Laporan praktikum Respirasi Praktikum Bio
RATIFICATION PAGE
Complete report of Basic Biology experiment the title “Respiratory”
arrenged by :
name
: Devi Putriana
ID
: 1314441006
class
: Biology ICP B
group
: VI (Six)
It has been checked and consult for an assistant and assistant coordinator then
stated acceptable.
Makassar, January
Assistant
Assistant Coordinator
Rachmayani Ardiansyah, S.Pd
Rukman Muslimin
ID: 101404157
Known by
Lecturer of Responsible
Dr. Muhiddin P. S.Pd, M.Pd
NIP : 19721231 19903 1 042
th
2013
CHAPTER I
INTRODUCTION
A. Background
In biology science, we cannot stop discussing about living creature. The
object study of biology is living creature. If we talk about the living creature,
then we will discuss about what living creature do for stay to live and how
about that work system. In their daily activity (human), that is so many
activity that always done, for example schooling, go to office, clean home and
many soon. But if thinking, there is an activity that we can call fundamental
activity to operating other activity like example above, this activity is
breathing. Did you know, living creature cannot live without breathing,
because if do this activity, everything work system that in into body can be
operating.
Not only human is breathing, plant, animal and other microorganism do
this activity, suitable with their method. But if there is a question what living
creature use to breathing, what will be answering for this question. Living
creature use air to breathing, so living creature can stay to live. But is there
different between living creature in breathing. Yes, there is some living
creature have different respiratory system with other living creature, this
respiratory system describe work in breathing. There is living creature use
their lung as respiration tool, but other use skin or trakea as respiration tool.
That is important as science to know how living creature to breathing. In
this experiment we will try to observe how living creature to breathing
specific about RQ or respiratory quotient. Or we can say we will try to
observe how many oxygen that needed for one living creature, is there
different between other living creature of oxygen needs. So to answer this
question, we will do this observation. This observation use living creature :
cockroach and grasshopper. We want to know too how many different oxygen
needs between them. So it is good time to this observe.
B. Purpose
1.
Proving that live organism need oxygen for their respiration.
2.
Comparing oxygen needs of some organism according to kind and size of
weight.
C. Benefit
After do this observation, apprentice can explain why the organism
need oxygen for stay to live. And apprentice can make comparison oxygen
needs between two different organism. This comparison will indicate which
organism that need much oxygen and it can explain what factor can influence
that.
CHAPTER II
PREVIEW OF LITERATURE
Respiration is oxidation process of material food or organic material which
happen inside cell that can done by aerob although anaerob. In aerob condition,
this respiration need free oxygen and release carbon dioxide and energy. If oxided
is sugar, then reaction happen is :
C6H12O6 + 6H2O 6CO2 + 6H2O + energy
Amount of CO2 produced and amount of O2 used in aerob respiration not always
same. This depends to kinds of material that used. Comparison between amount of
CO2 that released and amount of O2 that needed, called Respiratory Quotient
(RQ). For carbohydrate, its RQ value = 1. This RQ value can being varied
depends to material for respiration, perfect or not for respiration and others
condition (Tim Penyusun Biologi Umum, 2013).
One of the most important catabolic process is respiration. Respiration is the
means by which organic material such as carbohydrates, fats, and proteins are
broken down into carbon dioxide and water, with much of the energy released
being used in the synthesis of ATP. The biochemical reaction that accomplish this
break down are referred to collectively as respiratory metabolism. Respiratory
metabolism is of primary importance in both plant and animal cell since the ATP
produce essential to wide variety of reaction necessary for growth, development
and cellular work. The carbohydrates (sugar, for example) that are broken down in
respiration are produced originally via photosynthesis. In this model, the energy
of sunlight is used to drive the conveyor belt and raise water. Water that has been
lifted upward some distance has increased in potential energy. The utilization of
the energy in sunlight to raise water and increase it’s potential energy is analogous
to using light energy produce high-energy sugar from low – energy carbon
dioxide and water during photosynthesis. The high potential energy water can be
made to do work as it flows down the trough and drives the paddle wheel. As this
occurs, the water drops to a lower level and near for loses some potential energy.
This analogous to using the high – energy sugar produce by photosynthesis to
drive the energy requiring reaction of the cell. This is made possible by the
pathway of respiration (analogous to the trough), which harness the high – energy
sugar. As the high – energy sugar is respired, it is converted to the low – energy
products carbon dioxide and water. This is analogous to the water’s losing
potential energy as it flows down the trough to a lower level. Since the process of
respiration is of fundamental importance in all plant and animal cells, it will be
described in some detail in this chapter. However, it’s should be emphasized that
understanding the overall a accomplishment of respiratory metabolism and its
relationship to other cellular process is viewed as more important here then the
details of the individuals steps involved (Peter, 1989).
