Biochemistry enzyme kinetics lab report

Biochemistry Lab Report

Siavosh Naji-Talakar

siavosh@outlook.com

TITLE: Enzyme Kinetics

INTRODUCTION

Alkaline phosphatases are enzymes that are typically membrane-bound glycoproteins that catalyze the hydrolytic cleavage of monoesters at basic pH levels2. This enzyme is found in most advanced level eukaryotes and prokaryotes. When the hydrolysis of the monophosphate ester takes place at the basic pH levels an inorganic phosphate is released2. The enzyme can remove phosphate groups from several different types of molecules such as alkaloids, proteins, or nucleotides. In humans’ alkaline phosphatases play critical roles in the growth and development of teeth and bones, however, it can be found it other parts of the body such as in the liver and kidneys3. The phosphatases are essential for mineralization in humans to allow calcium and phosphorus to be deposited in bones and teeth3. The enzyme was reacted with 4-nitrophenyl phosphate as the substrate1. 4-nitrophenyl phosphate does not resemble a protein and is a non-specific substrate commonly used to assay alkaline phosphatases4. When the alkaline phosphatase performs hydrolysis on 4-nitrophenyl phosphate a highly colored phosphate free product is given1,4. This reaction releases the phosphate group on 4-nitrophenyl phosphate to give the product 4-nitrophenolate which has a molar absorptivity at 405nm under basic conditions with an extinction co-efficient of 18.8x103 M-1cm-1 1,4.

To experiment the effects of inhibition on the enzyme the inhibitor phenylalanine at 75mM and Na2HPO4 at 15mM was also used. Inhibition of enzymes may be carried out via irreversible pathways that work through covalent bonds or through reversible pathways. Reversible pathways include a competitive inhibition where the inhibitor binds to the active site of the enzyme or through a noncompetitive pathway where the enzyme binds to a side other than the active site, which may subsequently change the shape or conformation of the active site6. When a phosphate group PO43- is removed by hydrolysis from an organic compound it is referred to as dephosphorylation. This is the reaction in which the phosphatases operate. The reaction is important in a physiological setting because it enables the activation or deactivation of enzymes by removal of phosphoric esters; a prime example is the conversation of adenosine triphosphate to adenosine diphosphate through phosphorylation7.

The Michaelis-Menten kinetics model is used in biochemistry as a model for enzyme kinetics. The model uses an equation to explain the rate of the enzymatic reactions that occur . It does this by relating the reaction rate, Velocity or Vmax, to the substrate concentration, [S]. Vmax describes the constant values for each enzyme-substrate complex with the theoretical maximum velocity of the enzyme to turn over products at maximum saturation of the substrate concentration6. The Km value, known as the Michaelis constant, is the dissociation constant of the enzyme-substrate complex and measures the enzyme-substrate affinity6. Low Km values equate to a higher affinity of the enzyme-substrate6. The Michalis constant equal to the substrate concentration at half of the Vmax value6. The model best describes single substrate single substrate kinetics with an assumption that the back reaction of enzyme plus product being negligible6.

Inhibitors effect both the Vmax and Km values according to the type of inhibition produced on the enzyme or enzyme-substrate. Irreversible inhibitors create permanent inhibition through covalent bonds6. Reversible inhibitors work through intermolecular forces. Competitive inhibition prevents the substrate from binding active site of the enzyme, the Km appears to increase, and the Vmax does not change6. Noncompetitive inhibitors bind to any site other than the active site allowing for the enzyme-substrate complex to still form, no change to the Km, a decrease to the Vmax, with the enzyme conformational change restricting activity6. For each of these models a disassociation constant value of Ki can be calculated and is analogous to the Km for a substrate6. The Beer-Lambert Law, , is applied to determine what the concentration is in these biomolecules after samples have been prepared and analyzed through a platereader by measuring the attenuation of light8. The absorbance, or attenuation of light, is reliant on two assumptions, the value ‘A’ is directly proportional to the concentration ‘C’ and is also directly proportional to the light path ‘L’8. ‘E’ is a constant that is referred to as the molar extinction coefficient and is used to measure the probability of the electron transition8.

The Michaelis-Menten equation requires deeper calculations that need computer software to process in a timely manner1. Prior to the availability of computers or non-linear regression software the Lineweaver-Burk plot was used to determine the critical points in the enzyme kinetics such as Km and Vmax6,9. The Lineweaver-Burk equation, , linearizes the Michaelis-Menten equation and allows the Km and Vmax to be determined more accurately1,6,9. It is a the double reciprocal setup of the plot that allows for the y-intercept of the graph to be equal to the inverse of Vmax and the x intercept representing -1/Km9. The plot gives a visual impression of how enzymes are inhibited9.

