Category: 1st Semester

Catalase Peroxide Lab

 

 

Enzyme Rate of Catalase

 

 

Introduction:

Enzymes are molecular substances found in cells.  Enzymes act as catalysts and most are proteins.  Enzymes bind temporarily to one or more of the reactants of the reaction they catalyze. In doing so, they lower the amount of activation energy needed and thus speed up the reaction.
 Not only do enzymes economize energy usage, but also provide a variety of other functions. Cells uses an enzyme (catalase) to rid itself of a poisonous substance (hydrogen peroxide). The rate at which this occurs depends on the amount of catalase that is available. In this lab we are going to measure the time it takes for a disc of filter paper, soaked with different concentrations of enzyme, to make its way to the top of a plastic vial filled with peroxide.  Rate of enzyme activity = distance (depth of hydrogen peroxide in mm)/time (in sec).

Catalase catalyzes the decomposition of hydrogen peroxide into water and oxygen.  One molecule of catalase can break 40 million molecules of hydrogen peroxide each second.

catalase
2H2O2 —–> 2H2O + O2

Objectives:

Students will prepare various dilute solutions from a 100% enzyme solution.
Students will determine how enzyme concentration affects reaction rate.

 

Materials:

6 medicine cups for dilutions, *catalase stock solution, clear plastic vial, forceps, 18 filter paper disks, Hydrogen Peroxide (H2O2), paper towel, apron, safety glasses, watch with second hand, marker, metric ruler, calculator

* The enzyme has been prepared for you as follows: 50g of peeled potato was mixed with 50 ml cold distilled water and crushed ice and homogenized in a blender for 30 seconds. This extract was filtered through cheesecloth and cold distilled water was added to a total volume of 100 ml. Extract concentration is arbitrarily set at 100 units/ml. ENZYME SHOULD BE KEPT ON ICE AT ALL TIMES!!

Procedure:

  1. Make a series of dilutions of the enzyme catalase using the following table.

 

Final Quantity Needed Concentration of Final Solution mL of Catalase mL of Water
10 ml 100% c10 0
10 ml 80% 8 2
10 ml 60% 6 4
10 ml 40% 4 6
10 ml 20% 2 8
10 ml 0% 0 10

 

 

  1. Use a marking pencil and mark the enzyme solutions as follows: 100%, 80%, 60%, 40%, 20%, and 0%.
  2.  Fill a clear vial with 20 mL of hydrogen peroxide.
  3. Using your forceps, pick up one filter paper disk and submerge it in the 100% enzyme solution for 5 seconds. Continue to hold the disk with the forceps.
  4. Using your forceps, pick up one filter paper disk and submerge it in the 100% enzyme solution for 5 seconds. Continue to hold the disk with the forceps.
  5. Remove the disk from the solution and blot it dry, for five seconds, using your paper towel.
  6. Drop the disk in the hydrogen peroxide and measure the time it takes for the disk to rise up from the bottom. Begin timing as soon as the disk touches the surface of the hydrogen peroxide.
  7. Use the metric ruler to measure the distance the disk sinks into the hydrogen peroxide. multiply by two to determine the entire distance the disk traveled. Enter the time and distance the disk traveled in the column for Trial 1 in the data table below.

 

 

% Catalase

Time in seconds

Distance in millimeters

 Reaction Rate mm/s
Trial 1 Trial 2 Trial 3 Avg. Trial 1 Trial 2 Trial 3 Avg.
100
80
60
40
20
0

 

  1. Repeat the above steps for the remainder of the solutions. Remember to use clean filter paper each time you use a different solution. Enter the times and distances for trial 2 and 3 in their appropriate columns.

Analysis & Conclusion:

1. Which concentration of catalase had the fastest reaction time?

2. Which concentration of catalase had the slowest reaction time?

3. What is catalase & why is it important to cells in your body?

 

 

4. How did you know that catalase was present in the above compounds?

 

5. What 2 substances form when catalase breaks down hydrogen peroxide?

 

6. What type of organic compound is catalase?

7. Produce a line graph of the above data. Use the enzyme concentration as the independent variable and the reaction rate as the dependent variable.

Graph Title:__________________________________________________

Legend:

8.  Based on the graph and overall slope of the line, what can you conclude about the effect of enzyme concentration on reaction rate?

