AP Biology Exam Review: Lab Essays
At least one essay (FRQ) on the exam will be based on an AP laboratory. To prepare for
this question, review the objectives for all twelve laboratory exercises. The College
Board does not necessarily expect that you have completed that lab, but rather that you
have investigated the objectives of the lab.
You may be asked to “design an experiment to determine….” You don’t necessarily
need to create a new lab; if you have done an activity that would answer the question,
simply describe it.
For a good response, you should include the following.
1. State a hypothesis [as an “if…..(conditions), then….(results)” statement]
Be sure your hypothesis is testable.
2. Identify the variable factor.
3. Identify the control.
Be certain to explain the control for the experiment.
4. Hold all other variables constant.
5. Manipulate the variable.
6. State how you would measure the results.
7. Discuss the expected results.
Relate the results to your hypothesis.
8. Include steps to replicate or verify.
You may be asked to graph data. Be sure to use a graph that is appropriate for you
data. Bar graphs are used when data points are discrete (not related to one another),
while line graphs are used with the data are continuous. If there is a data point at zero,
be certain to extend your line to 0, but do not extend the line to 0 if there is no data
point at zero.
Other points to keep in mind:
1. Label a graph with a descriptive title.
2. Label the x- and y-axes. (Know which variable is dependent and which is
independent.)
3. Keep all measurement units constant. Each division on a graph must
represent the same value.
Lab 1: Diffusion and Osmosis
Know the terminology! This is one place where you will not get any credit if you
understand the concept but mess up the vocabulary.
diffusion
hypertonic
dynamic equilibrium
osmosis
isotonic
selective permeability
solvent /solute
hypotonic
water potential
Remember: Water moves from a region of higher water potential to a region of lower
water potential.
Ψ = Ψp + Ψs
In a plant cell, pressure exerted by the cell wall limits water uptake.
Pure water at atmospheric pressure has a solute potential of zero. As solute is added,
the solute potential value becomes more negative.
As solute is added, the water potential of a solution drops, and water will tend to move
into the solution.
The water potential of pure water in an open container is zero because there is no
solute and the pressure in the container is zero.
Lab 2: Enzyme Catalysis
This experiment investigates the rate at which the catalase (enzyme) converts H2O2
(substrate) to O2 and H2O (products). Catalase is allowed to react with H2O2 for varying
amounts of time and then the reactions are stopped by adding H2SO4. To determione
the amount of H2O2 that remains after the reaction you do a titration with KMnO4.
[Note that we also used catalase, but looked at the effect of pH, temperature, and
concentration on reaction rate.]
Remember: Change the shape of a protein and change its function.
Enzymatic activity is affected by pH and temperature because these affect the 3-D
shape. Extremes of pH or temperature result in protein denaturation. (Cold
temperatures don’t denature proteins, but they do decrease reaction rates.}
Rate =
Δy
Δx
Lab 3: Mitosis and Meiosis
Part I: Observing mitosis
You used collected data to calculate the relative time spent in each of the phases
of mitosis.
If a particular cell takes, on average, 24 hours (1440 minutes) to complete the cell cycle, then
multiply the percentage of cells in a particular stage by 1440 minutes to calculate the time for
each stage.
% of cells in a stage x 1440 minutes = number of minutes in the stage
Part II: Meiosis simulation and study of crossing over in Sordaria
When growing filaments of two haploid strains of Sordaria that produce spores
of different colors meet, fertilization occurs and zygotes form.
Meiosis occurs within fruiting bodies to form four haploid ascospores (spores
contained in special sacs called asci).
One mitotic division doubles the number of ascopores to eight.
The number of map units between two genes or between a gene and the
centromere is calculated by determining the percentage of recombinants that
result from crossing over. The greater the frequency of crossing over, the
greater the map distance.
Calculate the percent of crossovers by dividing the number of crossover asci by
the total number of asci x 100.
To calculate the map distance, divide the percent of crossover asci by 2 (because
only half the spores in each ascus are the result of crossing over.
Laboratory 4: Plant Pigments and Photosynthesis
There are two parts to this lab exercise – separate out chlorophyll pigments using paper
chromatography and measure the rate of photosynthesis by determining the change in
color that occurs when DPIP is reduced over time.
The migration of pigment relative to the migration of solvent relative to the migration of
solvent would not change. This migration of pigments relative to migration of solvent is
expressed as a constant Rf (rerference front). It is calculated by using the formula
Rf =
distance pigment migrated
distance solvent front migrated
When chloroplasts are exposed to light, electrons from each photosystem will be
boosted to a higher energy level and this energy is used to produce ATP and to reduce
NADP+ to NADPH. In this experiment, DPIP is substituted for NADP+ and will be
reduced when light hits the coloroplasts. DPIPox is blue; DPIPre is colorless.