Enzymes are the catalysts in living creatures and are a critical component in all chemical reactions that maintain homeostasis. This key role in life sustaining processing lends credence to the need to understand their behavior and chemical mechanisms. Pharmacological and medical understanding of enzyme kinetics are important in diagnostic assays or therapeutics such as medicines10. Enzymes are found throughout the human body and come in various forms. Oxidoreductases help in redox reactions, transferases transfer functional groups, hydrolases add water, lyases remove atoms to form double bonds or add atoms to double bonds, isomerases help in intramolecular rearrangement, and ligases help catalyze bond formations6. The plethora of different enzymes with varying functions impresses the importance of understanding how each work and the kinetics associated with them.

METHODS

Part 1: Effect of Enzyme Concentration upon Reaction Rate

Enzyme concentration’s effect on the reaction rate was analyzed using a 96-well microplate. The well B1-B6 and C1-C6 were filled with 240 µl of substrate solution. The substrate solution was 5mM 4-nitrophenyl phosphate in buffer. The multichannel pipette was used for ease of loading the substrate. After the substrate was loaded care was taken to ensure no air bubbles were present in the wells since that may lead to erroneous readings.

Enzyme dilutions were prepared on the same microplate. The dilutions were made in wells A1-A5 of the microplate. Alkaline phosphatase, stock concentration of 1mg/ml, was used to make 100 to 200 µl of each of the prescribed concentrations. The concentrations ranged from 250 µg/ml, 125 µg/ml, 50 µg/ml, 25 µg/ml and 10 µg/ml. The dilutions were done in the form of serial dilutions; taking the larger dilution to make the next and using that smaller dilution to make the following in subsequent order. Water was used as the diluting agent for each of the enzyme solutions. Well A6 held 200 µl of water and was used as a blank control with 0 µg/ml concentration.

Care was taken to ensure the enzyme was not added to the substrate solutions until the microplate was placed on the platereader and ready to be processed. The platereader was set up to read at λ = 405nm. Using the multichannel pipette tool 10 µl of the enzymes from row A1-6 were added to row B1-6. The solution was gently stirred with the tip of the pipette and the process repeated for row C1-C6. The platereader was initiated quickly thereafter. The data was collected and exported to Excel for further processing. The data was collected for 2 minutes and the actual increase in absorbance for each well was calculated by subtracting the time zero absorbance value from the reading. The change in absorbance per minute was translated to change in product concentration per minute using the extinction coefficient. The data was plotted on a graph of initial velocity vs. enzyme concentration. From the graph the optimal concentration of enzyme was determined that gave a linear plot of absorbance vs. time.

Part 2: Effect of Substrate Concentration and an Inhibitor

Part 2 was carried out very similar to part 1 with the difference being the enzyme concentration was held constant and the substrate concentration was varied. The best constant amount of enzyme derived from part 1 was used. Using the 96-well plate 36 wells were used. A1-12, B1-12, and C1-12 will be the only wells read by the platereader instrument. The substrate, 5 mM 4-nitrophenyl phosphate in buffer, was prepared to the final concentrations of 2.5 mM, 1.0 mM, 0.5 mM, 0.1 mM, and 0.05 mM. The total volume of each prepared substrate was about 1.5 ml and diluted using stock 0.05M amidol buffer. The instrument was programmed to scan row A to measure spontaneous hydrolysis of the sample, row B for the effect of the substrate concentrations, and row C for the effect of an inhibitor1. Table 1, Table 2, and Table 3 describe the setup of each well and row on the microplate. All samples were run in duplicate with column 1+2, 3+4, 5+6, 7+8, 9+10, and 11+12 being doubles of one another. The multichannel pipette was used to add the amounts of water or inhibitor. Stock concentration of the inhibitor phenylalanine was 75mM and Na2HPO4 was 15mM. The plate was taken to the microplate reader prior to the enzyme being added.

TABLE 1: Analysis of substrate hydrolysis (Row A)

Well

A1-2

A3-4

A5-6

A7-8

A9-10

A11-12

substrate concentration (mM)

0

0.05

0.1

0.5

1

2.5

substrate volume (µl)

230

230

230

230

230

230

water (µl)

20

20

20

20

20

20

TABLE 2: Kinetics of substrate concentration (Row B)

Well

B1-2

B3-4

B5-6

B7-8

B9-10

B11-12

substrate concentration (mM)

0

0.05

0.1

0.5

1

2.5

substrate volume (µl)

230

230

230

230

230

230

water (µl)

10

10

10

10

10

10

enzyme (µl)

10

10

10

10

10

10

TABLE 3: Kinetics with inhibitor (Row C)

