# Lab 1: Diffusion and Osmosis

W
Shared by:
Categories
Tags
-
Stats
views:
444
posted:
4/20/2012
language:
English
pages:
66
Document Sample

```							Lab 1: Diffusion and Osmosis

Introduction:

Atoms and molecules in cells have kinetic energy and are constantly in motion. This kinetic
energy causes molecules to bump into each other and move in new directions. One result of
this molecular motion is the process of diffusion.

Diffusion is the random movement of molecules from an area of higher concentration of those
molecules to an area of lower concentration. Eventually a dynamic equilibrium will be reached;
the concentration of the molecules will even out, and there will be no net movement from one
area to the other.

Osmosis is a special case of diffusion. Osmosis is the movement of water through a selectively
permeable membrane (a membrane that allows for diffusion of certain solutes and water) from a
region of higher water potential to a region of lower water potential. Water potential is the
measure of free energy of water in a solution.

When two solutions have the same concentration of solutes, they are isotonic to each other (iso
= same, ton = condition, ic = pertaining to). If the two solutions are separated by a selectively
permeable membrane, water will move between the two solutions, but there will be no net
change in the amount of water in either solution.

If two solutions differ in the concentration of solutes that each has, the one with more solute is
hypertonic to the one with less solute. The solution that has less solute is hypotonic to the one
with more solute. These words can only be used to compare solutions.

If two solutions are separated by a semipermeable membrane, the hypertonic solution has more
solute and less water. At standard atmospheric pressure, the water potential of the hypertonic
solution is less than the water potential of the hypotonic solution so the net movement of water
will be from the hypotonic solution into the hypertonic solution.

Water potential is a term used when predicting the movement of water into or out of cells.
Water potential is abbreviated by the Greek letter psi (Ψ) and it has two components: a physical
pressure component (pressure potential Ψp) and the effects of solutes (solute potential Ψs).
Water potential equals pressure potential plus solute potential.

Ψ = Ψp + Ψs
Water will always move from an area of higher water potential (higher free energy, more water
molecules) to an area of lower water potential (lower free energy, fewer water molecules).
Water potential measures the tendency of water to leave one place in favor of another place.
You can picture the water diffusing down a water potential gradient.

Water potential is affected by two physical factors. One factor is the addition of solute which
lowers the water potential. The other factor is pressure potential (physical pressure). An
increase in pressure raises the water potential. By convention, the water potential of pure water
at atmospheric pressure is defined as being zero.

Adapted from Biology Lab Manual for Students
The College Board, 2001
Movement of water into and out of a cell is influenced by the solute potential (relative
concentration of solute) on either side of the cell membrane. If water moves out of the cell, the
cell will shrink. If water moves into an animal cell, it will swell and may even burst. In plant
cells, the presence of a cell wall prevents cells from bursting as water enters the cells, but
pressure eventually builds up inside the cell and affects the net movement of water. As water
enters a dialysis bag or a cell with a cell wall, pressure will develop inside as water pushes
against the bag or cell wall. The addition of solutes lowers the water potential because water
always goes from high to low water potential. A higher solute concentration draws water in, so
the potential must initially be lower.

Movement of water into and out of a cell is also influenced by the pressure potential (physical
pressure) on either side of the cell membrane. In plant cells, this physical pressure can be
exerted by the cell pressing against the partially elastic cell wall. Pressure potential is usually
positive in living cells.

The water potential value can be positive, zero, or negative. Remember that water will move
across a membrane in the direction of the lower water potential. Pressure potential is positive if
it’s a pushing pressure (pushing water out) or negative if it’s a pulling pressure (sucking water
in). Solute potential is always negative. An increase in solute potential makes the water
potential more negative and an increase in pressure potential makes the water potential more
positive.

If enough solute is added to water outside of cells, water will leave the cells, moving from an
area of higher water potential to an area of lower water potential. The loss of water from the
cells will cause the cells to lose turgor. A continued loss of water will eventually cause the cell
membrane to shrink away from the cell wall. The movement of the membrane away from the
cell wall is called plasmolysis.

The solute potential of a solution can be calculated using the following formula:

Ψs = -iCRT
where i = ionization constant (1 for sucrose since it doesn’t ionize in water)
C = molar concentration (determined above)
R = pressure constant (R = 0.083 liter bars/mole K)
T = temperature in Kelvin (273 + oC of solution)

Solute potential will be measured in bars, a metric measure of pressure that’s about the same
as 1 atmosphere.

Exercise 1A: Diffusion

In this experiment, you will measure diffusion of small molecules through dialysis tubing, an
example of a selectively permeable membrane. Small solute molecules and water can move
freely through the tubing, but larger molecules will pass through more slowly or not at all. The
size of the pores in the dialysis tubing determines which substances can pass through the
membrane.

A solution of glucose and starch will be placed inside a bag of dialysis tubing. Distilled water will
be placed in a beaker outside the dialysis bag. After 30 minutes have passed, the solution

Adapted from Biology Lab Manual for Students
The College Board, 2001
inside the dialysis tubing and the solution in the beaker will be tested for glucose and starch.
The presence of glucose will be tested with glucose test strips. The presence of starch will be
tested with iodine.

Procedure:

1. Obtain a piece of dialysis tubing that has been soaking in water. Tie off one end of the
tubing to form a bag. Open the other end of the bag.

2. Test the glucose/starch solution for presence of glucose using a glucose test strip.

3. Make sure the glucose/starch solution is mixed well. Using a plastic pipette, transfer
some of the glucose/starch solution into the bag. Tie off the other end of the bag.
Record the color of the solution.

4. Fill a beaker with distilled water. Add iodine to the water and record the color of the
solution. Test for the presence of glucose and record the results.

5. Immerse the bag into the beaker of solution.

6. Allow your setup to stand for approximately 30 minutes. Record the final color of the
solution in the bag and the solution in the beaker.

7. Test the liquid in the beaker and in the bag for the presence of glucose. Record the
results.

Exercise 1C: Water Potential

In this experiment, you will use potato cores placed in different molar concentrations of sucrose
in order to determine the water potential of potato cells.

1. Pour enough of each solution into the appropriately labeled cups so that they will cover
the potato cores.

2. Use a cork borer to cut four potato cylinders for each cup. Do not include any skin on
the cylinders.

3. Determine the mass of four cylinders together and record the mass in the table. Put the
four cylinders into the appropriate sucrose solution.

4. Repeat for each molar concentration of sucrose.

5. Cover the cups with plastic wrap and let stand overnight.

6. Remove the cores from the cups, blot them gently on a paper towel, and determine the
total mass.

7. Record the final mass, and record class data. Calculate the percent change in mass for
your individual results and for the class average.

8. Graph both your individual data and the class average.

Adapted from Biology Lab Manual for Students
The College Board, 2001
Prelab Questions:
1. What is diffusion?

2. What is osmosis?

3. How does solute size affect diffusion across selectively permeable membranes?

4. What is water potential?

5. What is the relationship between solute concentration, pressure potential, and the water
potential of a solution?

6. What is molarity?

Exercise 1A Questions:

Initial Contents            Solution Color        Presence of glucose
Initial           Final   Initial         Final
Bag        Glucose, starch

Beaker        Water, iodine

1. Which substance(s) are entering the bag and which are leaving the bag? What

Adapted from Biology Lab Manual for Students
The College Board, 2001
2. Quantitative data uses numbers to measure observed changes. How could this
experiment be modified so that quantitative data could be collected to show that water
diffused into the dialysis bag?

3. Based on your observations, which molecules are smaller than the pores in the dialysis
tubing? Which are larger?

4. What results would you expect if the experiment started with a glucose and iodine
solution in the bag and only starch and water outside? Why?

Exercise 1B Questions: Interpreting Data

Dialysis bags were prepared with the following solutions. They were weighed and placed in
beakers of distilled water. After 30 minutes, they were removed, blotted dry, and weighed
again. All masses were recorded in the chart.

Contents in          Initial Mass        Final Mass   Mass Difference   Percent Change
Dialysis Bag                                                                in Mass
0 M Distilled            26 g              26.3 g
Water
0.2 M Sucrose            26.2 g              27 g

0.4 M Sucrose            26.1 g             28.1 g

0.6 M Sucrose            26.4 g             29.3 g

0.8 M Sucrose            26.3 g             30.2 g

1.0 M Sucrose            26.4 g             31.2 g

Fill in the rest of the chart and graph the results.

1. What is the independent variable

2. What is the dependent variable?

Adapted from Biology Lab Manual for Students
The College Board, 2001
1. How does the molarity of the sucrose in the bags affect the change in mass?

2. What would happen to the mass of each bag in this experiment if all of the bags were
placed in a 0.4 M sucrose solution instead of in distilled water? Explain your response.