Experimental Design
1. Set up four tubes with buffer
and distilled water.
2. Add DPIP and chloroplasts as
indicated in the picture.
3. Place a container of water
between the light source and
the tubes to serve as a heat
sink.
4. Expose the tubes to strong
lights.
5. Using a spectrophotometer,
read the % transmittance for
each sample.
noDPIP, unboiled
chloroplasts
DPIP, unboiled
chloroplasts
wrapped in foil
DPIP, boiled
chloroplasts
DPIP, unboiled
chloroplasts
Sample questions:
1. What is the role of DPIP in this experiment?
A. It mimics the action of chlorophyll by absorbing light energy,
B. It serves as an electron donor and blocks the formation of NADPH.
C. It is an electron acceptor and is reduced by electrons from chlorohyll.
D. It is bleached in the presence of light and can be used to measure light levels.
2. Some students were not able to get many data points in this experiment because
the solution went from blue to colorless in only 5 minutes for the unboiled
chloroplasts exposed to light. What modification to the experiment do you think
would be most likely to provide better results?
A. Increase the number of drops of chloroplasts used from three to five.
B. Double the volume of DPIP so that the solution has a lower initial
transmittance.
C. Modify the blank so that the initial transmittance is higter.
D. Use fresher spinach
and prepare the chloroplast solution during the laboratory.
E. Change the wavelength at which readings are taken.
3. If a student performed this experiment and got a flat curve, showing very little
change in percent transmittance over time, which of these would be the most
plausible explanation?
A. The DPIP was too pale at the beginning of the experiment.
B. The chloroplast solution was too concentrated.
C. The experimenter used chloroplasts that were damaged and could not
respond to light.
D. The blank was not properly used to calibrate the spectrophotometer.
Lab 5: Cell Respiration
In this lab, the rate of cellular respiration in both germinating and nongerminating peas
at two different temperatures is measured.
How can the rate of cellular respiration be measured?
1. measure the amount of glucose consumed
2. measure the amount of oxygen consumed
3. measure the amount of carbon dioxide produced
Pipette: fresh air equilibrates with the air in the vial
Cotton: protects the organism from caustic KOH
KOH: combines with CO2 to form a solid precipitate
Stopper: prevents gases and water from leaking
into the respirometer
peas: living organism that required O2 for respiration
The vial of beads acts as a control for the experiment,
compensating for any change in pressure of
temperature.
Understand the difference between control and
constant. Variables that are held constant in this
experiment include the volumes within the containers,
the number of peas, and the temperature of the water
for each set of three vials.
Lab 6: Molecular Biology
There are two different exercises in this AP lab, bacterial transformation and gel
electrophoresis.
Genetic transformation occurs when a host organism takes in foreign DNA and
expressed the foreign DNA. In genetic engineering, plasmids are one means used to
introduce foreign genes into a bacterial cell. Some plasmids have the ampR gene, which
confers resistance to the antibiotic ampicillin. E. coli cells containing this plasmid
(+ ampR cells) can survive and form colonies on LB agar that has been supplemented
with ampicillin.
In contrast, cells lacking the ampR plasmid (-ampR
cells) are sensitive to and killed by the antibiotic.
An antibiotic-sensitive cell can be transformed to
an ampicillin-resistant cell through
transformation. Competent cells are those most
likely to undergo transformation and take up
extracellular DNA.
-ampR
No ampR
plasmid
added
Control 1
Control 2
No ampicillin
in agar
Ampicillin
in agar
+ampR
Exp 1
No ampicillin
in agar
ampR
plasmid
added
Exp 2
Ampicillin
in agar
How do we know that transformation is successful?
1 Use a selection gene (antibiotic resistance)
2. Use a reporter gene (GFP)
Restriction enzyme digestion and gel electrophoresis
Lab 7: Genetics of Organisms
For this lab, it is important to be able to use data to determine the mode of transmission
(dominant, recessive, sex-linked, autosomal) and genetic make-up of the parents. You must also
be able to do chi-square analysis of data. Because the data we obtained was limited at best, we
were unable to perform chi-square analysis. It is, therefore, very important for you to go
through the exercise on the LabBench website
(http://www.phschool.com/science/biology_place/labbench/lab7/intro.html)
Chi square analysis is a way to determine if variance from expected results is statistically significant. One way to decide is to
test the validity of a nullI hypothesis, which states that there is no statistically significant difference between the observed
results and the expected results. To calculate chi-square you need two values provided by the chi-square critical value table,
the degrees of freedom (a value one less the total number of outcomes…for monohybrid crosses, the degree of freedom is 1;
for dihybrid crosses, there are three degrees of freedom) and the probability (many biologists agree that deviations having a
chance probability greater than 0.05 do not support the null hypothesis, to used the p value in the 0.05 row).
