3, and 0.6 mmol l -1. The hydrolysis rates obtained are listed in Table 3.3 [5]. Determine the inhibition mechanism and the kinetic parameters (Km, Vmax, KI) of this enzyme reaction.
SOLUTION: The results given in Table 3.3 are plotted as shown in Figure 3.9. This Lineweaver–Burk plot shows that the mechanism is competitive inhibition. From the line for the data without the inhibitor, K m and Vmax are obtained as 0.98 gmol m-3 and 9.1 mmol m-3 s-1, respectively. From the slopes of the lines, K I is evaluated as 0.6 gmol m-3.
Reference: Biochemical Engineering Katoh
Macalintal, Cristel M. Carbonic anhydrase catalyzes the hydration of CO 2. CO2 + H2O ↔ H2CO3 The Km of carbonic anhydrase for CO2 is 12 mM. Carbonic anhydrase gave an initial velocity vo=4.5 μmols H2CO3 formed/mL sec when [CO2]=36 mM. What is the Vmax for this enzyme? Solution:
Given:
For an enzyme that displays Michaelis- Menten kinetics, what is the reaction velocity, V (as a percentage of Vmax), observed at: a. [S]=Km b. [S]=0.5Km Solution:
a.
b.
Albizzia Falcataria, a specie, of plywood, after an initial thermochemical hydrolysis yielded 25% maltose, 3% sucrose, 12% cellibiose, 43% oligosaccharides and 17% non-carbohydrates residues. The resulting hydrolysate is passed through a column of immobilized enzyme systems so that all types of dissacharides are converted further to hexone units. In alcohol fermentation, the rule of thumb is 10% of the substrate is converted to biomass and 90% to alcohol. If one metric ton of pulpwood is processed daily, the mass of glucose that may be expected to be produced per six-day week is Solution: Dissacharides = maltose (25%), sucrose (3%), cellibiose (12%) Total % of disaccharides = 25% + 3% + 12% = 40% Dissacharide hydrolysis:
Expected glucose production per week:
Magboo, Michael S. 1. A 10 m3 chemostat operating at 75% capacity is producing biomass from a glucose feed at a volumetric rate of 46.9 L/min. The specific growth rate of the enzyme used is Q = () () () () = 0.5 m3 /hr 2. The hydrolysis of sucrose (S) catalyzed by the enzyme invertase has been studied by mea- surmg the initial rate, rpo, at a series of initial concentrations of sucrose (cS0). At a particular temperature and enzyme concentration, the following results were obtained (Chase et al., 1962): cs,/mol L-l
0.0292
0.0584
0.0876
0.117
0.146
0.175
0.234
rp,/mol L-1 s-l 0.182 0.371
0.265
0.311
0.330
0.349
0.372
Determine the values of Vmax and the Michaelis constant Km in the MichaelisMenten rate law. SOLUTION: Linear regression of the data given, results in Vmax = 0.46 mol L-l s-l, and Km = 0.043 mol L-l. Figure 10.2 shows the given experimental data plotted according to equation 10.3-2. The straight line is that from the linear regression; the intercept at l/cSo = 0 is l/Vmax = 2.17 mol-l L s, and the slope is Km/Vmax = 0.093 s.
3. The following mechanism relates to an enzyme E with two binding sites for the substrate S. Two complexes are formed: a reactive binary complex ES, and a nonreactive ternary complex ESS .
Derive the rate law for this mechanism. SOLUTION:
We apply the SSH to the complexes ES and ESS; in the latter case, this is equivalent to assumption of equilibrium with the dissociation constant for ESS given by K2 = k-,/k,:
Solving for cE and cESS in terms of cEs and cs from (1) and (2), respectively, and substituting the results in (3) and rearranging to obtain c ES in terms of cEO and cs, we have
where Km is the Michaelis constant (equation 10.2-17). Substituting the result for cus from (5) into (4) we obtain the rate law:
Marasigan, John Edrian T.
1 In a particular enzyme-catalyzed reaction, Vmax = 0.2 mol/sec and Km = 5 mM. Assume the enzyme shows standard Michaelis-Menten kinetics. What is the rate of the reaction when [S] = 10 mM? Solution:
2 An enzyme catalyzed reaction has a Km of 1 mM and a Vmax of 5 nM/sec. What is the reaction velocity when the substrate concentration is a) 0.25 mM; b) 1.5 mM; or c) 10 mM? a
b
c
3 To study the dependence of the rate of an enzyme-catalyzed reaction on the substrate concentration, a constant amount of enzyme is added to a series of reaction mixtures containing different concentrations of substrate (usually expressed in mol/L). The initial reaction rates are determined by measuring the number of moles (or µmoles) of substrate consumed (or product produced) per minute. Using theis experimental procedure, the data in the table below were obtained for an enzyme in 10-ml reaction mixtures. Use numerical (not graphical) calculations in answering the following questions.