3. Why did you calculate the percent change in mass rather than simply using the change
in mass?

4. A dialysis bag is filled with distilled water and then placed in a sucrose solution. The
bag’s initial mass is 20 g, and its final mass is 18 g. Calculate the percent change in
mass.

5. Would the sucrose solution in the beaker in the previous question have been hypertonic,
hypotonic, or isotonic to the distilled water in the bag?

Adapted from Biology Lab Manual for Students
The College Board, 2001
Exercise 1C Questions:

Potato Core Individual Data
Contents in         Initial Mass       Final Mass   Mass Difference Percent Change
Beaker                                                               in Mass
0 M Distilled
Water
0.2 M Sucrose

0.4 M Sucrose

0.6 M Sucrose

0.8 M Sucrose

Potato Core Results—Class Data
Contents            Percent Change in Mass of Potato Cores                  Total      Class
in                                                                                 Average
Group Group Group Group Group Group Group Group
Beaker
1     2      3       4        5        6       7 8
0M
Distilled
Water
0.2 M
Sucrose

0.4 M
Sucrose

0.6 M
Sucrose

0.8 M
Sucrose

Adapted from Biology Lab Manual for Students
The College Board, 2001
1. Determine the molar concentration of sucrose in the potato core. This would be the
sucrose molarity in which the mass of the potato core does not change. To find this,
draw the straight line on the graph that best fits your data. The point at which this line
crosses the x-axis represents the molar concentration of sucrose with a water potential
that is equal to the potato tissue water potential. What is this concentration of sucrose
in the potato?

Calculation of Water Potential from Experimental Data

1. Calculate the solute potential of the potato.

2. The pressure potential of the solution is zero. What is the water potential of the potato?

Adapted from Biology Lab Manual for Students
The College Board, 2001
3. If a potato core is allowed to dehydrate by sitting in the open air, would the water
potential of the potato cores increase of decrease? Why?

4. If a plant cell has a lower water potential than its surrounding environment and if
pressure is equal to zero, is the cell hypertonic or hypotonic to its environment? Will the
cell gain or lose water? Why?

5. In the figure above, the beaker is open to the atmosphere. What is the pressure
potential of the system?

6. Is the water potential greater in the beaker or the bag?

7. Will water diffuse into or out of the bag?

Exercise 1D Questions
1. Zucchini cores are placed in sucrose solutions at 27 oC and resulted in the following
percent changes after 24 hours.

% Change in Mass       Sucrose Molarity
20%                Distilled Water
10%                     0.2 M
-3%                    0.4 M
-17%                    0.6 M
-25%                    0.8 M
-30%                      1M

Adapted from Biology Lab Manual for Students
The College Board, 2001
Graph the results.

2. What is the molar concentration of solutes within the zucchini cells?

3. Calculate the solute potential of the sucrose solution in which the mass of the zucchini
cores does not change.

4. Calculate the water potential of the solutes within the zucchini cores.

5. What effect does adding solute have on the solute potential component (Ψs) of that
solution? Why?

Adapted from Biology Lab Manual for Students
The College Board, 2001
6. Consider what would happen to a red blood cell placed in distilled water.
a. Which would have the higher concentration of water, the distilled water or the red
blood cell?

b. Which would have the higher water potential, the distilled water or the red blood
cell?

c. What would happen to the red blood cell? Why?

Exercise 1E Questions: Plasmolysis

1. What is plasmolysis?

2. If onion cells are placed in a hypertonic solution, what happens?

3. In the winter, grass often dies near roads that have been salted to remove ice. What
causes this to happen?

Postlab Questions:

1. How can you measure the water potential of a solution in a controlled experiment?

2. How can you determine the osmotic (molar) concentration of living tissue or an unknown
solution from experimental data?

Adapted from Biology Lab Manual for Students
The College Board, 2001
3. Describe the effects of water gain and loss in animal cells.

4. Describe the effects of water gain and loss in plant cells.

5. How is solute potential related to solute concentration and water potential?

Adapted from Biology Lab Manual for Students
The College Board, 2001
Lab 2: Enzyme Catalysis

Introduction:

Enzymes are proteins produced in living cells; they act as catalysts in biochemical reactions. A
catalyst increases the rate of a chemical reaction. One benefit of enzyme activity is that cells
can carry out complex chemical activities at relatively low temperatures.

In an enzyme-catalyzed reaction, the substance to be acted on, the substrate, binds reversibly
to the active site of the enzyme. One result of this temporary union is a reduction in the energy
required to activate the reaction of the substrate molecule so that products of the reaction are
formed.

The enzyme is not changed in the reaction and can be recycled to break down additional
substrate molecules. Each enzyme is specific for a particular reaction because its amino acid
sequence is unique and causes it to have a unique three-dimensional structure. The active site
is the portion of the enzyme that interacts with the substrate. Any substance that blocks or
changes the shape of the active site affects the activity of the enzyme. Salt concentration, pH,
temperature, activators, and inhibitors affect enzymatic activity.

Catalase is an enzyme that has four polypeptide chains, each of which is composed of more
than 500 amino acids. The enzyme is found in all aerobic organisms and prevents the
accumulation of toxic levels of hydrogen peroxide formed as a byproduct of metabolic
processes. The primary reaction catalyzed by catalase is the decomposition of hydrogen
peroxide to form water and oxygen.

2H2O2  2H2O + O2

In the absence of catalase, this reaction occurs spontaneously but very slowly. Catalase
speeds up the reaction considerably. Much can be learned about enzymes by studying the
kinetics (particularly the changes in rate) of enzyme-catalyzed reactions. For example, it is
possible to measure the amount of product formed or the amount of substrate used from the
moment the reactants are brought together until the reaction has stopped. If the amount of
product formed is measured at regular intervals and this quantity is plotted on a graph, a curve
like the one in the figure below is obtained.

The rate of the reaction is the slope of the linear portion of the curve. To determine the rate,
pick any two points on the straight-line portion of the curve. Divide the difference in the amount
of product formed between these two points by the difference in time between them. The rate is
expressed as μmoles product/sec.

Adapted from Biology Lab Manual for Students
The College Board, 2001
During the early period of a reaction, the number of substrate molecules is usually so large
compared with the number of enzyme molecules that changing the substrate concentration
does not affect the rate. The enzyme is acting on substrate molecules at a nearly constant rate
(the initial rate). The initial rate of any enzyme-catalyzed reaction is determined by the
characteristics of the enzyme. It is always the same for any enzyme and its substrate at a given
temperature and pH. This also assumes that the substrate is in excess.

The rate of the chemical reaction can be studied in a number of ways including the following:

1. measuring the rate of disappearance of substrate
2. measuring the rate of appearance of product
3. measuring the heat released or absorbed during the reaction

In this experiment, we will measure the appearance of the product, O2 using an oxygen gas
probe. Your group will be assigned either concentration of enzyme, temperature, or pH as a
variable to test.

Exercise 2D: An Enzyme-Catalyzed Rate of H2O2 Decomposition

In this experiment, you will determine the rate at which a 1.5% solution decomposes when
catalyzed by the catalase extract. To do this, you will determine how much H2O2 has been
consumed after 0, 30, 60, 90, 120, 150, and 180 seconds.

Before starting, set up the Palm Pilot to collect data ever 30 seconds for 180 seconds.

Enzyme Concentration

1. Obtain 3 test tubes, and label them 5 drops, 10 drops, and 20 drops.
2. Add 5 ml of distilled water to each tube.
3. Add 5 ml of a 3% solution of H2O2 to each tube.
4. Add 5 drops of catalase to the first tube. Cover the top with parafilm, and invert quickly
to mix.
5. Pour the test tube contents into the plastic bottle, insert the oxygen gas probe, and
collect data for 180 seconds.
6. Clean out the bottle, and repeat with the other 2 test tubes, adding 10 drops of catalase
and 20 drops of catalase respectively.

Temperature

1. Obtain 4 test tubes, and label them cold, room, warm, and hot
2. Add 5 ml of distilled water to each tube.
3. Add 5 ml of a 3% solution of H2O2 to each tube
4. Place each tube except the room temperature tube in the appropriate water bath for at
least 5 minutes. Make sure to record the temperature of each water bath!
5. For the room temperature tube, add 10 drops of catalase. Cover the top with parafilm,
and invert quickly to mix.
7. Pour the test tube contents into the plastic bottle, insert the oxygen gas probe, and
collect data for 180 seconds.
6. Clean out the bottle, and repeat with the other 3 test tubes once they’ve been in the
water baths for at least 5 minutes.

