Download APBIO/GAYNOR SEM: 02/09 NAME: DATE: PERIOD: ______ AP

APBIO/GAYNOR
NAME: ________________________________
SEM: 02/09
DATE: ___________________ PERIOD: _________
AP BIOLOGY LAB REVIEW
INSTRUCTIONS: This is a helpful summation of each of the labs performed in class. The
OVERVIEW is good for looking at material that should be discussed in essays and the
OBJECTIVES are what you should know about the experiment (purpose!!). RESULTS
sections include some mock data that should be used to derive some conclusions. Think of this
review as a guide to essay writing and what should be included in the essays!
PART I: THE LABS
LAB #3: Mitosis & Meiosis
Overview
In this lab you will study plant mitosis using prepared slides of onion root tips and will calculate the relative
period of the phases of mitosis in the meristem of root tissue. You will also study the crossing over and
recombination that occurs during meiosis in a computer simulation.
Objectives
• Compare the events of mitosis in plant cells with those of animal cells
• Calculate the relative duration of mitosis (cell division) to interphase (no dividing), and compare the relative
duration of the stages of mitosis
Results
The relative lengths of the mitotic stages are: 53.4% prophase, 17.4% metaphase, 16.8% anaphase and 12.4%
telophase. Meiosis is important for sexual reproduction because it reduces the chromosome number by half and
it also results in new combinations of genes through independent assortment and crossing over, followed by the
random fertilization of eggs by sperm.
LAB #1: Diffusion and Osmosis
Overview
In this lab you will investigate the movement of water across semi-permeable membranes. You will also
examine the effect of solute concentrations on water potential as it relates to living tissues.
Objectives
• Describe the mechanisms of diffusion and osmosis
• Describe how solute size and molar concentration affect the process of diffusion through a selectively
permeable membrane
• Design an experiment to demonstrate and measure water potential
• Relate osmotic potential to solute concentration
• Describe how pressure affects the water potential of a solution
• Describe the effects of water gain or loss in animal or plant cells
• Calculate the water potential of living plant cells from experimental data
Results
When a solution such as that inside dialysis tubing is separated from pure water by a selectively-permeable
membrane water will move by osmosis from the surrounding area where the water potential is higher into the
cell where water potential is lower due to the presence of solute. The movement of water into the cell causes the
cell to swell and the cell membrane pushes against the cell wall to produce an increase in pressure (turgor). This
process will continue until the water potential of the cell equals the water potential of the pure water outside the
cell. At this point, a dynamic equilibrium is reached and net water movement will cease.
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APBIO/GAYNOR
NAME: ________________________________
SEM: 02/09
DATE: ___________________ PERIOD: _________
Sample Multiple-Choice Question for Diffusion and Osmosis
A dialysis bag is filled with a 3% starch solution. The bag is immersed in a beaker of water containing a 1% IKI
solution. All of the following observations are correct EXCEPT:
A. When the bag is first placed in the beaker, the water potential inside the bag is negative.
B. When the bag is first placed in the beaker, the solution in the beaker is yellow brown.
C. The starch solution inside the bag is hypertonic relative to the solution in the beaker.
D. After fifteen minutes, the solution in the bag turns blue.
E. After fifteen minutes, the mass of the dialysis bag has decreased.
Answer: (E) The solution inside the dialysis bag is more concentrated than the surrounding solution (the
solution in the bag is hypotonic to the surrounding solution. Water will diffuse into the bag (hypo = hippo!),
causing the bag to swell, i.e., the bag’s mass will increase.
Other Questions
You have two different sucrose solutions in two different sets of dialysis tubing. How might you determine
which solution is the more concentrated without removing the solutions from the tubing?
A cell placed in a hypertonic solution will shrink, swell, or stay the same? Explain.
A cell placed in a hypotonic solution will shrink, swell, or stay the same? Explain.
A cell placed in an isotonic solution will shrink, swell, or stay the same? Explain.
LAB # 2: Enzyme Catalysis
Overview
In this lab you will measure the rate of a reaction in the presence and absence of a catalyst. The catalyst,
(catalase or peroxidase), is an enzyme in cells that catalyzes the breakdown of toxic H202.