Well

C1-2

C3-4

C5-6

C7-8

C9-10

C11-12

substrate concentration (mM)

0

0.05

0.1

0.5

1

2.5

substrate volume (µl)

230

230

230

230

230

230

inhibitor (µl)

10

10

10

10

10

10

enzyme (µl)

10

10

10

10

10

10

Row A was initially measured. Once this row was complete the machine ejected the disk and 10 µl of established enzyme concentration from part A was added to row B. Once complete, the same was done for row C. The data in the platereader was not zeroed out and therefore the data was relative to the initial reading. The data was collected for 90 seconds and the actual increase in absorbance for each well was calculated by subtracting the time zero absorbance value from the absorbance reading at 90 seconds. This task was done by exporting the data to Excel for further processing. The data was converted from change in absorption per minute to change in product concentration per minute using the extinction coefficient. The graph of initial velocity vs. substrate concentration in the absence of inhibitor and in the presence of inhibitor was prepared. The value for maximum velocity (Vmax) and the Michaelis constant (Km) were generated from the plot of velocity vs. substrate concentration. The values for Vmax, Km, and the inhibitor constant (Ki) were generated from the data by using one of the linear forms of the Michaelis-Menten equation.

RESULTS

Part 1: Effect of Enzyme Concentration upon Reaction Rate

The absorbance data from Row B and C were averaged out. The blank column, C6, was subtracted from each row. Then, the zero-time value from each column (concentration) was subtracted from each time value. The value of L was calculated by taking the known pathlength in a 96-well plate, and solving for a 250 µl well volume. The pathlength L = 0.625cm. The molar extinction coefficient for the produced product 4-nitrophenolate is 18.8x103 M-1cm-1. Using the Beer-Lambert equation the value for concentration was derived from known absorbance’s. The data in Table 4 references calculated concentrations (µM) with corresponding time periods (sec). Figure 1 displays the results of the data and when a trend line is applied the point slope formula of each trend is displayed along with the R linearity value. The final values were multiplied by 106 to give the value in µM. The closer the R value to 1 the more linear the trend line. Plotting velocity vs enzyme concentration as in Figure 2 helps determine the best concentration of enzyme to use for part B. The velocity was determined by taking the slope of the trendline assigned per enzyme concentration as shown in Figure 1. The slope is defined as the product over time in units of µM per second. Each slope was referenced to their corresponding enzyme concentrations as seen in Table 5. The concentration of 250 µg/ml was inferred to be the best choice for enzyme concentration to use for part B based on velocity of the reaction.

TABLE 4: Absorbance values converted to concentrations in µg/ml listed by time in seconds.

Time (s)

250 µg/ml

125 µg/ml

50 µg/ml

25 µg/ml

10 µg/ml

0 µg/ml

0

0.0

0.0

0.0

0.0

0.0

0.0

20

16.3

6.9

3.7

1.8

0.8

0.0

40

33.1

14.3

6.2

3.4

1.8

0.0

60

48.9

20.4

9.1

4.9

2.5

0.0

80

59.1

26.6

12.3

6.7

3.1

0.0

100

70.6

34.0

15.5

8.2

3.7

0.0

120

83.1

41.2

18.6

9.6

4.4

0.0

TABLE 5: Enzyme concentration in µg/ml listed by velocity (slope) in µM/second.

[Enzyme] µg/ml

Velocity µM/second

250

0.685

125

0.339

50

0.153

25

0.080

10

0.036

0

0.000

Sample Calculations

Serial dilutions

Concentration from absorbance values

Part 2: Effect of Substrate Concentration and an Inhibitor

The 96-well plate had the rows A, B and C analyzed by the plate reader. Row A contained substrate only and was used as a blank. Row B contained the enzyme only and Row C contained the enzyme plus inhibitor. The stock 5mM 4-nitrophenyl phosphate substrate solutions were prepared into final concentrations of 2.5mM, 1.0mM, 0.5mM, 0.1mM, and 0.05mM by diluting with 0.05M amidol buffer to a volume of 1.5 ml. The dilutions were carried out by simply calculating what the dilution fold factor that was required for each requested dilution would be. For example, from stock 5.0mM substrate to be diluted to 2.5mM substrate was a 2x dilution. Therefore, to make a total volume of 1.5ml of 2.5mM substrate 0.75mL of substrate was added to 0.75ml of buffer. The substrate in each well was filled to the volume of 230 µl with the addition of 20 µl of water for Row A, 10 µl of water and 10 µl of enzyme for Row B, and 10 µl of enzyme and 10 µl of inhibitor for Row C. Therefore, the total concentration of the substrate was no longer the assigned values of 2.5mM, 1.0mM, 0.5mM, 0.1mM, or 0.05mM. The new calculated concentration of each substrate concentration was evaluated by using . For example, when adding 230 µl of the 2.5mM substrate solution and topping off to 250 µl with either water, enzyme, or inhibitor results in a 2.3mM concentration.