Respiration is the biologically controlled breakdown of energy-containing
substance, such as carbohydrates, fats, and protein, with the release of energy. As
such, it is reverse of photosynthesis, in which energy is used to synthesize
compounds. The major pathway of respiration in the cell can be broken into two
parts: ( 1 ) glycolysis and ( 2 ) the Krebs cycle. In the first part, glycolysis,
glucose is eventually split into two pyruvic acid molecules, each having three
carbon atoms. At remaining two-carbon compound enters the Krebs cycle, is
regenerated at end of the cycle and can accept another two-carbon piece. The two
parts of respiration differ in a number of important features. In the absence of
oxygen, glycolysis can occur but the Krebs cycle can not function. If no oxygen is
present, the pyruvic acid produced at the end of glycolysis is converted, with the
loss of one carbon atom, to ethyl alcohol. In this case, the process is fermentation
( Wiliam, 1967 ).
We have studied process of respiration, that process of oxidation of
metabolic by organism when there is oxygen to catch energy that had in metabolic
bondings. Respiration is not just produce energy, but other products, like carbon
dioxide and water. ( such as, glycolysis processes and fermentation break down
metabolic to get energy and produce other products ). So, gas exchange with one
way is needed to supplies oxygen for cells and remove carbon dioxide. It is rather
make us confused, but this process is known as respiration. ( Fried, 2003 ).
Insects require oxygen just as we do, and must "exhale" carbon dioxide, a
waste product of cellular respiration. Oxygen is delivered to the cells directly
through respiration, and not carried by blood as in vertebrates. Along the sides of
the thorax and abdomen, a row of small openings called spiracles (8) allow the
intake of oxygen from the air. Most insects have one pair of spiracles per body
segment. Small flaps or valves keep the spiracle closed until there is a need for
oxygen uptake and carbon dioxide discharge. When the muscles controlling the
valves relax, the valves open and the insect takes a breath. Once entering through
the spiracle, oxygen travels through the tracheal trunk (8), which divides into
smaller tracheal tubes. The tubes continue to divide, creating a branching network
that reaches each cell in the body. Carbon dioxide released from the cell follows
the same pathway back to the spiracles and out of the body. Most of the tracheal
tubes are reinforced by taenidia, ridges that run spirally around the tubes to keep
them from collapsing. In some areas, however, there are no taenidia, and the tube
functions as an air sac capable of storing air. In aquatic insects, the air sacs enable
them to "hold their breath" while under water. They simply store air until they
surface again. Insects in dry climates may also store air and keep their spiracles
closed, to prevent water in their bodies from evaporating. Some insects forcefully
blow air from the air sacs and out the spiracles when threatened, making a noise
loud enough to startle a potential predator or curious person (Debbie, 2013).
CHAPTER III
EXPERIMENT METHOD
A. Place and Date
Day/Date
: Saturday/January 4th 2013
Time
: 01.00 pm until 03.00 pm
Place
: Biology Laboratory 3nd floor of the west, faculty of
mathematic and science university of Makassar.
B. Tools and Materials
1. Tools
a. Respirometer
b. Spoit
c. stopwatch
2. Materials
a. Grasshopper(Dissoteria carolina)
b. Cockroach(Blatta orientalis)
c. Mung bean sprouts(Phaseolus radiatus)
d. Vaseline
e. Eosin solution
f. Cotton
g. KOH crystal
C. Work Procedure
1. Prepared the tools and materials that used in experiment.
2. Decided the object that used, what is a cockroach or grasshopper.
3. Wrapped KOH crystal with cotton.
4. Entered the cotton that contain KOH crystal into respirometer tube, then
determined the object which will used into respirometer tube.
5. Closed, then greased Vaseline in around mouth tube, so more session and
there is not oxygen entered in respirometer tube.
6. Then put a respirometer to prop it.
7. So, injected the iosin solution into scale glass ribbon. With first scale
0,0mL.
8. And then recorded the scale every one minute.
9. Repeated, the step 1 until 8 to the next object.
10. Recorded the experiment resulted into the table.
CHAPTER IV
RESULT AND DISCUSSION
A. Result observation
No.
Type
1.
2.
Times to-
Pariplanet
Big
1
0,01
a
Smal
0,02
0,02
0,02
0,02
0,02
americana
Dissosteria
l
Big
0,48
0,52
0,65
0,75
0,50
cardina
Smal
0,25
0,28
0,35
0,44
0,58
Vigna
l
Open
0,20
0,80
1,20
1,50
1,90
radiata
Close 0,80
1,80
2,50
3,00
3,10
3.
