bio 110 lab manual spring 2011

Bio 110

Laboratory Manual

Spring 2011

Dr. Maureen Knabb and Dr. Win Fairchild

Revised, Dr. Oné R. Pagán

Department of Biology, West Chester University

Bio 110 Laboratory Manual - Spring 2011

2

GENERAL BIOLOGY LABORATORY

MANUAL INDEX

PAGE

Laboratory safety 3

Adopt-a-bug 4

Avian evolution 14

Organic molecules 18

Microscopes & cells 20

Diffusion and osmosis 28

Data analysis & statistics 34

Fermentation and metabolism 42

Mitosis, meiosis and life cycles 49

Mendelian genetics 57

Duckweed population biology

Daphnia feeding ecology 65

Molecular genetics 81

LABORATORY EVALUATION:

Activity Points

Apply-your-knowledge 30

Lab practical exams 60 (2 x 30)

Quizzes 40

Total 130


3

Bio 110 - Lab Safety & General Rules

1. Read the experimental procedure ahead of lab.

2. Report all accidents to your lab instructor, regardless of how minor.

3. Use goggles and wear lab coats when instructed to do so. It is recommended (but not required) that

students wear laboratory coats to protect their clothing.

4. Due to the dangers of broken glass and corrosive liquid spills in the lab, open sandals or bare feet

are not permitted in the lab.

5. Learn the location and proper usage of the eyewash fountain, fire extinguisher, safety shower, fire

alarm box, evacuation routes, clean-up brush and dust pan, glass/chemical disposal can.

6. Use care whenever samples must be heated.

7. If you get any chemical in your eyes, immediately wash them with the eye-wash fountain and

notify the lab instructor.

8. Never look directly into a test tube. View the contents from the side.

9. Food and drink are prohibited in the lab. Avoid hand-to-mouth operations (e.g. chewing pencils,

licking labels, mouth pipetting).

10. Return all lab materials and equipment to their proper places after use.

11. Upon completion of work, wash and dry all equipment, your lab bench, and your clean-up area.

12. Wash your hands before you leave the lab.

13. Cell Phones: Not allowed in lab. Please silence them before the lab period starts. If you must

take a call, PLEASE take it outside.

14. Grades I will not be e-mailed!

15. Academic Integrity: Be familiar with the University policy on Academic Integrity, whose

principles will be adhered to in this course. Note that it is just as dishonest, and unfair to other

students, to knowingly allow someone to copy your work as it is to copy someone's work. All

instances of academic dishonesty will be dealt with according to University policies.

I have read and will abide by the syllabus and lab safety rules.

______________________________________________________

Print your name here ID#

______________________________________________________

Student signature date

Copies of this form will be available during our first lab.

Please print & sign your name and give it to your lab

instructor


4

Lab exercise 1: Adopt-a-Bug!

Milkweed Bug Growth Experiment

Large milkweed bug - Oncopletus fasciatus; Order Hemiptera, Family Lygaeidae

Congratulations! You’ve just adopted a cute and cuddly milkweed bug. The large milkweed bug is used

extensively in laboratory work by entomologists because it can be reared easily on a diet of the

cracked seeds of sunflower, watermelon, squash and almond. (Note: the seeds must be raw,

unsalted and cracked, as the insects can’t open the seeds!) Sunflower seeds will be used in this

experiment.

Host Plant

Milkweed bugs in the wild feed

on the seeds and tissue of a plant

that is common in old fields and

along roadways, the milkweed

plant (Asclepias sp.).

This species is often found in

small groups on milkweed leaves.

The 1 to 1.5 meter high milkweed

plants, which produce a distinctive

milky white sap when a leaf is

removed, bloom in sprays of small

white flowers in the summer.

In the fall, seedpods develop

which are about 10 mm long and 4

mm wide. When the seedpods

ripen they open up releasing seeds

that float on fluffy white parasols.

Milkweed bugs are often found

piercing the wall of the seedpods

to feed on seeds inside.

Life History

Milkweed bugs are hemimetabolous insects, which mean they exhibit gradual

metamorphosis with nymphs (immatures) that look like the adults without full wings.

Black wing pads appear early in development and their color pattern also changes

gradually with age. Nymphs have bright red abdomens.


5 Eggs are about 1 mm in diameter and change gradually from yellow to an orange color as

they incubate. First instar nymphs are about 1.5 mm long (tip of head to tip of

abdomen). As they grow they periodically molt (shed) their old exoskeleton. Milkweed

bugs usually molt 5 times which means they have 5 nymphal instars before becoming

an adult. Eggs take about 1 week to hatch and a month to become adults at room

temperature (75°F or 23°C).

The underside of the abdomen of the female has one black strip and two black dots

while the male has two thick black strips.

Aposematic Coloration to Avoid Predation

Milkweed bugs are one of a group of insects that are adapted to feed on poisonous

members of the milkweed family. The bugs have few predators because they sequester

(concentrate in their bodies) bad tasting compounds, cardiac glycosides, found in the

sap of milkweed plants.

The bugs use the bright (aposematic) coloration to advertise their bad taste.

Inexperienced birds that eat their first milkweed bug will vomit from the sequestered

glycosides and are unlikely to try to eat another orange and black insect. Some insect

species that do not taste bad use similar color patterns to fool birds. These are known as

mimics.

Interesting Behaviors

Milkweed bugs often gather in groups on the milkweed plant. This gregarious behavior

probably enhances their warning coloration. In the container you may see groups of

bugs forming at night.

During the day they are usually found feeding on seeds or walking around the

container. Sometimes groups form on the side of the container away from the seeds

when some of the bugs are molting.

Care

You will be provided an appropriate habitat for your bug: a plastic dish with some

sunflower seeds (a milkweed substitute) and a water supply, including a plastic tube

containing wet cotton.

The cotton should be kept moist but not saturated. If the cotton becomes dirty or moldy,

get a replacement form your lab instructor.

Be sure to keep the water away from the seeds so that they don't become moldy. If you

need to handle your bug, a soft camelhair brush can be used to move them.

Insects are ectothermic and their growth and development are very sensitive to

temperature. Try to find a warm place where the temperature is stable. Check your

adopted bug frequently, daily if possible.

Examine the habitat for discarded exoskeletons to see if it has molted (they look a little

like spiders).

Each week, measure the length of your bug (tip of head to tip of abdomen in mm) using

a piece or paper with 1 mm gridlines taped to the plastic cup.

Record the length and any notes on the datasheet provided. Later in the semester, you

will graph and analyze these results in the Data Analysis/ Statistics Lab.


6

References

Price, Peter W. 1977. Insect Ecology. John Wiley & Sons, NY. pp. 514.

http://insected.arizona.edu/milkinfo.htm

http://www.uwex.edu/ces/cty/milwaukee/urbanag/bugnet/outflowers/milkweedbug/milkweedb

ug.html

http://www.life.uiuc.edu/entomology/105/milkweedbug.html

http://entowww.tamu.edu/extension/youth/bug/bug027.html

http://pubpages.unh.edu/~pcj/adoptbug.html

http://www.cnr.berkeley.edu/citybugs/Taxa/Heteroptera/Lygaeidae/Oncopeltus/Oncopeltus_fas

ciatus/Oncopeltus_fasciatusPage.htm

http://everest.ento.vt.edu/Courses/Undergraduate/IHS/distance/html_files/buginacup/journal2.h

tml

Adopt a Bug! Datasheet

Name (your name, not the bug’s name): _________________ Lab section _______

Variable tested:____________________

Date Size

(mm)

Notes █ Date Size

(mm)

