BIOL 3380 Name:_____________________________________

Circle Session: T-PM W-AM W-PM R-AM R-PM F-AM F-PM

Experiment 9 – Pre-lab Homework
Enzyme Kinetics of LDH
This pre-lab homework assignment is due at the beginning of your lab session.

You are provided with the following portion of a protocol: • Determine concentration of enzyme stock solution, if unknown, by taking an A280 nm reading of a 1:100 dilution (in water). Use a total volume of 1 ml in the cuvette.

• Dilute some of the enzyme stock with buffer A to make a 4 mg/ml solution.

• Serially dilute the 4 mg/ml solution with buffer A to make working solutions of 400 µg/ml and 40 µg/ml.

• Prepare 30 µl of each working solution for every sample

The PI of the lab gives you a tube of enzyme and tells you the following before disappearing into the office to write more grant proposals: ➢ There is 50 µl of enzyme stock solution. The enzyme is expensive to purify, so follow the protocol exactly, using as little of the stock solution as possible.

➢ The concentration of the stock solution is currently not known, but a 1 mg/ml concentration of the pure enzyme has an A280 nm of 2.0.

➢ You’ll be performing the assay on 12 samples.

➢ Make enough of each working solution so that you have at least 400 ul to work with when you do the assay (to cover any waste and/or inefficiencies in pippetting).

Using the spectrophotometer to read the absorbance at 280 nm, you get a reading of 0.784.
1. (2 pts) What is the concentration of the solution in the cuvette? What is the concentration of the stock solution? (Show your work, box your answers.)

2. (1 pt) Calculate the minimum amount of each working solution would you need in order to do the number of assays requested.

How much excess has the PI instructed you to make? 3. (6 pts) Figure out your plan for making each solution required in the protocol. Be certain you FINISH with the required volumes of EACH solution! a. Calculate what is needed to make the 4 mg/ml solution

b. Calculate what is needed to make the 400 ug/ml working solution

c. Calculate what is needed to make the 40 ug/ml working solution

4. (1 pt) You compare the data from your assay to data obtained on a previous day by the lab’s star graduate student. You recorded a reading of 0.020 ΔA340/5 secs. The grad student recorded a reading of 0.160 ΔA340/min. Whose sample had more LDH activity? (Show your work for credit.)

5. (1pt) Looking at the “Reaction Rate as a Function of Enzyme Concentration” section in the lab protocol, which assay would you predict to have the highest LDH activity and why?

6. (1pt) Looking at the “Reaction Rate as a Function of Substrate Concentration” section in the lab protocol, which assay would you predict to have the highest LDH activity and why?

Experiment 9

Enzyme Kinetics of LDH

You are in a position to characterize the LDH enzyme as a catalyst. This is done by measuring the kinetics of enzyme reactions. (kinetics means the rate of change, or reaction rate). First, a review of enzyme basics:

I. Definitions

Enzymes increase the rate at which a reaction occurs. Enzymes are not consumed by the reaction. They do not shift the equilibrium of the reaction. Substrates are substances acted upon by an enzyme.

II. Properties of Enzymes A. The majority of known enzymes are proteins (some nucleic acids with catalytic function have been identified). This means that those factors that alter protein structure also influence enzyme activity. B. Catalysis occurs on a specific site on the enzyme (the active site). The active site is usually less than 5% of the surface area of the protein, and is always in a cleft. The rest of the molecule serves to present the active site in a three dimensional structure that is capable of binding substrate and catalyzing the reaction. Binding to a substrate is very specific, and involves ionic interactions, H bonds and van der Waals forces. C. Only amino acids with polar side chains participate directly in catalysis (Cys, His, Ser, Asp, Glu and Lys). Some reactions require electron acceptors, and since no side chains are good electron acceptors coenzymes and cofactors are required for catalysis of some reactions. D. Enzymes catalyze reactions by several means: ▪ by orienting the substrate molecules. ▪ by increasing the effective concentration of the substrate. ▪ by acid-base catalysis ▪ by covalent catalysis ▪ by using the energy of substrate binding to promote the enzyme catalyzed reaction.