2
0,06
3
0,11
4
0,13
5
0,17
B. Data analysis
s
Used the formula v = t
1. Pariplaneta Americana
a. Big :
scale
s
scale
V2= s/t = 0,06/120 = 0.0005 s
scale
V3= s/t = 0,11/180 = 0.00061 s
scale
V4= s/t = 0,13/240 = 0.00054 s
scale
V5= s/t = 0,17/300 = 0.00057 s
1) V1= s/t = 0,01/60 = 0.00017
2)
3)
4)
5)
Average speed
V=
=
( v 1+ v 2+ v 3+ v 4 +v 5 )
5
scale
s
( 0.00017 +0.0005+0.00061+0.00054+ 0.00057 )
5
scale
s
0.00239
=
5
= 0.00048
scale
s
scale
s
b. Small
scale
s
scale
V2= s/t = 0,02/120 = 0.00017 s
scale
V3= s/t = 0,02/180 = 0.00011 s
scale
V4= s/t = 0,02/240 = 0.00008 s
scale
V5= s/t = 0,02/300 = 0.00007 s
1) V1= s/t = 0,02/60 = 0.00033
2)
3)
4)
5)
Average speed
( v 1+ v 2+ v 3+ v 4 +v 5 )
V=
5
=
scale
s
( 0.00033+ 0.00017+0.00011+0.00008+ 0.00007 )
=
5
0.00076
5
= 0.00015
scale
s
scale
s
2. Dissosteria cardina
a. Big
scale
s
scale
V2= s/t = 0,52/120 = 0.00433 s
scale
V3= s/t = 0,65/180 = 0.00361 s
scale
V4= s/t = 0.72/240 = 0.003 s
scale
V5= s/t = 0,82/300 = 0.00273 s
1) V1= s/t = 0,48/60 = 0.008
2)
3)
4)
5)
scale
s
Average speed
V=
=
=
( v 1+ v 2+ v 3+ v 4 +v 5 )
5
scale
s
( 0.008+ 0.00433+0.00361+0.003+0.00273 )
5
0.02167
5
= 0.00433
scale
s
scale
s
scale
s
b. Small
scale
s
scale
V2= s/t = 0,28/120 = 0.00233 s
scale
V3= s/t = 0,35/180 = 0.00194 s
scale
V4= s/t = 0,44/240 = 0.00183 s
scale
V5= s/t = 0,50/300 = 0.00167 s
1) V1= s/t = 0,25/60 = 0.00417
2)
3)
4)
5)
Average speed
V=
=
=
( v 1+ v 2+ v 3+ v 4 +v 5 )
5
scale
s
( 0.00417 +0.00233+0.00194+ 0.00183+0.00167 )
5
0.01194
5
= 0.00239
scale
s
3. Vigna radiate
a. Open
1) V1= s/t = 0,20/60 = 0.00333
scale
s
scale
s
scale
s
scale
3) V3= s/t = 1,20/180 = 0.00667 s
scale
4) V4= s/t = 1,50/240 = 0.00625 s
scale
5) V5= s/t = 1,90/300 = 0.00633 s
2) V2= s/t = 0,80/120 = 0.00667
Average speed
V=
=
=
( v 1+ v 2+ v 3+ v 4 +v 5 )
5
scale
s
( 0.00333+ 0.00667+0.00667+0.00625+ 0.00633 )
5
0.02925
5
= 0.00585
scale
s
scale
s
scale
s
b. Close
scale
1) V1= s/t = 0,80/60 = 0.01333 s
scale
2) V2= s/t = 1,80/120 = 0.015 s
scale
3) V3= s/t = 2,50/180 = 0.01389 s
scale
4) V4= s/t = 3,00/240 = 0.0125 s
scale
5) V5= s/t = 3,10/300 = 0.01033 s
Average speed
V=
=
( v 1+ v 2+ v 3+ v 4 +v 5 )
5
scale
s
( 0.01333+ 0.015+0.01389+0.0125+0.01033 )
5
scale
s
=
0.06505
5
= 0.01301
scale
s
Scale
s
Chart between Big Periplaneta americana and Small Periplaneta americana
0.18
0.17
0.16
0.14
0.13
0.12
Big Periplaneta americana
Small Periplaneta
americana
0.11
0.1
0.08
0.06
0.06
0.04
0.02 0.02
0.01
0
0.02
0.02
0.02
0.02
Chart between Big Periplaneta americana and Small Periplaneta americana
graphic relation between Big Dissosteria cardina and Small Dissosteria cardina
0.9
0.82
0.8
0.72
0.7
0.65
0.6
0.5 0.48
0.52
0.44
0.4
0.3
0.2
0.5
Big Dissosteria
cardina
Small Dissosteria
cardina
0.35
0.25
0.28
0.1
0
graphic relation between Big Dissosteria cardina and Small Dissosteria cardina
graphic relation between Open Vigna radiata and Close Vigna radiata
3.5
3
3
3.1
2.5
2
1.5
1
1.9
1.8
1.5
1.2
0.8
Open Vigna radiata
Close Vigna radiata
1.5
0.8
0.5
0.2
0
minute 1minute 2minute 3minute 4minute 5
graphic relation between Open Vigna radiata and Close Vigna radiata
C. Discussion
Based on the experiment result above, we can see the respiration every
living things, especially Periplaneta americana, Dissosteria cardina and
Vigna radiata. The weight was inversely proportional with the used of oxigen,
because every living things required O2 in a large more than weight in the
body. Size of body organism affects the respiratory, if organism has small size
of body, the respiration occurred quickly.