The enzyme concentration determined from Part A, 250 µg/ml, was loaded at about 30 µl into row D of the plate and Rows A, B and C were loaded as described in Table 1, Table 2, and Table 3. The plate was taken to the plate reader where Row A was initially read by the instrument. Once complete the 10 µl of the enzyme was added to Row B and the plate read. Finally, enzyme was added to Row C and the plate read. The resulting data was exported to Excel.

The data from column 1+2, 3+4, 5+6, 7+8, 9+10, and 11+12 per respective Row A, B and C were averaged since the experiment was completed in duplicate. The data from Row A as the blank substrate only was subtracted from Row B and C individually taking into account any readings that the substrate only would reveal. Next the data from time zero at Row B and C were subtracted from all other readings within the respective row to blank time zero readings. The resulting absorbance optical density data points were then converted to concentrations using the Beer-Lambert equation. The same values for L = 0.625cm and €=18.8x103 M-1cm-1 were used from part A. Table 6 reviewed the calculated concentration values for the enzyme only and Table 7 reviewed the calculated concentrations values for the enzyme plus inhibitor. Values for Row A were no longer reviewed since they were simply blanks and post all calculations all equaled zero. The data points were plotted with concentration of product in µM vs time in seconds per substrate concentration as reviewed in Figure 3 for enzyme only and Figure 4 for enzyme plus inhibitor. Trendlines were added per concentration of substrate values as plotted to elucidate the given slopes per. The slopes per substrate concentration indicated the velocity in µM per second.

TABLE 6: Concentration of product against time for enzyme only.

Time (s)

0 mM

0.046 mM

0.092 mM

0.46 mM

0.92 mM

2.3 mM

0

0

0

0

0

0

0

20

-0.766

1.532

7.830

26.170

39.149

50.340

40

-0.766

2.340

11.021

41.447

61.106

86.298

60

-0.723

2.979

14.426

51.404

76.128

115.532

80

-0.723

3.319

17.702

63.362

93.957

143.191

100

-0.766

3.957

20.128

73.234

109.532

170.681

120

-0.766

4.213

22.511

83.745

123.574

197.362

TABLE 7: Concentration of product against time for enzyme plus inhibitor.

Time (s)

0 mM

0.046 mM

0.092 mM

0.46 mM

0.92 mM

2.3 mM

0

0

0

0

0

0

0

20

-0.043

2.043

8.723

0.809

13.957

21.489

40

-0.085

3.574

15.447

19.702

35.787

64.085

60

0.043

4.553

19.064

40.723

58.723

92.809

80

0.255

5.064

22.340

58.681

82.043

120.170

100

0.255

5.787

26.213

76.468

103.660

153.064

120

0.128

5.957

29.745

90.468

120.766

172.511

Plotting velocity, or slope, of each calculated substrate concentration for enzyme and enzyme plus inhibitor resulted in the Michalis-Menten plot as reviewed in Figure 5. These corresponding velocities to substrate concentration were listed in Table 8 for Row B, enzyme only, and Row C, enzyme plus inhibitor.

TABLE 8: Corresponding slopes per substrate concentration from Figure 3 and Figure 4 with substrate concentration in mM and enzyme in µM/second

[Substrate] mM

Enzyme

µM/s

Enzyme + Inhibitor µM/s

2.3

1.5887

1.4942

0.92

0.972

1.0499

0.46

0.6559

0.8245

0.092

0.1764

0.2341

0.046

0.033

0.0479

0

-0.004

0.0024

From Michalis-Menton plot in Figure 5 the values of Vmax, Km and Ki could have been determined. However, to ensure full accuracy the data was manipulated into the Lineweaver-Burk plot as in Figure 6. This was referred to as the double reciprocal plot with corresponding values listed in Table 9. The plot was generated by dividing 1 by the substrate concentration and velocity for both enzyme and enzyme plus inhibitor values.

TABLE 9: Evaluated Table 8 values after application of double reciprocal with substrate concentration in 1/mM and enzyme concentration in seconds/µM.

1/[Substrate] mM

1/Enzyme µM/s

1/Enzyme + Inhibitor µM/s

0.43

0.63

0.67

1.09

1.03

0.95

2.17

1.52

1.21

10.87

5.67

4.27

21.74

30.30

20.88

Sample Calculations

Preparation of substrate concentrations

New substrate concentration after addition of 230 µl to wells and topping off with water

Converting resulting O.D. data from Row B and C into concentrations

Converting Michaelis-Menton plot to Lineweaver-Burk plot by double reciprocal.

