By the end of this lab, you will be able to:
- Explain the structure and function of an enzyme using the terms substrate, active site, activation energy, and products
- State the equation for the specific reaction that is being investigated in this lab (name the substrate, enzyme, and products, and know the source of the enzyme)
- State the effect of enzyme concentration, pH, temperature, and inhibitors on enzyme activity; explain what is occurring at a molecular level
- Make a graph based on data given and be able to interpret a graph to draw conclusions
Required prior knowledge and textbook readings:
- Enzyme structure and function
- Enzyme inhibitors
- Chapter 4, Section 4.1: Energy and Metabolism in Concepts of Biology textbook
Introduction:
Enzymes are proteins that speed up reaction rates of specific chemical reactions (they are a type of catalyst). An enzyme must have a particular shape to combine with its substrate at the enzyme’s active site. If the shape of an enzyme molecule is altered (due to environmental conditions), the enzyme’s active site may change shape, resulting in a poor fit with the substrate and decreased catalytic activity. All enzymes have optimal environments in which they are most effective. Some enzymes will only function at low pH (think about enzymes in the stomach!) while others only function at particular temperatures.
In this lab, we will study the enzyme catalase. This enzyme speeds up the breakdown of hydrogen peroxide into water and oxygen gas, according to the following reaction equation:
2H2O2 (liq) → 2H2O (liq) + O2 (g)
[INSERT H5P ACTIVITY TO IDENTIFY WHICH IS THE SUBSTRATE]
Note that the progress of the chemical reaction can be monitored by either the disappearance of one of the reactants (on the left side of the reaction) or the accumulation of one of the products (on the right side of the reaction). The oxygen that is released during the above reaction provides a convenient way of measuring the progress of the reaction. More oxygen produced = more enzyme activity!
The source of our enzyme in this experiment will be a turnip (or some other root vegetable). Hydrogen peroxide (H2O2) is a by-product of aerobic cell activity (e.g. fatty acid metabolism). Virtually all cells that carry out oxidative respiration produce small amounts of hydrogen peroxide in organelles known as peroxisomes. Hydrogen peroxide is highly toxic to cells; their response to this is the evolved ability to produce enzymes such as catalase that converts hydrogen peroxide to harmless products. Most organisms therefore have catalase in very high amounts within their cells. Root vegetables such as turnip and carrot are full of them!
In the following exercise, you will qualitatively examine the effects of catalase (from a turnip) on the degradation of hydrogen peroxide into water and oxygen. You will then look at data from an enzyme lab experiment to quantitatively examine the same phenomenon.
Before beginning, ensure you have the following materials ready:
Materials:
- 1 turnip (if you don’t have a turnip, use another root vegetable such as a carrot or parsnip)
- 1 knife
- 1 cutting board
- 1 strainer
- 3 small bowls
- 4 Ziploc bags
- Measuring spoon (tablespoon)
- Tape
- Marker
- Stopwatch
- Tap water
- Tool to boil water (e.g. kettle, microwave)
- Small bottle of hydrogen peroxide (can be purchased at a pharmacy in the first aid section)
- Lemon juice (or lime juice)
Part 1: Qualitative Enzyme Experiment
Begin by considering the following hypothesis: The decomposition reaction of hydrogen peroxide into water and oxygen by the catalase enzyme is affected by environmental conditions.
You will then test this hypothesis using various kitchen supplies, a root vegetable, and hydrogen peroxide. You will cut your turnip into small pieces, place the pieces in various environmental conditions (e.g. high temperature, low pH), and then determine how effectively the catalase from the turnip is able to convert hydrogen peroxide into water and oxygen. This will be a qualitative investigation.
Method:
1. Use your knife and cutting board to finely chop the raw root vegetable into small pieces. Smaller pieces will increase the surface area, allowing for more catalase to be exposed. You should have at least a few tablespoons worth of chopped vegetable.
2. Using tape and a marker, label one bowl “+ Control”, one bowl “Low pH”, and one bowl “High Temp”.
3. Place 1 tablespoon of chopped vegetable into each bowl.
4. In your “+ Control” bowl, add enough tap water to cover the vegetable pieces. The water should be about room temperature (not hot, not ice cold). Let sit for 30 minutes.