1. Compare the oxygen demand Big Periplaneta Americana and Small
Periplaneta americana.
Firstly, we observed respiration of big Periplaneta americana.
After it, we observed respiration of small Periplaneta Americana. From
the result of experiment we obtain an the average speed on the big
Periplaneta americana and we got 0.00048
scale
. While the small
s
Periplaneta americana has average speeed is 0.00015
scale
s . At the time
of observing our small Periplaneta americana get inappropriate results
caused by the tube respirometer not clean properly so that there is water
on the scale that caused movement of eosin.
2. Compare the oxygen demand Big Dissosteria cardina and Small
Dissosteria cardina.
The next, we observed respiration of big Dissosteria cardina. After
it, we observed respiration of small Dissosteria cardina. There was
different scale that showed it mean that there was a different a respiration
speed between them. Based on the big Dissosteria cardina, we obtain an
the average speed,and we got 0.00433
scale
s , While the small Dissosteria
cardina, has average speeed is 0.00239
scale
s . We can observe that
respiration on a big Dissosteria cardina in more need of small Dissosteria
cardina oksygen. This proves that the larger the organism or the more
weight the more oxygen is needed. This proves that the organism's body
weight affect speed of respiration. It can be proved that this experiment in
accordance with the theory that the bigger or more weight fast then the
organism also the process of respiration.
3. Compare the oxygen demand open Vigna radiata and close Vigna radiata
And the last, we observed respiration of open Vigna radiata. After
it, we observed respiration of close Vigna radiata. There was different
scale that showed it mean that there was a different a respiration speed
between them. Based on the open Vigna radiata, we obtain an the average
speed and we got 0.00585
average speeed is 0.01301
scale
s , While the close Vigna radiata , has
Scale
s . In this experiment, we observed that the
sprouts are still closed need oksygen more to breathe compared to mung
bean that is already open. This happens because Sprouts make breathing
by involving oxygen gas (O2) as an ingredient is absorbed/required and
generate carbon dioxide gas (CO2), water (H2O) and a number of energy
so that it can be concluded that in the process of respiration sprouts,
require compliance with requirements level oksygen and also effected by
the weight of the size of the sprouts.
CHAPTER V
CONCLUTION AND SUGGESTION
A. Conclusion
Based on the experiment we can concluded that the size of the body a
affect the speed of respiration in animals, where the more size of animal, the
speed of it respiration will more quickly, even thought the less size of animal,
the speed of it respiration will more slowly.respiration was affected by type of
organism body and weight of organism.The animal with different kinds have
different speed of the respiration. The plants with different weight have
different speed of the respiraton
B. Suggestion
It is expected that the laboratory, so check carefully whether the goods
will be given to apprentice and goods returned apprentice. Expected assistant
companion kept an eye on the apprentice especially during the heating of
materials, and continue to give explanations and directions to the apprentice
.
BIBLIOGRAPHY
Debbie, Hadley. 2013. Internal Anatomy of an Insect. http://insects.about.com/od/
morphology/ss/internalanatomy_5.htm. accessed on 06.00 am
Fried, George. 2003. Biologi. Erlangga. Jakarta.
Kaufman, Peter. 1989. Their Biology and Importance. New York : Harper and
Row.
Knobloch, Irving Wiliam. 1967. Readings in Biological Science. Appleton
Century Crofts. United State of America.
Tim Penyusun Biologi Umum. 2013. Guide Book of Basic Biology.Makassar:
Biology Department Faculty Mathematic and science state university of
Makassar.
ANSWER THE QUESTION
1. What is the function KOH that covered by cotton?
Answer : To bound CO2 that came out by organism that exist in tube
respirometer.
2. Why eosin experience friction to direction of respirometer.?
Answer : Because mere are more less O that exist in this tube prove that
organism need oxygen to did it’s activity.