The Michaelis-Menten equation may be plotted in order to estimate the Km, Vmax, and Ki values. However, the Lineweaver-Burk equation as a double reciprocal plot allows a more accurate value for these figures. From this double reciprocal plot Figure 6 a trend line was applied to reveal the point slope formula associated with the linear line. The equation is in the form . The equation for the enzyme only was and the enzyme plus inhibitor was found to be . Therefore the Vmax can be solved for in the form of and then by inserting Vmax into to solve for Km. For the enzyme only the Vmax was 0.5638 and the Km was 0.7459. For the enzyme plus inhibitor the Vmax was 1.038 and the Ki was 0.9380. These values were derived from the Lineweaver-Burk plot.

Sample Calculation

Using enzyme only equation.

DISCUSSION

Prior to formulating the initial test calculations were carried out to understand the dilution factors and how to carry out the serial dilutions. The serial dilutions had to be applied correctly to the correct total volume of the initial step in Part A. The final concentration and volume divided by the initial concentration to get the volume of enzyme required to continue with each serial dilution. Running duplicate samples for Part A on two rows allowed for an average value to be derived for absorbance, which was converted to concentration using Beer-Lambert law. The same principle was applied in Part B but the concentrations were given in mM and the duplicate runs were completed on joined columns in the same row.

The results of the initial procedures conducted in Part A corroborated with predicted results. It was hypothesized that give an increase in enzyme concentrations that there would be more product. If more enzyme was added and it resulted in more product, then in sue the velocity would be increased with added enzyme. The experimental results corroborated this initial hypothesis. If there is excess substrate with small amounts of enzymes the velocity of the reaction will increase with the amount of enzyme. This occurs with varying concentrations of enzyme where all of the enzyme molecules catalyze reactions even though there is substrate that can be turned into product. Therefore, the velocity of the reaction can only increase if the concentration of enzyme is increased. Figure 1 and 2 match with this initial hypothesis. Moreover, the initial thought was that the larger increase in enzyme concentration would correlate to a stronger velocity. The more enzyme, the quicker the reaction. This was also confirmed from experimental results.

However, moving onto part B with the enzyme concentration being held constant the hypothesis did not match the experimental results fully. The initial findings when experimenting on concentration of product versus time for enzyme and enzyme + inhibitor did fall in line with the initial hypothesis; if more enzyme is used then there will be more product over a period of time given larger amounts of substrate. Based on Figure 3 and Figure 4 this initial hypothesis was confirmed.

It was hypothesized that when the enzyme concentration is kept constant and the concentration of substrate is increased it will cause the velocity to also increase quickly until half of the enzyme becomes saturated with substrate. Then, the velocity will begin to taper out. As it continues the velocity will approach an asymptote; at that point even when the substrate concentration is increased velocity will not. Experimental results from Figure 5 confirm this hypothesis as it can be noted that the velocity begins to slow down at the Km value around 1 and then reaches for the asymptote.

The variance begins when reviewing the double reciprocal or Lineweaver-Burk plot in Figure 6. Based on previous study and prelab work the hypothesis was that regardless of a competitive inhibitor or non-competitive inhibitor the slope of the inhibitor + enzyme line would be greater than that of the enzyme only line. However, the experiment showed that the slope of the line decreased with the inhibitor. This result does not match any real known reasoning. It was postulated that the reason for this weird result could be from two extenuating circumstances. The inhibitor could have simply not worked or been ‘stale’ in a sense. All chemicals have a shelf life and it is possible that the inhibitor provided no longer worked as intended with the enzyme. Further, the error could be a simple mix up of data points. If Row B and C were accidentally switched while conducting the experiment, then it would explain these results.

The calculated Km, Ki, and Vmax values did not match what was expected. There were no expectations on the actual numerical value of these fields but it was expected that the Vmax should be the same if the inhibitor was competitive or the km should be the same if it was a non-competitive inhibitor. The experimental data revealed that neither of these figures were corroborating with the hypothesis. Therefore, it was deduced that the inhibitor must be a mix of competitive and non-competitive. The initial data points as shown in Figure 6 have the slope of the line starting at a negative Y value. Either way though the Figure indicates almost the same Vmax and a change in Km. This data pushes the notion that the inhibitor is a mixed inhibitor. If we delete our 0.046mM data point from the double reciprocal plot the lines begin to look a little better. See below. With the deleted data point, the recalculated Vmax for the Enzyme Only is 2.129 with a Km of 1.019 and the Enzyme plus Inhibitor Vmax is 1.912 and a Ki of 0.6585. Based, especially on these new slopes derived with the deleted data point, with such a small change in Vmax it cannot be attributed to being a fully competitive or non-competitive. With the change in Km and an ever so slight change in Vmax the only conclusion that can be made is that it was a mixed inhibitor with more of a competitive inhibitor in the mix.

