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This manual is required of students. The experiments in the following pages will be recorded in your lab book that will be graded for accuracy, completeness, reproducibility and readability.

University of Minnesota, Crookston 2900 University Avenue

218-281-8380

TABLE OF CONTENTS

LABORATORY TOPIC PAGE

The Microscope ----------------------------------------------------------------------- - 1

Cell Structure and Function------------------------------------------------------------- 9

The Microbes around Us------------------------------------------------------------- 19

Ecology (Field Trip) -------------------------------------------------------------------- 27

Diffusion, Osmosis and Dialysis -------------------------------------------------------- 34

Enzymes and Catalysts -------------------------------------------------------------------- 43 Respiration Fermentation ------------------------------------------------------------ 50

Photosynthesis, oxygen production------------------------------------------------- 56

Mitosis ------------------------------------------------------------------------------------ 73

Corn and Human Genetics ----------------------------------------------------------- 80

Populations and Sampling---------- -------------------------------------------------- 99

Microbiology of Milk ------------------------------------------------------------------ 110

DNA Extraction ----------------------------------------------------------------------- 115

Gel Electrophoresis and Restriction Digest ---------------------------------------- 120

Human Physiology----------------------------------------------------------------------- 126

Vertebrate Anatomy and Physiology------------------------------------------------- 151

Animal Behavior and Plant Responsiveness ----------------------------------------- 155

Appendix A: Lab Equipment and Use ----------------------------------------------- 167

Appendix B: DNA goes to the Races---------------------------------------------- 172

Appendix C: DNA Scissors -------------------------------------------------------- 175

Appendix C: DNA Scissors -------------------------------------------------------- 178

THE MICROSCOPE

DESIRED LEARNER OUTCOMES:

Upon completion of this lab the student will be able to:

1. Identify the parts of compound and dissecting microscopes and state the function of each.

2. Compare and contrast the compound and the dissecting microscopes.

3. Describe and demonstrate the correct way to:

a. carry a microscope. clean the lenses.

b. focus with each objective. c. observe the available specimen with each

microscope. d. estimate the size of objects.

4. Convert from one metric length unit to another metric length unit.

KEY TERMS:

Body tube Micrometer Compound microscope Monocular/binocular lenses Condenser Nosepiece Dissecting microscope Objective lenses Field of view Reflection light source Fine and coarse adjustment Stage Iris and disc diaphragm Transmission light source Magnification Zoom knob

INTRODUCTION: The purpose of a microscope is to see inside organisms and cells, to see what is invisible to the

naked eye. The eye can be aided with a simple hand lens or magnifying glass or with a

compound microscope, which has two lenses at opposite ends of a tube. The ocular lens is the

one nearest the eye, and the objective lens is nearest the object or specimen.

Microbes (organisms too small to be seen with the naked eye) are less than a millimeter (1 mm =

1/1000 of a meter) in size and are usually measured in micrometers (1 pm = 1/1000 of a

millimeter). They could range in size from 1 to a 1000 µm. You should be able to see the

organism with your naked eye if it were about 100 µm or larger, but you wouldn't be able to see

much detail. Plant and animal cells are typically 5 to 40 µm in size, so they are not visible

unless magnified under the microscope.

For even smaller dimensions, such as parts of cells, the nanometer (nm) and the angstrom (A)

are appropriate. The nanometer is one one-thousandth of a micrometer (0.001 gm) and the

angstrom is one ten-thousandth of a micrometer (0.0001 pm) or one-tenth of a nanometer (0.1

nm). 1

The following table compares the sizes of these frequently used metric units of

length: TABLE I: Equivalent Metric Lengths

UNIT ABBREVIATION EQUIVALENT METRIC

Magnification is a measure of how big an object looks to your eye compared to "life size". Life size images are specified as 1. Magnification is usually written by a number followed by "X" which means "times life size". For example, 10X means 10 times life size. A magnifying glass magnified 3X to 10X, a dissecting microscope about 10X to 45X and a standard light microscope about 40X to 450X.

A microscope is a precision instrument and should be handled accordingly. It may appear to be indestructible, but in reality the slightest jar may damage its working parts. Handle the microscope carefully at all times and carry it by the arm.

MATERIALS NEEDED:

Dissecting microscope (one for each pair of students) Clean slides Cover slips Monocular standard microscope (one for each student) Clear plastic metric rulers Meter stick Prepared slides:

Flea Letter "e" slides Frog blood smears Human blood smears

Possible specimens: Insects Flowers Fungus Leaves pioneer species

2

Carry the dissecting microscope by its arm, place it on the lab desk and locate the following parts:

1. Illumination Control Switches:

a. Reflection: Light is directed down onto the specimen from which it is reflected up through the microscope.

b. Transmission:

Light travels up through the specimen. c. Reflection-Transmission:

Combines (a) and (b) the best possible illumination for a particular sample.

2. Oculars: NOTE: This is a binocular microscope and is really two microscopes so that objects can be viewed in binocular vision in three dimensions.

Move the oculars together or apart until you feel comfortable looking into the microscope. The two fields of view should overlap completely so that you see a single circle of light.

3

3. "Zoom" Knob:

Controls the objective lenses. It is located next to the oculars on the body of the microscope. This knob changes the magnification.

4. Focus Adjustment Knob:

Turn until specimen is in sharp focus.

PROCEDURE:

1. Place a clear plastic metric ruler on the stage of your dissecting microscope. 2. Turn the zoom knob until the indicator mark is on the lowest magnification.

3. Rotate the illumination knob to get the best illumination and turn the focus knob until the ruler is in sharp focus.

a. How big is the field of view (the area you see looking into the microscope) under the lowest

possible

magnification? __________ mm and in _____________ µm

a. _______ b. Turn the zoom knob to the highest magnification.

How big is the field of view? ______________ mm ____________ µm

c. Now look at your finger under the lowest magnification.

Is it right side up or upside down?

d. Move your finger to the right; which way does the image move?

Sketch two of the specimens provided for viewing using the dissecting microscope.

ALWAYS place the specimen on a glass slide or glass dish before viewing.

NEVER put specimens directly on the microscope stage.

If the organisms are in a liquid culture, make a wet mount as described in the section "Preparing a Wet Mount", or place a few drops in a depression slide or small dish. (SEE PAGE )

pm

_______µm _______mag. _______µm _______mag.

4

PROCEDURE:

Pick up your microscope by its arm, keeping it upright, and supporting it underneath with your free hand. Set it gently on your lab desk and locate the following parts. Label Figure 2 as you study each part.

1. Ocular lens:

Magnifies ten times (10X). This lens is often unattached, and thus can fall out unless the microscope is kept upright.

2. Objective lens:

Magnifies the object by the factor marked on the particular lens. Low power is considered the 1 OX lens and high power is the 40X (or 43X or 45X) lens. A very low power scan objective lens (5X) is present on some microscopes.

NOTE: Lens paper may be used to clean dirty lenses.

3. Nosepiece:

The revolving part to which the objectives are attached. It must be firmly clicked into position when the objective is changed. Rough treatment can cause it to snap off.

4. Body Tube:

Joins the nosepiece to the ocular lens.

5

5. Stage:

Supports the slide that is held onto it by stage clips, and has a hole so that light can shine up through the specimen. Always center the specimen over this hole.

6. Coarse Adjustment Focus Knob:

Moves the body tube or stage up and down, depending on the design of the microscope, to approximately the right position so that the specimen is in focus. This knob should only be used with low power or the scan objective.

7. Fine Adjustment Focus Knob: Moves the body tube or stage up and down to precisely the correct position so that the specimen is perfectly in focus. Use it to achieve fine focus with the low power objective and for all focusing with high power.

8. Light Source: Usually a small electric light beneath the stage that is controlled by a switch.

9. Iris or Disc Diaphragm:

Regulates how much light and lamp heat go through the specimen. Iris diaphragm is controlled by a lever that is moved back and forth. The disc diaphragm is a rotating plate with varying size holes.

10. Condenser:

A lens located above the diaphragm, which concentrates the light before it passes through the specimen.

Microscopy Skills:

PROCEDURE:

A. Measurement of real organism using ocular micrometer: It Is often necessary to measure the size of organisms or parts of organisms being observed under the microscope. Micrometers are used to measure objects seen through the microscope. On your compound microscopes the micrometers are located in your ocular or eye piece. Thus objects can be easily measured by comparing their size to the micrometers of the microscopes. Most microscopic measurements are made in terms of microns (1000 microns equals 1 mm). The width between the small marks is 25 microns (µm) with the 4X objective, 10 microns (µm) with the 10X objective, 2.5 microns (µm) with the 40X objective, 1 microns (µm) with the 100X objective.

B. Observation of the Flea:

1. A study of the effect of increased magnification. Obtain a slide of a flea and place it on the stage of the microscope. Observe using 1OX objective. When proper illumination of the field has been achieved, slowly raise the body tube or lower the stage until the flea comes into focus. It is helpful to move the slide slowly back and forth until you find the flea.

2. Once the flea comes into the field of view move the adjustment knobs to bring the flea into clear focus. 6

3. Now swing the high-power objective into position. Most modem microscopes are parfocal -- that is, once the image is brought into sharp focus under low power, will remain in focus when the high-power objective is turned into position. Continue to switch from low power to high power. Sketch the flea under low power and high power.

_______µm _______mag. _______µm _______mag.

Complete Table I by checking circling the correct response (increases or decreases) when you go from low to high power.

TABLE II: Effects of Increased Magnification When Going From Low Power to

High Power:

MICROSCOPE CHARACTERISTICS

FIELD OF VIEW (SIZE) [INCREASES or DECREASES]

AMOUNT OF LIGHT [INCREASES or DECREASES]

WORKING DISTANCE

(DISTANCE FROM TIP OF OBJECTIVE LENS TO THE STAGE) [INCREASES or DECREASES]

A. Observation of the Letter ("e"): Find a prepared slide with a letter mounted in it. Place the slide (oriented so that you can read the letter "e" properly) on the stage and position the letter over the hole in the stage. Make the proper light and focusing adjustments when looking at it under low pgwer until you have a sharp, clear image.

1. Does the object appear normal or upside down?______________________________________ 2. Move the slide a tiny bit away from you while observing the object in the

microscope.

Does it move away from you or toward you? ______________

3. Move the slide a tiny bit to the left while observing the object in the microscope.

Does it move to the left or to the right? ________________

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The ocular lens on your microscope gives 10X magnification of the image made by the objective lens. The objective lens magnifies the object 4X or more, so the total magnification is the magnification of the ocular lens times the magnification of the objective lens. So if the ocular lens is 10X and the objective lens is 4X the total magnification would be 40X. Fill in the magnification of the objective lenses on your microscope and the total magnification in Table II.

TABLE III: Computing Magnification

F. Observation of Human Blood Cells: Obtain a prepared slide of human blood and a frog blood cell. Both red blood cells and white blood cells are present. (The white blood cells are usually stained purple.) Sketch and measure the blood cells under high power.

Red blood cell White blood cell

x

_______µm _______mag. _______µm _______mag.

Frog blood cell

_______µm _______mag.

8

CELL STRUCTURE AND FUNCTION



1. Differentiate between the characteristics of Prokaryotic and Eukaryotic cells. 2. Make temporary wet mounts of organisms and cells where appropriate. 3. Draw, identify and give the function of the cell parts of the Prokaryotic cell

examples available. 4. Draw, identify and give the functions of the cell parts for the following Eukaryotic cells:

a. Specialized protists. b. Plant cells. c. Animal cells.

5. Estimate the size of various Prokaryotic and Eukaryotic cells. 6. Determine which cells or organisms studied were more specialized.

KEY TERMS:

Amyloplast Nucleolus Cell wall Nucleus Chloroplast Plasma membrane Cilia Prokaryote Cytoplasm Vacuole Eukaryote

INTRODUCTION:

All living things are composed of one cell or a number of cells. Whereas the original use of the term "cell" by Hooke applied only to the cell wall of cork, the term now has a broader meaning. In living cells, the living matter has been called protoplasm which, in 1868, Huxley termed "the physical basis of life." Every cell (with some highly specialized exceptions) has DNA in a nuclear area or in a nucleus, cytoplasm, and a plasma membrane. All cells also contain ribosomes, the structures that carry out protein synthesis. There are two major types of cells. Prokaryotic cells are small (0.5 to 5 mm) and less complex, usually exist as unicellular organisms, and have limited capabilities compared with Eukarvotic cells. Higher organisms, namely plants, animals, and fungi, are made up of highly integrated aggregations of large (10 to 50mm), specialized eukaryotic cells. Some sophisticated organisms, known as protists, consist of a single eukaryotic cell. These cells are large (5 to 200 mm or more), have a membrane bound nucleus, and have membrane bound organelles.

9

MATERIALS NEEDED: Standard microscopes (1/pair) Culture of yeast Prepared slides of bacterial types Culture of blue-green algae Cultures of protozoans:

Amoeba Euglena Paramecium

Prepared slides of mixed protozoans "Proto-Slo" Microscope slides and cover slips Onions Elodea Potatoes Aceto-carmine stain IKI Methylene blue stain Knife, razor blades Forceps

Preparing a Wet Mount:

PROCEDURE:

Wet mounts are used to study fresh, living material. They can be used only for a little while because they will soon dry out, but they are useful for observing qualities such as color, movement, or behavior that cannot be observed on dead, stained material. Follow the steps below to prepare your wet mount. 1. Obtain a slide, coverslip, and teasing needle from your instructor.

2. Place a drop of the culture on your slide.

3. Touch the coverslip to one edge of the drop, and gently lower it with the teasing needle, as shown.

If you have been careful, the slide will not have any bubbles. If not, you will see them as circles of various sizes with very dark edges when you look at your sample.

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Use lower or higher magnification to view your organisms, depending on their size.

FIGURE 2. Preparing a Wet Mount

Mount your living specimen or culture as shown to avoid

trapping air bubbles under the coverslip

1. Prokaryotic Cells:

Cell Darts to note: Cell Wall Flagella

PROCEDURE:

A. Bacteria:

1. Prepare a wet mount of one drop of the yeast culture.

2. Focus on the slide under low power and then switch to high power.

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a. Draw a sketch of the organisms you see.

b. What shape are the organisms you see?

c. Can you see any structures within the cells?

d. Are the organisms stationary or are they moving?

3. Obtain a prepared slide of bacteria and view the three main shapes of bacteria using high power.

a. What shapes can you see?

B. Blue-Green Algae - Oscillatoria:

1. Prepare a wet mount of a culture of filamentous Oscillatoria, a common pond inhabitant.

2. Focus on the slide under low power and then switch to high power.

a. What is the shape of the individual cells which make up the filament?

b. Can you see any organelles within the cells?

c. Does Oscillatoria contain chloroplasts?

d. Does Oscillatoria contain chlorophyll?

3. Notice the characteristics oscillations that give the organism its name and draw what you see.

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II. Specialized Protists:

Euglena Amoeba Paramecium

Protists are unicellular organisms not clearly related to a group of multicellular organisms. Cell parts to note: Cilia

Nucleus Cytoplasm Vacuole Flagellum

PROCEDURE:

1. Select one of the protozoa types available.

2. With a dropper take a sample from the lower surface of the container and place one drop on a slide. Make a wet mount and observe under low power. (If the organism you have chosen is very active, it may be necessary to add "Proto-Slo.")

3. Observe, sketch and label the protozoan you chose.

a. Can you see any structures within the cells?

b. Are the organisms stationary or mobile?

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III. Plant Cells (Eukarvotic):

MATERIALS NEEDED:

Onion Elodea

A. Onion Epidermis

Cell parts to note: Nucleus Nucleolus Cell wall Cytoplasm

PROCEDURE:

1. Break or cut off with a razor blade a piece of a single layer of onion from within the onion (the outermost layer may contain only dead cells).

2. Snap the piece in half and then use forceps to peel off a bit of tissue-like

transparent epidermis from the inner layer.

3. Mount the epidermis in tap water on a slide so that it is flat and not doubled over on itself. Add a coverslip.

4. View the slide under low power. -- What is the shape of the cells?

-- What is the thick layer that surrounds each of the cells?

- What does the clear part of each cell contain?

14

If you are looking at a confusing mix of overlapping cell parts, you probably do not have a good piece of epidermis. Ask your instructor for help. Some cells will not have a central vacuole, if they are young. Also, the plasma membrane, which lines the inner surface of the cell wall, is too thin to be seen in the light microscope, but you can see exactly where it must be.

5. Raise the coverslip and add a drop of 45% aceto-carmine dye; this will stain the cytoplasm and nucleus of your cells pink or red.

6. Distinguish between the vacuoles and the cytoplasm. The cytoplasm is within the cell membrane and is "living," whereas the contents of the vacuoles are often inert storage materials. 7. Locate the nucleus, which now should be red. It is a true, eukaryotic nucleus and is surrounded by a nuclear envelope.

-- What kinds of molecules are located in the nucleus?

-- What is the function of the nucleus?

8. Make a sketch of the onion cell.

B. Elodea Leaf:

Cell parts to note: Cell wall Central vacuole Cytoplasm Chloroplast Plasma membrane

PROCEDURE:

1. Obtain one of the leaves from the tip of a vigorous Elodea stem and make a wet mount of the entire leaf.

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2. Focus the leaf under low power and switch to high power. With the fine focus adjustment knob raise and lower the plane of focus, focusing at various depths of the leaf. An apparent shift in position of the cells indicates the passing vertically from one layer of cells to another.

How many cells in thickness is the elodea leaf?

3. Choose a cell in the first three rows from the margin to study. Distinguish the following parts of the cell: cell wall, a very thin layer enclosing the protoplast and vacuole; the cytoplasm, a thin layer just inside of the cell wall including certain obvious cell contents; central vacuole, the portion of the cell within the cytoplasm containing water and materials in solution. The cytoplasm contains two important and rather conspicuous parts: the numerous green bodies or chloroplasts, which are confined to the periphery of the cell, and the nucleus, an opaque oval body seen in many cells. The nucleus is usually most easily seen in the outer pointed cells near the tip of the leaf. The plasma membrane, present as the outer surface of the protoplast, is not visible because it is so thin and in direct contact with the cell wall. The plasma membrane is involved in the passage of cell materials into and out of the living cell.

The movement of the cytoplasm is usually evident because the chloroplasts in the cytoplasm are carried along like wooden discs in a current of water. Such movement is known as cvclosis or streaming and probably occurs in all living cells under suitable conditions. In which direction does the streaming occur?

4. Draw one of the cells, at least three inches in length. With arrows, indicate the directions of cyclosis and label the cell parts located in Step 3.

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IV. Animal Cell (Eukaryotic) :

A. Epithelium

Cell parts to note: Cell membrane Nucleus

PROCEDURE:

The first cell you will observe will be one that is part of a tissue or one of a group of cells that are performing a similar specific function. 1. Place a drop of stain (methylene blue) on a slide. Gently scrape the side of your cheek with

the flattened end of a toothpick. Mix the cells in the drop of stain. 2. Cover the mixture with a cover slip and observe under the microscope. 3. After locating the cells it will be necessary to go to high power.

4. Sketch the epithelial cell and observable contents.

It appears that epithelial cells are lost very readily. How does this affect you?

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V. QUESTIONS:

1. What structures are found in some plant cells but not in the animal cells?

2. What is a specialized cell? What cell type observed today was most specialized?

3. What evidence from today's work do you have that some of the plant cells were alive?

4. How would you describe the shape of the chloroplast?

5. What do we call the internal circulation of a cell?

6. Why was it important that all cell slides made today were wet mounts?

7. Do cells ever have more than one nucleus? If yes, example? 8. Were there any cells in which you were unable to locate nuclei? Do you think they

had any?

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THE MICROBES AROUND US


Upon completion of this lab the student will be able to

1. Demonstrate the occurrence of air borne bacteria in the environment by culturing samples from various locations.

2. Demonstrate the effectiveness of various mouth antiseptics on student mouth

bacterial cultures.

3. Demonstrate the effectiveness of various disinfectants on student mouth bacterial

cultures. 4. Demonstrate the effectiveness of various antibiotics on the inhibition of bacteria in

student mouth cultures.

KEY TERMS:

Agar medium

Antibiotic discs Antiseptics

Bacteria

Bacterial colony characteristics

Disinfectants

Incubator Sterile discs

Zone of Inhibition

INTRODUCTION:

Bacteria are microscopic organisms that are usually thought of as being bad or pathogenic. In

fact, most of these organisms are neither good nor bad and there are more beneficial bacteria

than harmful but they are not usually noticed.

The bacteria have the same needs to continue life, as do other organisms, that of water, food

and a suitable environment. The food needs of bacteria are often supplied to laboratory grown

organisms in the form of a medium. The medium usually contains nutrients and sufficient water for

rapid growth. It is the lack of food, water or suitable environment that prevents the bacterial

numbers from becoming very high in every part of our environment.

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Morphology:

Before lab, describe the following:

Description

A. Diplococcus

B.

C.

D.

E.

F.

G.

H.

I.

J.

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I. AIR BORNE BACTERIA: (Work in pairs)

The air usually has some microorganisms floating or moving in it but if they do not end up

on a suitable environment they go unnoticed. You will expose a petri dish with agar in it

to the air for 15 minutes and see what happens.

MATERIALS NEEDED:

1 petri dish with nutrient agar

PROCEDURE:

1. Select an area in the building that you want to test.

2. Open the dish in the area to be tested keeping the cover in the area free from

contamination.

3. After 15 minutes, replace the cover and place the dish in the 37° C incubator.

Location:

In describing the colonies use the following guides:

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Results:

Environmental Sampling Site Growth?

H. ORGANISMS OF THE BODY ENVIRONMENT:

The human body is constantly being exposed to many bacteria, but it only becomes

infected after the organisms gain entry to the moisture and nutrients under the body

coverings.

MATERIALS NEEDED:

2 agar plates per pair of students

2 sterile swabs

Bar of soap Wax pencils

PROCEDURE:

1. Turn the plates upside down and divide the dish into quarters with a marking pen.

2. Label the quarters #1, #2, #3, and #4.

3. Touch quarter #1 with your left thumb.

4. Touch quarter #2 with your left thumb after washing with water.

5. Touch quarter #3 with your left thumb after washing with water and soap.

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6. Leave the final quarter untouched (control).

7. Incubate the dish at 32° C in the incubator for several days.

Results:

Description of growth

Unwashed left thumb

Washed left thumb with water

Washed left thumb with water and soap

No touch

OUESTIONS:

1. How do you explain the hand-washing results?

2. Would you expect any bacteria to remain if you washed your hands a second time?

3. How do you explain the abundance of bacteria in your mouth? What is this source

of nutrients?

III. ANTISEPTICS AND DISINFECTANTS:

Many chemicals, both inorganic and organic, are toxic to microorganisms, and in man's

struggle to control microbial deterioration and disease thousands of these chemicals have

been discovered. The chemical agents either simply inhibit cell activities and growth or

are lethal and kill the microorganisms. Inhibitory chemicals are usually termed

antiseptics; and lethal, disinfectants. This exercise shows you how to compare the action of

antiseptics and disinfectants. 23

MATERIALS NEEDED:

1 agar plates per pair of students

Sterile blank discs

Antiseptics and disinfectants

PROCEDURE:

1. Turn the plate upside down and divide the plate bottom into sixths using a marking

pen.

2. Label # 1, #2, #3. #4, #5, and #6.

3. Using a cotton tip, swab a prepared broth culture of bacteria onto the agar plates.

4. Using sterile forceps, place the sterile discs in the provided antiseptics or

disinfectants. Place the disc in the center of each sixth of the seeded plate. Keep

a record of the antiseptics and disinfectants used.

5. Incubate (not inverted) at 37° C until the following lab.

6. Observe the plate and note the areas of inhibition of growth.

QUESTIONS:

1. Does this exercise differentiate between inhibition and killing? Explain.

2. Which antiseptics and disinfectants were effective inhibitors of Mouth bacteria?

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3. Does listerine "kill the germs that cause bad breath?" Explain.

IV. ANTIBIOTICS:

Certain living cells, especially fungi and bacteria, produce chemicals that inhibit the

growth of other organisms. These naturally occurring substances are called antibiotics.

MATERIALS NEEDED:

1 agar plate per pair of students

Antibiotic discs dispenser 2 sterile swabs

PROCEDURE:

1. Using a cotton tip, swab your mouth secretions on the surface of the hardened agar.

2. Using the antibiotic disc dispensers, place the different kinds of antibiotic discs on

the surface of the agar. Keep a record of the discs used. Invert the plates and

incubate for 2 or 3 days.

3. Record your results and collect class results (on class chart).

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Results

QUESTIONS:

1. Which of the antibiotics were effective against mouth bacteria?

2. Discuss the effectiveness of Penicillin.

3. What is an antibiotic?

4. How do antibiotics inhibit bacteria?

26

ECOLOGY (FIELD TRIP)



1. Observe examples of plant succession in several localities.

2. Identify examples of hydrosere species in various stages of succession.

3. Identify examples of dune plant species in various stages of succession.

4. Identify areas of SW-NE slope and determine various adaptations of plant life

suited to these environments.

5. Compare and contrast the environmental conditions among dune, grass and wooded

local areas.

KEY TERMS:

Ecology

Hydrosere succession

Primary succession

Secondary succession

Stability-diversity principle

SW-NE slope influence

INTRODUCTION:

Plant ecology is the study of the relationships that exist between the living plant and its

environment. Modern plant ecology is a composite science consisting of several interrelated

sciences. In order to understand fully the ecological relationships that exist between plants and

their environment, it becomes necessary to have a fair knowledge of many areas of science

including taxonomy, geology, geography, plant anatomy and physiology.

The effect on growth and distribution of plants that the living and nonliving elements of the

environment produce are better understood in the field rather than in the laboratory. Of course,

the nature of any ecological field work will depend largely upon the season and the geographical

location.

All plants bring about a change in their environment which may result in the dying off of some

species and the influx of new species. Such changes are gradual and occur over long periods of

time; they are hardly perceptible to the untrained observer.

This long-term sequence in plant communities is known as a succession. One example of a plant

succession would be a barren area free of plants which is gradually becoming inhabited by

certain migrating species and then eventually joined or replaced by other plants in sequential

fashion. The final stage in a plant succession is regarded as the climax. This stage does not

give way to a more advanced community and it is therefore relatively stable.

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The purpose of this exercise is to observe several examples of plant successions which may be

recognized in many localities.