 

 

 

Calorimetry lab

Calorimetry – Measuring the energy in Foods

Introduction:
There are two processes that organisms use to make usable energy. The process by which autotrophs convert sunlight to a usable form of energy is called photosynthesis. Photosynthesis supports all life on earth. Products from photosynthesis include food, textiles, fuel, wood, oils, and rubber. During photosynthesis, light energy is used to make organic compounds from inorganic water and carbon dioxide. Photosynthesis goes through light dependent reactions and the light independent reactions which include the Calvin cycle.
The process where heterotrophs break down food molecules to release energy for work is called cellular respiration. Cellular respiration is the reverse of photosynthesis; the reactants of one are the products of the other. The reactants of cellular respiration are glucose and oxygen, and the products are carbon dioxide, water, and energy.  Cellular respiration breaks down glucose to form carbon dioxide and water, while releasing energy usable by the cells. The first step, glycolysis is the process  that converts glucose to pyruvate and releases a small amount of cellular energy.  The second step may be aerobic or anaerobic depending on the amount of oxygen available.  Aerobic respiration is the breakdown of pyruvate in the presence of oxygen.  A larger amount of cellular energy or ATP is produced during the Kreb’s cycle and electron transport chain. Anaerobic respiration is the breakdown of food molecules in the absence of oxygen. Less ATP is produced by anaerobic respiration or fermentation.

Hypothesis:
If the heat given off by a burning pecan is measured by how much the temperature increases in a given amount of water, then the number of calories of energy stored in the nut during photosynthesis can be determined.

Materials:
Items needed for the lab included a large paper clip, a 100 ml graduated cylinder, thermometer, 2 soft drink cans, electronic balance, butane lighter, plastic tray, scissors, paper, and pencil.

Procedure:
Use a graduated cylinder to measure 100 ml of water and add this to an empty soft drink can. Cut holes on two sides of a second soft drink can so there is room to place a large bent paper clip.  Measure and record the mass of one pecan using the electronic balance. Bend a large paper clip to make a “nut stand” and measure and record  the mass of this clip. Place the pecan on the nut stand and put the stand inside the cut-out drink can.  Use a thermometer to measure and record the temperature of the water in the second can.  Place this can on top of the can with the nut. Use a butane lighter to ignite the nut. Record the temperature of the water when the nut is completely burned. Complete the data table by calculating the  the total calories in the pecan.

Data:    

Data Table 1
Before Burning After Burning Difference
Mass of Nut 1.7 g 0.1g 1.6g
Temperature of Water 20 40.1 20.1
Mass of Paper Clip 1.4g 1.4g 0g

 

Data Table 2
Mass of pecan 0.1 g
Temperature change of 100 ml of water 20.1
Calories required to produce temperature change in 100 ml water 2010
Calories per gram contained in the pecan 1182.4

Error Analysis:
Errors may have occurred in several ways during this experiment. One error that may have occurred is that some of the energy may have been lost during the burning. Some of the pecan’s energy was lost as light instead of heat energy. Also some of the heat measured in the water could have been due to the butane lighter used to ignite the pecan.

Conclusion:
The temperature of the 100 ml of water in the can above the burning pecan was changed by the energy given off by the pecan when it was burned.  The energy given off by the burning pecan was great enough to increase the water temperature by 20.1 degrees Celsius. The mass of the unburned pecan was 1.7g. It takes 100 calories to raise the temperature of 1 ml of water by 1 degree Celsius. The temperature of 100 ml of water was recorded to have increased by 20.1 degrees Celsius; therefore, the total number of calories in the pecan equals 20.1 x 100 or 2010 calories. Since the nut had a mass of 1.7g, the number of calories per gram equals 2010 divided by 1.7 or 1182.4 calories per gram.
The increase of temperature in the water showed that energy had been stored in the pecan. In this experiment, the amount of calories of heat energy stored in a pecan during photosynthesis was measured by a process known as calorimetry.

 

Campbell Problem 7

Molecular Genetics Problem 7
7. Using the information from problem 6, a further testcross was done using a heterozygote for height and nose morphology. The offspring were tall-upturned nose, 40; dwarf-upturned nose, 9; dwarf-downturned nose, 42; tall-downturned nose, 9. Calculate the recombination frequency from these data; then use your answer from problem 6 to determine the correct sequence of the three linked genes.

Experiment 3. (Frequency/Distance between T and S)

Determine the recombination frequency for the genes controlling Tallness and Snout:

40 tall-upturned snout = 40% expected
42 dwarf-downturned snout = 42% expected
9 dwarf-upturned snout = 9% recombinant
9 tall-downturned snout = 9% recombinant

Total = 100%

Therefore this recombination frequency between genes T and S is 18%

One can determine the relative frequency between genes using the percent frequencies as distances.