III. Criteria For Establishing The Presence Of Enzymes

Since most of the known biological catalysts are proteins two criteria are generally used for establishing the existence of enzymes. The first is that the rate of a reaction in the presence of an enzyme is greater than the rate in its absence. Because the uncatalyzed rates of most biologically important reactions are effectively zero, the mere demonstration of a reaction rate in a biological system is taken as evidence of the presence of an enzyme. The second criterion is that the increased rate is labile to treatments that destroy proteins, for example, heat denaturation and/or proteolysis. Furthermore, the rate of enzyme catalyzed reactions should be proportional to the amount of enzyme added. A word of caution, though. Some enzymes increase the rate of reaction only by a modest amount, others are heat stable, resist proteolysis, are inhibited by high concentrations of reactants, etc. The demonstration of the existence of an enzyme in a crude tissue extract sometimes can be complicated by the presence of other enzymes that compete for substrates or utilize products, the absence of activators or the presence of inhibitors. Thus preliminary fractionation (to remove competing enzymes), dialysis (to remove inhibitors), and additions (to achieve appropriate concentrations of reactants, cofactors and activators) may be required.

IV. Regulation of Enzyme Activity In Vivo In vivo, the activity of many enzymes is tightly regulated.. Among the mechanisms used by cells to regulate enzyme activity are: ▪ Control of transcription (both positive and negative). ▪ Synthesis as proenzymes and processing to active enzymes when needed. ▪ pH regulation (as in endocytic vesicles, secretory compartments, lysosomes). ▪ Inhibitors, modulators.

V. Factors That Influence Enzyme Catalyzed Reactions.
1. Enzyme concentration. If one enzyme molecule can transform a million substrate molecules each minute, two enzyme molecules will transform twice that many molecules in the same time (the reaction rate will double). As long as sufficient substrate is present, the reaction rate is proportional to the enzyme concentration.

Figure 1. Optical assay of a dehydrogenase. The change in absorption ((E) is plotted against time. With more enzyme used, the reaction runs faster. (From Introduction to Modern Biochemistry, by P. Karlson, 1965)

2. Substrate concentration. The velocity of an enzyme catalyzed reaction increases with the increase in substrate up to a certain point and then becomes constant. This point is known as the Vmax.. At this point the enzyme is said to be saturated, and further increases in the concentration of the substrate will not increase the rate of reaction. The enzyme can work no faster.

Figure 2. The effect of substrate concentration on the rate of an enzyme-catalyzed reaction.

3. Temperature. Reactions rates increase as the temperature increases, but only to a point. Most enzymes are denatured at temperatures above 60(C, with loss of their activity.

Figure 3. The maximum velocity of an enzyme-catalyzed reaction increases as a function of the temperature at which the assay is performed.

4. pH, ionic strength. The effect of pH involves titration of charged R groups that participate in catalysis. Enzymes have an optimum pH at which an enzyme’s activity is greatest. The optimum pH for pepsin, a proteolytic enzyme in the stomach is around 2. Trypsin, another proteolytic enzyme, has an optimum pH of around 8. If trypsin were placed in the highly acidic environment of the stomach, it would likely be denatured and lose its catalytic activity. Ionic strength affects noncovalent interactions of the active site.

Figure 4. At pHs above and below the optimum pH for an enzyme, the activity of the enzyme is reduced and the reactions rates are slower.

5. Cofactor concentration. Cofactors must be present for some enzymes; if absent these enzymes will have little or no activity. If cofactors are present but limiting, only a fraction of the enzyme concentration will be effective in catalyzing reactions.

6. Inhibitors. Inhibitors are substances that slow or completely stop enzyme-catalyzed reactions. Most enzymes can be inhibited by certain chemical reagents. Natural enzyme inhibitors occur in living systems to control the rates of specific reactions. Much of the current drug therapy is based on inhibition of specific enzymes with a substrate analog.

