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Finding total concentration of enzymes

Finding total concentration of enzymes


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Sorry if my question is very basic for biology majors because I am not. I am trying to build a mathematical model of a particular pathway using systems of differential equations and in order to reduce the number of parameters, I need to know the total concentration of multiple enzymes. How are enzyme concentrations usually reported in the literature so I can google it? Also, this does not seem to me to be a straightforward measurement for professional biologists, so how are enzyme concentrations actually measured or approximated? Thanks in advance!

I am interested in the following pathway:


Hi again TheLast Cipher,

Enzymes in a cell are usually semi-quantified by the: Western Blot technique.

But there are lesser known techniques.

E.g. For a Western Blot the unit of measurement is the color intensity of the enzyme band on the blot membrane.

If I were you I would look for journal articles covering your pathway of interest.

E.g. https://iubmb.onlinelibrary.wiley.com/doi/abs/10.1002/biof.5520320129

Covers the electron transport chain (Co-enzyme Q10)

If you can get through the pay-wall & read their results and you will find:

"Western blot analysis of membrane protein isolated from keratinocytes derived from one young and old donor. A representative picture out of five experiments is displayed."

Also they used and I quote:

"Arbitrary Units"

Maybe you could comment on which pathway and I can find some assisting material.


The BioNumbers site (https://bionumbers.hms.harvard.edu/search.aspx) is often useful for these types of questions, if you're interested in an order of magnitude. However, if you want precise concentration ratios (i.e. something like "is there twice as much Ras as ERK, or more like half as much"), you are unlikely to get a definitive answer.

These may be cell type-dependent and even vary from cell to cell due to noise. You probably want to simulate your network with a variety of absolute concentration, and then the most interesting behaviors are those that are reasonably robust to the exact numbers used (results that depend on, e.g., having almost exactly the same number of molecules of Ras as ERK are likely irrelevant as they would be too "brittle" inside a living cell). Once you find out what the necessary limits are for a given behavior that you find interesting, you could think about checking whether values reported by the literature indeed tend to fall within those bounds.


IB Biology Lab Report Example: Enzyme – Catalase

This is an biology lab report example about enzyme concentration and the activity of catalase. This biology lab report example can be used in order to figure out how to write a lab report for biology courses. The headings of the biology lab report example are given below.

Aim

To investigate the effect of enzyme concentration on the activity of catalase.

Research question

How does the concentration of catalase where Saccharomyces cerevisiae is the source of enzyme affects the rate of the process of the break down of hydrogen peroxide will be measured by Vernier oxygen gas sensor where the values of temperature, pH, medium, substrate concentration are kept constant?

Hypothesis

As the enzyme concentration is increased, the activity of catalase will increase until a point and then it will stay the same.

Introduction

Chemical reactions can be speeded up by a few ways, one of which is the use of catalysts. A catalyst is a substance that speeds up a reaction by decreasing the activation energy without being affected by the reaction or being used up. Enzymes are biological catalysts that speed up reactions.

A Substrate is a substance on which Enzymes act. Enzymes are spesific for particular substrates. Enzymes work best when some conditions such as pH, temperature and ion concentration are at optimum values. Every enzyme has a spesific optimum value that it works. If the optimum values are not provided, the reaction will proceed in a low rate or there will be no reaction at all.

Hydrogen Peroxide (H2O2) is a reactive chemical that is formed as a by-product in cellular reactions. It must be removed to prevent it from disturbing chemical reactions in the cell. Catalase is an enzyme that breaks down Hydrogen Peroxide to water and oxygen, which are harmless, and thus Hydrogen Peroxide is a substrate for catalase enzyme. Hydrogen Peroxide is broken down by catalase as in the following equation

The enzyme, Catalase can be obtained from assaccharomyces cerevisiae and in this experiment Catalase will be obtained from assaccharomyces cerevisiae.

Oxygen is a product of the break down of Hydrogen Peroxide and in this experiment, the rate of reaction will be calculated by measuring the Oxygen given off with a Vernier Oxygen Probe.