I. Hydrosere Aquatic Succession:

Plant successions beginning in a water habitat are called hydrarch and the series of stages

that evolve are collectively called hydrosere. The hydrosere begins when a body of water

is formed. This may occur naturally when landslides block a stream or artificially when

man builds a dam. In any case, the newly formed body of water does not remain free

of vegetation for any period of time and the pioneer submerged aquatic plants may soon

be recognized. These are usually replaced by floating plants which eventually become

so thick that the submerged forms are deprived of light and are soon replaced. Many

other types of plants may develop along the banks which will undergo a succession of

their own.

List and examine the types of plants and animals that are present from the pond edge to the

open water.

Sketch a cross-section of a pond environment beginning with the open water and extending to the grass of an adjacent field.

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QUESTIONS:

1. What region of the pond shows the greatest productivity? How do you know?

2. What would be an example of a food chain in this system?

3. Why is it better to think of the relationship as a food web?

II. Influences of Slope:

The vegetation of an area can be significantly different over a short distance because of

the slope of the land. The slopes in North America that exhibit this best are SW - NE.

Careful observation of these two slopes will be revealing.

A. Characteristics of SW vegetation:

1.

2.

3.

4.

Air Temperature___ Relative Humidity Soil Temperature

B. Characteristics of NE vegetation:

1.

2.

3.

4.

Air Temperature Relative Humidity Soil Temperature

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OUESTIONS:

1. What kind of reasons would you give for the lack of "hairy leafed" plants on the

NE slope?

2. How does "stability-diversity" enter into the slope concept?

III. Sand Dune Succession:

Examine designated areas in the sand dunes and list the organisms associated with the

three listed stages of succession.

Pioneers:

Serial Species:

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IV. Environmental Conditions of Three Adjacent Areas:

The effect of vegetation on an area is sometimes difficult to separate the effect of the

environment on the vegetation. The following experiment may produce evidence that will

clarify this apparent conflict.

PROCEDURE:

Three areas will be sampled: an open dune area where secondary succession is taking

place; a grass covered dune area in a later stage of succession; and the third area will be a

wooded area, a later stage of succession.

In each area the following parameters will be determined; at a height of 1 meter and

surface air temperature, relative humidity; light intensity, and soil temperature.

32

DIFFUSION, OSMOSIS AND DIALYSIS



1. Demonstrate diffusion of a liquid in a liquid.

2. Demonstrate diffusion of solids in a solid and calculate the rates of diffusion. 3. Demonstrate osmosis and dialysis in a model cell with solutions separated by a

selectively permeable membrane. 4. Observe osmosis microscopically in a living cell.

KEY TERMS:

Concentration Lysis

Dialysis Osmosis

Diffusion Plasmolysis

Entropy Rate of diffusion

Equilibrium Selectively permeable membrane Flaccid Solute

Hypertonic Solvent

Hypotonic Turgid Isotonic

INTRODUCTION:

Cell membranes are selectively permeable in that some substances may pass through them and

others may not. Molecules depending on their size and functional groups, may pass between

the membrane components, through channels surrounded by protein, or they may be escorted

across the lipid bilayer by specific transport proteins called carriers. The plasma membrane is

thus not only a physical container for the cell but is an important regulator of the cell contents.

If you put a high concentration of a solid substance, the solute, into a liquid, the solvent, the

randomly moving particles of the dissolving solute, will spread throughout the liquid. This

spontaneous or "downhill" process is called diffusion and it does not require any import of

energy. As they spread out, the molecules form a concentration or elect no chemical gradient

with the highest concentration near the dissolving solute and the lowest concentration farthest

away. The diffusing molecules will tend to move down the concentration gradient from higher to

lower concentration until they reach equilibrium. Then they will be spread out fairly evenly and

there will be no further net movement. The free energy of the solution will be lower than it was

before and the entropy (disorder) will be higher after diffusion has occurred.

I. Diffusion of a Liquid in a Liquid:

Since liquids are highly sensitive to convection currents, the success of the experiment

depends on not allowing your setup to be disturbed or jiggled once the experiment is

under way. 34

PROCEDURE:

1. Fill a small beaker with water and set it in a protected place. Wait 10 minutes.

2. Add a drop of red dye to the surface of the solution as gently as possible. Try not to disturb the surface, but hold the dropper as close to it as you can.

3. Watch the dye spread through the solution from time to time during the lab period.

At the end of the period complete the data sheet and answer the questions given

below:

QUESTIONS:

1. Is the concentration of the colored part of the solution greater at the beginning or end of the experiment?

2. Is the free energy of the solution greater at the beginning or at the end?

3. Is the entropy of the solution greater at the beginning or at the end?

4. What difference would you have seen if you had originally set up the beaker in the

refrigerator?

H. Diffusion in a Solid:

Background Information:

You are to determine the rate of diffusion of two substances that have different molecular weights: methylene blue (mol. wt. 320) and potassium permanganate (mol. wt. 158).

Both substances are soluble in water and readily diffuse through an agar gel that is about

98% water. As you might expect, the larger the ion is, the slower it will diffuse through

the agar. Which should move the slowest?

35

MATERIALS NEEDED:

Forceps

Petri dish

Crystals of:

Methylene blue

Potassium permanganate

PROCEDURE:

1. Place equal-sized crystals of potassium permanganate and methylene blue about 5

cm apart on an agar plate. Gently press the crystals into the agar with forceps to

assure good contact. Record the time.__________

2. After 1 hour, record the diameter of the colored circles.

-- Methylene blue ________________mm

-- Potassium permanganate________________mm

-- What relationship seems to exist between molecular weight and the rate of

diffusion?

I II. Movement of Substances Through Membranes:

Background Information:

When diffusion occurs across a plasma membrane, solutes may pass into or out of the cell

in the process called dialysis. Each solute will move from the side where its

concentration is higher to the side with a lower concentration, because the direction of

movement is always down the concentration gradient. Its movement will be independent of

the movement of the other solutes.

In the special case of diffusion in which the substance diffusing across the membrane is

water, the process is called osmosis. During osmosis water will flow from the side of

the membrane with the most water and lowest solute concentration, to the side with the

lowest water concentration and higher solute concentration. Therefore, osmosis is the

passive movement of water across a differentially permeable membrane in response to

solute concentration gradients (and/or pressure gradients). Osmotic movements across

cell membranes are affected by tonicity - that is, the relative concentrations of solutes in

two fluids. When solute concentrations are equal in both fluids, or isotonic, there is no

net osmotic movement of water in either direction. When the solute concentrations are

36

not equal, one fluid is hypotonic (has less solutes) and the other is hypertonic (has more solutes). Water molecules tend to move from a hypotonic fluid (more water-less solutes)

to a hypertonic fluid (less water-more solutes). Water will always flow from the region

of higher water concentration to the region of lower water concentration. Pure water has

the highest possible water concentration: 100%. The more solute a solution contains,

the less water it will have.

Osmosis in a Model Cell

You will use the following system to become familiar with osmosis: Two solutions will be

separated by a selectively permeable membrane that is perm i to water, glucose, and

small molecules, but impermeable to larger molecules. Water will diffuse from one to the

other in the process of osmosis, and small molecules will diffuse across the

membrane in the process of dialysis. This setup is shown in figure 2.

A selectively permeable membrane filled with solution A represents the cell. It is placed in

solution B to start the experiment.

PROCEDURE:

1. Obtain a short piece of dialysis tubing (12-15 cm), and soften it by soaking it in distilled water.

2. Fold one end over and tie it tightly with a piece of thin string or strong thread. 3. Fill the bag with solution A, a liquid "meal" containing:

25 % glucose

0.5% egg albumin

1 % starch

37

4. Fold over the top, squeeze out all the air, and tie tightly. The bag should be limp.

5. Weigh the bag. Record the weight. ___________grams.

6. Place the bag in a small beaker and add enough of solution B (distilled water

containing a very small amount of iodine) to cover it. Let it stand for an hour or

so while you go on with other experiments.

7. Test the "Meal" mixture with Iodine and "Test Tape" and record results on the data

sheet.

8. Weigh the bag at the end of the experiment. _____________grams.

9. Test the solution "B" for glucose and starch and record.

QUESTIONS:

1. Did the bag gain or lose weight? What caused the weight gain or

loss?

2. Which solution was hypertonic (A or B)?

3. Which solution gained water (A or B)?

4. Which substance inside the membrane reacted with the iodine?

5. Which substance(s) moved out through the membrane?

6. If you swallowed some of solution A, which substance(s) could be absorbed directly by the cells of your gut, assuming that they had the same permeability as the

dialysis tubing?

7. Which substance(s) would have to be further digested?

8. If your dialysis bag was flaccid or limp at the beginning of the experiment, at the

end was it very flaccid, flaccid, turgid, or very turgid?

38

9. Which of the following statements best describes the situation at equilibrium if you

let the system stand for a long time?

a. No molecules move across the membrane.

b. All molecules cross the membrane equally often in either direction.

c. Molecules to which the membrane is permeable cross equally often in either

direction.

d. Only water molecules cross the membrane equally often in either direction.

IV. Osmosis in a Living Cell:

To watch osmosis across a living membrane, see what happens when a plant cell comes in

contact with a hypertonic solution. Note that the plant is kept in an aquarium that

contains fresh water that is hypotonic to the plant cell. The cell swells and becomes r

tu gid but is prevented from undergoing lyaig (bursting) because its rigid cellulose walls

exert wall pressure on the cell contents. Many animal cells have no such protection

against lysis and can exist only in an isotonic solution.

Figure 3. Elodea Cell

39

PROCEDURE:

1. Make a wet mount of an Elodea leaf and focus on it under high power.

2. Add a drop or two of 10% NaCl solution to the edge of the coverslip.

3. Touch a piece of tissue to the opposite side of the coverslip to pull the solution

through and observe.

The cell is now (turgid, flaccid)?

The cell has undergone (lysis/plasmolysis/death)?

4. Now add a few drops of distilled water to the edge of the coverslip and pull it

through with tissue while watching the cell.

Is the process you have observed reversible?

DATA SHEET:

40

II. Size of spot from diffused solids after 1 hour.

41

41

IV. Initial bag wt.

Final (1 hr) bag wt.

Describe the color of solution "B" at the beginning of the experiment.

Explain.

42

ENZYMES AND CATALYSTS



1. Compare the rates of the reactions of organic and inorganic catalysts for the

breakdown of hydrogen peroxide.

2. Compare the various sources and effects of concentration of catalase on the rates of the breakdown of hydrogen peroxide.

3. Use the spectrophotometer to measure the absorbance of the enzyme-mediated

breakdown of catechol to compare the effects of:

a. reaction rate over time.

b. various enzyme concentrations.

c. various substrate concentrations.

d. various temperatures.

e. inhibitors.

KEY TERMS:

Peroxidase

Catalyst

Oxidation/reduction reaction

Enzyme

Enzyme-substrate complex

INTRODUCTION:

Without enzymes, most biochemical reactions would take place at a rate far too slow to keep

pace with the metabolic needs and other life functions of organisms. Enzymes are catalysts that

speed up chemical reactions but are not themselves consumed or changed by the reaction.

The cell's biological catalysts are proteins. These enzymes have a very complex three-

dimensional structure consisting of one or more polypeptide chains folded to form an active site

-- a special area into which the substrate (material to be acted on by the enzyme) will fit.

Changes in temperature, alternations in pH, the addition of ions or molecules, and the presence

of inhibitors all may affect the structure of an enzyme's active site and thus the activity of the

enzyme and the rate of the reaction in which it participates. The rate of an enzymatic reaction

can also be affected by the relative concentrations of enzyme and substrate in the reaction

mixture.

43

During this laboratory period, we will investigate how changes in pH, temperature, substrate

concentration, and enzyme concentration affect the enzymatic activity of catecholase. We will

also observe the action of rennin during the process of cheesemaking.

Investigating the Enzymatic Activity of Peroxidase

During this exercise we will study the activity of the enzyme peroxidase contained in turnips.

This enzyme acts to oxidize organic compounds in the presence of hydrogen peroxide

(H2O2), which is the specific substrate of this enzyme. The biological significance of this

peroxidase in cells is that it helps to remove hydrogen peroxide that is present in all cells

that use oxygen in metabolism (e.g., humans). If left alone the hydrogen peroxide would

build up in our cells and severly damage the cells.

We will be using a spectrophotometer to measure the production of a brown-colored,

oxidized product (tetraguaiacol) as it forms from the colorless, reduced form (guaiacol) in

the presence of hydrogen peroxide and the peroxidase enzyme. This reaction is considered

an oxidation reduction reaction. Guaiacol will be oxidized as it donates hydrogens,

hydrogen peroxide is reduced as it donates an oxygen atom to form water and the oxidized

tetraguaiacol compound.

I. Base experiment

PROCEDURE:

1. Grind about 2 grams of turnip (the peeled, inner portion) in 200 ml of distilled water,

using a blender for 1 minute.

2. Filter the extract through several layers of cheesecloth. This is your turnip peroxidase enzyme! Do not discard this liquid!

3. Number seven test tubes serially; fill each of the test tubes as shown in Table I.

44

TABLE I.

Tube #1 (control tube without H2O2) add the following:

1. 0.1 mL of guaiacol 2. 1.0 mL of turnip peroxidase enzyme

3. 8.9 mL of distilled water; mix well

Tube #2 (substrate without peroxidase enzyme) add the following:

1. 0.1 mL of guaiacol 2. 0.2 mL of 0.1% H2O2 (substrate)

3. 4.7 mL of distilled water

Tube #3 (peroxidase enzyme) add the following:

1. 1.0 mL of turnip peroxidase enzyme

2. 4.0 mL of distilled water

THE SPECTROPHOTOMETER:

A colored solution appears that way because some of the light entering the solution is absorbed

by the colored substance. A clear solution will allow almost all of the light to pass through.

The amount of absorbance can be determined by using a spectrophotometer, which measures

quantitatively what fraction of the light passes through a given solution, and indicates on the

absorbance scale the amount of light absorbed compared to that absorbed by a clear solution.

The darker the solution, the greater its absorbance.

Inside the machine there is a light that shines through a filter (which can be adjusted to control

the color, or wavelength, of light), then through the sample and onto a light-sensitive phototube.

The phototube produces an electric current proportional to the amount of light striking it. The

absorbance meter measures how much light has been blocked by the sample and thereby

prevented from striking the phototube. A clear tube of water or other solvent is the blank and

has zero absorbance. A solution that contains a small amount of a colored substance might show

an absorbance of 0.1, a solution with a moderate amount might show an absorbance of 0.4, and

so forth. In fact, in the lower portion of the absorbance scale, the amount of substance in

solution is directly proportional to the absorbance reading so that a graph of absorbance versus

concentration will give a straight line. This very useful relationship is known as Beer's Law.

When you are ready to read a sample, use figure 4 to follow these steps.

1. Turn on the instrument with the power switch knob. Allow 5 minutes for warm-up time.

2. Adjust to the desired wavelength (in this case use 470 nm). 45

3. With the sample chamber empty and the cover closed, use the power switch knob to set the meter needle to read infinity absorbance. (When the chamber is empty, a shutter

blocks all light from the phototube.)

Always read from the absorbance scale, not the transmission scale.

4. Fill a spectrophotometer tube or cuvette halfway with distilled water (or other solvent

when water is not the solvent in your experiment) to serve as the blank. Wipe it free of

moisture or fingerprints with a lint-free tissue (Kimwipe), and insert it into the sample

holder. Line up the etched mark with the raised line on the front of the sample holder,

and close the cover.

5. With the right-hand knob, adjust the meter to read zero absorbance. Remove the tube,

empty it, and shake it as dry as possible. This precaution is essential for accuracy.

6. Fill and insert the sample cuvette.

7. Read the absorbance directly from the meter. Rinse the cuvette with clean water and

shake it as dry as possible.

8. It is necessary to reset the machine to infinity absorbance and zero absorbance before

each set of readings because the settings drift a bit. Whenever the wavelength is changed,

the infinity and zero absorbance also must be reset.

The numbered controls correspond to steps in the operating instructions.

46

PROCEDURE:

1. Turn on the Spectronic 20 spectrophotometer to warm up for a few minutes while

you review the procedure for using this instrument described in the section "The

Spectrophotometer. " Make sure the wavelength is set to 470 nm.

3. Obtain a stopwatch to use in the timing of the reaction. Work in a group with one person

timing, one person mixing and one person handling the spectrophotometer.

4. Insert your blank, and adjust the needle to zero absorbance with the right-hand

knob.

5. Tube #2 and tube #3 will be mixed together by pouring the tubes back and forth between the

the two tubes. Once the tubes begin to be mixed the stopwatch is started. The mixed tube is

then poured into one of the tubes and placed in the spectrophotometer.

6. Record the absorbance at time zero and 30 seconds for 2 minutes. Use Table 2. as an example

for the results table.

Time (sec) Run 1 (baseline) Run 2 (2X) Run 3 (1/2X) Run 4 (?)

30

60

90

120

150

180

II. Effect of Enzyme Concentration 1. Use the following experimental design to test for the effect of doubling the enzyme

concentration. Make sure you start with clean test tubes!

Tube #4 (control) add the following:

0.1 mL of guaiacol 2.0 mL of turnip peroxidase enzyme

7.9 mL of distilled water; mix well


0.1 mL of guaiacol 0.2 mL of 0.1% H2O2 (substrate)

4.7 mL of distilled water


2.0 mL of turnip peroxidase enzyme


47

2. Insert your blank, and adjust the needle to zero absorbance with the right-hand knob.

3. Tube #5 and tube #6 will be mixed together by pouring the tubes back and forth between the

the two tubes. Once the tubes begin to be mixed the stopwatch is started. The mixed tube is

then poured into one of the tubes and placed in the spectrophotometer.

4. Record the data in your data table.

5. Use the following experimental design to test for the effect of cutting the enzyme

concentration in half. Make sure you start with clean test tubes!

Tube #7 (control) add the following: 0.1 mL of guaiacol 0.5 mL of turnip peroxidase enzyme

9.4 mL of distilled water; mix well


0.1 mL of guaiacol 0.2 mL of 0.1% H2O2 (substrate)



0.5 mL of turnip peroxidase enzyme


6. Repeat steps 2 through 4 for additional 2 minutes (using tube #8 and tube #9 instead of

using #5 and #6) and record this data in your data table. NOTE: Follow the directions below and as your instructor provides you with in the laboratory.

48

III. Effect of Substrate on Enzyme Activity

PROCEDURE:

1. Set up an experiment to test the effect of substrate on enzyme activity.

2. Use the same set-up as the baseline experiment but instead just using 0.1% H2O2,

run the experiment at the following substrate concentrations: 0.025%, 0.05%, 0.2%, 0.4%.

3. You will need to replace the 0.1% H2O2 in the baseline experiment with the specified

hydrogen peroxide solutions. Read the absorbance as before and record in your data table.

IV. Effect of Temperature on Enzyme Activity

PROCEDURE:

1. Set up an experiment to test the effect of temperature on enzyme activity.

1. Use the same set-up as the baseline experiment but instead of just room temperature

run the experiment at the following water temperatures: 4o, 21

o (ambient room

temperature), 37o, or 55

o.

2. You will need to place tubes #2 and #3 in the specified water bath for five minutes before

mixing. Return the mixed tube back to the water bath after recording the absorbance.

Read the absorbance as before and record in your data table.

V. Effect of pH on Enzyme Activity

PROCEDURE:

1. Set up an experiment to test the effect of pH on enzyme activity.

2. Use the same set-up as the baseline experiment but replace the distilled water in

test tube #3 with one of the following pH buffers: 4, 5, 7, or 9.

3. Read the absorbance as before and record in your data table.

49

RESPIRATION FERMENTATION



1. Demonstrate and measure the rate of fermentation by yeast cells on various carbohydrates.

2. Compare metabolic rates of various freshwater organisms by determining the rate

of carbon dioxide production using titration techniques.

3. Compare the metabolic rates of various organisms by measuring oxygen

consumption in a closed system.

KEY TERMS:

Aerobic respiration Anaerobic respiration Citric acid cycle

Coenzymes

Desiccator

End-point

Endergonic reactions

ETS-Electron transport system

Exergonic reactions Fermentation

Glycolysis

Manometer

Titrate

INTRODUCTION:

Heterotrophs ("other feeder") obtain energy by oxidizing organic food molecules and coupling these exergonic oxidation reactions to the synthesis of ATP, an endergonic process. The ATP is

then used to carry out metabolic reactions necessary to maintain the organisms's physical

integrity, and to support all its other activities.

Some organisms are able to exist in the absence of molecular oxygen, and may even be harmed by the presence of oxygen. They still carry out oxidation reactions but do not use molecular

oxygen itself.

In any case, the cytoplasm of all cells contains the enzymes needed in the ancient, central pathway of glycolysis, in which ATP is synthesized during the oxidation of energy-rich glucose to pyruvic acid.

50

Anaerobic respiration in the absence of oxygen is called fermentation. It begins with glucose,

and utilizes the enzymes of glycolysis to oxidize it to pyruvic acid in order to make ATP. It

then uses pyruvic acid as a sink for leftover hydrogen, producing ethanol and CO2, or a mixture

of organic acids and other compounds. There is a net gain of only two molecules of ATP for

each molecule of glucose oxidized in fermentation. These ATP molecules are formed during

glycolysis.

The process of alcohol fermentation in yeast may be summarized as:

fermentation

C6H12O6 ------------> 2C6 H5OH + 2CO2 + energy

glucose ethanol carbon dioxide ATP and heat

I. Alcohol Fermentation in Yeast

Yeasts are eukaryotic fungi that are commercially very important. They are necessary for

the production of beer, wine, bread, and industrial chemicals. In this experiment you will

study the production of CO2 during the fermentation of various carbohydrates by yeast

cells under anaerobic conditions.

MATERIALS NEEDED:

Fermentation tubes (4/group of four)

Actively growing yeast suspension

Sucrose solution

Galactose solution

Molasses solution

Distilled water

PROCEDURE: (Work in groups of four)

1. Obtain four clean, dry fermentation tubes.

2. Label the tubes #1, #2, #3, and #4.

3. Fill the tubes as follows:

Tube #1: 8 mls of sucrose + 8 mls of yeast suspension

Tube #2: 8 mls of galactose + 8 mls of yeast suspension

Tube #3: 8 mls of molasses + 8 mls of yeast suspension

Tube #4: 8 mls of water + 8 mls of yeast suspension

51

4. Mix the solution in each tube by placing your finger over the open end of the tube

and then tilt the tube back and forth. Then tilt the tube so that the fermentation end of

the tube is filled completely with solution.

5. Set the tubes on a counter where you can take readings from them without

disturbing them.

6. Measure the amount of displacement of the solution in the germination tubes by the

gas evolving from the alcohol fermentation reaction.

7. Record the exact level of gas in each tube at the intervals indicated on the Table.

TABLE I: Fermentation in Yeast:

Time (minutes) Tube #1 Tube #2 Tube #3 Tube #4

Substrate (control)

4

12

20

28

36

48

Total CO2 in ml/hr

QUESTIONS:

1. What is the gas that you are measuring in this experiment?

2. Did the type of carbohydrate being fermented affect the rate of gas formation?

3. Which carbohydrate supported the fastest rate of fermentation?

52

Most organisms use molecular oxygen in the process of cell respiration. Pyruvic acid, produced

by the reactions of glycolysis, loses a molecule of CO2, then enters another set of reactions in

the mitochondrion, the citric acid cycle in which the carbon skeleton is completely oxidized to

CO2. Certain coenzymes, NAD' and FAD, are reduced by the reactions of glycolysis and the

citric acid cycle and act as hydrogen carriers: NADH and FADH2 transfer their hydrogen atoms

to a third set of proteins called the electron transport system or electron transport chain. The

flow of electrons along this chain of protein carriers is coupled to the synthesis of ATP. The very last enzyme in the chain, reacts directly with molecular oxygen to produce water. Aerobic

respiration of a molecule of glucose gives a maximum yield of 38 molecules of ATP.

This is the overall process of cell respiration in summary:

respiration

C6H1206 + 602 ----------- > 6 CO2 + 6 H2O + energy

glucose oxygen Carbon metabolic ATP & Heat

dioxide water

II. Measuring Aerobic Respiration in Plants and Animals by Carbon Dioxide Production

In aerobic respiration the carbon skeleton of glucose is completely broken down to CO2.

The CO2 must be quickly removed via the transport and gas exchange systems, so that

it does not build up and make the body fluids so that the organism cannot function. In

this experiment you will measure the rate of CO2 production in several organisms. This

rate is an indicator of the respiration rate, or metabolic rate. During the experiment a

certain amount of CO2 will be put out into the beakers, making the water more acidic.

[H2O + CO2 -----> H2CO3 (carbonic acid)]

At the end of the experiment you will 'rat each solution by adding a base, NaOH, until

a certain, fixed pH is reached. When that pH is reached, an indicator that you will add

before starting to titrate turns color at its endpoint.

more NaOH will be required to get to the end point.

MATERIALS NEEDED:

Five 200 ml beakers per group of 4

Adjusted water

2 goldfish/group of 4

2 10 cm sprigs of Elodea/group of 4 2 snails/group of 4

Burets (1 per group of four)

Phenolphthalein (1 % stock) Indicator

NaOH (0.025 M) 1 bottle/group of four 50 ml beakers

25 ml graduated cylinders

53

The more CO2 in a solution, the

PROCEDURE: (Work in groups of 4)

1. Obtain five 200 ml beakers and label them #1, #2, #3, #4 and #5.

2. Set up the beakers as follows using adjusted aquarium water or a source of water specified by your instructor.

3. Cap beakers #1, #2, #3, and #5 with aluminum foil and note the time.

4. Completely cover beaker #4 with aluminum foil so that the Elodea is kept dark, and

do not open this beaker until the end of the experiment.

5. At the end of 30 minutes return the living organisms to the proper containers on

the front counter.

6. Obtain: a 250 ml Erlenmeyer

a 25 ml graduated cylinder

a biuret

a dropping bottle of phenolthalein (1 % stock, indicator)

a bottle of NaOH (0.025 M)

54

7. Add the contents of beaker #1 to the flask and place it on a white background.

8. Add 4 drops of phenolthalein solution (indicator) and swirl to mix thoroughly.

9. Fill your buret with NaOH exactly to the top mark.

10. Add NaOH drop by drop to the flask, swirling after each drop, until the solution

definitely turns color. [DO NOT DISCARD the remaining NaOH.]