Adapted from Biology Lab Manual for Students
The College Board, 2001
pH

1. Obtain 3 test tubes, and label them pH 4, 7, and 10
2. Add 5 ml of the appropriate pH buffer to each tube.
3. Add 5 ml of a 3% solution of H2O2 to each tube
4. For the pH 4 test tube, add 10 drops of catalase. Cover the top with parafilm, and invert
quickly to mix.
5. Pour the test tube contents into the plastic bottle, insert the oxygen gas probe, and
collect data for 180 seconds.
6. Clean out the bottle, and repeat with the other 2 test tubes.

The data sheet is in a separate Excel document. Be sure to print it out!

1. Graph the data for enzyme-catalyzed O2 production for each set of data (enzyme
concentration, temperature, and pH)—there will be 3 different graphs:

Adapted from Biology Lab Manual for Students
The College Board, 2001
2. What is the dependent variable for each set of data?

3. What is the independent variable for each set of data?

Adapted from Biology Lab Manual for Students
The College Board, 2001
4. Determine the initial rate of the reaction and the rates between each of the time points.
Record the rates in the table below.

Time intervals (seconds)
Initial: 0-30    30-60     60-90       90-120    120-150      150-180
Rates (ml
O2/s)

5. When is the rate the highest? Explain why.

6. When is the rate the lowest? Why is it lowest there?

Adapted from Biology Lab Manual for Students
The College Board, 2001
7. Explain the inhibiting effect of sulfuric acid on the function of catalase. Relate this to
enzyme structure and chemistry.

8. What effect does lowering the temperature would have on the rate of enzyme activity.

Adapted from Biology Lab Manual for Students
The College Board, 2001
Lab 3: Mitosis and Meiosis

Introduction

All new cells come from previously existing cells. New cells are formed by the process of cell
division, which involves both division of the cell’s nucleus (karyokinesis) and division of the
cytoplasm (cytokinesis).

There are two types of nuclear division: mitosis and meiosis. Mitosis typically results in new
somatic (body) cells. Formation of an adult organism from a fertilized egg, asexual
reproduction, regeneration, and maintenance and repair of body parts are accomplished
through mitotic cell division. Meiosis results in the formation of either gametes (in animals) or
spores (in plants). These cells have half the chromosome number of the parent cell.

In plants, new cells are formed in special growth regions called meristems. These regions
usually are found at the tips of stems or roots. In animals, cell division occurs anywhere new
cells are formed or as new cells replace old ones. However, some tissues in both plants and
animals rarely divide once the organism is mature.

To study the stages of mitosis, you need to look for tissues where there are many cells in the
process of mitosis. This restricts your search to the tips of growing plants, such as the onion
root tip, or, in the case of animals, to developing embryos, such as the whitefish blastula.

Exercise 3A: Observing Mitosis in Plant Cells Using Prepared Slides of the Onion Root
Tip

Examine prepared slides of the onion root tip. Locate the meristematic region of the onion with
the 10X objective, and then use the 40X and then 400X objective to study individual cells.
Identify one cell that clearly represents each phase. Sketch and label the cell in the space
provided. Write a description of the events that occur in each phase.

1. Interphase

Description:                                     Picture:

Adapted from Biology Lab Manual for Students
The College Board, 2001
2. Prophase

Description:                               Picture:

3. Metaphase

Description:                               Picture:

4. Anaphase

Description:                               Picture:

5. Telophase

Description:                               Picture:

Adapted from Biology Lab Manual for Students
The College Board, 2001
Questions:

1. How does mitosis differ in plant and animal cells? How does plant mitosis accommodate
a rigid, inflexible cell wall?

2. What is the role of the centrosome (area around the centrioles)? Is it necessary for

To estimate the relative length of time that a cell spends in the various stages of cell division,
you will examine sample data. All of the cells in each cell cycle phase in 3 fields of view from
one onion root tip have been counted (be glad I didn’t make you do the counting!). From this,
you can infer the percentage of time each cell spends in each phase. The length of the cell
cycle is approximately 24 hours for cells in actively dividing onion root tips. That’s 1,440
minutes.

Percentage of cells in phase x 1,440 minutes = ___ minutes of cell cycle spent in each phase

Number of Cells                       Percent of      Time in
Field 1         Field 2     Field 3         Total       Total Cells      Each
Counted        Phase
Interphase        1, 247          1, 820         4, 557
Prophase            67              88            101
Metaphase           39              67             53
Anaphase            26              36             45
Telophase           13               9             19
Total Cells Counted: ___________

Adapted from Biology Lab Manual for Students
The College Board, 2001
Questions:

1. If your observations had not been restricted to the area of the root tip that is actively
dividing, how would your results have been different?

2. Summarize the data in the table. What can you infer about the relative length of time an
onion root tip cell spends in each stage of cell division?

3. Draw and label a pie chart of the onion root tip cell cycle using the data from the table.

Title: ___________________________

Adapted from Biology Lab Manual for Students
The College Board, 2001
Meiosis Questions:

1. List 3 major differences between the events of mitosis and meiosis.

2. Compare mitosis and meiosis with respect to each of the following in the table below:

Mitosis                  Meiosis
Chromosome number of parent cells (human)
Number of DNA replications
Number of divisions
Number of daughter cells produced
Chromosome number of daughter cells (human)
Purpose/function

3. How are meiosis I and meiosis II different?

4. How do oogenesis and spermatogenesis differ? (Hint: look in Ch. 46!)

5. Why is meiosis important for sexual reproduction?

Adapted from Biology Lab Manual for Students
The College Board, 2001
Lab 4: Plant Pigments and Photosynthesis

Introduction:

Paper chromatography is a useful technique for separating and identifying pigments and other
molecules from cell extracts that contain a complex mixture of molecules. The solvent moves
up the paper by capillary action which occurs as a result of the attraction of solvent molecules to
the paper and to each other. As the solvent moves up the paper, it carries along any
substances dissolved in it. 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
fibers in the paper through the formation of intermolecular bonds such as hydrogen bonds.

Beta carotene, the most abundant carotene in plants, is carried along near the solvent front
because it is very soluble in the solvent being used and because it forms no hydrogen bonds
with cellulose. Xanthophyll is found further from the solvent front because it is less soluble in
the solvent and is slowed down by hydrogen bonding to the cellulose. Cholorophylls a and b
bind more tightly to the paper, so they run more slowly.

The relationship of the distance moved by a pigment to the distance moved by the solvent is a
constant called Rf. It can be calculated using the formula:

Rf = distance pigment migrated (mm)/distance solvent front migrated (mm)

Photosynthesis may be studied in a number of ways. For this experiment, a dye reduction
technique will be used. Light and chloroplasts are required for the light reactions to occur. In
place of the electron acceptor, NADP+, the compound DPIP (2,6-dichlorophenol-indophenol) will
be substituted. When light strikes the chloroplasts, electrons boosted to high energy levels will
reduce DPIP. It will change from blue to colorless.

Chloroplasts are extracted from spinach leaves and incubated in the presence and absence of
light. As the DPIP is reduced and becomes colorless, the resultant increase in light level
transmittance is measured over a period of time using a spectrophotometer.

Procedure:

Exercise 4A:

1. Obtain a vial that has 1 cm of solvent in the bottom. Keep the lid on as much as
possible because the solvent is volatile.

2. Cut one end of a piece of filter paper into a point and cut the top off the filter paper so
that it will fit into the vial. Draw a pencil line 1.5 cm above the point.

3. Use a coin to extract the pigments from spinach leaf cells. Place a small section on top
of the pencil line. Use the edge of the coin to crush the cells. Be sure that the pigment
line is on top of the pencil line. Repeat the procedure 8-10 times being sure to use a
new part of the leaf each time (but make the pigment line on the same line!).

4. Place the chromatography paper in the cylinder so that the pointed end is barely
immersed in the solvent. Do not allow the pigment to be in the solvent!

Adapted from Biology Lab Manual for Students
The College Board, 2001
5. Put the cap on the vial. When the solvent is about 1 cm from the top of the paper,
remove the paper and immediately mark the location of the solvent front before it
evaporates.

6. Mark the bottom of each pigment band. Measure the distance each pigment migrated
from the bottom of the pigment origin to the bottom of the separated pigment band.
Record the distance that each front, including the solvent front, moved. Depending on
the species of plant used, you may be able to observe 4 or 5 pigment bands.

Exercise 4B
1. Obtain 3 cuvettes. Label the lid of one D (dark), and cover the cuvette with
aluminum foil. Label one lid U (unboiled) and one lid B (boiled).
2. Plug the colorimeter into Channel 1 of the interface.
3. Prepare a blank by filling an empty cuvette ¾ full with distilled water. Seal the
cuvette with a lid. To correctly use a cuvette, remember:
 All cuvettes should be wiped clean and dry on the outside with a tissue.