Notes

█

█

█

█

█

█

█

█

█

█

█

█

█

http://insected.arizona.edu/milkinfo.htm

http://www.uwex.edu/ces/cty/milwaukee/urbanag/bugnet/outflowers/milkweedbug/milkweedbug.html

http://www.uwex.edu/ces/cty/milwaukee/urbanag/bugnet/outflowers/milkweedbug/milkweedbug.html

http://www.life.uiuc.edu/entomology/105/milkweedbug.html

http://entowww.tamu.edu/extension/youth/bug/bug027.html

http://pubpages.unh.edu/~pcj/adoptbug.html

http://www.cnr.berkeley.edu/citybugs/Taxa/Heteroptera/Lygaeidae/Oncopeltus/Oncopeltus_fasciatus/Oncopeltus_fasciatusPage.htm

http://www.cnr.berkeley.edu/citybugs/Taxa/Heteroptera/Lygaeidae/Oncopeltus/Oncopeltus_fasciatus/Oncopeltus_fasciatusPage.htm

http://everest.ento.vt.edu/Courses/Undergraduate/IHS/distance/html_files/buginacup/journal2.html

http://everest.ento.vt.edu/Courses/Undergraduate/IHS/distance/html_files/buginacup/journal2.html


7

Lab exercise 2: Avian Evolution

This exercise is intended to demonstrate the processes of natural selection, sexual selection,

speciation, and convergent evolution using specimens from the Biology Department's bird

collection. Because many of the birds are quite old and valuable, please refrain from handling the

specimens as you make your observations.

I. Evolutionary Adaptation

The physical characteristics and behaviors of birds illustrate a vast assortment of evolutionary

adaptations for a particular way of life, often termed the ecological "niche". For example, bill

size and shape reflect the diet of most birds.

Seed eating species, for example, have short, strong bills for cracking the seed coat. Predators

typically have hooked bills for tearing apart their prey. Birds such as swallows have bills

which, when opened, provide a wide gape for catching aerial insect prey.

Long bills may be used for spearing prey, probing in soft substrates, or hammering in wood.

Examine the specimens below and decide on their probable diet.

Bird Bill Description Probable Diet

Elf Owl

Dowitcher

Skimmer

Night hawk

Chimney swift

Hummingbird

Evening

Grosbeak

Woodcock

Least Bittern

II. The Process of Natural Selection

Adaptation typically occurs within populations by the following microevolutionary processes,

termed natural selection:

a. Individuals in the population differ in their genetic composition (genotypes).

b. The different genotypes produce different phenotypes, typically visible as slight

variations in height, weight, or the size/shape of particular structures.


8 c. These phenotypic differences influence survival and reproductive success.

d. Differential reproductive success leads to a different mix of genotypes (and thus phenotypes) in the

next generation.

Shown is a Crossbill, a bird with an unusual bill that is inserted into pinecones to extract the pine

seeds. The Crossbill is thus a dietary specialist. Also shown is a bird of about the same size, a Pine

Grosbeak, which is a dietary generalist capable of feeding on a wide variety of seeds.

Using the information on the previous page, describe a process of natural selection on a Grosbeak-

like bill, which, repeated over enough generations, would cause such a bill to evolve to the

Crossbill-like shape.

The theory of natural selection, formally proposed by Charles Darwin, replaced an earlier theory

espoused by Lamarck that adaptations acquired during an individual’s lifetime were slowly

heritable.

For example, Lamarck proposed that the giraffe stretched its neck over time to reach leaves on trees

and that these adaptations were passed to the offspring. Lamarck's theory slowly lost favor for lack

of supporting evidence.

For example, the crow shown here survived an apparent accident, but its offspring would not have

crossed bills.

III. Phenotypic Variation within Populations

In order for natural selection to produce evolutionary change, there must be genetically

determined phenotypic variation among individuals within a population. An extreme

example of such variation is that of the Eastern Screech Owl Otus asio, which has two

distinct color morphs: a ―grey phase‖ and a ―red phase‖.

These two phases are heritable and are also associated with behavioral differences. Grey

phase owls typically roost (sleep during the day) next to tree trunks, whereas red phase

owls more commonly roost in the outer foliage.

Why might roosting next to a tree trunk decrease mortality and thereby increase the

reproductive success of a grey phase owl?

The sparrows shown are all members of the common house sparrow, Passer domesticus,

but differ slightly in body size. In a very early, now classic study of the effects of natural

selection on sparrow body size, Hermon Bumpus at Brown University collected sparrows

incapacitated by a severe ice storm that hit Rhode Island on February 1, 1898 and brought

them back to the lab. Of the 136 sparrows collected, 72 survived while 64 died. Bumpus

recorded 9 size measurements on each bird. Body weight and total length data for the 59

adult males collected are shown in the graph below. Based on the graph, were smaller or

larger birds favored by natural selection in this instance?


9

IV. Sexual Selection

Males and females of many species (humans included) lead slightly different lives and have

differing roles in courtship. Particularly in polygynous species (where a dominant male may

mate with more than one female and some males are left without mates), competition among

males for mates is fierce and natural selection favors traits that lead to winning battles (e.g.,

larger size).

Natural selection may also favor traits that "advertise" males (e.g., bright coloration) to

females, even if those same traits increase their risk of getting eaten by predators. Selection

on the two sexes may thus be different, and lead to sexual dimorphism (males and females

look different). Describe the dimorphism in color pattern of the bird species shown.

Male

Female

Cardinal

Scarlet Tanager

Red-bellied

Woodpecker

Bobolink

Total l ength (mm)

170165160155150

Bod

y w

eigh

t (g

)

32

30

28

26

24

22

SURVIVE?

no

yes


10

The males of these species molt immediately after the breeding season to less ostentatious winter plumage.

Suggest why this may be an advantage:

______________________________________________________________________

______________________________________________________________________

______________________________________________________________________

V. Speciation and Divergent Evolution

Populations of a species that become isolated from one another often evolve along separate

trajectories to the point where, even if the two populations once more came into contact,

reproduction is impossible. When the two populations are no longer capable of interbreeding, they

are considered distinct species. This separation of one species into two distinct species is called

divergent evolution.

An example occurs in Chester County, which is situated along a zone where the ranges of the

Black-capped Chickadee (Parus atricapillus) and Carolina Chickadee (P. carolinensis) overlap

slightly (see the maps).

The laptop at this station has pictures and songs of both the Black-capped Chickadee and Carolina

Chickadee. Double click on each file to view and hear them.

Based on the specimens shown, and the information provided, how do the two species appear to

differ in their size, coloration and calls?

______________________________________________________________________

______________________________________________________________________

______________________________________________________________________

Black-capped Chickadee Carolina Chickadee

Size

Coloration

Calls/ Songs

VI. Convergent Evolution

Most recently evolved species continue to grow more dissimilar over time, showing divergent

evolution (see above). Occasionally, however, species of very different ancestry exploit similar

niches, and their adaptations show convergent evolution.

An example is shown by Grebes and Ducks, both evolved from very different ancestors, but both

using aquatic animals and plants as food. What trait do both species share which indicates

convergent evolution?


11

___________________________________________________________________________________

___________________________________________________________________________________

___________________________________________________________________________________

References:

http://www.aquatic.uoguelph.ca/birds/morphevol/main.htm

http://www.chickadee-web.com/

http://research.amnh.org/ornithology/crossbills/

http://research.amnh.org/ornithology/crossbills/nathist.html

The beaks of Finches and natural Selection

For this part of the lab you will work in pairs. You have been provided with a tool to simulate the

beak of a finch. The different tools represent different beak types within a population of a single

finch species.

At your table there are 4 petri dishes containing different seeds: corn, sunflower, caraway and

mixed seeds.

Your goal is to determine how many seeds of each type you can successfully pick up with your tool

(the finch’s beak) in 30 seconds. Success is defined as the transfer of the seeds to a plastic cup (the

bird’s belly).