There were potentially several sources of error added to the experiment. First and foremost, since the volumes that were required were so small and necessitated exactness it is certainly possible that any little pipetting error was magnified throughout the experiment. Either of the sample Rows from section B of the experiment were susceptible to being contaminated. It is possible that even 1 drop of the inhibitor solution into Row B (Enzyme only row) would give extreme erroneous readings on the plate reader. The platereaders read the absorbance of light through the solution given its length. However, it should be noted that for the enzyme only test that Column B11 was run on the plate reader with a bubble in it. Bubbles in the solution make the light travel a different distance compared to the other wells. This can cause a huge error in readings. For this experiment the data from Column B11 was expunged from the data readings and only Column B12 was used, therefore these two columns were not averaged. Perhaps most importantly, when part A of the experiment was carried out it was concluded that the enzyme with 250 µg/ml was the best concentration to use for part B. However, a lapse in concentration occurred and the followed lab session when part B was carried out the enzyme was not diluted to 250 µg/ml from the original 1 mg/ml. This in theory should only increase the velocity of the reaction. However, it is certainly an experimental error.

To improve this experiment, it may be very beneficial to the students to know prior to conducting the experiment which inhibitor was given to them. Relying on the data to determine the inhibitor type is fine but it leaves a sense of the unknown still. It is much better learning to know what the end result should look like and then to work backwards to see why the data attained did fit or did not fit. Sometimes students can be left still feeling anxious not knowing for certain if all the time and effort spent conducting the experiment, understanding and applying the logic learned to it, and writing an extensive lab report were correct or not. The learning opportunity is lost especially if over a week is spent reviewing wrong data or wrong answers. Providing this information to the student certainly does not take away from the learning objectives of the lab. If it was known what the inhibitor was then as the experiment is being carried out errors can be identified much more quickly. Rather than, going back and forth between TA and student trying to see what went wrong where and for what possible reason. This can let students work through errors themselves and may even provide a better learning outcome. Additionally, if the inhibitors were given out and known, the lab can be improved by given a competitive, non-competitive, and mixed inhibitor to students and allowing additional tests to be run on the 96 well plate. There were certainly more rows open for additional test of inhibitors to be conducted. In this way students can see the effects of no enzyme, enzyme only, enzyme + competitive, enzyme + noncompetitive, and enzyme + mixed. Instead, we are left with only an enzyme + unknown inhibitor.

QUESTIONS

1. Did you observe a direct proportionality between enzyme concentration and initial velocity?

Yes, we did. Based on Figure 2 when the velocity, or slope, of the varying enzyme concentrations were graphed the trend line was y = 0.0027x + 0.0084 and was therefore directly proportional. As enzyme concentration was increased so was the initial velocity.

2. Does the velocity of the reaction change with time? If so, why?

If there is excess substrate with small amounts of enzymes the velocity of the reaction will increase with the amount of enzyme i.e. the velocity of the reaction should increase in direct proportion to the concentration of enzyme. This occurs with varying concentrations of enzyme where all of the enzyme molecules catalyze reactions even though there is substrate that can be turned into product. Therefore, the velocity of the reaction can only increase if the concentration of enzyme is increased.

When we keep the enzyme concentration constant and the concentration of substrate is increased it will cause the velocity to also increase quickly until half of the enzyme becomes saturated with substrate. Then, the velocity will begin to taper out. As it continues the velocity will approach an asymptote; at that point even when the substrate concentration is increased velocity will not. This is called the Vmax. At half of the Vmax we see the Km, Michaelis-Menten constant.

The reaction rate is proportional to the concentration; the proportionality constant is the rate constant.

Rate = k[R]X where x = order of reaction.

So over time, the reactant concentration will reduce and so will the rate.

3. What type of inhibition is indicated by your data? Justify your answer.

The initial data points as shown in Figure 6 have the slope of the line starting at a negative Y value. Either way though the Figure indicates almost the same Vmax and a change in Km. This data pushes the notion that the inhibitor is a mixed inhibitor. If we delete our 0.046mM data point from the double reciprocal plot the lines begin to look a little better. See below. With the deleted data point, the recalculated Vmax for the Enzyme Only is 2.129 with a Km of 1.019 and the Enzyme plus Inhibitor Vmax is 1.912 and a Ki of 0.6585. Based, especially on these new slopes I derived with the deleted data point, with such a small change in Vmax I cannot attribute it to being a fully competitive or non-competitive. With the change in Km and an ever so slight change in Vmax the only conclusion that can be made is that it was a mixed inhibitor with more of a competitive inhibitor in the mix.

