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Lab 4: Enzymes (Team 5) (FINAL)

Page history last edited by Jean-Pierre Jacob 10 years, 5 months ago

A.  Learning Objectives 

In this lab, students will:

• analyze the effect of catechol oxidase on the production of benzoquinone.

• design and conduct experiments to study how physical conditions affect enzyme activity.

• plot data graphically.


B. Textbook Correlation  
Please review Section 6.2 Enzymes and Ribozymes in Chapter 6: An Introduction to Energy, Enzymes, and Metabolism to assist in writing the intrroduction and researching the experiment.


C.  Introduction

Please write a two paragraph introduction to enzymes;


 Enzymes are found in all living cells. Enzymes are essentially proteins so they would be made up of a polypeptide is fold and twisted upon itself to form a tertiary structure of a protein that has a specific 3-D shape. Enzymes have a location on it that is called the active site. The active site is where the chemical reaction takes place. Substrates, the reacting molecules, bind to the active site in order to participate in the chemical reaction. The active site has a specific shape that only one type of substrate can bind to. Enzymes also have a allosteric site, which is a location outside the active site where noncompetitive inhibitors can bind to. When the binding of an enzyme and its substrate occurs, it results in the production of an enzyme-substrate complex.A very unique feature of enzymes is their ability to bind to their substrates with a high level of specificity. This means that only certain substrates can bind to the enzyme's active site. When substrates bind to an enzyme, the enzyme undergoes a conformational change. These conformational changes cause the substrates to bind more tightly to the substrates. This is referred to as the induced fit theory. After this occurs in an enzyme, only then will the reaction take place.

     Enzymes function as catalysts in the cell. Catalysts speed up a chemical reaction by lowering the activation energy. The activation energy is the initial input of energy that allows for a reaction to overcome the electron repulsion between the reacting molecules. Enzymes are able to lower the activation energy because enzymes can bind small reactants, or substrates, to its active site. When the substrates are bound to the enzyme, the bonds in the reacting molecules are strained which helps them achieve the transition state. The Transition state is where the original bonds have been stretched to their limit. An enzymes' active site also positions the substrates very close to each other. This favors the formation of new bonds and lowers the activation energy of a chemical reaction. Another way to lower the activation energy is by changing the local environment of the reactants through the course of a chemical reaction. Sometimes, enzymes also lower the activation energy by directly participating in a chemical reaction. The Induced fit also helps to lower the activation energy as the conformational change of the enzyme allows it to bind to the substrates even tighter. Overall, Enzymes speed up the rate of chemical reaction by lowering the activation energy that is initially required to get the reaction started. Enzyme-catalyzed reactions occurs millions of times faster than uncatalyzed reactions as the activation energy is higher without an enzyme.



Paragraph #1:  Discuss the structure/function of enzymes.  In your discssion, address the make-up of most enzymes, the role of the active site and its impact on specificity, and the idea behind the induced fit theory,  Also discuss activation energy and how enzymes speed up chemical reactions by impacting the activation energy. 







This is an illustration of a typical enzyme, containing a multitude of proteins that work together to form a functional, 3-dimensional molecule that catalyzes many chemical reactions in the cell.


This is an illustration of an enzyme's impact on activation energy. Activation energy is the amount of energy needed to initiate a chemical reactions. Enzymes will lower activation energy, allowing chemical reactions to occur more efficiently and quicker. For example, an enzyme might pull two molecules together, allowing them to form covalent bonds, which would speed up the combination of those two molecules, as well as lower the amount of activation energy that would've been required without the assistance of the enzyme. As the picture above emphasizes, without enzymes, chemical reactions require a higher amount of activation energy.


Above is a chemical reaction known as a catabolic reactions. This type of reaction involves the breaking down of a complex molecule into simpler molecules. The breakage of complex molecules is usually completed through a process known as hydrolysis, which involves the addition of a water molecule to break a complex molecule into its building blocks. Enzymes will assist in catabolic reactions by holding the complex molecule and allowing the h20 molecule to bind and break apart the complex molecule into simpler parts.





Paragraph #2:  Discuss how enzyme activity is regulated in a cell.  Include in the discussion the idea of enzyme saturation, how saturation is overcome, the physical requirements for optimal enzyme activity, and the role of inhibitors (both competitive and non-competitive).  When discussing inhibitors, include the idea behind allosteric regulation.