Your professor will provide direction and serve as the ―timekeeper‖ for this experiment. One of the

members of your team will be the ―bird‖ and the second one will count the seeds. Only 1 seed per

transfer!!!! We will collect the data in the tables below.

Which tool are you? _____________________________________

Which tool do you predict will work best for each seed type?

Corn: ___________________ Sunflower: __________________ Caraway: ________________

Tool

Small forceps

Medium forceps

Long forceps

Pliers

Wire cutters

Wrench


12

Seed type = Corn

Tool Number of seeds Average

Small forceps

Medium forceps

Long forceps

Pliers

Wire cutters

Wrench

Seed type = Sunflower seeds


Small forceps

Medium forceps

Long forceps

Pliers

Wire cutters

Wrench

Seed type = Caraway seeds


Small forceps

Medium forceps

Long forceps

Pliers

Wire cutters

Wrench

Which tool (if any) worked better for the:

Corn seeds? _____________ Sunflower seeds?_______________ Caraway seeds?________________

Were your initial predictions correct? ___________

Based on your predictions, which beak type would be an advantage if the environment changed so that the

predominant seed type transitioned from corn to caraway seeds?

______________________________________________________________________________________

______________________________________________________________________________________

______________________________________________________________________________________


13

In the remaining Petri dish, there is a mixture of seeds. Based on what you learned about the efficiency of

each tool for the different sizes of seeds, do you think that any tool would be advantageous for survival

(obtaining sufficient seeds in a limited time)? _____________________________________

Seed type = Mixed

10 seconds

Tool Number of seeds Average Rate (seeds/second)

Small forceps

Medium forceps

Long forceps

Pliers

Wire cutters

Wrench

Suppose the finches needed to consume seeds at a rate of about 0.5 seeds/sec per feeding to survive long

enough to reproduce. Which finches would survive?

_______________________________________________________

Suppose the finches needed to consume seeds at a rate of about 1 seed/sec per feeding to survive long


_______________________________________________________

Suppose the finches needed to consume seeds at a rate of about 2 seeds/sec per feeding to survive long


_______________________________________________________

Explain how this activity demonstrates natural selection.

______________________________________________________________________________________

______________________________________________________________________________________

______________________________________________________________________________________

______________________________________________________________________________________


14

Lab exercise 3: Organic Molecules

Most organic molecules

found in living systems can

be classified as

carbohydrates, fats,

proteins, or nucleotides.

Each of these classes of

molecules has specific

properties that can be

identified by simple

chemical tests called

assays.

Assays can be either quantitative or qualitative. A qualitative assay simply shows whether or not a

specific molecule is present in a sample at detectable levels. A quantitative assay shows the

presence and amount of a molecule in a sample.

In this laboratory you will use qualitative assays and learn to detect the presence of the organic

molecules: carbohydrates, fats, and proteins.

Exercise 1: Testing for Carbohydrates

The basic structural unit of carbohydrates is the monosaccharide (or single sugar).

Monosaccharides are classified by the number of carbons they contain: for example, trioses have

three carbons, pentoses have five carbons, and hexoses have six carbons. They may contain as few

as three or as many as ten carbons.

Monosaccharides are also characterized by the presence of a terminal aldehyde group (CHO) or an

internal ketone (C=O) group. Both of these groups contain a double-bonded oxygen atom that

reacts with Benedict's reagent to form a colored precipitate that will change the color of the

original solution.

When two monosaccharides are joined together, they form a disaccharide. If the reactive aldehyde

or ketone groups are involved in the bond between the monosaccharide units (as in sucrose), the

disaccharide will not react with Benedict's reagent.

If only one group is involved in the bond (as in maltose), the other is free to react with the reagent.

Sugars with free aldehyde or ketone groups, whether monosaccharides or disaccharides, are called

―reducing sugars.'' In this exercise, you will use Benedict's reagent to test for the presence of

reducing sugars.

C

C

C

OC

O

H

KETONE ALDEHYDE


15

Monosaccharides may join together to form long chains (polysaccharides) that may be either

straight or branched.

Starch is an example of a polysaccharide formed entirely of' glucose units. Starch does not show a

reaction with Benedict's reagent because the number of free aldehyde groups (found only at the end

of each chain) is small in proportion to the rest of the molecule. Therefore, we will test for the

presence of starch with Lugol's reagent (iodine/potassium iodide, I2KI).

Part 1: Benedict’s test

When Benedict's reagent is heated with a reducing sugar, the color of the reagent changes

depending on the amount of sugar present.

Orange and red indicate the highest proportion of these sugars. Green or yellow indicate lesser

amounts of reducing sugar. (Benedict's test will show a positive reaction for starch only if

excessive heating has broken down the starch.)

Procedure:

1. To each large glass test tube, add approximately 1 ml (about 20 drops) of each solution to be tested.

2. Add approximately 5 drops of Benedict's reagent to each tube.

3. Mix the reagent and the solution by agitating the solution in each tube from side to side.

4. Record the original color of each tube's contents in Table 1.

5. Heat the test tubes in a warm water bath for 10 minutes. Be careful when placing tubes into or

removing them from the water bath. Record any color changes.

Part 2: Lugol's Test

When an iodine solution comes in contact with starch, the yellow-brown color of the iodine changes

to dark brown/black.

Monosaccharides and disaccharides are too small to interact with iodine. Therefore, the Lugol's test

is specific for a polysaccharide, starch.


16

Procedure:

1. To each small plastic test tube, add approximately 1 ml of each solution to be tested.

2. Add approximately 2 drops of Lugol’s reagent to each tube (bottles may be labeled iodine solution).

3. Mix the reagent and the solution by agitating the solution in each tube from side to side.

4. Record any color changes (DO NOT HEAT) in the table below.

SUBSTANCE TYPE OF SUGAR

MOLECULE

BENEDICT’S

TEST

BENEDICT’

S TEST

LUGOL’S

TEST

control, mono-, di-, poly- before boiling after boiling after addition

1. Water

2. Glucose

3. Maltose

4. Sucrose

5. Starch

Questions:

a. Why did you test water with Benedict's and Lugol’s reagent? ____________________

b. What color change resulted from a positive Benedict’s test? _________________

Lugol’s test? ____________________

c. Did maltose cause Benedict's reagent to change color?___________

Lugol’s reagent? ___________ Why or why not? ____________________

d. Did glucose cause Benedict's reagent to change color? ___________

Lugol’s reagent? ___________ Why or why not?_____________________

e. Did starch cause Benedict's reagent to change color? ___________

Lugol’s reagent? ___________ Why or why not? _____________________

f. Did sucrose cause Benedict's reagent to change color? ___________

Lugol’s reagent? ___________ Why or why not? _____________________

Exercise 2: Testing for Lipids

The word lipid refers to any of the members of a rather heterogeneous group of organic molecules

that are soluble in nonpolar solvents such as chloroform (CHCl3), but are insoluble in water (polar).

Although lipids include fats, steroids, and phospholipids, this exercise will focus primarily on fats.


17 Triglycerides, a popular topic in discussions of diet and nutrition, are the most common form

of fat. They consist of three fatty acids attached to a glycerol molecule. Triglycerides are found

predominantly in adipose (fat) tissue and store more energy per gram than any other molecule.

At room temperature, some fats are solid (generally those found in animals) and are referred to as

fats, while others are liquid (generally those found in plants) and are referred to as "oils." Vegetable

oil, a liquid fat, is a mixture of triglycerides.

Since both solid and liquid fats are nonpolar, we will test for their presence by using Sudan IV, a

nonpolar dye that preferentially dissolves in nonpolar substances like fats and oils, resulting in red

droplets on the surface of a liquid containing lipid.

Procedure

1. Get two small plastic test tubes and add the same amount of water to both of them.

2. Add one small drop of vegetable oil to one tube.

3. Add one drop of Sudan IV to both tubes.

What is the result of the water versus water + oil reaction with Sudan IV?