 Handle cuvettes only by the top edge of the ribbed sides.
 All solutions should be free of bubbles.
 Always position the cuvette with its reference mark facing the white reference
mark at the right of the cuvette slot on the colorimeter.
4. Set up the interface for the colorimeter and calibrate it.
a. Place the blank in the cuvette slot of the colorimeter and close the lid.
b. Make sure the interface displaces ABSORBANCE in Channel 1 and that the
colorimeter is set to a wavelength of 635 nm.
c. For the first calibration point, turn the wavelength knob of the colorimeter to
the 0%T position. In the Value field, enter “0” as the percent transmittance.
You can enter this information usin the onscreen keyboard (tap “123”, or by
using the Graffiti writing area. When the voltage reading stabilizes, tap “Keep
Pt 1”.
d. For the second calibration point, turn the wavelength knob of the colorimeter
to the Red LED position (635 nm). Enter “100” as the percent transmittance.
When the voltage reading stabilizes, tap “Keep Pt 2”.
5. You will be given a beaker or fish bowl with water and a flood lamp. The container of
water will act as a heat shield, protecting the chloroplasts from warming by the flood
lamp.
6. Add 2.5 ml of DPIP/phosphate buffer solution to each of the 3 cuvettes (U, D, B).
Then perform the following steps as quickly as possible.
a. Cuvette U: Add 3 drops of unboiled chloroplasts. Place the lid on the cuvette
and gently mix.
b. Cuvette D: Add 3 drops of unboiled chloroplasts. Place the lid on the cuvette
and gently mix. Make sure that the foil is wrapped around the cuvette so that
light won’t penetrate it.
c. Cuvette B: Add 3 drops of boiled chloroplasts. Place the lid on the cuvette
and gently mix.

Adapted from Biology Lab Manual for Students
The College Board, 2001
7. Take absorbance readings for each cuvette. Invert each cuvette twice to resuspend
the chloroplasts before taking a reading. To take a reading, place the cuvette in the
slot of the colorimeter and close the lid. Allow 10 seconds for the readings displayed
on the screen to stabilize. Record the absorbance value in Table 1. Remove the
cuvette.
8. Place the cuvettes in front of the lamp (with the heat sink in between). Turn on the
lamp.
9. Repeat Step 7 when 5 minutes have elapsed. Repeat again at 10, 15, and 20
minutes. Be sure to invert the cuvettes before taking the readings and to put them
back in front of the lamp as soon as each reading is taken.

Exercise 4A Questions:

Distance Moved By Pigment Band (mm)

Band Number                           Distance (mm)                Band Color
1
2
3
4

Distance Solvent Front Moved: _________ mm

Rf for beta carotene (yellow to yellow orange): _________

Rf for xanthophyll (yellow): _________

Rf for chlorophyll a (bright green to blue green): __________

Rf for chlorophyll b (yellow green to olive green): _________

1. What factors are involved in the separation of the pigments?

2. Would you expect the Rf value of a pigment to be the same if a different solvent were
used? Why or why not?

Adapted from Biology Lab Manual for Students
The College Board, 2001
Exercise 4B Questions:
Time (min)           Absorbance Unboiled       Absorbance in Dark        Absorbance Boiled
0
5
10
15
20

Chloroplast          Rate of Photosynthesis
Unboiled
Dark
Boiled
Plot the absorbance for the three cuvettes on the graph, and label each plotted line.

Adapted from Biology Lab Manual for Students
The College Board, 2001
1. What is the independent variable?

2. What is the dependent variable?

3. What is the function of DPIP in this experiment?

4. What molecule found in chloroplasts does DPIP replace?

5. Where do the electrons that reduce DPIP come from?

6. What was measured with the colorimeter?

7. What is the effect of darkness on the reduction of DPIP?

8. What is the effect of boiling the chloroplasts on the reduction of DPIP?

9. What were the controls in this experiment?

Adapted from Biology Lab Manual for Students
The College Board, 2001
Lab 5: Cell Respiration

Introduction:

Aerobic cellular respiration is the release of energy from organic compounds by metabolic
oxidation in the mitochondria within each cell. Cell respiration involves a series of enzyme-
mediated reactions. The equation below shows the complete oxidation of glucose. Oxygen is
required for the energy releasing process to occur.

C6H12O6 + 6O2  6CO2 + 6H2O + 686 kcal

There are three possible ways to measure cell respiration. You could measure the consumption
of O2, the production of CO2, or the release of energy.

In this experiment, the relative volume of CO2 produced by germinating and nongerminating
(dry) peas at two different temperatures will be measured.

Procedure:
Part 1
1. Set the CO2 gas sensor to the Low (0-10,000 ppm) setting. Connect the CO2 gas sensor
to Channel 1 of the interface.
2. Set the program to collect data every 30 seconds for 10 minutes (600 seconds).
3. Obtain 25 germinating peas, and blot them dry between 2 pieces of paper towel. Use
the thermometer to measure the room temperature. Record the temperature.
4. Place the germinating peas in the respiration chamber.
5. Place the shaft of the CO2 gas sensor in the opening of the respiration chamber.
6. Wait one minute, and then begin measuring carbon dioxide concentration by clicking
“start”.
7. Record the data in your data table.
8. Remove the CO2 gas sensor from the respiration chamber. Place the peas in a 100 ml
beaker filled with cold water and ice. The cold water will prepare the peas for Part 2 of
the experiment.
9. Use a notebook or notepad to fan air across the openings in the probe shaft of the CO2
gas sensor for one minute.
10. Fill the respiration chamber with water, and then empty it. Thoroughly dry the inside of
the respiration chamber with a paper towel.
11. Obtain 25 non-germinating peas, and place them in the respiration chamber.
12. Repeat steps 5-10 for the non-germinating peas.

Part 2
1. Remove the peas from the cold water, and blot them dry between two paper towels.
2. Repeat steps 5-10 to collect data with the cold germinating peas.
3. Calculate rates of carbon dioxide production for all three samples.

Adapted from Biology Lab Manual for Students
The College Board, 2001
Data Table
Germinating Peas    Non-germinating        Germinating Peas
Room Temp (      )       Peas                Cold (     )
Room Temp (      )
0 sec
30 sec
60 sec
90 sec
120 sec
150 sec
180 sec
210 sec
240 sec
270 sec
300 sec
330 sec
360 sec
390 sec
420 sec
450 sec
480 sec
510 sec
540 sec
570 sec
600 sec
Rate

1. This experiment uses a number of controls. What conditions must remain constant?
Why?

Adapted from Biology Lab Manual for Students
The College Board, 2001
2. Graph the carbon dioxide produced by the germinating peas and the dry peas at room
temperature and for the germinating peas in the cold. (You’ll have 3 lines)

3. What is the independent variable?

4. What is the dependent variable?

Adapted from Biology Lab Manual for Students
The College Board, 2001
5. Describe and explain the relationship between the amount of carbon dioxide produced
and time.

6. Explain the effect of germination (vs. nongermination) on pea respiration.

7. The graph below is a sample graph of possible data obtained for oxygen consumption by
germinating peas up to about 8oC. Draw in predicted results through 45oC. Explain your
prediction.

Adapted from Biology Lab Manual for Students
The College Board, 2001
Lab 8: Population Genetics and Evolution

Introduction

In 1908, Hardy and Weinberg independently suggested a scheme by which evolution could be
viewed as changes in the frequency of alleles in a population of organisms. In this scheme, if A
and a are alleles for a particular gene locus,

p = frequency of the A allele
q = frequency of the a allele

These are referred to allele frequencies. Because there are only 2 alleles, p + q = 1.

The frequencies of the genotypes are expressed as

p2 = AA
2pq = Aa
q2 = aa

Since these are the only 3 genotypes, p2 + 2pq + q2 = 1.
Hardy and Weinberg said that if five conditions are met, the population’s allele and genotype
frequencies will remain constant from one generation to the next. These conditions are:

1. Large population size (the effect of chance on changes in allele frequencies is greatly
reduced)
2. Random mating (individuals show no mating preference for a particular phenotype)
3. No mutation (no alteration in the DNA sequence of alleles)
4. No gene flow (no immigration or emigration)
5. No natural selection (all genotypes have an equal chance of surviving and reproducing)

This provides a benchmark by which changes in allele frequency, and therefore evolution, can
be measured.

Exercise 8A:

Using the class as a sample population, the allele frequency of a gene controlling the ability to
taste the chemical PTC (phenylthiocarbamide) could be estimated. A bitter-taste reaction to the
PTC is evidence of the presence of a dominant allele. The inability of the chemical at all
depends on the presence of homozygous recessive alleles.