4. Given the following reaction and equation for the initial velocity of the reaction:

(where kcat is the rate constant for the reaction which forms the product from the ES complex), explain in words why the velocity is directly proportional to the amount of enzyme added in the presence of saturating substrate levels.

If there is excess substrate with small amounts of enzymes the velocity of the reaction will increase with the amount of enzyme i.e. the velocity of the reaction should increase in direct proportion to the concentration of enzyme. This occurs with varying concentrations of enzyme where all of the enzyme molecules catalyze reactions even though there is substrate that can be turned into product. Therefore, the velocity of the reaction can only increase if the concentration of enzyme is increased.

For nearly all forward, irreversible reactions, the rate is proportional to the product of the concentrations of the reactants, each raised to some power.

The rate is proportional to [A]m[B]n that is:

rate = k[A]m[B]n

here the rate of the reaction is proportional to concentration of ES.

This reaction is of first order, so it is directly proportional

5. Consider the Michaelis-Menten equation. Under what conditions will Km be a measure of substrate binding?

Km shows the affinity between the substrate and the enzyme as it is a substrate concentration at which half of the Vmax is achieved. So if the Km value increases substrate binding decreases. Km = [S] at half Vmax

REFERENCES

1. Biochemistry Lab Manual, Pages 1-2. Summer 2016

2. Sharma, U. Deeksha, P. Rajendra, P. 2014. Alkaline Phosphatase: An Overview. Indian J Clin Biochem. Jul; 29(3) 269-278. Retrieved from http://www.ncbi.nlm.nih.gov/pmc/articles/PMC4062654/?tool=pmcentrez

3. National Institutes of Health. 2016. Alkaline phosphatase, liver/bone/kidney. Retrieved from https://ghr.nlm.nih.gov/gene/ALPL

4. New England Biolabs Inc. 2016. P-Nitrophenyl Phosphate (PNPP). Retrieved from https://www.neb.com/products/p0757-p-nitrophenyl-phosphate-pnpp

5. Fernley, HN. Walker, PG. 1970. Inhibition of alkaline phosphatase by L-phenylalanine. Biochem J. Feb; 116(3): 543-544. Retrieved from http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1185393/

6. Tymoczko JL, Berg JM, Stryer L. 2013. Biochemistry: A Short Course Second Edition. W.H. Freeman and Company. New York.

7. Casiday, R. Herman, C. Frey, R. 2008. Energy for the Body: Oxidative Phosphorylation. Washington University in St. Louis. Retrieved form http://www.chemistry.wustl.edu/~edudev/LabTutorials/Cytochromes/cytochromes.html

8. Gooch, Jan. 2007. Encyclopedic Dictionary of Polymers. SpringerLink. Retrieved from http://link.springer.com/referencework/10.1007%2F978-0-387-30160-0

9. Lineweaver, H. Burk, D. 1934. The Determination of Enzyme Dissociation Constants. J. Am. Chem. Soc. 56(3), pp 658-666. Retrieved form http://pubs.acs.org/doi/abs/10.1021/ja01318a036

10. King, MW. 2016. Enzyme Classifications. The Medical Biochemistry Page. Retrieved from http://themedicalbiochemistrypage.org/enzyme-kinetics.php

Figure 3: Concentration of [Product] vs Time (enzyme only)

0 mM
0 20 40 60 80 100 120 0 -0.76595744680851019 -0.76595744680851019 -0.72340425531914831 -0.72340425531914831 -0.76595744680851019 -0.76595744680851019 0.046 mM

0 20 40 60 80 100 120 0 1.5319148936170219 2.3404255319148937 2.9787234042553199 3.3191489361702144 3.9574468085106389 4.2127659574468099 0.092 mM

0 20 40 60 80 100 120 0 7.8297872340425538 11.021276595744681 14.425531914893622 17.702127659574472 20.12765957446809 22.51063829787234 0.46 mM

0 20 40 60 80 100 120 0 26.170212765957444 41.446808510638299 51.40425531914893 63.361702127659569 73.234042553191472 83.744680851063819 0.92 mM

0 20 40 60 80 100 120 0 39.148936170212764 61.10638297872341 76.127659574468098 93.957446808510625 109.53191489361701 123.57446808510636 2.3 mM

0 20 40 60 80 100 120 0 50.340425531914896 86.297872340425542 115.53191489361701 143.19148936170214 170.68085106382983 197.36170212765958

Time (s)

[Product] (µM)