Objectives
• Graph data from an enzyme experiment
• Determine the rates of enzymatically catalyzed reactions
• Discuss the method for determining enzyme activity
• Discuss the relationship between dependent and independent variables
• Discuss the effect of initial reaction rates produced by changes in temperature, pH, enzyme concentrations,
and substrate concentrations
• Design an experiment to measure the effect of enzyme activity produced by changes in temperature, pH,
enzyme concentrations, and substrate concentrations
Results
In the first few minutes of an enzymatic reaction, the number of substrate molecules is usually so large
compared to the number of enzyme molecules that changing the substrate concentration does not (for a short
period at least) affect the number of successful collisions between substrate and enzyme. During this early
period, the enzyme is acting on substrate molecules at a constant rate. The slope of the graph line during this
early period is called the initial velocity of the reaction. The initial velocity, or rate, of any enzyme catalyzed
reaction is determined by the characteristics of the enzyme molecule. It is always the same for an enzyme and
its substrate as long as temperature and pH are constant and substrate is present in excess. Also, in this
experiment the disappearance of the substrate is essential in this reaction. Eventually, the rate of the reaction
levels off.
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APBIO/GAYNOR
NAME: ________________________________
SEM: 02/09
DATE: ___________________ PERIOD: _________
Sample Multiple - Choice Question
Which of the following is LEAST likely to increase the forward rate of an enzyme-mediated reaction?
A. An increase in the substrate concentration
B. An increase in the enzyme concentration
C. An increase in the product concentration
D. An increase in pH
E. An increase in the temperature
Answer: (C) Since enzyme-mediated reactions are reversible (they convert product back
to substrate), increasing the concentration of the product will slow the forward direction
of the reaction and accelerate the reverse reaction. Conversely, and increase in the
substrate concentration will increase the forward rate of the reaction. Increasing the
enzyme concentration will not slow the reaction rate but may increase it if the substrate
concentration is high enough to utilize additional enzyme. An increase in pH or
temperature may change the rate of reaction, but the nature of the enzyme must be known
in order to determine whether the rate is increased or decreased
LAB #5: Cell Respiration
Overview
In this lab you will measure oxygen consumption during respiration as a change in gas volume in germinating
and non-germinating peas at two different temperatures.
Objectives
• Discuss the gas laws as they apply to the function of a respirometer
• Interpret data related to the effects of temperature on cell respiration
• Interpret data related to the effects of germination or non-germination on cell respiration
• Explain or determine the significance of a control
• Explain the relationship between dependent and independent variables
• Calculate a rate of cell respiration by utilizing graphed data
• Design an experiment to use a respirometer to measure cellular respiration
Results
Examples: O2 Graph and CO2 Graph
Germinating peas respire and need to consume oxygen in order to continue the growing process. Pea seeds are
non-germinating and do not respire actively. These seeds are no longer the site of growth and thus do not need
oxygen for growth. In consideration to temperature, at higher temperatures more oxygen is consumed which
means more respiration is occurring. 686 kilocalories are released during respiration. When temperature
decreases molecular motion slows down and respiration decreases because less energy is made available.
LAB #4: Plant Pigments and Photosynthesis
Part A: Chromatography
Part B: Relationship Between Respiration & Photosynthesis in Elodea
Part C: DPIP Reduction
Overview
In this lab you will separate plant pigments using paper chromatography. You will also measure the rate of
photosynthesis in isolated chloroplasts using a measurement technique that involves the reduction of the dye,
DPIP. The transfer of electrons during the light-dependent reactions of photosynthesis reduces DPIP and
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APBIO/GAYNOR
NAME: ________________________________
SEM: 02/09
DATE: ___________________ PERIOD: _________
changes its color from blue to colorless.