5. In your “Low pH” bowl, add enough lemon or lime juice to cover the vegetable pieces. Let sit for 30 minutes.
6. In your “High Temp” bowl, add enough boiling water to cover the vegetable pieces. Let sit for 10 minutes. **BE CAREFUL HANDLING BOILING WATER. WEAR PROPER PROTECTIVE EQUIPMENT!**
7. After 10 minutes, drain the water from the “High Temp” bowl and add more boiling water. Let sit for 20 more minutes.
8. While the vegetable pieces are sitting in their experimental conditions, label your Ziploc bags as “+ Control”, “Low pH”, “High Temp”, and “- Control”.
9. After a total of 30 minutes in each experimental condition, drain and transfer the vegetable pieces from each bowl to the corresponding labelled Ziploc bags. NO vegetable pieces will be added to the “- Control” bag.
10. Before proceeding, predict what will happen in each Ziploc bag once hydrogen peroxide is added. Consider the function of the catalase enzyme in the different experimental conditions. Make a note of your predictions.
[INSERT H5P ACTIVITY ABOUT THE PREDICTIONS]
11. Add two tablespoons of hydrogen peroxide to each Ziploc bag (enough to cover the vegetable pieces). Immediately seal the bags, trying to expel as much air as possible.
12. Observe the chemical reactions taking place (or lack thereof) in each bag over the course of 10 minutes. Focus on the bubbles forming – this is oxygen being produced as part of our chemical reaction! Record your observations regarding the bubble formation in each bag.
[INSERT H5P ACTIVITY THAT RELATES TO RESULTS AND ENZYME ACTIVITY]
13. Take a photo of your Ziploc bags 10 minutes after adding the hydrogen peroxide.
Part 2: Quantitative Enzyme Data Analysis
A quantitative version of the experiment in Part 1 can also be performed in a laboratory setting. This involves using turnip homogenate (made by pureeing turnip in a blender) as the source of catalase. Hydrogen peroxide can be added to the turnip homogenate in different environmental conditions, with the amount of oxygen production measured numerically over 4 minutes. As before, oxygen is a product of the reaction and therefore indicates the activity of the catalase enzyme in each experimental condition. The following environmental conditions can be varied to determine the effects on enzyme activity:
- Enzyme concentration (how much turnip homogenate is added)
- Substrate concentration (how much hydrogen peroxide is added)
- Temperature
- pH
- Presence or absence of copper sulfate (an enzyme inhibitor)
Below are five tables of quantitative data showing the average oxygen produced (in mL) after 4 minutes in each experimental condition. Review the data and answer the questions that follow.
Table 1. Volume of oxygen produced (mL) after 4 minutes of various amounts of turnip homogenate reacting with hydrogen peroxide.
Volume of Turnip Homogenate (mL) | Volume of O2 (mL) |
0 | 0 |
0.5 | 3 |
1 | 7 |
1.5 | 18 |
2 | 25 |
[INSERT H5P ACTIVITY ABOUT INTERPRETING THESE RESULTS]
Table 2. Volume of oxygen produced (mL) after 4 minutes of turnip homogenate reacting with various amounts of hydrogen peroxide.
Volume of H2O2 (mL) | Volume of O2 (mL) |
0 | 0 |
0.5 | 4 |
1 | 9 |
2 | 17 |
5 | 30 |
[INSERT H5P ACTIVITY ABOUT INTERPRETING THESE RESULTS]
Table 3. Volume of oxygen produced (mL) after 4 minutes of turnip homogenate reacting with hydrogen peroxide at various temperatures.
Temperature (°C) | Volume of O2 (mL) |
0 | 11 |
20 | 25 |
37 | 28 |
80 | 0 |
[INSERT H5P ACTIVITY ABOUT INTERPRETING THESE RESULTS]
Table 4. Volume of oxygen produced (mL) after 4 minutes of turnip homogenate reacting with hydrogen peroxide at various pH levels.
pH | Volume of O2 (mL) |
3 | 0 |
5 | 7 |
7 | 19 |
9 | 22 |
11 | 20 |
13 | 0 |
[INSERT H5P ACTIVITY ABOUT INTERPRETING THESE RESULTS]
Table 5. Volume of oxygen produced (mL) after 4 minutes of turnip homogenate reacting with hydrogen peroxide in the presence and absence of copper sulfate inhibitor.
Copper Sulfate | Volume of O2 (mL) |
Absent | 35 |
Present | 0 |
[INSERT H5P ACTIVITY ABOUT INTERPRETING THESE RESULTS]
Assignment:
If required by your instructor, you can download the assignment for the Enzymes lab here. Note that the file download works best using Microsoft Edge or Mozilla Firefox web browsers (not Google Chrome). Once you have completed the assignment, submit the file to your instructor as per their instructions.