Complete report of Basic Biology experiment the title “Respiratory”
arrenged by :
name
: Devi Putriana
ID
: 1314441006
class
: Biology ICP B
group
: VI (Six)
It has been checked and consult for an assistant and assistant coordinator then
stated acceptable.
Makassar, January
Assistant
Assistant Coordinator
Rachmayani Ardiansyah, S.Pd
Rukman Muslimin
ID: 101404157
Known by
Lecturer of Responsible
Dr. Muhiddin P. S.Pd, M.Pd
NIP : 19721231 19903 1 042
th
2013
CHAPTER I
INTRODUCTION
A. Background
In biology science, we cannot stop discussing about living creature. The
object study of biology is living creature. If we talk about the living creature,
then we will discuss about what living creature do for stay to live and how
about that work system. In their daily activity (human), that is so many
activity that always done, for example schooling, go to office, clean home and
many soon. But if thinking, there is an activity that we can call fundamental
activity to operating other activity like example above, this activity is
breathing. Did you know, living creature cannot live without breathing,
because if do this activity, everything work system that in into body can be
operating.
Not only human is breathing, plant, animal and other microorganism do
this activity, suitable with their method. But if there is a question what living
creature use to breathing, what will be answering for this question. Living
creature use air to breathing, so living creature can stay to live. But is there
different between living creature in breathing. Yes, there is some living
creature have different respiratory system with other living creature, this
respiratory system describe work in breathing. There is living creature use
their lung as respiration tool, but other use skin or trakea as respiration tool.
That is important as science to know how living creature to breathing. In
this experiment we will try to observe how living creature to breathing
specific about RQ or respiratory quotient. Or we can say we will try to
observe how many oxygen that needed for one living creature, is there
different between other living creature of oxygen needs. So to answer this
question, we will do this observation. This observation use living creature :
cockroach and grasshopper. We want to know too how many different oxygen
needs between them. So it is good time to this observe.
B. Purpose
1.
Proving that live organism need oxygen for their respiration.
2.
Comparing oxygen needs of some organism according to kind and size of
weight.
C. Benefit
After do this observation, apprentice can explain why the organism
need oxygen for stay to live. And apprentice can make comparison oxygen
needs between two different organism. This comparison will indicate which
organism that need much oxygen and it can explain what factor can influence
that.
CHAPTER II
PREVIEW OF LITERATURE
Respiration is oxidation process of material food or organic material which
happen inside cell that can done by aerob although anaerob. In aerob condition,
this respiration need free oxygen and release carbon dioxide and energy. If oxided
is sugar, then reaction happen is :
C6H12O6 + 6H2O 6CO2 + 6H2O + energy
Amount of CO2 produced and amount of O2 used in aerob respiration not always
same. This depends to kinds of material that used. Comparison between amount of
CO2 that released and amount of O2 that needed, called Respiratory Quotient
(RQ). For carbohydrate, its RQ value = 1. This RQ value can being varied
depends to material for respiration, perfect or not for respiration and others
condition (Tim Penyusun Biologi Umum, 2013).
One of the most important catabolic process is respiration. Respiration is the
means by which organic material such as carbohydrates, fats, and proteins are
broken down into carbon dioxide and water, with much of the energy released
being used in the synthesis of ATP. The biochemical reaction that accomplish this
break down are referred to collectively as respiratory metabolism. Respiratory
metabolism is of primary importance in both plant and animal cell since the ATP
produce essential to wide variety of reaction necessary for growth, development
and cellular work. The carbohydrates (sugar, for example) that are broken down in
respiration are produced originally via photosynthesis. In this model, the energy
of sunlight is used to drive the conveyor belt and raise water. Water that has been
lifted upward some distance has increased in potential energy. The utilization of
the energy in sunlight to raise water and increase it’s potential energy is analogous
to using light energy produce high-energy sugar from low – energy carbon
dioxide and water during photosynthesis. The high potential energy water can be
made to do work as it flows down the trough and drives the paddle wheel. As this
occurs, the water drops to a lower level and near for loses some potential energy.
This analogous to using the high – energy sugar produce by photosynthesis to
drive the energy requiring reaction of the cell. This is made possible by the
pathway of respiration (analogous to the trough), which harness the high – energy
sugar. As the high – energy sugar is respired, it is converted to the low – energy
products carbon dioxide and water. This is analogous to the water’s losing
potential energy as it flows down the trough to a lower level. Since the process of
respiration is of fundamental importance in all plant and animal cells, it will be
described in some detail in this chapter. However, it’s should be emphasized that
understanding the overall a accomplishment of respiratory metabolism and its
relationship to other cellular process is viewed as more important here then the
details of the individuals steps involved (Peter, 1989).