______________________________________________________________________

Why do the droplets float, rather than mix with the water?

______________________________________________________________________

Exercise 3: Testing for proteins and amino acids

Proteins are made up of one

or more polypeptides, which

are linear polymers of smaller

molecules called amino acids.

Amino acids derive their name

from the amino group and the

carboxyl group (acidic) that

each possesses.

Polypeptides are formed when

amino acids are joined

together by peptide bonds

between the amino group of

one amino acid and the

carboxyl group of a second

amino acid.

The biuret reagent reacts

with peptide bonds and will,

therefore, react with proteins,

like egg albumin, but will not react with free amino acids, like glycine and alanine.


18 On the other hand, the reagent ninhydrin will react with the amino group of free amino acids,

but will not react with polypeptides.

In this exercise you will distinguish between free amino acids and proteins (polypeptides) on the basis of

their ability to react with either biuret reagent or ninhydrin.

Part 1: Biuret Test Procedure

1. Obtain two small plastic test tubes and add approximately 1 ml of water to one tube.

2. Add approximately 1 ml of albumin to the second tube.

3. Add about 5 drops of biuret reagent to both tubes.

4. Allow the contents of the tubes to react for about 2 minutes at room temperature.

Record your results. ________________________________________________

What is the result of the water versus albumin reaction with Biuret reagent? ________

Part 2: Ninhydrin Test Procedure

1. Obtain a piece of filter paper and divide it into six pieces

with a pencil. Label the areas A, B, C, D, E, and F (see picture at

right).

2. Place one drop of each solution (labeled A, B, C, D) onto the

filter paper in the area with the corresponding letter. Only add a

small drop of each solution so that the sample does not spread

into the adjacent areas. Allow the spots to dry. Place a drop of

water in area E. Press your index finger into quadrant F.

3. Apply one drop of ninhydrin to each spot. Caution:

Ninhydrin is poisonous; avoid contact with your skin. Allow the

paper to dry at room temperature for 10 to 15 minutes. (The

reaction will occur more quickly if the paper is heated using the lamp on your table.) All amino acids,

except proline, will turn purple in the presence of ninhydrin. Proline will turn yellow.

Record your results in the table below.

SAMPLE FINAL COLOR TYPE OF MOLECULE

A

B

C

D

E

F


19

Exercise 4: Analyzing unknowns

Using reagents and methods from the previous exercises, identify the types of molecules contained in the

unknown solutions. Fill out the table below.

SAMPLE BENEDICT’S

TEST

LUGOL’S

TEST

BIURET

TEST

NINHYDRIN

TEST

SUDAN

IV TEST

1.

2.

3.

What are the chemical components in each of the unknowns tested?

1. ____________________________________________________________________________

2. ____________________________________________________________________________

3. ____________________________________________________________________________

References:

http://www.sidwell.edu/sidwell.resources/bio/VirtualLB/Macro/macro.html

http://www.surry.cc.nc.us/ayers_bio/BIO%20111/labexcs/chemistry%20lab/chemistry%20for%20life.htm

http://www.science.mcmaster.ca/Biology/1A03/Laboratories/Lab_One/index.html


20

Lab exercise 4: The Microscope,

Measurement, and Cells

Understanding and properly applying the methods of scientific inquiry require that you become

proficient at observing and recording data accurately. To do this, you need to be familiar with the

types of instruments used for experimental work and with proper sampling techniques.

During this laboratory period you will learn about the use and care of the compound microscope.

You will also prepare living materials for observation.

In the late 1500s, two Dutch spectacle makers (The Jannsens) developed the compound microscope.

Their device had two convex lenses placed at either end of a tube and was capable of magnifying an

object to 10 times (10x) its actual size.

Today, developments in microscopy provide scientists with a wide selection of instruments with

which to view the smallest organisms and even the components of individual cells. These

microscopes range in complexity from the relatively simple models you will use in the laboratory

today to highly sophisticated scanning and transmission electron microscopes.

Exercise 1: Identifying the parts of the compound microscope

To use the microscope properly, you must first be familiar with the care of this expensive and delicate

instrument. Keep the following precautions in mind:

Always carry the microscope in an upright position. Use one hand to grasp the arm of the microscope;

use the other to support the base. The eyepiece (ocular lens) slides into the body tube and could fall out if

the microscope is tilted too much.

Never place the microscope close to the edge of the table. Keep the electrical cord out of the way.

Use only lens paper for cleaning the lenses. Using your fingers, handkerchief, or other materials could

smudge or damage the lenses.

When you are finished with your observations, turn off the illuminator and rotate the low-power

objective into viewing position. Never put a microscope away with the high-power objective in the

viewing position.

Obtain a compound microscope from your instructor and familiarize yourself with the location and function

of each part.

Light source: May be built into the base with a lens that focuses light onto the lower condenser lens

or may be a separate light that is focused onto the condenser lens by a mirror.

Condenser: Contains a system of lenses that focuses light on the object. Locate the knob that raises

and lowers the condenser.

Iris diaphragm: Used to control contrast. It can be opened or closed using the lever protruding

from the lower front edge of the condenser. Greatest contrast is achieved when the iris is fully

closed. Closing the iris diaphragm reduces the resolving power of the microscope, the ability to

distinguish the smallest structures. As the iris is opened, resolving power increases but contrast

decreases.

Objective lenses are mounted on a revolving nosepiece or turret (e). Most new microscopes are

parfocal, that is, when an object is in focus with one lens, the lenses can be changed without


21 completely losing focus. Each objective contains a complex lens system. Magnification is

indicated on the side of the objective. The nosepiece usually holds the objectives listed below.

Check to see which of these are present on your microscope: scanning objective (4 x

magnification), low-power objective (10 x magnification), and two high- power objectives (40 x

and 60 x magnification).The objective is also responsible for determining the resolving power.

Resolving power is the ability to reveal detail, to distinguish two closely spaced objects as being

two rather than one. The smaller the distance between two objects that can be distinguished from

one another, the greater the resolving power of the instrument used to view the objects. The unaided

human eye can distinguish (resolve) two objects when they are at least 0.1 mm apart, whereas with

the light microscope, the human eye can distinguish objects as separate when they are up to 1,000

times closer than that!

Ocular lens or eyepiece: The lens you look through. It will magnify the image from the objective

an additional 10 times (10 x). The body tube (g) holds the ocular lens. Your microscope has two

oculars and is said to be binocular.

Stage: Holds the slide to be viewed. The stage can be moved vertically by turning the coarse

adjustment (i) and the fine adjustment (j) knobs. These are located in different places on different

types of microscopes, either separately or together. Coarse adjustment is used for initial focusing of

specimens at low power. Fine adjustment makes very slight changes, allowing precise focusing at

high power.

Mechanical stage: Your microscope is equipped with a mechanical stage. Adjustment knobs are

used to move the slide in the horizontal plane, that is, side to side and toward and away from you.

Base (l) and arm (m): Important structural parts of the microscope.

Exercise 2: Using the compound microscope

Procedure

To get maximum performance from the microscope, you will need to adjust the illumination and

focus properly.

1. Place the microscope on the table with the oculars pointing toward you.

2. Revolve the nosepiece until the scanning objective (4x) is in line with the body tube. A click will be

heard (or felt) when the objective is properly engaged (otherwise you will see only your eyelashes against a

black background). Raise the stage as close to the objective as it will go. You should notice that with this

objective there is considerable space between the end of the objective and the stage. The space between the

end of the objective and the specimen is referred to as the working distance and it will decrease with the

higher power objectives and increased magnification.