The Recombinant relationships from experiments 1-3 are:

Exp. 1 T-A = 12 map units Exp. 2 A-S = 5 map units Exp. 3 T-S = 18 map units

An arrangement that fits the data would be:

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Carbon Dioxide Use in Plants

 

 

Do Plants Consume or Release CO2?

 

Introduction

The rate of photosynthesis can be determined by measuring the rate of production of sugar or oxygen or by measuring the rate of decrease in carbon dioxide concentration. A common aquarium plant called, Elodea,  can be used to show fast carbon dioxide is being removed from the water in which the Elodea is submerged.

6CO2 + 12H2O + light energy —> C6H12O6 + 6O2 + 6H2O

 

In this lab, you will use phenol red as an indicator to show whether CO2 is being consumed or produced in a reaction. It is well known that in the presence of light, plants perform photosynthesis. At the same time, plants are also performing cell respiration. To demonstrate this, we will determine whether CO2 is consumed or produced as Elodea is placed in either a light or dark environment. The change in CO2 will be detected by the pH indicator phenol red. Phenol red is yellow under acidic conditions (high H+ ion concentration), pink to magenta under basic or alkaline conditions (low H+ ion concentration), and orange under neutral conditions. A change in the amount of CO2 will cause a directly proportional change in H+ ion.

If the CO2 concentration decreases, the H+ ion concentration will also decrease, and the solution will change to pink, becoming basic.

If the CO2 concentration increases, the H+ ion concentration will also increase, and the solution will change to yellow, becoming acidic.

Neutral solutions of phenol red will be orange.

Materials:

phenol red solution,  4 sprigs of Elodea, soda straw, 4 test tubes, labeling marker, 100 ml graduated cylinder, beaker, aluminum foil

Procedure:

  1. Create a solution of phenol red by adding concentrated phenol red to about 100 ml of water in a beaker. The phenol red may change color as a result of adding water (depending on how acidic your tap water is). Your goal is to make your solution a neutral orange color. You can do this by gently blowing into the solution with a straw.
  2. Label 4 test tubes 1, 2, 3, and 4.
  3. Once you have the solution at an orange color, transfer it to 4 test tubes (they should be filled about 2/3 full with your orange solution).
  4. Place a cut piece of Elodea (cut end up) into tubes 2 and 4 and tightly cap.
  5. Test tubes 3 and 4 will not have Elodea. Cap and then cover these tubes with aluminum foil so no light can enter.
  6. Place tubes 1 and 2 in bright light.
  7. Place tubes 2 and 4 in the dark.
  8. After 24 hours, uncover and examine all 4 test tubes and record the results.

Data:

 

Test Tube # Contents of Tube Initial Color Final Color
1
2
3
4

 

Conclusion:

1. What test tubes served as the controls in this experiment. Why?

 

 

2. What was the dependent variable?

 

3. Do you think there would have been any change in any of the test tubes if they were left for 48 or 72 hours? Explain.

 

 

4. Describe and explain what happened in the test tubes.

 

 

 

5. Why did the color change occur?

 

6. Where does the carbon dioxide that is removed from the solution go?

 

7. What other process goes on in plant cells that requires oxygen and produces carbon dioxide?

 

8. What was the purpose in tightly capping all four test tubes?

 

 

Campbell Problem 10

Molecular Genetics Problem 10
10. An aneuploid person is obviously female, but her cells have two Barr bodies. What is the probable complement of sex chromosomes in this individual?

This individual probably is XXX.

The individual is a female. Nondisjunction of sex chromosomes produces a variety of aneuploid conditions in humans. Most of these conditions appear to upset genetic balance less than aneuploid conditions involving autosomes. Extra copies of the X chromosome are deactivated as Barr bodies in the somatic cells. Females with trisomy of the X chromosome (XXX), which occurs about once in approximately 1000 live births, are healthy and cannot be distinguished from XX females except by karyotype.

An Example of nondisjunction:

Klinefelter’s syndrome

49 ,XXXXY

This karyotype shows a variant of Klinefelter’s syndrome.

Individuals with this syndrome are male, typically with the karyotype 47,XXY.

Individuals with Klinefelter’s syndrome exhibit a characteristic phenotype including tall stature, infertility, gynecomastia and hypogonadism.

Aneuploidy above one extra chromosome is usually fatal but because of X-inactivation, which “turns off” all but one X chromosome per cell, the effects of 3 extra chromosomes are reduced.

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