Method

  1. 0.5, 0.75, 1.00, 1.25, 1.50, 1.75, 2.00, 2.25, 2.50 grams of yeast are measured with a digital balance.
  2. Each of the different amounts of yeast are poured into different beakers and the beakers are named from 1 to 9 with respect to the amount of yeast it contains.
  3. Take a syringe and fill the syringe with a 10% Hydrogen Peroxide solution.
  4. Yeast is put into the special flask of the Vernier gas sensor, which has a neck that fits with the gas sensor so that there is no gas escaping from the flash when the reaction occurs.
  5. Add the Hydrogen Peroxide solution into the flask that contains yeast. After adding the Hydrogen Peroxide solution, place the vernier gas sensor so that no oxygen gas escapes.
  6. Repeat the experiment with all the different amounts yeast and take three trials for each amount of yeast.
  7. Record your data in a table.
  8. Calculate the rate of Oxygen production for each trial and plot the graph of your data.

Practical relationships between [R], [L], Kd and [RL]

For the same concentration of added ligand and receptor, more complex formation will occur with a smaller value of Kd

Receptor and ligand are added to a total concentration of 1.0 x 10 -4 M. The Kd for the RL complex is 1.0 x 10 -4 M. What is the concentration of [R], [L] and [RL] at equilibrium?

The added ligand, L, will be partitioned between free and bound forms:

Ltotal = L + RL = 1.0 x 10 -4 M

Similarly, the added receptor, R, will be partitioned between free and bound forms:

Rtotal = R + RL = 1.0 x 10 -4 M

Substituting these values of L and R into the expression for Kd yields:

This is a quadratic with values a = 1, b = -3.0 x 10 -4 and c = 1.0 x 10 -8 . This yields two possible values for RL:

Since the maximum possible value for [RL] is 1.0 x 10 -4 (i.e. given the starting concentrations of R and L, this is the most amount of RL that can be formed), the first result is not possible and [RL] = 3.82 X 10 -5 M

Thus, 38.2% of the added receptor is in the complex formation (i.e. 3.82 X 10 -5 M/1.0 x 10 -4 )x100%

Same amount of added R and L, but this time the Kd for the RL complex is 1.0 x 10 -6 M. What is the concentration of [R], [L] and [RL] at equilibrium?

his is a quadratic with values a = 1, b = -2.01 x 10 -4 and c = 1.0 x 10 -8 . This yields two possible values for RL:

Since the maximum possible value for [RL] is 1.0 x 10 -4 (i.e. given the starting concentrations of R and L, this is the most amount of RL that can be formed), the first result is not possible and [RL] = 9.05 X 10 -5 M

Thus, 90.5% of the added receptor is in the complex formation (i.e. 9.05 X 10 -5 M/1.0 x 10 -4 )x100%

For the same concentration of added ligand and receptor, more complex formation will occur with a smaller value of Kd


8.4.1. The Significance of KM and Vmax Values

Conceptual Insights, Steady-State Enzyme Kinetics

Learn how the kinetic parameters KM and Vmax can be determined experimentally using the enzyme kinetics lab simulation in this media module.

The Michaelis constant, KM, and the maximal rate, Vmax, can be readily derived from rates of catalysis measured at a variety of substrate concentrations if an enzyme operates according to the simple scheme given in equation 23. The derivation of KM and Vmax is most commonly achieved with the use of curve-fitting programs on a computer (see the appendix to this chapter for alternative means of determining KM and Vmax). The KM values of enzymes range widely (Table 8.5). For most enzymes, KM lies between 10 -1 and 10 -7 M. The KM value for an enzyme depends on the particular substrate and on environmental conditions such as pH, temperature, and ionic strength. The Michaelis constant, KM, has two meanings. First, KM is the concentration of substrate at which half the active sites are filled. Thus, KM provides a measure of the substrate concentration required for significant catalysis to occur. In fact, for many enzymes, experimental evidence suggests that KM provides an approximation of substrate concentration in vivo. When the KM is known, the fraction of sites filled, fES, at any substrate concentration can be calculated from

Table 8.5

KM values of some enzymes.

