Document 318947

 Advanced Biology Laboratory Manual 2010-­‐2011 Name:_______________________ 1
2
Biology Table of Contents
CB#
Suggested
Unit
Dissolved Oxygen Probe Calibration
Dissolved Oxygen
Primary Productivity
Animal Behavior
Transpiration
Circulatory: Blood Pressure
Evolution: Hardy Weinberg
Enzymes
Osmosis and Diffusion
12A
12B
11
9
10
8
2
1
Ecology
Ecology
Animals
Plants
Circulation
Evolution
Biochemistry
Cells
Mitosis
Meiosis
Cellular Respiration
3A
3B
5
Plant Pigments
4A
Photosynthesis
4B
Electrophoresis
Transformation
Chi Square Pre-Lab
Corn and Genetics
Extra Graph Paper
6B
6A
7
7
Cells
Cells
Cell
Energetics
Cell
Energetics
Cell
Energetics
Biotechnology
Biotechnology
Genetics
Genetics
Page Numbers
5-7 & 15-17
9-14
19-24
25-29
31-38
39-43
45-54
55-63
65-86
89-95
97-101
103-108
109-112
113-122
123-133
135-139
141-142
143-147
149-152
4
PRE-LAB DISSOLVED OXYGEN PROBE PROCEDURE
**Important: Prior to each use, the Dissolved Oxygen Probe must warm up
for a period of 15–20 minutes as described below. If the probe is not
warmed up properly, inaccurate readings will result. Perform the following
steps to prepare the Dissolved Oxygen Probe.**
You have two bottles of fluid in your dissolved oxygen probe box: DO
electrode filling solution (used to lubricate the membrane inside the probe)
AND Sodium Sulfite Calibration Solution used to calibrate the sensor.
1. Prepare the Dissolved Oxygen Probe for use.
a.
b.
c.
d.
e.
Remove the protective cap.
Unscrew the membrane cap from the tip of the probe.
Using a pipet, fill the membrane cap with 1 mL of DO Electrode Filling Solution.
Carefully thread the membrane cap back onto the electrode.
Place the probe into a 250 mL beaker containing distilled water.
Figure 1
2. Connect the Dissolved Oxygen Probe to CH 1 and the Temperature Probe to CH 2 on
LabQuest. Choose New from the File menu. If you have older sensors that do not
auto-ID, manually set up the sensors.
3. It is necessary to warm up the Dissolved Oxygen Probe for 5–10 minutes before
taking readings. With the probe still in the distilled water beaker, wait while the probe
warms up. The probe must stay connected at all times to keep it warmed up.
4. On the Meter screen, tap Mode. Change the mode to Selected Events. Select OK.
5. Calibrate the Dissolved Oxygen Probe.
If your instructor directs you to use the calibration stored in the experiment file,
continue directly to the Procedure.
• If your instructor directs you to manually enter the calibration values, choose
Calibrate►Dissolved Oxygen from the Sensors menu. Tap Equation. Enter the
values for the Slope and the Intercept. Select Apply. Select OK and continue
directly to the Procedure.
• If your instructor directs you to perform a new calibration, use the procedure at the
end of this lab.
•
5
CALIBRATION PROCEDURE
1. Perform the calibration.
a. Choose Calibrate►Dissolved Oxygen from the Sensors menu.
b. Select Calibrate Now.
Zero-Oxygen Calibration Point
c. Remove the probe from the water bath and place the
tip of the probe into the Sodium Sulfite Calibration
Solution. Important: No air bubbles can be trapped
below the tip of the probe or the probe will sense an
inaccurate dissolved oxygen level. If the voltage does
not rapidly decrease, tap the side of the bottle with the
probe to dislodge the bubble. The readings should be
in the 0.2 to 0.6 V range.
d. Enter 0 as the known value in mg/L for Reading 1.
e. When the voltage stabilizes (~1 minute), tap Keep.
Figure 3
Saturated DO Calibration Point
f. Rinse the probe with distilled water and gently blot dry.
g. Unscrew the lid of the calibration bottle provided with the probe. Slide the lid and
the grommet about 1/2 inch onto the probe body.
h.
i.
j.
k.
l.
Figure 4
Add water to the bottle to a depth of about 1/4 inch and screw the bottle into the
cap, as shown. Important: Do not touch the membrane or get it wet during this
step.
In the Reading 2 field, enter the correct saturated dissolved-oxygen value
(in mg/L) from Table 3 (for example, 8.66) using the current barometric pressure
and air temperature values. If you do not have the current air pressure, use Table 4
to estimate the air pressure at your altitude.
Keep the probe in this position for about a minute. The readings should be above
2.0 V. When the voltage reading stabilizes, tap Keep.
Return the Dissolved Oxygen Probe to the distilled water beaker.
Select OK
6
CALIBRATION TABLES
0°C
1°C
2°C
3°C
4°C
5°C
6°C
7°C
8°C
9°C
10°
C
11°
C
12°
C
13°
C
14°
C
15°
C
16°
C
17°
C
18°
C
19°
C
20°
C
21°
C
22°
C
23°
C
24°
C
25°
C
26°
C
27°
C
28°
C
29°
C
30°
C
770
mm
14.7
14.36
14.08
13.61
13.35
12.91
12.67
12.36
12.05
11.75
11.57
11.20
10.94
10.78
10.54
10.21
10.09
9.867
9.67
9.47
9.29
9.11
8.94
8.78
8.62
8.47
8.32
8.17
8.04
7.90
7.77
760
mm
14.5
14.17
13.89
13.42
13.17
12.83
12.41
12.19
11.99
11.60
11.32
11.05
10.89
10.64
10.30
10.17
9.945
9.74
9.54
9.35
9.17
9.00
8.83
8.66
8.51
8.36
8.21
8.07
7.93
7.80
7.67
Table 3: 100% Dissolved Oxygen Capacity (mg/L)
750
740
730
720
710
700
690
680
mm 14.1
mm 13.9
mm 13.8
mm 13.6
mm 13.4
mm 13.2
mm 13.0
mm
14.3
14.08 13.89 13.69 13.40 13.21 13.02 12.83 12.74
13.60 13.42 13.23 13.14 12.96 12.77 12.58 12.30
13.24 13.16 12.98 12.70 12.52 12.43 12.25 12.07
12.99 12.72 12.64 12.46 12.29 12.11 11.93 11.75
12.66 12.49 12.31 12.14 11.97 11.80 11.62 11.45
12.34 12.17 12.00 11.83 11.66 11.50 11.33 11.16
12.03 11.86 11.70 11.53 11.37 11.21 11.04 10.98
11.73 11.57 11.41 11.25 11.19 10.93 10.87 10.61
11.44 11.38 11.13 11.07 10.81 10.76 10.50 10.35
11.26 11.01 10.96 10.71 10.65 10.40 10.35 10.19
10.90 10.85 10.60 10.55 10.30 10.25 10.00 9.925
10.74 10.50 10.45 10.21 10.16 9.991 9.847 9.70
10.40 10.36 10.11 10.07 9.903 9.77 9.63 9.49
10.26 10.12 9.968 9.834 9.69 9.55 9.42 9.28
10.04 9.880 9.75 9.62 9.48 9.35 9.22 9.08
9.812 9.68 9.55 9.42 9.29 9.15 9.02 8.89
9.61 9.48 9.35 9.22 9.10 8.97 8.84 8.71
9.41 9.29 9.16 9.04 8.91 8.79 8.66 8.54
9.23 9.11 8.98 8.86 8.74 8.61 8.49 8.37
9.05 8.93 8.81 8.69 8.57 8.45 8.33 8.20
8.88 8.76 8.64 8.52 8.40 8.28 8.17 8.05
8.71 8.59 8.48 8.36 8.25 8.13 8.01 7.90
8.55 8.44 8.32 8.21 8.09 7.98 7.87 7.75
8.40 8.28 8.17 8.06 7.95 7.84 7.72 7.61
8.25 8.14 8.03 7.92 7.81 7.70 7.59 7.48
8.10 7.99 7.89 7.78 7.67 7.56 7.45 7.35
7.96 7.86 7.75 7.64 7.54 7.43 7.33 7.22
7.83 7.72 7.62 7.51 7.41 7.30 7.20 7.10
7.69 7.59 7.49 7.39 7.28 7.18 7.08 6.98
7.57 7.47 7.36 7.26 7.16 7.06 6.96 6.86
670
mm
12.8
12.54
12.11
11.89
11.58
11.28
11.09
10.71
10.45
10.29
10.04
9.780
9.56
9.35
9.14
8.95
8.76
8.58
8.41
8.24
8.08
7.93
7.78
7.64
7.50
7.37
7.24
7.11
6.99
6.87
6.76
660
mm
12.6
12.35
12.02
11.71
11.40
11.10
10.82
10.55
10.39
10.03
9.869
9.63
9.41
9.21
9.01
8.82
8.63
8.45
8.28
8.12
7.96
7.81
7.67
7.52
7.39
7.26
7.13
7.01
6.89
6.77
6.66
Table 4: Approximate Barometric Pressure at Different Elevations
Elevation Pressure
Elevation Pressure
Elevation Pressure
(mm Hg)
(mm Hg)
(mm Hg)
(m)
(m)
(m)
0
760
800
693
1600
628
100
748
900
685
1700
620
200
741
1000
676
1800
612
300
733
1100
669
1900
604
400
725
1200
661
2000
596
500
717
1300
652
2100
588
600
709
1400
643
2200
580
700
701
1500
636
2300
571
7
8
Dissolved Oxygen and Water Temperature
Although water is composed of oxygen and hydrogen atoms, biological life in water depends upon
another form of oxygen-dissolved oxygen (or oxygen saturation). Oxygen is used by organisms in
aerobic respiration where energy is released by the combustion of sugar in the mitochondria. This form
of oxygen can fit into the spaces between water molecules and is available to aquatic organisms. Water
obtains oxygen from two major sources: the atmosphere and from aquatic plants.
Aquatic animals depend upon the oxygen dissolved in water. Without this oxygen, they would
suffocate. Fish have different tolerances for temperature and dissolved oxygen. The ecological quality
of the water depends largely upon the amount of oxygen the water can hold. Table 1 indicates the
normal tolerance of selected animals to temperature and oxygen levels. By observing the aquatic
animal populations in a stream, The quality of the water can be assessed with fair accuracy.
Table 1: Temperature and Dissolved Oxygen Ranges for Fish
Animal
Trout
Smallmouth bass
Caddisfly larvae
Mayfly larvae
Stonefly larvae
Catfish
Carp
Mosquito
Water boatmen
Temperature range
(°C)
5–20
5–28
10–25
10–25
10–25
20–25
10–25
10–25
10–25
Minimum dissolved
oxygen (mg/L)
6.5
6.5
4.0
4.0
4.0
2.5
2.0
1.0
2.0
OBJECTIVES:
In this experiment, you will
• Study the effect of temperature on the amount of dissolved oxygen in water.
• Predict the effect of water temperature on aquatic life.
MATERIALS:
LabQuest
Temperature Probe
Vernier Dissolved Oxygen Probe
D.O. calibration bottle and cap
Adapted from Advanced Biology with Vernier
two 250 mL beakers
hot and cold water
1 gallon plastic milk container
Styrofoam cup
9
PROCEDURE
1. Start data collection.
2. Obtain two 250 mL beakers. Fill one beaker with ice and water. Fill the second beaker with warm
water about 40–50°C.
3. Place approximately 100 mL of cold water and a couple small pieces of ice, from the beaker filled
with ice, into a clean plastic one-gallon milk container. Seal the container and vigorously shake the
water for a period of 2 minutes. This will allow the air inside the container to dissolve into the
water sample. Pour the water into a clean Styrofoam cup.
4. Place the Temperature Probe in the Styrofoam cup as shown
in Figure 2. Place the shaft of the Dissolved Oxygen Probe
into the water and gently stir. Avoid hitting the edge of the
cup with the probe.
5. Monitor the dissolved oxygen readings displayed on the
screen. Give the dissolved oxygen readings ample time to
stabilize (90–120 seconds). At colder temperatures the probe
will require a greater amount of time to output stable
readings. When the readings have stabilized, tap Keep.
6. Place the Dissolved Oxygen Probe back into the distilled
water beaker.
Figure 2
7. Pour the water from the Styrofoam cup back into the milk
container. Seal the container and shake the water vigorously for 1 minute. Pour the water back into
the Styrofoam cup.
8. Repeat Steps 4–7 until the water sample reaches room temperature.
9. When room temperature has been reached, begin adding about 25 mL of very warm water (40–
50°C) prior to shaking the water sample. This will allow you to take warmer water readings.
Repeat Steps 4–7 until the water sample reaches 35°C.
10. When all samples have been taken, stop data collection.
11. Create a single graph of dissolved oxygen vs. temperature.
a. Choose Show Graph ►Graph 1 from the Graph menu.
b. Tap the x-axis label and select Temperature.
c. To examine the data pairs on the displayed graph, tap any data point. As you tap each data point,
the dissolved oxygen and temperature values are displayed to the right of the graph.
d. Record the data values in Table 2.
12. Use the graph of dissolved oxygen concentration vs. temperature to help you answer the questions
below.
Adapted from Advanced Biology with Vernier
10
DATA
TABLE 2
Temperature
(°C)
Dissolved oxygen
(mg/L)
QUESTIONS
1. At what temperature was the dissolved oxygen concentration the highest? Lowest?
2. Does your data indicate how the amount of dissolved oxygen in the water is affected by the
temperature of water? Explain.
Adapted from Advanced Biology with Vernier
11
3. If you analyzed the invertebrates in a stream and found an abundant supply of caddisflies, mayflies,
dragonfly larvae, and trout, what minimum concentration of dissolved oxygen would be present in
the stream? What maximum temperature would you expect the stream to sustain?
4. Mosquito larvae can tolerate extremely low dissolved oxygen concentrations, yet cannot survive at
temperatures above approximately 25°C. How might you account for dissolved oxygen
concentrations of such a low value at a temperature of 25°C? Explain.
5. Why might trout be found in pools of water shaded by trees and shrubs more commonly than in
water where the trees have been cleared?
Adapted from Advanced Biology with Vernier
12
13
14
PRE-LAB DISSOLVED OXYGEN PROBE PROCEDURE
**Important: Prior to each use, the Dissolved Oxygen Probe must warm up
for a period of 15–20 minutes as described below. If the probe is not
warmed up properly, inaccurate readings will result. Perform the following
steps to prepare the Dissolved Oxygen Probe.**
You have two bottles of fluid in your dissolved oxygen probe box: DO
electrode filling solution (used to lubricate the membrane inside the probe)
AND Sodium Sulfite Calibration Solution used to calibrate the sensor.
1. Prepare the Dissolved Oxygen Probe for use.
a.
b.
c.
d.
e.
Remove the protective cap.
Unscrew the membrane cap from the tip of the probe.
Using a pipet, fill the membrane cap with 1 mL of DO Electrode Filling Solution.
Carefully thread the membrane cap back onto the electrode.
Place the probe into a 250 mL beaker containing distilled water.
Figure 1
2. Connect the Dissolved Oxygen Probe to CH 1 and the Temperature Probe to CH 2 on
LabQuest. Choose New from the File menu. If you have older sensors that do not
auto-ID, manually set up the sensors.
3. It is necessary to warm up the Dissolved Oxygen Probe for 5–10 minutes before
taking readings. With the probe still in the distilled water beaker, wait while the probe
warms up. The probe must stay connected at all times to keep it warmed up.
4. On the Meter screen, tap Mode. Change the mode to Selected Events. Select OK.
5. Calibrate the Dissolved Oxygen Probe.
If your instructor directs you to use the calibration stored in the experiment file,
continue directly to the Procedure.
• If your instructor directs you to manually enter the calibration values, choose
Calibrate►Dissolved Oxygen from the Sensors menu. Tap Equation. Enter the
values for the Slope and the Intercept. Select Apply. Select OK and continue
directly to the Procedure.
• If your instructor directs you to perform a new calibration, use the procedure at the
end of this lab.
•
15
CALIBRATION PROCEDURE
1. Perform the calibration.
a. Choose Calibrate►Dissolved Oxygen from the Sensors menu.
b. Select Calibrate Now.
Zero-Oxygen Calibration Point
c. Remove the probe from the water bath and place the
tip of the probe into the Sodium Sulfite Calibration
Solution. Important: No air bubbles can be trapped
below the tip of the probe or the probe will sense an
inaccurate dissolved oxygen level. If the voltage does
not rapidly decrease, tap the side of the bottle with the
probe to dislodge the bubble. The readings should be
in the 0.2 to 0.6 V range.
d. Enter 0 as the known value in mg/L for Reading 1.
e. When the voltage stabilizes (~1 minute), tap Keep.
Figure 3
Saturated DO Calibration Point
f. Rinse the probe with distilled water and gently blot dry.
g. Unscrew the lid of the calibration bottle provided with the probe. Slide the lid and
the grommet about 1/2 inch onto the probe body.
h.
i.
j.
k.
l.
Figure 4
Add water to the bottle to a depth of about 1/4 inch and screw the bottle into the
cap, as shown. Important: Do not touch the membrane or get it wet during this
step.
In the Reading 2 field, enter the correct saturated dissolved-oxygen value
(in mg/L) from Table 3 (for example, 8.66) using the current barometric pressure
and air temperature values. If you do not have the current air pressure, use Table 4
to estimate the air pressure at your altitude.
Keep the probe in this position for about a minute. The readings should be above
2.0 V. When the voltage reading stabilizes, tap Keep.
Return the Dissolved Oxygen Probe to the distilled water beaker.