3. Turn on the light switch and adjust the illumination. Higher power objectives require more illumination.

4. Adjust the condenser to its highest position. To truly optimize the condenser position, after focusing on

the subject, gently place a sliver of paper on the top of the illuminator lens. Now raise or lower the

condenser so that the edge of the paper is in sharp focus with the specimen. The iris diaphragm should be

adjusted so that the field of view is brightly and evenly lit.

5. Obtain a slide of the letter "e". With the 4x objective in position, place the slide on the microscope

stage, making sure that the coverslip is facing upward toward the objective. Hold the slide in place with the

mechanical stage. Position the slide so that the specimen is directly over the hole in the stage and confirm

that the stage is as close to the objective as possible.

6. Since the stage is as close to the 4x objective as it can get, focus must occur at a lower stage position.

Looking through the oculars, slowly lower the stage by turning the coarse adjustment focus knob. As the


22 specimen comes into view, continue lowering the stage until the image becomes clear. At first

continue past focus, then reverse direction and determine the point at which the image is in perfect focus.

7. Once you are sure that the specimen is in sharp focus and in the center of your field of view, rotate the

next higher power objective into place. The specimen should appear larger, but perhaps blurred. Turn the

fine adjustment knob slowly to sharpen the focus. Your microscope is parfocal, so you should be able to

bring the specimen into focus by using only the fine adjustment. If you need to use the coarse adjustment,

always return to a lower power objective with a long working distance, check focus, and make sure the

specimen is centered. Never turn the coarse adjustment while using an objective with very little working

distance.

8. You can determine how much your specimen is magnified by multiplying the power of the objective

lens by the power of the ocular lens. For example, a specimen is magnified 600 times when using the 60x


23 objective lens and the 10x ocular lens. What is the magnification when using the 4x objective?

_______________________

9. Move the slide to the right. Which way does the letter "e" move (as viewed through the microscope)?

_______________________________________.

10. Prepare a slide of Elodea leaf and repeat steps 1-8. Place a drop of water in the center of a clean

microscope slide. To the drop, add a torn piece of an Elodea leaf. Place one edge of a coverslip at the edge

of the water drop and gently lower it so that the water containing the specimen completely spreads out

under the coverslip. Take care not to trap bubbles under or around the specimen. Do not press down on the

coverslip. If there is too much water present, draw off the excess by touching the corner of a paper towel to

the edge of the coverslip.

11. Rotate the nosepiece so that the 10x objective clicks into place. The working distance (the space

between the objective lens and the slide) decreases with higher-power objectives and increased

magnification. The size of the field of view (the area you can see) varies inversely with magnification. The

greater the magnification, the smaller the field of view.

12. Examine your slide of Elodea. Count the number of cells you can see in one row at 4x and 10x. What

happened to your field of view as you changed from the 4x to the 10x objective?

________________________. What is the relationship between magnification and field of view?

____________________________________________________________________

13. When you can operate the microscope successfully using 10x magnification, change to the 40 x high-

power objective. When using high power, the object to be viewed must be at the center of the field because

the high-power objective magnifies only a small portion of the field of view observed under low power.

You may need to use the fine adjustment knob to focus the specimen. Remember, never use the coarse

adjustment knob with high power. What is the total magnification of your specimen with the 40 x objective

in place? ________________________.

14. Below, draw what you observe using the 10x and 40x objectives. Drawings should routinely be done in

pencil and be labeled to indicate the name of the specimen. A scale bar should be included to indicate the

approximate size of the specimen (see exercise 5).

A few additional suggestions for using a microscope.

a. Be certain that you have the proper amount of light. With an unstained specimen, you can find the right

focus by first focusing on the edge of the cover slip.

b. Try to keep both eyes open. This will take some practice on your part, but will be less tiring for your

eyes.

c. Always clean ocular and objective lenses with lens paper before use.

d. Always begin by using a low-power objective lens to find the specimen. You can then turn to a

higher power to make your observations.

e. If you are having difficulty locating the specimen, use a systematic pattern to search the slide.

f. If all else fails, ask your instructor for help.

Exercise 3: Observing wet mount slides

Procedure

1. Examine your wet mount of Elodea under low power and then under high power. Use the 10x objective

to focus on the torn edge of the Elodea leaf. Notice that only part of the whole thickness of the leaf is in

focus. Using the fine adjustment knob, focus on various planes throughout the thickness of the leaf. Note

that the leaf is more than one cell-layer thick. How many cell layers can you discern?

_______________________


24 2. Now switch to your 40x objective. You should notice that even less of the leaf’s thickness is in

focus at any one setting. What you have just accomplished is a demonstration of a principle of

magnification that was discussed earlier: the higher the magnification, the shallower the depth of field

(depth of the area that is clearly focused). To a microscopist, this means that your ―DEPTH

DISCRIMINATION‖ has been improved. Depth discrimination is the ability to make clear optical sections

unblurred by under- and over-lying material.

3. Notice the movement of little green bodies inside each cell. These are chloroplasts, organelles

responsible for photosynthesis in plant cells. The movement you observe is called cyclosis, or cytoplasmic

streaming. As the cytoplasm moves around the large central vacuole, it carries with it dissolved substances

as well as suspended organelles. Does cyclosis occur in the same direction in all cells?

__________________________

Exercise 4: Measuring the size of objects using the compound microscope

The microscope can be used as a tool to gather quantitative data in addition to serving as an instrument for

making qualitative observations.

Procedure

The size of objects viewed with the compound microscope can be estimated by first determining the

diameter of the field of view for a particular microscope objective. It is then possible to estimate the size of

the specimen by comparing it with the total field.

1. Place a transparent mm grid across the field of view under scanning power and record the diameter in

millimeters: ________

2. What is the diameter in micrometers (µm)? ________

3. What is the diameter in cm? ____________

4. What is the diameter in nm? __________

The diameter of the field of view using the scanning objective (A) can be used to calculate the diameter

using any other objective (B):

magnification A x diameter A = magnification B x diameter B

5. Calculate the diameters of the fields of view using the other objectives on your microscope.

Objective Diameter (µm)

10 X

40 X

60 X

6. Using the slide of Elodea, estimate the length of one cell in micrometers. Hint: Use the diameter of the

field of view to determine the length of a row of cells, then divide by the number of cells: ___________µm

Exercise 5: Comparing prokaryotic and eukaryotic cells

Understanding the nature of cell structure and function is important to an understanding of

organisms. All organisms are composed of cells whether they exist as single cells, colonies of

cells, or in the multicellular form.

Cells are usually very small, and for this reason, a thorough understanding of their subcellular

structure and function has been possible only through advances in electron microscopy and cell

biochemistry.


25 There are two general types of cells: prokaryotic and eukaryotic. These two words have their

root in the Greek word karyon (nut), which refers to a cell's nucleus. The prefix pro- means

"before" or "prior to." Thus the word prokaryotic literally means "before having a nucleus."

Prokaryotic cells do not have a membrane-bound nucleus and their genetic material (DNA) is

only loosely confined to a nuclear area within the cell. Bacteria, including the cyanobacteria

(formerly known as "blue-green algae"), are prokaryotes. All other organisms are eukaryotes.

The prefix eu- means "true." The cells of eukaryotes have true, membrane-bound nuclei

containing their genetic material.

There are other distinctions between prokaryotic and eukaryotic cells. Eukaryotic cells are

generally larger and contain additional specialized compartments (membrane-bound organelles)

in which cell functions such as energy production may occur.

Prokaryotic cells lack membrane-bound organelles, their cell functions are carried out in the

cytoplasm.

Observing Prokaryotic cells

Bacteria are extremely small (approximately 1 to 2 µm). Most are heterotrophic, depending on

organic molecules for food, but some are photosynthetic and make their own food.

Morphologically, they are either round (cocci), rod-shaped (bacilli), or spiral-shaped (spirilli).