Second, KM is related to the rate constants of the individual steps in the catalytic scheme given in equation 9. In equation 15, KM is defined as (k-1 + k2)/k1. Consider a limiting case in which k-1 is much greater than k2. Under such circumstances, the ES complex dissociates to E and S much more rapidly than product is formed. Under these conditions (k-1 >> k2),

The dissociation constant of the ES complex is given by

In other words, KM is equal to the dissociation constant of the ES complex if k2 is much smaller than k-1. When this condition is met, KM is a measure of the strength of the ES complex: a high KM indicates weak binding a low KM indicates strong binding. It must be stressed that KM indicates the affinity of the ES complex only when k-1 is much greater than k2.

The maximal rate, Vmax, reveals the turnover number of an enzyme, which is the number of substrate molecules converted into product by an enzyme molecule in a unit time when the enzyme is fully saturated with substrate. It is equal to the kinetic constant k2, which is also called kcat. The maximal rate, Vmax, reveals the turnover number of an enzyme if the concentration of active sites [E]T is known, because

For example, a 10 -6 M solution of carbonic anhydrase catalyzes the formation of 0.6 M H2CO3 per second when it is fully saturated with substrate. Hence, k2 is 6 × 10 5 s -1 . This turnover number is one of the largest known. Each catalyzed reaction takes place in a time equal to 1/k2, which is 1.7 μs for carbonic anhydrase. The turnover numbers of most enzymes with their physiological substrates fall in the range from 1 to 10 4 per second (Table 8.6).

Table 8.6

Maximum turnover numbers of some enzymes.


CoEnzymes & CoFactors

Kinetics of Enzyme-Substrate Chemistry

Homogeneous Catalysts

Homogeneous catalysts interact with the reactants in the same phase (i.e: turning a substrate into a product at a faster rate). The homogeneous catalysts do not change their current states, unlike heterogeneous catalysts. By "states" we mean the phase state, which indicates either solid, liquid, or gas. If homogeneous catalysts are solid, then they will remain solid after the reaction is completed the same is true for liquid and gas homogeneous catalysts. Although many biological enzymes are heterogeneous, there are some homogeneous enzymes that remain in their same state after the reaction, such as in an immunoassay (EIA).

Heterogeneous Catalysts

Heterogeneous catalysts are catalysts that speed up the rate of reactions by allowing them to occur on a solid surface. An example of a heterogeneous catalyst is a clay DNA polymer scaffolding, where the DNA's individual purines and pyrimidines link together on a clay surface to enable more secure bonding.


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Exploring Enzymes

Introduction
Have you ever wondered how all the food that you eat gets digested? It is not only the acid in your stomach that breaks down your food&mdashmany little molecules in your body, called enzymes, help with that too. Enzymes are special types of proteins that speed up chemical reactions, such as the digestion of food in your stomach. In fact, there are thousands of different enzymes in your body that work around-the-clock to keep you healthy and active. In this science activity you will investigate one of these enzymes, called catalase, to find out how it helps to protect your body from damage.

Background
Enzymes are essential for our survival. These proteins, made by our cells, help transform chemicals in our body, functioning as a catalyst. A catalyst gets reactions started and makes them happen faster, by increasing the rate of a reaction that otherwise might not happen at all, or would take too long to sustain life. However, a catalyst does not take part in the reaction itself&mdashso how does this work? Each chemical reaction needs a minimum amount of energy to make it happen. This energy is called the activation energy. The lower the activation energy of a reaction, the faster it takes place. If the activation energy is too high, the reaction does not occur.

Enzymes have the ability to lower the activation energy of a chemical reaction by interacting with its reactants (the chemicals doing the reacting). Each enzyme has an active site, which is where the reaction takes place. These sites are like special pockets that are able to bind a chemical molecule. The compounds or molecules the enzyme reacts with are called their substrates. The enzyme pocket has a special shape so that only one specific substrate is able to bind to it, just like only one key fits into a specific lock. Once the molecule is bound to the enzyme, the chemical reaction takes place. Then, the reaction products are released from the pocket, and the enzyme is ready to start all over again with another substrate molecule.