11. Wait 10 seconds. If there is still a faint, but definite color, you have reached the

end point. If the color disappears, you are very near the end point and should

cautiously continue adding drops of NaOH.

12. Read off the level of NaOH remaining and record data as in Table II the volume of NaOH

used from the biuret.

13. Refill the biuret with NaOH and repeat steps 10-12 for beakers #2, #3, #4, and #5.

(Reminder: do not uncover beaker #4 until you are ready to do the titration.

Exposure to light prior to using the solution could cause improper results.)

14. Multiply your values by 4 for the experiments where the total volume was 100 ml

(beakers #1, #3, #4 and #5).

TABLE II: CO2 Production in Plants and Animals

Record the number of drops or ml (20 drops = 1 ml) needed to change color. Calculate CO2

production in each case.

55

100

QUESTIONS:

1. Which organism had the greatest CO2 production?

2. Which one had the least production of CO2 ?

3. Did any of the experiments have a decrease in CO2 (as compared to the control)?

Why?

4. Explain your results with the Elodea in the dark versus the light.

5. Which Elodea experiment was a better measure of the amount of carbohydrate

metabolism alone (aerobic respiration)? Why?

6. In the reactions of aerobic respiration, which ones actually release CO2? (Refer to

the reactions described in your text.)

56

PHOTOSYNTHESIS



1. Identify and measure the Rf values of photosynthetic plant pigments using paper chromatography.

2. Identify the absorption spectrum of photosynthetic plant pigments using the

spectrophotometer.

3. Demonstrate the test used for identification of oxygen produced during

photosynthesis.

4. Identify the absorption spectrum of chloroplast pigments using a spectroscope.

5. Compare the effects of various wavelengths of light on photosynthesis.

6. Demonstrate carbon fixation in plants leaves using tests for starch and glucose.

KEY TERMS OR TOPICS:

Absorbance scale Matrix

Absorption spectrum for photosynthetic Oxygen Test

pigments PAR

Accessory pigments Rf value

Carbon fixation-Calvin cycle Solvent-solvent front

Chlorophylls a and b Spectroscope Chromatography-Chromatogram Starch test

Electron transport Sugar test Light intensity

Light wavelengths (color)

INTRODUCTION:

The overall reaction of photosynthesis can be understood as two closely linked processes. Light

driven electron transport produces ATP and NADPH, which feed into the assembly line of the

Calvin cycle where enzymes carry out carbon fixation by converting CO2 to glucose. A diagram

showing how these processes are linked is shown in Figure 1. Reactions on the left side take

place in the h 1 oid membranes of the chloroplast; those on the right occur in intervening fluid

regions of r m

The first step in photosynthesis is light absorption by pigments. This drives the flow of

electrons from water to NADP' along a chain of carriers. The flow also sets up a H+

electrochemical gradient that is used for chemiosmotic ATP synthesis. This entire process, by

which light energy has been converted to chemical energy in the form of NADPH and ATP, can

be summarized as non-cyclic photophosphorylation. Electrons flow in a non-cyclic fashion from

water to NADP+, which is reduced to NADPH. This electron transport depends directly on light

57

(photo-) and permits the coupled synthesis of ATP from ADP (.phosphorylation) through

chemiosmosis. Sometimes ATP alone is made by cyclic photophosphorylation, but this process is

much harder to study because O2 and NADPH are not produced.

In the final part of photosynthesis, carbon fixation, the energy of NADPH and ATP is used to

reduce M to glucose in a series of reactions known as the Calvin cycle. These reactions are

light-independent in that they can occur in the absence of light as long as NADPH and ATP are

provided. In intact photosynthetic systems, however, light is an absolute necessity, since

sufficient NADPH and ATP cannot be provided without it.

THYLAKOID MEMBRANE STROMA

H2O C6H1206

glucose

NON-CYCLIC PHOTOPHOSPHORYLATION CARBON FIXATION

Figure]. Photosynthesis

ATP synthesis by chemiosmosis depends on a hydrogen ion electrochemical gradient set up as a

result of electron transport. The Calvin cycle is tightly coupled to electron flow and quickly stops

if light is not available.

The overall balance chemical equation for the synthesis of 1 glucose molecule from 6 carbon

dioxide and 12 water molecules is shown in Figure 2.

58

carbon +

dioxide

water

light oxygen

green plants

organic + + water

nutrients

Expressed chemically and showing the reaction of the atoms, the equation appears as follo�vs:

i---------------= ------1

r

C CO, + 12 H2O light 6 02

green plants

t +

;'6t11206

+ 611,0

LI--------------------J

T I -----------L------J

Figure 2. The Photosynthesis Reaction

Answer the following questions before lab.

1. What is the source of oxygen produced in photosynthesis?

2. Where does the carbon finally become a part of?

3. What is the energy source for the breakup of water?

4. What compound is essential for the "splitting" of water?

5. Does the oxygen of carbon dioxide become the molecular oxygen released in the process.

59

I. Photosynthetic Pigments: Chromatography:

If you touch a pen containing washable black ink to a piece of tissue, the ink will spread

out on the paper and will separate into red, green, and blue bands. As the liquid in

which the colored pigments are dissolved is absorbed by the cellulose fibers of the tissue,

the pigments can't keep up: Some will move almost as fast as the solvent, but some will

move more slowly. That is because each pigment has a tendency to stick or absorb to

the cellulose fibers, and those that stick more strongly will be slowed down the most.

As a result, each substance will have its characteristic rate of movement, and the

pigments will separate from each other. This is the essential principle of

chromatography.

Chromatography has been adapted to separate all kinds of mixtures into their individual

components. The substances do not have to be colored, so long as they can be identified

somehow when the chromatography is finished. To make the chromatography

reproducible, the mixture is applied in a tiny spot to the matrix, or stationary phase,

which might be cellulose, silica gel, alumina, or some other inert substance. One end of

the matrix is then dipped into a solvent, which it will absorb. As the solvent rises up the

matrix by capillary action, it dissolves the substances in the mixture, and allows them to

migrate along the matrix. Different substances will migrate different distances. The

solvent is carefully formulated so that the substances to be separated are not completely

soluble in it, or they would move as rapidly as the solvent itself. Nor must they be

completely insoluble or they would not migrate at all. As they move they tend to absorb

or stick to the matrix to different degrees, so adsorption also helps to separate them.

When the solvent has almost reached the edge of the matrix, the chromatography is

stopped, and the finished chromatogram is dried after noting the exact position of the

solvent front, the leading edge of the area wet by the solvent. Finally the individual

substances are identified, and the distance moved by each one from the origin is carefully

measured. This distance is compared with the distance travelled by the solvent itself:

The ratio of the two is the Rf and is always constant for a given substance in a particular

solvent-matrix system:

Rf = distance travelled by substance (cm)

distance travelled by solvent (cm)

In this experiment the matrix will be the cellulose of Whatman #1 filter paper, and the

solvent is petroleum ether: acetone (9:1). Although substances do not have to be colored to

be separated by chromatography, it is a little more dramatic when you can actually see the

different substances as they pull apart to form distance spots. You will begin by

applying a green mixture of PIGMENT containing chlorophylls and accessory pigments to

the paper, and will obtain a separation of the mixture into several distinct components at

the end of the chromatography. The separation depends on the size, solubility, and

affinity for paper or adsorption of the various pigment molecules.

60

MATERIALS NEEDED:

1 inch wide Whatman #1 filter paper

Solvent [petroleum ether: acetone (9:1)] Chromatography tube and cork Chlorophyll extract

Pasteur pipet

PROCEDURE:

1. Obtain an eight inch strip of Whatman #1 filter paper and notch it at one end as

shown in figure 3. `J Place drop

of pigment

here

L _. Figure 3.

2. With a Pasteur pipet place one drop of chlorophyll extract at spot indicated in

figure 3 and let dry. Repeat this procedure 10 times, letting it dry between

applications.

3. Pour petroleum ether: acetone solvent into a clean chromatography tube to a depth

of 0.5 cm.

4. Immerse the tip of the paper strip (but not the dot of pigment) in the solvent. It should stand up straight and not touch the sides.

5. Cover the tube with a cork, place it in a large beaker and leave it undisturbed during the chromatography.

6. Check the tube every few minutes, and note the progress of the solvent as it wets

the paper and moves toward the top. Do not let the solvent reach the top of the paper.

7. Once the solvent is close to the top of the paper, stop the chromatography by

removing the chromatogram (use forceps) from the tube. Replace the cork in the

tube.

61

8. Immediately mark the solvent front with a pencil. You will not be able to calculate

Rf values if you fail to do this before the rapidly drying solvent dries.

9. Allow your chromatogram to dry.

10. Outline each visible spot with a pencil and note its color on the chromatogram.

Make a dot with your pencil in the center of each band. (Use your judgement.)

11. Carefully measure and record in Table I the distance from the origin to the solvent

front. Measure and record the distance from the origin to the dot in the of each

spot observed.

12. Calculate the Rf for each spot and describe its appearance.

13. If you can, identify the major pigments and record your Rf values for them in Table I.

14. Tape the chromatogram into your lab book.

Note: The number of spots that you observe will depend on the exact condition of the

extract that you used. These are the major pigments in the extract:

Chlorophyll a: a blue-green pigment

Chlorophyll b: a yellow-green pigment

Carotenes: yellow-orange accessory pigments

Xanthophylls: yellow accessory pigments

Pigment molecules in the faster spots will be more soluble in the solvent; they will be

smaller and will have less affinity for the paper.

TABLE I: Data From Chromatography:

R,

(cm/cm

S2Qt cm moved )v nt Color Identification

Solvent

Front

Fastest

SpQt

lowest

62

Chromatogram:

Rf Values: carotene (yellow-orange) _

xanthophyll,,(yellow) _

chlorophyll a (blue-green) _

chlorophyll b (yellow-green) _

THE SPECTROPHOTOMETER:

A colored solution appears that way because some of the light entering the solution is absorbed

by the colored substance. A clear solution will allow almost all of the light to pass through.

The amount of absorbance can be determined by using a spectrophotometer, which measures

quantitatively what fraction of the light passes through a given solution, and indicates on the

absorbance scale the amount of light absorbed compared to that absorbed by a clear solution.

The darker the solution, the greater its absorbance.

Inside the machine there is a light that shines through a filter (which can be adjusted to control

the color, or wavelength, of light), then through the sample and onto a light-sensitive phototube.

The phototube produces an electric current proportional to the amount of light striking it. The

absorbance meter measures how much light has been blocked by the sample and thereby

prevented from striking the phototube. A clear tube of water or other solvent is the blank and

has zero absorbance. A solution that contains a small amount of a colored substance might show

an absorbance of 0.1, a solution with a moderate amount might show an absorbance of 0.4, and

so forth. In fact, in the lower portion of the absorbance scale, the amount of substance in

solution is directly proportional to the absorbance reading so that a graph of absorbance versus

concentration will give a straight line. This very useful relationship is known as Beer's Law.

When you are ready to read a sample, use figure 4 to follow these steps.

1. Turn on the instrument with the power switch knob. Allow 5 minutes for warm-up time.

2. Adjust to the desired wavelength (in this case use 470 nm).

63

3. With the sample chamber empty and the cover closed, use the power switch knob to set the meter needle to read infinity absorbance. (When the chamber is empty, a shutter

blocks all light from the phototube.)

Always read from the absorbance scale, not the transmission scale.

4. Fill a spectrophotometer tube or cuvette halfway with distilled water (or other solvent

when water is not the solvent in your experiment) to serve as the blank. Wipe it free of

moisture or fingerprints with a lint-free tissue (Kimwipe), and insert it into the sample

holder. Line up the etched mark with the raised line on the front of the sample holder,

and close the cover.

5. With the right-hand knob, adjust the meter to read zero absorbance. Remove the tube,

empty it, and shake it as dry as possible. This precaution is essential for accuracy.

6. Fill and insert the sample cuvette.

7. Read the absorbance directly from the meter. Rinse the cuvette with clean water and

shake it as dry as possible.

8. It is necessary to reset the machine to infinity absorbance and zero absorbance before

each set of readings because the settings drift a bit. Whenever the wavelength is changed,

the infinity and zero absorbance also must be reset.

Figure 4. Spectronic 20 spectrophotometer

The numbered controls correspond to steps in the operating instructions.

64

II. Photosynthetic Pigments: Absorption Spectrum:

Only the light that is absorbed by chloroplasts can be used in photosynthesis. As you

have already seen, the pigments of a green plant look green to the eye because they

permit green light to pass through, but absorb the red and blue light. The particular

wavelengths of light that are absorbed by a certain substance form a pattern called its

absorption spectrum. The spectrum is determined by illuminating a solution of the

substance with each wavelength of light in turn, and measuring the absorption in each

case. Using visible light, you will determine the visible spectrum for a mixture of

pigments extracted from chloroplasts. The spectrum can then be plotted as a graph of

absorbance versus the wavelength or color of light used.

PROCEDURE:

1. Turn on the Spectronic 20 spectrophotometer to warm up for a few minutes while

you review the procedure for using this instrument described in the section "The

Spectrophotometer. "

2. Fill a spectrophotometer cuvette halfway with 80% acetone to use as your blank.

3. Fill a second tube halfway with diluted Chlorophyll Extract.

4. Set the wavelength at 400 nm.

5. Insert your blank, and adjust the needle to zero absorbance with the right-hand

knob.

6. Insert the tube containing chlorophyll, and take your first reading. It should be in

the vicinity of 0.8. If it is not, ask your instructor for help. Record the

absorbance in Table II. The zero and absorbance will change each time you

adjust the wavelength, so you will have to reset the spectrophotometer with and without the blank for each of your readings.

7. Change the wavelength to 405 nm, reset the spectrophotometer, and take your

second reading; enter the absorbance in Table II.

8. Continue your readings every 5 nm until the readings have become quite low. Then switch to every 10 or 20 nm.

9. When the absorbance starts to rise, switch back to taking readings every 5 nm, and

continue until you reach 700 nm.

65

The purpose of taking the spectrum is to get the shape as accurately as possible,

especially around the regions of greatest absorbance (low 400s and high 600s). When the

absorbance rises and then falls again, the peak is called the absorption maximum, and the

wavelength at which it occurs is used as a characteristic for identifying a compound. If

you are not sure where at least two maxima are in your spectrum, go back to the regions of

high absorbance and take some more readings.

10. Record your data in a table such as Table II. Plot your data in Microsoft Excel and tape

into your lab book.

66

III. Wavelengths Used in Photosynthesis:

When a beam of white light passes through a glass prism (or strikes a diffraction grating),

as in a spectroscope, it emerges as a spectrum of colors (like a rainbow) of different

wavelengths. Plants are able to use only certain wavelengths (colors) of this total

spectrum. Most of this useful light energy is within the visible portion of the solar

spectrum, and is often referred to as the "photosynthetically active radiations", or PAR.

If a colored solution is placed between the light source and the spectroscope, the solution

will absorb some of the wavelengths and will transmit others. As seen through the

spectroscope, therefore, regions in the spectrum where wavelengths have been absorbed

will appear as dark bands. This spectrum, from which specific wavelengths have been

absorbed, is known as an absorption spectrum and is characteristic of the solution being examined. Thus, the absorption spectrum of each substance that absorbs light can be used to identify the substance.

The color of a solution depends not upon the wavelengths of light absorbed but upon

those that pass through or are transmitted by it. Thus chlorophyll pigments in solution

appear green in transmitted light because the green wavelengths of the visible spectrum

pass through them, while the red and blue regions of the spectrum are largely absorbed.

1. Observe the electromagnetic spectrum shown in figure 5.

0

Gamma rays

4000 L

X-rays U tra Violet isible

light

Violet Indigo Blue Green

Shorter waves- higher potential

67

2. A spectroscope has been set up on the laboratory.

a. Observe the spectra formed by visible light.

b. Move the tube containing chloroplast extracts between the light and the

spectroscope.

c. Observe the absorption spectra of chloroplast pigments.

3. Sketch the absorption spectra in figure 6 in your lab book. (NOTE: If colors shown in figure 5 are not present, indicate a blank space.)

M µ

400

450

500

550

600

650

700

(Blue) (Green) (Yellow) (Red)

Figure 6. Absorption Spectrum of Chloroplast Pigments

QUESTIONS:

1. What colors (wavelengths) were absorbed?

2. What wavelengths are used in photosynthesis? Explain.

68

IV. Carbon Fixation:

Demonstration

Although carbon fixation can occur in the absence of light, it will quickly slow down and stop because the ATP and NADPH will soon be used up. A leaf or part of the plant that

is not illuminated will obtain food from the other parts of the plant, but will not be able

to carry on photosynthesis itself. Roots, tree trunks, and the other non-green parts of plants must routinely obtain carbohydrate from the green photosynthetic parts. In this experiment you will study the effect of depriving part of a green leaf of light.

Small pieces of the black paper were folded and attached to geranium leaves and then the

plants were placed in growth chambers for at least 48 hours prior to this laboratory.

Some variegated plants (coleus or geranium) have also been exposed to light for at least 48

hours. One plant has been placed in complete darkness for 48 hours.

A. Starch Test:

The extent of carbon fixation can be measured as the amount of the storage

carbohydrate, starch, that is formed.

PROCEDURE:

1. Remove a leaf from a plant (remove black paper if appropriate).

a. from a plant with leaves covered with black paper

b. variegated leaf

c. from a plant that has been in complete darkness

2. Drop the leaf into boiling water for 2 minutes to rupture cells.

3. Transfer the leaf into a beaker of boiling 95% alcohol (beware of fumes),

and boil it until it turns white.

4. Transfer leaf to dish of iodine solution and let it absorb the iodine for a few

minutes.

5. Rinse the leaf in tap water and observe which parts stain darkly. Record

your results in Table V.

69

TABLE V: Results of Carbohydrate Fixation Experiments

(Indicate a positive (+) or negative (- response

TiSsue Portion Starch Test

Covered leaf in light

a. covered portion

b. uncovered portion

Variegated leaf

a. green portion

b. non-green portion

Leaf in the light

Leaf in the dark

tiolated seedling

lbino seedling

QUESTIONS:

1. What part of the leaf stained dark?

2. Was glucose present in the leaf in the dark or the light?

3. Was glucose present in the etiolated or albino seedlings? Explain.

70

4. What substances are necessary for photosynthesis on the basis of your experiments?

5. In the photosynthesis reaction what is the source of:

a. the carbon atoms in glucose?

b. the oxygen molecules?

c. the hydrogen atoms in glucose?

71

PHOTOSYNTHESIS LAB OXYGEN PRODUCTION

Introduction:

Photosynthesis produces oxygen as one of the products of the light reaction. The measuring of the oxygen produced can be accomplished in several ways, some very complex and others quite simple. In this experiment you will use a simple method. First you will remove the existing oxygen gas from spinach leaves, then determine the length of time required to fill the intercellular spaces of leaf mesophyll tissue with oxygen.

MATERIALS: 1. Spinach leaves 2. Filter Flasks, 2 3. Sodium bicarbonate solution, 0.1%, 500 ml. 4. Paper punch

5. Petri dishes, 5 6. Colored cellophane (red, green, blue or yellow) 7. Light Meter 8. Glass stirring rod 9. Lamps and growth chambers 10. Aspirator 11. Growth chambers

PROCEDURE:

Create an experimental design and corresponding data sheet for an experiment in which conclusions can be drawn regarding oxygen production during photosynthesis. The following steps will be helpful in your design and plan. 1. Soak several fresh spinach leaves in R 0 water at 4 degrees Centigrade for 24 hours. 2. Punch out 70 discs with a paper punch, allow them to drop into filter flasks that

contain 100 ml of the bicarbonate solution (0.1%). 3. Attach the aspirator hose to the filter flask and turn on the water. 4. Place a rubber stopper in the mouth of the flask and aspirate for 30 seconds. 5. Remove the stopper; swirl if the discs do not sink repeat the process up to five times. 6. When most of the discs have "sunk", pour the solution into a large beaker. 7. Pour sodium bicarbonate solution into petri dishes to a depth of 5 mm. 8. Remove 10 discs from the beaker with a glass rod and transfer to a petri dish. 9. Repeat #8 until enough dishes are prepared for the experiment designed. 10. Setting up the control and test dishes:

• Using a lamp or growth chamber light, illuminate the dish #1 (control) 640 foot-candles.

• If colored cellophane is used on dishes, measure the illumination between the cellophane and the dish.

• If light intensity is a variable, measure the illumination and record. (If an illumination increase Is likely to increase the temperature of the leaf discs, a water filter should be considered.

• If temperature is the variable. Acclimate the solutions in the dark prior to beginning of the experiment.

11. At pre-determined times count the number of discs that have risen to the surface of the solution in each petri dish. 12. Develop a written report, including procedure, data table and/or graph and conclusion.

72

MITOSIS

DESIRED LEARNER OUTCOMES: Upon completion of this lab the student will be able to

1. Identify and describe the chromosomal activity during mitosis of the

cell cycle.

2. Identify the various stages of mitosis in the onion root tip and

determine the length of time spent in each stage.

KEY TERMS:

Cell cycle

Mitotic cycle

Stages of Mitosis -- Prophase, Metaphase, Anaphase, Telophase

INTRODUCTION:

The internal control of the cell is achieved through direct control of enzyme formation by genes.

In order to ensure that each cell has the genetic factors responsible for each enzyme, a regulated

means of duplication and equal separation of these factors must exist when new cells are formed.

Since these genetic factors are on the chromosomes, the chromosomes must undergo duplication

and orderly segregation. The process by which they do so is known as the mitotic cycle or

mitosis. In unicellular organisms, growth and reproduction are often accomplished by this

means. In multicellular organisms, multiplication of cells by the mitotic process is accompanied

by maturing phases, during which time cells undergo conversion to specialized types.

In unicellular organisms, each cell is capable of division and is therefore able to give rise to two

daughter cells, resulting in an increase in the size of the population. In multicellular organisms,

special embryonic cells are capable of division and thereby produce a sufficient number of cells

necessary for development. Highly differentiated cells are usually incapable of division. As the

cells that undergo division are performing their specialized functions, they divide many times.

Therefore, the term mitotic cycle is preferred in describing their activity, because mitosis

conveys the impression of a single and noncontinuing event. (You may also recall that this

nuclear process is usually accompanied by a division or partitioning of the cytoplasm; this is

usually referred to as cytokinesis.)

73

1. Trace one turn of the mitotic cycle in Figure 1 cytokinesis

mitosis

Figure 1. The Cell Cycle

The decision to divide occurs in G„ the chromosomes are replicated in S, the preparations for cell division are made in G2, the nucleus divides in mitosis, and the cytoplasm divides in cytokinesis.

II. Description of the Stages of the Mitotic Cycle

The mitotic cycle is made up of a series of stages that are arbitrarily established for

convenience of observation. Since the genetic factor are contained on the chromosomes, the

stages are based on chromosomal behavior.

Complete Table I before coming to lab. Consult your textbook.

74

TABLE 1: Summary of Mitotic Cycle:

DESCRIPTION OF

STAGE CHROMOSOME ACTIVITY

Interphase:

Prophase:

Metaphase:

Anaphase:

Telophase:

III. Examination of Root Sections Showing Cells in Division:

PROCEDURE:

1. Obtain a slice of Allium (onion) root tip.

2. Observe the slide under low power and find the mitotic area. (Cell division occurs

here.) See Figure 2.

75

Figure 2

Onion Root Tip

QUESTIONS:

1. In which portion of the root do you find the youngest cells?

2. How do young cells differ from older cells?

3. Do all cells have nuclei? Explain.

76

PROCEDURE. continued:

1. Focus the high power objective on the area of cell division.

2. Locate and sketch the various stages of mitosis.

Figure 3

A

Interphase Prophase Metaphase

r

I J

Early Telophase LateTelophase

Anaphase Telophase

77

IV. Allium (Onion) Root Tip Slide:

In plants, the root tip contains tissue that is forever dividing and producing new cells.

Examine the prepared slide of onion root tip cells undergoing mitotic cell division. Try to

find the stages that correspond to those shown in figure #3.

Duration of Mitotic Stages

On basic assumption of this exercise is that the more cells there are in a particular stage, the

longer the duration of that stage. A second assumption is that the duration from start to finish (including all five stages) is 24 hours.

1. Select a region of the root tip that is one cell thick and that seems to have dividing

cells scattered regularly throughout it. Once a field has been chosen the slide

should not be moved, and only the cells within that field should be examined.

2. In this field, count the total number of cells in your field and the number of nuclei

in prophase, metaphase, anaphase, and telophase. Record these numbers in table

#1.

3. Count the number of cells undergoing mitotic cell division and subtract this number

from the total number of cells. This gives you the number of nuclei in interphase.

Record this number in the table.

Total mitotic figures = _________________

Total number of cells - total mitotic figures = ___________________nuclei in

interphase

4. The duration of each mitotic stage may now be estimated using the following

equation:

number of cells in a stage 60 min

Duration of mitotic stage = total number of cells X 24 hr X 1 hr

5. Using a calculator, your instructor will pool the information for the class and give

you the data to complete the second half of the table. Compare your personal data and class data to the reported data in the literature: prophase (71 min); metaphase

(6.5 min); anaphase (2.4 min.); telophase (3.8 min.); interphase (1,356.3 min.).

Which is closer to the data in the literature, your data or that of the class?

_____________________ Can you suggest a reason for this?

78

TABLE 1: Duration of Mitotic Stages:

STUDENT CLASS

Number of CellsDurationNumber of Cells Duration (Min.) (Min.)

1. Total number

nuclei

2. Number in

prophase

3. Number in metaphase

Number in

ananhase

Number in

telophase

Number in

interohase

79

CORN AND HUMAN GENETICS



1. Identify a variety of traits used in monohybrid and dihybrid corn plant crosses. 2. Determine phenotypic ratios of various one and two trait corn crosses.

3. Determine your own phenotypes and possible genotypes for several human traits

controlled by single genes.

4. Identify the genotypes of various human blood phenotypes. 5. Determine your phenotype for colorblindness.

6. Identify various chromosomal characteristics associated with defective karyotypes.

7. Identify and compare various student fingerprint ridge patterns.

KEY TERMS:

Alleles

Dominant-Recessive

F1 and F2 progeny

Genotype-Phenotype

Homozygous-Heterozygous

Karyotypes

Monohybrid-Dihybrid cross

Multiple alleles Phenotypic ratios

Polygenes

Sex-linked gene traits

Test Cross

INTRODUCTION:

Genetics is the study of the mode of transmission of genes from one generation to the next.

