Category: 1st Semester

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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Catalase Lab Sample 1

 

 

Enzyme Rate of Reaction for Catalase

 

Introduction:
Enzymes are an important part of life that regulate chemical reactions with in the body. Enzymes speed up chemical reactions in four different ways, one way is heat, another is increasing the rate of reactants, the third way is decreasing the amount of products and the fourth way is enzymes, which speed up reaction without themselves being used up. Enzymes are also involved in digestion, respiration, reproduction, vision, movement, thought, and also in the productions of other enzymes. Simple cells may have as many as 2000 enzymes with each one catalyzing a different reaction. An enzyme can speed up a reaction making it 10, 000,000,000 times faster. An enzyme is a catalyst. A catalyst is a chemical that reduces the amount of activation interim needed for a reaction. Without enzymes a reaction would take much longer than if it had and enzyme. Enzymes also the control the rate and direction of the reaction.
Without catalysts chemical reactions would take much longer that the average human life expectancy. So that would mean that in 76 years only a couple chemical reactions would take place. Since our bodies have enzymes though hundreds of chemical reactions a day. If our bodies didn’t have catalysts our bodily cells couldn’t function. Some bacteria, however, possess a defense mechanism which can minimize the harm done by the two compounds. These resistant bacteria use two enzymes to catalyze the conversion of hydrogen peroxide  back into diatomic oxygen and water. One of these enzymes is catalase and its presence can be detected by a simple test. The catalase test involves adding hydrogen peroxide to a cultures sample or an agar slant.

 

Hypothesis:
The reaction rate of catalase splitting hydrogen peroxide into water and oxygen will increase over time.

Materials:
The materials used consisted of 100 paper H2O2 molecules, a data table, paper, pencil, calculator, scissors, watch with a second hand, and an enzyme rate of reaction catalase worksheet.

Methods:
Cut out 100 hydrogen peroxide paper molecules. Double check to make sure there are only 100 paper molecules and place them in an envelope. Then one person will keep track of the time while another person acts as a catalase and tears the paper hydrogen peroxide molecules in half. The torn paper molecules should be returned to the envelope each time. Another person times the person acting as the catalase.  The time intervals in which the paper molecules are to be ripped are 10 seconds, 20 seconds, 30 seconds, and two different 60second periods of time. The results should be  recorded in a data table. The reaction rate for catalase is figured using the formula:
M2 – M1 = Reaction Rate
T2  –  T1

 

Results:

 

 

Time in Seconds Ripped H2O2 Molecules Rate of Reaction
0-10 5 .5
10-30 13 .4
30-60 31 .6
60-120 61 .5
120-180 91 .5

 

1. What is an enzyme? What are its functions in living things?
chemicals that reduce the amount of activation energy needed for reactions to occur; they are proteins in cells that control metabolic reactions

 

2. Name several things that can affect the functioning of an enzyme.
temperature, pH, and the amount of reactant or product

 

3. Write the chemical equation for the breakdown of hydrogen peroxide by the enzyme catalase.
H2O2   +   Catalase –>  H2O  + O2

 

4. An enzyme’s efficiency increases with greater substrate concentration, but only up to a point. Why?
once all active sites are filled, the enzyme’s reaction rate won’t continue increasing

 

5. If you were allowed to continue this lab and rip hydrogen peroxide molecules for 240 and 300 seconds, what would happen to the reaction rate and why would this happen?
there would be more molecules ripped because of the increased amount of time

 

6. What can you say about the length of time and the reaction rate?
The more time available, the faster the reaction will occur.

 

7. What would happen to the reaction rate if you removed the water and oxygen molecules as soon as they were produced?
The rate of reaction would go even faster

 

Error Analysis:
The counting of the time  may have been off a couple of seconds.

 

Discussion and Conclusion:
The data shows that the more time there is, the more hydrogen peroxide molecules will be ripped. The catalase in the lab ripped about 6 molecules every 5 seconds. The same thing occurs in a cell as more hydrogen peroxide is produced, catalase speeds up breaking down this waste into water and oxygen.

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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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