Phenotypes              Allele Frequency Based
on the H-W Equation
Tasters       Nontasters          p            q
North             .55            .45
American
Population

Adapted from Biology Lab Manual for Students
The College Board, 2001
Question:

1. What percentage of the North American population is heterozygous for the taster trait?

Exercise 8B:

Case I—Ideal Hardy-Weinberg Population

The class will simulate a population of randomly mating heterozygous individuals with an initial
gene frequency of .5 for the dominant allele A and .5 for the recessive allele a. Your initial
genotype is Aa. Record this on the data page. Each member of the class will receive 2 cards.
One will have A written on it and the other will have a. These will represent the products of
meiosis. Remember that in reality, 4 gametes are produced from each original cell in meiosis!

1. Turn the cards over so that you can’t see them. Shuffle your cards and pick one to
contribute to the production of the first offspring. Your partner should do the same. The
two alleles represent alleles from the first offspring. One of you should record the
genotype of this offspring.

2. Return the cards to the original owners, and pick cards for a second offspring. The other
partner should record the genotype of the second offspring on his or her data sheet.
One person should assume the identity of each offspring!

3. Assume the identity of one of your offspring, and find a new mater. Follow the same
procedure for 5 generations

Questions:

1. What does the Hardy-Weinberg equation predict for the new p and q?

2. Do the results in the simulation agree with the prediction?

3. What requirements for Hardy-Weinberg equilibrium were not met in this simulation?

Case II—Selection

1. Start as Aa again and repeat the procedure.

Adapted from Biology Lab Manual for Students
The College Board, 2001
2. HOWEVER, aa is dead! If you have an aa offspring, try again. You must end up with
two viable offspring!

3. Proceed through 5 generations.

Questions:

1. How do the frequencies of p and q at the end compare to the initial frequencies?

2. What requirements were not strictly followed in this simulation?

3. Predict what will happen to the frequencies of p and q if you simulated another 5
generations.

4. In a large population, would it be possible to completely eliminate a deleterious
recessive allele? Explain.

coin. Heads= Dead, Tails = Alive Again, you need 2 viable offspring!

2. Proceed through 10 generations.

Questions:

1. Explain how the changes in p and q frequencies in this simulation compare with Case I
and Case II.

2. Do you think the recessive allele will be completely eliminated in either Case II or Case
III?

Adapted from Biology Lab Manual for Students
The College Board, 2001
3. What is the importance of heterozygotes in maintaining genetic variation in populations?

Case IV—Genetic Drift

1. Same as Case I, but we’ll divide into 3 small populations. Make sure to write down how
many people are in your populations.

2. Proceed for 5 generations.

Questions:

1. What do your results indicate about the importance of population size as an evolutionary
force?

Hardy-Weinberg Problems

1. In Drosophila, the allele for normal-length wings is dominant over the allele for vestigial
wings. In a population of 1000 individuals, 360 show the recessive phenotype. How
many individuals would you expect to be homozygous dominant and heterozygous for
this trait?

2. The allele for unattached earlobes is dominant over the allele for attached earlobes. In a
population of 500 individuals, 25% show the recessive phenotype. How many
individuals would you expect to be homozygous dominant and heterozygous for this
trait?

Adapted from Biology Lab Manual for Students
The College Board, 2001
3. The allele for the hair pattern called widow’s peak is dominant over the allele for no
widow’s peak. In a population of 1000 individuals, 510 show the dominant phenotype.
How many individuals would you expect of each of the possible three genotypes for the
trait?

4. In the US, about 16% of the population is Rh negative. The allele for Rh negative is
recessive to the allele for Rh positive. If the student population of a high school in the
US is 2000, how many students would you expect for each of the three possible
genotypes?

5. In certain African countries, 4% of the newborn babies have sickle-cell anemia, which is
a recessive trait. Out of a random population of 1000 newborn babies, how many would
you expect for each of the three possible genotypes?

6. In a certain population, the dominant phenotype of a certain trait occurs 91% of the time.
What is the frequency of the dominant allele?

Adapted from Biology Lab Manual for Students
The College Board, 2001
Data Page

Case I                                         Case IIII

Initial Class Frequencies:                     Initial Class Frequencies:

AA_____ Aa_____ aa_____                        AA_____ Aa_____ aa_____

p______ q ______                               p______ q ______

Initial genotype: Aa                           Initial genotype: Aa

F1 Genotype _____                              F1 Genotype _____ F6 Genotype _____
F2 Genotype _____                              F2 Genotype _____ F7 Genotype _____
F3 Genotype _____                              F3 Genotype _____ F8 Genotype _____
F4 Genotype _____                              F4 Genotype _____ F9 Genotype _____
F5 Genotype _____                              F5 Genotype _____ F10 Genotype _____
Final Class Frequencies:                       Final Class Frequencies:

AA_____ Aa_____ aa_____                        AA_____ Aa_____ aa_____

p______ q ______                               p______ q ______

Case II                                        Case IV
Selection                                      Genetic Drift

Initial Class Frequencies:                     Initial Class Frequencies:

AA_____ Aa_____ aa_____                        AA_____ Aa_____ aa_____

p______ q ______                               p______ q ______

Initial genotype: Aa                           Initial genotype: Aa

F1 Genotype _____                              F1 Genotype _____
F2 Genotype _____                              F2 Genotype _____
F3 Genotype _____                              F3 Genotype _____
F4 Genotype _____                              F4 Genotype _____
F5 Genotype _____                              F5 Genotype _____
Final Class Frequencies:                       Final Class Frequencies:

AA_____ Aa_____ aa_____                        AA_____ Aa_____ aa_____

p______ q ______                               p______ q ______

Adapted from Biology Lab Manual for Students
The College Board, 2001
Lab 9: Transpiration

Introduction

The amount of water needed daily by plants for the growth and maintenance of tissues is small
in comparison to the amount that is lost through the process of transpiration (the evaporation of
water from the plant surface) and guttation (the loss of liquids from the ends of vascular tissues
at the margins of leaves). If this water is not replaced, the plant will wilt and may die.

The transport of wter up from the roots in the xylem is governed by differences in water
potential. These differences account for water movement from cell to cell and over long
distances in the plant. Gravity, pressure, and solute concentration all contribute to water
potential, and water always moves from an area of high water potential to an area of low water
potential. The movement itself is facilitated by osmosis, root pressure, and adhesion and
cohesion of water molecules.

Minerals actively transported into the root accumulate in the xylem, increasing solute
concentration and decreasing water potential (making it more negative). Water moves in by
osmosis. As water enters the xylem, it forces fluid up the xylem due to hydrostatic root
pressure. But this pressure can only move fluid a short distance. The most significant force
moving the water and dissolved minerals in the xylem is upward pull as a result of transpiration,
which creates tension. The “pull”on the water from transpiration results from the cohesion and

Transpiration begins with the evaporation of water through the stomata, small openings in the
leaf surface that open into air spaces that surround the mesophyll cells of the leaf. The moist air
in these spaces has a higher water potential than the outside air, and water tends to evaporate
from the leaf surface (moving from high to low water potential). The moisture in the air spaces
is replaced by water from the adjacent mesophyll cells, lowering their water potential (since the
cytosol becomes more concentrated). Water will then move into the mesophyll cells by osmosis
from surrounding cells with higher water potentials, including the xylem. As each water
molecule moves into a mesophyll cell, it exerts a pull on the column of water molecules existing
in the xylem all the way from the leaves to the roots. This transpirational pull occurs because of
the cohesion of water molecules to each other (H bonds) and the adhesion of water molecules
to the walls of the xylem cells.

The upward transpirational pull on the fluid in the xylem causes a tension (negative pulling
pressure) to form in the xylem, pulling the walls inward. This contributes to lowering the water
potential of the xylem which allows water to move inward from the soil, across the cortex of the
root, and into the xylem.

Evaporation through the open stomata is a major route of water loss in plants. However, the
stomata must open to allow the entry of CO2 used in photosynthesis. Therefore, a balance must
be maintained between the gain of CO2 and the loss of water by regulating the opening and
closing of stomata on the leaf surface. Temperature, light intensity, air currents, and humidity
influence the opening and closing of the stomata and affect the transpiration rate. Different
plants also vary in the rate of transpiration and stomatal regulation.

Adapted from Biology Lab Manual for Students
The College Board, 2001
Calculate leaf surface area by taking a large leaf and tracing it on the grid below. Count all of
the squares that are completely within the tracing, and estimate the number of squares that lie
partially within the tracing. 4 squares = 1 cm2. The total surface area can be calculated by
dividing the total number of squares covered by 4. Record this value here.

____________ = Leaf Surface Area (cm2)

Calculate the water loss per square centimeter of leaf surface by dividing the water loss at each
reading in the table below by the leaf surface you calculated.

