biology lab

1. NAME: ____________________________
ID# _______________________
Lab 6
RESPIRATION: ENERGY CONVERSION
Objectives:
After completing this exercise, you will be able to
1. Define metabolism, reaction, metabolic pathway, respiration, ATP, aerobic respiration, alcoholic
fermentation.
2. Give the overall balanced equation for aerobic respiration and alcoholic fermentation.
3. Distinguish among the inputs, products, and efficiency of aerobic respiration and those of
fermentation.
Introduction:
The first law of thermodynamics states that energy can neither be created nor destroyed only
converted from one form to another. Because all living organisms have a constant energy
requirement, they have mechanisms to gather, store, and use energy. Collectively, these
mechanisms are called metabolism. A single, specific reaction that starts with one compound and
ends up with another compound is a reaction, and a sequence of such reactions is a metabolic
pathway.
Carbohydrates are temporary energy stores. The process by which energy stored in carbohydrates is
released to the cell is respiration.
Both autotrophs and heterotrophs undergo respiration. Photoautotrophs such as plants utilize the
carbohydrates they have produced by photosynthesis to build new cells and maintain cellular
machinery. Heterotrophic organisms may obtain materials for respiration in two ways, by digesting
plant material or by digesting the tissues of animals that have previously digested plants.
Several different forms of respiration have evolved. The specific respiration pathway used depends
on the specific organism and/or environmental conditions. In this exercise, we will consider two
alternative pathways: (1) aerobic respiration, an oxygen-dependent pathway common in most
organisms; and (2) alcoholic fermentation, an ethanol – producing process occurring in some yeast.
Perhaps the most important aspect to remember about these two processes is that aerobic
respiration is by far the most energy – efficient. Efficiency refers to the amount of energy captured
in the form of ATP relative to the amount available within the bonds of the carbohydrate. ATP,
adenosine triphosphate, is the so-called universal energy currency of the cell. Energy contained
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2. within the bonds of carbohydrates is transferred to ATP during respiration. This stored energy can
be released later to power a wide variety of cellular reactions.
For aerobic respiration, the general equation is
C6H12O6 + 6O2 6CO2 +6H2O + 36 ATP+
Glucose + oxygen carbon dioxide +water+ chemical energy
If glucose is completely broken down to CO 2 and H2O about 686,000 calories of energy are released.
Each ATP molecule produced represents about 7500 calories of usable energy. The 36 ATP represent
270,000 calories of energy (36 x 7500 calories). Thus, aerobic respiration is about 39% efficient
[(270,000/686,000) x 100%].
By contrast, fermentation yields only 2 ATP. Thus, these processes are only about 2% efficient
[(2x7500/686,000) x 100%]. Obviously, breaking down carbohydrates by aerobic respiration gives a
bigger payback than the other means.
During the process of aerobic respiration, relatively high- energy carbohydrates are broken down in
stepwise fashion, ultimately producing the low-energy products of carbon dioxide and water and
transferring released energy into ATP. But what is the role of oxygen?
During aerobic respiration, the carbohydrate undergoes a series of oxidation-reduction reactions.
Whenever one substance is oxidized (loses electrons), another must be reduced (accept, or gain,
those electrons), another must be reduced (accept, or gain, those electrons). The final electron
acceptor in aerobic respiration is oxygen. Tagging along with the electrons as they pass through the
electron transport process are protons (H +). When the electrons and protons are captured by
oxygen, water (H2O) is formed:
2H+ + 2e- + ½ O2 H2O
In the following experiment, we examine aerobic respiration in two sets of seeds.
Oxygen Consumption:
One set of pea seeds has been soaked in water for the past 48 hours to initiate germination. In this
section, you will measure oxygen consumption to answer the question, “Do germinating peas have a
higher respiratory rate than ungerminated peas?”
Materials
•
Volumeter
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3. •
China marker
•
80 germinating pea seeds
•
80 ungerminated (dry)pea seeds
•
Glass beads
•
Non-absorbent cotton
•
Metric ruler
•
Bottle of potassium hydroxide (KOH) pellets
•
¼ teaspoon measure
•
Marker fluid in dropping bottle
Procedure:
1. Obtain a volumeter set up as in figure 6-2. Skip to step 6 if your instructor has already assembled the
volumeter as described by steps 2-5.
2. Remove the test tubes from the volumeter. With a china marker, number the tubes and then fill as
follows:
Tube 1: 80 germinating (soaked) pea seeds
Tube 2: 80 ungerminated (dry) pea seeds plus enough glass beads to bring the total volume equal to
that in tube 1.
Tube 3: enough glass beads to equal the volume of tube 1
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4. Both temperature and pressure affect gas pressure within a closed tube. Tube 3 serves as
thermobarometer and is used as a control to correct experimental readings to account for changes in
temperature and barometric pressure taking place during this experiment.
3. Pack cotton loosely into each tube to a thickness of about 1.5 cm above the peas/beads.
4. Measure out 1 cubic centimetre (cm3) (about 1/4 teaspoon) of KOH pellets and pour them atop the
cotton.
Potassium hydroxide absorbs CO2 given off during aerobic respiration. Since the volumeter measures
change in gas volume, any gas given off during respiration must be removed from the tube so an
accurate measure of O2 consumption can be made.
5. Insert the stopper-syringe assembly in place.
6. Add a drop of marker fluid to each side arm pipet by touching the dropper to the end of each. The
drop should be taken into the side arm by capillary action. Gently withdraw the plunger of each
syringe and adjust the position of the drop so it is between 0.80 and 0.90 cm 3 on the scale of the
graduated pipet.