[S] (mol/L) 0.25 0.25 0.25 0.20 0.071 0.0096
v (µmol/min) 0.25 0.25 0.25 0.20 0.071 0.0096
a What is Vmax for this concentration of enzyme? Looking at the table, it is clear that b What is the Km of this enzyme? For a reaction obeying Michaelis-Menten kinetics, Vmax and Km are simply constants relating v to [S]. Km can be calculated by substituting Vmax and any pair of v and [S] values at v < Vmax. For example, at [S] = 5.0 x 10-5 M and v = 0.20 µmol/min the equation becomes
Matibag, Cherriellyn B. 1) For an enzyme (5 μM) , the following initial velocities have been reported depending on the substrate concentration: [Substrate] mM 0.02 0.04 0.07 0.1 0.15 0.2 0.3 0.5 0.7 Draw a Michaelis-Menten plot for this enzyme.
V0,mM, s-1 10.83 18.57 26.76 32.50 39.00 43.33 48.75 54.17 56.88
2. You are trying to reproduce experimental data from the previous student in the lab. He/she had reported that the enzyme under investigation has a kcat of 1875 s-1 and a catalytic efficiency of 7.5 107 M-1 s-1. What is the KM of this enzyme? From kcat/KM and kcat, you can calculate KM as Solution:
Km =
3. In repeating the measurement, you want to have a vmax of 10 mM s-1 (as this is easy to measure). How much enzyme should you use? Provide your result in μM.
Solution:
Bernadeth R. Pagcaliwagan 1. Exploring the Michaelis-Menten equation I. What is the v/Vmax ratio when [S] = 4Km?
2. Given a specie of a microorganism that doubles every 2.5 hours, the mass of biomass that may be expected from 500 liters of seed if each liter contains 5 grams biomass and the fermentation culture was maintained for 12 hours.
3. A solution is measured for alcohol dehydrogenase activity by adding excess ethanol and NAD+, and monitoring the formation of NADH2+ (which absorbs UV light at 340 nm with a molar extinction coefficient ε of 6.22 Abs Units·L/mmol). A 100 L solution containing 33.5 mg/L protein is mixed with 900 L buffer, NAD+ and ethanol. After 1 minute, the absorbance of the mixture has increased by 0.46 AU. a) What is the activity of alcohol dehydrogenase in the original protein solution? b) What is the specific activity of alcohol dehydrogenase in the original protein solution? Solution:
a. Note that the “original protein solution” has been diluted by a factor of 10 as a result of the assay, and there are 1000 mol in 1 mmol.
b.
Reyes, Kristine Joy L. The κ for alkaline phosphatase-catalyzed hydrolysis of methyl phosphate is approximately 14/sec at pH 8 and 25°C. The rate constant for the uncatalyzed hydrolysis of methyl phosphate under the same conditions is approximately 1x1015/sec. What is the difference in the activation free energies of these two reactions? Solution:
Where KS is the dissociation constants for enzyme-substrate complex, and K T is the dissociation constants for enzyme transition state complex. In general,
and
or
Acetyl cholinesterase acetylcholine:
catalyzes
the
hydrolysis
of
the
neurotransmitter
Acetyl choline + H2O yields acetate + choline The Km of acetyl cholinesterase for its substrate acetylcholine is 9x10 5 M. In a reaction mixture containing 5 nanomoles/mL of acetyl cholinesterase and 150 µm acetylcholine, a velocity vo = 40 µmol/mL·sec was observed for the acetyl cholinesterase reaction. Calculate the Vmax for this amount of enzyme. Solution:
The enzyme catalase catalyzes the decomposition of hydrogen peroxide: 2H2O2 = 2H2O + O2 The turn over number (kcat) for catalase is 40 000 000 / sec. The Km of catalase for its substrate H2O2 is 0.11 M. What is the catalytic efficiency of catalase? Solution: The catalytic efficiency is the turnover number divided by known.
, both of which is
Rivera, Lourelane H. 1. The concentration-velocity data shown below were obtained for an enzyme catalyzing a reaction S P. Calculate Km and Vmax. (S) V M nmoles x liter -1 x min-1 2.5 x 10-6 24 3.33 x 10-6 30 -6 4.0 x 10 34 5 x 10-6 40 -5 1 x 10 60 2 x 10-5 80 4 x 10-5 96 -4 1 x 10 109 2 x 10-3 119 -2 1 x 10 120 Linearizing the Michaelis-Menten equation:
Solution:
Let x = ; y = By linear regression using the calculator: y-intercept = A = = 8343317.435 Therefore, M/min Slope = B = = 83.4440 Therefore, /min 2. Estimate k, the first-order rate constant, for an enzyme preparation with a Vmax of 4.6 under the given experimental conditions. M. Solution: 3. The velocity of an enzyme-catalyzed reaction was measured at several substrate concentrations. Calculate Km and Vmax for the reaction.
[S] (M) 0.25 0.5 1.0 2.0 4.0
V (mM/sec) 0.75 1.20 1.71 2.18 2.53
Solution: By linearizing Michaelis-Menten equation:
Let x = ; y = By linear regression using the calculator: y-intercept = A = = 333.6629 Therefore, M/sec Slope = B = = Therefore, /sec