Select OK
16
CALIBRATION TABLES
0°C
1°C
2°C
3°C
4°C
5°C
6°C
7°C
8°C
9°C
10°
C
11°
C
12°
C
13°
C
14°
C
15°
C
16°
C
17°
C
18°
C
19°
C
20°
C
21°
C
22°
C
23°
C
24°
C
25°
C
26°
C
27°
C
28°
C
29°
C
30°
C
770
mm
14.7
14.36
14.08
13.61
13.35
12.91
12.67
12.36
12.05
11.75
11.57
11.20
10.94
10.78
10.54
10.21
10.09
9.867
9.67
9.47
9.29
9.11
8.94
8.78
8.62
8.47
8.32
8.17
8.04
7.90
7.77
760
mm
14.5
14.17
13.89
13.42
13.17
12.83
12.41
12.19
11.99
11.60
11.32
11.05
10.89
10.64
10.30
10.17
9.945
9.74
9.54
9.35
9.17
9.00
8.83
8.66
8.51
8.36
8.21
8.07
7.93
7.80
7.67
Table 3: 100% Dissolved Oxygen Capacity (mg/L)
750
740
730
720
710
700
690
680
mm 14.1
mm 13.9
mm 13.8
mm 13.6
mm 13.4
mm 13.2
mm 13.0
mm
14.3
14.08 13.89 13.69 13.40 13.21 13.02 12.83 12.74
13.60 13.42 13.23 13.14 12.96 12.77 12.58 12.30
13.24 13.16 12.98 12.70 12.52 12.43 12.25 12.07
12.99 12.72 12.64 12.46 12.29 12.11 11.93 11.75
12.66 12.49 12.31 12.14 11.97 11.80 11.62 11.45
12.34 12.17 12.00 11.83 11.66 11.50 11.33 11.16
12.03 11.86 11.70 11.53 11.37 11.21 11.04 10.98
11.73 11.57 11.41 11.25 11.19 10.93 10.87 10.61
11.44 11.38 11.13 11.07 10.81 10.76 10.50 10.35
11.26 11.01 10.96 10.71 10.65 10.40 10.35 10.19
10.90 10.85 10.60 10.55 10.30 10.25 10.00 9.925
10.74 10.50 10.45 10.21 10.16 9.991 9.847 9.70
10.40 10.36 10.11 10.07 9.903 9.77 9.63 9.49
10.26 10.12 9.968 9.834 9.69 9.55 9.42 9.28
10.04 9.880 9.75 9.62 9.48 9.35 9.22 9.08
9.812 9.68 9.55 9.42 9.29 9.15 9.02 8.89
9.61 9.48 9.35 9.22 9.10 8.97 8.84 8.71
9.41 9.29 9.16 9.04 8.91 8.79 8.66 8.54
9.23 9.11 8.98 8.86 8.74 8.61 8.49 8.37
9.05 8.93 8.81 8.69 8.57 8.45 8.33 8.20
8.88 8.76 8.64 8.52 8.40 8.28 8.17 8.05
8.71 8.59 8.48 8.36 8.25 8.13 8.01 7.90
8.55 8.44 8.32 8.21 8.09 7.98 7.87 7.75
8.40 8.28 8.17 8.06 7.95 7.84 7.72 7.61
8.25 8.14 8.03 7.92 7.81 7.70 7.59 7.48
8.10 7.99 7.89 7.78 7.67 7.56 7.45 7.35
7.96 7.86 7.75 7.64 7.54 7.43 7.33 7.22
7.83 7.72 7.62 7.51 7.41 7.30 7.20 7.10
7.69 7.59 7.49 7.39 7.28 7.18 7.08 6.98
7.57 7.47 7.36 7.26 7.16 7.06 6.96 6.86
670
mm
12.8
12.54
12.11
11.89
11.58
11.28
11.09
10.71
10.45
10.29
10.04
9.780
9.56
9.35
9.14
8.95
8.76
8.58
8.41
8.24
8.08
7.93
7.78
7.64
7.50
7.37
7.24
7.11
6.99
6.87
6.76
660
mm
12.6
12.35
12.02
11.71
11.40
11.10
10.82
10.55
10.39
10.03
9.869
9.63
9.41
9.21
9.01
8.82
8.63
8.45
8.28
8.12
7.96
7.81
7.67
7.52
7.39
7.26
7.13
7.01
6.89
6.77
6.66
Table 4: Approximate Barometric Pressure at Different Elevations
Elevation Pressure
Elevation Pressure
Elevation Pressure
(mm Hg)
(mm Hg)
(mm Hg)
(m)
(m)
(m)
0
760
800
693
1600
628
100
748
900
685
1700
620
200
741
1000
676
1800
612
300
733
1100
669
1900
604
400
725
1200
661
2000
596
500
717
1300
652
2100
588
600
709
1400
643
2200
580
700
701
1500
636
2300
571
17
18
Primary Productivity
Oxygen is vital to life. In the atmosphere, oxygen comprises over 20% of the available gases. In
aquatic ecosystems, however, oxygen is scarce. To be useful to aquatic organisms, oxygen must
be in the form of molecular oxygen, O2. The concentration of oxygen in water can be affected by
many physical and biological factors. Respiration by plants and animals reduces oxygen
concentrations, while the photosynthetic activity of plants increases it. In photosynthesis, carbon
is assimilated into the biosphere and oxygen is made available, as follows:
6 H2O + 6 CO2(g) + energy ® C6H12O6 + 6O2(g)
The rate of assimilation of carbon in water depends on the type and quantity of plants within the
water. Primary productivity is the measure of this rate of carbon assimilation. As the above
equation indicates, the production of oxygen can be used to monitor the primary productivity of
an aquatic ecosystem. A measure of oxygen production over time provides a means of
calculating the amount of carbon that has been bound in organic compounds during that period
of time. Primary productivity can also be measured by determining the rate of carbon dioxide
utilization or the rate of formation of organic compounds.
One method of measuring the production of oxygen is the light and dark bottle method. In this
method, a sample of water is placed into two bottles. One bottle is stored in the dark and the
other in a lighted area. Only respiration can occur in the bottle stored in the dark. The decrease in
dissolved oxygen (DO) in the dark bottle over time is a measure of the rate of respiration. Both
photosynthesis and respiration can occur in the bottle exposed to light, however. The difference
between the amount of oxygen produced through photosynthesis and that consumed through
aerobic respiration is the net productivity. The difference in dissolved oxygen over time between
the bottles stored in the light and in the dark is a measure of the total amount of oxygen produced
by photosynthesis. The total amount of oxygen produced is called the gross productivity.
The measurement of the DO concentration of a body of water is often used to determine whether
the biological activities requiring oxygen are occurring and is an important indicator of pollution.
OBJECTIVES
In this experiment, you will
•
•
Measure the rate of respiration in an aquatic environment using a Dissolved Oxygen Probe.
Determine the net and gross productivity in an aquatic environment.
MATERIALS
LabQuest
Vernier Dissolved Oxygen Probe
1.5L of pond or lake water per group
Deep buckets
Distilled water
250 mL beaker
Aquatic Bottles
19
PROCEDURE
Day 1
1. Obtain a set of aquatic bottles.
2. Fill each of the containers with the water sample from the same bucket. There must be NO
air bubbles in each container. Be sure no air is in the container.
3. Use a marker to label the cap and the bottom of each container.
4. Mix the contents of each container. Be sure that there are no air bubbles present in any of the
containers. Fill with more sample water if necessary.
5.
Remove the Dissolved Oxygen Probe from the beaker. Place the probe into a container, so
that it is submerged half the depth of the water. Gently and continuously move the probe up
and down a distance of about 1 cm in the container. This allows water to move past the
probe’s tip (if you have a Vernier microstirrer, you can attach it to the bottom of the probe
and place on a magnetic stirrer). Note: Do not agitate the water, or oxygen from the
atmosphere will mix into the water and cause erroneous readings.
6.
After 60 seconds, or when the dissolved oxygen reading stabilizes, record the reading in
Table 2. Replace the displaced water in the container removing all air bubbles and cap.
7.
Rinse the Dissolved Oxygen Probe with distilled water and place it back in the distilled water
beaker. The probe should remain in the beaker overnight, so that measurements can be made
the following day.
8.
Place the containers in the aquatic chamber and place near the light source, as directed by
your instructor.
Day 2
1. Repeat the pre-lab procedure to polarize the Dissolved Oxygen Probe and manually enter the
calibration values.
2. Place the probe into the 100% light container. Gently and continuously move the probe up
and down a distance of about 1 cm in the container. After 60 seconds, or when the dissolved
oxygen reading stabilizes, record the reading in Table 2.
3. Repeat Step #2 (Day 2) for the remaining containers.
4. Clean your containers as directed by your instructor. Rinse the Dissolved Oxygen Probe with
distilled water and place it back in the distilled water beaker.
20
DATA
Table 2
Test tube
% Light
1
Initial
2
100%
3
75%
4
50%
5
25%
6
0%
DO (mg/L)
PROCESSING THE DATA
1. Determine the number of hours that have passed since the onset of this experiment. Subtract
the DO value in test tube 1 (the initial DO value) from that of test tube 7 (the dark test tube’s
DO value). Divide the DO value by the time in hours. Record the resulting value as the
respiration rate in Table 3.
Respiration rate = (dark DO – initial DO) / time
2. Determine the gross productivity in each test tube. To do this, subtract the DO in test tube 7
(the dark test tube’s DO value) from that of test tubes 2–6 (the light test tubes’ DO value).
Divide each DO value by the length of the experiment in hours. Record each resulting value
as the gross productivity in Table 4.
Gross productivity = (DO of test tube – dark DO) / time
3. Determine the net productivity in each test tube. To do this, subtract the DO in test tubes 2–6
(the light test tube’s DO value) from that of test tube 1 (the initial DO value). Divide the
result by the length of the experiment in hours. Record each resulting value as the net
productivity in Table 4.
Net Productivity = (DO of test tube – initial DO) / time
Table 3
Respiration (mg O2/L/hr)
21
Table 4
Test Tube
% Light
2
100%
3
65%
4
25%
5
10%
6
2%
Gross productivity
(mg/L/hr)
Net productivity
(mg/L/hr)
4. Prepare a graph of gross productivity and net productivity as a function of light intensity.
Graph both types of productivity on the same piece of graph paper.
QUESTIONS
1. Is there evidence that photosynthetic activity added oxygen to the water? Explain.
2. Is there evidence that aerobic respiration occurred in the water? If so, what kind of organisms
might be responsible for this (autotrophs, heterotrophs)? Explain.
3. What effect did light have on the primary productivity? Explain.
4. Refer to your graph of productivity and light intensity. At what light intensity do you expect
there to be no net productivity? no gross productivity?
22
5. How would turbidity affect the primary productivity of a pond?
EXTENSIONS
1. Determine the effects of nitrogen and phosphates on primary productivity. Why would the
presence of phosphates and nitrogen, in the form of nitrates and ammonium ions, be
important to an aquatic ecosystem during the spring season? How do they accumulate in the
watershed? What is eutrophication?
2. Measure the dissolved oxygen of a pond at different temperatures. What is the effect of
temperature on the primary productivity of a pond?
3. Calculate the amount of carbon that was fixed in each of the tubes. Use the following
conversion factors to do the calculations.
mg O2/L X 0.698 = mL O2/L
mL O2/L X .536 = mg carbon fixed/L
23
24
Animal Behavior: Pillbugs Biotic and abiotic factors are limiting factors in an ecosystem because they can regulate the maximum size of the population based on their availability. Organisms are able to exist in a relatively narrow range of environmental conditions that favor them and their offspring. Since most organisms cannot change the nature of their environment, they must position themselves in an environment with favorable conditions Many organisms exhibit a tactic response to these environmental factors. Tactic responses may be positive, toward a favorable environment, or negative, away from an unfavorable environment. These static responses enable an organism to locate prey, avoid predators, seek food or shelter, or avoid a negative environment. There are two types of behavior movements: taxis and kinesis. Taxis is a deliberate movement toward or away from a stimulus. Kinesis, on the other hand, is a random movement that is not oriented toward or away from a stimulus. A movement classified as taxis defines the physiological needs of the organism, its evolutionary history, its’ nervous system, etc. Objectives: • Design an experiment to determine how pillbugs respond to environmental changes. • Observe behavior characteristics of pillbugs. Materials pillbugs (10 per group) Masking tape 1% hydrochloric acid solution Magnifier or stereoscope 2% potassium hydroxide solution filter papers (2 per group) Animal behavior tray Camel’s hair paint brush Procedure I. General Observations 1. Place 10 pillbugs in the behavior tray and carefully observe them for at least 10 minutes. 2. In Table 1, document any behaviors you see. Remember to document every behavior in chronological order. Note: Do not disturb the pillbugs during your observation. 3. In Table 2, sketch a pillbug and label its structures. Adapted from Advanced Biology with Vernier 25
Table 1: Pillbug Observations Table 2: Pillbug Anatomy Adapted from Advanced Biology with Vernier 26
II. Kinesis 1. Label one chamber of the tray A and one chamber B. 2. Place 5 pillbugs in each chamber of the tray 3. For 10 minutes, count the number of pillbugs in each tray each minute and record your data in table 3. 4. Calculate the average number of pillbugs in each chamber at the end of the 10-­‐minute period. 5. Collect class data for the average number of pillbugs in each chamber. Table 3: Pillbug Kinesis Time (min) Side A (# pillbugs) Side B (# pillbugs) 0 1 2 3 4 5 6 7 8 9 10 Group Avg. Class Avg. III. Experiment 1. Your instructor will assign you a variable to use as your environmental factor that will elicit either: chemotaxis, hydrotaxis, or Phototaxis. a. Chemotaxis-­‐ the orientation of an organism when exposed to a chemical b. Hydrotaxis-­‐ the orientation of an organism when exposed to water. c. Phototaxis-­‐ the orientation of an organism when exposed to light. 2. Formulate a hypothesis about how your pillbugs may react to the environmental variable that you are assigned. 3. Design an experiment to test your variable. 4. Once you have designed your experiment, check with your instructor before preceding. 5. Graph your results from part II and part III on the same graph. Adapted from Advanced Biology with Vernier 27
Hypothesis:___________________________________________________________________________________
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_________________________________________________________________________________________________ Table 4: Condition Tested: ____________________________________________ Time (min) Side A (# pillbugs) Side B (# pillbugs) 0 1 2 3 4 5 6 7 8 9 10 Group Avg. Adapted from Advanced Biology with Vernier 28
29
30
Whole Plant Transpiration
Background Information
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 (evaporation of water from the
stomata of leaves) and guttation (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.
Water transport from the roots to the leaves occurs in the xylem and is governed by differences in water
potential (the potential energy of water molecules). These differences account for water movement from
high to low water potential, working cell to cell 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.
The Overall Process
Minerals actively transported into the root accumulate in the xylem, increase solute concentration and
decrease water potential. 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 a negative tension. The "pull" on the water from transpiration is increased as
a result of cohesion and adhesion of water molecules.
The Details
Transpiration begins with evaporation of water through the stomates (stomata), small openings in the leaf
surface which 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.
The moisture in the air spaces is replaced by water from the adjacent mesophyll cells, lowering their
water potential. Water will then move into the mesophyll cells by osmosis from surrounding cells with the
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 is caused by (1) the cohesion of water molecules to one another due to
hydrogen bond formation, (2) by adhesion of water molecules to the walls of the xylem cells which aids
in offsetting the downward pull of gravity. The upward transpirational pull on the fluid in the xylem
causes a tension (negative pressure) to form in the xylem, pulling the xylem walls inward. The tension
also contributes to the lowering of the water potential in the xylem. This decrease in water potential,
transmitted all the way from the leaf to the roots, causes water to move inward from the soil, across the
cortex of the root, and into the xylem. Evaporation through the open stomates is a major route of water
loss in the plant. However, the stomates 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 stomates on the leaf surface. Many environmental conditions influence the
opening and closing of the stomates and also affect the rate of transpiration. Temperature, light intensity,
air currents, and humidity are some of these factors. Different plants also vary in the rate of transpiration
and in the regulation of stomatal opening.
Importance of the Transpiration “Engine”
•
•
•
supply photosynthesis (1%-2% of the total)
bring minerals from the roots for biosynthesis within the leaf
cool the leaf
Modification of Margaret Bahe’s Transpiration Lab and AP Biology Lab 9
31
OBJECTIVES
In this experiment, you will:
1. Observe how transpiration affects and correlates with the overall
process of water movement in plants
2. Understand the effects of temperature, humidity, air movement, and
light intensity on transpiration through experimentation
3. Understand the overall structure of a leaf (including cells,
transportation, tissues)
MATERIALS per group:
Incandescent lamp
Labeling tape and sharpie
4 basil plants~equal size
5 plastic bags and twist ties
small fan
cup for balance
spray bottle
balance
PROCEDURE
1. Select 4 plants that are about the same size. Remove any blossoms.
2. Water and drain the plants well. Carefully remove the plant from the plastic container by gently
squeezing the bottom and lifting the plant. Don’t force the plant out.
3. Wrap the root ball of each plant with a plastic bag. Tie the bag close to the stem with string or
twist tie. Label with your group name and experimental treatment.
4. Measure and record the masses (g) of each of the plants. Place them into an appropriate container.
5. Place the plants in different environmental conditions in the room:
a. CONTROL -The control is placed out of a draft or excessive heat.
b. WIND - 2 feet away from a fan on low (leaves shouldn’t rustle, can cause stomata to
close)
c. LIGHT- 2 feet away from a floodlight.
d. HUMID - The plant covered with bags on top and bottom or plant under an inverted
aquarium (heavily mist each day after massing)
e. DARK- in a closed cabinet.
6. Measure and record the masses of each plant every day for one week. Calculate the percent mass
change each day as compared to initial mass on Day 1. Enter data on a group data Table A.
7. Graph your plants’ daily masses over time.
Modification of Margaret Bahe’s Transpiration Lab and AP Biology Lab 9
32
Transpiration Lab Data
Group Data Table A
Day 1
Day 2
Day 3
Day 4
Day 5
Mass
Control
% change
0*
Mass
Light
% change
0*
Mass
High
Humidity
% change
0*
Mass
Fan (wind)
% change
0*
Mass
Dark
% change
0*
Class Data Table B (Averages)
Type
Group 1
Group 2
Group 3
Percent Mass Change
Group 4
Group 5
Group 6
Group 7
Group 8
Control
Light
High
Humidity
Fan
(Wind)
Dark
Modification of Margaret Bahe’s Transpiration Lab and AP Biology Lab 9
33
Average
PROCESSING THE DATA
1. Determine the surface area of all the leaves on your plant using the following method:
a.
b.
c.
d.
Cut one leaf (not stem) off your plant.
Estimate the surface area of the leaf in cm2.
Find the total leaf surface area by multiplying this by the number of leaves that size on your plant.
Repeat steps a-c for smaller or larger leaves and add up the total surface area of your plant’s
leaves.
e. Record the calculated surface area in Table 1.
2. Calculate the rate of transpiration (grams of water lost per day)/surface area. To do this, divide the
rate of transpiration by the surface area for each plant. These rate values can be expressed as
g/day/cm2. Record the rate/area in Table 1.
3. Subtract the control (rate/area) value from the experimental value. Record this adjusted rate in the
last column of Table 1.
4. Record the adjusted rate for your experimental test on the board to share with the class. Record the
class results in Table 2 for each of the environmental conditions tested. If a condition was tested by
more than one group, take the average of the values and record in Table 2.
5. Graph the class’ data for Transpiration Rate
Table 1: Individual Group Data
Test
Slope
(g/day)
Surface area
(cm2)
Rate/area
(g/day/cm2)
Adjusted rate
(g/day/cm2)
Experimental ______________
Control
Table 2: Class Data Transpiration Rate
Test
Adjusted rate (g/day/cm2)
Light
Humidity
Wind
Temperature
Modification of Margaret Bahe’s Transpiration Lab and AP Biology Lab 9
34
PRELAB QUESTIONS
1. What is transpiration? How does transpiration affect the size and shape of guard cells?
2. How does cohesion and adhesion relate to the movement of water by evaporation?
3. What is the independent variable? What is the dependent variable?
4. What are the controlled variables (constants)?
5. Describe and explain the impact of each of the following factors on the rate of transpiration.
a. Light Intensity
b. Temperature
c. Humidity
d. Air Movement or Wind
e. Soil water
6. Predict which environmental variable has the greatest effect on the rate of transpiration.
Modification of Margaret Bahe’s Transpiration Lab and AP Biology Lab 9
35
7. The data below were collected using a potometer which measures the volume of water lost during
transpiration (mL/m2). Graph the data below and calculate the rates of transpiration for each
condition.
Show your work!!
Transpiration in mL/m2
Condition
Wind
Control
Light
Humidity
0 min
0
0
0
0
3min
3.70
0.67
2.15
1.59
6min
7.77
1.34
4.27
1.70
9min
9.87
1.94
5.89
1.85
12min
12.34
2.61
8.15
1.97
15min
14.19
3.13
9.41
2.21
18min
16.29
3.63
10.96
2.46
21min
17.90
4.33
11.53
2.70
24min
19.87
4.78
12.76
2.83
27min
21.60
5.30
13.90
3.13
30 min
23.45
5.75
16.42
3.39
Rates of Transpiration:
Rate of Transpiration
Condition
(mL/m2/min)
Wind
Control
Light
Humidity
POST LAB: ANALYSIS QUESTIONS
1. Calculate the average rate of water loss per day for each of the treatments (humidity, light, fan,
dark or control). Show your work.
Modification of Margaret Bahe’s Transpiration Lab and AP Biology Lab 9
36
2. Explain why each of the conditions causes an increase or decrease in transpiration compared to
the control. Be sure to discuss the activities of the stomata.
3. Why is it advantageous for a plant to close its stomata when water is in short supply? What are
the disadvantages?
4. Describe several adaptations that enable plants to reduce water loss from their leaves. Include
both structural and physiological adaptations. What environment would you expect a plant to
have the adaptations mentioned above? (You may need your book to answer this question).
5. Why did you need to calculate leaf surface area in tabulating your results?
6. Which variable resulted in the greatest rate of water loss? Explain why this factor might increase
water loss when compared to the others.