There are two copies of each gene called alleles in every diploid cell, with one on each

homologous chromosome. Since the alleles must follow the movement of chromosomes during

meiosis, they will show segregation during meiosis I, with one allele going to each daughter, so

that a gamete will have only one allele for each gene. The alleles of genes on different pair of

homologous chromosomes will move independently during meiosis and show independent

assortment in the gametes and the offspring.

In order to observe what is happening to the genes during reproduction, there must be a

difference between the alleles. In most of the experiments you will conduct today, one allele

will be considered the dominant (masks the expression of the other allele and is symbolized with

a capital letter "A") and the other the recessive (can be masked by the other allele and is

80

symbolized by a small case letter "a"). For experimental purposes the recessive allele must

cause an observable change in appearance or pheno than that of the dominant allele. An

organism's gay is its genetic makeup, and its genotype will determine its phenotype.

Genetic experiments usually begin with parents that are true-breeding or homozygous in that both of their alleles for the gene in question look alike: AA or aa. Thus homozygous parents are

crossed to produce the F, hybrid generation, which is heterozygous and contains an allele from

each parent:

P

(dominant) (recessive)

F, all

These F, offspring can be crossed among themselves to give the F generation. If such a cross is

analyzed for one gene, it is called a monohybrid cross (Aa X Aa); if it is analyzed for two

genes at the same time, it is called a dihybrid cross (AaBb X AaBb). Alternatively, the F,

offspring can be crossed to an individual homozygous recessive for the trait in question in a test

cross. Data from the test cross can be used to verify that Mendelian inheritance is involved, to

detect linkage between two genes, or to determine the genotype (AA or Aa) of an individual that

shows the dominant phenotype.

1. Corn Genetics:

Corn plants are diploid (2N = 20) and produce both male and female gametes. Large

numbers of offspring can be obtained. Some traits appear as altered phenotypes in the

corn kernels, and for these it is not necessary to wait a whole generation to find out the

results of a cross.

Many traits are available for genetic studies. Possible traits to be-used:

Kernel color: purple (purple pigment in outer layer of cells of the kernel)

yellow (no pigment in outer layer of cells of kernel and carotenoids in

inner layers of kernel)

white (no pigments present)

Kernel shape: smooth (starch present in kernel)

wrinkled (sugar present in kernel)

81

Plant color: green (normal photosynthetic pigments present)

white (no photosynthetic pigments present - lethal in seedlings)

Seedling

height: normal

dwarf (shortened internodes - shorter, compact plants)

A. Corn Crosses

You will be asked to work with a partner to study one or more of the following

crosses in which homozygous parents (P) differing in one or more traits were

crossed and the resulting F2 progeny were again crossed to give the F2. In each of

the following possible crosses give the genotypes of the parents (P) and the phenotypic ratios expected in the F2 progeny.

Possible crosses: [where A is dominant to a]

1. Monohybrid cross of F, progeny:

F, ----- > Aa X Aa

F2 1AA:2Aa: laa

QUESTIONS:

1. What are the possible genotypes of the parents?_________X__________

2. What is the expected F2 phenotypic ratio?________________

Activity 1. Obtain an ear of corn and carefully examine the color of the kernels. Record the

results in Table I. Examine 100 kernels.

TABLE I.

Phenotype Hypothesized Ratio No. of Kernels Observedrv No. of Kernels Expected

Dark

Yellow

82

Activity 2. Dihybrid Inheritance in Maize (Work in Pairs)

Obtain an ear of maize (corn) from your instructor and carefully examine the

texture and color of the kernels. Many of the kernels are dark colored due to the

presence of a dominant gene (C). The lighter yellow kernels result when the

recessive allele (c) is present in the homozygous condition. Kernel texture is

controlled by a dominant gene (W) that produces starchy, smooth-textured kernels.

The recessive allele for texture (w) produces sugary kernels that appear wrinkled.

To decide whether kernel color and kernel texture are inherited in accordance with the

expectations of the principle of independent assortment, hypothesize a

phenotypic ratio such as 9:3:3:1. Then count and classify all kernels on one ear of

maize and enter the results in Table II.

TABLE U. Number of Kernels on One Ear of Maize

PHENOTYPEHYPOTHESIZED NUMBER OF NUMBER OF

RATIO F2 KERNELS F2 KERNELS

(OBSERVED) (EXPECTED)

Dark/Smooth

D Wri kl

Yellow /Smooth

Yellow/Wrinkled

I _L Total Assuming a 9:3:3:1 hypothesis, values for the "expected" column in Table II are obtained as

follows:

DARK/SMOOTH Multiply total kernels counted X 9/16

DARK/WRINKLED Multiply total X 3/16

YELLOW/SMOOTH Multiply total X 3/16

YELLOW/WRINKLED Multiply total X 1/16

2. Exercise 2: Crosses Involving Cord Seedling Traits:

MATERIALS NEEDED:

Flats of F2 corn seedling progeny

Paper towels

83

1. Obtain a flat of Fz seedlings and a paper towel. Record the label

number of your cross in Table H.

2. Inspect the progeny to determine how many phenotypic classes there are in the progeny.

3. Decide which of the possible expected phenotypic ratios (3:1 or 9:3:3:1)

for that number of classes is more likely to apply. Record your decision

in Table H.

4. Count all the seedlings in the flat. Use the paper towels as dividers to

divide the flat into sections to make them easier to count.

NOTE: Do not remove or damage the seedlings.

5. Record the number of progeny in each phenotypic class in Table II and

calculate ratios. Return flat to the front counter.

TABLE II: Data from Crosses Involving Corn Seedling Traits:

Total number of seeds planted (see flat)

Label of cross

Expected phenotypic ratio (you chose in step 3)

Phenotype # of Individuals Ratio

Total

84

Total seeds planted by team

Total seeds germinated

Number of green plants

Number of albino plants

Total seeds planted by class Total seeds germinated

Number of green plants

Number of albino plants

GENERAL QUESTIONS:

1. Would you expect your calculated (or observed) ratios to match the

expected ratios for these types of crosses when you have counted one

ear of corn or one flat of seedlings?

2. What is the role of statistical analysis in genetics?

85

II. Human Genetics:

Humans may be of much more interest to the average person than corn, but it is not easy to

study variation in human genes by conventional methods. Most traits are not

controlled by a single gene, and few would volunteer to be mutanized to help find a single

gene trait. There are also problems in doing traditional mating schemes, sibling matings,

backcrosses of offspring to parent. So much of the early work in human genetics relied on

the painstaking analysis of agrees in which the pattern of appearance of a certain trait

in several generations of a family showed how the allele for that trait was inherited. Figure 1

shows the pedigree for free vs. attached ear lobes.

1 2 3 4 5

Figure 1. Pedigree for free (normal) vs. attached ear lobes

Males are indicated by squares and females by circles. Open symbols are for individuals with

"normal" phenotype and filled symbols are for individuals who show the phenotype under

investigation. Roman numerals I-IV designate generations.

A. Exercise 3: Human Traits Controlled by a Single Gene

Several characteristics are known to be controlled by single gene differences. Some of these are listed below. Determine your own phenotypes and possible genotypes for each of the following traits and record them in Table III.

1. Pigmented iris: The P allele for pigmented iris (green, hazel, brown or black

eyes) is dominant over the p allele for lack of pigment (gray or blue eyes).

2. Tongue rolling: The R allele for the ability to roll the tongue into a U shape

is dominant over the r allele for the lack of this ability.

86

3. Bent little finger: The B allele for bent little finger is dominant over the b

allele for a straight little finger.

4. Widow's Deak: The W allele for a widow's peak is dominant over the w

allele for a straight hairline.

5. Thumb crossing: The C allele for crossing your left thumb over the right

thumb when you interlace your fingers is dominant over the c allele for

crossing the right thumb over the left.

6. Attached ear lobes: the a allele for attached ear lobes is recessive to the A

allele for unattached ear lobes.

7. Hitchhiker's Thumb: The h allele for the ability to bend the last joint of the thumb back at an angle of 60° or more is recessive to the H allele for the

lack of the ability to bend the thumb back this far.

8. Freckles: The F allele for freckles is dominant over the f allele for normal

pigment distribution.

9. PTC tasting: The T allele for the ability to sense this bitter taste is dominant

over the nontaster.

TABLE III: Human Traits Determined by Single Genes:

Trait Dominant Allele Your Phenotype Your Genotype(s)

Pigmented Iris P=pigment present

roneue Roller R=ability to roll

Bent little fingerB=bent fin r

Widow's W =wi present

Thumb in C=left over right Attached

87

QUESTIONS:

1. What is your genotype taking all of these traits into account? (Use A if

you are uncertain if you are AA or Aa)

B. Exercise 4: Human Colorblindness - A sex-linked System:

Some human traits such as color vision are carried on the sex or X or Y

chromosomes. The Y chromosome is small by comparison and carries very few

genes for traits other than traits for "maleness." The X chromosome is much

larger and does carry other genes besides those genes for traits for "femaleness."

The gene for color vision is carried on the X chromosome. Therefore,. females

(XX) have two alleles controlling vision and males (XY) have one allele. The

allele for normal color vision (C) is dominant to the allele for colorblindness (c).

NOTE: There are many types of colorblindness.

PROCEDURE:

1. Obtain a set of colored charts for colorblindness and test yourself.

2. Complete Table V summarizing your observations.

TABLE V: Colorblindness Test:

Chart NumberWhat do you see? Interpretation

1

2

3

4

5

6

7

Q

10

88

QUESTIONS:

1. Are you colorblind?

2. What is your genotype? (Check with your parents and family.)

3. Could you be a carrier?

4. If a colorblind female mates with a normal male, what types of offspring will they have in reference to color vision?

D. Exercise 5: Human Karyotvpes:

A Karyotype is the display of a persons chromosomes. Karyotypes can easily be

made from cheek cells or white blood cells or from amniotic fluid of a fetus

extracted by amniocentesis. After the cells have been stained and squashed, the

individual chromosomes are apparent and can be photographed. The chromosomes

are cut out, matched in pairs, and arranged in a chart according to size and shape.

The Karyotype of a normal male is shown in figure 2. A special staining technique

is used to produce the banding patterns that you see in which each band

corresponds to many genes. Such patterns help to identify the individual

chromosomes and to detect and locate large chromosomal defects.

89

1 1 11

1 2

A

II 11

13 14

D

19 20

11

3

11

15 16

11 11

4 5

\-B---/

11 11 17 18

E

F 1 �z G

Figure 2. Human Male Karyotype

These are the chromosomes found in each cell of a normal male diagrammed to show the

banding pattern after staining. Length and centromere position are important in sorting

the chromosomes. (Redrawn from Jorge J. Ynis. i n 191:1269, 1976 @ American

Association for the Advancement of Science.)

90

PROCEDURE:

1. Refer to Table VI during the discussion of human characteristics associated

with defective karyotypes.

2. Determine the number of chromosomes that you would find in cells of

persons with the chromosomal defects listed (Table VI) and enter the

numbers in Table VI.

TABLE VI: Characteristics Associated with Defective Karyotvpes:

Chromosome

Syndrome K Characteristics Num er

Klinefelters's Unusual body proportions and

Male XXY sterility-, subnormal mental ability Tall, prone to acne, impaired fertility;

"XYY" male XYY mentally normal Short stature; webbing of the neck;

Turner's may have low mental ability and

Female X sterility Super

Female XXX M have low mental ability* fertile #5 defective Catlike cry; severe physical and mental

"Cri chat" in short arms abnormalities: nonlethal Patau's Physical abnormalities; lethal soon

Syndrome Extra #13 after birth

Edward's Unusual features of the head and

Syndrome Extra #18 fin r• f n dies in infancy

Characteristic facial features; low

Down's mental ability; stocky build; some-

Syndrome Extra #21 im heart defects

E. Exercise 7: Fingerprint Ridge Count - A Polygenic System:

Most human traits are not controlled by a single gene. Traits such as height, skin

color, and intelligence (measured by I.Q. tests) are polygenic. Polygenic traits, in

contrast to single gene traits and chromosome abnormalities, exhibit a wide and continuous range of expression which is measurable.

91

Polygenic Model of Inheritance:

the trait is controlled by many independently assorting gene loci each gene locus has one active allele (contributing to the trait) or an

inactive allele (no contribution to the trait) all alleles at each gene locus lack dominance

phenotype is determined by the sum total of all the active alleles present

in the individual

Classification of fingerprints: (see figure 3)

Fingerprint patterns of dermal ridges can be classified into three major groups.

1. Arches: simplest and least frequent pattern.

2. Loops: has a triradius and a core. Triradium is a point at which three

groups of ridges coming from three directions meet at angles of about 120 degrees. Core is a ridge that is surrounded by fields of ridges which turn

back on themselves at 180 degrees.

a. Radial loop: if triradius is on the side of the little finger for the hand in

question and the loop opens toward the thumb.

b. Ulnar loop: if triradius is on the side of the thumb for that hand and the loop opens toward the little finger.

3. Whorl: has two triradii with ridges forming various patterns inside.

Ridge Count:

The focus of this exercise is the polygenic trait called total ridge count (TRC), the

sum of the ridge counts for all 10 fingers.

The ridge count on a finger with a loon is determined by counting the number of

ridges between the triradius and the center core of the pattern. For an arch, the

ridge count is zero. For a whorl a ridge count is made from each triradius to the

center of the fingerprint, but only the higher of the two possible counts is used (see

figure 3).

92

F�1 c Figure 3. Fingerprint Patterns

Three principal types of fingerprint patterns: (a) arch with no triradius and a ridge

count of 0; (b) loop with one triradius and a ridge count of 12 and (c) whorl with two triradii and a ridge count of 15 (the higher of the two possible counts).

MATERIALS NEEDED:

#2 lead pencil

Sheet of paper

3/4 inch scotch brand magic tape

Hand lenses or dissecting microscope

PROCEDURE:

1. Using a number 2 lead pencil, on a piece of paper shade in a square having

sides three centimeters in length.

2. Rub one of your fingers on the graphite square, making certain you have

covered all the triradii on the fingerprint.

3. Carefully place a piece of scotch tape onto your finger so that tape comes in

contact with the entire print.

4. Peal away the tape and affix it to the appropriate place on your data sheet

(Table VII).

93

5. Repeat steps 1-4, preparing a print of each of your ten fingers.

6. Examine each print carefully. (You may want to use a hand lens or

dissecting microscope.) Classify the pattern and determine the ridge count

for each print. Record in Table VII.

7. Calculate your total ridge count (TRC) and record in Table VII. 8. Record your fingerprint pattern data, total ridge count (TRC) and sex on the

table on the chalkboard, as directed by the instructor.

9. Use the class data to answer the following questions and to construct a

histogram. (See Figure 4.)

TABLE VII: Data Sheet for Fingerprints:

Right Hand

Thumb Second Third Fourth Fifth

Pattern

Ridge

Count

Place

Prints in

This Space

Left Hand

Thumb Second Third Fourth Fifth

Pattern

Ridge

Count

Place

Prints in

This Space

94

Number of types of p

Loop

Whorl

Arch

Total

%W

0

v

Total Ridge Count (TRC) of all ten fingers

10

20t

18+

16+

14-�-

12--

10--

8

z 6

2 4-

20 40 60 80 100 120 140 160 180 200 220 240 260 280 300

Total Ridge Count

QUESTIONS:

1. What was the average TRC for the class?

95

2. What was the average TRC for the males in the class? For the females?

3. How does your TRC compare to the average for your class?

4. How does your TRC compare to the average for your sex in the class?

5. Is there a difference between male and female average TRC's?

6. How does the class data compare to the averages of the general population

averages (Holt 1968) 145 for males and 126 for females?

7. How does the class percentages of fingerprint types compare to the general

population averages (Holt 1968), loop 68.9%; whorl 26.1 %; and arch 5.0%?

8. Summarize and describe the histogram you produced from the class data.

96

Genetics

Taste Test:

In actual populations it is usually not possible to tell the genotypes of all the individuals.

If the "A" allele for a certain trait is dominant over the "a" allele, the AA and Aa

genotypes will have the same phenotype. The phenotype is what can be easily observed

or measured. If the frequency of the recessive phenotype (aa) is measured (aa = q2), q2

is known. If the frequency of the dominant phenotype is known (AA + Aa = p2 + 2pq),

q2 can be calculated because the dominant and recessive phenotypes must add up to 1:

q2 = 1 - (p2 + 2pq) aa=1-(AA+Aa)

This means that the recessive phenotype frequency is equal to 1 minus the dominant

phenotype frequency.

In this experiment you will determine your phenotype for a certain trait and use the

results from the entire class to calculate the approximate number of individuals in the

class who are homozygous dominant or heterozygous for the trait, assuming the class is a

population meeting the Hardy-Weinberg conditions.

Phenylthiocarbamide (PTC) is a chemical that tastes bitter to some people (tasters) but not to

others (nontasters).

1. Place a piece of control paper on your tongue and chew it a bit. This is the taste

of paper alone.

2. Now place a piece of PTC-impregnated paper on your tongue and moisten it.

Do you detect a bitter taste other than the taste of the paper itself?

3. Chew the paper a bit and roll it around on your tongue.

4. Classify yourself as a taster or nontaster, and record the results of this test in Table

IV at the end of this topic.

5. Also record the class results and the total number of students in the class.

The ability to taste PTC is genetically determined by the dominant alleles r (early tasters

and T' (late tasters). (Late tasters can detect PTC only after chewing the paper slightly.)

These dominant alleles will be lumped together as T. Those people homozygous for the

recessive allele "t" are nontasters. Assume that your class is a reasonably accurate

sample of an ideal randomly mating population, and use the Hardy-Weinberg Law to

calculate the frequencies of "t" and "T" in the class. Once you know "t" and "T",

calculate the percentage of TT and Tt genotypes in the class.

97

1. How many individuals are expected to be homozygous dominant (TT)?

2. How many are heterzogous (Tt)?

Record your results in Table IV.

Sodium benzoate (SB) is another chemical that only some people can taste. Its' taste may

seem sour, sweet, or bitter.

1. Repeat the test with a piece of paper impregnated with sodium benzoate.

Are you a taster for sodium benzoate?

2. Which type of taste did you detect, if you are a taster?

Record your observation and the class results in Table IV.

3. What is the frequency of the "t" allele, assuming "t" is a recessive allele for

nontaster?

4. Is this frequency different from that of "t" for PTC tasting?

Variation in the ability to taste PTC and sodium benzoate has no obvious survival value

today, and yet there is ethnic variation in allele frequencies between different groups. For

instance, 63 % of Arabs are tasters for PTC, whereas 98 % of Native Americans can taste it.

At least one condition for the Hardy-Weinberg Law was not met with respect to this gene

in the world population.

QUESTIONS:

1. How do you think this ethnic variation came about?

2. How could this variation in allele frequency be associated with survival potential

for the ability to detect a bitter taste?

3. Assuming there was no selective advantage for this trait, what other factor might

account for the ethnic differences?

98

Populations Census and Sampling

Objectives When you have completed this exercise and related reading, you should be able to:

1. Define and describe what a population is.

2. Describe the difference between determining population density by sampling and census.

3. Define natality, mortality, emigration and immigration.

4. Explain population density.

5. Define population distribution.

6. Carrying capacity.

Materials and information needed for this exercise

1. Normal color vision.

2. Campus map including parking lots.

3. Data sheets.

4. Sampling and census designation: Run-1 (20 min. after beginning of lab) Run-2 (20

min. before end of lab)

5. Vehicle color choices: (The color of the upper part of the multiple-colored vehicles will be

the recorded

• White (Off-whites included) (W)

• Black (B)

• Green(G)

• Blue (B)

• Red (R)

• Yellow or Gold (Y)

• Gray or silver (S)

• Tan or Brown (T)

• Other (0)

Introduction In this exercise you and two partners are going to conduct a census and sample a population of

vehicles by color in University parking lots; then compare the population density by color of the two

methods. Since some populations often undergo cyclic changes over time; you will complete a census

and sample the same lot twice in a two hour period.

99

EXPERIMENTAL ROCEDURE

The first sampling and census will begin at 20 minutes after the beginning of lab period; the 2"d census and sampling will begin 20 minutes before the end of the scheduled lab.

1. Form a team of 3 students. a. Team-member #1 will be the vehicle-color assigner.

b. Team-member #2 will record the colors and totals of all vehicles in the

lot.(census taker) c. Team-member #3 will record the color and number of every 5`h vehicle and 10t''

vehicle. (The sampler)

2. Select a parking lot. ( Your instructor will coordinate the selections)

3. Develop a data sheet that can be used in collecting data accurately. 4. Go to the selected lot and decide where you will begin the census and sampling. (This is

usually a corner or end of the lot; do not begin sample with the first vehicle)

5. Collect information for "Run" number 1.

6. Complete the data summary tables for Run #1 7. At 20 minutes before the end of the scheduled lab repeat the census and sampling.(Run-

2)

8. Complete the data summary 9. Answer the following questions as a team and turn in to the lab instructor at the beginning

of the next lab.

Questions

1. Are the vehicles in this population exercise more like a population of animals or plants?

Explain your reasons for that choice.

2. Is the distribution of the colors of the vehicles in the lot density-dependent or density

independent?

3. In a population of nocturnal animals would 2 runs be necessary? Explain your answer.

4. What part of your team results would you expect to be different if you ran this exercise in

a Wal-mart lot? 5. What information your team gathered, least represented the campus-wide vehicle

population? Explain.

6. How would design an experiment in which you would relate vehicle color to gender?

Age?

100

Data Tables - Census -Run-I Lot letter Start time End time Total Vehicle # Team name

Color distribution of census: (actual numbers)

White (Off-whites included) (W)

Black (B)

Green (G)

Blue (B)

Red (R)

Yellow or Gold (Y)

Gray or silver (S)

Tan or Brown (T)

Other (0)

Color distribution percentage: (based on numbers data)

White (Off-whites included) (W) Black (B)

Green (G)

Blue (B)

Red (R)

Yellow or Gold (Y) Gray or silver (S) Tan or Brown (T) Other (0)

Data Tables - Census -Run-2 Lot letter Start time End time Total Vehicle # Team name



Green (G)

Blue (B)

Red (R)

Yellow or Gold (Y)

Gray or silver (S)

Tan or Brown (T)

Other (0)

Color distribution percentage: (based on numbers data) White (Off-whites included) (W)

Black (B)

Green (G)

Blue (B)

Red (R)

Yellow or Gold (Y) Gray or silver (S) Tan or Brown (T) Other (0)

101

Data Tables -Sampling (1 of 5) Run-I Lot letter Start time End time Total Vehicle #



Black (B)

Green (G)

Blue (B) Red (R)

Yellow or Gold (Y)

Gray or silver (S)

Tan or Brown (T)

Other (0)

# of vehicles in sample

Color distribution percentage:


Black (B)

Green (G) Blue (B)

Red (R)

Yellow or Gold (Y)

Gray or silver (S)

Tan or Brown (T)

Other (0

Data Tables -Sampling (I of 5) Run-2

Lot letter Start time End time

(based on numbers data)

Total Vehicle # # of vehicles in sample



Green (G)

Blue (B)

Red (R)

Yellow or Gold (Y)

Gray or silver (S)

Tan or Brown (T)

Other (0)




Black (B)

Green (G)

Blue (B)

Red (R)

Yellow or Gold (Y)

Gray or silver (S)

Tan or Brown (T)

Other (0)

102

Data Tables -Sampling (1 of 10) Run-1

Lot letter Start time End time Total Vehicle #



Black (B)

Green (G)

Blue (B)

Red (R)

Yellow or Gold (Y)

Gray or silver (S)

Tan or Brown (T)

Other (0)




Black (B)

Green (G)

Blue (B)

Red (R)

Yellow or Gold (Y)

Gray or silver (S)

Tan or Brown (T)

Other (0


Data Tables -Sampling (1 of 10) Run-2 Lot letter Start time End time Total Vehicle #



Black (B)

Green (G)

Blue (B)

Red (R)

Yellow or Gold (Y)

Gray or silver (S)

Tan or Brown (T)

Other (0)




Black (B)

Green (G)

Blue (B)

Red (R)

Yellow or Gold (Y)

Gray or silver (S)

Tan or Brown (T)

Other (0)


103

I

EXERCISE: POPULATIONS

"POPULATION GROWTH: A MODEL"

INTRODUCT10N

Just as we need tools like the microscope to help us extend our powers of observation, so we need mental "tools" to help us extend our thinking. One such mental tool is called a model. Here the word is used in a sense that is somewhat

different from the sense in which it is generally

used. The model we are discussing is not an object; it is a mental image. This kind of model simplifies a complex, real situation so that we can more easily understand it. Should the model give results similar to those given in the real situation, we can have confidence that it "works" in the same way the real situation does. Of course, the two will never match exactly, but the degree of match will determine the extent of our confidence in the model.

Because the model is a simplification, it differs in some respects from the real situation. The simplifications we make are called

assumptions. To

simplify, we assume certain things that may be only approximately true. We must keep these assumptions in mind whenever we use the model to try to understand a real situation.

PURPOSE:

The purpose of this experiment is to observe the way in which a "model" population might grow. In the exercise we will set up an experiment to see how closely a real population fits our model.

MATERIALS:

1 Ordinary graph paper, one sheet per student.

2. Semilogarithmic graph paper, 5 cycle, I sheet per person.

PROCEDURE:

A. SETTING UP THE MODEL

We can build our model, our mental tool, around a real organism, the house sparrow.

We will begin on an island, in the spring of 1981 with a population of 10 house sparrows---5

mating pairs, that is 5 males and 5 females. Assumption 1 Each year each pair of sparrows produces 10 young, always 5 males and 5 females Assumption 2: Each year all the breeding (parent) birds die, before the next

spring breeding season.

Assumption 3: Each year all offspring live through to breed during the next breeding

(Note that in most real situations, some parents would live

to breed a second time, and some of the offspring would die

before breeding. But taken together, assumptions 2 & 3 tend to balance each other, thus reducing the difference between our model and a real situation.)

Assumption 4: During the time of our study, not other sparrows arrive on the island, and none leave.