Respiration is the biologically controlled breakdown of energy-containing
substance, such as carbohydrates, fats, and protein, with the release of energy. As
such, it is reverse of photosynthesis, in which energy is used to synthesize
compounds. The major pathway of respiration in the cell can be broken into two
parts: ( 1 ) glycolysis and ( 2 ) the Krebs cycle. In the first part, glycolysis,
glucose is eventually split into two pyruvic acid molecules, each having three
carbon atoms. At remaining two-carbon compound enters the Krebs cycle, is
regenerated at end of the cycle and can accept another two-carbon piece. The two
parts of respiration differ in a number of important features. In the absence of
oxygen, glycolysis can occur but the Krebs cycle can not function. If no oxygen is
present, the pyruvic acid produced at the end of glycolysis is converted, with the
loss of one carbon atom, to ethyl alcohol. In this case, the process is fermentation
( Wiliam, 1967 ).
We have studied process of respiration, that process of oxidation of
metabolic by organism when there is oxygen to catch energy that had in metabolic
bondings. Respiration is not just produce energy, but other products, like carbon
dioxide and water. ( such as, glycolysis processes and fermentation break down
metabolic to get energy and produce other products ). So, gas exchange with one
way is needed to supplies oxygen for cells and remove carbon dioxide. It is rather
make us confused, but this process is known as respiration. ( Fried, 2003 ).
Insects require oxygen just as we do, and must "exhale" carbon dioxide, a
waste product of cellular respiration. Oxygen is delivered to the cells directly
through respiration, and not carried by blood as in vertebrates. Along the sides of
the thorax and abdomen, a row of small openings called spiracles (8) allow the
intake of oxygen from the air. Most insects have one pair of spiracles per body
segment. Small flaps or valves keep the spiracle closed until there is a need for
oxygen uptake and carbon dioxide discharge. When the muscles controlling the
valves relax, the valves open and the insect takes a breath. Once entering through
the spiracle, oxygen travels through the tracheal trunk (8), which divides into
smaller tracheal tubes. The tubes continue to divide, creating a branching network
that reaches each cell in the body. Carbon dioxide released from the cell follows
the same pathway back to the spiracles and out of the body. Most of the tracheal
tubes are reinforced by taenidia, ridges that run spirally around the tubes to keep
them from collapsing. In some areas, however, there are no taenidia, and the tube
functions as an air sac capable of storing air. In aquatic insects, the air sacs enable
them to "hold their breath" while under water. They simply store air until they
surface again. Insects in dry climates may also store air and keep their spiracles
closed, to prevent water in their bodies from evaporating. Some insects forcefully
blow air from the air sacs and out the spiracles when threatened, making a noise
loud enough to startle a potential predator or curious person (Debbie, 2013).
CHAPTER III
EXPERIMENT METHOD
A. Place and Date
Day/Date
: Saturday/January 4th 2013
Time
: 01.00 pm until 03.00 pm
Place
: Biology Laboratory 3nd floor of the west, faculty of
mathematic and science university of Makassar.
B. Tools and Materials
1. Tools
a. Respirometer
b. Spoit
c. stopwatch
2. Materials
a. Grasshopper(Dissoteria carolina)
b. Cockroach(Blatta orientalis)
c. Mung bean sprouts(Phaseolus radiatus)
d. Vaseline
e. Eosin solution
f. Cotton
g. KOH crystal
C. Work Procedure
1. Prepared the tools and materials that used in experiment.
2. Decided the object that used, what is a cockroach or grasshopper.
3. Wrapped KOH crystal with cotton.
4. Entered the cotton that contain KOH crystal into respirometer tube, then
determined the object which will used into respirometer tube.
5. Closed, then greased Vaseline in around mouth tube, so more session and
there is not oxygen entered in respirometer tube.
6. Then put a respirometer to prop it.
7. So, injected the iosin solution into scale glass ribbon. With first scale
0,0mL.
8. And then recorded the scale every one minute.
9. Repeated, the step 1 until 8 to the next object.
10. Recorded the experiment resulted into the table.
CHAPTER IV
RESULT AND DISCUSSION
A. Result observation
No.
Type
1.
2.
Times to-
Pariplanet
Big
1
0,01
a
Smal
0,02
0,02
0,02
0,02
0,02
americana
Dissosteria
l
Big
0,48
0,52
0,65
0,75
0,50
cardina
Smal
0,25
0,28
0,35
0,44
0,58
Vigna
l
Open
0,20
0,80
1,20
1,50
1,90
radiata
Close 0,80
1,80
2,50
3,00
3,10
3.