7. What combination of experimental situations will create the most water loss? The least?
8. How does the concept of hydrogen bonding relate to transpirational pull? Be specific in your
explanation, using a drawing to assist your explanation.
Modification of Margaret Bahe’s Transpiration Lab and AP Biology Lab 9
37
Modification of Margaret Bahe’s Transpiration Lab and AP Biology Lab 9
38
Blood Pressure as a Vital Sign
Blood pressure is a measure of the changing fluid pressure within the circulatory system. It
varies from a peak pressure produced by contraction of the left ventricle, to a low pressure,
which is maintained by closure of the aortic valve and elastic recoil of the arterial system. The
peak pressure is called systole, and the pressure that is maintained even while the left ventricle is
relaxing is called diastole (see Figure 1).
Mean arterial pressure (MAP) is not a simple average of the two pressures, because the duration
of diastole is twice that of systole. MAP is used by emergency room and intensive care unit
personnel as a measure of the adequacy of blood supplied to vital tissues (such as the brain,
heart, and kidneys) when the blood pressure is dangerously low.
Blood pressure is traditionally reported with the systolic pressure stated first and the diastolic
pressure stated second. In adults, 120/80 and below is considered normal blood pressure. High
blood pressure is 140/90 or above. The seriousness of low blood pressure, as well as the health
risks of high blood pressure (also called hypertension), has been elucidated over the past several
decades. High blood pressure is a major risk factor for a number of health problems including
strokes and congestive heart failure. Diet and exercise are beneficial, but many people require
medication for optimal blood pressure control.
In this experiment, you will examine your blood pressure using the Vernier Blood Pressure
Sensor. You will compare blood pressures taken before and after exposure to cold. The cold
stimulus activates the sympathetic nervous system, resulting in hemodynamic changes that
prepare the body for a “fight or flight” response (i.e., when fighting or running from danger).
The sensitivity of blood pressure to harmful external or internal injuries makes it useful as a vital
sign, an indicator of health, disease, excitement, and stress.
Figure 1
Adapted from Biology with Vernier
39
OBJECTIVES
In this experiment, you will
Compare blood pressure before and after exposure to cold stimulus.
• Observe an example of sympathetic nervous system activation (“fight or flight” response).
•
MATERIALS
LabQuest
LabQuest App
Vernier Blood Pressure Sensor
ice water bath
towel (paper or cloth)
PROCEDURE
Select one or more persons from your lab group to be the subject(s).
Part I Baseline Blood Pressure
1. Connect the Blood Pressure Sensor to LabQuest. There are two rubber tubes connected to the
pressure cuff. One tube has a black Luer-lock connector at the end and the other tube has a
bulb pump attached. Connect the Luer-lock connector to the stem on the Blood Pressure
Sensor with a gentle half turn.
2. Choose New from the File menu.
3. On the Meter screen, tap Rate. Change the data-collection rate to 60 samples/second and the
data-collection length to 100 seconds. Select OK.
4. Attach the Blood Pressure cuff firmly around the upper arm, approximately
2 cm above the elbow. The two rubber hoses from the cuff should be
positioned over the biceps muscle (brachial artery) and not under the arm
(see Figure 2).
5. Have the subject sit quietly in a chair with his or her forearm resting on a
table surface. The person having his or her blood pressure measured must
remain still during data collection; there should be no movement of the arm
or hand during measurements.
6. Start data collection. Immediately pump the bulb pump until the cuff
Figure 2
pressure reaches at least 160 mm Hg. Stop pumping. The cuff will slowly
deflate and the pressure will fall. During this time, the systolic, diastolic, mean arterial
pressures, and pulse will be calculated by the software. These values will be displayed on the
screen. When the cuff pressure drops below 50 mm Hg, the program will stop calculating
blood pressure. At this point, you can stop data collection. Release the pressure from the cuff,
but do not remove it. Data collection will stop automatically after 100 seconds. If the final
pressure value recorded was not below 50 mm Hg, repeat this step to collect another run.
7. Tap the Meter tab. Enter the pulse and the systolic, diastolic, and mean arterial pressures in
Table 1.
Adapted from Biology with Vernier
r
40
Blood Pressure as a Vital Sign
Part II Blood Pressure Response to Cold
8. Prepare an ice water bath for use in the next step. The subject will be instructed to place his
or her opposite hand (the one to which the Blood Pressure cuff is not attached) in the ice
water bath for 15 s.
9. Collect data to examine the body’s response to cold.
a. With the cuff still attached, have the subject from Part I put the hand of his or her
non-cuffed arm in the ice water bath.
b. As soon as the subject’s hand enters the ice water bath, start data collection.
c. Pump the bulb until the cuff pressure reaches at least 160 mm Hg, then stop pumping.
d. When data have been collected for 15 s, have the subject remove his or her hand from the
ice water bath.
e. When the blood pressure readings have stabilized (after the pressure drops to 50 mm Hg),
the program will stop calculating blood pressure. At this point, you can stop data
collection. Release the pressure from the cuff, and remove the cuff from the subject’s arm.
10. Tap the Meter tab. Record the systolic, diastolic, and mean arterial pressures, and the pulse in
Table 2.
DATA
Table 1–Baseline Blood Pressure
Systolic pressure
(mm Hg)
Diastolic pressure
(mm Hg)
Mean arterial pressure
(mm Hg)
Pulse
(beats/minute)
Table 2–Blood Pressure Response to Cold
Systolic pressure
(mm Hg)
Diastolic pressure
(mm Hg)
Mean arterial pressure
(mm Hg)
Pulse
(beats/minute)
DATA ANALYSIS
1. Describe the trends that occurred in the systolic pressure, diastolic pressure, mean arterial
pressure, and pulse with cold stimulus. How might these be useful in a “fight or flight”
response?
Adapted from Biology with Vernier
41
2. Vasovagal syncope is a condition in which severe pain or fright activates the parasympathetic
nervous system instead of the sympathetic nervous system, resulting in fainting. Keeping in
mind that the parasympathetic system causes a response opposite to that of the sympathetic
system, describe the hemodynamic changes that would explain this.
3. As a vital sign, blood pressure is an indicator of general health. A high blood pressure
(140/90 or higher) increases the risk of cardiovascular disease and strokes. Collect the systolic
and diastolic pressures for the class and calculate the average for each. Rate the class average
blood pressure using the following scale:
Blood Pressure
Category
140/90 or higher
High
120–139/80–89
Pre-hypertension
119/79 or below
Normal
4. Why does blood pressure differ among individuals?
Adapted from Biology with Vernier
r
42
Blood Pressure as a Vital Sign
5. Why do some athletes have a low blood pressure?
6. How would the blood pressure differ between an endothermic organism and an ectothermic
organism with a change in temperature?
EXTENSION
Blood pressure is traditionally obtained by using a stethoscope to listen to the brachial artery.
The pumping of air into the blood pressure cuff acts to stop the blood flow through this artery.
As the pressure is released, the blood again is allowed to flow. When the blood begins to flow,
pulsations can be heard through the stethoscope. The pressure in the cuff at that time can be
noted, and corresponds closely to the systolic blood pressure. As pressure continues to be
released from the cuff, the pulsations of the artery become less audible. The pressure at which
they disappear has been found to approximate the diastolic pressure. These sounds are known as
Korotkoff Sounds.
With a stethoscope, obtain the blood pressure of a classmate by listening for the appearance and
disappearance of pulsations as the pressure in the cuff is released. Compare this to the blood
pressure you obtained with the Vernier Blood Pressure Sensor.
Adapted from Biology with Vernier
43
44
AP Lab 8: Population Genetics & Evolution
OVERVIEW
In this lab you will:
•
Learn about the Hardy-Weinberg law of genetic equilibrium, and
•
Study the relationship between evolution and changes in allele frequency by using your class as a
sample population.
OBJECTIVES
Before doing this lab you should understand:
•
How natural selection can alter allele frequencies in a population
•
The Hardy-Weinberg equation and its use in determining the frequency of alleles in a population;
and,
•
The effects of allelic frequencies of selection against the homozygous recessive or other
genotypes.
After doing this lab you should be able to:
•
Calculate the frequencies of alleles and genotypes in the gene pool of a population using the
Hardy-Weinberg formula, and
•
Discuss natural selection and other causes of microevolution as deviations from the conditions
required to maintain Hardy-Weinberg equilibrium.
INTRODUCTION
In 1908, G. H. Harding and W. Weinberg independently suggested a scheme whereby evolution could be
viewed as changes in the frequency of alleles in a population of organisms. They reasoned that if A and
a are alleles for a particular gene locus and each diploid individual has two such loci, then p can be
designated as the frequency of the A allele and q as the frequency of the a allele. Thus, in a population
of 100 individuals (each with two loci) in which 40% of the alleles are A, p would be 0.40. The rest of
the alleles (60%) would be a, and q would equal 0.60. Hardy and Weinberg also argued that if five
conditions are met, the population’s allele and genotype frequencies will remain constant from
generation to generation. These conditions are as follows.
1. The population is very large. The effects of chance on changes in allele frequencies is thereby greatly
reduced.
2. Individuals show no mating preference for A or a, i.e., mating is random.
3. There is no mutation of alleles.
4. No differential migration occurs (no immigration or emigration).
5. All genotypes have an equal chance of surviving and reproducing, i.e., there is no selection.
Basically, the Hardy-Weinberg equation describes the status quo. If the five conditions are met, then no
change will occur in either allele or genotype frequencies in the population. Of what value is such a rule?
It provides a yardstick by which changes in allele frequency, and therefore evolution, can be measured.
One can look at a population and ask: Is evolution occurring with respect to a particular gene locus?
The purpose of this laboratory is to examine the conditions necessary for maintaining the status quo and
see how selection changes allele frequency. This will be done by simulating the evolutionary process
using your class as a sample population.
45
AP Lab 8: Population Genetics and Evolution
PROCEDURE
Exercise 8A: Estimating Allele Frequencies for a Specific Trait within a Sample Population
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 PTC is evidence of
the presence of a dominant allele in either the homozygous condition (AA) or the heterozygous condition
(Aa). The inability to taste the chemical at all depends on the presence of homozygous recessive alleles
(aa). (Instead of PTC tasting, other traits, such as tongue-rolling, may be used.) To estimate the
frequency of the PTC-tasting allele in the population, one must find p. To find p, one must first
determine q (the frequency of the nontasting PTC allele), because only the genotype of the homozygous
recessive individuals is known for sure (i.e., those that show the dominant trait could be AA or Aa).
1. Using the PTC taste test papers provided, tear off a short strip and press it to your tongue tip. PTC
tasters will sense a bitter taste. For the purposes of this exercise these individuals are considered to
be tasters.
2. A decimal number representing the frequency of tasters (p2 + 2pq) should be calculated by dividing
the number of tasters in the class by the total number of students in the class. A decimal number
representing the frequency of nontasters (q2) can be obtained by dividing the number of nontasters
by the total number of students. You should then record these numbers in Table 8.1.
3. Use the Hardy-Weinberg equation to determine the frequencies (p and q) of the two alleles. The
frequency q can be calculated by taking the square root of q2. Once q has been determined, p can
be determined because 1 - q = p. Record these values in Table 8.1.
Table 8.1: Phenotypic Proportions of Tasters and Nontasters and Frequencies of the Determining
Alleles
Phenotypes
Class
Population
North
American
Population
% Tasters
(p2 + 2pq)
% Nontasters
(q2)
0.55
0.45
Allele Frequency Based
on the H-W Equation
p
q
Exercise 8B: A Test of an Ideal Hardy-Weinberg Population – NO SELECTION
It will be assumed that the class is initially a population of randomly mating heterozygous individuals
each of genotype Aa. Therefore, the initial allele frequency is 0.5 for the dominant allele A and for
the recessive allele a, that is p = 0.5 and q = 0.5.
Each member of the class will receive four cards. Two cards will have the letter A printed on them
and the other two the letter a. A represents the dominant allele and a is a recessive allele. The four
cards represent the products of meiosis. Each student should work with a partner.
Each student (parent) contributes a haploid set of chromosomes in the next generation. So for each
trait, each partner donates one allele for each offspring.
46
AP Lab 8: Population Genetics and Evolution
1. Turn the four cards over so the letters are not showing, shuffle them, and take the card on top to
contribute to the production of the first offspring. Your partner should do the same. Put the two
cards together. You are now the proud parents of the first offspring. One of you should record
the genotype of this offspring in the Generation 1 row of Table 8.2. Each student pair must
produce two offspring, so all four cards must be reshuffled and the process repeated to produce a
second offspring.
2. Your partner should then record the genotype of the second offspring in his or her Table 8.2. The
very short reproductive career of this generation is over. You and your partner now become the
next generation by assuming the genotypes of the two offspring. That is, student 1 assumes the
genotype of the first offspring and student 2 asumes the genotype of the second offspring.
3. Each student should obtain, if necessary, new cards representing the alleles in his or her
respective gametes after the process of meiosis. Each student should randomly seek out another
person with whom to mate in order to produce the offspring of the next generation. The sex of
your mate does not matter, nor does the genotype. You should follow the same mating process as
for the first generation, being sure to record your new genotype after each generation in the table.
Class data should be collected after each generation for five generations. At the end of each
generation, remember to record the genotype that you have assumed. Your teacher will collect
class data after each generation by asking you to raise your hand to report your genotype. Record
these data in Table 8.2.
Table 8.2: Data
Generations
Offspring’s Genotype
(AA, Aa, or aa)
Class Totals For Each Genotype
AA
Aa
aa
Total
1
2
3
4
5
47
AP Lab 8: Population Genetics and Evolution
Exercise 8B: Analysis of Results
1.
Genotype Frequency: From the genotypes recorded for Generation 5 in Table 8.2, genotype
frequencies for the class should be computed using the following equations:
Frequency of AA =
Total AA
Total AA + Total Aa + Total aa
Frequency of Aa =
Total Aa
Total AA + Total Aa + Total aa
Frequency of aa =
Total aa
Total AA + Total Aa + Total aa
What are the theoretical genotype frequencies of the beginning population where p= 0.5 and q= 0.5?
Record below.
P2 (AA)_______
2pq (Aa )_______
q2 (aa )_______
What are genotype frequencies after five generations of mating? Calculate from the class data and
record below.
P2 (AA)_______
2.
2pq (Aa)________
q2(aa)________
The allele frequencies, p and q, should be calculated for the population after five generations of the
random mating.
Number of A alleles present at the fifth generation
Number of offspring with genotype AA _____x (2) = ______ A alleles
Number of offspring with genotype Aa _____x (1) = ______ A alleles
Total = ______ A alleles
p=
Total number of A alleles = _______
Total number of alleles in the population
Number of a alleles present at the fifth generation
Number of offspring with genotype aa _____x (2) = ______ a alleles
Number of offspring with genotype Aa _____x (1) = ______ a alleles
Total = ______ a alleles
q=
Total number of a alleles = _______
Total number of alleles in the population
Exercise 8B: A Test of Ideal Hardy-Weinberg Equilibrium
p
Original Population
q
Fifth Generation
48
AP Lab 8: Population Genetics and Evolution
Exercise 8C: Selection
In humans, several genetic diseases have been well characterized. Some of these diseases are
controlled by a single allele where the homozygous recessive gentotype has a high probability of not
reaching reproductive maturity, but both the homozygous dominant and the heterozygous individual
survive. At this point, both the homozygous dominant (AA) individuals and the heterozygous individuals
(Aa) will be considered to be phenotypically identical. The selection will be made against the
homozygous recessive individuals (aa) 100% of the time.
Procedure
1. The procedure is similar to that in the preceding exercise. You should start with the initial
genotype of Aa and, with another student, determine the genotypes of your two offspring. This
time, however, there is one important difference. Every time an offspring is aa, it dies. To
maintain the constant population size, the two parents of an aa ofspring must try again until
they produce two surviving offspring (AA or Aa).
2. The process should proceed through five generations with selection against the homozygous
recessive offspring. It is important to remember that an aa individual cannot mate. After each
generation, you should record the genotype that you assume. Also, class data should be
collected after each generation and recorded in Table 8.3.
Data Table 8.3
Generation
Offspring’s
Genotype
(AA or Aa)
Class Totals
AA
Aa
Total
1
2
3
4
5
Exercise 8C: Analysis of Results
Allele Frequency: The allele frequencies, p and q, should be calculated for the population after five
generations of random mating.
Number of A alleles present at the fifth generation
Number of offspring of AA ______ x (2) = _______ A alleles
Number of offspring of Aa __ ___ x (1) = _______ A alleles
Total = _______ A alleles
p = ________
49
AP Lab 8: Population Genetics and Evolution
Number of a alleles present at the fifth generation
Number of offspring of aa ______ x (2) = _______ a alleles
Number of offspring of Aa ______x (1) = _______ a alleles
Total = _______ a alleles
q = ________
Exercise 8C: Selection
p
q
Original Population
Fifth Generation
Exercise 8D: Heterozygote Advantage
From exercise 8C, it is easy to see what happens to the lethal recessive allele in the population.
However, data from many human populations show an unexpectedly high frequency of the sickle-cell
allele (a) in some populations. Thus our simulation does not accurately reflect the real situation; this is
because individuals who are heterozygous are slightly more resistant to a deadly form of malaria than
homozygous dominant individuals. In other words, there is a slight selection against homozygous
dominant individuals as compared to heterozygotes. This fact is easily incorporated into our simulation.
Procedure
1. In this round, keep everything the same as it was in Exercise 8C, except that if your offspring is
AA, flip a coin. If the coin lands heads-up, the individual does not survive; if it lands tails-up,
the individual does survive.
2. Simulate five generations, starting again with the initial genotype from Exercise 8B. The
genotype aa never survives, and the genotype AA only survives if the coin toss comes up tails.
Since we want to maintain a constant population size , the same two parents must try again until
they produce two surviving offspring. Get new “allele” cards from the pool as needed. Total the
class genotypes and calculate the values of p and q.
3. Starting with the F5 genotype, go through five more generations and again total the class
genotypes and calculate the values of p and q.
Data Table 8.4
Generation
Offspring’s
Genotype
(AA or Aa)
Class Totals
AA
Aa
Total
1
2
3
4
5
50
AP Lab 8: Population Genetics and Evolution
Exercise 8D: Analysis of Results (Part 1)
Allele Frequency: The allele frequencies, p and q, should be calculated for the population after five
generations of random mating.
Number of A alleles present at the generation
Number of offspring of AA ______ x (2) = _______ A alleles
Number of offspring of Aa __ ___ x (1) = _______ A alleles
Total = _______ A alleles
p = ________
Number of a alleles present at the fifth generation
Number of offspring of aa ______ x (2) = _______ a alleles
Number of offspring of Aa ______x (1) = _______ a alleles
Total = _______ a alleles
q = ________
Exercise 8D: Heterozygote Advantage (Generations 1 – 5)
p
Original Population
q
Fifth Generation
Data Table 8.5
Generation
Offspring’s
Genotype
(AA or Aa)
Class Totals
AA
Aa
Total
6
7
8
9
10
51
AP Lab 8: Population Genetics and Evolution
Exercise 8D: Analysis of Results (Part 2)
Allele Frequency: The allele frequencies, p and q, should be calculated for the population after ten
generations of random mating.