GROWTH OF THE MODEL POPULATION

Now we want to see how this model population (we can call it a hypothetical population) will grow. To do this , we must calculate the size of the population at the beginning of each breeding season. The first spring, g, a according olof 50 to assumption one, we have 5 pairs, each producing 10 offspring, According to assumption 2, the 10 breeding birds of 1981, will die before the next

104

spring. According to assumption 3, all of the 50 offspring live to breed during the

season of 1 982. Thus at the start of the 1982 breeding season, there will be 50 sparrows, 25 mating pairs, each of which produce 10 offspring. Go on with this kind

of reasoning to calculate at the island's sparrow population will be at the

beginning of the breeding seasons in 1983-1987. Show work on data sheet at end of this exercise.

You should now have a series of numbers, but they probably do not give any

clear idea of the way the population grows. A graph will show us more. Construct the graph so that the years are shown along the horizontal axis (abscissa) and the number of birds along the vertical axis (ordinate). You will need to make the

vertical scale large enough to show the small 1981 population. Plot as many generations as you can.

The difficulty we meet in plotting our data on population growth Is the choice of a scale large enough to show small gains, but not so large that later generations will go beyond the height of the ordinate or vertical scale. This difficulty can be overcome using another tool, a cyclic semilogarithmic (called "semi-log") type of graph paper. It Is not necessary to fully understand the mathematics of logarithms to appreciate our present use of this tool. Your instructor will explain

at you need to know to be able to plot your data on this type of paper.

Basically, however, this paper is set up in such a way that all data values

between 0-10 are plotted on the first cycle; any value between 10 and 100 if plotted

on the second cycle; 100 - 1 , 000 values go on the third cycle; 1 ,000 - 10,000 values

are in the fourth; and values between 10,000 - 100,000 go into the fifth cycle.

Now construct your semi-log graph with the same data you used before, remembering to put points In proper cycle.

1 What advantage(s) does the semi-log graph have over the regular graph paper for plotting data on population growth?

2. Place the two graphs in front of you. How does the slope of the line connecting the plotted points change as you go from left to right (year to year) across the, ordinary graph paper?

3. What does this mean in terms of rate of population growth?

4. What kind of line shows the same thing on the semi-log paper?

5. Finally, we must relate our results to the purpose of the exercise. In on or two sentences, describe the growth of a hypothetical population that is limited by the assumptions stated In this exercise.

FOR FURTHER INVESTIG

ATION:

We can examine the effects that changes in our assumptions will have on

population growth. By doing this, we can expect to gain a better understanding of the factors involved In population changes.

1 Chan a Assumption 2 as follows: Each year all of the breeding birds (equal y males and females) live to breed again through all the years of the study. ALL OTHER ASSUMPTIONS REMAIN THE SAFE. Calculate and compare the population size of each generation with results of the original assumption by drawing In a different color on the original graphs.

2. Change Assumption 3 as follows: Each year 40 percent of the offspring die (males & females equally) before the next breeding season. ALL OTHER

ASSUMPTIONS REMAIN UNCHANGED FROM ORIGINAL. As before, calculate the new population numbers, and draw a new colored graph on the same graph papers.

3. Change Assumption 4 as follows: Each year 20 new birds, or 10 new breeding pair, arrive on the Island from elsewhere. None leave. ALL OTHER ASSUMPTIONS REMAIN UNCHANGED FROM ORIGINAL. Calculate the populations and draw a comparative graph.

DEVISE OTHER PROBLEMS FOR YOURSELF, BY CHANGING OTHER ASSUMPTIONS.

105

BIRD POPULATION MODEL DATA SHEET

STARTING TOTAL 1981 1982 1983 1984 1985 1988 1987

BREEDING PAIRS

OFFSPRING/PAIR

TOTAL OFFSPRING

SURVIVING PARENTS

TOTAL BIRDS

DEATHS

SURVIVORS

IMMIGRANTS

START NEW YEAR

MODEL ASSUMPTIONS:

STARTING TOTAL

BREEDING PAIRS

OFFSPRING/PAIR

TOTAL OFFSPRING

SURVIVING PARENTS

TOTAL BIRDS

DEATHS

SURVIVORS

IMMIGRANTS

START NEW YEAR

YEARS BASIC PROBLEM MODIFICATION I MODIFICATION II MODIFICATION III

1981

1982

1983

1984

1985

1988

1987

106

BIRD POPULATION MODEL DATA SHEET

STARTING TOTAL 1981 1982 1983 1984 1985 1988 1987

BREEDING PAIRS

OFFSPRING/PAIR

TOTAL OFFSPRING

SURVIVING PARENTS

TOTAL BIRDS DEATHS

SURVIVORS

IMMIGRANTS

START NEW YEAR

MODEL ASSUMPTIONS:

STARTING TOTAL

BREEDING PAIRS

OFFSPRING/PAIR

TOTAL OFFSPRING

SURVIVING PARENTS

TOTAL BIRDS DEATHS

SURVIVORS IMMIGRANTS

START NEW YEAR

YEARS BASIC PROBLEM MODIFICATION I MODIFICATION II MODIFICATION III

1081 1982

1083

1984

1085

1088

1987

107

TABLE IV. Data on PTC

Data from the PTC Test

Individual results: Taster

Class results:

Number of tasters _

Number of nontasters

Total

q2 =

Nontaster

% tasters (TT + Tt)

% nontasters (tt)

; q = ; frequency of t allele =

1 - q = p = frequency of T allele =

p2 = genotype of TT = % of the class

2pq = genotype of Tt = % of the class

If a class member is known to be a taster, what is the chance that he or she is homozygous fo the T

allele?

% IT X 100/(%TT = %Tt) _ %

108

TABLE V. PTC Tasters in Human Populations

Population SampleTasters(%) Gene Fr uencies

Sampled Size R 9

Welch 237 57.7 0.36 0.64

Eskimo (unmixed) 130

EsIdmo (mixed) 49 69.4 0.45

Whites (Ohio) 70.2 0.45

Blacks (Alabama) 76.5 0.52 0.48

Blacks (Ohio) 3156 0.70 .030

Blacks (Kenya) 91.9 0.72 0.2

Blacks (Sudan) 805 95.8 0.80 0.20

Navaho Indians 1 269 1 98.2 0.87 0.13

109

Standard Plate Counts of Raw and Pasteurized Milk

Microbiology Laboratory

Background

Milk is one our most nutritionally complete foods adding protein, milk sugar, fat, and minerals and vitamins

to our diet. In addition, the milk industry utilizes milk for the production of other food products such as

butter, sour cream, yogurt, cottage cheese among others. It is therefore very important for the industry to

test for and prevent/reduce the transmission of microbial diseases that can be found in contaminated milk.

There is a normal flora of bacteria that exists in milk that is nonpathogenic (harmless) but milk if not treated

can carry pathogenic organisms that can cause problems. Most outbreaks of disease occur due to the

consumption of raw milk. For example, Salmonellosis (Salmonella), listeriosis (Lysteria monocytogenes),

tuberculosis (Mycobacterium) and gastrointestinal problems (Campylobacter jejuni, and Escherichia coli

0157:H7) all have caused problems in recent years due to contaminated raw milk.

The aforementioned organisms can all be killed or at least inactivated by a heat process referred to as

pasteurization. This process uses heat at a specific temperature for a specified time period to reduce or kill

the microorganisms. The following methods of heat treatment are designed to kill Mycobacterium

tuberculosis due to its higher tolerance of heat compared to other bacteria. Several methods are employed

in pasteurization but they involve either low temperature for a long period or time (LTLT) or high

temperature for a short time (HTST) and ultra-high temperature (UHT). The following temperatures and

times are usually associated with these methods:

LTLT: 62.8oC for 30 minutes

HTST: 71.7oC for 15 seconds or

UHT: 138oC for 2 seconds.

The Milk industry and the American Public Health Association monitor the quality of the pasteurization

process by following a protocol to ensure proper treatment and prevention of microbial contamination. We

can test the microbial flora of raw versus pasteurized milk in a very similar way. Simple agar plate counts

are used to monitor the quality of pasteurized milk and we will compare plate counts from both raw and

pasteurized milk samples in this laboratory. The State of Minnesota has set the following standards for raw

and pasteurized Grade A milk: 100,000 bacteria/ml and 20,000 bacteria/ml, respectively. Another

technique associated with these agar plate counts is a standard serial dilution procedure. This technique

has the effect that there are numerically less bacterial colonies found on an agar plate as you increase the

amount of dilution or dilution factor.

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Procedures

Dilution of Raw Milk and Pasteurized milk

1. Obtain a test tube of raw milk, 4 test tubes, test tube rack, 8 sterile Petri plates and transfer pipettes from the instructor.

2. Label the test tubes as follows: 10-1, 10-2, 10-3, 10-4

3. Prepare water blanks as shown in figure 1 by transferring 9.0 ml into the labeled test tubes, exactly as shown.

4. Thoroughly mix the raw milk.

5. Make serial dilutions of the raw milk. The raw milk will be serially diluted from 10-1 through 10-4 as shown in Figure 1-1

i. Aseptically transfer 1.0 ml of the suspension into the first test tube labeled 10-1. Discard the pipette.

ii. Mix using a vortex and transfer 1.0 ml of the 10-1 dilution into the second test tube labeled 10-2. Save the pipette for step 5 by placing the pipette back into its original sleeve (label the pipettes to avoid confusion later).

iii. Mix using a vortex l and transfer 1.0 ml of the 10-2 dilution into the third test tube labeled 10-3. Save the pipette as before.

iv. Mix using a vortex and transfer 1.0 ml of the 10-4 dilution into the fourth test tube labeled 10-4. Save the pipette as before.

6. Pipette 1 ml from the 10-2 test tube into each of 2 Petri dishes labeled 10-2. Discard the pipette.

7. Repeat step 5 for the remaining dilutions (10-3 and 10-4).

8. Pour 10-12 ml of cooled plate (45oC) agar into the Petri dishes and mix thoroughly but gently (figure-8 motion).

9. After allowing the agar to harden, invert and place the plates in an incubator at 32oC for 48 hours.

10. After incubation, select the plating dilution that has between 30-300 colonies. Count the colonies and record the result in the attached data table.

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Figure 1-1. Dilution series for Analysis of Raw Milk

1 ml

10-2 10-2

10-3 10-3 10-4

10-4

10-1

(1:10)

10-2

(1:100)

10-3

(1:1000)

10-4

(1:10000)

9.0 ml 9.0 ml 9.0 ml 9.0 ml

100

(1:1)

1

1.0 ml 1.0 ml 1.0 ml 1.0 ml

9.0 ml

11. Repeat steps 1 through 9 for the pasteurized milk with the following exceptions:

a. Serially dilute the pasteurized milk from 10-1 through 10-3. There is no need to dilute the milk any further than 10-3. b. In step 5, instead of plating 10-2 through 10-4 you should plate the dilutions 100 through 10-3.

12. Use the following formula to calculate the colony forming units per ml of milk: number of colonies x the reciprocal of the dilution = colony forming units/ml

Results

Data table:

Type of Milk Dilution 1/ Dilution Average number

of colonies/plate

Colony forming

units/ml

Pasteurized

Raw

Questions

1. How many bacteria were in your raw milk sample? How many bacteria were in your pasteurized milk sample?

2. According to the American Public Health Association standards, does the milk you tested meet the allowable bacterial counts/ml for the respective samples?

3. Did you see any bacterial growth in the pasteurized milk? If so, explain why there may be bacteria in milk after the pasteurization procedure.

4. Define pasteurization, in your own words, as it relates to the milk industry.

5. Compare/contrast the three main methods of pasteurization mentioned in this laboratory.

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DNA Extraction


Upon the completion of this lab the student will be able to:

1. extract a visible mass of DNA;

2. describe the physical and chemical properties of DNA; and,

3. explain the universal function of DNA.

Applications and Practices

Biotechnology has the potential to greatly affect plant agriculture of the future. The word biotechnology can be divided into its two root words: bio which means life and technology which means applying science to solve a problem. Biotechnology is the collection of techniques that use living organisms to make products or solve problems. The most common techniques include genetic engineering, diagnostics, and cell/tissue culture.

Genetic engineering has great potential for improving both the quality and quantity of the crops we raise. Plants are now available with special genes that act as herbicides and insecticides. Other genes have been introduced into plants which produce frost resistance or delay the ripening process. Plants that are resistant to drought or can be irrigated with salt water are also being tested. One day, plants may be used to manufacture vaccines. Immunity to a disease could be obtained by simply eating the plant which manufactured the medicine as it grew. This new way of health care would be particularly valuable in developing countries. Scientists are also working on changing the color of.fruits and vegetables that we eat to more easily' identify a food which contains a special vaccine.

Science Connections - Questions for Investigation

1. What is the basic structure of DNA? 2. How is information stored within the DNA structure? 3. How can DNA be isolated?

4. What is the basic function of DNA?

DNA Extraction from an Onion

MATERIALS: • two 4-cup measuring cups (1000 ml) with ml markings • one 1-cup measuring cup (250 ml) with ml markings • measuring spoons • sharp knife for cutting onion • large spoon for mixing

• food processor or blender

• thermometer that will measure 60° C (140° F), such as a candy thermometer

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• strainer or funnel that will fit in a 4-cup measuring cup •

• #6 coffee filter or cheese cloth • hot tap water bath (60° C) • ice water bath • distilled water • light-colored dishwashing liquid

• large onion • table salt • meat tenderizer that contains papain • 1 test tube for each student, preferably with a cap

• Pasteur pipettes or medicine droppers

• 95% ethanol (grain alcohol)

PROCEDURE:

1. Set up hot water bath at 55-60° C and an ice water bath.

2. For each onion, make a solution consisting of one tablespoon (10 ml) of liquid dishwashing detergent or shampoo and one level 114 teaspoon (1.5 g) of table salt. Put in a 1-cup

measuring cup (250 ml beaker). Add distilled water to make a final volume of 100 ml. Dissolve the salt by stirring slowly to avoid foaming.

3. Coarsely chop one large onion with a food processor or blender (may be done by hand if neither is available) and put into a 4-cup measuring cup (1000 ml). For best results, do not chop the onion too finely. The size of the pieces should be like those used in making spaghetti. It is better to have the pieces too large than too small.

4. Cover chopped onion with the 100 ml of solution from Step 1.

5. Put the measuring cup in a hot water bath at 55-60° C for 10-12 minutes. During this time,

press the chopped onion mixture against the side of the measuring cup with the back of the spoon. (Do not keep the mixture in the hot water bath for more than 15 minutes because the DNA will begin to break down.)

6. Cool the mixture in an ice water bath for 5 minutes. During this time, press the chopped

onion mixture against the side of the measuring cup with the back of the spoon.

7. Filter the mixture through a #6 coffee filter or four layers of cheese cloth placed in a

strainer over a 4-cup measuring cup.

8. Dispense the onion solution into test tubes, one for each student. The test tube should contain about 1 teaspoon of solution or be about 1/3 full. For most uniform results among test tubes, stir the solution frequently when dispensing it into the tubes. There is not an advantage to dispensing more than one teaspoon of solution into a test tube. The solution

can be stored in a refrigerator for about a day before it is used for the laboratory exercise. When the solution is removed from the refrigerator, it should be gently mixed before the test tubes are filled.

9. Add two toothpicks full of meat tenderizer to the onion solution, cap the tube, and mix gently

to avoid foaming.

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10. Add cold alcohol to the test tube to create an alcohol layer on top of about 1 cm. For best

results, the alcohol should be as cold as possible. The alcohol can be added to the solution in at least three ways. (a) Fill a Pasteur pipette with alcohol, put it to bottom of the test tube, and release the alcohol. (b) Put about 1 cm of alcohol into the bottom of a test tube and add the onion solution.

11. (c) Slowly pour the alcohol down the inside of the test tube with a Pasteur pipette or medicine dropper.

12. Let the solution sit for 2-3 minutes without disturbing it. It is important not to shake the test tube. You can watch the white DNA precipitate out into the alcohol layer. When good

results are obtained, there will be enough DNA to spool on to a glass rod, a Pasteur pipette

that has been heated at the tip to form a hook, or similar device. DNA has the appearance of

white mucus.

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DNA Extraction from a Banana

MATERIALS:

One blender per class:

Tissue homogenate -take 1 banana + 75 ml water in blender - strain

through cheesecloth

95% ethanol - keep cold in freezer

SDS solution - 11 g SDS in 100 ml water

PROCEDURE:

1. Place 1.0 ml tissue homogenate in test tube

2. Add 1.0 ml SDS solution Mix well

3. Add 10 ml cold ethanol

Mix well, swirl but do not shake tube.

4. Place tube on ice for 5-10 minutes to precipitate DNA

Swirl lightly.

Find DNA in the interface between the ethanol and the homogenate

5. Spool onto a glass rod

What does the DNA look like on the glass rod?

What do you see under the microscope?

Does banana DNA look like mice DNA or human DNA under the microscope or

spooled on a glass rod?

What is the purpose of using a blender?

What does SDS do to the banana mixture?

What does "to precipitate DNA" mean when discussing the action of the ethanol?

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Data Summary and Analysis

Students should record the procedures for extracting DNA from the E. Coli and record observations on the properties of the spooled DNA.

Laboratory reports should contain answers to the following questions: 1.

1. What effect did the detergent have on the cell membrane?

2. What are three properties of DNA demonstrated by the lab?

3. Why does DNA appear flexible when it is actually a very rigid structure?

Anticipated Findings The liquid detergent causes the cell membrane to break down and dissolves the lipids and proteins of the cell by disrupting the bonds that hold the cell membrane together. The detergent causes lipids and proteins to precipitate out of the solution. NaCI enables nucleic acids to precipitate out of an alcohol solution because it shields the negative phosphate end of DNA, causing the DNA strands to come closer together and coalesce.

The heat treatment softens the phospholipids in the cell membrane and denatures the DNA's enzymes which, if present, would cut the DNA into small fragments so that it could not be extracted.

DNA is not soluble in alcohol. When alcohol is added to the mixture, all the components of

the mixture, except for DNA, stay in solution while the DNA precipitates out into the alcohol layer.

The DNA molecule is extremely long and acidic in nature. The strands can bend without breaking.

The stiff property of DNA allows it to spool on the rod.

Ideas for Other Experiments

Similar procedures can be used to extract DNA from other organisms, such as E.coli.

Data Summary and Analysis Students should record the procedures for extracting DNA from the E. Coli and record observations on the properties of the spooled DNA.

Laboratory reports should contain answers to the following questions:

1. What effect did the detergent have on the cell membrane?

2. What are three properties of DNA demonstrated by the lab? 3. Why does DNA appear flexible when it is actually a very rigid structure?

119

GEL ELECTROPHORESIS

DESIRED LEARNER OUTCOMES: Upon completion of this lab the student will be able to:

1. explain what restriction enzymes are and how they are used;

2. explain what gel electrophoresis is and how it is used; and,

3. demonstrate how to properly load and run DNA samples on an agarose gel.

In this lab, gel electrophoresis will be used to separate DNA fragments after restriction diges using specified

restriction enzymes. In a normal research setting, samples of various DNA, RNA, or proteins would be run on

this type of system.

This experiment introduces the student to the basics of gel electrophoresis. Background information on gel

electrophoresis is included in the introduction.

INTRODUCTION:

Electrophoresis means literally to carry with electricity. It is a common

biotechnology technique used to separate charged molecules such as DNA, RNA, and proteins.

Electrophoresis is frequently performed with an agarose gel. Agarose is a polysaccharide like agar or pectin that

dissolves in boiling water and then gels as it cools. In electrophoresis the sample is applied to a slab of gelled

agarose and then an electric current is applied across the gel. Agarose gels must be prepared and run in a

buffer.

The buffer is necessary because ions would otherwise cause the anode to become alkaline and the cathode to

become acidic. The buffer you have used is tris-borate-EDTA (TBE). Remember that a buffer is a mixture of a

weak acid or base and its salt. Tris is a weak base and the Tris salt is made by adding boric acid. The metal

chelator EDTA (ethylenediaminetetra acetic acid) is also added. By chelating calcium ions, EDTA inhibits

RNases and DNases, which could degrade RNA and DNA. It is important to have the proper concentration of

buffer. In the absence of ions (say, if the gel were mistakenly run in water) electrical conductance is minimal.

On the other hand, if the buffer is too concentrated (say, if 10x buffer were used by mistake) electrical

conductance is too efficient and heat is generated. Too much heat can melt the gel.

When an electric field is present, a negatively charged sample (such as DNA) migrates through the gel toward

the positive electrode. The rate of migration is specific to the properties of the sample. For a DNA molecule,

which is negatively charged, the rate of migration depends on the size of the DNA fragment and also on whether

the DNA is circular (like a plasmid) or linear. The voltage applied to the agarose gel also determines how quickly

the sample moves through the gel. The higher the voltage, the more quickly the sample moves. However, there

is a trade-off because at higher voltages samples do not separate with as much resolution.

120

PRE-LAB PROCEDURE:

Mix (TBE) Buffer

Because tris-borate-EDTA (TBE) buffer solution is stable, it can be made

ahead of time and stored in a carboy or other container at room temperature until

ready for use. Dilute 20 mL (entire bottle) of the 20x TBE buffer mix in 380 mL of

distilled or deionized water. Stir until solution is completely mixed. Rinse any

residue from the buffer container with a portion of the 380 mL of distilled or

deionized water.

Prepare Agarose Solution

Before class on Lab Day, prepare 1.0% agarose solution. Add 4 g of agarose to 400

mL 1x TB E. Heat in microwave until solution becomes clear as agarose dissolves.

Swirl and observe bottom to insure that no undissolved agarose remain.

Cast Agarose Gel

1. Seal ends of gel-casting tray and insert well-forming comb into the proper slot. Place gel-casting

tray out of the way on lab bench so that agarose poured in the next step can set undisturbed.

2. Carefully pour enough agarose solution into casting tray to fill to a depth of about 5 mm. Gel should

cover only about half the height of the comb teeth. Use the tip of a transfer pipet to move large bubbles

or solid debris to sides or ends of tray, while gel is still liquid.

3. Gel will become cloudy as it solidifies (about 10 min). Do not move or jar casting tray while agarose is solidifying.

PROCEDURE:

DNA Digest by Enzymes

1 In a floating rack, place one of each colored tube containing enzymes (blue is

BamHl, pink is EcoRl, green is HindIII, and yellow is the control which contains no

enzyme).

2. Using the micropipette (yellow color), transfer 20 µL of DNA to one of the tubes.

Mix the DNA and enzyme by pipetting up and down several times. There should

be no concentration of the blue color in the bottom of the tube if the DNA and

enzyme are mixed sufficiently. Change pipette tips (to avoid cross

contamination) and repeat the process for the other tubes.

3. When all tubes are mixed well, cap the tubes and place the floating rack in the

water bath for incubation. 120

Remove the casting tray and gel from the electrophoresis chamber.

I Unplug the power supply and remove the leads. Slide the cover of the

electrophoresis chamber open making sure not to bend the poles.

2. Carefully lift the casting tray from the electrophoresis chamber and place the tray

on the white paper. Remove the gel from the casting tray by sliding it into a

disposable staining tray.

3. Pour Carolina Blue Stain over the gel, using enough to just cover the gel and

begin timing. After 10 minutes, pour the stain into the "Used Stain" container.

Next, cover the gel with deionized water and allow it to stand. These gels will be

viewed later.

Load Gels and Start Run

1. Obtain a floating rack with tubes of digested DNA from the water bath containing

the previous lab's digestions.

2. Using the smaller micropipette (gray color) set it to deliver 2 µL. Using a clean

tip each time, deliver 2 µL loading dye to each enzyme tube. Mix the tubes by

tapping the tube on the counter several times.

3. Obtain a casting tray that fits your electrophoresis chamber. Unseal both ends of

the casting tray and place it into the chamber so that the comb is at the negative

(black) end.

4. Gently remove the comb. Pull straight up trying not to rip the wells. Adjust the

level of the buffer (1X TBE buffer) by making sure that the buffer just covers

the gel entirely. If you see any ripples around the wells, add more buffer.

5. Load the gel with 20 µL of the enzyme/DNA samples, using a fresh pipette tip for

each sample.

• Steady the micropipette over the well using both hands

• Be certain to expel any air in the micropipette tip end before loading the

gel. If an air bubble forms a "cap" over the well, the sample will flow into the

buffer around the edges of the well. r

• Make sure the micropipette tip extends through the surface of the buffer

and is positioned over the well. Slowly expel the sample into the well

allowing it to sink to the bottom. Be careful not to punch the pipette tip

too deeply so the sample enters the gel itself.

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Close the top of the electrophoresis chamber by sliding the top, matching up the red

and black electrodes. Plug the leads into the power supply, again matching red-to-

red and black-to-black. Also, make sure that both^ leads are connected to the same

channel.

7. Turn on the power supply and set the voltage to 45. The chambers will run

through the rest of the class period.

RESULTS AND DISCUSSION:

1. Examine your stained gel on a light box or overhead projector. Compare your gel with the ideal gel shown in Figure 1, and try to account for the fragments of lambda DNA in each lane.

2. How can you account for differences in separation and band intensity between your gel and

the ideal gel? 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? Why would it work? 4. Linear DNA fragments migrate at rates inversely proportional to the loglo of their molecular weights. For simplicity's sake, base-pair length is substituted for molecular weight.

a. The matrix below gives the actual size in base pairs (Act. bp) of lambda DNA fragments

generated by a HindIll digest:

HindI1l EcoRI

Distance Act. bp Distance Measured bp Act. bp

*27,491

*23,130

9,416

6,682

4,361

2,322

2,027

**564

**125

* Pair appears as single band. ** Does not appear on this gel.

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b. Using your gel or the ideal gel shown in Figure 1, carefully measure the distance (in mm) each HindIII and EcoRI fragment migrated from the origin. Measure from the front edge of well to front edge of each band. Enter distances into matrix.

c. Match base-pair sizes of Hindlll fragments with bands that appear in the ideal digest. Label each band with kilobase-pair (kbp) size. For example, 27,491 bp equals 27.5 kbp. d. Set up semilog paper with distance migrated as the x (arithmetic) axis and the base-pair length as the y (logarithmic) axis. Then, plot distance migrated versus base-pair length for each Hindlll fragment. e. Connect data points with a best-fit line.

f. Locate on x axis the distance migrated by the first EcoRI fragment. Using a ruler, draw

a vertical line from this point to its intersection with a best-fit data line. g. Now extend a horizontal line from this point to the y axis. This gives the base-pair size of this EcoRI fragment.

h. Repeat steps f and g for each EcoRI fragment and place answer under Calc. bp after completion, your teacher will provide data for Act. bp for comparison. i. For which fragment sizes was your graph most accurate? For which fragment sizes was it least accurate? What does this tell you about the resolving ability of agarose-gel electrophoresis?