Objectives
• Understand the principles of chromatography
• Calculate Rf values
• Design an experiment in which chromatography is used as a separation technique
• Describe a technique for determining photosynthetic rate
• Understand the relationship between dependent and independent variables
• Describe how light intensity, light wavelength, and temperature can affect photosynthesis
• Design an experiment to measure how light intensity, light wavelength, and temperature can affect
photosynthetic rates
Results
The solvent moves up the paper by capillary action, which occurs as a result of the attraction of solvent
molecules to the paper and the attraction of solvent molecules to one another. As the solvent moves up the
paper, it carries along any substances dissolved in it, in this case pigments. The pigments are carried along at
different rates because they are not equally soluble in the solvent and because they are attracted, to different
degrees, to the cellulose in the paper through the formation of hydrogen bonds. Also, as the DPIP is reduced and
becomes colorless, the resultant increase in light transmittance is measured over a time course using a
spectrophotometer.
Sample Multiple - Choice Question
In a paper chromatography procedure, molecules with which of the following characteristics migrate the fastest
up the chromatography paper?
A. High solubility in solvent and weak hydrogen bonding to cellulose.
B. High solubility in solvent and strong hydrogen bonding to cellulose.
C. Low solubility in solvent and strong hydrogen bonding to cellulose.
D. Low solubility in solvent and weak hydrogen bonding to cellulose.
E. Insoluble in solvent.
Answer: (A)
Other Questions
How does chromatography work?
Using the data below, calculate the Rf values for the pigments (the solvent front moved 100 mm)
Band 1 = 10 mm, Rf =
Band 2 = 35 mm, Rf =
Band 3 = 42 mm, Rf =
Band 4 = 66 mm, Rf =
LAB #7: Genetics Of Drosophila
Computer Simulation of Drosophila Genetics
Overview
In this lab you will use the fruit fly Drosophila melanogaster to do genetic crosses. You will learn how to
collect and manipulate fruit flies, collect data from F1 and F2 generations, and analyze the results from a
monohybrid, dihybrid or sex link cross.
Objectives
• Conduct a genetics experiment for a number of generations
• Compare predicted results with actual results
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APBIO/GAYNOR
NAME: ________________________________
SEM: 02/09
DATE: ___________________ PERIOD: _________
• Explain the importance of Chi-Square analysis
• Design genetic crosses to illustrate independent assortment and sex linkage
• Discuss the life cycle of the fruit fly, recognize the sex of fruit flies, and recognize several types of classic
mutations
Results
From this lab, you will be able to find genotypes and phenotypic expression within a fruit fly. Also, recessive
genes and mutations will be revealed as the student crosses a variety of Drosophila alleles. For example, if a
female carrier for an x-linked, recessive trait, was crossed with a male without the recessive trait the results
would be:
½ males with x-linked trait ½ males without
½ female carriers ½ females without
0 females express sex linked traits
Sample Multiple - Choice Question
A population consists of 20 individuals of which 64% are homozygous dominant for a
particular trait and the remaining individuals are all heterozygous. All of the following
can explain the situation except
A. Genetic drift is occurring
B. The recessive allele is deleterious
C. All homozygous recessive individuals emigrate
D. The population is very small
E. Only heterozygous individuals mate
Answer: (E)
Other Questions
What does Chi-Square measure?
What are the chances of getting the following from a cross of AaBbCC x AaBBCc?
Genotype = AaBBCC (½ x ½ x ½ = 1/8)
Complete the following data for the following monohybrid cross of incomplete dominance: Bb x Bb where BB
= black, Bb = brown and bb = albino
Actual (observed) phenotypes: 30 black, 40 brown, and 30 albino
Answer
Expected phenotypes: 25 black, 50 brown, and 25 albino
Chi-Square
At the 0.5 probability level, are the results significantly different from what was expected? Explain.
LAB #6: Molecular Biology
Part A: DNA Restriction
Part B: Bacterial Transformation
Overview
In this lab you will investigate some basic principles of genetic engineering. Plasmids containing specific
fragments of foreign DNA will be used to transform E. coli cells, conferring antibiotic (ampicillin) resistance
and the Lac + phenotype (ability to metabolize lactose). Restriction enzyme digests of phage lambda DNA will
also be used to demonstrate techniques for separating and identifying DNA fragments using gel electrophoresis.