To view them with the light microscope, the highest power objective lens (60x) must be used. Even

then, not much more than their basic shape will be visible. Both a cell membrane and a thicker

polysaccharide cell wall surround the bacterium. Some bacteria have flagella, threadlike structures

used in locomotion. Bacterial flagella are composed of the protein, flagellin. We will not observe

bacteria in this lab.

Cyanobacteria, formerly called blue-green algae, are photosynthetic prokaryotes. However, unlike

photosynthetic bacteria, which contain bacteriochlorophyll, cyanobacteria contain chlorophyll a--

the same type of chlorophyll found in eukaryotic green algae. The chlorophyll molecules are not

located within chloroplasts, as in higher plants, but are found, instead, within photosynthetic

thylakoid membranes dispersed throughout the cytoplasm. In addition to chlorophyll a,

cyanobacteria contain other accessory pigments including the yellow and orange carotenoids and

the phycobilins (reddish phycoerythrins and bluish allophycocyanins).

1. Make a wet mount of the cyanobacteria species, Anabaena,

available in the laboratory.

2. Draw your observations at right.

The cells of protists, fungi, plants, and animals are

eukaryotic, containing both a membrane-bound nucleus and

membrane-bound organelles.

Magnification?______

Procedure

1. Prepare a wet-mount slide of the following samples available in the lab and draw a picture of your

observations.


26

Kingdom Protista

Stentor: a ciliated, single-celled eukaryote Volvox: a colony of biflagellated green algal

cells (NO COVER SLIP ON VOLVOX)

Magnification?______ Magnification?______

Plant cells: All plant cells possess a cell wall, cell membrane, cytoplasm, nucleus, and plastids. Plastids

are membrane-bound organelles unique to plants -chloroplasts (containing chlorophyll) and leucoplasts

(containing starch). Chromoplasts contain several types of pigment including carotenoids, which give

plants an orange or yellow color. Add the dye Lugols reagent to the plant samples in order to visualize

starch granules in the plants (if they are present).

Kingdom Plantae

Elodea: Onion skin: Potato:

Magnification?______ Magnification?______ Magnification?______

How are the Elodea, onion cells, and potato cells similar to each other? ______________

___________________________


27 How are they different? ________________________

_______________________________________________________________________

Animal cells: Unlike plant cells, animal cells do not possess a cell wall or plastids. View the prepared

slides of animal cells using the 10 x and 40 x objectives.

Kingdom Animalia

Nerve cells: Oral epithelial cells

Magnification?______ Magnification?______

References:

http://www.science.mcmaster.ca/Biology/1A03/Laboratories/Lab_One/Microscopy/Microscopy

http://www.science.mcmaster.ca/Biology/1A03/Laboratories/Lab_One/Microscopy/microscopy.html

http://www.wisc.edu/botit/Botany_130/Microscope/Parts_microscope.html

http://www.wisc.edu/botit/Botany_130/Microscope/Microscope.html


28

Lab exercise 5: Diffusion and Osmosis

The law of diffusion is of particular importance to our understanding of the movement of molecules

into and out of cells.

The law of diffusion states that molecules tend to move from areas of high chemical potential

(high concentration) to areas of low chemical potential (low concentration).

Osmosis, a special case of diffusion with special relevance for cells, is the movement of water

molecules from regions of high water potential (low concentration of solutes) to regions of low

water potential (high concentration of solutes) across a selectively permeable membrane.

The more solute dissolved in water, the lower the water potential. Thus, pure water with no

dissolved solutes has the highest water potential.

Exercise 1: Dialysis

The process of movement (diffusion) of solutes through a selectively permeable membrane is

called dialysis. The cell membrane is a living example of a selectively permeable membrane. Small

hydrophobic solute molecules, water, and other very small polar, but uncharged, molecules can

move freely through the cell membrane. But, larger molecules may pass more slowly or sometimes

not at all.

In this experiment, dialysis tubing, made of a material with a specific pore size, will be used to

simulate a cell membrane.

1. Work in groups of four. Obtain a piece of dialysis tubing, soak it in water for a minute, and gently rub

the untied end to open the bag. Fill the bag with water to make sure that there are no holes in the bag.

Empty the water from the bag.

2. Add 4 pipettefuls of 15% glucose to the bag and then add 4 pipettefuls of 1% starch solution to the bag.

3. Hold the bag closed and mix its contents. Carefully rinse off the outside of the bag in tap water. Place

the bag in a beaker so that the untied end of the bag hangs over the edge of the beaker.

4. This is important! Fill the beaker with only enough water to cover the liquid within the bag. Add

about 1 dropperful of Lugol's reagent (I2KI) to the water in the beaker. Record the color of the solution in

the bag and the beaker.

5. Allow the setup to stand until you see a distinct color change in the bag or in the beaker. After about 1

hr, record the final color of both the solution in the bag and the solution in the beaker.

6. Remove about 1 ml of the solution from the beaker, and test with Benedict's reagent (see Organic

Molecules lab #2 for the procedure if you have forgotten this test). Record your results in the table below.

ORIGINAL

CONTENTS

ORIGINAL

COLOR

FINAL

COLOR

COLOR AFTER

BENEDICT’S TEST

Bag Glucose and

starch

not performed

Beaker water and

Lugol’s


29

How would you explain the results of your experiment?

_________________________________________________________________________

_________________________________________________________________________

Which substance or substances entered the bag; which ones left the bag? Give evidence for your answer in

terms of the color changes that occurred.

_________________________________________________________________________

_________________________________________________________________________

Exercise 2: Osmosis (Diffusion of water)

The movement of water molecules through a selectively permeable membrane is a special case

of diffusion known as osmosis. Osmosis is a passive process, that is, it requires no metabolic

energy. The principle factor driving osmosis in animal cells is osmotic potential.

The osmotic potential is a function of solute concentration.

The addition of solute to water lowers the osmotic potential of a solution. Other things being equal,

if two unlike solutions are separated by a membrane that allows water to pass through but not

solutes, water molecules will move across the membrane towards the solution that has more solute.

Thus, osmosis results in movement of water from an area of low solute concentration to an area of

higher solute concentration.

Consider red blood cells that are normally suspended in plasma. Plasma contains salts, proteins and

other solutes. The osmotic potential of red blood cells and the surrounding plasma are equal so that

they are in equilibrium. If a drop of blood is placed on a slide and pure water is added to it, the

blood cells will quickly swell and burst because the solution surrounding the cells has a higher

osmotic potential than the cells.

Now let’s consider a plant cell. When plant cells are submerged in pure water, they will also take

up water because the intracellular contents are more concentrated than the surrounding solution. In

contrast to animal cells, plant cells do not burst because of the presence of a rigid cells wall located

outside the plasma membrane.

As water is taken up into the plant cell, pressure builds up within the cell. This pressure is called

turgor pressure. Turgor pressure will not increase indefinitely. Eventually, the turgor pressure

will prevent any further uptake of water even if the osmotic potentials are different on the two sides

of the membrane. Hence the combined effects of osmotic potential and turgor pressure drive

osmosis into and out of plant cells.

In this experiment, one group at your table will place several dialysis bags containing solutions of

different sucrose concentrations into beakers containing distilled water.

The direction of water movement can be determined by weighing the bags before and after placing

them in distilled water: the dialysis bags are permeable to water, but not to sucrose, and can gain or

lose water. The water potential of the sucrose solutions in the dialysis bags will be negative (recall

that the addition of solute to pure water decreases osmotic potential). Can you predict which way

the water will move?

___________________________________________________________________


30

The other group at your table will investigate the osmotic potential in a living system; the deshelled

egg. Share your results with your group when you are done your portion of the osmosis experiment.

Osmosis in an artificial system (for ½ of table)

1) Obtain five presoaked dialysis tubes. Use a pipette to put approximately 5 ml of each of the solutions

listed below into separate bags. After adding the solution, remove most of the air from the bag by drawing

the unfilled portion between two fingers. Tie a knot near the open end to seal the solution within the tube.