V. Reaction Rates As A Function Of Enzyme Concentration.

Michaelis-Menton analysis of kinetic data requires measurement of the initial velocities of a series of enzyme catalyzed reactions. The initial reaction rate reflects test conditions where amounts of substrate are unlimited (saturating) and there is negligible back reaction during the time of reaction. Under these conditions there is a linear production of product in time. This initial experiment is to establish conditions that allow accurate determination of the initial velocity.

Figure 5. Time course of four enzyme catalyzed reations with increasing enzyme concentrations. The initial rate of a reaction is where there is a linear generation of product as a function of time. Assay 2 uses an optimal enzyme concentration; the assay is in the linear range at the time of assay, and activity is high enough to give a reproducible reading.

By considering only the initial velocity of a reaction, you only look at the reaction where the instantaneous velocity at any time depends on the concentration of ES: k1 k p E + S ========== E S ------------- E + P k-1 This is contrasted to the reaction after a significant concentration of product (P) has accumulated: k1 k2 k3 E + S ===== E S ====== E P ======== E + P k-1 k-2 k-3

To estimate the initial velocity of a reaction, look at the graph of A340 nm versus time (sec). The initial velocity of the reaction is the tangent drawn as accurately as possible to the earliest reliable part of the curve. The progress curve may be non-linear for several reasons: the reaction may reach equilibrium, substrate might be depleted, or the product might be inhibitory.

Your data will be very similar to data published by Michaelis and Menten in 1913. They were able to mathematically describe their data by postulating the existence of a transient enzyme-substrate complex (ES, sometimes called the Michaelis complex) and the achievement of maximal velocities of reactions (Vmax ) when all the enzyme is complexed with substrate.

Michaelis Menton Equation v - initial velocity of the reaction Vmax [ S ] Vmax - the maximum velocity v = (((((( Km - the Michaelis-Menten constant [ S ] + Km [ S ] - substrate concentration (molar)

Assumptions of the Michaelis-Menten Equation: ▪ The enzyme concentration is constant. ▪ An enzyme-substrate complex is formed. ▪ There is only one substrate. ▪ The reaction proceeds in one direction (the product is not converted back into substrate.) In simple terms, the velocity of an enzyme catalyzed reaction increases with the increase in substrate concentration up to a certain point and then becomes constant and reaches a maximum velocity, Vmax. (This maximum velocity is obtained when all the enzyme is complexed with substrate.)

Figure 6. The relationship of substrate concentration and initial velocity in enzymatic reactions.

The Michaelis constant (Km; moles/liter) is obtained by extrapolation from the curve at ½ of the Vmax. It is usually difficult to obtain Km from this curve. The Km is an indication of the relative affinity of the enzyme for the substrate; the lower the Michaelis constant, the higher the affinity.

B) Lineweaver-Burk plots. Lineweaver and Burk pointed out that rearranging the Michaelis-Menten equation by taking the reciprocal of both sides gives an equation of the straight line form y = mx + b. (Recall that this describes a straight line of slope m with the y intercept at (0, b).) Thus,

1 Km 1 1 ((( = ((( ( ( + ((( v Vmax S Vmax

Lineweaver Burk plots are more convenient. To construct these plots, 1/v (Y-axis) is graphed versus 1/ [S} (on the X-axis).

Figure 7. A Lineweaver-Burk plot.

On the Lineweaver-Burke plot, the Km (Michaelis constant) is determined by ascertaining the x-intercept (-1/ Km). It is one of the most fundamental numbers used in characterizing an enzyme.

The Vmax (maximal velocity) is dependent on the enzyme concentration. Is used to determine the turnover number of an enzyme. The turnover number is the number of molecules of substrate modified per molecule of enzyme per minute. For example, a preparation of catalase which has a turnover number of 5.6 x 106 can pick up and decompose 5.6 million H2O2 molecules each minute. This is a measure of an enzyme's efficiency.

Experiment 9 – Protocol

Enzyme Kinetics of LDH

You will be given an aliquot (100-200 (l) of LDH stock solution. You need to determine the protein concentration of this stock solution and then perform a serial dilution to end up with two working solution concentrations (250ug/ml and 25ug/ml).