Time interval (min)
0        3         6        9      12       15      18    21       24      27       30
Water
loss
(ml)
Water
loss
per

Adapted from Biology Lab Manual for Students
The College Board, 2001
cm2

The following data shows water loss in ml/cm2 over time from the same type of plant under
different laboratory conditions: normal room conditions, light shining on the plant, a fan blowing
on the plant, and misting of air around the plant.

Time (minutes)
Treatment                   0                     10             20                   30
Room                      0                   2.19            4.56                 6.57

Light                    0                   4.16            7.57                11.73

Fan                     0                   4.50            7.58                11.00

Mist                    0                   1.30            2.44                 3.65

On the same set of axes, for each treatment, graph the data for each time interval.

Adapted from Biology Lab Manual for Students
The College Board, 2001
1. What is the dependent variable?

2. What is the independent variable?

3. Calculate the rate (average amount of water loss per square centimeter) for each of the
treatments.

Room:

Light:

Adapted from Biology Lab Manual for Students
The College Board, 2001
Fan:

Mist:

4. Explain why each of the conditions causes an increase or decrease in transpiration
compared with the control.

Condition                           Effect                  Explanation of Effect
Room                                           Control                      -----------
Light

Fan

Mist

5. Explain the role of water potential in the movement of water from soil through the plant
and into the air.

6. What is the advantage of closed stomata to a plant when water is in short supply? What

7. Describe several adaptations that enable plants to reduce water loss from their leaves.
Include both structural and physiological adaptations.

Adapted from Biology Lab Manual for Students
The College Board, 2001
Structure of the stem and root:

8. Describe characteristics of the following cell types:

a. Parenchyma cells

b. Sclerenchyma cells

c. Collenchyma cells

9. Describe the following tissue types:

a. Xylem

b. Phloem

c. Epidermis

10. Draw pictures of cross sections of the stems and roots of monocots and dicots.

Adapted from Biology Lab Manual for Students
The College Board, 2001
Monocot Stem                 Monocot Root

Dicot Stem                  Dicot Root

Adapted from Biology Lab Manual for Students
The College Board, 2001
Lab 10: Physiology of the Circulatory System

Introduction

The cardiovascular (circulatory) system functions to deliver oxygen and nutrients to tissues for
growth and metabolism and to remove metabolic wastes. The heart pumps blood through a
circuit that includes arteries, arterioles, capillaries, venules, and veins. One important circuit is
the pulmonary circuit, where there is an exchange of gases within the alveoli of the lungs. The
right side of the human heart receives deoxygenated blood from body tissues and pumps it to
the lungs. The left side of the heart receives oxygenated blood from the lungs and pumps it to
the tissues.

With increased exercise, several changes occur within the circulatory system, thus increasing
the delivery of oxygen to actively respiring muscle cells. These changes include increased
heart rate, increased blood flow to muscular tissue, decreased blood flow to nonmuscular
tissue, increased arterial pressure, increased body temperature, and increased breathing rate.

An important measurable aspect of the circulatory system is blood pressure. When the
ventricles of the heart contract, pressure is increased throughout all the arteries. Arterial
pressure is directly dependent on the amount of blood pumped by the heart per minute and the
resistance to blood flow through the arterioles. The arterial blood pressure is measured using a
sphygmomanometer. This device consists of an inflatable cuff connected by rubber hoses to a
hand pump and to a pressure gauge graduated in millimeters of mercury. The cuff is wrapped
around the upper arm and inflated to a pressure that will shut off the brachial artery. The
examiner listens for the sounds of blood flow in the brachial artery by placing the bell of a
stethoscope in the inside of the elbow below the biceps.

At rest, the blood normally goes through the arteries so that the blood in the central part of the
artery moves faster than the blood in the peripheral part. Under these conditions, the artery is
silent when one listens. When the sphygmomanometer cuff is inflated to a pressure above the
systolic pressure, the flow of blood is stopped, and the artery is again silent. As the pressure in
the cuff gradually drops to levels between the systolic and diastolic pressure of the artery, the
blood is pushed through the compressed walls of the artery in a turbulent flow. Under these
conditions, the blood is mixed, and the turbulence sets up vibrations in the artery that are heard
as sounds in the stethoscope. These sounds are known as the heart sounds or sounds of
Korotkoff. The sounds are divided into five phases based on the loudness and quality of the
sounds.

Phase 1: A loud, clear, snapping sound is evident which increased in intensity as the cuff is
deflated. In the example in the figure, this phase begins at a cuff pressure of 120 mm Hg and
ends at a pressure of 106 mm Hg.

Phase 2: A succession of murmurs can be heard. Sometimes the sounds seem to disappear
during this time, which may be a result of inflating or deflating the cuff too slowly. In the
example, this phase begins at a cuff pressure of 106 mm Hg and ends at a pressure of 86 mm
Hg.

Phase 3: A loud, thumping sound, similar to that in Phase 1, but less clear, replaces the
murmurs. In the example, this phase begins at a cuff pressure of 86 mm Hg and ends at a
pressure of 81 mm Hg.

Adapted from Biology Lab Manual for Students
The College Board, 2001
Phase 4: A muffled sound abruptly replaces the thumping sounds of Phase 3. In the example,
this phase begins at a cuff pressure of 81 mm Hg and ends at a pressure of 76 mm Hg.

Phase 5: All sounds disappear.

The cuff pressure at which the first sound is heard (beginning of Phase 1) is taken as the
systolic pressure. The cuff pressure at which the muffled sound of Phase 4 disappears
(beginning of Phase 5) is taken as the diastolic pressure. In the example, the pressure would
be recorded as 120/76. A normal blood pressure measurement for a given individual depends
on the person’s age, sex, heredity, and environment. When these factors are taken into
account, blood pressure measurements that are chronically elevated may indicate hypertension,
a major contributing factor in heart disease and stroke. Typical blood pressure for men and
women varies with age and fitness. For high school students, the typical range is usually 100-
120/70-90.

Exercise 10A: Measuring Blood Pressure

Note: These labs are only for experimental and not diagnostic purposes!

The earpieces of the stethoscope should be cleaned with alcohol swabs before and after each
use!

1. Work in pairs. Those who are to have their blood pressure measured should be seated
with both shirt sleeves rolled up.

2. Attach the cuff of the sphygmomanometer snugly around the upper arm.

3. Place the stethoscope directly below the cuff in the bend of the elbow joint.

Adapted from Biology Lab Manual for Students
The College Board, 2001
4. Close the valve of the bulb by turning it clockwise. Pump air into the cuff until the
pressure gauge reaches 180 mm Hg.

5. Turn the valve of the bulb counterclockwise and slowly release air from the cuff. Listen
for a pulse.

6. When you first hear the heart sounds, note the pressure on the gauge. This is the
systolic pressure.

7. Continue to slowly release air and listen until the clear thumping sound of the pulse
becomes strong and then fades. When you last hear the full heart beat, note the
pressure. This is the diastolic pressure.

8. Repeat the measurement two more times and determine the systolic and diastolic
pressure, and record the data in the box below.

Measurement                    1                 2                   3                Average
Systolic

Diastolic

Exercise 10B: A Test of Fitness

The point scores on these tests provide an evaluation of fitness based not only on cardiac
muscular development but also on the ability of the cardiovascular system to respond to sudden
change in demand. CAUTION: Make sure that you do not attempt this exercise if strenuous
activity will aggravate a health problem!

Work in pairs. Determine the fitness level for one member (Tests 1-5), and then repeat the
process for the other member.

Test 1: Standing Systolic Compared with Reclining Systolic

1. The subject should recline on a table for at least 5 minutes. At the end of this time,
measure the systolic and diastolic pressure and record the values.

Reclining systolic pressure: ______ mm Hg            Reclining diastolic pressure: ______ mm Hg

2. Remain reclining for two minutes, then stand, and immediately repeat measurements on
the same subject. Record the values below.

Standing systolic pressure: ______ mm Hg             Standing diastolic pressure: ______ mm Hg

3. Determine the change in systolic pressure from reclining to standing by subtracting the
standing measurement from the reclining measurement. Assign fitness points based on
the table below, and record in the fitness data box at the end.

Adapted from Biology Lab Manual for Students
The College Board, 2001
Cardiac Rate and Physical Fitness

During physical exertion, the cardiac rate (beats per minute) increases. This increase can be
measured as an increase in pulse rate. Although the maximum cardiac rate is generally the
same in people of the same age group, those who are physically fit have a higher stroke volume
(ml/beat) than more sedentary individuals. A person who is in poor physical condition,
therefore, reaches his or her maximum cardiac rate at a lower work level than a person of
comparable age who is in better shape. Individuals who are in good physical condition can
deliver more oxygen to their muscles (have a higher aerobic capacity) before reaching
maximum cardiac rate than can those in poor condition. Thus, the physically fit have a slower
increase in cardiac rate with exercise and a faster return to resting cardiac rate after exercise.