Objectives
• Discuss the principles of bacterial transformation
• Describe how to prepare competent E. coli cells
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APBIO/GAYNOR
NAME: ________________________________
SEM: 02/09
DATE: ___________________ PERIOD: _________
• Discuss the mechanisms of gene transfer using plasmid vectors
• Discuss the transfer of antibiotic resistance genes and tell how to select positively for transformed cells that
are antibiotic resistant
• Discuss the mechanisms of action for restriction endonucleases
• Discuss how a plasmid can be engineered to include a piece of foreign DNA that alters the phenotype of the
transformed cells
• Understand and be able to explain the principles of electrophoresis as they pertain to separating and
identifying DNA fragments
Results
Bacterial Transformation-Ampicillin Resistance: In this exercise, we will introduce competent E. Coli cells to
take up the plasmid pAMP, which contains a gene for ampicillin resistance. Normally, E. Coli cells are
destroyed by the antibiotic ampicillin, but E. Coli cells that have been transformed will be able to grow on agar
plates containing ampicillin. Thus, we can select for transformants; those cells that are not transformed will be
killed by ampicillin; those that have been transformed will survive.
Restriction Enzyme Cleavage of DNA: Restriction endonuclease recognizes specific DNA sequences in doublestranded DNA and digests the DNA at these sites. The result is the production of fragments of DNA of various
lengths corresponding to the distance between identical DNA sequences within the chromosome. By taking
DNA fragments and systematically reinserting the fragments into an organism with minimal genetic material, it
is possible to determine the function of particular gene sequences
Electrophoresis: Fragments of DNA can be separated by gel electrophoresis when any molecule enters the
electrical field, the mobility or speed at which it will move is influenced by the charge (negative charges travel
to positive/top pole of gel), the density of the molecule, (the smaller the molecule, the faster it travels), the
strength of the electrical field, and the density of the medium (gel) which it is migrating.
Sample Multiple - Choice Question
After growth on ampicillin to select bacteria transformed with a mixture of recombinant
DNA containing plasmids, you must identify a clone containing a specific gene sequence.
You would:
A. Blot transfer clones to membranes, and screen using a radioactive probe complementary to the gene.
B. Re-grow bacteria in ampicillin where only transformants containing the gene of interest can grow.
C. Culture bacteria in both ampicillin and tetracycline to select for bacteria containing the gene of interest.
D. Digest DNA from the plasmid to isolate the gene fragment.
E. Do a restriction map of plasmid DNA to identify the correct clone.
Other Questions
How does a restriction enzyme work?
At what location in the cell do endonucleases work?
What is a plasmid?
Which DNA fragments migrate further along a gel in electrophoresis, long or short?
What happened to each of the colonies below and why?
LB without plasmid
LB with plasmid
LB with plasmid & ampicillin
LB without plasmid, but with ampicillin
Population Genetics and Evolution
Overview
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APBIO/GAYNOR
NAME: ________________________________
SEM: 02/09
DATE: ___________________ PERIOD: _________
In this lab you will learn about Hardy-Weinberg law of genetic equilibrium and study the relationship between
evolution and changes in allele frequency by using your class as a sample population.
Objectives
• Calculate allele and genotype frequencies using the Hardy-Weinberg theorem
• Discuss the effect of natural selection on allelic frequencies
• Explain and predict the effect of allelic frequencies of selection against the homozygous recessive
• Discuss the relationship between evolution and changes in allele frequencies, as measured by deviations
from the Hardy-Weinberg law of genetic equilibrium
Results
Assuming that Hardy-Weinberg equilibrium is maintained allele and genotype frequencies should remain
constant from generation to generation. For this to happen the five following situations must all occur:
1. Population is very large. The effects of chance on changes in allele frequencies is
thereby greatly reduced.
2. Individuals show no mating preference, i.e. random mating.
3.There is no mutation of alleles.
4. No differential migration occurs, (no immigration or emigration).
5. All genotypes have an equal chance of surviving and reproducing, i.e. there is no natural selection.
In humans, several genetic diseases have been well characterized. Some of these diseases are controlled by a
single allele where the homozygous recessive genotype has a high probability of not reaching reproductive
maturity. If this were to occur both the homozygous dominant and heterozygous individuals will survive while
the homozygous recessive will become extinct.