When you get your LDH stock mix it for 2-3 seconds using a vortexer. Throughout the experiment, keep your LDH on ice.

A. Determining the concentration of the stock solution:

We learned the proper way to handle and clean cuvettes in a previous lab. If you have forgotten please ask the TA or Instructor for a refresher demonstration!

1. Make sure that there are no cuvettes in the spectrophotometer turret. Close the chamber lid.

2. Turn on your spectrophotometer (switch below paper roll next to power cord). After the machine has initialized (~2minutes) a list of programs should appear on the screen.

3. Using the arrow keys (bottom left side of keypad), scroll down to the “A280” program and then press <enter>. A list of program parameters should appear.

4. Press the <triangle> key (upper right side of keypad) that corresponds to the “run test” command. A box should appear that instructs you to place your blank and sample cuvettes into the machine.

5. Prepare a blank: Make 1 ml of a 1:100 dilution (using water) of the provided Tris buffer in a centrifuge tube, vortex, and then transfer the dilution to a cuvette. Do not perform a serial dilution for this step. Place the cuvette into the sample “B” holder of the spectrophotometer.

6. Prepare your sample: Make 1 ml of a 1:100 dilution (using water) of your LDH stock in a centrifuge tube, vortex, and then transfer the dilution into a new cuvette. Do not perform a serial dilution for this step. Place the cuvette into the sample “1” holder.

7. Shut the lid and press <enter> and record the A280 value.

8. Given the conversion factor that 1mg/ml of pure LDH gives an A280 of 1.5, calculate the concentration of the stock LDH. Show your work to the TA/Instructor.

B. Making the 250ug/ml and 25ug/ml working solutions:

Perform all calculations and dilutions (using Tris buffer) so that you end up with 500ul of each working LDH solution of 250ug/ml and 25ug/ml.

Have the Instructor/TA check your calculations before actually diluting the LDH stock!

If the concentration of your stock LDH is greater than 2.5mg/ml, then dilute the stock LDH with Tris buffer to make a 2.5 mg/ml solution. Then serially dilute the 2.5 mg/ml LDH to make the 250ug/ml and a 25ug/ml LDH working solutions.

If the concentration of your stock LDH is less than 2.5 mg/ml, then dilute the stock LDH with Tris buffer directly to 250 (g/ml. Then using the 250ug/ml working solution, perform a 1:10 dilution to make the 25ug/ml LDH solution.

C. Reaction Rates as a Function of Enzyme Concentration.
Measure the rate of NADH formation as a function of LDH concentration. Reaction mixtures:

Assay 1 2 3 4 5

LDH ((g/assay) 6.25 2.5 1.25 0.625 0.25

1.5 M Tris pH 8.8 915 ul 930ul 890ul 915 ul 930ul
300 mM Lactate 30 30 30 30 30
38 mM NAD 30 30 30 30 30
LDH 25 10 50 25 10 (250 (g/ml) (250 (g/ml) (25 (g/ml) (25 (g/ml) (25 (g/ml)

|A340nm Data |Assay 1 |Assay 2 |Assay 3 |Assay 4 |Assay 5 |
|0 seconds | | | | | |
|5 seconds | | | | | |
|∆ A340nm/min | | | | | |

Pre-mix all assay components except LDH in a 1.5ml centrifuge tubes using the table above.

1. While still in the A280 program, press the <Esc> key twice.

2. Check to make sure that there are NO cuvettes in the spectrophotometer turret. Close the lid.

3. Using the arrow keys (bottom left side of keypad), scroll down to the “LDH” program and then press <enter>. A list of program parameters should appear.

4. Press the <triangle> key (upper right side of keypad) that corresponds to the “run test” command. A blank graph (absorbance versus time) should appear on the screen. This program measures the change in A340nm over a period of 60 seconds.

5. Add 1 ml of A0 to a cuvette and place it into the designated sample “B” holder of the spectrophotometer. This blank cuvette remains in the machine during the entire experiment.