Number of A alleles present at the 10th generation
Number of offspring of AA ______ x (2) = _______ A alleles
Number of offspring of Aa __ ___ x (1) = _______ A alleles
Total = _______ A alleles
p = ________
Number of a alleles present at the 10th generation
Number of offspring of aa ______ x (2) = _______ a alleles
Number of offspring of Aa ______x (1) = _______ a alleles
Total = _______ a alleles
q = ________
Exercise 8D: Heterozygote Advantage (Generations 6 – 10)
p
Original Population
(Generation 1)
Tenth Generation
q
ANALYSIS QUESTIONS
Case 8A: Allele Frequency
1. What percentage are heterozygotes? Show your work
2. What percentage are homozygotes? Show your work
Case 8B: A Test of Ideal Hardy-Weinberg Equilibrium
1. What does the Hardy-Weinberg equation predict for the new p and q?
2. Do the results you obtained in this simulation agree? If not, why?
3. What major assumption(s) were NOT followed in this simulation?
52
AP Lab 8: Population Genetics and Evolution
Case 8C: Selection
4. How do the new frequencies (after 5 generations) of p and q compare to the initial frequencies?
5. What major assumption(s) were NOT followed in this simulation?
6. Predict what would happen to the frequencies of p and qif you simulated another 5
generations.
7. In a large population, would it be possible to completely eliminate a deleterious recessive
allele? Why or why not?
Case 8D: Heterozygote Advantage
8. Explain how (and why) the changes in p and q frequencies in Case 8C compare with both Case
8B and Case 8D.
9. What is the importance of heterozygotes in maintaining genetic variation in populations?
53
AP Lab 8: Population Genetics and Evolution
HARDY-WEINBERG PROBLEMS
1. In Drosophila, the allele for normal-length wings is dominant over the allele for vestigial wings
(vestigial wings are stubby little curls that cannot be used for flight). In a population of 1,000
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?
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 1,000 individuals, 510 show the dominant phenotype. How many
individuals would you expect of each of the possible three genotypes for this trait?
4. In the United States 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 U.S. is
2,000, 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 1,000 newborn babies, how many would you
expect for each of the three possible phenotypes?
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?
54
Enzyme Action: Testing Catalase Activity
Enzymes are globular proteins, responsible for most of the chemical activities of living organisms.
Enzymes act as catalysts, substances that speed up chemical reactions without being destroyed or
altered during the process. One enzyme may catalyze thousands of reactions every second and can be
used over and over again. Temperature, pH, and concentration can effect how enzymes perform. Most
organisms have a preferred temperature and pH range that they perform best. Enzymes have a specific
shape. When they are altered their shape (also called conformation) changes and makes them
unusable. For example, If the environment of the enzyme is too acidic, or too basic, the enzyme may
irreversibly denature, or unravel, until it no longer has the shape necessary for proper functioning.
Hydrogen Peroxide (H2O2) is toxic to most living organisms. Many organisms are capable of
enzymatically destroying the H2O2 before it can do much damage. H2O2 can be converted to oxygen
and water, in the following reaction:
2 H2 O 2
2 H2 O + O 2
This reaction will occurs spontaneously, but adding an enzymes increases the rate at which the reaction
will occur. At least two different enzymes are known to catalyze this reaction: catalase, found in
animals and protists, and peroxidase, found in plants. The rate of a chemical reaction may be studied in
a number of ways including:
• measuring the rate of appearance of a product (in this case, O2, which is given off as a gas)
• measuring the rate of disappearance of substrate (in this case, H2O2)
• measuring the pressure of the product as it appears (in this case, O2).
In this experiment, you will measure the rate of enzyme activity under various conditions, such as
different enzyme concentrations, pH values, and temperatures. It is possible to measure the
concentration of oxygen gas formed as H2O2 is destroyed using an O2 Gas Sensor. At the start of the
reaction, there is no product, and the concentration is the same as the atmosphere. After a short time,
oxygen accumulates at a rather constant rate. The slope of the curve at this initial time is constant and
is called the initial rate. As the peroxide is destroyed, less of it is available to react and the O2 is
produced at lower rates. When no more peroxide is left, O2 is no longer produced.
OBJECTIVES
In this experiment, you will
• Use an Oxygen Gas Sensor to measure the production of oxygen gas as hydrogen peroxide is
destroyed by the enzyme catalase or peroxidase at various enzyme concentrations.
• Measure and compare the initial rates of reaction for this enzyme when different concentrations
of enzyme react with H2O2.
• Measure the production of oxygen gas as hydrogen peroxide is destroyed by the enzyme catalase
or peroxidase at various temperatures.
• Measure and compare the initial rates of reaction for the enzyme at each temperature.
• Measure the production of oxygen gas as hydrogen peroxide is destroyed by the enzyme catalase
or peroxidase at various pH values.
• Measure and compare the initial rates of reaction for the enzyme at each pH value.
• To maximize time, we will do two variables (as assigned on your card) simultaneously since you
have access to the Lab Quest and 2 Oxygen probes.
Adapted from Advanced Biology with Vernier
55
Figure 1
Materials
LabQuest Interface
2 Vernier O2 Gas Sensors
Water baths ( hot, cold, room)
Disposable pipets
Distilled water (50mL)
3.0% H2O2 (30mL)
Thermometer (Vernier temp probe)
2- 250 mL Nalgene bottles per group
3 buffers (7mL) (buffers 4,7,10)
*Nalgene bottle goes directly in the bath
5 ml syringe (qty 4) for H2O2 and Buffers
Procedure
1. Obtain and wear goggles.
2. Connect the O2 Gas Sensor to LabQuest and choose New from the File menu. If you have an older
sensor that does not auto-ID, manually set up the sensor.
3. On the Meter screen, tap Rate. Change the data-collection rate to 0.2 samples/second and the datacollection length to 180 seconds.
4. You will be using a Nalgene Bottle for all three parts.
Part I: Testing the Effect of Enzyme Concentration
*****You will be running two levels At the same time in TWO different Nalgene bottles*****
5. Fill each Nalgene bottle with 5 mL of 3.0% H2O2 and 5 mL of water.
6. Add the number of enzyme suspension indicated to each bottle. For example: If your card says
to add 5 drops and 20 drops. Add 5 drops to one Nalgene Bottle AND then add 20 drops to the
other Nalgene Bottle.
7. Place the sensor in the Nalgene bottle and wait 30 seconds.
8. Select the Record key on your LabQuest to begin taking measurements. While the measurements
are being taken, you can toggle between the screens by selecting the icons at the top of the screen.
9. Record your measurements in Table 2
10. When data collection is complete, a graph of O2 gas vs. time will be displayed. Remove the O2 Gas
Sensor from the Nalgene bottle. Rinse the bottle with water and dry with a paper towel.
11. Store the data from the first run by tapping the File Cabinet icon.
56
Part II Testing the Effect of Temperature
Your teacher will assign a temperature range for your lab group to test. Depending on your assigned
temperature range, set up your water bath as described below. Place a thermometer in your water bath
to assist in maintaining the proper temperature throughout the experiment.
•
•
•
0–5°C:
600 mL beaker filled with ice and water.
20–25°C: No water bath needed to maintain room temperature.
50–55°C: 600 mL beaker filled hot water.
1. Fill the Nalgene bottle with 5 mL of 3.0% H2O2 and 5 mL of water then place the Nalgene bottle in
the appropriate water bath (look at the two assigned values on your card in the water bath. The
Nalgene bottle should be in the water bath for 5 minutes before proceeding to Step 2. Record the
temperature of the water bath, as indicated on the thermometer, in the space provided in Table 3.
2. Add 10 drops of enzyme solution to the Nalgene bottle.
3. Calculate the average rate for the three trials you tested. Record the average in Table 3.
4. Record the average rate and the temperature of your water bath from Table 3 on the class
chalkboard. When the entire class has reported their data on the chalkboard, record the class data in
Table 5.
Part III Testing the Effect of pH
*For each buffer, you will perform each of the steps below. Clear all runs from your LabQuest before
you begin.
1. Add 5 mL of 3% H2O2 and 5 mL of pH 3 buffer to a Nalgene bottle
2. Tap Table. Choose Clear All Data from the Table menu.
3. Tap Graph to display the graph.
4. Add 10 drops of enzyme solution to the Nalgene bottle.
5. Place the Oxygen sensor in the top of the Nalgene bottle.
6. Record your measurements in table 4
7. Graph all three runs of data on a single graph.
a. Tap Run 3 and select All Runs. All three runs will now be displayed on the same graph axes.
b. Use the displayed graph and the data in Table 5 to answer the questions for Part III.
57
DATA
Table 2: Part I Effect of Enzyme Concentration
Table 2
Bottle
Slope (rate= % / s)
Trial 1
Trial 2
Trial 3
Trial 4
Mean Rate
5 drops
10 drops
20 drops
Table 3: Part II Effect of Temperature
Table 3
Slope (rate= % / s)
Temperature
°C
Trial 1
Trial 2
Trial 3
Trial 4
Mean Rate
cold
room temp
hot
Table 4: Part III Effect of pH
Table 4
pH
Slope (rate= % / s)
Trial 1
Trial 2
Trial 3
Trial 4
Slope
Mean Rate
(rate= % / s)
pH 4
pH 7
pH 10
Table 5: Class Data
pH
G1
G2
G3
G4
G5
G6
G7
G8
Avg.
Slope
Mean Rate
(rate= % / s)
pH 4
pH 7
pH 10
Cold
Room
Temp
5 drops
10 drops
20 drops
58
PROCESSING THE DATA
1. For Part II of this experiment, make a graph of the rate of enzyme activity vs. temperature by
hand. Plot the rate values for the class data in Table 3 on the y-axis and the temperature on the
x-axis. Use this graph to answer the questions for Part II.
Analysis
Part I Effect of Enzyme Concentration
1. How does changing the concentration of enzyme affect the rate of decomposition of H2O2?
2. If one increases the concentration of enzyme to thirty drops, what do you think will happen to the
rate of reaction? Predict what the rate would be for 30 drops.
3. Are there other factors in the experiment that may prevent the rate from reaching the previously
predicted rate from # 2? Explain your answer.
Part II Effect of Temperature
4. At what temperature is the rate of enzyme activity the highest? Lowest? Explain.
5. How does changing the temperature affect the rate of enzyme activity? Does this follow a pattern
you anticipated?
6. Why might the enzyme activity decrease at very high temperatures?
59
Part III Effect of pH
7. At what pH is the rate of enzyme activity the highest? Lowest?
8. How does changing the pH affect the rate of the enzyme activity?
9. Choose one variable and perform a linear regression to calculate the rate of reaction. What does a
linear regression show? Explain
a. Choose Curve Fit from the Analyze menu.
b. Select Linear for the Fit Equation. The linear-regression statistics for these two data columns
are displayed for the equation in the form
y = mx + b
c. Enter the absolute value of the slope, m, as the reaction rate in Table 2.
d. Select OK.
10. Graph all three of your runs of data and the class average on a single graph for each variable. (You
will have 3 graphs).
60
61
62
63
64
Diffusion and Osmosis Across Membranes
BACKGROUND
In nature, things tend to move from areas of high concentration to low concentration. To survive, all
organisms must exchange molecules (gases, nutrients, wastes, and water) into and out of their cells.
Diffusion is a process by which ions or molecules move through a medium from high to low
concentrations. For instance, oxygen from the atmosphere and photosynthetic organisms diffuses
through pond water for use by fish and other aquatic animals. When animals use oxygen, more oxygen
will diffuse from the neighboring environment to replace it. Waste products released by these animals
are diluted by diffusion and dispersed throughout the pond.
At the cellular level, ions or molecules diffuse through the intracellular and extracellular fluids to cross
the cell membrane.
The cell membrane is said to be selectively permeable (semi-permeable),
because it allows some molecules to move across but not others. Due to their
small size, water molecules and certain gases like CO2 and O2 can move
freely back and forth across the membrane. However, larger molecules
(proteins, sugars, DNA, RNA) and charged particles (ions) are not allowed to
pass through the membrane without the help of transport proteins. On a
larger scale, think of using strainer to sift a large cup of wet beach sand.
Certain particles will fit through the holes (water, small sand particles) and
other larger particles won’t (pebbles, shells).
A concentration gradient exists when concentrations inside and outside the cell are not equal. When a
concentration gradient exists, diffusion continues until the molecules are equally distributed.
There are two ways that particles move across the cell membrane: passive and active transport.
Passive transport (diffusion) occurs spontaneously and requires no energy to move molecules from
high to low concentrations. In passive transport, no energy is needed because the particles are small
enough to pass through the membrane. When determining what will be moving across the membrane,
we have to think about the particle’s size and charge. If it’s too big or charged, it is more difficult to
go through. Depending on the composition of a solution, water may be the only thing small enough to
move across the membrane.
Osmosis is a special form of diffusion that involves the movement of water across a membrane.
Facilitated diffusion describes situations where transport channel proteins are used to help
molecules that are too large to move freely through the membrane. In the case of larger
molecules (protein, starch, DNA, RNA) transport proteins act as gateways to help particles pass
from high to low concentrations.
Active transport uses protein pumps to push molecules against the concentration gradient (from low
to high concentrations). This requires the cell to use chemical energy to move substances across the
cell membrane. For example, neurons use sodium-potassium pumps to transmit nerve impulses.
The rate of diffusion can be influenced by the temperature of the environment, pressure, and
concentration gradient of molecules (solutes).
65
OBJECTIVES
In this experiment, you will
• Determine if the diffusion rate for a molecule is affected by the presence of a second molecule.
• Study the effect of concentration gradients on the rate of diffusion.
• Determine the molar concentration of carrot or potato cells.
• Determine the water potential of carrot or potato cells.
• Observe the effect of hypotonic and hypertonic solutions on the cell membrane of elodea or onion cells.
PART I: DIFFUSION
MATERIALS
plastic cups
distilled water
15% glucose/1% starch
Small funnel (optional)
Dialysis tubing
Lugol’s solution
Glucose test strips
balance
PROCEDURE
1.
2.
3.
4.
5.
6.
7.
8.
9.
Wear your goggles
Obtain the 15 mL of a 15% glucose/1%starch solution
Test the solution for the presence of glucose and starch. Record in Table 1.
Obtain a piece of dialysis tubing (25-30cm long). Tie one end off in a knot or with string.
Using your fingers, carefully open the tubing and pour 15ml of the solution into the open end of
the dialysis tubing using a funnel (be sure not to get any solution on the outside of the tubing).
Tie off the tubing (be sure to leave room in the bag for the volume of liquid to expand).
Fill the plastic cup (or 250 ml beaker) 2/3 of way with distilled water. Add 4ml of Lugol’s solution
to the water. Record the color of the solution in Table 1.
Immerse the tubing (tied on both ends) in the cup (dialysis bag should be completely submerged)
Let stand for 30 minutes. Record the colors of the water and the bag at the end of 30 minutes.
PART II: OSMOSIS
BACKGROUND
Isotonic, Hypertonic, and Hypotonic are terms used when predicting the net movement of water by
comparing concentrations of solutions inside a cell and in its environment. A more precise measurement
of the net flow of water can be calculated using the concept of water potential. Think of water potential
like potential energy, moving from high to low (outlined in Part IV). For the following examples, keep in
mind that salt is an ionic compound that can’t move across the membrane. Therefore water will be the
substance (molecule) moving across the membrane from an area of high to low concentration of water.
A salt water fish living in its normal habitat, the ocean, is in an isotonic solution or environment. This is
because the concentration of salt (and other solutes) and water are the same inside and outside of the
organism’s cells (no concentration gradient). At equilibrium there is no net movement of water across the
membrane because the water potential is equal on both sides.
Placing the salt water fish in a freshwater environment creates a concentration gradient because the fish
has more dissolved solutes than the environment. The salt water fish is in a hypotonic solution because
the water potential is greater outside the cell than inside. Because the water has less dissolved particles
(higher water potential) than the fish’s cells, water moves into the cells, from an area of high to a low
66
Diffusion Through Membranes
water potential. As more water enters the cells, cytolysis occurs, eventually causing them to swell and
rupture. Like a balloon, if too much water enters, the membrane can burst, killing the cell. Some cells,
paramecia, prevent too much water from entering, by pumping excess water out via the contractile
vacuole. Fungi, plants, and bacteria have a cell wall that will not allow the cell to fill beyond capacity.
Conversely, placing a freshwater fish in a salt water environment, hypertonic solution, creates a gradient
because the environment has more dissolved particles (lower water potential) than the fish’s cells.
Plasmolysis, cell shrinkage, occurs when the concentration of water inside the cell is greater than outside,
more water moves out than in. If a cell becomes too dehydrated, it may not be able to survive.
In this investigation, you will study the effect of concentration gradients on the rate of diffusion.
MATERIALS
plastic cups/beaker
distilled water
string
5 lb bag sugar
Small funnel (optional)
Dialysis tubing (6 pieces)
balance
PROCEDURE
1. Obtain 6 pieces of dialysis tubing (approximately 20-25cm long)
2. Time off one end of each piece of tubing to make 6 bags
3. Using a funnel, pour 15-20 mL of each solution into its own dialysis bag.
(distilled water, sucrose solutions 0.2M, 0.4M, 0.6M, 0.8M, and 1.0M)
4. Remove most of the air from the bag and tie off the end (leave space for expansion of the contents
in the bag)
5. Rinse each bag with distilled water and pat dry.
6. Mass each bag and record on Table 2.
7. Put each bag in a separate cup. Fill 2/3 full with distilled water. Bag must be submerged.
8. Let stand 30 minutes
9. At the end of 30 minutes, remove the bags, pat dry, mass them, and record in Table 2.
10. Add your lab group’s data to the class data chart.
11. Graph the results for Table 2 and Table 3.
PART III: WATER POTENTIAL
BACKGROUND
Through the collaborative work of scientists, instead of predicting, we can quantitatively measure the
movement of water by determining the water potential.
Water potential is a term used when predicting the movement of water into or out of cells. 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, then, measures the
tendency of water to leave one place in favor of another place. The symbol for water potential is the
Greek letter Psi, Ψ. Water potential is impacted by two physical factors:
1. pressure component called pressure potential Ψp, which increases water potential.
2. the effects of solutes called solute potential, Ψs, which reduces water potential.
Ψ
=
Ψp
+
ΨS
Water potential
=
Pressure potential
+
Solute potential
67
There is a numerical relationship between water potential and its components (pressure potential and
solute potential). The water potential value can be positive, zero, or negative. Water can move freely
across the membrane, but will always move in the direction of the lower water potential. Water potential
depends on pressure and solutes.
Pressure Potential
Increase = more positive Ψ value
Decrease (tension/pulling) = more negative Ψ value, as in the case of dead xylem cells
Solute Potential
Is always negative
Distilled water in an open beaker (at atmospheric pressure) has a water potential of zero (Ψ= 0). A water
potential value can be positive, negative, or zero. Water potential is usually measured in bars, which is a
metric unit of pressure. One bar is approximately equal to 1 atmosphere. Another measure of pressure is
the megapascal (MPa). (1 MPa = 10 bars, 1 kPa = 0.1 bar)
Water movement is directly proportional to the pressure on a system. For example, pressing on the
plunger of a water- filled syringe causes the water to exit via any opening. In plant cells this physical
pressure can be exerted by the cell pressing against the partially elastic cell wall.
It is important to realize that water potential and solute concentration are inversely related. More
dissolved particles would yield a lower water potential (less water per volume). Less dissolved particles
or solute equates to a higher water potential (more water per volume). In other words, the solute potential
is the effect that solutes have on a solution's overall water potential.
The following figures (potato tissue in distilled water) will help to illustrate the relationship between water
potential and its components.
68
Diffusion Through Membranes
Due to the tissue’s higher solute potential (Ψs), the
water potential is lower inside the cells. Water will
move (by osmosis) from the surrounding water
where water potential is higher, into the cell where
water potential is lower (more negative). In Figure
a, the pure water potential (Ψ= 0) and the solute
potential (Ψs = -3). For this example, assume that
the solute will not diffuse out of the cell.
By the end of the observation, the movement of
water into the cell causes the cell to swell and the
cell contents push against the cell wall to produce
an increase in pressure potential (turgor) (Ψp = 3).
Eventually, enough turgor pressure builds up to
balance the negative solute potential of the cell.
When the water potential of the cell equals the
water potential of the pure water outside the cell
(Ψof cell = Ψof pure water = 0), a dynamic
equilibrium is reached and there will be no NET
water movement.