Sample Multiple - Choice Questions
1. All of the following are examples of evolution, except:
A. Mutations in an individual
B. Changes in an allele frequency in a population
C. Changes in an allele frequency in a species
D. Divergence of a species into two species
E. Adaptive Radiation
Answer: (A) Only populations evolve.
Other Questions
1. In a Hardy-Weinberg population, the frequency of the a allele is 0.4. What is the frequency of
individuals with genotype Aa?
2. In a population with two alleles, A and a, the frequency of a is 0.6. What is the population of
heterozygotes in the population in H-W equilibrium?
3. Why is the heterozygote for sickle-cell anemia favorable against malaria?
LAB #11: Behavior/Habitat Selection
Overview
In this lab, you will examine the habitat preferences of the isopod (roly poly). You will use controlled
experimentation to determine the thermal, pH, and light environments selected by this isopod. Based on your
experience with this lab, you will design an experiment that could be used to survey other variables and other
organisms.
7
APBIO/GAYNOR
NAME: ________________________________
SEM: 02/09
DATE: ___________________ PERIOD: _________
Objectives
• Describe the relationship between dependent and independent variables
• Discuss the value of comparing experimental results with control results
• Graph an interpret histogram data
• Measure the volumes, distances, and temperature using metric scales
• Design and conduct an experiment to measure the effect of environmental variables on habitat selection
Results
When conducting this experiment, a couple of things should be understood.
A. Three variables are tested: light, temperature and pH. The control is exposed to room
light, room temperature, and neutral pH is also prepared. For each of the variables, a
gradient is established providing a continuous variation from weak to strong intensities.
B. The histograms are prepared showing the number of isopods in each of the four
intensities. A histogram is prepared for each of the three variables and the control. From
the data in the histograms, conclusions cab be made describing the habitat preferences of
the isopod.
LAB #12: Dissolved Oxygen and Aquatic Primary Productivity
Overview
You will measure and analyze the dissolved oxygen concentration in water samples. You will also measure and
analyze the primary productivity of natural waters or lab cultures.
Objectives
• Describe the physiological importance of carbon and oxygen in an ecosystem
• Understand the physical and biological factors that affect the solubility of dissolved gases in aquatic
ecosystems
• Describe a technique for measuring dissolved oxygen
• Define primary productivity
• Describe the relationship between dissolved oxygen and the processes of photosynthesis and respiration as
they affect primary productivity in an ecosystem
• Design an experiment to measure primary productivity in an aquatic ecosystem
• Understand the effect of light and nutrients on photosynthesis
Results
The amount of oxygen dissolved in natural water samples is measured and analyzed to determine the primary
productivity of the sample. The amount of dissolved oxygen is dependent upon many factors.
A. Temperature
B. Salinity
C. Photosynthesis
D. Respiration
Primary productivity is a measure of the amount of biomass produced by autotrophs through photosynthesis per
unit time. It can be examined by the following factors
A. Gross Primary Productivity
B. Net Primary Productivity
C. Respiratory Rate
8
APBIO/GAYNOR
NAME: ________________________________
SEM: 02/09
DATE: ___________________ PERIOD: _________
Sample Multiple - Choice Questions
The net primary productivity for a temperate forest was measured at 2000 mg C/m2/day. The respiratory rate of
the community was determined to be 1000 mg C/m2/day. The gross primary productivity for this community is
A. 1000 mg C/m2/day
B. 2000 mg C/m2/day
C. 3000 mg C/m2/day
D. 4000 mg C/m2/day
E. 5000 mg C/m2/day
Answer: (C) The gross primary productivity is the sum of the net primary productivity and the respiratory rate.
Other Questions
1. Create a graph showing the relationship between temperature and the percent saturation of dissolved oxygen.
2. Create a graph showing the relationship between the amount of light received and the amount of productivity.
3. Create a graph showing the relationship between the amount of oxygen consumed by photosynthetic
organisms and the carbon fixed.
4. Why is net primary productivity lower than gross primary productivity?
LAB #9: Transpiration
Overview
In this lab you will apply what you learned about water potential from Lab 1 (Diffusion and Osmosis) to the
movement of water within the plant. You will study the organization of the plant stem as it relates to these
processes by observing sections of fresh tissue.