2
0,06
3
0,11
4
0,13
5
0,17
B. Data analysis
s
Used the formula v = t
1. Pariplaneta Americana
a. Big :
scale
s
scale
V2= s/t = 0,06/120 = 0.0005 s
scale
V3= s/t = 0,11/180 = 0.00061 s
scale
V4= s/t = 0,13/240 = 0.00054 s
scale
V5= s/t = 0,17/300 = 0.00057 s
1) V1= s/t = 0,01/60 = 0.00017
2)
3)
4)
5)
Average speed
V=
=
( v 1+ v 2+ v 3+ v 4 +v 5 )
5
scale
s
( 0.00017 +0.0005+0.00061+0.00054+ 0.00057 )
5
scale
s
0.00239
=
5
= 0.00048
scale
s
scale
s
b. Small
scale
s
scale
V2= s/t = 0,02/120 = 0.00017 s
scale
V3= s/t = 0,02/180 = 0.00011 s
scale
V4= s/t = 0,02/240 = 0.00008 s
scale
V5= s/t = 0,02/300 = 0.00007 s
1) V1= s/t = 0,02/60 = 0.00033
2)
3)
4)
5)
Average speed
( v 1+ v 2+ v 3+ v 4 +v 5 )
V=
5
=
scale
s
( 0.00033+ 0.00017+0.00011+0.00008+ 0.00007 )
=
5
0.00076
5
= 0.00015
scale
s
scale
s
2. Dissosteria cardina
a. Big
scale
s
scale
V2= s/t = 0,52/120 = 0.00433 s
scale
V3= s/t = 0,65/180 = 0.00361 s
scale
V4= s/t = 0.72/240 = 0.003 s
scale
V5= s/t = 0,82/300 = 0.00273 s
1) V1= s/t = 0,48/60 = 0.008
2)
3)
4)
5)
scale
s
Average speed
V=
=
=
( v 1+ v 2+ v 3+ v 4 +v 5 )
5
scale
s
( 0.008+ 0.00433+0.00361+0.003+0.00273 )
5
0.02167
5
= 0.00433
scale
s
scale
s
scale
s
b. Small
scale
s
scale
V2= s/t = 0,28/120 = 0.00233 s
scale
V3= s/t = 0,35/180 = 0.00194 s
scale
V4= s/t = 0,44/240 = 0.00183 s
scale
V5= s/t = 0,50/300 = 0.00167 s
1) V1= s/t = 0,25/60 = 0.00417
2)
3)
4)
5)
Average speed
V=
=
=
( v 1+ v 2+ v 3+ v 4 +v 5 )
5
scale
s
( 0.00417 +0.00233+0.00194+ 0.00183+0.00167 )
5
0.01194
5
= 0.00239
scale
s
3. Vigna radiate
a. Open
1) V1= s/t = 0,20/60 = 0.00333
scale
s
scale
s
scale
s
scale
3) V3= s/t = 1,20/180 = 0.00667 s
scale
4) V4= s/t = 1,50/240 = 0.00625 s
scale
5) V5= s/t = 1,90/300 = 0.00633 s
2) V2= s/t = 0,80/120 = 0.00667
Average speed
V=
=
=
( v 1+ v 2+ v 3+ v 4 +v 5 )
5
scale
s
( 0.00333+ 0.00667+0.00667+0.00625+ 0.00633 )
5
0.02925
5
= 0.00585
scale
s
scale
s
scale
s
b. Close
scale
1) V1= s/t = 0,80/60 = 0.01333 s
scale
2) V2= s/t = 1,80/120 = 0.015 s
scale
3) V3= s/t = 2,50/180 = 0.01389 s
scale
4) V4= s/t = 3,00/240 = 0.0125 s
scale
5) V5= s/t = 3,10/300 = 0.01033 s
Average speed
V=
=
( v 1+ v 2+ v 3+ v 4 +v 5 )
5
scale
s
( 0.01333+ 0.015+0.01389+0.0125+0.01033 )
5
scale
s
=
0.06505
5
= 0.01301
scale
s
Scale
s
Chart between Big Periplaneta americana and Small Periplaneta americana
0.18
0.17
0.16
0.14
0.13
0.12
Big Periplaneta americana
Small Periplaneta
americana
0.11
0.1
0.08
0.06
0.06
0.04
0.02 0.02
0.01
0
0.02
0.02
0.02
0.02
Chart between Big Periplaneta americana and Small Periplaneta americana
graphic relation between Big Dissosteria cardina and Small Dissosteria cardina
0.9
0.82
0.8
0.72
0.7
0.65
0.6
0.5 0.48
0.52
0.44
0.4
0.3
0.2
0.5
Big Dissosteria
cardina
Small Dissosteria
cardina
0.35
0.25
0.28
0.1
0
graphic relation between Big Dissosteria cardina and Small Dissosteria cardina
graphic relation between Open Vigna radiata and Close Vigna radiata
3.5
3
3
3.1
2.5
2
1.5
1
1.9
1.8
1.5
1.2
0.8
Open Vigna radiata
Close Vigna radiata
1.5
0.8
0.5
0.2
0
minute 1minute 2minute 3minute 4minute 5
graphic relation between Open Vigna radiata and Close Vigna radiata
C. Discussion
Based on the experiment result above, we can see the respiration every
living things, especially Periplaneta americana, Dissosteria cardina and
Vigna radiata. The weight was inversely proportional with the used of oxigen,
because every living things required O2 in a large more than weight in the
body. Size of body organism affects the respiratory, if organism has small size
of body, the respiration occurred quickly.
