AP Biology Lab Activities
Laboratory activities are an integral component of any college level biology course.
These activities allow students to process data, formulate hypotheses, and apply what
they have learned from a lecture or reading assignment. The College Board has
developed 12 exercises that are both hands-on and high-level, to provide students with an
opportunity to gain experiences consistent with those offered at the college level.
Through out the year we will conduct all of these labs (either as demonstration, computer
simulation or hands-on), as well as other labs and activities to prepare you for the AP
Exam in May. These labs require that you analyze and interpret both qualitative and
quantitative data, as well as use proper procedures and techniques in order to be
successful. Successful completion of these exercises will prepare you for the FreeResponse portion of the AP exam.
Below is a description of each of the 12 required laboratory exercises (and the
introductory scientific method simulation lab), the expected outcomes for each exercise,
expected results at the completion of each lab, sample multiple choice questions related
to the exercise, and links to sites with additional information about the topic being
studied.
Introductory Lab: Scientific Method/Cricket Chirps
Computer Simulations (All 12 College Board/AP Biology Labs)
Lab Bench Review of Labs
Laboratory 2: Mitosis
Part A: Mitosis/Microscope Lab
Part B: Computer Simulation of Meiosis (later in the term)
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
• Demonstrate the procedure to stain tissue for the identification of cells in the various
stages of mitosis
• Calculate the relative duration of mitosis
Results
The relative length of mitotic stages are, 53.4% of prophase, 17.4% of metaphase, 16.8%
of anaphase and 12.4% of 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.
Laboratory 3: 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 and water potential
• 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 a potato cell 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.
Sample Multiple - Choice Question
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.
Laboratory 4: 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.
Sample Multiple - Choice Question
Which of the following is LEAST likely to increase the forward rate of an enzymemediated 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
Laboratory 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.
Laboratory 6: Plant Pigments and Photosynthesis
Part A: Chromatography
Part B: Computer Simulation of 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 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.
Laboratory 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
• 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
Laboratory 8: 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
• 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
ampicllin 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 double-stranded 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.
Laboratory 9: Population Genetics and Evolution
Overview
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. Which of the following generates the formation of adaptations?
A. Genetic drift
D. Sexual reproduction
B. Mutations
E. Natural selection
C. Gene flow
2. 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
Laboratory 10: 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.
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.
Laboratory 11: 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.
Laboratory 12: 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
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
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
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
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
Laboratory 12: 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, after the baroreceptor reflex, 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.
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
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.