Objectives
• Describe how differences in water potential affect the transport of water from roots to stems to leaves
• Relate transpiration to the overall process of water transport in plants
• Discuss the importance of properties of water - including hydrogen bonding, adhesion, and cohesion - to the
transport of water in plants
• Quantitatively demonstrate the effects of different environmental conditions on the rate of transpiration in
plants
• Identify the vascular tissues of the plant stem and describe their functions
Results
Conditions that cause a decreased rate of water loss from leaves result in a decreased water potential gradient
from stem to leaf and therefore in a decreased rate of water movement up the stem to the leaves. Conditions that
cause an increased rate of water loss from leaves result in an increase in the water potential gradient from stem
to leaf and therefore in an increase in the rate of water movement up the stem to the leaves.
Normal Room Conditions (CONTROL)
When you expose a plant to room conditions nothing is supposed to happen. The
reasoning for this is room conditions don’t cause drastic changes in the plants
environment for major transpiration or even water gain to occur. The plant under room
conditions is considered to be your control.
Floodlight
When light is absorbed by the leaf, some of the light energy is converted to heat and
remember that transpiration rate increases with temperature. We learned in Unit One of
the Campbell’s edition that when the temperature of liquid water rises, kinetic energy of
the water molecules increases. As a result, the rate at which liquid water is converted to
water vapor increases. When the water is turned into water vapor, it easily passes out
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APBIO/GAYNOR
NAME: ________________________________
SEM: 02/09
DATE: ___________________ PERIOD: _________
through stomata out into the outer atmosphere.
The floodlight is an example of a plant near the sun (which is why the plant is one meter
away from the light). Due to the aforementioned properties of plants you should see a
loss of water.
Fan (WIND)
An increase in wind speed results in an increase in the rate of leaf water loss because
increased wind decreases the boundary layer of still air at the leaf surface. This boundary
layer acts to slow leaf water loss. Increased wind also causes the rapid removal of
evaporating water molecules from the leaf surface. This results in a low water potential
in the air immediately and the water level should drop.
Mist (HIGH HUMIDITY)
Increased humidity in the air surrounding the leaf decreases the water potential gradient
between the saturated air in the leaf air spaces and the air surrounding the leaf, resulting
in a decreased rate of leaf water loss. However, when the humidity of the air surrounding
the leaf if very low, the water potential of the air is low. Therefore, the water potential
gradient between the air spaces of the leaf and the surrounding air is high, and the rate of
leaf water loss increases.
Or maybe this explanation is better
When there is a great amount of humidity, transpiration decreases because of water
potential. When the humidity is at a low or normal, the mesophyll cells in the plant are
much higher in water potential than the relatively drier surrounding air. Due to the
properties of water potential, which states that water tends to evaporate from the leaf
surface moving from an area of higher water potential to an area of lower water potential,
transpiration occurs. But, because of the high humidity, the surrounding air has a higher
water potential than the mesophyll cells and water loss is at a minimum.
Adaptations to reduce leaf water loss include a reduced number of stomates, an increase in the thickness of the
leaf cuticle, a decrease in leaf surface area, and adaptations that decrease air movements around stomates, such
as dense hairs and sunken stomates. Because leaves are all different in size, reporting the water loss without
considering a unit area would provide non-comparable data.
Sample Multiple - Choice Question
Which of the following series of terms correctly indicates the gradient of water potential from lowest water
potential to highest water potential?
A. Air, leaf, stem, root, soil
B. Soil, root, stem, leaf, air
C. Root, leaf, stem, air, soil
D. Air, soil, root, leaf, stem
E. Stem, leaf, root, soil, air
Answer: (A) Water potential is highest in soil, decreases from root to leaf, and is lowest in the air. Water moves
from the soil into the roots and through the plant an transpires from the leaf because water moves from the area
of greatest water potential to the area of lowest water potential
Other Questions
Using the following conditions, create a graph showing the relationship between the amount of transpiration and
time. Then write a short explanation of the physiological responses due to the environmental conditions.