Another website resource for practicing
chi-square analysis is
http://cas.bellarmine.edu/tietjen/Stats/
Chi-square.htm .
If the chi-square value is greater than the critical values, the null
hypothesis is rejected.
Hint from the graders:
There was a chi-square problem on the AP Biology Exam a few years
ago, and we found that a common error was the some students took
the square root of the value they got for Χ2. Don’t make this
mistake! Χ2 is just the shorthand for the name of this mathematical
technique, chi-square.
Fred and Theresa Holtzclaw
Lab 8: Population genetics and evolution
This is the (in)famous “grebe” lab. You are expected to know the Hardy-Weinberg
equation and use it to determine the frequency of alleles in a population. You also
should know the conditions required for a population to remain in genetic equilibrium
and be able to explain how genetic drift, selection, and the heterozygote advantage
affect Hardy-Weinberg equilibrium.
Hint from the graders:
There are a couple of ways you can get confused when doing these problems. Here’s
what to look for:
1. If you are given the number of individuals showing the dominant
trait, remember this is NOT p2 because it includes heterozygotes.
However, you can use it to determine q2 and from that get a value
for q.
2. You also may be given a problem where you are told the
frequency of an allele. You are being directly given p or q!
Fred and Theresa Holtzclaw
Lab 9: Transpiration
In this lab (which we will be doing possibly next week), you investigate factors that
influence the rate of transpiration. Transpiration is a major mechanism that drives the
movement of water through a plant.
Be sure to review hydrogen bonding! Relate hydrogen bonding to the unique properties
of water, including adhesion and cohesion.
Know the path water takes through a plant, entering through the root hairs, passing
through root tissues into the xylem, traveling up through the xylem into the leaves.
Understand the roles of stomata and guard cells.
Relate transpiration rates to light intensity, to water potential, and to windy or calm
conditions.
Lab 10: Physiology of the Circulatory System
In this lab, you take pulse and blood pressure readings of a classmate and gather data
for a fitness index. You also measure the effect of temperature on heart rate.
You should be able to discuss how exercise changes heart rate and relate this to fitness.
In addition, you should be able to describe how changes in temperature affect the rate
of respiration or circulation.
The metabolic rate of ectothermic animals increases as environmental temperature
increases. This occurs because the reactants in the cell have greater thermal energy and
many enzymes are more active as temperature increases. This is noticeable in a range
from approximately 5⁰C and 35⁰C; at higher temperatures, enzymes are often
denatured.
The relationship between termperature and metabolic rate is often measured as Q10.
For most ectotherms, each 10⁰C increase in temperature results ins a doubling of the
metabolic rate. If the metabolic rate doubles with a 10⁰C increase in termperature,
then Q10 = 2.
Lab 11: Animal Behavior
In this lab, detailed observations of an organism’s behavior are made. This serves as the
basis for formulating a hypothesis and designing an experiment to test the hypothesis.
This is good practice for designing a controlled experiment.
In one exercise, pillbugs (roly-polies) are placed in a “choice chamber” with one half
lined with dry filter paper and the other half lined with wet filter paper. Many students
hypothesize that because pillbugs are crustaceans respire through gills, they will be
found more often in the moist chamber.
Realize that this experiment is not controlled. Refer to the beginning of the packet to
see how to design a controlled experiment.
Lab 12: Dissolved Oxygen and Aquatic Primary Productivity
You should understand:
 the biological importance of carbon and oxygen cycling in ecosystems
 how primary productivity relates to the metabolism of organisms in an
ecosystem
 the physical and biological factors that affect the solubility of gasses in aquatic
ecosystems
 the relationship between dissolved oxygen and the process of photosynthesis
and respiration as they affect primary productivity
You should be able to:
 measure primary productivity based on changes in dissolved oxygen in a
controlled experiment
 investigate the effects of changing light intensity and/or inorganic nutrient
concentrations on primary productivity in a controlled experiment
Primary productivity is a term used to describe the rate at which plants and other
photosynthetic organisms produce organic compounds in an ecosystem. Gross
productivity (the total photosynthetic production of organic compounds in an
ecosystem) and net productivity (the organic material that remain after some has
been used for respiration) are two aspects of primary productivity.
Net productivity = Gross productivity – Respiration
Two cultures are prepared in the lab as follows: one has
little phytoplankton but is high in zooplankton, and has
a low DO to start. The other is rich in phytoplankton
and zooplankton, and has a high dissolved oxygen to
start.
Which should have the highest net productivity?
Which should have the highest gross productivity?
Which should have the highest respiration rate?