Biochem Enzyme Kinetics

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 (