In this experiment, you will measure the percent change in mass of carrot cells after they have soaked in
various concentrations of sugar solutions for a 24 hour period. You will use this data to calculate the the
molar concentration and water potential of carrot or potato cells.
MATERIALS
Computer
Logger Pro
4 baby carrot slices
250mL beaker
plastic wrap
balance
paper towels
assigned sucrose solutions
(0, 0.2, 0.4, 0.6, 0.8, or 1M)
PROCEDURE
You will be assigned one or more of the sugar solutions in which to soak your carrots.
1. Pour 100 mL of the assigned sugar solution into a 250 mL beaker.
2. Measure and record the mass of the four carrot slices together.
3. Put the four carrot slices into the beaker of sugar solution.
4. Cover the beaker with plastic wrap and allow it to stand for a 24 hour period.
5. Remove the carrots from the beaker, blot with a paper towel, and determine the mass of the four
carrots together after soaking.
6. Calculate the percent change in mass and record data for the sugar concentration tested in Table 4 as
well as on the class data sheet Table 5.
7. Repeat Steps 1–6 for any additional assigned sugar solutions.
69
PROCESSING THE DATA
1. Plot the percent change in mass on the y-axis and concentration of sugar on the x-axis.
2. The point at which the line of best fit crosses the x-axis represents the molar concentration of
sucrose with a water potential that is equal to the carrot tissue water potential. At this concentration,
there is no net gain or loss of water from the tissue.
3. Perform a linear regression to determine the molar concentration of sugar solution at which the
mass of the carrot cells does not change.
a. Move the mouse pointer to the first data point. Press the left mouse button. Drag the pointer
to the last data point.
b. Select the Regression option from the Analyze menu. A floating box will appear with the
formula for a best-fit line.
c. Choose Interpolate from the Analyze menu. Move the mouse pointer along the regression
line to the point where the line crosses the x-axis. This point represents the molar
concentration of sugar with a water potential equal to the carrot cells water potential.
Meaning, there is no net movement of water in or out. Record this concentration in Table 6.
4. Use the sugar molar concentration from the previous step to calculate the solute potential of the sugar
solution.
a. Calculate the solute potential of the sugar solution using the equation
Ψs = - iCRT
where i
C
R
T
=
=
=
=
ionization constant (1.0 for sugar since it doesn’t ionize in water)
sugar molar concentration (determined from the graph)
pressure constant (0.0831 liter bars/mol-K OR 8.314 kPa L/mol K)
temperature (295 K)
b. Record this value with units in Table 6.
5. Calculate the water potential of the solution using the equation
Ψ =
Ψp
+
Ψs
The pressure potential of the solution in this case is zero because the solution is at equilibrium. Record the
water potential value with units in Table 6.
70
Diffusion Through Membranes
PART IV PLASMOLYSIS
In this investigation you will observe the effect of hypotonic and hypertonic solutions on the cell membrane of
elodea or onion cells.
MATERIALS
paper towels
eye dropper
blank slides
cover slip
15% NaCl solution
elodea or onion
microscope
PROCEDURE
1. Make a wet mount slide of the elodea or onion epidermis. Observe and sketch at 100x magnification
2. Add 2 to 3 drops of 15% NACl to the edge of the coverslip. On the opposite side of the coverslip
where you added the salt, use a piece paper towel to draw the fluid across the plant. Observe and
sketch at 100x magnification.
3. Remove the coverslip and flood the elodea or onion with water. Observe and sketch at 100x
magnification.
EXTENSIONS
1. Make a plot of the rate of diffusion vs. the salt concentration. Using your plot, estimate the rate of
diffusion of a 3% salt solution.
2. If the results of the experiments in Part I can be extrapolated to diffusion in living systems, how would
a single-celled organism respond in an oxygen rich pond compared to an oxygen-poor pond? Explain.
3. Design an experiment to determine the effect of temperature on the diffusion of salt. Perform the
experiment you designed.
4. Ectotherms are organisms whose body temperatures vary with the surrounding environment. On the
basis of your data from Extension Question 3, how do you expect the oxygen consumption of
ectotherms to vary as the temperature varies? Explain.
5. If waste products of an aquatic single-celled organism were released into a pond, how would that affect
the organism’s ability to obtain oxygen from the pond water? Explain how your data from Part II
supports your answer.
71
DATA
Table 1: Diffusion Data with 15% glucose/1% Starch solution
Diffusion
Data
Table
Dialysis
Tubing
Initial Contents
Solution Color
Initial
Final
Presence of Glucose (+/-)
Initial
Final
Presence of Starch (+/-)
Initial
Final
Beaker
Table 2: Lab Group Data on Osmosis through Dialysis Bag
Dialysis Bag
Contents
Initial Mass
(g)
Mass
Difference
(g)
Final Mass
(g)
Percent
Change in
Mass
0.0 M Distilled H2O
0.2 M Sucrose
0.4 M Sucrose
0.6 M Sucrose
0.8 M Sucrose
1.0 M Sucrose
Table 3: Class Data on Osmosis through Dialysis Bag
Dialysis
Bag
Contents
Percent Change in Mass
Group
1
Group
2
Group
3
Group
4
Group
5
Group
6
Group
7
Group
8
TOTAL
Class
Avg
0.0 M Distilled
H 2O
0.2 M Sucrose
0.4 M Sucrose
0.6 M Sucrose
0.8 M Sucrose
1.0 M Sucrose
72
Diffusion Through Membranes
Table 4: Lab Data for Carrot Cells
Sugar solution
concentration
Initial mass
(g)
Final mass
(g)
Mass difference
(g)
Percent change
in mass
0.00 M
0.2 M
0.4 M
0.6 M
0.8 M
1.0 M
Table 5: Class Data for Carrot Cells (% change)
Sugar solution Group
1
concentration
Group
2
Group
3
Group
4
Group
5
Group
6
Group
7
Group
8
Total
Class
average
0.0 M
0.2 M
0.4 M
0.6 M
0.8 M
1.0 M
Table 6: Group Data for Water Potential
Sugar molar concentration in carrot
=
Solute potential of solution
=
Water potential of solution
=
73
PRE-LAB QUESTIONS (PARTS I &II)
1. What makes dialysis tubing an ideal model for the plasma membrane?
2. Predict the impact of temperature, concentration gradient, and pressure on the rate of diffusion.
3. Explain the role of integral proteins in diffusion.
4. Explain the role of cholesterol in the plasma membrane.
5. Explain how small molecules versus large molecules pass through the plasma membrane.
6. Explain the difference between plant and animal cells response to a hypertonic solution.
7. Explain the difference between plant and animal cells response to a hypotonic solution.
8. Predict how you think the following cells will behave in distilled water.
Animal cell_____________________
fungus cell_____________________
Elodea__________________________
carrot stick____________________
Yeast cell_______________________
blood cell______________________
74
Diffusion Through Membranes
POST LAB ANALYSIS QUESTIONS (PARTS I & II)
1.
Discuss the sizes of the substance molecules relative to each other and the dialysis membrane. How
did you determine their relative sizes?
2.
Which solutions, if any, produced a positive slope? Was water moving in or out of the cell (dialysis
tubing) under these circumstances? Explain.
3.
Which solutions, if any, produced a negative slope? Was water moving in or out of the cell under
these circumstances? Explain.
4.
Does sucrose move in or out of the cell? Explain.
5.
Examine the graph of the rate of pressure change vs. the sucrose concentration. Describe any trends in
the data.
6.
Use the graph to determine (estimate) the concentration of sucrose that would yield no change in
pressure. Why is this biologically significant?
75
7.
When wilted plants are watered, they tend to become rigid. Explain how this might happen.
8.
Explain the strengths and weakness of this dialysis model with respect to animal and plant cells.
9.
Discuss and explain potential reasons to account for variation in class average mass change at specific
sucrose concentrations.
10.
In the winter, people pour salt on roads to breakdown the ice. The runoff often leaves dead plants in
its wake. Explain why plants die when exposed to salt.
11.
Indicator solutions are used to test for various substances. What indicators are used to test for the
presence of the following organic compounds?
Simple carbohydrates _____________________ Starch _____________________________
Proteins________________________________ Lipids_____________________________
76
Diffusion Through Membranes
12. Using the data from the table, answer the questions.
Sample Data:
Contents in Dialysis
bag (20mL)
0.0 M Distilled H20
0.2 M Sucrose
0.4 M Sucrose
0.6 M Sucrose
0.8 M Sucrose
1.0 M Sucrose
Percent Change
in Mass
12.5
3.5
-4.0
-6.0
-14.0
-14.0
a. What is the molar concentration of the carrot cells? Show your work.
b. Identify each of the bags (in the chart above) in the beakers below as isotonic, hypertonic, or
hypotonic.
Contents in
Beaker
Is the water in the Are the carrot cells
beaker Isotonic,
Isotonic,
Hypertonic, or
Hypertonic, or
Hypotonic to the
Hypotonic to the
carrot cells
water in the beaker
0.0 M Dist H20
0.2 M Sucrose
0.4 M Sucrose
0.6 M Sucrose
0.8 M Sucrose
1.0 M Sucrose
13. Predict the effect of increased and decreased initial temperature on rates of mass change for each of
these different solution concentrations.
77
14. Predict the rates of pressure change if the dialysis tubing is placed in an insulated cup holding 1000
mL of 37°C water. Explain your reasoning.
15. Predict the rate of mass change if the dialyisis tubing were filled with 10 mL of 0.8 M sucrose solution
and 5 mL of 1.0 M sodium chloride solution. Explain your reasoning.
78
Diffusion Through Membranes
POST LAB ANALYSIS QUESTIONS (PART III)
16. Describe the trends in your lab group data and compare/contrast to the class’ data. Explain any
differences.
17. Whether you used whole carrots or slices, explain how your data may have changed if you had used
the other type. What would explain the differences?
18. What may happen to an animal cell if water moves into it? How does this differ from what would
happen in a plant cell?
19. What factors affect water potential?
20. If a plant cell has a higher water potential than its surrounding environment and the pressure is equal
to zero, will water move in or out of the cell? Explain why.
21. When you increase the concentration of a solute in a solution, the concentration gradient of water
changes. Does the water increase its water potential or decrease its water potential?
79
80
81
82
83
84
85
86
88
Mitosis
In 1858, Rudolf Virchow discovered that new cells arise only from previously existing cells.
Cells can arise in two ways: mitosis and meiosis. Somatic (body) cells divide by mitosis
followed by cytokinesis. Germ cells produce gametes by the process of meiosis. Plant cells grow
by enlargement, essentially by taking up water. When they reach a certain size, they divide,
forming two identical daughter cells via mitosis. The various parts of the cell are divided in such
a way that the new daughter cell is identical to the parent cell. Mitosis is one part of the cell
cycle.
Figure 1: The Cell Cycle
Mitosis is the division of the nucleus, and is therefore distinct from cell division, in which the
cytoplasm is divided. In prokaryotes, the DNA is replicated before division. In eukaryotes, the
DNA is part of the chromosomes. Division requires a specific method for replication, separation,
and divided between the daughter cells.
Mitosis, or nuclear division, ensures the equal division of the nuclear material between
the daughter cells in eukaryotic organisms. During mitosis the chromosomes condense,
and move to the center of the cell where they fully contract. They then split longitudinally
into two identical halves that appear to be pulled to opposite poles of the cell by a series
of microtubules. In these two genetically identical groups, the coiling of the
chromosomes relaxes again, and they are reconstituted into the nuclei of the two
daughter cells. It is a continuous process that can be divided into four major phases:
prophase, metaphase, anaphase, and telophase.
89
Interphase: The chromatin, if visible at all, can only be seen as small grains or threads.
Interphase is generally considered to be a “resting phase.” However, the cell is replicating the
genetic material, preparing for mitosis.
Mitosis
Prophase: The beginning of mitosis is illustrated by the chromosomes gradually becoming
visible. They start out as elongated threads that shorten and thicken. Chromosomes become more
condensed and undergo spiral contractions, like a thin wire being turned into a coiled spring.
This coiling involves the entire DNA–protein complex. Each chromosome is composed of two
longitudinal halves, called chromatids, joined in a narrow area known as the centromere, where
the chromatids are not coiled. The centromere, located on each chromosome, divides the
chromosomes into two arms of varying lengths. As prophase progresses, the nucleoli grow
smaller and finally disappear. Shortly after, in most cell types, the nuclear envelope breaks down,
putting the contracted chromosomes into direct contact with the cytoplasm; this marks the end of
prophase.
Metaphase: The chromosomes, still doubled, become arranged so that each centromere is on the
equatorial region of the spindle. Each chromosome is attracted to the spindle fibers by its
centromere; often the arms of the chromosome point toward one of the two poles. Some of the
spindle fibers pass from one pole to the other and have no chromosome attached.
Anaphase: The chromatids separate from one another and become individual chromosomes.
First, the centromere divides and the two daughter chromosomes move away from the equator
toward opposite poles. Their centromeres, which are still attached to the spindle fiber, move first,
and the arms drag behind. The two daughter chromosomes pull apart; the tips of the longer arms
separate last. The spindle fibers attached to the chromosomes shorten as the chromatids divide
and the chromosomes separate. The fibers appear to move, but in fact the microtubules are
continuously formed at one end of the spindle fiber and disassembled at the other. In the process,
it appears as if the spindle fibers were tugging the chromosomes toward the poles by their
centromeres. By the end of anaphase, the two identical sets of chromosomes have separated and
moved to opposite poles.
Telophase: The separation is made final; the nuclear envelopes are organized around the two
identical sets of chromosomes. The spindle apparatus disappears. Nucleoli also reform at this
time. The chromosomes become increasingly indistinct, uncoiling to become slender threads
again.
Post Mitosis (Cell Division)
Cytokinesis: As mitosis ends, cytokinesis begins, resulting in the formation of two daughter
cells. The cleaved membrane slowly draws together, forming a narrow bridge, then separates the
cell into two daughter cells. The cells now enter interphase.
In order to investigate the process of mitosis, plant and animal tissues where cells are dividing
rapidly must be examined. In animals, the most rapidly growing and dividing tissues are found in
the embryonic stages of development. Although most animal tissues continue to undergo mitosis
throughout the life cycle of the organism, they do so very slowly when compared to their
embryos. Some animal cells, like most plant tissues, rarely replicate after the organism reaches
maturity.
90
In plants, these tissues are primarily found in the tips of stems and roots. The root tip plants are
exceptionally good places to look for cells undergoing mitosis. Plant root tips consist of several
different zones where various developmental and functional processes of the root are performed.
The primary region for the formation of new cells is the apical meristem. The root cap offers
protection for the rest of the root, the region of elongation is the area where the bulk of cell
growth occurs, and the region of maturation is where tissue differentiation occurs.
OBJECTIVES
In this experiment, you will
• Examine and compare the phases of mitosis in animal and plants cells.
• Determine the relative time cells spend in each phase of mitosis.
• Follow the processes of mitosis in the life cycle of Sordaria.
• Examine the arrangement of Sordaria ascospore microscopically to determine the
frequency of crossing over.
• Calculate the distance, in map units, between a specific gene and the chromosome
centromere.
MATERIALS
whitefish mitosis slide
onion mitosis slide
compound microscope
PROCEDURE
Part A: Observing Mitosis in Plant and Animal Cells
1. Observe the prepared microscope slide of onion root tip mitosis, first at 100X, then 400X.
Using the Plant Mitosis Chart as a guide, identify cells that represent each mitotic phase.
2. In the Analysis section, draw each phase of plant cell mitosis that you see. Write a brief
description of each phase below each drawing.
3. Observe the prepared microscope slide of whitefish blastula. Using the Animal Mitosis Chart
as a guide, identify each phase of animal cell mitosis.
4. In the Analysis section, draw each phase of plant or animal cell mitosis that you see. Write a
brief description of each phase below each drawing.
Part B: Relative Lengths of Phases of Mitosis
5. Examine at least three fields of view of the apical meristem of the onion root tip at 400X. In
each view, count the number of cells in the various stages of mitosis. Record this data in
Table 1.
6. Calculate the total number of cells counted and the percentage of total cells counted for each
stage of mitosis. Record this data in Table 1. Record the percentages in Table 2, as well.
91
7. Assuming that it takes an average of 24 hours (1,440 minutes) for onion root tip cells to
complete the cell cycle, calculate the amount of time cells spent in each phase of the cycle.
Use the formula provided below. Enter your results in Table 1.
Percent of Cells in Phase × 1,440 minutes = _________ minutes cell spent in phase
ANALYSIS
Stage: ___________
Description: ______
_________________
_________________
_________________
_________________
Stage: ___________
Description: ______
_________________
_________________
_________________
_________________
Stage: ___________
Description: ______
_________________
_________________
_________________
_________________
Stage: ___________
Description: ______
_________________
_________________
_________________
_________________
Stage: ___________
Description: ______
_________________
_________________
_________________
_________________
92
Table 1
Stage
# of cells in
Field 1
# of cells in
Field 2
# of cells in
Field 3
Total # of
cells
% of Total
# of cells
Time of Each
stage (min)
Interphase
Prophase
Metaphase
Anaphase
Telophase
Total number of cells counted __________
Table 2:
Table 2
Percentage of cells in each stage of mitosis
Stage
% of Total # of cells
Interphase
Prophase
Metaphase
Anaphase
Telophase
Total
QUESTIONS
1. Referring to the percentage of total cells counted in each phase of mitosis, determine which
phase takes the longest for the cell to complete, and explain why. Sketch a pie graph of the
percentage of cells in each phase to illustrate. Be sure to title your graph and include a key.
2. What is the relationship between the processes of mitosis and cytokinesis?
3. Describe the phase that is different between plant and animal cell mitosis.
4. Explain the difference between chromatin, chromosomes, and chromatid.
93
ANIMAL CELL MITOSIS
Interphase
Prophase
Metaphase
Anaphase
Telophase
Cytokinesis
94
PLANT CELL MITOSIS
Interphase
Prophase
Metaphase
Anaphase
Telophase
Cytokinesis
95
96
Meiosis
Crossing Over during Meiosis in Sordaria
Sordaria fimicola is an ascomycete fungus that can be used to
demonstrate the results of crossing over during meiosis. Sordaria is a haploid
organism for most of its life cycle. It becomes diploid only when the fusion of the
mycelia (very small filaments) of two different strains results in the fusion of the
two different types of haploid nuclei to form a diploid nucleus. The diploid nucleus
must then undergo meiosis to resume its haploid state.
Meiosis, followed by mitosis, in
Sordaria results in the formation of
eight haploid ascospores contained
within a sac called an ascus (plural,
asci). Many asci are contained within a
fruiting body called a perithecium.
When ascospores are mature the
ascus ruptures, releasing the
ascospores. Each ascospore can
develop into a new haploid fungus.
The life cycle of Sordaria fimicola is
shown in Figure 1.
Figure 1: Life Cycle
To observe crossing over in
Sordaria, one must make hybrids
between wild-type and mutant strains
of Sordaria. Wild-type (+) Sordaria
have black ascospores. One mutant
strain has tan spores (tn). When
mycelia of these two different strains
come together and undergo meiosis,
the asci that develop will contain four
black ascospores and four tan
ascospores. The arrangement of the
spores directly reflects whether or not crossing over has occurred. In Figure 2, no
crossing over has occurred. Figure 3 shows the results of crossing over between
the centromere of the chromosome and the gene for ascospore color.
Figure 2:
97
Two homologous chromosomes line up at metaphase I of meiosis. The
two chromatids of one chromosome each carry the gene for tan spore color (tn)
and the two chromatids of the other chromosome carry the gene for wild-type
spore color (+).
The first meiotic division (MI) results in two cells each containing just one
type of spore color gene (either tan or wild-type). Therefore, segregation of
these genes has occurred at the first meiotic division (MI). The second meiotic
division (MII) results in four cells, each with the haploid number of chromosomes
(lN). A mitotic division simply duplicates these cells, resulting in 8 spores. They
are arranged in the 4:4 pattern.