Catalase is a very common enzyme that is present in almost all organisms that are exposed to oxygen. The purpose of catalase in living cells is to protect them from oxidative damage, which can occur when cells or other molecules in the body come into contact with oxidative compounds. This damage is a natural result of reactions happening inside your cells. The reactions can include by-products such as hydrogen peroxide, which can be harmful to the body, just as how a by-product of a nice bonfire can be unwanted smoke that makes you cough or stings your eyes. To prevent such damage, the catalase enzyme helps getting rid of these compounds by breaking up hydrogen peroxide (H2O2) into harmless water and oxygen. Do you want to see the catalyze enzyme in action? In this activity you will disarm hydrogen peroxide with the help of catalase from yeast.

  • Safety goggles or protective glasses
  • Five teaspoons of dish soap
  • One package of dry yeast
  • Hydrogen peroxide, 3 percent (at least 100 mL)
  • Three tablespoons
  • One teaspoon
  • Five 16-ounce disposable plastic cups
  • Tap water
  • Measuring cup
  • Permanent marker
  • Paper towel
  • Workspace that can get wet (and won't be damaged by any spilled hydrogen peroxide or food-colored water)
  • Food coloring (optional)

Preparation

  • Take one cup and dissolve the dry yeast in about one-half cup of warm tap water. The water shouldn't be too hot but close to body temperature (37 Celsius). Let the dissolved yeast rest for at least five minutes.
  • Use the permanent marker to label the remaining four cups from one to four.
  • To all the labeled cups, add one teaspoon of dish soap.
  • To cup one no further additions are made at this point.
  • Before using the hydrogen peroxide, put on your safety goggles to protect your eyes. In case you spill hydrogen peroxide, clean it up with a wet paper towel. If you get it on your skin, make sure to rinse the affected area with plenty of water.
  • To cup two, add one tablespoon of 3 percent hydrogen peroxide solution. Use a fresh spoon for the hydrogen peroxide.
  • To cup three, add two tablespoons of the hydrogen peroxide.
  • To cup four, add three tablespoons of the hydrogen peroxide.
  • Optionally, you can add a drop of food color to each of the labeled cups. (You can choose a different color for each one for easy identification)
  • Take cup number one and place it in front of you on the work area. With a fresh tablespoon, add one tablespoon of the dissolved yeast solution to the cup and swirl it slightly. What happens after you add the yeast? Do you see a reaction happening?
  • Place cup number two in front of you and again add one tablespoon of yeast solution to the cup. Once you add the enzyme,does the catalase react with the hydrogen peroxide? Can you see the reaction products being formed?
  • Add one tablespoon of yeast solution to cup number three. Do you see the same reaction taking place? Is the result different or the same compared to cup number two?
  • Finally, add one tablespoon of yeast solution to cup number four. Do you see more or less reaction products compared to your previous results? Can you explain the difference?
  • Place all four cups next to each other in front of you and observe your results. Did the enzymatic reaction take place in all of the cups or was there an exception? How do the results in each cup look different? Why do you think this is the case?
  • Now, take cup number one and add one additional tablespoon of 3 percent hydrogen peroxide to the cup. Swirl the cup slightly to mix the solution. What happens now? Looking at all your results, what do you think is the limiting factor for the catalase reaction in your cups?
  • Extra: Repeat this activity, but this time do not add dish soap to all of the reactions. What is different once you remove the dish soap? Do you still see foam formation?
  • Extra: So far you have observed the effect of substrate (H2O2) concentration on the catalase reaction. What happens if you keep the substrate concentration constant but change the concentration of the enzyme? Try adding different amounts of yeast solution to three tablespoons of hydrogen peroxide, starting with one teaspoon. Do you observe any differences, or does the concentration of catalase not matter in your reaction?
  • Extra: What happens if the environmental conditions for the enzyme are changed? Repeat the catalase reaction but this time vary conditions such as the pH by adding vinegar (an acid) or baking soda (a base), or change the reaction temperature by heating the solution in the microwave. Can you identify which conditions are optimal for the catalase reaction? Are there any conditions that eliminate the catalase activity?
  • Extra: Can you find other sources of catalase enzyme that you could use in this activity? Research what other organisms, plants or cells contain catalase and try using these for your reaction. Do they work as well as yeast?