Figure 4: Concentration of [Product] vs Time (enzyme+inhibitor)

0 mM
0 20 40 60 80 100 120 0 -4.2553191489361743E-2 -8.5106382978723485E-2 4.2553191489361743E-2 0.25531914893617041 0.25531914893617041 0.12765957446808521 0.046 mM

0 20 40 60 80 100 120 0 2.0425531914893615 3.5744680851063833 4.5531914893617023 5.0638297872340434 5.7872340425531927 5.9574468085106398 0.092 mM

0 20 40 60 80 100 120 0 8.7234042553191493 15.446808510638295 19.063829787234042 22.340425531914889 26.212765957446809 29.744680851063833 0.46 mM

0 20 40 60 80 100 120 0 0.80851063829787306 19.702127659574472 40.723404255319153 58.680851063829792 76.468085106383 90.468085106382986 0.92 mM

0 20 40 60 80 100 120 0 13.957446808510637 35.787234042553187 58.723404255319146 82.042553191489347 103.6595744680851 120.76595744680851 2.3 mM

0 20 40 60 80 100 120 0 21.489361702127663 64.085106382978694 92.808510638297847 120.17021276595743 153.06382978723403 172.51063829787236

Time (s)

[Product] (µM)

Figure 5: Velocity vs Substrate Concentration

(enzyme & enzyme + inhibitor)

Enzyme 2.2999999999999998 0.92 0.46 9.1999999999999998E-2 4.5999999999999999E-2 0 1.5887 0.97199999999999998 0.65590000000000004 0.1764 3.3000000000000002E-2 -4.0000000000000001E-3 Enzyme + Inhibitor 2.2999999999999998 0.92 0.46 9.1999999999999998E-2 4.5999999999999999E-2 0 1.4942 1.0499000000000001 0.82450000000000001 0.2341 4.7899999999999998E-2 2.3999999999999998E-3
[Substrate] (mM)

Verlocity (µM/sec)

Figure 6: Lineweaver-Burk Plot Velocity vs Substrate Concentration (enzyme & enzyme+inhibitor)

1/Enzyme

0.43478260869565222 1.0869565217391304 2.1739130434782608 10.869565217391305 21.739130434782609 0.62944545855101652 1.0288065843621399 1.5246226558926665 5.6689342403628116 30.303030303030301 1/Enzyme + Inhibitor

0.43478260869565222 1.0869565217391304 2.1739130434782608 10.869565217391305 21.739130434782609 0.66925445054209609 0.95247166396799687 1.212856276531231 4.2716787697565142 20.876826722338205

1/[Substrate] (1/mM)

1/Verlocity (sec/µM)

Deleted datapoint 0.046mM: Double Reciprocal Plot Velocity vs Substrate Concentration (enzyme & enzyme+inhibitor)

1/Enzyme

0.43478260869565222 1.0869565217391304 2.1739130434782608 10.869565217391305 0.62944545855101652 1.0288065843621399 1.5246226558926665 5.6689342403628116 1/Enzyme + Inhibitor

0.43478260869565222 1.0869565217391304 2.1739130434782608 10.869565217391305 0.66925445054209609 0.95247166396799698 1.212856276531231 4.2716787697565142

1/[Substrate] (1/mM)

1/Verlocity (sec/µM)

Figure 1: Concentration of Product vs Time

250 ug/ml
0 20 40 60 80 100 120 0 16.340425531914896 33.063829787234049 48.893617021276597 59.063829787234049 70.595744680851055 83.106382978723417 125 ug/ml

0 20 40 60 80 100 120 0 6.8936170212765973 14.340425531914892 20.382978723404257 26.638297872340427 33.957446808510639 41.191489361702132 50 ug/ml

0 20 40 60 80 100 120 0 3.6595744680851072 6.212765957446809 9.1063829787234081 12.297872340425533 15.489361702127658 18.595744680851062 25 ug/ml

0 20 40 60 80 100 120 0 1.8297872340425547 3.4468085106382986 4.9361702127659566 6.6808510638297882 8.2127659574468108 9.617021276595743 10 ug/ml

0 20 40 60 80 100 120 0 0.80851063829787073 1.7872340425531907 2.4680851063829783 3.063829787234043 3.7021276595744688 4.425531914893619 0 ug/ml

0 20 40 60 80 100 120 0 0 0 0 0 0 0

Time (s)

[Product] (µM)

Figure 2: Velocity vs. Enzyme Concentration

Velocity
250 125 50 25 10 0 0.68540000000000001 0.33929999999999999 0.1527 8.0100000000000005E-2 3.6299999999999999E-2 0

Enzyme Concentration (µg/ml)

Velocity (µM/sec)