In this example, crossing over has occurred in the region between the
gene for spore color and the centromere. The homologous chromosomes
separate during meiosis I. This time, the MI results in two cells, each containing
both genes (1 tan, 1 wild-type); therefore, the genes for spore color have not yet
segregated. Meiosis II (MII) results in segregation of the two types of genes for
spore color. A mitotic division results in 8 spores arranged in the 2:2:2:2 or 2:4:2
pattern. Any one of these spore arrangements would indicate that crossing over
has occurred
between the
gene for
spore coat
color and the
centromere.
Two strains of Sordaria (wild-type and tan
mutant) have been inoculated on a plate of agar.
Where the mycelia of the two strains meet (Figure 4),
fruiting bodies called perithecia develop. Meiosis occurs
within the perithecia during the formation of asci. A
slide has been prepared of some perithecia (the black
dots in figure 4).
Procedure
Using the 40x objective, view the 10 slides and
locate a group of hybrid asci (those containing both tan and black ascospores).
Count at least 10 hybrid asci on each and enter your data in Table 1.
The frequency of crossing over appears to be governed largely by the
distance between genes, or in this case, between the gene for spore coat color
and the centromere. The probability of a crossover occurring between two
particular genes on the same chromosome (linked genes) increases as the
98
distance between those genes becomes larger. The frequency of crossover,
therefore, appears to be directly proportional to the distance between genes.
A map unit is an arbitrary unit of measure used to describe relative
distances between linked genes. The number of map units between two genes or
between a gene and the centromere is equal to the percentage of
recombinants. Customary units cannot be used because we cannot directly
visualize genes with the light microscope. However, due to the relationship
between distance and crossover frequency, we may use the map unit.
2:2:2:2
Asci Spores
with crossovers
2:4:2
2:2:2:2
4:4
4:4
Here is another picture of spore formation.
99
Pre-Lab Questions
1. What type of fungus is Sodaria fimicola?
2. What are the following? ASCUS, ASCOSPORES, PERITHECIUM
3. Why are there 8 spores in an ascus?
Table 1
Slide #
Noncrossover
Asci 4:4
Crossover
Asci 2:2:2:2
OR 2:4:2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
TOTAL
100
Analysis of Results
1. Using the data in Table 1, determine the distance between the gene for
spore color and the centromere. Calculate the percent of crossovers by
dividing the number of crossover asci (2:2:2:2 or 2:4:2) by the total
number of asci x 100. To calculate the map distance, divide the
percentage of crossover asci by 2. The percentage of crossover asci is
divided by 2 because only half of the spores in each ascus are the result
of a crossover event (Figure 3). Record your results in Table 2.
Percent crossovers = crossover asci/ total asci X 100
Table 2
Number
of 4:4
Number of
Crossovers
Total
Asci
%Asci
Showing
crossover
Gene to
Centromere
Distance
(map Units)
2. Draw a pair of chromosomes in MI and MII, and show how you would get
a 2:4:2 arrangement of ascospores by crossing over. (Hint: refer to Figure
3).
3. Make a Venn diagram showing the similarities and differences between
mitosis and meiosis.
101
102
Cell Respiration
Cell respiration refers to the process of converting the chemical energy of organic molecules into
a form immediately usable by organisms. Glucose may be oxidized completely if sufficient
oxygen is available according to the following equation:
C6H12O6 + 6O2(g) → 6 H2O + 6 CO2(g) + energy
All organisms, including plants and animals, oxidize glucose for energy. Often, this energy is
used to convert ADP and phosphate into ATP. Peas undergo cell respiration during germination.
Using the CO2 Gas Sensor and O2 Gas Sensor, you will monitor the carbon dioxide produced and
the oxygen consumed by peas during cell respiration. Both germinating and non-germinating
peas will be tested. Additionally, cell respiration of germinating peas at different temperatures
will be investigated.
OBJECTIVES
In this experiment, you will
Study the effect of temperature on cell respiration.
• Determine whether germinating peas and non-germinating peas respire.
• Compare the rates of cell respiration in germinating and non-germinating peas.
•
The Oxygen Probe must always be
UPRIGHT- when stored & when being used
Figure 1
Adapted from Advanced Biology with Vernier and College Board Lab # 5 Cellular Respiration
103
MATERIALS
LabQuest
Vernier O2 Gas Sensor
Vernier CO2 Gas Sensor
25 non-germinated peas
25 germinated peas
BioChamber 250 (2 hole chambers)
ice cubes
two 100 mL beakers
thermometer
PROCEDURE
1. If your CO2 Gas Sensor has a switch, set it to the Low (0–10,000 ppm) setting. Connect the
O2 Gas Sensor and the CO2 Gas Sensor to LabQuest.
2. Choose New from the File menu. If you have older sensors that do not auto-ID, manually set
up the sensors.
3. Measure the room temperature using a thermometer and record the temperature in Table 1.
4. Obtain 25 germinated peas and blot them dry between two pieces of paper towel.
5. Place the germinated peas into the respiration chamber.
6. Place the O2 Gas Sensor into the BioChamber 250 as shown in Figure 1. Insert the sensor
snugly. The O2 Gas Sensor should remain vertical throughout the experiment. Place the CO2
Gas Sensor into the neck of the BioChamber 250.
7. Wait two minutes, then start data collection.
8. When data collection has finished, remove the sensors from the respiration chamber. Place
the peas in a 100 mL beaker filled with cold water and an ice cube.
9. Fill the respiration chamber with water and then empty it. Thoroughly dry the inside of the
respiration chamber with a paper towel.
10. Perform a linear regression to calculate the rate of respiration.
a. Choose Curve Fit from the Analyze menu and select CO2 Gas.
b. Select Linear as the Fit Equation. The linear-regression statistics for these two data
columns are displayed for the equation in the form
c. Enter the absolute value of the slope, m, as the rate of respiration for the CO2 Gas Sensor
in Table 2.
d. Select OK.
11. Calculate the rate of respiration for the O2 Gas Sensor.
a. Choose Curve Fit from the Analyze menu and select O2 Gas.
b. Select Linear as the Fit Equation. The linear-regression statistics are displayed for the
equation in the form
Adapted from Advanced Biology with Vernier and College Board Lab # 5 Cellular Respiration
104
c. Enter the absolute value of the slope, m, as the rate of respiration for the O2 Gas Sensor in
Table 2.
d. Select OK.
12. Repeat Steps 5–11 substituting the germinated peas with non-germinated peas. In Step 8
place the non-germinated peas on a paper towel and not in the ice bath.
Part II Germinated peas and Temperatures
13. Remove the peas from the cold water and gently blot them dry between two paper towels.
14. Repeat Steps 5–11 using the cold germinated peas.
15. Graph all of your data on one graph.
DATA
Table 1: Temperatures
Condition
Temperature (°C)
Room
Cold
Table 2: Respiration Rates
Peas
O2
Rate of respiration
(ppt/s)
CO2
Rate of respiration
(ppt/s)
Germinating, room temperature
Non-germinating, room temperature
Germinating, cool temperature
Table 3: Class Data Oxygen
Peas
Germinating,
Room Temp
NonGerminating,
Room Temp.
Germinating,
Cold Temp
G1
G2
G3
G4
G5
G6
G7
G8
G9
Adapted from Advanced Biology with Vernier and College Board Lab # 5 Cellular Respiration
G10
Avg.
105
Table 4: Class Data Carbon Dioxide
Peas
Germinating,
Room Temp
NonGerminating,
Room Temp.
Germinating,
Cold Temp
G1
G2
G3
G4
G5
G6
G7
G8
G9
G10
Avg.
QUESTIONS
1. Do you have evidence that cell respiration occurred in peas? Explain.
2. What is the effect of germination on the rate of cell respiration in peas?
3. What is the effect of temperature on the rate of cell respiration in peas?
4. Why do germinating peas undergo cell respiration?
5. Compare the class average Oxygen in all of the trials. Explain the significance of each.
EXTENSIONS
1. Compare the respiration rate among various types of seeds.
2. Compare the respiration rate among seeds that have germinated for different time periods,
such as 1, 3, and 5 days.
3. Compare the respiration rate among various types of small animals, such as insects or
earthworms.
Adapted from Advanced Biology with Vernier and College Board Lab # 5 Cellular Respiration
106
107
108
Plant Pigment Chromatography
Paper chromatography is a technique used to separate and identify substances in a mixture based
on the movement of the different substances up a piece of paper by capillary action. Pigments
extracted from plant cells contain a variety of molecules, such as chlorophylls, beta carotene, and
xanthophyll that can be separated using paper chromatography. A small sample of plant pigment
placed on chromatography paper travels up the paper due to capillary action. Beta carotene is
carried the furthest because it is highly soluble in the solvent and because it forms no hydrogen
bonds with the chromatography paper fibers. Xanthophyll contains oxygen and does not travel
quite as far with the solvent because it is less soluble than beta carotene and forms some
hydrogen bonds with the paper. Chlorophylls are bound more tightly to the paper than the other
two, so they travel the shortest distance.
The ratio of the distance moved by a pigment to the distance moved by the solvent is a constant,
Rf. Each type of molecule has its own Rf value.
Rf =
distance traveled by pigment
distance traveled by solvent
Chlorophyll a is the primary photosynthetic pigment in plants. A molecule of chlorophyll a is
located at the reaction center of photosystems. Other chlorophyll a molecules, chlorophyll b, and
the carotenoids (that is, carotenes and xanthophylls) capture light energy and transfer it to the
chlorophyll a at the reaction center. Carotenoids also protect the photosynthetic system from the
damaging effects of ultraviolet light.
OBJECTIVES
In this experiment, you will
•
•
Separate plant pigments.
Calculate the Rf values of the pigments.
MATERIALS
chromatography jar with lid
chromatography paper
spinach leaves
Coin
goggles
ruler
pencil
scissors
10-15 ml chromatography solvent
PROCEDURE
Obtain and wear goggles! Caution: The solvent in this experiment is flammable and poisonous.
Be sure there are no open flames in the lab during this experiment. Avoid inhaling fumes. Wear
goggles at all times. Notify your teacher immediately if an accident occurs.
1. Obtain a chromatography jar that has 10-15 ml of solvent in the bottom. Keep the jar closed
as much as possible because the solvent is volatile.
2. Obtain an 8-cm square piece of chromatography paper and one fresh spinach leaf.
Adapted from Advanced Biology with Vernier
109
3. Make two pencil marks 1.5 cm from one edge of the chromatography paper as shown in the
figure below.
4. Lay the leaf on the chromatography paper. Using the marks as a guide lay a ruler on top of
the leaf so that the edge of the ruler is on the paper 1.5 cm from, and parallel to, the edge.
5. Using the ruler as a guide, roll a coin over the leaf so that you drive the leaf pigments into the
paper in a straight line 1.5 cm from the edge of the paper. Repeat the procedure 10 times
making sure to use a different part of the leaf each time.
6. Fold the chromatography paper in half vertically so that it will stand up by itself. Place the
chromatography paper in the jar so that the end of the paper is in the solvent but the pigment
streak itself is not in the solvent. Do not let the sides of the paper touch the sides of the jar.
7. Close the lid and do not disturb the jar.
8. Wait until the solvent is approximately 1 cm from the top of the paper. Remove the
chromatography paper and mark the solvent front before it evaporates. Recap your jar.
9. Allow the paper to dry. Mark the bottom of each pigment band. Measure the distance each
pigment moved from the starting line to the bottom of the pigment band. Record the distance
that each of the pigments and the solvent moved, in millimeters in Table 1. Band 1 is the
pigment band nearest the origin, etc.
10. Identify each of the bands and label them on the chromatography paper.
•
beta carotene:
yellow to yellow orange
•
xanthophyll:
yellow
•
chlorophyll a:
bright green to blue green
•
chlorophyll b:
yellow green to olive green
11. Cut your chromatogram along the fold line. Staple half of the chromatogram to the front of
your lab sheet. Write your name on the back of the other half and give it to your teacher. The
pigments from this half may be separated from each other, extracted from the paper, and
analyzed.
Adapted from Advanced Biology with Vernier
110
DATA
Table 1
Band number
Distance traveled
(mm)
Band color
Identity
1
2
3
4
5*
Distance solvent front moved =
mm
* The fifth band may not appear.
PROCESSING THE DATA
Calculate the Rf values and record in Table 2.
Table 2
Molecule
Rf
beta carotene
xanthophyll
chlorophyll a
chlorophyll b
Adapted from Advanced Biology with Vernier
111
PRE LAB QUESTIONS
1. In this experiment we are separating pigments by paper chromatography. Other than
chlorophyll a, what pigments might we find in the spinach leaves?
2. Why do you mark the chromatography paper with pencil, not pen?
3. Jessica didn’t read the instructions and allowed the liquid level to be above the baseline.
Will she still get valid data? Explain.
4. Which pigment, chlorophyll a, chlorophyll b, and/or carotenoids, will travel the farthest
on the chromatography paper? Explain.
QUESTIONS
1. What factors are involved in the separation of the pigments?
2. Would you expect the Rf value to be different with a different solvent? Explain.
3. Why do the pigments become separated during the development of the chromatogram?
4. What type of chlorophyll does the reaction center contain? What are the roles of the other
pigments?
Adapted from Advanced Biology with Vernier
112
Photosynthesis
The human eye responds to a certain range of wavelengths of electromagnetic radiation. We call
radiation within this range “visible light.” The shorter the wavelength, the greater the energy of
the radiation. Ultraviolet radiation, X-rays, and gamma rays possess shorter wavelengths and
more energy than visible light. These high-energy radiations can harm living tissues.
Electromagnetic radiations with lower energies (longer wavelengths) than visible light are called
infrared radiation and radio waves.
Plant cells contain pigments that absorb electromagnetic radiation of wavelengths within the
visible range. The pigment molecules are part of complexes called photosystems, which are
found in the chloroplasts of palisade mesophyll leaf cells.
The process of photosynthesis involves the use of pigment molecules to absorb light energy
which will be used to convert carbon dioxide and water into sugar, oxygen, and other organic
compounds. This process is often summarized by the following reaction:
6 H2O + 6 CO2 + light energy → C6H12O6 + 6 O2
This process is an extremely complex one, occurring in two stages. The first stage, called the
light reactions of photosynthesis, requires light energy. The products of the light reactions are
then used to produce glucose from carbon dioxide and water. Because the reactions in the second
stage do not require the direct use of light energy, they are called the light independent or dark
reactions of photosynthesis.
In the light reactions, electrons derived from water are “excited” (raised to higher energy levels)
in several steps using photosystems I and II. In both steps, chlorophyll absorbs light energy that
is used to excite the electrons. Normally, these electrons are passed to a cytochrome-containing
electron transport chain. In the first photosystem, these electrons are used to generate ATP. In the
second photosystem, excited electrons are used to produce the reduced coenzyme nicotinamide
adenine dinucleotide phosphate (NADPH). Both ATP and NADPH are then used in the light
independent reactions to produce glucose.
In this experiment, a blue dye (2,6-dichlorophenol-indophenol, or DPIP) will be used to replace
NADP+ in the light reactions. When the dye is oxidized, it is blue. When reduced, however, it
turns colorless. Since DPIP replaces NADP+ in the light reactions, it will turn from blue to
colorless when reduced during photosynthesis. In this activity, you will add an extract of
chloroplasts from spinach leaves to a DPIP solution and incubate the mixture in the presence of
light. As the DPIP is reduced and becomes colorless, the resultant decrease in light absorbance is
measured over time.
To measure the change in color more precisely, you will use a device called a
Spectrophotometer (Vernier SpectroVis). Inside the spectrophotometer, there is a light bulb
which can be made to shine a beam of light of just one wavelength through your sample tubes.
(Remember that normal white light is a mixture of all the colors of the rainbow. Each of these
colors is different because it has a different wavelength.) There is also a detector which will
measure how much light passes through your sample. You will adjust the bulb to emit light that
Adapted from Advanced Biology with Vernier
113
is absorbed best by the unreduced DPIP. The wavelength of this light is known as λmax (lambda
max). At the beginning, when the DPIP is blue, it will absorb all the light and will not allow any
to pass through. As the DPIP is reduced to DPIPH and becomes colorless, more and more light
will pass through the sample. You will measure this change at different time intervals. The
amount of light that passes through the sample is known as transmittance. The amount of light
that is ‘caught’ by the molecules of DPIP is known as absorbance.
When you use DPIP in this experiment, it will be part of a mixture. The other parts of the
mixture are water, buffer, and chloroplasts. They also transmit and absorb light. If you want the
spectrophotometer to read the concentration of just the DPIP, and not the other substances, you
must adjust the meter so that it will ‘‘ignore’’ everything except the DPIP. You will do this by
using a blank. A blank is a sample that contains all the substances in the experimental sample
except the one you will be measuring. The blank is inserted into the spectrophotometer before
any experimental readings are taken. When the blank is in the chamber, the meter will be
adjusted to read 0% absorbance. In other words, the meter will ‘ignore’’ buffer, water, and
chloroplasts. When an experimental sample is placed in the chamber, the meter will record only
the amount of light absorbed by the DPIP.
This experiment is designed to test the hypothesis that light and active chloroplasts are required
for the light reactions of photosynthesis to occur.
OBJECTIVES
In this experiment, you will
• Use a spectrophotometer to measure color changes due to photosynthesis.
• Study the effect of light on photosynthesis.
• Study the effect that the boiling of plant cells has on photosynthesis.
• Compare the rates of photosynthesis for plants in different light conditions.
Figure 1
MATERIALS
LabQuest
Vernier SpectroVis
Adapted from Advanced Biology with Vernier
two small test tubes
4 graduated transfer pipets (or syringes)
114
5 cuvettes with lids (test tubes)
aluminum foil
100 watt floodlight
Stopwatch
600 mL beaker (or fish bowl)
250 mL beaker (ice bath)
distilled water
10 mL DPIP
phosphate buffer solution
vial with unboiled chloroplast suspension
vial with boiled chloroplast suspension
ice
marker
PROCEDURE
1. Obtain and wear goggles.
2. Your teacher will provide you with vials containing boiled and unboiled chloroplasts.
3. Fill the 250 mL beaker with ice and place the vials in the beaker.
4. Mark the five cuvette lids with with a BK (blank), a UD (unboiled/dark), a
UL(unboiled/light), a BL (boiled/light) and a NL (no chloroplasts/light).
5. Cover all four sides and the bottom of the UD cuvette with aluminum foil.
6. Use the information in Table 1 to add the phosphate buffer, distilled H2O, and DPIP to each
cuvette. Important: Do not add chloroplasts at this time.
Note: You have four plastic transfer pipets for setting up the experiment. Keep track of them
carefully and use them as directed, so that you will not cross-contaminate reagents. The first
is to be used for the H2O and then later for the buffer solution, the second for DPIP, the third
for unboiled chloroplasts suspension, and the fourth for boiled chloroplasts suspension. Be
aware that the pipets have a 1-mL graduation mark at the top of the neck, near the bulb.
Table 1
BK
UD
UL
BL
Blank
(no DPIP)
Unboiled
Dark
Unboiled
Light
Boiled
Light
NL
No
Chloroplasts
Light
Phosphate buffer
1 mL
1 mL
1 mL
1 mL
1 mL
Distilled H2O
4 mL
3 mL
3 mL
3 mL
3 mL + 3
drops
—
1 mL
1 mL
1 mL
1 mL
3 drops
3 drops
3 drops
—
—
—
—
—
3 drops
—
DPIP
Unboiled chloroplasts
Boiled chloroplasts
7. Connect the spectrophotometer to LabQuest and choose New from the File menu. Tap Mode.
From the dropdown menu select Selected Events then select OK. Tap on the red box and
select Change Wavelength. Enter 605 into the wavelength box, put a check in the box to
Adapted from Advanced Biology with Vernier
115
Report average of wavelength band, then select OK. (605 nm was selected because it is the
λmax for DPIP).