Observations and results
You probably saw lots of bubbles and foam in this activity. What made the foam appear? When the enzyme catalase comes into contact with its substrate, hydrogen peroxide, it starts breaking it down into water and oxygen. Oxygen is a gas and therefore wants to escape the liquid. However, the dish soap that you added to all your solutions is able to trap the gas bubbles, which results in the formation of a stable foam. As long as there is enzyme and hydrogen peroxide present in the solution, the reaction continues and foam is produced. Once one of both compounds is depleted, the product formation stops. If you do not add dish soap to the reaction, you will see bubbles generated but no stable foam formation.

If there is no hydrogen peroxide present, the catalase cannot function, which is why in cup one you shouldn't have seen any bubble or foam production. Only when hydrogen peroxide is available, the catalase reaction can take place as you probably observed in the other cups. In fact, the catalase reaction is dependent on the substrate concentration. If you have an excess of enzyme but not enough substrate, the reaction will be limited by the substrate availability. Once you add more hydrogen peroxide to the solution, the reaction rate will increase as more substrate molecules can collide with the enzyme, forming more product. The result is an increasing amount of foam produced in your cup as you increase the amount of H2O2 in your reaction. You should have seen more foam being produced once you added another tablespoon of hydrogen peroxide to cup one, which should have resulted in a similar amount of foam as in cup two. However, at some point you will reach a substrate concentration at which the enzyme gets saturated and becomes the limiting factor. In this case you have to add more enzyme to speed up the reaction again.

Many other factors affect the activity of enzymes as well. Most enzymes only function under optimal environmental conditions. If the pH or temperature deviates from these conditions too much, the enzyme reaction slows down significantly or does not work at all. You might have noticed that when doing the extra steps in the procedure.

Cleanup
Pour all the solutions into the sink and clean all the spoons with warm water and dish soap. Wipe your work area with a wet paper towel and wash your hands with water and soap.

This activity brought to you in partnership with Science Buddies


How to Calculate Units of Concentration

Once you have identified the solute and solvent in a solution, you are ready to determine its concentration. Concentration may be expressed several different ways, using percent composition by mass, volume percent, mole fraction, molarity, molality, or normality.

Percent Composition by Mass (%)

This is the mass of the solute divided by the mass of the solution (mass of solute plus mass of solvent), multiplied by 100.
Example:

Determine the percent composition by mass of a 100 g salt solution which contains 20 g salt.
Solution:

20 g NaCl / 100 g solution x 100 = 20% NaCl solution

Volume Percent (% v/v)

Volume percent or volume/volume percent most often is used when preparing solutions of liquids. Volume percent is defined as:
v/v % = [(volume of solute)/(volume of solution)] x 100%
Note that volume percent is relative to the volume of the solution, not the volume of solvent. For example, wine is about 12% v/v ethanol. This means there is 12 ml ethanol for every 100 ml of wine. It is important to realize liquid and gas volumes are not necessarily additive. If you mix 12 ml of ethanol and 100 ml of wine, you will get less than 112 ml of solution.
As another example, 70% v/v rubbing alcohol may be prepared by taking 700 ml of isopropyl alcohol and adding sufficient water to obtain 1000 ml of solution (which will not be 300 ml).

Mole Fraction (X)

This is the number of moles of a compound divided by the total number of moles of all chemical species in the solution. Keep in mind, the sum of all mole fractions in a solution always equals 1.
Example:
What are the mole fractions of the components of the solution formed when 92 g glycerol is mixed with 90 g water? (molecular weight water = 18 molecular weight of glycerol = 92)
Solution:

90 g water = 90 g x 1 mol / 18 g = 5 mol water
92 g glycerol = 92 g x 1 mol / 92 g = 1 mol glycerol
total mol = 5 + 1 = 6 mol
xwater = 5 mol / 6 mol = 0.833
x glycerol = 1 mol / 6 mol = 0.167
It's a good idea to check your math by making sure the mole fractions add up to 1:
xwater + xglycerol = .833 + 0.167 = 1.000

Molarity (M)

Molarity is probably the most commonly used unit of concentration. It is the number of moles of solute per liter of solution (not necessarily the same as the volume of solvent!).
Example:

What is the molarity of a solution made when water is added to 11 g CaCl2 to make 100 mL of solution? (The molecular weight of CaCl2 = 110)
Solution:

11 g CaCl2 / (110 g CaCl2 / mol CaCl2) = 0.10 mol CaCl2
100 mL x 1 L / 1000 mL = 0.10 L
molarity = 0.10 mol / 0.10 L
molarity = 1.0 M

Molality (m)

Molality is the number of moles of solute per kilogram of solvent. Because the density of water at 25°C is about 1 kilogram per liter, molality is approximately equal to molarity for dilute aqueous solutions at this temperature. This is a useful approximation, but remember that it is only an approximation and doesn't apply when the solution is at a different temperature, isn't dilute, or uses a solvent other than water.
Example:
What is the molality of a solution of 10 g NaOH in 500 g water? (Molecular weight of NaOH is 40)
Solution:

10 g NaOH / (40 g NaOH / 1 mol NaOH) = 0.25 mol NaOH
500 g water x 1 kg / 1000 g = 0.50 kg water
molality = 0.25 mol / 0.50 kg
molality = 0.05 M / kg
molality = 0.50 m

Normality (N)

Normality is equal to the gram equivalent weight of a solute per liter of solution. A gram equivalent weight or equivalent is a measure of the reactive capacity of a given molecule. Normality is the only concentration unit that is reaction dependent.
Example:

1 M sulfuric acid (H2SO4) is 2 N for acid-base reactions because each mole of sulfuric acid provides 2 moles of H + ions. On the other hand, 1 M sulfuric acid is 1 N for sulfate precipitation, since 1 mole of sulfuric acid provides 1 mole of sulfate ions.

  1. Grams per Liter (g/L)
    This is a simple method of preparing a solution based on grams of solute per liter of solution.
  2. Formality (F)
    A formal solution is expressed regarding formula weight units per liter of solution.
  3. Parts per Million (ppm) and Parts per Billion (ppb)Used for extremely dilute solutions, these units express the ratio of parts of solute per either 1 million parts of the solution or 1 billion parts of a solution.
    Example:

    A sample of water is found to contain 2 ppm lead. This means that for every million parts, two of them are lead. So, in a one gram sample of water, two-millionths of a gram would be lead. For aqueous solutions, the density of water is assumed to be 1.00 g/ml for these units of concentration.

Finding total concentration of enzymes - Biology

Which of the following graphs shows the results of reaction rate vs substrate concentration for an allosteric enzyme in the absence and presence of an allosteric inhibitor?

Allosteric enzymes

Binding of effectors to regulatory subunits

Allosteric enzymes may also have regulatory subunits that bind either activators or inhibitors. Activators and inhibitors are termed "effectors." Inhibitors cause the allosteric enzyme to adopt the inactive shape. Activators promote the active shape.

An equilibrium exists between the active and inactive shapes. The amount of active and inactive enzyme is dependent on the relative concentrations of substrate and inhibitor, as suggested by the diagram:

The binding of an allosteric inhibitor causes the enzyme to adopt the inactive conformation, and promotes the cooperative binding of a second inhibitor.

An excess of substrate can overcome the inhibitor effect. Substrate binding causes the enzyme to assume the active conformation, and promotes the cooperative binding of additional substrate, leading to product formation.


D. INFLUENCE OF IONIC CONCENTRATION ON CATALASE ACTIVITY

Repeat steps 1-6 (using three catalase-soaked disks) using different concentrations of NaCl in the substrate solution.

1.1.5% H 2 O 2 solution containing 10% NaCl.

2.1.5% H 2 O 2 solution containing 2% NaCl.

3.1.5% H 2 O 2 solution containing 0% NaCl.

Graph your results as ionic concentration (x-axis) vs. Velocity (Vol of O 2 /min) and explain the relationship which appears to exist between the concentration of sodium chloride ions and catalase activity.

E. INFLUENCE OF TEMPERATURE ON CATALASE ACTIVITY

Repeat steps 1-6 (using three catalase-soaked disks and 10 mls of 3% H 2 O 2 ) in each of the different temperature water baths.

1.Three disks soaked in the "boiled" catalase extract (100C) in a room temperature water-bath and reaction vessel with 3% H 2 O 2 . Allow equipment to stabilize at room temperature for 2-3 minutes before making run. This simulates a boiling water bath, without the danger.