6. Add your pre-mixed assay components into a clean sample cuvette. Working quickly, add your LDH, pipet up and down three times, cover the cuvette with a small square of Parafilm, mix by quickly inverting three times, place the cuvette into the sample “1” holder, shut the lid, and press the <triangle> key that corresponds to “Measure Samples”. The scan will take 60 seconds.

7. Press the <triangle > key that corresponds to “tabular data” and record the A340nm for the 0 and 5 second data points. Calculate the ∆A340/min. Record this information in table above.

8. Clean the sample cuvette and repeat the assay starting at step 6 above.

D. Reaction Rate as a Function of Substrate Concentration

The initial velocities in reactions with varying lactate concentrations will be measured. These measurements can then be graphed to determine the Km and Vmax of the enzyme-substrate reaction. From this the turnover number of LDH can be determined.

The assay is performed as in section C above, except the LDH concentration in the assay is held constant and the lactate concentration is varied (Use 15 mM lactate for this experiment!). Set up the reactions as given below and measure the initial reaction rate.

Reaction mixture:

Assay 6 7 8 9 10 11
|Lactate (mM) |9 |3.15 |1.8 |1.35 |0.9 |0.45 |
|1.5 M Tris pH 8.8 |345 (l |735 (l |825 (l |855 (l |885 (l |915 (l |
|15 mM Lactate |600 |210 |120 |90 |60 |30 |
|38 mM NAD |30 |30 |30 |30 |30 |30 |
|250 (g/ml LDH |25 |25 |25 |25 |25 |25 |

|A340nm Data |Assay 6 |Assay 7 |Assay 8 |Assay 9 |Assay 10 |Assay 11 |
|0 seconds | | | | | | |
|5 seconds | | | | | | |
|∆ A340nm/min | | | | | | |

At the end of class: turn off the spectrophotometer return the buffer and water tubes throw all your solutions and cuvettes into the trash
BIOL 3380
Experiment 9 - Lab Report

Enzyme Kinetics of LDH

Name:

Circle your lab section:

| |W-AM |R-AM |F-AM |
|T–PM |W-PM |R-PM |F-PM |

Instructor:

Graduate TA:

Date:

Partner:

Experiment 9 – Lab Report

Enzyme Kinetics of LDH

This report is due at the beginning of your next scheduled laboratory session. This report must be typed. Questions should be included with your answer. All calculations and graphs may be handwritten. Attach a copy of your raw data to the back of your report. Use the provided cover page (page 9-13). (You will be penalized up to 4 points if you do not follow these directions.)

Based upon the Beer-Lambert equation (A=bc() you can convert the (A340nm/min to (NADH/min. Remember that "c" is a concentration and the units will be M/min. You will need to convert this answer to nM/min and nmol/min.

All figures/graphs/tables must include appropriate titles, legends, and labeled/ scaled axes.

1. (2pts) Present a table showing initial velocity data ((A340nm/min, (NADH[nM]/min, and ∆NADH[nmol]/min) as a function of LDH concentration.

2. (3pts) Present a graph of (NADH[nmol]/min as a function of LDH concentration.

3. (2pts) Present a table showing initial velocity data ((A340nm/min, (NADH[nM]/min, and ∆NADH[nmol]/min) as a function of lactate concentration.

4. (5pts) Present a graph of (NADH[nmole]/min as a function of lactate concentration (Michaelis-Menten plot).

5. (5pts) Convert the Michaelis-Menten plot into a Lineweaver-Burke plot.

6. (4pts) What is the Vmax as determined by the Michaelis-Menton plot? Vmax as determined by the Lineweaver-Burke plot? Why the difference (if any)? Show your work.

7. (4pts) What is the Km as determined by Michaelis-Menten plot? Km as determined by Lineweaver-Burke plot? Why the difference (if any)? Show your work.

8. (3pts) Using the data from the Lineweaver-Burke plot, what is the calculated turnover rate for a LDH monomer? An LDH monomer is composed of a 331 amino acids. An average amino acid weighs 110 Daltons. Show your work. (1 Dalton = 1g/mole)

9. Make a complete copy of your lab report before turning it in!