123

HUMAN PHYSIOLOGY



1. Record and compare student heart rate after various activities. 2. Record and compare student blood pressure and breathing rates under varying

conditions. 3. Determine various lung capacities of students using hand-held spirometers.

KEY TERMS:

Chemoreceptors

Diastole

SA node

Sphygmomanometer

Spirometry

Systole

Tidal volume

Vital capacity

1. Human Heart Rate:

The beating pattern of the heart is not due to nervous stimulation, but is rather due to

spontaneous electrical events starting with the pacemaker of the sinoatrial (SA) node in

the right atrium. The electrical-chemical impulse spreads quickly and smoothly through

the cardiac muscle of the atria to trigger the atrioventricular (AV) node, which in turn

stimulates the ventricles to contract. The rate of the intrinsic heartbeat, however, is very

responsive to changes in conditions within the body and to external stimuli. It speeds up

when the demand for oxygen is high and also under the influence of the hormone

epinephrine (adrenalin).

During exercise, epinephrine is released and acts to stimulate the heart rate. Exercise is

supposed to benefit your circulatory system because the capillaries in the muscles,

especially those of the heart, dilate in order to increase the supply of oxygen to the active

tissues. After the exercise is over, the capillaries remain a little bit more open than

before so that oxygenation of the tissues is improved all the time. An athlete in top

condition will often have a very low heart rate of 50 beats/min. or less because the

pumping heart meets very little resistance in the peripheral capillaries. In order for the

exercise to help significantly, however, the heart rate must reach 150 beats/min. for at

least a few minutes during each exercise session.

126

PROCEDURE:

1. Place your fingertips of your left hand in the hollow of your wrist at the base of

your right thumb. If you have trouble detecting your pulse, use the carotid artery

alongside your larynx instead.

2. Count your heartbeats for 3 minutes, calculate the rate in beats per minute, and

record the results in Table I.

3. Jump up and down until you feel out of breath or tired, and again record your heart

rate.

4. After your heart rate is back to normal, lie down for 5 minutes, and measure the

rate while resting. Try to relax as fully as possible.

5. Try other activities and then test your heart rate. (Climbing stairs, sit-ups, etc.)

TABLE I. Human Heart Rate:

Condition f Time Interyal Beats/Minute

Initial Normal

Jumping

Normal

Lying Down

Other Activities

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H. Blood Pressure:

Blood pressure varies throughout the circulatory system, so it is usually measured in the

upper arm for comparison of different individuals. The pressure is the hydrostatic

pressure exerted on the walls of the blood vessels by the blood under the force of the

contracting heart. It is greatest in the vessels adjacent to the heart, it drops off in the

arterioles and capillaries due to the friction caused by the blood moving along the walls,

and it is very low in the venules and veins leading back to the heart. Muscular

contraction is required to squeeze the blood in the veins from the extremities to the heart

because the pressure in these vessels is less than the force of gravity. Valves in the veins

are extremely important in aiding the return of the blood to the heart, because its pressure

is so low.

The blood pressure in the arteries follows a rhythmic cycle with each beat of the heart.

When the ventricles contract, exerting the strongest pressure, the resulting high-pressure

phase is termed systole. While the ventricles of the heart relax, the pressure in the

arteries decreases in a low-pressure phase called diastole. Blood pressure is always

measured at both systole and diastole, and the result is reported as the systolic

value/diastolic value. The blood pressure reading is generally higher in older people and

lower in people in good physical condition or in athletic training.

Work in pairs.

MATERIALS NEEDED:

Sphygmomanometers

PROCEDURE:

1. Make sure that the subject is seated.

2. Wrap the blood pressure cuff around your partner's upper arm so that it is neither

tight nor loose, and secure it with the Velcro fasteners.

3. Squeeze the bulb several, times while watching the pressure gauge until the

pressure reaches 180-200 mm Hg (760 mm is 1 atm).

4. Place the stethoscope in the hollow at the inner side of the elbow, and be ready to

listen for the sound of the pulse.

5. Slowly release the pressure while watching the gauge, and make a mental note of

the pressure at which you first hear a single sound with each heartbeat; this is the

systolic pressure.

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6. Continue to release the pressure until you hear the double sound of a normal

heartbeat (lub-dup . . lub-dup . .); this is the diastolic pressure. 7. Record the pressure values for systole and diastole in Table II, and repeat the

measurement twice more to be sure your value is reasonably accurate.

8. Next, ask your partner to lie down and rest for a few minutes. Take the blood

pressure three times, and record the results.

Did the value for the blood pressure depend on the position of the subject?

9. Repeat the measurement of blood pressure under other conditions. Record in Table II.

TABLE II. Blood Pressure:

Blood Pressure

Condition Reading Systolic/diastolic

/

Initial Pressure 2

3

Resting Pressure

Other Condition 1 /

2

2

QUESTIONS:

1. What effect would smoking a cigarette have on your blood pressure?

2. What effect would a cup of coffee or a can of coke have on your blood pressure?

129

3. What is hypertension?

4. What can hypertension lead to?

During exercise more blood is sent to the heart and muscles because their blood vessels

dilate. At the same time, vessels in other areas (lungs, abdominal organs) constrict so

that blood pressure does not fall and may even increase. If a person loses blood through

injury or during blood donation, veins will automatically constrict to reduce the volume

available for the circulating blood and thereby maintain normal blood pressure.

III. Human Breathing:

A. Vertebrae lungs can be ventilated in two ways. In mammals the lungs are enclosed

in the airtight thoracic cavity: Enlargement of this cavity creates negative pressure

outside the lungs so that the air rushes into them. Negative pressure is the same

as suction. Air is pulled into the lungs. The frog, on the other hand, uses pgsitiv

pressure from contraction of the oral cavity to "push" air into the lungs. Negative

pressure inflation of the lungs is like sucking in on your bubble gum so that it

forms a bubble inside your mouth. Positive pressure inflation is like blowing a

regular bubble.

Inspiration in mammals is due to expansion of the chest cavity when the rib cage

lifts and the diaphragm contracts. Muscles between the ribs and radial muscles in

the diaphragm must contract for this to happen. Negative pressure breathing has

the advantage that the animal need not use the muscles of the mouth at the same

time.

PROCEDURE:

1. Place one hand on your stomach just below your ribs to feel the effect of

your diaphragm contracting, and place the other hand on your rib cage.

2. Take a few normal breaths.

-- Does the rib cage move?

-- Does the stomach move?

3. Try to breathe without moving the rib cage at all.

130

4. Then try not to move your diaphragm.

-- Can you get enough air using only one set of muscles?

B. How the Rate of Breathing is Controlled:

The rate at which you breathe is automatically controlled so that there will be

enough oxygen and so that CO2 will be promptly removed. Chemoreceptors in the

medulla (part of the spinal cord just below the brain) respond to the pH of the

blood. Blood pH will decrease as the CO2 level increases due to the formation of

carbonic acid:

H2O + CO2 -----> H2CO3 -----> H+ + HC03-

As soon as the CO2 level rises, the receptors detect carbonic acid and the

inspiratory center in the medulla speeds up the breathing rate to get rid of the CO2.

Breathing rate is normally determined by the level of CO2 rather than that of

oxygen, probably because CO2 is easier for the body to detect. The inspiratory

center shuts itself off by sending a message to an expiratory center, also in the

medulla. The expiratory center works by inhibiting the inspiratory center, allowing

relaxation of all the muscles involved in inspiration. If the blood oxygen level ever

becomes abnormally low, oxygen chemoreceptors in the aorta and carotid arteries

stimulate the inspiratory center, but oxygen concentration does not play a role in

normal breathing. The control of breathing is not perfect, and apnea, a

spontaneous lapse in breathing may occur, particularly during sleep or under the

influence of drugs.

PROCEDURE:

In these experiments, stop at once if you begin to feel faint. If you have any

medical problem with your heart or lungs, be a timekeeper, not a subject.

Work with a partner to make the following measurements:

1. While sitting down and breathing normally, measure the number of breaths

per minute three times. Record you results to the nearest half breath in

Table III.

2. Hyperventilate by breathing as deeply and as fast as you can for 20 breaths.

Then breathe normally.

STOP IF YOU FEEL FAINT.

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3. Measure your breathing rate for the first minute after you resume normal

breathing and record it in Table III.

QUESTIONS:

a. Did you feel an urge to breathe right after hyperventilating?

b. Was your blood pH higher or lower?

c. Was you blood CO2 higher or lower?

d. As a result, were your breaths faster or slower than normal?

e. Does increased oxygen in your blood affect your breathing rate?

4. Wait until you are breathing normally before continuing.

5. Breathe into a plastic bag for 2.5 minutes, and record in Table III your

breathing rate for each half-minute (30 second) interval.

6. Calculate breaths per minute for each interval.

STOP IF YOU FEEL FAINT. DO NOT EXCEED 3 MINUTES.

QUESTIONS:

a. Did your breathing become faster or slower after breathing into the bag?

b. Why?

c. There was still oxygen left in the bag. What would have happened if the

CO2 had been removed from each breath that you exhaled?

132

TABLE III. Rate of Human Breathing:

Condition Time Interval Breaths Br h Per Minute

1 m i n .

Normal 2nd min,

3rd min

After

rvnil in

Ist half min,

2nd min.

3rd half min,

Breaths into bag 4th half in

5th half min.

half mi,n,

IV. Spirometry:

The word respiration means one inspiration plus one expiration. A normal adult has 14 t o

18 respirations in a minute, during which the lungs exchange specific volumes of air with

the atmosphere. Pulmonary malfunction usually produces lower-than-normal

exchange volumes. A respirometer (spirometer) is the instrument commonly used to

measure volumes of air exchanged in breathing. Two types are available: the hand-held

type or the tank-type or Collin's recording spirometer.

The Collins respirometer consists of a weighted drum, containing air, inverted over a

chamber of water. The air-filled chamber is connected to the subject's mouth by a tube.

When the subject inspires, air is removed from the chamber, causing the drum to sink and

producing an upward deflection. This deflection is recorded by the stylus on the graph

paper on the kymograph (rotating drum). When the subject expires, air is added, causing the

drum to rise and producing a downward deflection. These deflections are recorded as a

spiroeram. These spirometric studies measure lung capacities and rates and depths of

ventilation for diagnostic purposes.

133

As the following respiratory volumes and capacities are discussed, keep in mind that the

values given vary with age, height, and sex. Each inspiration of normal, quiet breathing

pulls about 500 ml (cc) of air into the respiratory passageways. The same amount moves

out with each expiration, and this volume of air inspired (or expired) is called tidal

volumel(figure 3). Only about 350 ml of this tidal volume reaches the alveoli. The other

150 ml is called dead air volume because it remains in the dead spaces of the nose,

pharynx, larynx, trachea, and bronchi.

If we take a very deep breath, we can inspire much more than 500 ml. The additional

inhaled air, called the inspiratory reserve volume, averages 3100 ml above the 500 ml of

tidal volume. Thus, our respiratory system can pull in as much as 3600 ml of air. If we

inspire normally and then expire as forcibly as possible, we can push out 1200 ml of air

in addition to the 500-ml tidal volume. This extra 1200 ml is called expiratory reserve

volume. Even after the expiratory reserve volume is expelled, a considerable amount of

air still remains in the lungs because the lower intrapleural pressure keeps the alveoli

slightly inflated. This air, the residual volume, amounts to about 1200 ml. When the

chest cavity is opened, the intrapleural pressure equals the atmospheric pressure, which

forces out the residual volume. The lungs still contain a small amount of air called

minimal volume, which can be demonstrated by placing a piece of lung in water and

watching it float.

Lung capacity can be calculated by combining various lung volumes. Inspiratory

capacity, the total inspiratory ability of the lungs, is the sum of tidal volume plus

inspiratory reserve volume (3600 ml). Functional residual capacity is the sum of residual

volume plus expiratory reserve volume (2400 ml). Vital capacity is the sum of

inspiratory reserve volume, tidal volume, and expiratory reserve volume (4800 ml).

Finally, total lung capacity is the sum of all volumes (6000 ml). 6,000 ml

5,000 ml

4,000 ml

3,000 ml

2,000 ml

1 000 m1

6,000ml

- 5,000 ml - INSPIRATORY

RESERVE

VOLUME INSPIRATORY 3,100 rN CAPACITY

- 4,000 ml - 3.600 rru

VITAL

CAPACITY

4.800 m

TOTAL

- 3,000 ml - LUNG

TIDAL CAPACITY

VOLUME 500 nv 6.000 IN

EXPIRATORY - 2000 m RESERVE

VOLUME

1,200 ml FUNCTIONAL V - R RESIDUAL ESIDUAL

- 1 000 m1 - �Ir 2,400m]

RESIDUAL

VOLUME

1,200 mI

L

Figure 3. Spirogram of Lung Volumes and Capacities

134

6,000 ml

5,000 ml

4.000 ml

3.000 ml

2.000 ml

1,000 ml

A. Respiratory Sounds:

Air flowing through the respiratory tract creates characteristic sounds that can be

detected through the use of a stethoscope. Perform the following exercises.

MATERIALS NEEDED:

Stethoscope

PROCEDURE:

1. Place the bell portion of the stethoscope just below the larynx and listen for

bronchial sounds during both inspiration and expiration.

2. Move the stethoscope slowly downward toward the bronchial tubes until the

sounds are no longer heard.

3. Place the stethoscope under the scapula, under the clavicle, and over different

intercostal spaces on the chest, and listen for any sound during inspiration and expiration.

QUESTIONS:

1. What does the air movement in the bronchial tubes sound like?

2. Does it sound different when you place the bell of the stethoscope under the

scapula?

B. Use of a Hand-Held Spirometer:

Although this device cannot measure inhalation volumes, it is possible to determine most of the essential lung capacities.

MATERIALS NEEDED:

Hand-held spirometers

Disposable mouthpieces

70 % alcohol

Cotton 135

‘

1.

PROCEDURE:

Tidal Volume:

The amount of air that moves in and out of the lungs during a normal

respiratory cycle is called the tidal volume. Although sex, age, and weight

determine this and other capacities, the average normal tidal volume is

around 500 ml. Proceed as follows to determine your tidal volume.

a. Swab the stem of the spirometer with 70% alcohol and place a

disposable mouthpiece over the stem.

b. Rotate the dial of the spirometer to zero.

c. After three normal breaths, expire three times into the spirometer while inhaling through the nose. Do not exhale forcibly. Always hold the

spirometer with the dial upward.

d. Divide the total volume of the three breaths by 3. This is your tidal

volume. Record your tidal volume in Table IV. (Remember the so-

called norm is about 500 mls.)

2. Expiratory Reserve Volume (ERV):

The amount of air that one can expire beyond the tidal volume is called the

expiratory reserve volume. It is usually around 1,100 ml.

To determine this volume, set the spirometer dial on 1,000 first. After

making three normal expirations, expel all the air you can from your lungs

through the spirometer. Subtract 1,000 from the reading on the dial to

determine the exact volume. Record in Table IV.

3. Vital Capacity (VC):

If we add the tidal, expiratory reserve, and inspiratory reserve volumes, we

arrive at the total functional or vital capacity of the lungs. This value is

determined by directing an individual to take as deep a breath as possible and exhaling all the air possible. Although the average vital capacity for men and

women is around 4,500 ml, age, height, and sex do affect this volume

appreciably. Even the established norms can vary as much as 20% and still

be considered normal. Tables of normal values for men and women are

given in Tables V and VI.

136

Set the spirometer dial on zero. After taking two or three deep breaths and

exhaling completely after each inspiration, take one final deep breath and

exhale all the air through the spirometer. A slow, even, forced exhalation

is optimum.

Repeat two or more times to see if you get approximately the same readings. Your VC should be within 100 ml each time. Record the average in Table IV. Consult Tables V and VI for predicted (normal) values.

4. Inspiratory Capacity (IC):

If you take a deep breath to your maximum capacity after emptying your lungs of tidal air, you will have reached your maximum inspiratory capacity.

This volume is usually around 3,000 ml. Note from the spirogram that this

volume is the sum of the tidal and inspiratory reserve volumes.

Since this type of spirometer cannot record inhalations, it will be necessary to

calculate this volume, using the following formula:

IC=VC - ERV

Record in Table IV.

5. Inspiratory Reserve Volume (IRV):

This is the amount of air that can be drawn into the lungs in a maximal

inspiration after filling the lungs with tidal air. Since the IRV is the

inspiratory capacity less the tidal volume, make this subtraction and record

in Table IV.

6. Residual Volume (RV):

The volume of air in the lungs that cannot be forcibly expelled is the residual

volume. No matter how hard one attempts to empty one's lungs, a certain

amount, usually around 1,200 ml, will remain trapped in the tissues. The

magnitude of the residual volume is often significant in the diagnosis of

pulmonary impairment disorders. Although it cannot be determined by

simple ordinary spirometric methods, it can be done by washing all the

nitrogen from the lungs with pure oxygen and measuring the volume of

nitrogen expelled. We won't be measuring residual volume in this

laboratory. A residual volume of 1200 ml has been recorded for you.

137

TABLE IV: Lung Capacities Using a Hand-Held Spirometer:

Ts of QUacifics Volume ml "Norms"

Tidal Volume 500 ml

Expiratgry Reserve m

Vital Inspiratory

Inspirato1y Reserve V

Residual

Total Lung CaMity

1200 ml

1200 ml

4800 ml

ml

3100 ml

1200 ml

6000 ml

Sum of all Volumes)

138

TABLE V: Predicted Vital Capacities for Females:

HEIGHT IN CENTIMETERS AND INCHES

cm 152 154 156 158 160 162 164 166 168 170 172 174 176 178 180 182 184 186 188 Agein. 59.8 60.6 61.4 62.2 63.0 63.7 64.6 65.4 66.1 66.9 67.7 68.5 69.3 70.1 70.9 71.7 72.4 73.2 74.0

16 3,070 3,110 3,150 3,190 3,230 3,270 3,310 3,350 3,390 3,430 3,470 3,510 3,550 3,590 3,630 3,670 3,715 3,755 3.800

1 7 3,055 3,095 3,135 3,175 3,215 3,255 3,295 3,335 3,375 3,415 3,455 3,495 3,535 3,575 3,615 3,655 3.695 3,740 3.780

1 8 3,040 3,080 3,120 3.160 3,200 3,240 3,280 3,320 3,360 3,400 3,440 3,480 3,520 3,560 3,600 3,640 3,680 3,720 3,760

20 3,010 3,050 3.090 3,130 3,170 3,210 3,250 3,290 3,330 3,370 3,410 3,450 3,490 3,525 3,565 3,605 3.645 3,695 3,720

22 2,980 3,020 3,060 3,095 3,135 3,175 3,215 3,255 3,290 3,330 3,370 3,410 3,450 3,490 3,530 3,570 3,610 3,650 3,685

24 2.950 2,985 3,025 3,065 3,100 3,140 3,180 3.220 3.260 3.300 3,335 3,375 3.415 3,455 3,490 3,530 3,570 3.610 3.650

26 2,920 2,960 3.000 3,035 3,070 3,1 10 3,150 3,190 3,230 3,265 3,300 3,340 3,380 3,420 3,455 3,495 3,530 3,570 3,610

28 2,890 2,930 2.965 3,000 3,040 3,070 3,115 3,155 3,190 3,230 3,270 3,305 3,345 3,380 3,420 3,460 3,495 3,535 3,570

30 2,860 2,895 2,935 2,970 3.010 3,045 3,085 3,120 3,160 3,195 3,235 3,270 3,310 3,345 3,385 3.420 3,460 3,495 3,535

32 2,825 2,865 2,900 2,940 2,975 3,015 3,050.3,090 3,125 3,160 3,200 3,235 3,275 3,310 3,350 3,385 3,425 3.460 3.495 34 2,795 2,835 2,870 2,910 2,945 2,980 3,020 3,055 3,090 3,130 3,165 3,200 3,240 3,275 3,310 3,350 3,385 3,425 3,460

36 2,765 2,805 2,840 2,875 2,910 2,950 2,985 3,020 3,060 3,095 3,130 3,165 3,205 3.240 3.275 3,310 3.350 3,385 3.420

38 2.735 2.770 2.810 2,845 2.880 2,915 2,950 2,990 3,025 3,060 3,095 3,130 3,170 3,205 3,240 3,275 3,310 3.350 3,385

40 2,705 2,740 2.775 2,810 2,850 2,885 2,920 2,955 2,990 3,025 3,060 3,095 3,135 3,170 3,205 3,240 3,275 3,310 3,345

42 2,675 2,710 2.745 2,780 2,815 2,850 2,885 2,920 2,955 2,990 3.025 3060 3,100 3,135 3,170 3.205 3,240 3,275 3,310

44 2,645 2,680 2,715 2,750 2,785 2,820 2,855 2,890 2,925 2,960 2,995 3,030 3.060 3,095 3,130 3.165 3,200 3,235 3,270

46 2,615 2,650 2,685 2,715 2,750 2,785 2,820 2,855 2,890 2.925 2.960 2,995 3,030 3,060 3,095 3,130 3,165 3,200 3,235

48 2,585 2.620 2.650 2,685 2,715 2,750 2,785 2,820 2,855 2,890 2.925 2.960 2,995 3.030 3,060 3.095 3.130 3,160 3.195

50 2,555 2,590 2,625 2,655 2.690 2,720 2,755 2,785 2,820 2,855 2.890 2,925 2,955 2,990 3,025 3.060 3.090 3,125 3.155

52 2,525 2,555 2,590 2,625 2,655 2,690 2,720 2,755 2,790 2,820 2,855 2,890 2,925 2,955 2,990 3,020 3,055 3,090 3,125

54 2,495 2,530 2,560 2,590 2,625 2,655 2,690 2,720 2,755 2,790 2,820 2,855 2,885 2,920 2,950 2,985 3,020 3.050 3.085

56 2,460 2,495 2,525 2,560 2,590 2,625 2,655 2,690 2,720 2,755 2,790 2,820 2,855 2,885 2,920 2,950 2,980 3,015 3.045

58 2,430 2,460 2.495 2,525 2,560 2,590 2,625 2,655 2,690 2,720 2,750 2,785 2,815 2,850 2,880 2,920 2.945 2,975 3,010

60 2,400 2,430 2,460 2,495 2,525 2,560 2,590 2,625 2,655 2,685 2,720 2,750 2,780 2,810 2,845 2,875 2,915 2,940 2,970

62 2,370 2,405 2,435 2,465 2,495 2,525 2,560 2,590 2,620 2,655 2,685 2,715 2,745 2,775 2,810 2,840 2.870 2,900 2,935

64 2,340 2,370 2,400 2,430 2,465 2,495 2,525 3,555 2,585 2,620 2,650 2,680 2,710 2,740 2,770 2,805 2,835 2,865 2,895

66 2.310 2,340 2,370 2,400 2,430 2,460 2,495 2.525 2,555 2,585 2,615 2,645 2,675 2,705 2,735 2,765 2,800 2,825 2.860

68 2.280 2,310 2.340 2,370 2.400 2,430 2,460 2,490 2,520 2,550 2,580 2,610 2,640 2,670 2,700 2,730 2,760 2,795 2,820

70 2,250 2,280 2,310 2,340 2,370 2,400 2,425 2,455 2,485 2,515 2,545 2,575 2,605 2,635 2,665 2,695 2,725 2,755 2,780

72 2.220 2,250 2,280 2,310 2,335 2,365 2,395 2,425 2,455 2,480 2,510 2,540 2,570 2,600 2,630 2,660 2,685 2,715 2,745

74 2,190 2,220 2,245 2,275 2,305 2,335 2,360 2,390 2,420 2,450 2,475 2,505 2,535 2,565 2,590 2,620 2,650 2,680 2,710

From: Archives of Environmental Health, February, 1966, Vol. 12, pp 146-189, E. A. Gaensler, MD and G. W. Wright, MD

139

TABLE VI• Predicted Vital Capacities for Males: HEICUHT IN CENTIMETERS AND INCHES

3.665 3,710 166 168 170 172 65.4 66.1 66.9 67.7

4,285 4,335 4,385 4,440

4,250 4,300 4,350 4,405

4,215 4,265 4,320 4,370

4,185 4,235 4,285 4,335

4,135 4.185 4,235 4,285

4,100 4,150 4,200 4,250

4,070 4,1 1 5 4,165 4,215

4,035 4,080 4,130 4,180

4,000 4,050 4.095 4,145

3.950 4,000 4,045 4,095

3,920 3,965 4,010 4,060

3.885 3,930 3,980 4,025

3,850 3,900 3,945 3,990

3,820 3.865 3,910 3,955

3,770 3,815 3.860 3,905

3,735 3.780 3.825 3,870

3.700 3,745 3,790 3,835

3,650 3,695 3,740 3,785

3,620 3,660 3,705 3,750

3,585 3,630 3,670 3,715

3,550 3,595 3,640 3,680

3,500 3.545 3,585 3,630

3,470 3,500 3,555 3,595

3,440 3,480 3,520 3,560

3,400 3,440 3,490 3,530

3,350 3,390 3,430 3,470

3,320 3.360 3,400 3,440

3, 29 0 33 3 0 3 37 0

174 176 178 180

68.5 69.3 70.1 70.9

4,490 4,540 4,590 4,645

4,455 4,505 4,555 4,610

4,420 4,470 4,520 4,570

4,385 4,435 4,485 4,535

4,330 4,380 4,430 4,480

4,300 4,350 4,395 4,445

4,265 4,310 4,360 4,410

4,230 4,275 4,325 4,375

4,195 4,240 4,290 4.340

4,140 4,190 4,225 4,285

4,105 4,155 4,200 4,250

4,070 4,120 4,165 4,210

4.035 4,085 4,130 4,175

4,000 4,050 4,095 4,140

3,950 3,995 4.040 4.085

3,915 3,960 4,005 4,050

3,880 3,925 3,970 4,015

3,830 3,870 3,915 3,960

3,795 3,835 3,880 3,925

3,760 3,800 3,845 3.890

3,725 3,765 3,810 3,850

3.670 3,715 3,755 3,800

3.635 3.680 3,720 3,760

3.600 3,640 3,680 3,730

3,570 3,610 3,650 3,690

3,510 3,550 3,600 3,640

3,480 3,520 3,560 3,6

182 184 186 188 71.7 72.4 73.2 74.0

4,695 4.745 4.800 4.85

4.660 4,710 4.760 4.81

4,625 4.675 4.725 4.77.