Physical fitness involves not only muscular development but also the ability of the
cardiovascular system to respond to sudden changes in demand.

Test 2: Standing Pulse Rate

1. The subject should stand at ease for 2 minutes after Test 1.

2. After the 2 minutes, determine the subject’s pulse.

3. Count the number of beats for 30 seconds, and multiply by 2. The pulse rate is the
number of beats per minute. Record them in the fitness data box at the end, and assign
fitness points based on the table below.

Adapted from Biology Lab Manual for Students
The College Board, 2001
Test 3: Reclining Pulse Rate

1. The subject should recline for 5 minutes on a table.

2. Determine the subject’s resting pulse rate.

3. Count the number of beats for 30 seconds and multiply by 2. (Note: the subject should
remain reclining for the next test). Record the pulse, and assign fitness points based on
the table below.

Test 4: Baroreceptor Reflex (Pulse Rate Increase from Reclining to Standing)

1. The reclining subject should now stand up.

2. Immediately take the subject’s pulse. Record this value.

Pulse immediately upon standing = ______ beats per minute

Baroreceptors in the carotid artery and the aortic arch detect a drop in blood pressure upon
standing, and they signal the medulla of the brain to increase the heartbeat.

3. Subtract the reclining pulse rate (from Test 3) from the pulse rate immediately upon
standing (Test 4) to determine the increase upon standing. Record in the fitness data
box, and assign fitness points based on the table below.

Adapted from Biology Lab Manual for Students
The College Board, 2001
Test 5: Step Test—Endurance

1. The subject should do the following: Place your right foot on an 18 inch high stool.
food to the original position. Repeat this exercise 5 times, allowing 3 seconds for each
step up.

2. Immediately after the completion of this exercise, measure the subject’s pulse for 15
seconds, and record; continue taking the subject’s pulse and recording the rates at 60,
90, and 120 seconds.

Number of beats in the 0-15 second interval: _____ x 4 = ______ beats per minute
Number of beats in the 16-30 second interval: _____ x 4 = ______ beats per minute
Number of beats in the 31-60 second interval: _____ x 2 = ______ beats per minute
Number of beats in the 61-90 second interval: _____ x 2 = ______ beats per minute
Number of beats in the 91-120 second interval: _____ x 2 = ______ beats per minute

3. Observe the time that it takes for the subject’s pulse to return to approximately the level
that was recorded in Test 2. Assign fitness points based on the table below, and record
them in the fitness data box.

Adapted from Biology Lab Manual for Students
The College Board, 2001
4. Subtract the subject’s normal standing pulse rate (recorded in Test 2) from his/her pulse
rate immediately after exercise (0-15 second interval) to obtain pulse rate increase.
Record in the fitness data box, and assign fitness points based on the table below.

Fitness Data Box
Measurement                      Points
Test 1: Change in systolic
pressure from reclining to                                   mm Hg
standing
Test 2: Standing pulse rate
beats/min

Test 3: Reclining pulse rate
beats/min

Test 4: Baroreceptor reflex
Pulse rate increase on                                     beats/min
standing
Test 5: Step Test
Return of pulse to standing                                seconds
rate after exercise

Pulse rate increase                                        beats/min
immediately after exercise
TOTAL SCORE

Total Score                                    Relative Cardiac Fitness
18-17                                                Excellent
16-14                                                 Good
13-8                                                  Fair
7 or less                                               Poor

Adapted from Biology Lab Manual for Students
The College Board, 2001
Questions:

1. Explain why blood pressure and heart rate differ when measured in a reclining position
and in a standing position.

2. Explain why high blood pressure is a health concern.

3. Explain why an athlete must exercise harder or longer to achieve a maximum heart rate
than a person who is not as physically fit.

4. Do some research, and explain why smoking causes a rise in blood pressure.

Adapted from Biology Lab Manual for Students
The College Board, 2001
Exercise 10 C: Heart Rate and Temperature

In ectothermic animals, there is a direct relationship between the rate of many physiological
activities and environmental temperature. The rate of metabolism in these animals increases as
environmental temperatures increase from approximately 5oC to 35oC. Increasing the
temperature by approximately 10oC results in a doubling of the metabolic rate. That is why a
snake or lizard can hardly move when it is cold but becomes quite active after warming in the
sun.

The following table shows heart rate of the water flea, Daphnia, at different temperatures.

Reading                            Temperature oC        Heart rate (beats/minute)
1                                    5                           108
2                                    10                          152
3                                    15                          211
4                                    20                          290

Graph the temperature and heart rate data.

Adapted from Biology Lab Manual for Students
The College Board, 2001
1. What is the independent variable?

2. What is the dependent variable?

3. Why does temperature affect heart rate in ectothermic organisms?

4. Discuss what results you might obtain if you repeated this experiment using an
endothermic organism.

5. Describe at least four ways an ectothermic organism’s behavior helps it regulate its
temperature.

Adapted from Biology Lab Manual for Students
The College Board, 2001
Lab 11: Animal Behavior

Introduction:

Ethology is the study of animal behavior. Behavior is an animal’s response to sensory input and
falls into two basic categories: learned and innate (inherited).

Orientation behaviors place the animal in its most favorable environment. Taxis is when an
animal moves either toward or away from a stimulus. Taxis is often exhibited when the stimulus
is light, heat, moisture, sound, or chemicals. Kinesis is a movement that is random and does
not result in orientation with respect to a stimulus. If an organism responds to bright light by
moving away, that is taxis. If an animal responds to bright light by random movements in all
directions, that is kinesis.

In this lab, you will be working with terrestrial isopods commonly known as pillbugs, sowbugs, or
roly-polies. These organisms are members of the phylum Arthropoda, Class Crustacea, which
also includes shrimp and crabs. Most members of this group respire through gills. Isopods
locate their appropriate environments by using their compound eyes and their antennae.

Exercise 11A: General Observation of Behaviors

Note: Try to make your observations without disturbing the animals in any way. Do not prod,
poke, or shake the dish, make loud sounds, or subject them to bright lights. You want to
observe their behavior, not influence or interfere with it!

Procedure:

1. Place 10 pillbugs in a Petri dish or small plastic container. Pillbugs generally do not
climb, but if they do, you may cover the dish with plastic wrap or the Petri dish cover.

2. Observe the pillbugs for 10 minutes. Make notes on their general appearance,
movements around the dish, and interactions with each other. Notice if they seem to
prefer one area over another, if they keep moving, settle down, or move sporadically.
Note any behaviors that involve 2 or more pillbugs.

Exercise 11B: Kinesis in Pillbugs:

1. Prepare a choice chamber as illustrated in the figure on the next page. The choice
chamber consists of two large, plastic Petri dishes (or two small recyclable plastic
containers) taped together with an opening cut between them. Cut the opening with
scissors and use tape to hold the dishes together. Line one chamber with a moist piece
of filter paper (or paper towel) and the other with a dry piece of filter paper (or paper
towel).

Adapted from Biology Lab Manual for Students
The College Board, 2001
2. Place 5 pillbugs in each side of the choice chamber so that there are 10 pillbugs total.
Cover the chambers.

3. Count how many pillbugs are on each side of the choice chamber every 30 seconds for
10 minutes and record your data. Continue to record even if they all move to one side or
stop moving.

4. Graph both the number of pillbugs in the wet chamber and the number in the dry
chamber.

Exercise 11C: Student-Designed Experiment to Investigate Pillbugs’ Response to
Temperature, pH, Background Color, Light, or Another Variable

1. Select one of the variable factors listed above.

2. Design an experiment to test how the pillbugs will respond to that factor.

3. Decide what data you will collect and design your data sheet.

the data.

6. Write a conclusion based on your experimental results.

7. Set your pillbugs free! 

Adapted from Biology Lab Manual for Students
The College Board, 2001
Prelab Questions:

1. What is kinesis?

2. What is taxis?

3. What variable will you select for Exercise 11C?

4. Develop a hypothesis concerning the pillbugs’ response to that variable.

5. What materials will you use?

7. On a separate sheet of paper, design your data sheet.

Exercise 11A Questions:

1. What do you observe about the pillbugs’ appearance, movements, and interactions with
each other?