1. Compare the oxygen demand Big Periplaneta Americana and Small
Periplaneta americana.
Firstly, we observed respiration of big Periplaneta americana.
After it, we observed respiration of small Periplaneta Americana. From
the result of experiment we obtain an the average speed on the big
Periplaneta americana and we got 0.00048
scale
. While the small
s
Periplaneta americana has average speeed is 0.00015
scale
s . At the time
of observing our small Periplaneta americana get inappropriate results
caused by the tube respirometer not clean properly so that there is water
on the scale that caused movement of eosin.
2. Compare the oxygen demand Big Dissosteria cardina and Small
Dissosteria cardina.
The next, we observed respiration of big Dissosteria cardina. After
it, we observed respiration of small Dissosteria cardina. There was
different scale that showed it mean that there was a different a respiration
speed between them. Based on the big Dissosteria cardina, we obtain an
the average speed,and we got 0.00433
scale
s , While the small Dissosteria
cardina, has average speeed is 0.00239
scale
s . We can observe that
respiration on a big Dissosteria cardina in more need of small Dissosteria
cardina oksygen. This proves that the larger the organism or the more
weight the more oxygen is needed. This proves that the organism's body
weight affect speed of respiration. It can be proved that this experiment in
accordance with the theory that the bigger or more weight fast then the
organism also the process of respiration.
3. Compare the oxygen demand open Vigna radiata and close Vigna radiata
And the last, we observed respiration of open Vigna radiata. After
it, we observed respiration of close Vigna radiata. There was different
scale that showed it mean that there was a different a respiration speed
between them. Based on the open Vigna radiata, we obtain an the average
speed and we got 0.00585
average speeed is 0.01301
scale
s , While the close Vigna radiata , has
Scale
s . In this experiment, we observed that the
sprouts are still closed need oksygen more to breathe compared to mung
bean that is already open. This happens because Sprouts make breathing
by involving oxygen gas (O2) as an ingredient is absorbed/required and
generate carbon dioxide gas (CO2), water (H2O) and a number of energy
so that it can be concluded that in the process of respiration sprouts,
require compliance with requirements level oksygen and also effected by
the weight of the size of the sprouts.
CHAPTER V
CONCLUTION AND SUGGESTION
A. Conclusion
Based on the experiment we can concluded that the size of the body a
affect the speed of respiration in animals, where the more size of animal, the
speed of it respiration will more quickly, even thought the less size of animal,
the speed of it respiration will more slowly.respiration was affected by type of
organism body and weight of organism.The animal with different kinds have
different speed of the respiration. The plants with different weight have
different speed of the respiraton
B. Suggestion
It is expected that the laboratory, so check carefully whether the goods
will be given to apprentice and goods returned apprentice. Expected assistant
companion kept an eye on the apprentice especially during the heating of
materials, and continue to give explanations and directions to the apprentice
.
BIBLIOGRAPHY
Debbie, Hadley. 2013. Internal Anatomy of an Insect. http://insects.about.com/od/
morphology/ss/internalanatomy_5.htm. accessed on 06.00 am
Fried, George. 2003. Biologi. Erlangga. Jakarta.
Kaufman, Peter. 1989. Their Biology and Importance. New York : Harper and
Row.
Knobloch, Irving Wiliam. 1967. Readings in Biological Science. Appleton
Century Crofts. United State of America.
Tim Penyusun Biologi Umum. 2013. Guide Book of Basic Biology.Makassar:
Biology Department Faculty Mathematic and science state university of
Makassar.
ANSWER THE QUESTION
1. What is the function KOH that covered by cotton?
Answer : To bound CO2 that came out by organism that exist in tube
respirometer.
2. Why eosin experience friction to direction of respirometer.?
Answer : Because mere are more less O that exist in this tube prove that
organism need oxygen to did it’s activity.