4,585 4.635 4,685 4.735

4,530 4,580 4.630 4.68

4,495 4.545 4.595 4.645

4.460 4.510 4,555 4.605

4.425 4.470 4.520 4,570

4,385 4,435 4.485 4.530

4.330 4.380 4,425 4,475

4,295 4,340 4.390 4,435

4,260 4.305 4.350 4.400

4,220 4.270 4.315 4,360

4,185 4.230 4.280 4.325

4.130 4,175 4.220 4.270

4.095 4. 40 4.185 4,230

4.060 4.105 4,150 4,190

4,005 4,050 4.090 4,135

3.970 4,010 4,055 4,100

3,930 3,975 4.020 4.060

3.895 3,940 3,980 4.025

3.840 3,880 3.925 3,965

3.805 3,845 3.885 3.930

3,770 3,810 3,850 3.890

3.730

140

Experiment

Heart Rate, Blood Pressure, and Exercise

11

The adaptability of the heart can be observed during exercise, when the metabolic activity of

skeletal muscles increases. The cardiovascular system, consisting of the heart and blood vessels,

responds to exercise with an increase in heart rate and strength of contraction with each beat,

resulting in a higher cardiac output (cardiac output = quantity of blood pumped through the heart

per unit of time) and blood pressure. Positive pressure is created by forceful contraction of the

left ventricle of the heart, measured as systole. It is maintained during relaxation of the ventricle

by closure of the aortic valve and recoil of arteries, measured as diastole (see Figure 1).

Mean arterial pressure (MAP) is a useful measure of the adequacy of tissue perfusion, and is not a

simple average of systolic and diastolic blood pressures. This is because diastole continues for twice as long as systole. MAP can be reasonably approximated using the equation:

(systole + 2(diastole)) =MAP

3

The mean arterial'pressure is directly proportional to cardiac output and inversely proportional to total

peripheral resistance, where:

Cardiac output is the amount of blood pumped out of the heart with each beat (called the stroke

volume), multiplied by the number of beats per minute.

Total peripheral resistance depends on blood viscosity, length of the arterial system, diameter and elasticity of the blood vessels, and the pressure entering versus leaving the arterial system (systolic

pressure minus the pressure in the venous system).

Ventricular Contraction Ventricular Relaxation (systole) (diastole)

Figure 1

Human Physiology with Vernier 141 11 - 1

Experiment 11

In this experiment, you will observe how the heart responds to the increased metabolic demand

of muscles during exercise. You will compare heart rate and blood pressure readings taken before

and after exercise and measure changes in systolic, diastolic and mean arterial pressures. You

will also consider the effect that exercise has on cardiac output and peripheral vascular

resistance.

Important: Do not attempt this experiment if physical exertion will aggravate a health problem.

Inform your instructor of any possible health problems that might be exacerbated if you participate in this exercise.

OBJECTIVES

In this experiment, you will

• Obtain graphic representation of heart rate and blood pressure.

• Determine the effect of exercise on heart rate, and systolic, diastolic and mean arterial pressures.

• Use blood pressure readings and pulse to infer changes in cardiac output and peripheral vascular resistance with exercise.

• Correlate the fitness level of individuals with amount of daily exercise.

MATERIALS

computer Vernier computer interface Logger Pro Vernier Blood Pressure Sensor

PROCEDURE

Part I Baseline Blood Pressure

Vernier Hand-Grip Heart Rate Monitor or Vernier Exercise Heart Rate Monitor

saline solution in dropper bottle

(only for use with the Exercise HR Monitor)

1. Connect the Blood Pressure Sensor to Channel 1 of the Vernier computer interface. Open the file "11 a Heart Rate BP Exercise" from the Human Physiology with Vernier folder.

2. Attach the Blood Pressure Sensor to the blood pressure cuff if it is not

already attached. There are two rubber tubes connected to the 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.

3. Attach the Blood Pressure cuff to 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).

Figure 2 4. The subject should sit quietly in a chair and avoid moving his or her arm or

hand during blood pressure measurements.

11-2 142

Human Physiology with Vernier

Heart Rate, Blood Pressure, and Exercise

5. Click to begin data collection. Immediately begin to pump until the cuff pressure reaches at least 160 mm Hg. Stop pumping.

6. During this time the systolic, diastolic, and mean arterial pressures will be calculated by the software. These values will be displayed on the computer screen. 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 terminate data collection by clicking

_. Release the pressure from the cuff, but do not remove it.

7. Enter the pulse and the systolic, diastolic, and mean arterial pressures in Table 1.

Part II Heart Rate and Blood Pressure after Exercise

8. Connect the receiver module of the Heart Rate Monitor to Channel 2 of the Vernier computer interface. Open the file "1 la Heart Rate BP Exercise" from the Human Physiology with Vernier folder.

9. Set up the Heart Rate Monitor. Follow the directions for your type of Heart Rate Monitor.

Using a Hand-Grip Heart Rate Monitor

a. Grasp the handles of the Hand-Grip Heart Rate Monitor. Place the fingertips of each hand on the reference areas of the handles (see Figure 3).

b. The left hand grip and the receiver are both marked with an alignment arrow. When collecting data, be sure that the arrow labels on each of these devices are in alignment (see Figure 3) and that they are not too far apart. The reception range of the plug-in receiver is 80-100 cm, or 3 feet.

10. Stand quietly facing your table or lab bench. Figure 3

1 1. Click to begin data collection. There will be a 15 second delay while data are collected before the first point is plotted on the heart rate graph. Thereafter, a point will be plotted every 5 s.

12. Determine that the Heart Rate Monitor is functioning correctly. The readings should be consistent and within the normal range of the individual, usually between 55 and 100 beats per minute. If readings are stable, click ! stop and continue to the next step.

13. Click to begin data collection. If the baseline appears stable, begin to run in place at 40 s. Continue data collection while running in place for the next 2 minutes.

14. At approximately 160 s, stop running. Stand still. Do not move during blood pressure measurement.

15. Immediately begin to pump the blood pressure cuff until the cuff pressure reaches at least 160 mm Hg. Stop pumping.

16. During this time the systolic, diastolic, and mean arterial pressures will be calculated by the software. These values will be displayed on the computer screen. When the blood pressure

Human Physiology with Vernier 143

11-3

Experiment 11

readings have stabilized (after the pressure drops to 50 mm Hg), the program will stop calculating blood pressure. At this point, release the pressure from the cuff.

17. Enter the systolic, diastolic, and mean arterial pressures in Table 2.

18. The subject should continue to stand in place while his/her heart rate slows toward its resting pre-exercise value. Data will be collected for 280 s.

19. Click and drag over the area of the graph where the resting heart rate is displayed (from 0 to approximately 40 s). This will highlight the region of interest.

20. Click the Statistics button, Record the mean resting heart rate in Table 3.

21. Drag the right hand bracket to the right edge of the graph, until all the data points are highlighted. The values in the Statistics box will be adjusted based on the data within the brackets. Record the maximum heart rate in Table 2 (under "pulse") and in Table 3.

22. Move the statistics brackets to highlight the area of the graph beginning with the maximum heart rate and ending with the first data point that matches the initial baseline value (or the last point graphed, if baseline is not achieved). Record the Ox value displayed at the lower left corner of the graph as the recovery time in Table 3.

11-4 144 Human Physiology with Vernier

Experiment

Lung Volumes and Capacities 19

Measurement of lung volumes provides a tool for understanding normal function of the lungs as well as disease states. The breathing cycle is initiated by expansion of the chest. Contraction of the diaphragm causes it to flatten downward. If chest muscles are used, the ribs expand outward. The resulting increase in chest volume creates a negative pressure that draws air in through the nose and mouth. Normal exhalation is passive, resulting from "recoil" of the chest wall, diaphragm, and lung tissue. In normal breathing at rest, approximately one-tenth of the total lung capacity is used. Greater

amounts are used as needed (i.e., with exercise). The following terms are used to describe lung volumes (see Figure 1):

Tidal Volume (TV):

Inspirator y Reserve Volume (IRV):

Expiratory Reserve Volume (ERV):

Vital Capacity (VC):

Residual Volume (RV):

Total Lung Capacity (TLC):

Minute Ventilation:

The volume of air breathed in and out without

conscious effort

The additional volume of air that can be inhaled with maximum effort after a normal inspiration

The additional volume of air that can be forcibly exhaled after normal exhalation

The total volume of air that can be exhaled after a maximum inhalation: VC = TV + IRV + ERV

The volume of air remaining in the lungs after maximum exhalation (the lungs can never be completely emptied)

=VC+RV

The volume of air breathed in 1 minute: (TV)(breaths/minute)

inspiratory

reserve volume

!IRV)

expiratory

reserve volume

(ERV)

f residual volume (RV)

Figure I

In this experiment, you will measure lung volumes during normal breathing and with maximum effort. You will correlate lung volumes with a variety of clinical scenarios.

Human Physiology with Vernier 145 9-1

Experiment 19

OBJECTIVES


Obtain graphical representation of lung capacities and volumes.

Compare lung volumes between males and females.

Correlate lung volumes with clinical conditions.

MATERIALS

computer disposable mouthpiece Vernier computer interface disposable bacterial filter Logger Pro nose clip Vernier Spirometer

PROCEDURE

Important: Do not attempt this experiment if you are currently suffering from a respiratory

ailment such as the cold or flu.

1. Connect the Spirometer to the Vernier computer interface. Open the file "19 Lung Volumes" from the Human Physiology with Vernier folder.

2. Attach the larger diameter side of a bacterial filter to the "Inlet" side of the Spirometer. Attach a gray disposable mouthpiece to the other end of the bacterial filter (see Figure 2).

.i V-

IT Figure 2

3. Hold the Spirometer in one or both hands. Brace your arm(s) against a solid surface, such as a table, and click to zero the sensor. Note: The Spirometer must be held straight up and down, as in Figure 2, and not moved during data collection.

4. Collect inhalation and exhalation data.

a. Put on the nose plug.

b. Click to begin data collection.

c. Taking normal breaths, begin data collection with an inhalation and continue to breathe in and out. After 4 cycles of normal inspirations and expirations fill your lungs as deeply as possible (maximum inspiration) and exhale as fully as possible (maximum expiration). It is essential that maximum effort be expended when performing tests of lung volumes.

d. Follow this with at least one additional recovery breath.

5. Click to end data collection. 19-2 146 Human Physiology with Vernier

Luny Volumes and Capacities

6. Click the Next Page button, E, to see the lung volume data. If the baseline on your graph has drifted, use the Baseline Adjustment feature to bring the baseline volumes closer to zero, as in Figure 3.

7. Select a representative peak and valley in the Tidal Volume portion of your graph. Place the cursor on the peak and click and drag down to the valley that follows it. Enter the y value displayed in the lower left comer of the graph to the

nearest 0.1 L as Tidal Volume in Table 1. Fig ore 3

8. Move the cursor to the peak that represents your maximum inspiration. Click and drag down the side of the peak until you reach the level of the peaks graphed during normal breathing. Enter the y value displayed in the lower left corner of the graph to the nearest 0.1 L as Inspiratory Reserve Volume in Table 1.

9. Move the cursor to the valley that represents your maximum expiration. Click and drag up the side of the peak until you reach the level of the valleys graphed during normal breathing. Enter the y value displayed in the lower left corner of the graph to the nearest 0.1 L as Expiratory Reserve Volume in Table 1.

10. Calculate the Vital Capacity and enter the total to the nearest 0.1 L in Table 1.

VC=TV+ IRV+ ERV

1 1. Calculate the Total Lung Capacity and enter the total to the nearest 0.1 L in Table 1. (Use the value of 1.5 L for the RV.)

TLC=VC+RV

12. Share your data with your classmates and complete the Class Average columns in Table 1.

147

Experiment 12

Analyzing the Heart with EKG An electrocardiogram (ECG or EKG) is a graphical recording of the electrical events occurring

within the heart. In a healthy heart there is a natural pacemaker in the right atrium (the sinoatrial

node) which initiates an electrical sequence. This impulse then passes down natural conduction

pathways between the atria to the atrioventricular node and from there to both ventricles. The

natural conduction pathways facilitate orderly spread of the impulse and coordinated contraction

of first the atria and then the ventricles. The electrical journey creates unique deflections in the

EKG that tell a story about heart function and health (Figure 1). Even more information is

obtained by looking at the story from different angles, which is accomplished by placing

electrodes in various positions on the chest and extremities. A positive deflection in an EKG

tracing represents electrical activity moving toward the active lead (the green lead in this

experiment).

Five components of a single beat are

traditionally recognized and labeled P, Q, R, S, and T. The P wave represents the start of the electrical journey as the impulse spreads from the sinoatrial node downward from the atria through the atrioventricular node and to the ventricles. Ventricular activation is

represented by the QRS complex. The T wave results from ventricular repolarization, which

is a recovery of the ventricular muscle tissue

to its resting state. By looking at several beats you can also calculate the rate for each component.

Doctors and other trained personnel can look

at an EKG tracing and see evidence for disorders of the heart such as abnormal slowing, speeding, irregular rhythms, injury to muscle tissue (angina), and death of muscle tissue (myocardial infarction). The length of

an interval indicates whether an impulse is following its normal pathway. A long interval

P = Atrial

Contraction

QRS = Ventricular

Contraction

= Ventricular Repclarization

Figure I

reveals that an impulse has been slowed or has taken a longer route. A short interval reflects an

impulse which followed a shorter route. If a complex is absent, the electrical impulse did not rise normally, or was blocked at that part of the heart. Lack of normal depolarization of the atria leads

to an absent P wave. An absent QRS complex after a normal P wave indicates the electrical impulse was blocked before it reached the ventricles. Abnormally shaped complexes result from

abnormal spread of the impulse through the muscle tissue, such as in myocardial infarction where the impulse cannot follow its normal pathway because of tissue death or injury. Electrical

patterns may also be changed by metabolic abnormalities and by various medicines.

In this experiment, you will use the EKG sensor to make a five second graphical recording of

your heart's electrical activity, and then switch the red and green leads to simulate the change in

electrical activity that can occur with a myocardial infarction (heart attack). You will identify the

different components of the waveforms and use them to determine your heart rate. You will also

determine the direction of electrical activity for the QRS complex.

Human Physiology with Vernier 148 12-1

Experiment 12

OBJECTIVES


Obtain graphical representation of the electrical activity of the heart over a period of time.

Learn to recognize the different wave forms seen in an EKG, and associate these wave forms

with activity of the heart.

Determine the heart rate by determining the rate of individual wave forms in the EKG.

Compare wave fonns generated by alternate EKG lead placements.

MATERIALS

computer Vernier EKG Sensor Vernier computer interface electrode tabs Logger Pro

PROCEDURE

Part I Standard limb lead EKG

1 Connect the EKG Sensor to the Vernier computer interface. Open the file "12 Analyzing

Heart EKG" r fom the Hannan Physiology with Vernier folder.

2. Attach three electrode tabs to your arms, as shown in Figure 2. Place a single patch on the inside of the right wrist, on the inside of the right upper forearm (distal to the elbow), and on the inside of the left upper forearm (distal to elbow).

3. Connect the EKG clips to the electrode tabs as shown in Figure 2. Sit in a relaxed position in a chair, with your forearms resting on your legs or on the arms of the chair. When you are properly positioned, have someone

click to begin data collection. Green negative)

4. Once data collection is finished, click and drag to highlight each interval listed in Table I. Use Figure 3 as your guide when determining these intervals. Enter the x. value of each highlighted area to the nearest 0.01 s in Table 1. This value can be found in the lower left corner of the graph.

5. Calculate the heart rate in beats/min using the EKG

data. Record the heart rate to the nearest whole number in Table 1.

BIa kl,groundl

Figure 2

6. Store this run by choosing Store Latest Run from the Experiment menu.

Part 11 Alternate limb lead EKG

7. Exchange the red and green EKG clips so that the green clip is now attached to the electrode tab on the left arm and the red clip is on the right arm. Sit in a relaxed position in a chair, with your forearms resting on your legs or on the arms of the chair. When you are properly positioned, have someone click to begin data collection.

12-2 149 Human Physiology with Vernier

Analyzing the Heart with EKG

8. Print or sketch the tracing for alternate limb lead placement only.

Figure 3

P-R interval: time from the beginning of P wave to the start of the QRS complex

QRS complex: time from Q deflection to S deflection

Q-T interval: time from Q deflection to the end of the T

150

VERTEBRATE ANATOMY AND PHYSIOLOGY



1. Identify various parts of the external anatomy of the grass frog and demonstrate an

understanding of the function of the specified structures.

2. Identify various parts of the circulatory, respiratory, digestive, reproductive, and

2. excretory systems during the dissection of the grass frog and demonstrate an

understanding of the function of the specified structures.

KEY TERMS:

Blood smear

External anatomy of frog Internal anatomy of frog

INTRODUCTION:

The purpose of this laboratory is to give you an introduction to vertebrate systems. The

dissection of the frog will demonstrate parts of the circulatory, respiratory, digestive,

reproductive and excretory systems of vertebrate animals.

MATERIALS NEEDED:

Grass frogs

Dissecting pans and tools

Standard microscopes

Mounting boards

151

I. Exercise 1. Dissection of the Frog:

The frog is in many respects one of the ideal vertebrates to dissect in an unpreserved

condition. There is some advantage to this in that initially the frog's heart can still be

seen to be beating, and in that the tissues and organs appear and feel as they naturally do,

rather than in the discolored and hardened condition in which they are found in a

preserved animal.

The genus Rana includes several hundred species distributed all over the world, some of

which are quite striking in their coloration and behavior. The frog most used in the

laboratory is the North American grass frog, Rana pipiens. In common with all frogs,

this species is found in damp locations, usually in long grass close to water. Individuals

of this species frequently show their presence by leaping into water for protection, when

nature lovers or small boys armed with collecting nets, tin cans, or milk bottles walk

along the banks of ponds or streams. Sometimes they may be seen sitting comfortably

on lily pads, catching the flies that happen past. Frogs are carnivorous and feed on small

animals, which are swallowed whole. Except where the weather is mild enough, that they

need not, Rana pipiens spends its winters in hibernation in mud holes usually burrowing

so deeply as to become completely covered. In the spring those that have survived come

out of hibernation and congregate in shallow ponds and swamps, where mating takes

place. At this time the males croak loudly, and the choral sounds of mating and courtship

take place.

Obtain a frog from the preparation area, as well as a dissecting tray and equipment.

Wash off the frog to eliminate the slime which may cover the frog. Then place your frog in

the dissecting tray and begin the laboratory.

A.Observation of the external anatomy of a frog:

Pick up the frog in your hand and examine it externally. It consists of a head and

mink -- there is no neck or tail. The mouth is extremely wide and extends all the

way from the front to the back of the head. On the dorsal surface of the head in

front of the eyes, are two small holes called the external nares. These each lead

into two small nasal sacs and then through internal nares into the space inside the

mouth called the buccal cavity. The large eyes almost protrude above the dorsal

surface of the head but are directed laterally. They can be retracted into the head

and can be covered with a nictitating membrane that moves up from the lower

margin of the eye. Immediately posterior to the eyes are two circular areas of

rather flat-appearing tissue. These are the tympanic membranes or ear drums.

Posterior to the head the trunk widens, deepens, and then narrows in all dimensions

to almost a blunt point at the posterior end where the lc oaca is located.

152

Label the underlined structures (see above.)

Label External Anatomy of Frog

153

B. Internal Anatomy of the Frog:

Identify and sketch the internal structures of the frog in the space on the next page. A

large class chart on demonstration will help you identify various internal

structures. Check off each on the list below as you find and sketch them. Consult your

text and give the function of each.

Heart

Lungs

Liver

Gall Bladder

Stomach

Small intestine

Large intestine

Spleen

Pancreas

Kidney

Fat bodies

Gonads

(ovaries or testes)

154

ANIMAL BEHAVIOR AND PLANT RESPONSIVENESS



1. Compare several plant responses resulting from a variety of environmental stimuli. 2. Describe isopod response to moisture and light stimuli.

3. Gather and compare class data on the response of planaria to light stimuli.

4. Gather individual and class data on various human behaviors.

KEY TERMS:

Behavior

Conditioned response

Habits

Kinesis

Nastic movement

Reflexes

Stimulus-Response

Taxis

Trial and Error

Tropism

I. BEHAVIOR:

One of the basic characteristics of life is its capacity for response to environmental

change. An organism reacts to a specific change in its environment, a stimulus, by

carrying out some activity that we call a response. The repertoire of responses that

characterizes an organism is called its behavior. Living organisms have evolved a

remarkable diversity of behaviors which, for convenience of discussion, we may group

into two categories: learned behavior, that array of responses acquired by an organism

in the course of its experience, and inherited or innate behavior, which is a part of the

organism's genetic endowment. In this laboratory, we shall study a few examples of

inherited behavior.

Like the nervous and effector systems from which it is molded, behavior is subject to

evolutionary adaptation. As you watch different behaviors today, try to think about them

as adaptive mechanisms for the organisms in question, and ask yourself how they may be

useful to the organism in its natural situation. Think also about how the stimuli you are

using may differ from those to which the organism is normally subject. Discussion with your

neighbors and comparison of results is encouraged.

155

Several shorthand terms will be helpful to you in describing the behaviors you see in this

laboratory, and also in thinking about their adaptive value. Locomotor behavior, that

involving movement of an organism from one place to another, is described by two terms,

taxis and kinesis. A taxis is an oriented movement, toward or away from the stimulus

that brought about the movement. A kinesis, on the other hand, is random movement,

caused by a stimulus but not necessarily oriented by it. Some insects, for example, select

a dark habitat in preference to a lighted one. If we can demonstrate that an animal

actually avoids light by moving directly away from it, we can describe its behavior as a

taxis (in this case, a phototaxis, i.e., one caused by light). It is perfectly possible,

however, that light, or the "absence of darkness," may merely initiate random movements

which eventually chance to carry the animal into the darkness, where the movement

"turns off." This sort of behavior would be called kinesis. The terms taxis and kinesis

are most often used to describe animal behavior, but they apply equally well to any

organism that shows locomotion, for example, the motile algae.

Two other terms we shall find useful are nastic movement and tropism; we shall use them

to describe nonlocomotor behavioral movements in plants. A nastic movement is a

change in the posture of a plant or plant part, and it is the same no matter what the

stimulus direction. It does not orient the plant toward or away from the stimulus, as

could the locomotor behaviors we discussed above. A tropism, in sharp contrast, is a

growth curvature, a postural change oriented by the direction of the stimulus.

We may add a prefix, as shown in the following list, to any of the terms discussed above, to

describe the nature of the stimulus. We may add the word positive or negative to

describe the direction of the movement relative to the stimulus. For example, the bending of a sunflower toward the sun would be called a positive phototropism.

photo- reaction to light

thermo- reaction to temperature geo- reaction to gravity hydro- reaction to water

hygro- reaction to moisture or humidity thigmo- reaction to contact

chemo- reaction to chemicals

galvano- reaction to electricity

opto- reaction to change in the visual field

seismo- reaction to shock

Do the following experiments in groups, as directed by your instructor. Your group may be asked to give a report on one of the experiments. In the report, you are expected to pool the

results obtained by the rest of the class and to summarize information regarding the adaptive

value of the behavior you discuss.

156

I I Plant Movements

A. Mimosa

Obtain a potted specimen of the "sensitive plant," Mimosa and WITHOUT TOUCHING the leaves, move it carefully to your desk.

1. Stimulate one of the end leaflets by gently touching it; if nothing happens, stimulate it again by touching it with a little more force.

QUESTIONS:

a. What is the plant's response?

b) How would you describe this response, using the terminology in the

introduction?

c) What benefit might the plant derive from this behavior?

d) Note the approximate time when the response was given.

e) The movement you saw was due to a rapid change in the turgor cells at

the base of the leaflet. Can you locate in this region any structure that

might help to transmit the pressure of your touch to sensitive cell?

2. Time the recovery of the leaflets to their normal posture. Design and carry

out an experiment to test the effect of increased severity of stimulus, or try using other stimuli. You might wish to try to find out whether the plant is

particularly sensitive to touch in certain regions of the leaflets. While you are waiting for the plant to recover, go on to the next part.

157

B. Pbotonasty in Oxalis

Obtain an Oxali plant that has been kept in sunlight, and examine the leaves and

stalk. In the space provided, sketch the angle of the leaves relative to the stalk.

1. Place a cardboard box over the plant, and leave it for about an hour. In the

meantime, go on to other laboratory activities. At the end of that time, remove the box just long enough (less than a minute) to glance at the plant so

that you can again sketch the angle of the leaves relative to the stalk.

Immediately replace the cardboard box, and sketch the angle in.

Angle of OXALIS before Angle of OXALIS leaves after experiment

2. With the box still covering the plant, design and carry out an experiment to

prove that this behavior is a nastic movement (that its direction is not

determined by the stimulus).

C. Phototaxis in Flagellated Protists

You will be supplied with a petri dish containing Eu len , a flagellated unicellular

protist.

1. Stir the water in the dish gently, until the Euglena are equally dispersed

throughout the container. Cover the dish completely with aluminum foil,

making certain that no light can enter. After half an hour, remove the foil and

record the distribution of the protists in a sketch.

158

2. Cut a small circle (5 mm diameter) in the aluminum foil cover, and put it back on the container. Illuminate the Euglena through this hole, using a

microscope lamp or other source positioned to deliver a vertical beam of light

into the container, and set up far enough away so that the water will not

become heated. After half an hour, remove the foil cover, and record the

distribution of the Euglena.

QUESTIONS:

a. What would you call this response?

b) How would you design an experiment or observations to determine

whether these were taxis or kinesis?

3. Design and carry out an experiment to demonstrate negative taxis. How

would you design an experiment that would demonstrate negative phototaxis? Eliminate any possible effects of a temperature stimulus.

D. Geotropism

1. To demonstrate geotropism place a sheet of wet paper toweling around the inside of a 400 ml beaker, then crumble up another paper towel, dampen and place in the center.

2. Select eight corn seeds, placing them between the glass and the paper towel.

Place the seeds in such a way so that the seed tips point up, down, left and the last one, right.

3. Wrap in aluminum foil.

4. Place in a dark drawer and observe in two days.

5. Explain the direction taken by the coleoptiles at the end of the observation period, also explain root response to gravity.

E. Phototropism

(Observe film loop.)