Adapted from Biology Lab Manual for Students
The College Board, 2001
2. Make a detailed sketch of a pillbug.

Exercise 11B Questions:

Time     # in Wet # in Dry                   Other Notes
(mins)    Chamber Chamber
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
6.5
7
7.5
8
8.5
9
9.5
10

Adapted from Biology Lab Manual for Students
The College Board, 2001
1. What is the independent variable?

2. What is the dependent variable?

3. What conclusions do you draw from your data? Explain the physiological reasons for
the behavior observed in this activity?

Adapted from Biology Lab Manual for Students
The College Board, 2001
4. Obtain results from all of the lab groups in your class. What types of environments do
isopods prefer? How do the data support these conclusions? Give specific examples.

5. Is your isopod’s response to moisture best classified as taxis or kinesis? Explain your
response.

Exercise 11C Questions:

1. On a separate sheet of paper, make a graph of your data.

2. What can you conclude from your experimental results?

Adapted from Biology Lab Manual for Students
The College Board, 2001
Lab 12: Dissolved Oxygen and Aquatic Primary Productivity

Introduction:

Oxygen is necessary for the metabolic processes of virtually all organisms, including aquatic
life. Dissolved oxygen, therefore, is an important indicator of water quality.

Aquatic and terrestrial environments do not have the same ability to hold oxygen. Air contains
95% more oxygen than an equal volume of cold water. Water’s ability to hold oxygen
decreases as the temperature of the water increases. Because water does not hold oxygen as
efficiently as air does, respiration and organic degradation easily deplete its dissolved oxygen
concentration. Oxygen must be replenished from the atmosphere and from photosynthesis in
the aquatic environment.

Wind increases dissolved oxygen because oxygen is mixed into the water as wind blows across
the surface. Turbulence in the water as it flows through a stream or river has a similar effect
and increases dissolved oxygen.

A eutrophic body of water is one which has a fluctuating dissolved oxygen content from varying
amounts of activity of the life in the water and is always rich in nutrients such as calcium or
nitrates. An oligotrophic body of water is always rich in oxygen but poor in plant nutrients. The
oxygen content is constant because there isn’t much variation in life activity that would cause
serious depletion.

A concentration of dissolved oxygen less than 4 ppm is stressful to most forms of aquatic life.
The ideal range for an adequate fish population is about 8-15 ppm.

Energy accumulated by plants is termed primary production. All production in an ecosystem
stems from the energy in organic substances that autotrophs (primary producers) create from
inorganic raw materials. The flow of energy through a community starts with photosynthesis,
which, in itself, requires energy. All of the sun’s energy that is assimilated is called gross
primary production. The energy that plants need to do photosynthesis is provided by the
reverse process of photosynthesis, cell respiration. The energy remaining after respiration and
stored as organic material is net primary production.

Primary production can be measured by measuring oxygen production. A given concentration
of phytoplankton is added to two bottles. The “dark” bottle is wrapped in aluminum foil to
exclude light. The “light” bottle is clear. A quantity of oxygen proportional to the total organic
matter fixed (gross production) is produced by photosynthesis in the light bottle. At the same
time, some of the oxygen is being used in cell respiration. The amount of oxygen left is
proportional to the amount of fixed organic matter remaining after respiration (net production).
The quantity of oxygen in the light bottle indicates the net photosynthesis or net primary
production.

In the dark bottle, oxygen is used but not produced. Subtracting the amount of oxygen at the
start, determined by a measurement taken from a control “initial” bottle from the amount left at
the end of the run (usually 24 hours), determines the quantity of oxygen used (respiration). The
amount of oxygen in the light bottle added to the amount used in the dark bottle provides an
estimate of total photosynthesis or gross production.

Respiration = Initial – Dark

Adapted from Biology Lab Manual for Students
The College Board, 2001
Net Primary Production = Light – Initial
Gross Production = Light – Dark

Ecologists prefer to express primary production in terms of carbon fixed rather than oxygen
produced. Oxygen values are often converted to carbon. One method assumes that 1 mole of
oxygen is released for each mole of carbon dioxide that is fixed as implied by the simple
photosynthetic equation: 6CO2 + 6H2O + light  6 O2 + C6H12O6

The molecular weights, 44 for carbon dioxide and 32 for oxygen, are used to convert oxygen
produced into carbon dioxide consumed. For each milliliter of oxygen produced, approximately
0.536 milligrams of carbon has been fixed. (It’s actually pretty fun to figure this out. You can
convert ml oxygen into mg carbon dioxide fairly easily, and then just use the percent of carbon
in carbon dioxide to figure out how much carbon was used.)

There are two common ways to determine the amount of oxygen dissolved in water. One is by
using a dissolved oxygen probe. You stick it in the water and get a digital readout of how much
dissolved oxygen there is. Don’t you love technology? The other way is by the Winkler titration
method. You add a bunch of chemicals to react with the oxygen to produce iodine. The
quantity of iodine produced is equivalent to the amount of oxygen in the sample. The amount of
iodine is quantified by titration with sodium thiosulfate until it turns from purplish black to clear.
The volume of sodium thiosulfate (in ml) used to titrate the 20 ml sample is approximately
equivalent to the concentration of dissolved oxygen (mg/l) in the original sample. You can
determine the concentration of dissolved oxygen by observing how much sodium thiosulfate
working solution was required to convert free iodine.

ml titrant used = mg/l dissolved oxygen

You can convert the mg/l of dissolved oxygen to ml/l using the following formula:

mg O2/l x 0.698 = ml O2/l

Using a nomograph (below) and a ruler, you can estimate the percent saturation of dissolved
oxygen in your sample. Align the ruler so that it passes through both the water temperature and
the dissolved oxygen content. Then read the percent saturation.

Adapted from Biology Lab Manual for Students
The College Board, 2001
From the data in the following table, use the nomograph to estimate the percent oxygen
saturation at each of the three temperatures used in the experiment. Record the data in the
table.

Temperature oC               Mean Dissolved Oxygen (mg/l)               % Dissolved Oxygen
0                                    6.9                                 46
20                                    6.3                                 68
30                                    5.9                                 79

Measurement of Primary Productivity

Procedure:

1. Obtain seven water sample bottles. There will be an initial sample bottle, one in the dark
(covered with foil), and seven with different light intensities (100%, 65%, 25%, 10%, and
2%). The different light intensities will be simulated by covering the bottles with different
numbers of mesh screens.

2. Fill each bottle with a water sample, and put the appropriate number of screens or foil on
the bottles.

3. Measure the dissolved oxygen concentration of the initial sample.

4. Place the covered bottles under a light source for 24 hours.

5. After 24 hours, measure the dissolved oxygen concentration of each of the samples.

6. Determine the respiration rate, the gross productivity, and the net productivity for each of
the samples.

7. Convert the gross productivity data (ml O2) for your samples to carbon productivity (mg
C/m3) using the following formulas:

ml O2/l = 0.698 x mg O2/l
mg C/l = 0.536 x ml O2/l

1 m3 = 1000 l, so you can convert mg C/l to mg C/m3 by multiplying by 1000.

Adapted from Biology Lab Manual for Students
The College Board, 2001
Fill in the following table:

Percent Light           Dissolved               Gross       Net Productivity        Gross
Oxygen              Productivity      (mg O2/l)         Productivity
(mg O2/l)             (mg O2/l)                          (mg C/m3)
Initial                6.3                 ----------       ----------         ----------
Dark                    4.1               ----------       ----------         ----------
100%                     7.95
65%                     7.85
25%                     7.3
10%                     6.55
2%                     5.6

Questions:

1. How does temperature affect the solubility of oxygen in water?

2. Would you expect to find a higher dissolved oxygen content in a body of water in winter
or summer? Why?

3. Which samples in the above table were limited in their productivity by lack of light?

Adapted from Biology Lab Manual for Students
The College Board, 2001
Productivity Simulation:

Productivity of a body of water can be estimated by measuring the productivity of water samples
and then plotting the rates on a depth profile graph. To create a depth profile, the degree to
which the body of water reduces light must be known. Data was collected from 2 lakes to show
how much light was available at different depths.

Lake 1

Percent of Incident Light        Depth (meters)
100%                          0
65%                          0.5
25%                          1.5
10%                          2.5
2%                           4

Lake 2

Percent of Incident Light        Depth (meters)
100%                          0
65%                          1.5
25%                           4
10%                           7
2%                          11

Analysis:

Using the gross productivity data (mg C/m3) that you calculated before, simulate primary
productivity for the two data sets above.

On graph paper, plot this converted data at the depth which they could occur in each of these
lakes. Use axes that look like the sample graph below:

1. Based on your analysis, which lake is more productive?

Adapted from Biology Lab Manual for Students
The College Board, 2001

```
Related docs
Other docs by Yz242Sd
Fiche de revision 14 18 pour le site
Insert School name and logo here
PARTS OF AN ON-FICTION BOOK