A plant at room temperature
A plant in humid conditions
A plant in high light conditions
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APBIO/GAYNOR
NAME: ________________________________
SEM: 02/09
DATE: ___________________ PERIOD: _________
A plant in very dry conditions
LAB #10: Physiology of the Circulatory System
Part A: Cardiovascular Fitness
Part B: Computer Simulation of Heart Rate in Daphnia
Overview
You will learn how to measure blood pressure and measure pulse rate under different physiological conditions:
standing, reclining, and during and immediately after exercise. The blood pressure and pulse rate will be
analyzed and related to a relative fitness index. You will also measure the effect of temperature on the heart rate
of the Daphnia magna, and calculate a Q10 for the relationship between temperature and heart rate.
Objectives
• Measure pulse rate
• Measure blood pressure
• Describe the relationship between the changes in heart rate and blood pressure relative to changes in body
position
• Describe the relationship between changes in heart rate and exercise
• Determine the "fitness index" for an adult human
• Perform statistical analysis on class data
• Define Q10
• Determine the Q10 of heart rate in a living organism such as Daphnia
Results
The sphygmomanometer measures the blood pressure. The blood pressure cuff is inflated so that blood flow
stops to through the brachial artery in the upper arm. A stethoscope is used to listen to blood flow entering the
brachial artery. When blood first enters the artery, snapping sounds called the sounds of Korotkoff are
generated.
A. Blood pressure and heart rate increase when you move from a reclining to a standing
position counteracting gravitational pull on the blood
B. Elevated arterial blood pressure indicates increased arterial resistance to blood flow
C. Fit individuals can pump a larger volume of blood with each contraction and deliver
more oxygen to muscle tissue than the hearts of unfit individuals. As a result, blood
pressure and heart rate increases are smaller for fit individuals, and the time required to
return to normal conditions is shorter for fit individuals than unfit individuals.
D. For the Daphnia, remember that ectothermic animals use behavior to regulate their
body temperatures and that Q10 cannot be determined for endothermic animals because
body temperatures remain constant regardless of environmental temperatures.
Sample Multiple - Choice Question
A Q10 value of 3 in an ectothermic animal means that the metabolic rate
A. Triples when body temperature triples
B. Triples when body temperature increases by 10° C
C. Doubles when the body temperature increases by 3° C
D. Doubles when the body temperature increases by 10° C
E. Triples when the body temperature decreases by 10° C
Answer: (B) The Q10 is the ratio of the metabolic rate at one temperature to the metabolic rate at a temperature
10° colder. A Q10 equal to 3 indicates that he metabolic rate triples when the body temperature increases by
10° C.
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APBIO/GAYNOR
NAME: ________________________________
SEM: 02/09
DATE: ___________________ PERIOD: _________
Other Questions
What is systolic pressure?
What is diastolic pressure?
What does it mean to have high blood pressure?
Why does your heart rate go up when you exercise?
Why does your heart rate go up when you are vertical vs. horizontal?
What is Q10?
Calculate the Q10 for the following conditions: rate at lower temperature = 76, rate at higher temperature = 145.
Create a graph showing the relationship between temperature and heart rate.
PART II: LAB FREE RESPONSE QUESTIONS YOU MIGHT SEE!!
A. POINTS OF EMPHASIS
1. State a hypothesis, i.e., an expectation of results based on the known effects
of the independent variable.
2. Design and identify a control group for comparison.
3. Tell how you will hold at least two other experimental variables constant.
4. Indicate how the independent variable will be manipulated/varied/changed.
5. Describe how the dependent variable will be measured quantitatively.
6. Verify your results through multiple trials (repetition on the same procedure).
7. Analyze the results (take the average, etc.).
8. If a rate is derived, indicate how it is calculated (slope, etc.).
9. Explain why you are doing the various procedural steps.
10. It is a good idea to describe a range of expected results across the range of
biological diversity (dark to light, 0-100 degrees, etc.).
B. POSSIBLE PHRASES
Explain the purpose of each step of your procedure.
Describe how you could determine whether…
Summarize the pattern
Choose one of the variables that you identified and design a controlled
experiment to test
Describe the essential features of a device that could be used to measure
Explain how the concept of _ is used to account for _
Construct and label a graph using the data
Identify three different environmental variables that could account for the
differences
Design an experiment!
Design an experiment!
Design an experiment!
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