7. Adjust each side arm pipet so it is parallel to the table top. Wait five minutes before starting data
collection.
8. In table 6-2 at time 0, record the position of the marker droplet within each pipet. Record readings
for each tube at 5-minute intervals for the next 60 minutes, keeping track of whether changes are
positive (movement toward the test tube) or negative (movement away from the test tube).
If respiration is rapid and the marker drop moves to near the end of the scale to read, carefully use
the syringe to readjust its position so it is between 0.80 and 0.90 cm 3 on the scale of the graduated
pipet again. Note this in table6-2 and continue to record readings every five minutes.
9. To determine change in volume of gas within each tube at each sampling, subtract each subsequent
reading from the previous reading.
10. Determine cumulative volume change (cumulative oxygen consumption) by adding each volume
change to the previous volume-change measurement. The final figure represents the total oxygen
consumption (in ml) in that tube.
11. At the end of the experiment, correct for any volume changes caused by changes in temperature or
barometric by using the reading obtained from the thermobarometer. If the thermobarometric
marker moves toward the test tube (decrease in volume), subtract the volume change from the total
oxygen consumption measurement of tubes 1 and 2. If the marker droplet moves away from the test
tube (increase the volume), add the volume change to the last total oxygen consumption
measurement for tubes 1 and 2.
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5. 12. Plot a graph of oxygen consumption by germinating and non-germinating seeds against time. Use a
(+) for data points of germinating peas and a (.) for dry peas.
Time
(min.)
0
5
10
15
20
25
30
35
40
45
50
55
60
Tube 3:
Thermobarometer
Reading
Total
change
in
volume
0
Tube 1:
Tube2:
Germinating Peas
Dry Peas
Reading Total
Total oxygen Reading Total
Total oxygen
change consumption
change consumption
in
in
volume
volume
0
0
0
0
Fermentation:
Despite relatively low energy yield, fermentation provides sufficient energy for certain organisms to
survive. Alcoholic fermentation by yeast is the basis for the baking, wine-making, and brewing
industries. It’s been said the yeast and alcoholic fermentation made Milwaukee famous.
The chemical equation for this process is
C6H12O6 2CH3CH2OH +CO2+2ATP
GLUCOSE-> ETHANOL+CARBON DIOXIDE+ENERGY
Starch (amylase), a common storage carbohydrate in plants is a polymer consisting of a chain of
repeating glucose units. The polymer has the chemical formula (C 6H12O6)n; where n is a large number.
Starch is broken down by the enzyme amylase into individual glucose units. To summarize:
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6. (C6H12O6)n C6H12O6 + C6H12O6 +***
This section demonstrates the action of yeast cells on carbohydrates.
Materials:
•
3 50-ml beakers
•
25-ml measuring cylinders
•
10% glucose
•
1% starch
•
0.5% amylase in bottle fitted with graduated pipet
•
3 glass stirring rods
•
¼ teaspoon measure
•
3 fermentation tubes
•
15-cm ruler
•
0.5g pieces of yeast bread
•
Scale
•
Incubator
Procedure:
1. Number three 50-ml beakers
2. Measure 15-ml of the following solutions into each beaker:
Beaker 1: 15ml of 10% glucose
Beaker 2: 15ml of 1% starch
Beaker 3: 15-ml of 1% starch; add 5 ml of 0.5% amylase.
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7. 3. Wait 5 minutes and then to each beaker add 0.5g of bread yeast. Stir with separate glass stirring
rods.
4. When each is thoroughly mixed, pour the contents into three corresponding numbered
fermentation tubes. Cover the opening of the fermentation tube so the tail portion is filled with
solution.
5. Write a prediction about gas production in each tube.
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6. Place the tubes in a 370C incubator.
7. At intervals of 20, 40 and 60 minutes after the start of the experiment, remove the tubes and,
using a metric ruler, measure the distance from the tip of the tail to the fluid level. Record your
results in table 6-3. Calculate the volume of gas evolved using the formula below table 6-3.
Tube
Solution
1
Distance from tip of tube to fluid level (mm)
20 min.
40 min.
60 min.
10% glucose
+ yeast
1% starch +
yeast
1% starch +
2
3
Volume of
gas evolved
(mm2)
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8. yeast
+amylase
To calculate the volume of gas evolved, use the following equation: V = πr 2h, where π =3.14, r=
radius of tail of fermentation tube [r=1/2d (diameter)], h=distance from top of tail to level of
solution.
Did your results conform to your predictions? If not, speculate on reasons why this might be so?
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What gas accumulates in the tail portion of the fermentation tube?
Questions:
1. If you performed the experiment on oxygen consumption without adding KOH pellets to the test
tubes, what results would you predict? Why?
2. Sucrose (table sugar) is a disaccharide composed of glucose and fructose. Glycogen is a
polysaccharide composed of many glucose subunits. Which of the following fermentation tubes
would you expect to produce the greatest gas volume over a 1-hour period? Why?
Tube 1: glucose plus yeast
Tube 2: sucrose plus yeast
Tube 3: glycogen plus yeast
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9. 3. Bread is made by mixing flour, water, sugar, and yeast to form dense dough. Why does the dough
rise? What gas is responsible for the holes in bread?
4. Oxygen is used during aerobic respiration. What biological process is the source of the oxygen?
5. Compare aerobic respiration and fermentation in terms of
a. Efficiency of obtaining energy from glucose
b. End products
6. How would you explain this statement: “the ultimate source of our energy is the sun?”
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10. 7. The first law of thermodynamics seems to conflict with what we know about ourselves. For
example, after strenuous exercise we run out of energy. We must eat to replenish our energy
stores. Where has that energy gone? What form has it taken?
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