You should have 2 times as much empty space in the tube as that taken up by the volume of the solution.

This will leave enough unfilled space within the tube to accommodate the possible accumulation of water.

Mark the beakers to keep track of which tube contains which solution.

a. distilled water b. 20% sucrose c. 40% sucrose

d. 60% sucrose e. 80% sucrose

2) Carefully blot the outside of each bag. Tare the balance before weighing each bag, determine the initial

mass of each bag, and record the value in the table below.

3) Fill five 250-ml beakers three-quarters full with distilled water.

4) Place each bag in one of the beakers of water and label the beaker to indicate the % of the solution in

the dialysis bag. Make sure that all parts of the bag are completely covered by water.

5) Let stand for 1 hour. At the end of the required time, remove the bags from the water and carefully

blot them. Tare the balance, determine the mass of each bag, and record your data in the Table below.

Calculate % change = [(final mass- initial mass)/ initial mass] x 100

CONTENTS IN

BAG

INITIAL WEIGHT

(g)

FINAL WEIGHT

(g)

% CHANGE IN

WEIGHT

distilled water

20 % sucrose

40 % sucrose

60 % sucrose

80 % sucrose

a. What is the relationship between the increase in weight and the % sucrose within the dialysis bags?

_______________________________________________

b. How does the increase in sucrose concentration affect the water potential of the solutions inside the

dialysis bags? _________________________________

c. A solution (A) that contains more solute than another solution (B) is said to be

hyper-osmotic (hyper = more than; solution A is hyperosmotic to solution B).

Likewise, solution B can be described as being hypo-osmotic (hypo = less than) to solution A.

Solutions with equal solute concentration are called what? __________________.


31

Osmosis in a living system (for ½ of table)

A chicken egg is composed of a large cell surrounded by albumin and encased in a shell membrane

and a shell. The shell can be removed by soaking the egg in vinegar or acetic acid. The egg will serve as a

living model system for studying osmosis.

1) Place the following solutions in 3- 250 ml beakers:

a. distilled water

b. 20 % sucrose

c. 60 % sucrose

2) Carefully blot the outside of each egg. Tare the balance, determine the initial mass of each egg, and

record it in the table on the next page.

3) Place each egg in one of the beakers and label the beaker to indicate the % of the solution in the beaker.

Make sure that there is an equal amount of solution in each beaker.

4) Weigh the egg every 15 min for 1 hour. At the end of the required time, remove the eggs from the

beakers and carefully blot them. Tare the balance, determine the mass of each egg, and record your data in

the table below.

5) Calculate % change = [(final weight- initial weight)/ initial weight] x 100

Calculate rate of osmosis = (final weight- initial weight)/ 60 min

TIME

(MIN)

EGG IN WATER (g) EGG IN 20%

SUCROSE (g)

EGG IN 60%

SUCROSE (g)

0

15

30

45

60

% change

(at t=60 mins)

rate of osmosis

(% / minute)

Which of the solutions are iso-osmotic, hypo-osmotic, or hyper-osmotic compared to the chicken eggs?

_________________________________________________________________________

_________________________________________________________________________


32 Exercise 3: Diffusion of a liquid in a liquid

Done by your instructor!

We can make some observations about the rate of diffusion and changes in concentration of a

diffusing substance by observing the diffusion of a colored liquid in water.

Procedure

1. Fill a 4000 ml graduated cylinder to the top with water. Position the cylinder so that you can see the

milliliter scale. Allow the container to stand undisturbed for a few minutes to be sure that all water

convection has ceased.

2. Gently add a dropperful of methylene blue to the surface of the water. Take care to avoid disturbing the

surface of the water. Cover the container with a Petri dish lid to prevent disturbance by air currents.

Done by you!

3. Observe the diffusion of dye in the liquid and answer the following questions

a. What happens to the leading edge of the dye molecules?

b. What can you say about the chemical potential of the dye molecules at different levels of the cylinder?

c. Do you think that the dye molecules will become evenly distributed with more time? How long do

you think that it will take?

d. What do you think would happen if you used a smaller cylinder?

Exercise 4: Plasmolysis

If a plant is immersed in a hyper-osmotic solution, water will leave the cell. Microscopically,

increased loss of water and loss of turgor becomes visible as a withdrawal of the protoplast from the

cell wall (plasmolysis) and as a decrease in the size of' the vacuole.

Obtain a leaf from the tip of an Elodea plant. Place it in a drop of water on a slide, cover it with a

coverslip, and examine the material first at low power (10x) and then at higher power (40x). Locate

a region of healthy cells and sketch the location of the chloroplasts.


33 While touching one corner of the coverslip with a torn piece of paper towel to draw off the

water, add a drop of 40% sucrose solution to the opposite corner of the coverslip. Be sure that the

sugar solution moves under the coverslip. Wait about 5 minutes, and then draw your observations in

the space below.

Before After

The 40 % sucrose solution is _______(hyper-, iso-, or hypo-) osmotic to the Elodea cells.

References:

http://www.umanitoba.ca/faculties/science/biological_sciences/lab3/biolab3_3.html

http://biog-101-104.bio.cornell.edu/BioG101_104/tutorials/cell_division.html

http://arbl.cvmbs.colostate.edu/hbooks/cmb/cells/pmemb/transport.html

http://www.trinity.edu/slibs/OsmosisWeb/default.html

Barbara Cocanour and Alease Bruce in "Osmosis and The Marvelous Membrane" JCST November 1985,

pp127-130.


34

Lab exercise 6: Data* Analysis and

Statistics *Data is the plural form, datum is singular.

The Nature of Science

A critical part of the methods of learning about the universe that we call science is to gather

data.

Scientists then analyze the data and try to explain the conditions that caused them.

Statistical analysis is simply a collection of methods for evaluating questions or predictions

based on data. This lab has two objectives:

1) You should become familiar with the use of Excel spreadsheets to organize and summarize data, and to

perform routine mathematical calculations.

2) You will perform a statistical comparison of mean growth rates in milkweed bugs grown under two

experimental conditions.

Organizing and Summarizing Data

Data are usually a collection of either categories or numbers, corresponding to

measurements or “observations” of particular ―variables”.

Variables are called variables because the observations for a particular variable usually vary

(duh …!).

Observations are the individual measurements of a variable. Tables usually organize the

data into columns of variables and rows of observations under each variable title.

Recall from lecture that 1) the species Homo sapiens, to which we belong, varies widely in

terms of height and brain size, 2) brain size seems to be generally related to height when

comparing humans with some of the species from which we evolved (see figure below), and

3) brain size does not determine intelligence in humans.


35 Suppose you wished to determine whether head size depends in part on height in humans. We

will begin by collecting head lengths (in cm), head widths (in cm) and heights (in cm) in a table on

the whiteboard.

Following the procedure explained by your lab instructor, use the plastic calipers provided on each

lab bench to measure head length (in cm; from the point directly above the nose to the bony bump

at the back of the head), head width (in cm; with the caliper positioned directly in front of each

ear), and head height (in cm; estimated from right under your ears vertically to the crown of the

head). All estimates should be to the nearest half centimeter.

Two of the dimensions are shown in the photo above. Finally, estimate your body height (in cm) by

multiplying your height in inches by 2.54. Record the class data in Table 1.

Table 1. Morphometric measurements of head size and body height.

Head Length

(cm)

Head Width

(cm)

Head Height

(cm)

Head Volume

(cm3)**

Body Height

(cm)

**HV = LWH/6. Calculated as shown below.

Turn on your laptop, and download the file ―Brains and Bugs.xls‖ from the Blackboard website to

your desktop (make sure the ―desktop‖ is in fact the screen and not a directory within the hard

drive).