2.A water bath that has been chilled with ice to 5C. (Keep adding ice to keep temperature close to 5C. Allow reaction vessel to stabilize in water bath for 2-3 minutes before making the run. Record the temperature.

3.A water bath that has been warmed to 33C. Allow temperature of the reaction vessel to stabilize in the water bath for 2-3 minutes at 33C before making the run. This temperature is about 91.4F, which in the body, is equivalent to a state of hypothermia .

4.A water bath that has been warmed to 37C. Allow temperature of reaction vessel to stabilize in the water bath for 2-3 minutes before making the run. This temperature is about 98.6F, which is equivalent to standard body temperature.

5.A water bath that has been warmed to 41.1C. Allow temperature of reaction vessel to stabilize for 2-3 minutes in the water bath before making the run. This temperature is about 106F, which in the body, is equivalent to a state of hyperthermia .

6. Use the results of the trial from part A, step1 (the catalase reaction with 3 disks at room temperature) for the Room temperature data point requested on the class spreadsheet.

Graph your results as temperature (x-axis) vs. Velocity (Vol of O 2 /min) and explain the relationship which appears to exist between temperature and catalase activity.

F. INFLUENCE OF ENZYME INHIBITOR ON CATALASE ACTIVITY

Repeat steps 1-6 using 3 disks soaked with catalase at different concentration of copper sulfate. You will be given five tubes containing the catalase solution, five beakers containing copper sulfate (CuSO 4 ) solution at various concentrations and five weigh boats.

  • Weigh boat 1: Pour all 5 ml of catalase solution, then add 5 drops of 0.1 M CuSO 4 solution. Mix and wait exactly 5 mintues before making run.
  • Weigh boat 2: Pour all 5 ml of catalase solution, then add 5 drops of 0.25 M CuSO 4 solution. Mix and wait exactly 5 mintues before making run.
  • Weigh boat 3: Pour all 5 ml of catalase solution, then add 5 drops of 0.5 M CuSO 4 solution. Mix and wait exactly 5 mintues before making run.
  • Weigh boat 4: Pour all 5 ml of catalase solution, then add 5 drops of 0.75 M CuSO 4 solution. Mix and wait exactly 5 mintues before making run.
  • Weigh boat 5: Pour all 5 ml of catalase solution, then add 5 drops of 1.0 M CuSO 4 solution. Mix and wait exactly 5 mintues before making run.

Plan your experiment CAREFULLY.

Graph your results as copper sulfate concentration (M) (x-axis) vs. Velocity (Vol of O 2 /min) and explain the relationship, which appears to exist between copper sulfate concentration and catalase activity.

The 0 M copper sulfate data point is the value generated in Part A #1: three disks soaked in catalase.

G. INSTRUCTIONS FOR ENZYME PAPER - rough draft is due the week of February 21.

As a group, complete a full scientific paper. This paper will include parts A-F, even though your group only completed part of the experiment.

Here are a couple of internet resources that will help you to write a scientific paper on the experiment.

Writing a Lab Report: a link to an Introductory Biology course at UNCG dealing with writing a lab report.

Writing Lab reports and scientific papers: by Warren D. Dolphin at Iowa State University

Use the Long Island University library website for the proper MLA format for citing your references in your report.

Note 2: Use your text and the internet resources at the beginning of this lab to discuss the basic principles of enzyme function. Also include in your background information about catalase activity. Also, in discussing your results, in the discussion section you will also need to do some research to explain the effect of the various conditions have on catalase activity.

Abramoff, P. & Thomson, R. G. 1976. An Experimental Approach to Biology, pp. 65-73. W. H. Freeman and Co.

Starr, C. & Taggart, R. 1981. Biology: The Unity and Diversity of Life, pp. 84-94. Wadsworth Publishing Company.

Weisz, P. D., & Tarp, F. H. 1967. Laboratory Manual for Elements of Biology, pp. 25-32. McGraw-Hill Book Co.


Watch the video: Molarity, Molality, Volume u0026 Mass Percent, Mole Fraction u0026 Density - Solution Concentration Problems (January 2023).