159

I II. Animal Movements:

A. Reactions to Moisture and to Light

We will use animals known as is s. These terrestrial crustaceans are sometimes

called pill bugs, sow bugs, or volkswagens. While most crustaceans are aquatic,

such as lobsters, crayfishes, shrimps, etc., the isopods have also established

themselves on land. Their survival on land depends on a very moist habitat, and

terrestrial isopods are usually found only in very moist places, such as under

stones, rotting logs, leaf mold, etc.

Figure 1. Isopod chamber with waxed paper dividing wet and dry paper toweling

1. Prepare two glass bowls by putting a three-inch square of moistened (not

soaking) paper towel into one, and an identical unmoistened square into the

other. Obtain ten isopods from the stock container, and introduce five of

them into each bowl. Cover each bowl with a box to exclude the possibility of

light acting as a stimulus to activity.

QUESTIONS:

a) After ten minutes, remove the boxes and immediately observe the degree

of activity in each bowl. How would you describe the rate of

locomotion in each bowl?

b) Is there any directedness to movement in either of the bowls?

c) Using the terms previously introduced, how would you describe this

reaction?

d) Can you tell, from this experiment, whether the stimulus is presence or

absence of water?

160

e) Can you be certain that the stimulus has anything to do with direct

sensing of liquid water?

f) Could it be a response to presence or absence of moisture or humidity?

g) How could you set up experiments to test the answers to these questions?

2. Obtain a piece of waxed paper and a pan from the supply desk. Line the

bottom of the pan with two pieces of paper toweling, separated in the middle

and overlapped by a two-inch strip of waxed paper (figure 1). Secure the

waxed paper with tape. The paper should fit the pan snugly. With distilled

water, moisten one of the pieces of paper towel (n dripping wet) and leave

the other dry. (The pan may be prepared for you.) Introduce ten isopods to

the waxed-paper strip, and cover the pan with aluminum foil. After ten

minutes, remove the foil and observe the position and activity of the isopods.

QUESTIONS:

a) How many Isopods are on the wet paper?

b) How many are on the dry paper?

c) What is the state of activity. of the isopods on the wet paper?

d) On the dry paper?

e) Repeat the experiment several times. What conclusions can be drawn?

161

In interpreting these experiments, it is important to recognize the existence

of problems in experimental design that may greatly affect the outcome of

your experimental procedures. One such problem is the matter of where the

animals are introduced in the pan. If your introduction chances to direct

more of them toward the moistened area, more will naturally be counted

there; and it is possible that they might never have been counted there if the

introduction of animals had been altogether random. The use of the waxed-

paper strip is a step toward the attainment of random introduction, but there

are many better methods of solving the problem. Discuss the experiment

with others in your group, and be prepared to suggest other ways in which

the problem of attaining randomness in the introduction of animals could be

solved. The simpler your solutions, the better. Your instructor may be able

to provide you with some equipment to carry out your improved experiments,

if there is time.

B. Responsiveness in Animals

The development of a complex system of receptors, effectors, and modulators in

animals results in more complex responses at the organismic level than those

responses observed in plants. Responses may be immediate and may involve

locomotion; e.g., taxis and kinesis. However, sessile animals may show tropistic

movements.

Light

1. Planarians. These primitive worms are especially favorable for the study of

responses to light because their photoreceptors are simple and the variety of

response is limited. Because sensory adaption to light can occur, the

planarians are kept in the dark between experiments.

a) Using a lateral source of light, observe the creeping of planarians. Shift

the light source 90 ° and observe the turning reaction. Is the turning

response one of "trial and error" or is it a directed response? Record

your observations.

b) Place a dish which has one half of the bottom white and the other half

of the bottom black under a light. Place ten planarians in the center of

the dish and leave them undisturbed for 1 hour. Then count the number

that have come to rest on the dark and on the backgrounds. Record you

results in the table on the chalkboard and in Table I in the row

"individual results." After all individual results are recorded, copy them

into your table and sum them to obtain the total class results.

162

TABLE I:

Number on White Number on Black % on Individual

Results

Total

Class Results

QUESTIONS:

1. Is there any difference in the number of planaria on the white and black background?

2. Was the difference the same in all individual tests?

3. Calculate the percent of animals on black for each individual test and for the

class totals. What comments can you make concerning the advisability of

using large numbers of animals or of repeating experiments?

163

IV. BEHAVIOR

1. The Simple Reflex

Work in pairs.

A. Experiment

One student holds a sheet of window glass directly in front of his face with his

nose touching the glass. The second student throws a wadded piece of paper

toweling at the glass.

B. Observation

C. Conclusions

2. The Conditioned Response

Instructions will be given orally.

A. Experiment

B. Observation

C. Conclusions

D. Ouestions

1) What was the unconditioned stimulus?

2) What the conditioned stimulus?

164

3) What was the conditioned response

3. Habits


A. Experiment

B. Observations

C. Conclusions

4. The Conditioned Response


A. Experiment

B. Observations

C. Conclusions

5. Learning by Trial and Error

Work in pairs; instructions will be given orally.

A. Experiment

165

B. Observations

Trial Time i Trial Time 1 7 13 2 14

3 15 4 10 16

11 17

16 1 12 1

Plot a graph showing elapsed time (ordinate) as a function of the number of

the trial. Use the graph paper provided.

C. Conclusions

6. Learning: Meaningful vs. Meaningless Material

A. Experiment

Each student will be handed a list of words (face down) and a sheet of clean

paper. When your instructor gives the signal, turn the list over and try to

memorize as many words as you can before given the signal to stop. When

told to stop, turn the list back over and exchange papers with your neighbor

and score them. Record your results. The entire procedure will then be

repeated with a second list of words.

B. Observations

TEST A TEST B

Number Correct:

X 100 = % X 100 = %

Total Number:

166

APPENDIX A

MEASURING VOLUME To determine the concentration of dissolved substances, it is necessary to be able to measure

volume accurately. There are a number of devices available to measure volume, but the most

common are the graduated cylinder, volumetric flask, and pipette.

OBJECTIVES:

1. Measure volume and report data accurately and precisely.

2. Be able to select and use appropriate measuring devices.

1. THE PIPETTE

Pipettes are usually used to measure liquid volumes of 10 ml or less, although some types of

pipettes can measure up to 100 ml or more. There are two different types of pipettes, to

deliver (TD) and to contain (TC). The label near the upper end of the pipette shaft

indicates the type of pipette. Most pipettes are the TD type. There are two varieties of

TD pipettes: "delivery" and "blow out." Examples are shown below. Notice that on the

delivery pipette the scale stops before the pipette narrows. When using this type of

pipette, it is important to deliver liquid only down to this line. 10 ml in 1/10 TD

Types of pipettes: (a) to deliver (TD) (a)

delivery pipette, (b) to deliver

(TD) blow-out pipette, 10 ml in 1110 TD L m m -+ m rn a as N 1

(b)

Whenever using a pipette, use a propipette or bulb, never a mouth-pipette in this class.

Your instructor will demonstrate the proper method of using a plunger type of propipette.

167

H. THE GRADUATED CYLINDER

For measuring larger amounts of fluid, graduated cylinders are used. Two types are in

common use: glass cylinders and cylinders made from organic polymers such as

polycarbonates. Water adheres more strongly to glass than to polymers, and thus the

water at the top of a column will "climb" the sides of a glass cylinder to form a bowl-like

meniscus. To obtain a precise measurement, volume must be read at the bottom of the

"bowl."

PROCEDURE:

1. Partially fill a 100 ml graduated cylinder with water and set it on a table.

2. View the cylinder and estimate volume:

a. from a standing position

b. at eye level

c. from below (see figure below)

How do volumes estimated from these various positions compare?

3. Repeat the procedure using a graduated cylinder made from an organic polymer.

You will notice that there is much less of a meniscus and that the volume can be

read more precisely from any eye level.

168

4. Add water to the 10 ml level in a 100 ml glass graduated cylinder. Pour the 10 ml

of water into a 10 ml graduated cylinder. What does the measurement read on this

cylinder?

Which cylinder would be more accurate for measuring 5 ml of solution: the 100

ml cylinder or the 10 ml cylinder?

5. Fill an Erlenmeyer flask to the 50 ml mark with water. Pour this into a 100 ml

graduated cylinder.

Describe the accuracy of the Erlenmeyer flask.

169

METRIC SYSTEM The International System of Units (SI) has recognized seven base units of measurement, five of

which are of interest to general biology students: meter (length), kilogram (mass), second

(time), degree Celsius (temperature), and mole (amount of substance).

Length

Unit Value Symbol kilometer 1,000 meters km

meter base unit m

decimeter 0.1 meter dm

centimeter 0.01 meter cm

millimeter 0.001 meter mm

Volume

Volume is the three-dimensional space occupied by a material; therefore, volume is measured

by these units:

Unit Value Symbol

cubic meter derived unit m'

cubic decimeter 0.001 cubic meterdm'

cubic centimeter 0.000 001 cubic metercm'

cubic millimeter 0.000 000 001 cu metermm'

Very often, however, the unit called a liter is used for liquid measurement in science. A liter

(1) is equal to a cubic decimeter and a milliliter (ml) is equal to a cubic millimeter.

Mass

Unit Value Symbol

kilogram base unit kg

gram 0.001 kg g

decigram 0. l g dg

centigram 0.01 g cg

milligram 0.001 g mg

Time

Unit Value Symbol second base unit s

minute 60s min

hour 3,600s h

day 86,400 s d 170

Temperature

To change degrees Fahrenheit to degrees Celsius, subtract 32 from the Fahrenheit temperature

and multiply by 5/9. To convert from Celsius to Fahrenheit, multiply the Celsius temperature by

9/5 and add 32.

r-,

1 00°

800

600

40°

20°

00

-0°

40°

W-W

Water

boils

Water 0-0

freezes

Two scales

200°

1 80°

1 60°

140°

1 20°

1 000 80°

60°

40°

320

20° 00

--20°

-40° -5')0

equivalent -50°

'J Celsius (°C)

0

Fahrenheit (°F)

The term Centigrade is very often used instead of Celsius.

Mole

Mass refers to the total quantity of substance and amount of substance; mole refers to the

concentration of substance. the official SI definition of mole (mol) is that it is "the amount of

substance of a system which contains as many elementary entities as there are atoms in 0.012

kilogram of carbon-12." Mole is the unit of measurement to be used when comparing the

concentration of atoms or chemicals.

171

APPENDIX B: DNA Goes to the Races

Student Activity

You have already learned about restriction enzymes and

how they cut DNA into fragments. You may have even

looked at some DNA restriction maps and figured out how

many pieces a particular enzyme would produce from that

DNA. But when you actually perform a restriction digest,

you put the DNA and the enzyme into a small tube and let

the enzyme do its work. Before the reaction starts, the

mixture in the tube looks like a clear fluid. Guess what!

After the reaction is finished, it still looks like a clear fluid!

Just by looking at it, you can't tell that anything happened.

In order for restriction digestion to mean much, you have to

be able somehow to see the different DNA fragments that

are produced. There are chemical dyes that stain DNA, but

obviously it doesn't do much good to add them to the

mixture in the test tube. In the laboratory, scientists

separate DNA fragments so that they can look at the

results of restriction digests (and other procedures) by a

process called gel electrophoresis.

Gel electrophoresis takes advantage of the chemistry of

DNA to separate fragments. Under normal circumstances,

the phosphate groups in the backbone of DNA are

negatively charged. In electrical society, opposites do

attract, so DNA molecules are very much attracted to

anything that is positively charged. In gel electrophoresis,

DNA molecules are placed in an electric field (which has a

positive and a negative pole) so that they will migrate

towards the positive pole.

The electric field makes the DNA molecules move, but to

cause them to separate and be easy to look at later on, the

whole process is carried out in a gel (obviously the source

of the name gel electrophoresis). If you have ever eaten

Jell-O', you have had experience with a gel. The gel

material in Jell-O° is gelatin; different gel materials are used

to separate DNA. One gel material often used for

electrophoresis of DNA is called agarose, and it behaves

much like Jell-O'ID but lacks the sugar and color. To make a

gel for DNA (called pouring or casting a gel), you dissolve

agarose powder in boiling water, pour it into the desired

dish, and let it cool. As it cools, it hardens (sound familiar?).

172

Since the plan for agarose gels is usually to add DNA to

them, scientists place a device called a comb in the liquid

agarose after it has been poured into the desired dish, and

let the agarose harden around the comb. Imagine what

would happen if you stuck the teeth of a comb into liquid

Jell-O® and let it harden. Afterwards, when you pulled the

comb out, you would have a row of tiny holes in the solid

Jell-O® where the teeth had been. This is exactly what

happens with laboratory combs. When the comb is

removed from the hardened agarose gel, a row of holes in

the gel. The holes are called sample wells. DNA samples

are placed into the wells before electrophoresis is begun.

For electrophoresis, the entire gel is placed in a tank of salt

water (not table salt). An electric current is applied across

the tank, so that it flows through the salt water and the gel.

When the current is applied, the DNA molecules begin to

migrate through the gel towards the positive pole of the

electric field.

At this point the gel does its most important work. All of the

DNA in the gel migrates through the gel towards the positive

pole, but the gel material makes it more difficult for larger

DNA molecules to move than smaller ones. So in the same

amount of time, a small DNA fragment can migrate much

further than a large one. You can therefore think of gel

electrophoresis like a DNA footrace, where the "runners"

(the molecules being separated) separate just like runners in a

real race.The smaller the molecule, the faster it runs. Two

molecules the same size run exactly together.

After a time, the electric current is turned off and the entire

gel is placed into a DNA staining solution. After staining, the

DNA can be seen. The resulting pattern looks like a series of

stripes in the gel called bands, where each separate band is

composed of one size of DNA molecule. There are millions

of actual molecules in the band, but they are all the same

size (or very close to it). At any rate, after a restriction digest,

there should be one band in the gel for each different size

fragment produced in the digest. The smallest fragment will

be the one that has migrated furthest from the sample well,

and the largest will be closest to the well.

Activity

You have three representations of a DNA molecule and an

outline of an electrophoresis gel. The representations show

the cut sites of three different restriction enzymes on the

same DNA molecule. You will simulate the digestion of this

DNA with each of the three enzymes, then simulate

agarose gel electrophoresis of the restriction fragments.

1. Cut out the three pictures of the DNA molecule.

2. Simulate the activity of the restriction enzyme EcoRl

on the DNA molecule that shows the EcoRl sites by

cutting across the strip at the vertical lines

representing EcoRI sites. You have now digested the

molecule with EcoRl. Put your "restriction fragments"

in a pile apart from the other two DNA strips.

3. "Digest" the second DNA strip with BamHl. Put the

BamHl fragments in a separate pile.

4. Now digest the remaining DNA molecule with Hindlll.

Put these fragments in a third pile.

5. In our imaginary gel electrophoresis, you will separate

the EcoRl, BamHl, and Hind Ill fragments as if you

loaded the three sets of fragments into separate but

adjacent sample wells. Arrange your fragments as they

would be separated by agarose gel electrophoresis.

Designate an area on your desk as the end of the gel

with the sample wells. Starting with the EcoRl

fragments, arrange them from longest to shortest,

with the longest one closest to the well.

Restriction Maps for DNA Goes to the Races

6. Next, separate the BamHl fragments adjacent to the

EcoRl fragments. Be sure to order the fragments

correctly by size with respect to the other BamHI

fragments and to the EcoRl fragments you have

already laid out.

7. Repeat the same procedure for the Hindlll fragments.

You should now have all three of your sets of fragments

arranged in order in front of you.

8. Look at the outline of the electrophoresis gel. Notice

that it has a size scale in base pairs on the left-hand

side, and that sample wells are drawn in. Use the

outline and draw the pattern your restriction digest

would make in the gel, using the size scale as a guide

for where to draw your fragments. Use the EcoR!

sample well for the EcoRl fragments, and so on.

9. After you draw the bands representing the restriction

fragments, use the size information on the paper DNA

strips to label the bands on the gel with the sizes of

the fragments in base pairs.

10. Use the actual fragment sizes as a check for your work

Are all the smaller fragments across all the gel "lanes" in

front of all the larger fragments?

Did you notice that the size scale doesn't seem to have

regular intervals? The size scale looks the way it does

because agarose gels separate fragments that way.

Below are three representations of a 15,000-base-pair DNA molecule. Each representation shows the locations of

different types of restriction site, with vertical lines representing the cut sites. The numbers between the cut sites

show the sizes (in base pairs) of the fragments that would be generated by digesting the DNA with that enzyme.

EcoRl sites

4,000 1 3,500 2,500 5,00

6,000 4,000 3.000 2,000

Hindlll sites

8,000 4,500 2,500

173

Gel Outline for DNA Goes to the Races EcoRl Hindlll BamHl

Sample wells

Size scale in base pairs

8,000

6,000

4,000

3,000

2,000

174

APPENDIX C: DNA Scissors: Introduction to Restriction Enzymes Background Reading

Genetic engineering is possible because of special enzymes

that cut DNA. These enzymes are called restriction

enzymes, or restriction endonucleases. Restriction enzymes

are proteins produced by bacteria to prevent or restrict

invasion by foreign DNA. They act as DNA scissors, cutting

the foreign DNA into pieces so that it cannot function.

Restriction enzymes recognize and cut at specific places

along the DNA molecule called restriction sites. Each

different restriction enzyme (and there are hundreds, made

by many different bacteria) has its own type of site. In

general, a restriction site is a 4- or 6-base-pair sequence

that is a palindrome. A DNA palindrome is a sequence in

which the "top" strand read from 5' to 3' is the same as

the "bottom" strand read from 5' to 3.' For example:

5' GAATTC 3'

3' CTTAAG 5'

is a DNA palindrome. To verify this, read the sequence of

the top strand and the bottom strand from the 5' end to

the 3' end. This sequence is also a restriction site for the

restriction enzyme called EcoRl. Its name comes from the

bacterium in which it was discovered: Escherichia coli

strain RY 13 (EcoR), and "I" because it was the first

restriction enzyme found in this organism.

EcoRI makes one cut between the G and A in each of the

Drl;-. strands (see below). After the cuts are made, the

DNA is held together only by the hydrogen bonds between

the four bases in the middle. Hydrogen bonds are weak,

and the DNA comes apart.

I cut sites: 5' GAATTC 3'

3' CTTAAG 5'

T cut DNA: 5' G AATTC 3'

3' CTTAA G 5'

The EcoRl cut sites are not directly across from each other

on the DNA molecule. When EcoRl cuts a DNA molecule.

it therefore leaves single-stranded "tails" on the new ends

(see above example). This type of end has been called a

sticky end because it is easy to rejoin it to complementary

sticky ends Not all restriction enzymes make sticky ends,

some cut the two strands of DNA directly across from one

another, producing a blunt end.

When scientists study a DNA molecule, one of the first

things they do is to figure out where many restriction sites

are. They then create a restriction map, showing the

location of cleavage sites for many different enzymes. 175

These maps are used like road maps to the DNA molecule

Below are the restriction sites of several different

restriction enzymes, with the cut sites shown.

I I EcoRl: 5' GAATiC 3' Hindlll: 5' AAGGCTT 3'

3' CTTAAG 5' 3' TTCGAA 5'

T T

I 1 BamHl: 5' GGATCC 3' Alul: 5' ACC-l' 3'

3' CCTAGG 5' 3' TCGA 5'

T T

I I Smal: 5' CCCGGG 3' Hhal: 5' GCGC 3'

3' GGGCCC 5' 3' CGCG 5'

T T

Which ones of these enzymes would leave blunt ends?

Which ones would leave sticky ends?

Refer to the above list of enzyme cut sites as you do the

activity.

Exercises and Questions

Exercise I

Take the figure with the DNA sequence strips and cut out the

strips along the borders. These strips represent double-

stranded DNA molecules. Each chain of letters represents

the phosphodiester backbone, and the vertical lines between

the base pairs represent hydrogen bonds between the bases.

1. You will now simulate the activity of EcoRl. Scan along

the DNA sequence of Strip 1 until you find the EcoRF

site (refer to the list above for the sequence). Make

cuts through the phosphodiester backbone by cutting

just between the G and first A of the restriction site on

both strands. Do not cut all the way through the strip.

Remember that EcoRl cuts the backbone of each DNA

strand separately.

2. Now separate the hydrogen bonds between the cut sites

by cutting through the vertical lines. Separate the two

pieces of DNA. Look at the new DNA ends produced by

EcoRl. Are they sticky or blunt? Write "EcoRl" on the cut

ends. Keep the cut fragments on your desk.

3. Repeat the procedure with Strip 2, this time simulating

the activity of Smal. Find the Smal site and cut through

the phosphodiester backbones at the cut sites indicated

above. Are there any hydrogen bonds between the cut

sites? Are the new ends sticky or blunt? Label the new

ends "Smal" and keep the DNA fragments on your desk.

4. Simulate :he activity of H ndl ll with Strip 3. Are these

ends sticky or blunt? Label the new ends "Hindlll" and

keep the fragments.

5. Repeat the procedure once more with Strip 4, again

simulating EcoRl.

6. Pick up the "front end" DNA fragment from Strip 4 (an

EcoRl fragment) and the "back end" Hindlll fragment

from Strip 3. Both fragments have single-stranded tails

of 4 bases. Write down the base sequence of the two

tails and label them " EcoRl" and "Hindlll." Label the

5' and 3' ends. Are the base sequences of the HindIll

and EcoRl "tails" complementary?

7. Put down the Hindlll fragment and pick up the back end

DNA fragment from Strip 1 (Strip 1 cut with EcoRl).

Compare the single-stranded tails of the EcoRl fragment

from Strip 1 and the EcoRl fragment from Strip 4. Write

down the base sequences of the single-stranded tails

and label the 3' and 5' ends. Are they complementary?

8. Imagine that you cut a completely unknown DNA

fragment with EcoRl. Do you think that the single-

stranded tails of these fragments would be

complementary to the single-stranded tails of the

fragments from Strip 1 and Strip 4?

9. There is an enzyme called DNA ligase that re-forms

phosphodiester bonds between nucleotides. For DNA

Restriction Map of YIP5 DNA

5541 base pairs 176

1

9

2

ligase to \\0-k. t\\0 nucleotides must come ckisc

together in the proper orientation for a bond (the 5 silo

of one must be next to the 3' side of the other) Do you

think it would be easier for DNA ligase to reconnect

two fragments cut by EcoRl or one fragment cut by

EcoRl with one cut by Hindlll? What is your reason?

Exercise ll

Examine the figure that is the restriction map of the circular

plasmid YIP5. This plasmid contains 5,541 base pars.

There is an EcoRl site at base pair 1. The locations of other

restriction sites are shown on the map. The numbers after

the enzyme names tell at what base pair that enzyme

cleaves the DNA. If you digest YIP5 with EcoRl, you will

get a linear piece of DNA that is 5,541 base pairs in length

1. What would be the products of a digestion with the

two enzymes EcoRl and Eagl?

2. What would be the products of a digestion with the

two enzymes Hindlll and Apal?

3. What would be the products of a digestion witn the

three enzymes Hindlll, Apal, and Pvul?

4. If you took the digestion products from question 10

and digested them with Pvull, what would the

products be?

DNA Sequence Strips for DNA Scissors r - - - - - - - - -

5 ' –TAGACTGAATTCAAGTCA- 3-' I I I I I I I I I I I I I I I I I I

3'- ATCTGACTTAAGTTCAGT-5' L------------------------------------------------------------------------------------------

r-------------------------------------------------------------------------------------------

2 2I

5'- ATACGCCCGGGTTCTAAA-3' _ I I I I _ 1 I =1 -I I I I I

3 ' - TATGCGGGCCCAAGATTT - 5 ' L

r

5 ' - CAGGATCGAAGC.TTATGC - 3 ' I ::.',,I : I 41 I .I -:

3 GTCCTAGCTTCGAATACG-5' L------------------------------------------------------------------------------------------ J

r-------------------------------------------------------------------------------------------

4

5 -"-,AATAGAATTCCGATCCGA- 3 ' I I I I I I I I I I I I I I I I I I

3'- TTATCTTAAGGCTAGGCT-5' L-------------------------------------------------------------------------------------------

177

LAB-AIDS" #1-S SIMULATED BLOOD TYPING KIT

Student Worksheet and Guide

The ABO system is a means of classifying blood according to the antigens located on the surface of the erythrocytes (or red blood

cells). There are four blood groups in humans. The erythrocytes of a human may carry an A antigen, a B antigen, both A and B

antigens or no antigens at all. People with type A blood have A antigens on their red blood cells; type B have B antigens;type AB have

both A and B antigens; and type 0 produce neither A nor B antigens. The simulated tests you are about to do are very similar to

those done with real blood.

All materials used in this lab activity are SIMULATED. No human or animal blood products are used. These materials

cannot be used to type actual human blood.

PROCEDURE:

1. Each team of two students should have 4 Slide-Guides"', one for each of the four simulated blood types your team will test.

Label each Slide-Guide with one of the following titles:

1 -Type A 2-Type B 3 -Type AB and 4 - Type 0

I Prepare each of your Slide-Guides as follows:

Slide-Guide 1 -Type A

Carefully place 2 drops of simulated Type A blood on each of the circles.

Slide-Guide 2 -Type B

Carefully place 2 drops of simulated Type B blood on each of the circles. Slide-Guide 3 -Type AB

Carefully place 2 drops of simulated Type AB blood on each of the circles.

Slide-Guide 4 - Type 0

Carefully place 2 drops of simulated Type 0 blood on each of the circles.

3. On your Slide-Guide labeled 1- Type A, add 1-2 drops of simulated Anti-A serum in the circle marked Anti-A and 1-2 drops of

simulated Anti-B serum in the circle marked Anti-B.

4. Use a blue toothpick to gently mix the liquids in the circle marked Anti-A for 30 seconds and use a yellow toothpick to gently

mix the liquids in the circle marked Anti-B for 30 seconds. Observe for precipitate formation, which simulates blood agglutination.

5. Record your observations in the chart below using + for a positive reaction (agglutination) and - for no agglutination (or

precipitate being formed.)

Simulated Blood Antiens on g

Group the red blood cell

A A

B B

AB A and B

0 none

Reactions with: Anti-A Anti-B Serum Serum

6. Rinse your dirty toothpicks and repeat steps 3-5 for your Slide-Guides labeled 2-Type B, 3-Type AB, and 4-Type O.

178

biol 1009 biology lab manual 2010

Documents

appendix c

dna scissors

dissecting microscopes

lab equipment

lab book

appendix b

dna extraction

biology lab manual