8. Finish preparing the Blank cuvette by adding three drops of unboiled chloroplasts. (Be sure
to always gently shake the vial to resuspend chloroplasts before removing sample.) Place the
lid marked with BK on the cuvette and gently invert three times to mix.
Note: 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.
9. Calibrate the SpectroVis.
a. Choose Calibrate from the Sensors menu.
b. When the warm-up period is complete, place the Blank (BK) in the spectrometer. Make
sure to align the cuvette so that the clear sides are facing the light source of the
spectrometer.
c. Tap Finish Calibration, and then select OK.
10. Obtain a 600 mL beaker filled with water and a flood lamp. Arrange the lamp and beaker as
shown in figure 2. The beaker will act as a heat shield, protecting the chloroplasts from
warming by the flood lamp. Do not turn on the lamp until Step 13.
Figure 2
11. Finish preparing the cuvettes. Important: Perform the following steps as quickly as possible
and proceed directly to Step 12.
a. Cuvette UD: Add 3 drops of unboiled chloroplasts. Place the lid on the cuvette and gently
mix; try not to introduce bubbles in the solution. Make sure that no light can penetrate the
cuvette.
b. Cuvette UL: Add 3 drops of unboiled chloroplasts. Place the lid on the cuvette and gently
mix; try not to introduce bubbles in the solution.
c. Cuvette BL: Add 3 drops of boiled chloroplasts. Place the lid on the cuvette and gently
mix; try not to introduce bubbles in the solution.
d. Cuvette NL: Do not add chloroplasts to this cuvette.
Place the cuvettes in front of the lamp as shown in Figure 2. Mark the cuvettes’ positions so
that they can be placed back in the same spots.
12. Take absorbance readings for each cuvette. Invert each cuvette two times to resuspend the
chloroplast before taking a reading. If any air bubbles form, gently tap on the cuvette lid to
Adapted from Advanced Biology with Vernier
116
knock them loose.
a. Cuvette UD: Remove the cuvette from the foil sleeve and place it in the cuvette slot of the
SpectroVis. Allow 10 seconds for the readings displayed in the meter to stabilize. Record
the absorbance value in Table 2. Remove the cuvette and place it back into the foil sleeve.
Place the cuvette in its original position in front of the lamp.
b. Cuvette UL: Place the cuvette in the SpectroVis. Allow 10 seconds for the readings
displayed in the meter to stabilize. Record the absorbance value in Table 2. Remove the
cuvette and place it in its original position in front of the lamp.
c. Cuvette BL: Place the cuvette in the SpectroVis. Allow 10 seconds for the readings
displayed in the meter to stabilize. Record the absorbance value in Table 2. Place the
cuvette in its original position in front of the lamp.
d. Cuvette NL: Place the cuvette in the cuvette slot of the SpectroVis. Allow 10 seconds for
the readings displayed in the meter to stabilize. Record the absorbance value in Table 2.
Remove the cuvette and place it in its original position in front of the lamp.
13. Turn on the lamp and note the time.
14. Repeat Step 13 after 5, 10, 15, and 20 minutes have elapsed.
PROCESSING THE DATA
Calculate the rate of photosynthesis for each of the three cuvettes tested by performing a linear
regression on each of the data sets.
1. Prepare LabQuest for data entry.
a. Disconnect the SpectroVis and choose New from the File menu to prepare the program for
manual entry.
b. Tap Table to display the data table.
c. Tap Run 1 to change the name of the table to Absorbance.
d. Tap X to rename the column.
e. Enter the Name (Time) and Units (min). Select OK.
f. Tap Y to rename the column.
g. Enter the Name (Unboiled/Dark), leave the Units field blank, and select 3 decimal places.
Select OK.
h. Tap Table and select New Manual Column.
i. Enter the Name (Unboiled/Light), leave the Units field blank, and select 3 decimal places.
Select OK.
j. Repeat steps h-i to add columns for (Boiled/Light) and (No Chloroplast/Light).
2. Enter the absorbance and time data pairs from Table 1.
a. Tap the first cell in the Time column and enter the 0 for the initial time at which data was
taken.
Adapted from Advanced Biology with Vernier
117
b. Move to the first cell in the Unboiled/Dark column and enter the initial absorbance value
from the Unboiled/Dark column of Table 1.
c. Continue to enter the data pairs using the same process until you have entered the 5 time
and absorbance data pairs for your data in Table 1.
d. When all the data has been entered, tap Graph to view a graph of absorbance vs. time.
3. Perform a linear regression to calculate the rate of photosynthetic activity.
a. Choose Curve Fit from the Analyze menu.
b. Select the first set of data you wish to analyze.
c. Select Linear as the Fit Equation. The linear-regression statistics for these two data
columns are displayed for the equation in the form
y = mx + b
where x is concentration, y is absorbance, m is the slope, and b is the y-intercept.
c. Enter the absolute value of the slope, m, as the rate of photosynthetic activity in Table 2.
d. Select OK.
4. Repeat Step 3 of Processing the Data for the rest of your data sets.
DATA
Table 2 – Absorbance Readings
Time
(min)
Table 3 - Rates
UD
UL
BL
NL
unboiled
dark
unboiled
light
boiled light
no
chloroplast
light
Chloroplast
0
Unboiled/Dark
5
Unboiled/Light
10
Boiled/Light
15
None/Light
Rate of
photosynthesis
20
Adapted from Advanced Biology with Vernier
118
PRE LAB QUESTIONS
1.
How is red light different from green light? 2.
Which has more energy, short or long electromagnetic waves? 3.
If DPIP is a dark blue color, has light been absorbed by the chloroplast? Explain. 4.
Which instrument will be used to measure the absorbance of light so that we can measure the amount of photosynthesis occurring? 5.
Do you expect to see more or less absorbance of light if photosynthesis is actually occurring? 6.
Why must you use a blank when you make measurements with a SpectroVis?
7.
What is the purpose of the water container in step 10?
8.
Which cuvette do you expect to end up with the lighter color, the one that has boiled
chloroplasts or the one with unboiled chloroplasts? Explain.
Adapted from Advanced Biology with Vernier
119
QUESTIONS:
1. What is the function of DPIP in this experiment?
2. What molecule found in chloroplasts does DPIP “replace” in the experiment?
3. What is the source of the electrons that will reduce DPIP?
4. What was measured with the SpectroVis in this experiment?
5. What is the effect of darkness on the reduction of DPIP? Explain.
Adapted from Advanced Biology with Vernier
120
6. What is the effect of boiling the chloroplasts on the subsequent reduction of DPIP?
Explain.
7. Identity the function of each of the cuvettes:
BK:
_________________________________________________________________
_
_________________________________________________________________
_
UD:
_________________________________________________________________
_
_________________________________________________________________
_
UL:
_________________________________________________________________
_
_________________________________________________________________
_
BL:
_________________________________________________________________
_
_________________________________________________________________
_
NL:
_________________________________________________________________
_
_________________________________________________________________
_
Adapted from Advanced Biology with Vernier
121
122
DNA Detectives or “Who Dunnit?”
Introduction:
Many of the revolutionary changes that have occurred in biology over the past fifteen years can be
attributed to the ability to manipulate DNA in defined ways. The principal tools for the recombinant
DNA technology are enzymes that can “cut and paste” DNA. Restriction enzymes are the “chemical
scissors” of the molecular biologist; these enzymes cut DNA at specific nucleotide sequences. A
sample of someone’s DNA, incubated with restriction enzymes, is reduced to millions of DNA
fragments of varying sizes. A DNA sample from a different person would have a different nucleotide
sequence and would thus be enzymatically “chopped up” into a very different collection of
fragments. Because no two people (except identical twins) have exactly the same DNA, a person’s
DNA fingerprint is unique and can be used for purposes of identification. We have been asked to
apply DNA fingerprinting to determine which suspect should be charged with a crime perpetrated in
our city.
CONDUCTING ELECTROPHORESIS:
BACKGROUND INFORMATION
DNA is made up of a series of base pairs (guanine-cytosine, adenine-thymine, cytosine-guanine,
thymine-adenine).
G-C
A-T
C-G
T-A
Every individual has a unique series of base pairs in their DNA.
The DNA samples used in this lab have been treated with restriction enzymes which seek out
specific DNA base pair (bp) sequences and cut the DNA at that point. Since the DNA samples were
all different, they were all cut at different spots which resulted in different sized fragments of DNA
for each sample.
ATCCTGCCGGAAGTCCGATCCGGTA
TAGGACGGCCTTCAGGCTAGGCCAT
The number of fragments and the sizes of the fragments depend on the restriction enzyme used and
the size of the original DNA molecule. Restriction enzymes are microbial products, and their names
are derived from the names of the organisms in which they were found. There are currently about
2500 restriction enzymes known, with 200 different recognition sequences. It is believed that the
native function of enzymes was to digest foreign DNA, in other words, to protect the microorganism
from invading viral DNA.
Specific restriction enzymes may leave blunt ends on the DNA while others leave “sticky ends”
where one strand of the DNA overhangs the other. Sticky ends are useful in recombinant technology.
A new genetic characteristic can be transferred from organism to another by joining the sticky ends
of two different strands of DNA.
In order to determine what the DNA fragment sizes are, it is necessary to: (1) separate the fragments
by size; (2) have some way to visualize the DNA; and (3) have a standard to which the fragments
can be compared. The first is accomplished by separating the DNA using agarose gel
123
electrophoresis.
The term ‘electrophoresis’ literally means “to carry with electricity.” It is a technique for separating
and analyzing mixtures of charged molecules. When placed in an electric field, pieces of DNA
(because they are ionized and negatively charged) migrate toward the positive electrode (anode);
small pieces of DNA experience less resistance and move faster (farther) than larger pieces.
It is necessary to add a matrix such as agarose or acrylamide to act as a sieve and separate the DNA
molecules based on their size. The choice of matrix, agarose or acrylamide, is determined by the
sizes of the molecules to be separated. Acrylamide is used primarily to separate proteins and small
DNA molecules (under 1000 base pairs). Agarose is the matrix used to separate most DNA
molecules.
Agarose is a polysaccharide (from algae) that can be dissolved in hot water. As the agarose solution
cools, it solidifies to form a matrix of gelatin-like consistency. The matrix contains pores through
which the DNA molecules must pass. The size of the pores, and hence the sizes of the DNA
molecules that can be separated on the gel, is determined by the concentration of the agarose
solution. For example, large DNA molecules (>10,000 base pairs) can best be separated on a 0.3%
agarose gels (e.g., larger pores), whereas small DNA molecules (100-3000 base pairs) would
separate with better resolution on a 2.0% agarose gel (e.g., smaller pores). In the experiments in this
kit, 0.9% agarose gels will be used to separate the DNA molecules.
As these 0.9% gels are prepared, a comb is placed in the gel at the end closest to the cathode
(negative electrode). After the agarose solution has solidified, the comb can be removed, leaving
small holes or wells in the gel into which the samples will be loaded. The DNA samples are mixed
with a loading buffer that contains glycerol and a tracking dye. The glycerol adds density to the
samples, assuring that they will stay in the wells when loaded. The tracking dye usually contains a
dye like bromphenol blue, a small molecule that migrates through the gel at a position approximately
equivalent to a DNA fragment of 300 base pairs, or Orange G, which migrates through the gel at a
position approximately equivalent to a DNA fragment of 50 base pairs. The dyes serve two
functions. They makes it easier to see the samples while the wells are being loaded and, since the
dye can be seen as it migrates through the gel, it can be used to estimate how far the DNA has
migrated in the gel.
When it is time to load and run the gel, the gel is covered in buffer, the comb carefully removed, and
the samples loaded into the wells. A standard solution consisting of DNA fragments of known sizes
is loaded into an adjacent well. The lid is placed on the gel box, the gel box is connected to a power
supply, and an electrical current is passed through the gel. The DNA molecules immediately begin to
migrate toward the anode, with smaller molecules migrating more rapidly than larger DNA
molecules.
It is necessary to have some method for visualizing the DNA in the agarose gel. In the following
procedures, methylene blue is used as a post-stain for DNA. The DNA bands appear blue on a clear
background and the migration of the fragments can be measured. A DNA ladder is a mixture of
DNA fragments of known lengths. Using the data from the DNA fragments of known sizes, a
standard curve can be constructed and used to calculate the sizes of unknown DNA fragments (the
samples from the crime scene and suspects.)
124
Who Dunnit?
A murder has been committed, and police discover evidence of a struggle and blood traces at the
scene of the crime. Ian, a UPS delivery man is found dead in his truck on 221 just west of Forest
Middle School. Autopsy has shown that Ian was strangled to death but there is blood on the scene.
Packages are missing from the truck, and no witnesses an be found. Suspects X, Y, and Z are
arrested and will go through DNA tests to determine if they were at the scene of the crime. All of the
suspects proclaim their innocence adamantly, and want to see their lawyers. At their indictments, it
is learned that
1.
Suspect X - Bob Smith is a man in his middle thirties with prior convictions for armed
robbery. Bob was apprehended shortly after the murder in Bedford driving recklessly on an
expired license. No contraband was found in his possession but his hands are cut in several
places. He says it’s because he works construction.
2.
Suspect Y - Jim Dale is a man in his late forties. He is suspected of being romantically
involved with Ian’s wife, Pam. Unexplained scratches were found on the back of his neck.
3.
Suspect Z - Pam, wife of Ian. She says she was with Jim the entire day. Several cuts on both
hands are suspicious. She claims she got them while picking blackberries with Ian.
You are the lab worker who has been handed the DNA samples from the three suspects involved
plus the DNA from the blood at the crime scene. Using molecular biology techniques, your job is to
determine which of the suspects might have been at the crime scene. The court awaits your findings.
Purpose: To prepare and analyze a DNA fingerprint, the student will:
1.
2.
3.
4.
Prepare and load an agarose gel with enzyme cut DNA samples.
Conduct gel electrophoresis to sort out the DNA fragments in the samples.
Stain the gel to visualize the DNA fragments.
Analyze the resulting banding pattern or “DNA fingerprint” to solve a crime.
Materials:












Electrophoresis chamber
casting tray
comb
agarose
power supply
plastic tray for staining and storing gel
DNA from suspects and crime scene
(pre-cut)
micropipette and pipette tips
electrophoresis buffer [1X SB]
racks for 1.5 ml microcentrifuge tubes
1.5 ml microcentrifuge tubes
1X staining solution (Methylene Blue)












gloves
light box
plastic wrap
125 mL Erlenmeyer flask
scoopula
weighing boat
balance
thermometer
practice pipetting station & dye
parafilm
plastic wrap and transparency
marker
Procedures:
125
Preparing the gel
1.
Prepare the casting tray by placing it into the electrophoresis chamber and inserting the metal
buffer dams at each end of the tray. Insert the comb into the tray so that it is nearest the black
electrode.
2.
Measure out 30 mL of 1X SB buffer solution and pour it into a 125 mL flask.
3.
Mass out 0.27 g of agarose powder into a weighing boat. Add it to the flask with the buffer.
DO NOT SWIRL the solution, as the undissolved agarose will stick to the sides of the flask.
4.
Mass the flask with its contents. Record the mass on the flask’s label along with your name.
5.
Heat the solution in a microwave oven for 1 minute to dissolve the agarose.
a.
Do not seal the container. Watch carefully that the solution does not boil over.
b.
When the hot agarose solution is removed from the microwave, it may be
superheated. Carefully remove the flask from the microwave by holding onto the
neck of the flask pointing the flask away from yourself and others, swirl gently. The
superheated solution may bubble briefly at this point. (Be careful, as it may boil over,
out of the flask)
c.
As the agarose dissolves, the solution will become clear. Swirl the flask to be certain
that all the agarose is dissolved. If it is not, you will see little clear flecks. Hold the
flask up to the light and examine the solution carefully to be sure that the agarose is
all dissolved. If not completely dissolved place flask back in the microwave and heat
for a short time.
6.
Slightly cool the flask containing dissolved agarose then place it on the balance and add
distilled water to bring the mass back up to the original mass marked on your flask. Some of
the volume will be lost to evaporation during the process of heating and must be replaced to
maintain the proper agarose concentration.
7.
Place parafilm over the mouth of the flask to prevent further evaporation while cooling.
8.
Note: Use a thermometer to check the temperature of the agarose! If the agarose is
hotter than 55 – 60oC when the gel is poured, the gel tray can be warped by the heat.
Be Careful!
Carefully pour the agarose into the casting tray. DO NOT jar or move the casting tray as the
gel solidifies. As the agarose polymerizes (about 10 min), it changes from clear to slightly
opaque.
126
9.
After gel has set, carefully remove the comb and casting gates.
10.
Pour enough buffer into the gel box so that the gel is completely covered, with no ‘dimpling’
above the wells.
Loading and Running the Gels
11.
Load entire contents (10ul) of each sample tube into separate wells in the gel. Be sure the
micro-pipette tip is below the surface of the buffer and just above the center of each well that
you load. CHANGE PIPET TIPS BETWEEN SAMPLES TO AVOID
CONTAMINATION!
Leaving an empty lane on both sides, load in the order shown below:
X
Yellow tube
Green tube
Blue tube
Red tube
Clear tube
Y
Z
- Suspect X
- Suspect Y
- Suspect Z
-E
-L
E
L
= Bob Smith, former thief
= Jim Dale, boyfriend
= Pam, wife of victim
= evidence DNA found at crime scene
= ladder DNA (standardized control sample)
12.
Once the wells are loaded, put the top on the gel box and connect it to the power supply (red
to read, black to black.) Plug in the power supply and turn the unit to the desired voltage
(check with teacher for voltage settings). Check the electrodes to be sure that bubbles are
rising from the wires. If you don’t see bubbles, check all your connections.
13.
Run until loading dye nears the bottom of the gel. At this point the current may be shut off
and the leads (wires) disconnected.
Staining and Viewing the Gel
14.
Remove the casting tray from the gel box. CAREFULLY slide your gel off the casting tray
into its plastic container.
15.
Wearing gloves, pour the 1X methylene blue staining solution into the plastic container and
allow it to sit for 15-20 minutes, rocking the container periodically.
16.
To destain your gel, pour the stain carefully back into the beaker (do not throw away) and
gently rinse your gel with water for 5 minutes. You may need to let your gel soak in several
changes of water to increase contrast. DO NOT use large volumes of water, the water should
just cover the gel. It is better to use small volumes of water and change it frequently than to
flood the gel in a large volume!
17.
Place a piece of plastic wrap on your light box and examine your stained, rinsed gel by
placing it on the plastic wrap. Gently place a transparency over your gel and trace the bands
with a permanent marker.
127
18.
Store your gel in a labeled plastic bag in the refrigerator.
Upon completion of the lab:

dispose of designated materials in the appropriate places.

Gently rinse gel boxes in water (do not scrub or wipe inside the box to dry it)

leave equipment as you found it.

check that your workstation is in order.

wash your hands.
Analysis:
Compare the fingerprints of all the suspects in this case to the profile of the DNA isolated from the
blood droplets at the crime scene. By comparing the banding patterns of the DNA samples, you should
be able to determine who the murderer was. Which suspect’s blood was found at the site of the murder?
Unfortunately, most courts will not accept your preliminary results as being conclusive and will expect
a more detailed analysis. By calculating the size of the fragments from different samples, it is possible
to more definitively determine “guilt”.
1.
Look at the lane containing the 1 kb (kilobase) DNA ladder. To figure out which band is which,
find the two bands that represent 1650 and 2000 base pairs (bp). These two bands are separated
from the other bands, and rather easy to find.
2.
Find these two bands on the stained gel and measure their migration from their point of origin
in the gel. Measure from the bottom of the well to the foremost edge of the stained band. Be
certain to measure each from the same point, e.g., from the bottom of the well each time, not the
bottom one time and the top of the well the next. Record the base pair size of the band and its
migration distance (in mm) on your data sheet.