     Enzyme activity is regulated in a cell in a variety of ways. Enzyme activity may be regulated by substrate availability, environmental factors (temperature, pH, salinity), and inhibition. As the number of substrates increases, activity of the enzyme increases. In other words, if a certain enzyme accepts h20, the more h20 present in the enzyme's surroundings, the higher enzymatic activity. Enzyme saturation is reached when there is as many, or more, substrates (h20) as there are enzymes. Thus, the enzymes are "saturated," meaning little if any enzymes are free to accept substrates at a given time. Increasing the substrate levels will hardly change this, so the only way to overcome this is to either reduce substrate concentration or increase enzyme concentration. Enzymes are proteins, meaning at certain temperatures, salinity concentrations and pH levels they may denature and become dysfunctional. Enzymes require a specific temperatures (usually body temp. ~37C), and as temperature increases, enzyme activity increases (up to a certain point at which the enzyme will denature). As salinity increases, enzyme activity increases because the salinity may promote the attachment of hydrophobic substances to the enzyme. Lastly, optimal pH varies depending on the enzyme's environment (e.g. enzymes in the stomach thrive in a pH of 2). However, if an environment's environment is not appropriate, the protein will become dysfunctional due to denaturation. Inhibition also regulates enzyme activity. Non-competitive inhibitors are substrates that will bind to the allosteric site and affect a substrate's ability to bind to the enzyme's active site. When a molecule binds to the allosteric site, it will make the enzyme unusable by its designated substrate, decreasing overall enzyme activity in the cell. Competitive inhibition involve molecules competing with designated substrates for binding to the active site of the enzyme. If a competitive inhibitor binds to the enzyme before the substrate, enzyme activity will be reduced. A way to decrease competitive inhibition would be increasing substrate concentration so that it is more likely for the substrate to bind than the less numerous competitive inhibitors. During allosteric regulation, certain molecules are able to bind to the allosteric site of the enzyme. These molecules are called allosteric regulators. When allosteric regulators bind to the allosteric site, they can either activate the enzyme so it can carry out with its reaction or it can inhibit the enzyme and prevent any reactions from taking place.







This illustration shows that as the substrate levels increase, the substrate levels starts to approach the V max. The V max indicates the point at which the substrate concentrations are saturated. At this point, few if not any, enzymes are empty to allow for the binding of the substrate to the enzyme's active site.


This image shows the difference between the mechanisms of competitive inhibition and noncompetitive inhibition. During competitive inhibition, the competitive inhibitor binds to the enzyme's active site. This prevents the binding of substrates to the enzyme, ultimately making the enzyme useless. During noncompetitive inhibition, the noncompetitive inhibitor will bind to the allosteric site of the enzyme. This will cause the enzyme to undergo a conformational change that changes the shape of the active site. Ultimately preventing the substrates from binding to the active site and rendering the enzyme as useless.


This image shows the different parts of allosteric regulation. During allosteric regulation, molecules bind to the allosteric site of the enzyme. These molecules are called an allosteric regulator. Some allosteric regulators activate the enzyme while others inhibit it. 


In today’s exercise you will first observe the actions of the enzyme catechol oxidase. After this exercise you will be ready to design two experiments on your own to test the physical requirements for optimal enzyme activity.


D.  Catechol Oxidase Activity

In today’s exercise the enzyme you will use is catechol oxidase.  In plants this copper-containing enzyme creates brown pigment when exposed to air (specifically oxygen), and it is the reason fruits turn brown after they are sliced.  The brown color is due to the production of the product benzoquinone, a substance that is toxic to food-spoiling bacteria.   When the peel is damaged, oxygen can then react with the catechol, protecting the fruit.


In this experiment, we will test catechol oxidase activity.   The enzyme is extracted from potatoes using a blender and is referred to as potato extract in the subsequent experiments.


Experimental Procedure:

1. Label 3 test tubes 1–3.

2. Pipette the amount of catechol and water into the appropriate test tube as outlined in Table 1. Do not add the  catechol oxidase to all tubes until just before starting the incubation in step 3.



mL of Catechol

mL of Water

mL of Catechol Oxidase




1 mL (20 drops)




1 mL (20 drops)




0 mL (0 drops)


3.  Place the test tubes in the 37⁰C water bath for 10 minutes. 

4. Record your results in the table above. Use the following scale: 

0       no color change

1       little color change

2       more color change

3      dark color change






Benzoquinone was produced. 


Based off the results, benzoquinone was produced, even though it is not supposed to. 


During the time of the experiment, benzoquinone was not produced. 



1. Which tube is the negative control?  Which tube is your positive control? 


 Tube 2 is the negative control because no catechol (substrate) is present for the catechol oxidase (enzyme) to act on. This means that no benzoquinone will be produced and no reaction will take place. Since we expect no benzoquinone to be produced, test tube 2 is the negative control. Tube 1 is the positive control because catechol (substrate) and catechol oxidase (enzyme) are both present in the test tube. This means that the reaction will take place for sure and benzoquinone will be produced. Since we expect benzoquinone to be produced, test tube 1 is the positive control. 


2. What would it mean if tube 2 turned brown?


If test tube 2 turned brown, like it did a little with our experiment, some type of contamination must have occurred. Some form of contamination occurred that resulted in catechol being present in the tube and for the enzyme catechol oxidase activity to act on the catechol. The catechol might have entered the test tube through a contaminated test tube.



E.  Design an Experiment to Study Enzyme Activity Under Different Physical Conditions. 

Protein activity is highly dependent on its three-dimensional structure.  Conditions that cause a protein to denature (unfold) results in the loss of protein activity.  Environmental deviations from optimal cause an enzyme to lose activity.  Question:  What is the optimal temperature for catechol oxidase activity?  What is the ideal pH for catechol oxidase activity?  What is the optimal salinity for catechol oxidase activity?  Use the experiment from section D as a template.  Remember to include positive and negative controls when applicable.  Make sure you take photographic images of your results and a video of your procedure explaining how you designed the experiment.