3.
Working up (toward larger DNA fragments) and down (toward smaller DNA fragments) from
the 1650 and 2000 bp bands, measure and record the migration of the other bands in the DNA
ladder.
Note: The large bands will be too close together to be measured accurately, and the
128
smaller bands may have migrated off the bottom of the gel. Remember, if you have run
the gel until the dye has reached the bottom of the gel, then anything smaller than dye
about 400 bp will have run off the bottom of the gel.
4.
Measure the migration of the bands in the experimental lanes and record the migration distances
on your data sheet.
Graphing – Method One - Paper
5.
Create a standard curve using the data from the 1 kb DNA ladder. Graph the migration distance
of the DNA fragments (x-axis) against the size of the DNA fragments (y-axis) on semi-log
graph paper.
6.
Draw a straight “line of best fit” which comes as close as possible to each point.
7.
To determine the size of an enzyme-digested DNA fragment, find where the migration distance
of the DNA fragment intersects the standard curve. Draw a line from this point to the y-axis.
Where this line meets the y-axis is the size of the fragment. Record the size of the fragments on
your data sheet.
Graphing – Method Two – TI-83
8.
Create a standard curve using the data from the 1 kb DNA ladder.
9.
To clear your calculator of old information:

press 2nd, MEM, select 4:ClrAllLists, and press ENTER, ENTER

press Y=, then CLEAR for each entry

press 2nd, STATPLOT, then 4:PLOTSOFF, ENTER
10.
On your calculator select STAT, EDIT, then press ENTER
11.
Enter the migration distances from your ladder into the L1 column
12.
Enter the corresponding DNA fragment sizes (bp) into the L2 column
13.
Press 2nd, STAT PLOT, and select Plot 1, ENTER
14.
Select ON and press ENTER
15.
Select scatter plot, Xlist: L1, Ylist: L2, and press GRAPH
16.
To fit your graph to your data press ZOOM, select 9:ZoomStat, and press ENTER. You
should see a graph of your data points.
17.
To draw a regression curve to fit your data points, press STAT, select CALC, 0:ExpReg, and
press ENTER
18.
To tell the calculator where the data is that you want analyzed, press 2nd L1,2nd L2 then press
ENTER
129
19.
To tell the calculator to graph this curve, press Y=, select Y1=, press VARS, select 5:Statistics,
select EQ, and press ENTER (make sure all other Y equations are deleted)
20.
Press GRAPH. You should see a graph of your data points with the curve of best fit.
21.
To use this graph to determine the sizes of an enzyme-digested DNA fragment, press 2nd,
CALC, then select 1:Value, and press ENTER
22.
Enter the migration distance of a DNA band and press ENTER. The Y value showing on the
screen will be the size of the DNA fragment (in bp). Record this information in your data table.
23.
Repeat steps 20-21 for the rest of the DNA fragments.
24.
Using your calculated bp values, determine which suspect should be charged with the crime.
130
DATA TABLES
1 kb DNA ladder
Fragment size
(in bp)
12,000
Migration
distance (mm)
DNA sample: Suspect X
Migration
Distance
Calculated
Size (bp)
11.000
8,800
8,200
7,000
5,000
2000 
1650 
DNA sample: Suspect Y
1000
850
650
Migration
Distance
Calculated
Size (bp)
500
400
300
 Use these bands for orientation.
DNA sample: Evidence
Migration
Calculated
Distance
Size (bp)
DNA sample: Suspect Z
Migration
Calculated
Distance
Size (bp)
131
Prelab Questions for DNA Gel Electrophoresis
1.
What is the purpose of electrophoresis?
2.
Towards which electrode does DNA move? Why?
3.
What function do restriction enzymes serve in bacterial cells?
4.
Why is the DNA in this experiment cut with restriction enzymes?
5.
What is agarose?
6.
What function does the agarose serve?
7.
Why is a comb inserted into the gel before it solidifies?
8.
Why is glycerol added to the DNA samples?
9.
How can you tell the gel is solidified and ready?
10.
Which DNA molecules move faster through the gel?
11.
What is the purpose of the Methylene Blue?
12.
Why must the gel be destained?
13.
What is a DNA ladder?
132
Lab Questions
1. Restriction enzymes are primarily isolated from bacteria. Why would bacteria have enzymes
that cut DNA?
2. Discuss how each of the following factors would affect the results of electrophoresis:
a. Voltage used:
b. Running time:
c. Amount of DNA used:
d. Reversal of polarity:
3. Two small restriction fragments of nearly the same base pair size appear as a single band,
even when the sample is run to the very end of the gel. What could be done to resolve the
fragments?
4. A certain restriction enzyme digest results in DNA fragments of the following sizes: 4,000 bp,
2,400 bp, 2,000 bp, 400 bp. Sketch the resulting separation by electrophoresis. Show starting
point, positive and negative electrodes, and the resulting bands.
5. How can a mutation that alters a recognition site be detected by gel electrophoresis?
6. Restriction enzymes make one of two types of cuts in DNA being digested. Some of these
enzymes produce sticky ends at the cut site while others produce blunt cuts. What are sticky
ends and what makes them important to recombinant DNA studies?
7. Who committed the crime? Justify your answer using your data.
133
134
BACKGROUND
Advanced Placement (AP) is a
registered trademark of the College Entrance Examination
Board. The materials provided
in this lab activity have been
prepared by WARD’S according
to the specifications outlined in
the AP Laboratory Manual, Edition D. To perform the investigation, you may either follow
the directions in the AP manual
or, if one is not available, you
may follow the directions in this
guide.
DID YOU KNOW?
Transformation was discovered in the late 1920s by Fred
Griffith, an English medical
officer, while he was studying
the bacteria responsible for a
pneumonia epidemic in
London.
The ability to exchange genes within a population is a nearly
universal attribute of all living things. Among prokaryotes, there is no
known case where genetic exchange is an obligatory step (as it often
is in eukaryotes) in the completion of an organism’s life cycle.
Rather, genetic exchange in prokaryotes seems to be an occasional
process that occurs through three different mechanisms in various
prokaryotes. These three mechanisms are transformation,
transduction, and conjugation.
In transformation, DNA is released from a cell into the surrounding
medium. Recipient cells are then able to incorporate it into themselves from the medium. Transduction occurs when a phage viron
attaches to a bacterial cell and transfers some or all of its DNA into
the bacterium. Conjugation is controlled by plasmid-borne genes and
occurs between cells that are in direct contact with one another. Usually only the plasmid itself is transferred, although sometimes chromosomal genes can be transferred as well.
Although the existence of plasmids was inferred from genetic studies
in the 1950s, it has only been during recent decades, with the advent
of more accurate means of detection, that the impact of plasmids on
the biology and ecology of prokaryotes has been fully appreciated.
Plasmids are circular molecules of double-stranded DNA and generally function as small chromosomes. They are self-replicating and
can encode for a variety of cellular functions. However, unlike chromosomes, they are dispensable. The types of functions they encode
only benefit the cell in a limited set of environments, and none are
known to encode for essential cellular functions. Many, but not all,
bacteria may contain anywhere from one to several dozen plasmids
and although they are capable of autonomous replication, the number
of plasmids remains fairly constant from one generation to the next.
These bits of DNA vary in size from a few to several hundred Kb
(kilobase) pairs in length.
Many plasmids, called R factors, carry genes that confer resistance to
antibiotics on the host cell. First discovered in 1955, R factors have
spread rapidly among pathogenic bacteria in recent years, profoundly
affecting medical science by causing many strains of pathogenic bacteria to be highly resistant to antibiotics. The number of inhibitory
substances for R-mediated resistance has grown to almost all antibiotics, many other chemotherapeutic agents, and a variety of heavy metals. The mechanisms of resistance conferred by these genes tend to be
different from those that are chromosomally determined. Plasmid
genes often encode enzymes that chemically inactivate the drug or
eliminate it from the cell by active transport. In contrast, chromosomal mutations usually modify the cellular target of the drug, rendering the cell resistant to the drug’s action.
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4
135
The first mechanism of bacterial genetic exchange to be discovered
was transformation. In 1928, a now-famous experiment demonstrated
that injecting mice with an non-pathogenic (not capable of causing
disease) strain of Streptococcus pneumoniae, together with heat-killed
cells of a pathogenic (disease-causing) strain killed mice, while injecting these strains separately did not. This and subsequent experiments established that the surviving cells were recombinant, meaning
that they exhibited certain properties (including the ability to cause
disease) that were typical of the killed cells and others that were typical of the non-pathogenic culture. A genetic exchange of the DNA
dissolved in the external medium had occurred between the dead cells
and the live ones. At the time, it was thought that a particular substance, a “transforming principle”, caused the exchange to take place.
Thus the word “transformation”
came to be used to describe genetic
exchange among prokaryotes.
When cells are able to be transformed by DNA in their environments they are called “competent”.
In a significant number of bacteria,
entry into a competent state is encoded by chromosomal genes and
signaled by certain environmental
conditions. Such bacteria are said to
be capable of undergoing natural
transformation. Many other bacteria
do not become competent under ordinary conditions but can be made
competent by exposing them to a
variety of artificial treatments, such
as exposure to high concentrations
of divalent cations.
E. coli cells, which do not possess a
natural system for transformation,
are capable of being artificially
transformed. They become competent only after the cultured cells are
exposed to calcium chloride solution. These newly-competent cells
are now receptive to an insertion of
foreign DNA contained in a plasmid.
5
© 2005 WARD’S Natural Science Establishment, Inc.
All Rights Reserved
136
MATERIALS
MATERIALS NEEDED PER GROUP
2
2
2
1
2
4
1
1
Luria agar plates
Luria agar plates with ampicillin
Microcentrifuge tubes
Inoculating loop
Bacti-spreaders
Sterile graduated pipets
Rubber pipet bulb
Capillary micropipet
DID YOU KNOW?
Streptococcus pneumoniae
infections cause 3,000 cases
of meningitis, 50,000 blood
infections, and 100,000 150,000 hospitalizations for
pneumonia each year.
SHARED MATERIALS
Calcium chloride
Luria broth
Plasmid pUC8
Waterbath
Starter plate of E. coli
PROCEDURE
Figure 1
NOTE
Prior to conducting the experiment, make sure all
materials are present and ready to use. A 42ºC waterbath should be available and the calcium chloride should be in an ice bath and kept cold throughout the experiment.
.7
1. Obtain two microcentrifuge tubes and mark one tube “+”, the
other “–”. The “+” tube will have the plasmid added to it.
2. Using a sterile pipet, add 0.25 ml (250 µl) ice cold calcium chloride to each tube (Figure 1).
3. Obtain a starter plate. Use a sterile inoculating loop to transfer a
large colony of bacteria from the starter plate to each tube of
cold calcium chloride. Be sure not to transfer any agar to the
tube.
.8
.9
4. To remove the bacteria from the transfer loop, place the loop
into the calcium chloride and twirl rapidly. Dispose of the loop
according to your instructor.
NOTE
Gently tapping the loop against the side of the tube
may help dislodge the bacteria.
9
© 2005 WARD’S Natural Science Establishment, Inc.
All Rights Reserved
137
5. Using the provided capillary micropipets and plungers, add 10 µl
(Figure 2) of the plasmid pUC8 solution, with the antibiotic resistance gene to the “+” tube.
Figure 2
Plunger
6. Gently tap the tube with your finger to mix the plasmid into the
solution.
7. Incubate both tubes on ice for 15 minutes.
8. While the tubes are incubating, obtain two Luria agar plates and
two Luria agar plates with ampicillin. Label one Luria agar plate
“+”, the other “–”. Do the same for the Luria agar plates with ampicillin. Be sure to label all four plates with your group name.
NOTE
Pipet
Both time and temperature are critical in the following heat-shock protocol. Be sure your waterbath is at
42ºC and do not exceed 90 seconds in the waterbath.
9. The bacterial cells must be heat shocked to allow the plasmid to
enter the cells. Remove the tubes from ice and immediately place
in a 42°C hot waterbath for 60 to 90 seconds.
Figure 3
.7
10. Remove the tubes from the 42ºC waterbath and immediately
place on ice for two minutes.
11. Remove the tubes from the ice bath and add 0.25 ml (250 µl) of
room temperature Luria broth to each tube with a sterile disposable pipet. Gently tap the tube with your finger to mix the solution. The tubes may now be kept at room temperature.
12. Add 0.1 ml (100 µl) (Figure 3) of the “+” solution to the two “+”
plates with a sterile disposable pipet. Add 0.1 ml (100 µl) of the
“–” solution to the two “–” plates with a different sterile disposable pipet.
.8
13. Using a sterile Bacti-spreader, spread the cells over the entire surface of the Luria agar “-“ plate. Then, using the same Bactispreader, spread the liquid on the Luria agar w/ampicillin “-“
plate.
.9
14. Using a new Bacti-spreader, repeat the procedure for both of the
“+” plates. Spread the liquid on the Luria agar “+” plate first followed by the Luria agar w/ampicillin “+” plate. Dispose of the
Bacti-spreaders according to your instructor.
15. Let the plates sit for five minutes to absorb the liquid. Place the
plates in a 37°C incubator, inverted, overnight.
16. The next day, remove the plates from the incubator. Count and
record the number of colonies on each plate. If the bacteria has
grown over the entire surface so that individual colonies cannot
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10
138
Analysis
Luria agar(+) __________________
Luria agar (-)_______________________
Luria agar w/ampicillin(+)_____________ Luria agar w/ampicillin (-)______________
1. Based on your experimental results, did transformation occur? Why or why not?
2. What other methods can be used to verify that transformation occurred? Explain.
3. Transformation is one type of genetic exchange among bacteria. Name and explain
another type of genetic exchange.
4. Explain the four plates and the results. Provide an explanation of each plate and the
results.
5. You repeat the experiment and obtain the results below. Explain what did or did not
occur.
139
140
Chi Square (X 2 ) Modeling Using M & M’s
Introduction:
The Chi Square test (X2) is often used in science to test if data you observe from
an experiment his the same as the data that you would predict from the experiment.
Calculating X2 values allow you to determine if test results can be attributed to
randomness or not. If the data differs greatly and it is not due to randomness, other
factors must be influencing your results. This investigation will help you to use the Chi
Square test by allowing you to practice it with a population of familiar objects, M & M
candies.
Objectives: Before you start this investigation, you should be able to:
• Determine the degrees of freedom (df) for an investigation;
• Calculate the X2 value for a given set of data;
• Use the critical values table to determine if the calculated value is equal to or less than the
critical value;
• Determine if the Chi Square value exceeds the critical value and if the null hypothesis is
accepted or rejected.
Biologists generally accept p = 0.05 as the cutoff for accepting or rejecting a hypothesis. If the
difference between your observed data and your expected data would occur due to chance alone
fewer than 1 time in 20 (p = 0.05) then the acceptability of your hypothesis may be questioned.
Biologists consider a p value of 0.05 or less to be a “statistically significant” difference.
Procedure:
1. Record the different colors (classes) of M & M’s in Table 1 and Table 2.
2. Count the number of each color of candy and record the number in Table 1 under
“Number Observed.”
3. Calculate the number of each color expected in Table 1 and record under “Number
Expected.”
HINT: You must count all the colors and add the total number of M & M’s before you can
calculate the number expected of each color.
Table 1
Color of Candy
Number Observed (o)
Percentage
Expected
Number Expected (e)
(Total num ber of all
pieces of candy x
percentage expected)
Total # Candies =
4. Record the numbers expected and the numbers observed in Table 2.
5. Complete the calculations and determine the Chi Square value.
Table 2
Classes
(Colors)
Expected
(e)
Observed
(o)
o – e
(o – e) 2
(o – e) 2
e
141
Degrees of freedom =
(number of classes – 1)
∑=
Analysis Questions:
1. What is the X2 value for your data?
2. What is the critical value for your data?
3. Given the critical value (p = 0.05), is the null hypothesis accepted or rejected? Explain why or
why not.
4. If the null hypothesis is rejected, propose an alternative hypothesis.
5. How do you think the results would differ if you were to have used a “fun size” bag of M & M’s? A
1 pound bag? Explain.
6. Suppose you were to obtain a Chi-square value of 7.82 or greater in your data analysis with 2
degrees of freedom. What would this indicate?
142
AP Lab 7: Genetics of Organisms Using Corn
OVERVIEW
In this lab you will use living organisms to do genetic crosses. You will learn how to collect and
manipulate the organisms, collect data from F1 and F2 generations, and analyze the results from a
monohybrid, dihybrid, or sex-linked cross. The procedures that follow apply to fruit flies; your
teacher may substitute other procedures using other organisms.
OBJECTIVES
•
•
•
Investigate the independent assortment of two genes and determine whether the two
genes are autosomal or sex-linked using a multi-generation experiment.
Calculate Chi Square
Analyze the data from your genetic crosses using chi-square analysis techniques.
PROCEDURE
Part A: Monohybrid Cross
In corn plants, starchy kernels (T) are dominant over sweet kernels (t). You will analyze the
results of the F2 offspring between a cross of a homozygous starchy corn plant and a
homozygous sweet corn plant.
1. Count 20 rows of corn and record your data in table1. These are your observed values
2. Calculate the expected phenotypic ratio:
Parental Genotypes (P generation):
x
First Filial Generation Genotypes (F1 generation):
Expected F2 generation Phenotypic Ratio:
Expected F2 generation Genotypic Ratio:
Punnett Square Showing the F2 generation
3. Calculate Chi Square by filling in table 2.
modified from College Board Lab Manual
143
Table 1: Number of Each Phenotype Observed in Each Row of the corn
Row
Phenotype 1:
Phenotype 2:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
TOTAL
(individual)
TOTAL
(class data)
Table 2: Determining Chi Square Value Using Class Data
Phenotype
Expected (e) Observed (o)
o–e
Degrees of freedom =
(number of phenotypes – 1
(o – e)2
(o – e)2
e
∑=
modified from College Board Lab Manual
144
Chi Square Critical Values Table
Degrees of Freedom (df)
Probability
1
2
3
4
5
0.05
3.84
5.99
7.82
9.49
11.1
0.01
6.64
9.21
11.3
13.2
15.1
0.001
10.8
13.8
16.3
18.5
20.5
Part B: Dihybrid Cross
In corn plants, starchy kernels (T) are dominant over sweet kernels (t) and purple kernels (R) are
dominant over yellow kernels (r). You will analyze the F2 generation results of a cross between a
homozygous starchy, homozygous purple corn plant and a homozygous sweet, homozygous
yellow corn plant.
1. Count 20 rows of corn and record your data in table 3. These are your observed values
2. Calculate the expected phenotypic ratio:
Parental Genotypes (P generation):
x
First Filial Generation Genotypes (F1 generation):
3. Show your work in the punnett square below for the F2 generation.
Expected F2 generation Phenotypic Ratio:
modified from College Board Lab Manual
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Data Table 3: Number of Each Phenotype Observed in Each Row
Row
Phenotype 1:
Phenotype 2:
Phenotype 3:
Phenotype 4:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
TOTAL
(individual)
TOTAL
(class data)
Data Table 4: Determining Chi Square Value Using Class Data
Phenotype
Expected (e) Observed (o)
o–e
Degrees of freedom =
(number of phenotypes – 1)
(o – e)2
(o – e)2
e
∑=
modified from College Board Lab Manual
146
Analysis:
Part I: Monohybrid
1. What is your null hypothesis?
2. Refer to the Chi Square Critical Values Table. With the available data, is the null
hypothesis accepted or rejected? Explain your answer.
Part II: Dihybrid
3. Refer to the Chi Square Critical Values Table. With the available data, is the null
hypothesis accepted or rejected?
4. Why would you want to perform a Chi-Square?
modified from College Board Lab Manual
147
148
149
150
151
152
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