Materials Provided:

  • test tubes
  • plastic pipettes
  • catechol
  • potato extract
  • water baths (3) and hot plate
  • ice
  • thermometers 
  • phosphate buffers ranging from pH of 2 to 12 (actual buffers available (in pH units): 2, 4, 6, 7, 8, 10, 12) 
  • distilled water 
  • 10% NaCl stock solution


Experiment #1: Temperature 

1.  Hypothesis: 


The optimal temperature of catechol oxidase activity will be at 37 degrees celsius because 37 degrees celsius is the body temperature and if the temperature gets too high or cold in the body, then the enzyme activity will not function at its optimal temperature. Thus, 37 degrees celsius will be the optimal temperature for catechol oxidase activity.



  • 8 test tubes

  • ice

  • 3 water baths

  • hot plate

  • plastic pipettes

  • potato extract

  • catechol

  • thermometers


Posititive and Negative Controls are not applicable in this experiment




2.  Experimental Design:


  1. Label 6 tests tubes with numbers from 1-6

  2. Prepare 3 water baths

    1. At 40, 60, and 80 degrees Celsius

  3. Prepare the 3 beakers with water (using ice or a hot plate when necessary)

    1. At 0, 20, and 100 degrees Celsius

  4. Add 1 mL of potato extract to each of the 6 tubes.

  5. Place test tube 1 in the 0 degrees Celsius beaker

  6. Place test tube 2 in the 20 degrees Celsius beaker

  7. Place test tube 3 in the 40 degrees Celsius water bath

  8. Place test tube 4 in the 60 degrees Celsius water bath

  9. Place test tube 5 in the 80 degrees Celsius water bath

  10. Place test tube 6 in the 100 degrees Celsius beaker

  11. Let them sit for 5 minutes

  12. After about 5 minutes add 1 mL of catechol to each of the 6 test tubes

  13. Place the 6 test tubes back into their appropriate water baths or beakers

  14. Let them sit for 10 minutes

  15. Record the intensities of the color of the solution in the test tube

  16. Record the intensity of the color based on a number system0-5  (0 being no color and 5 being a very dark brown)

    1. The test tube with the darkest color provides the enzyme with the optimal temperature


Test Tube  Temperature  Intensity of Color 
0 degrees Celsius 
2 20 degrees Celsius 
40 degrees Celsius 

60 degrees Celsius 
5 80 degrees Celsius
6 100 degrees Celsius




Experiment #2: pH 

1.  Hypothesis: 


 It will function best at a pH of 7 because any pH other than neutral will cause changes on a molecular level within the enzyme. For example, if there is more H+ in the environment, the enzyme’s functional groups may respond a certain way, altering the enzyme’s structure. Thus, the optimal pH must be 7 (neutral)



test tubes

plastic pipettes


potato extract

phosphate buffers ranging from pH of 2 to 12 (actual buffers available (in pH units): 2, 4, 6, 7, 8, 10, 12)



2.  Experimental design:  


1. Take 7 test tubes, filling each of them with 40 drops of the 7 different pH buffers (2,4,6,7,8,10,12) with the plastic pipettes.

2. Add 10 drops of potato extract (which contains Catechol Oxidase) to each test tube.

3. Add 10 drops of Catechol to each test tube.

4. Shake the mixture.

5. Observe the intensity of color every 5 minutes for 20 minutes.

6. Record the intensity of color based on a number system 0-5, 0 being no color and 5 being a very dark brown.


(The use of positive and negative controls is not applicable in this experiment).


Interpretation of Results:

The pH that yields the most intense color (which should be the darkest brown) represents the pH in which Catechol Oxidase functions optimally.  We know that the test tube with the darkest brown is the one that catechol oxidase functions optimal in because when catechol oxidase oxidizes catechol it produces a brown pigment, benzoquinon. The more benzoquinone (brown coloring), the more oxidation that occurred.


Test Tube pH Intensity of Color







Experiment #3: Salt 

1.  Hypothesis: 


My hypothesis is that the optimal salinity is 1ml of the stock solution because it doesn’t have enough salt concentration to denature the proteins but it has enough to keep them together.


Postivive and Negative controls are not applicable in this experiment.


2.  Experimental design:  


  1. Put 6 test tubes on the rack (label them 1-6)

  2. Fill each test tube with 5 ml of deionized water

  3. In test tubes 1, 2, 3, 4, and 5 add the 10% NaCl in amounts of 1ml, 2ml, 3ml, 4ml, 5ml, respectively.

  4. In test tube 6, do not put any NaCl as this our negative control.

  5. Add catechol to all 6 test tubes

  6. Next add the potato extract 

  7. Monitor the tubes for 5 minutes  

  8. Record the results



Please embed your presentation below.


Please watch the voicethreads in order. Our presentation was too big to fit onto one voicethread so we had to split it up into two voicethread. Please also click on the youtube links below. These links show some aspects of our pH experiment. Thank you.





Here is pH video 1 where we are swirling the contents in the beakers




Here is pH video 2 where we show which test tubes have which pH 








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