36 Open the file and note a first spreadsheet marked ―Brains‖ (see the tab at the lower left) with

the same variables you just measured (head length, head width, head height, and body height), and

with a fifth variable, head volume, that has yet to be calculated.

Enter the data for head length, head width and height under the correct column headings in the

Excel worksheet. Note that, once the measurements have been entered, Excel automatically

calculates an estimate of head volume.

To find out what formula was used, click on the cell E3 in the spreadsheet and read the formula next

to the fx button at the top of the worksheet. What is the formula? =____________________

In effect, you have just calculated head volume (HV) as HV = LWH/6, where L, W and H are the

head length, head width and head height. This assumes that heads are approximately ovoid, which

is only a very crude estimate of head volume, and is approximately twice the brain volume.

To determine brain volume, the brain would usually be extracted, and the empty cranial cavity

stuffed with metal shot, rice grains or some other non-compressible material. The material could

then be poured out into a graduated cylinder to estimate its volume. That procedure is not

performed in this lab …. ☺

You now wish to determine the degree to which head size is influenced by body height. Which is

the independent variable?__________ Which is the dependent variable? __________________.

Graphing and Visualizing data

Figures are graphs that provide (with a little practice) a visual summary of relationships

among variables. One way to visually evaluate the effect of one variable on another is a

scatterplot.

The horizontal axis (the X-axis) is used to plot the observations for the independent

variable, while the Y-axis is used to plot the observations for the dependent variable.

Construct a scatterplot showing the effect of height on head size, using the independent

variable as the X-axis and the dependent variable as the Y-axis, in Figure 1 below.

To speed up the process, group members can each place a portion of the data in the figure,

then copy the remaining data points from other group members.

Label both axes and include the units of measurement in parentheses after each axis title

(e.g., Height (cm)). Each point in the scatterplot actually represents a pair of observations,

one for the independent variable and a corresponding observation for the dependent variable.

How many points should there be in the scatterplot?


37

To create a scatterplot in Excel, use your mouse to highlight the two columns Head Volume and

Body Ht, including the column titles, then pull down under Insert to Chart.

In the Chart Wizard – Step 1, select XY (Scatter), and click on the sub-type without lines (upper

left). Next. In the Chart Wizard – Step 2, make sure you indicate that the series are in columns,

and click Next. In the Chart Wizard – Step 3, add appropriate chart and axis titles (including

measurement units in parentheses at the end of each axis title), then click on Legend in the toolbar

at the top of the window. Remove the legend. Next. Finally, in Chart Wizard – Step 4, include the

graph as an object in the current spreadsheet. Finish. If you have been successful, the resulting

graph should resemble the hand-drawn plot in Figure 1 above.

To obtain a line that expresses the overall effect of your independent variable on the dependent

variable, click once on the chart to highlight it, then pull down under Chart to Add Trendline.

In the Add Trendline window, make sure the linear option is specified, and click OK. A ―best fit‖

line summarizing the relationship should now appear. When you are done, Save the file to your

desktop where you can locate it easily later.

Based on the data points and trendline, does the graph show a very ―tight‖ relationship, with most of

the data points close to the trendline and a clear ―slope‖ to the trendline, or is the relationship very

―loose‖, with lots of scatter and a nearly flat trendline?

Based on your analysis, does height appear to influence head size?________________________

Graphing changes in body length of your milkweed bug over time

Here you’ll use a second kind of scatterplot, in which successive observations are connected

by straight lines. Using the data set compiled from your Adopt-a-bug measurements (mm),

make a graph below to visualize your bug's length relative to time (days from the beginning

of experiment).

What is the independent variable in your milkweed bug growth experiment?

______________________What is the dependent variable? ____________________. Label

both axes in below, including the units of measurement.


38 Now create a similar graph in Excel. Reopen the file ―Brains and Bugs.xls‖ if

necessary, and click on the tab for the second worksheet ―Your Bug‖ at bottom of the

screen. Enter your data in the two columns Time (d) and Length (mm)

The data in the first column should be the number of days since you started taking data

(include all days as integers (0,1,2,3,…), even if you missed a measurement on a particular

day). Then enter the length data for your bug in a second column (opposite the appropriate

days), and a third column with your partner’s bug lengths further to the right (make sure all

three columns are adjacent and have titles).

Now highlight all three columns including column titles using your mouse. As before, pull

down under Insert to Chart, and specify XY (Scatter). This time, however, choose the sub-

type with straight lines and markers (lower left option). Next. In Step 3, add axis titles

(with units in parentheses). Next. In Step 4, add the Chart as an object in your current

Sheet 2. Again save the file.

Your bug probably showed a ―stair-step‖ rather than linear progression of body lengths.

Insects, unlike vertebrates, grow by molting. Size remains approximately constant between

molts. Shedding the old exoskeleton during molting allows a brief period of rapid increase

in size until the new exoskeleton hardens.

How many times did your bug molt during the time you observed it? _______ A milkweed

bug undergoes 5 molts from the time it hatches from an egg to the time it becomes a

reproductive adult. How much does body length change between molts? ____ mm.

Was the amount of the increase the same each time? _____________________.

Variation in Growth Rate within Populations

In order to compare the growth of bugs in the class, you need to calculate ―growth rates‖ for

each. Discuss how you would represent growth rate numerically as a class.

Growth Rate =

The milkweed bugs in your class all have slightly different genetic composition, differed in

gender, differed slightly in age when you received them, and were likely subjected to

slightly different living conditions.

Thus, it is expected that there will be some variation both in the growth rate of growth and

the final size. Despite all this variation among individual bugs, you are hoping to be able to

discern the effects of one variable that was deliberately manipulated, dividing the bugs into

two groups.

The (independent) variable, defining the two groups, that was chosen by the class was

_______________________________. The two values (treatments) of that variable were

__________ and _________.

In Table 2 below, list the value (treatment) of the independent variable tested (at the top of

each column), and the growth rate for each bug in each treatment in the rows below the

treatment titles.

Calculating Descriptive Statistics

Using your bug data you will perform two sorts of statistical analyses: (1) the calculation of

descriptive statistics and (2) hypothesis testing. ―Descriptive‖ statistics don’t directly test


39 predictions, but are useful in summarizing groups of data. Two kinds of descriptive

statistics are the mean and standard deviation.

The mean describes the ―average value‖ for a group of observations. The standard

deviation describes the ―amount of variation among values‖ within the group of

observations.

The mean (X) is the average of the data as defined by the formula:

X = X/n

where is the sum of the individual data values (in this case the Xs are growth rates for the bugs). Thus,

X represents the combined growth rates of all the bugs. The value for n is the number of observations (in

this case the number of bugs).

The standard deviation describes how the individual observations within the group of

data are spread out around the mean. Large differences among observations (in this case,

large differences in growth rates among individual bugs) will result in a larger value for

the standard deviation.

Leave the mean and standard deviations at the bottom of Table 2 blank for now. Excel

will make the necessary calculations for you, but it is important that you know the

meaning of both the mean and standard deviation.

Growth rate in two groups of milkweed bugs

Treatmentt 1 = Treatmentt 2 =

Growth Rate (units= ) Growth Rate (units= )

Mean for trt 1: Mean for trt 2:

Std. Dev. for trt 1: Std. Dev. for trt 2:

You are now going to analyze the milkweed bug growth data in Excel. Reopen the file you saved

earlier if necessary, and click on the third worksheet ―Growth Rates”.


40 Now enter the data for your lab section in columns in the same format as shown in Table 3

above (with 2 columns, including column titles for each treatment and using one row for each bug

in the cells below).

If you need to, you can widen your columns by clicking on the line between letters (e.g., between A

and B) and dragging the column wider. Save the file once more. Now you're ready to analyze your

results.

Computing Descriptive Statistics

Click on cell B25 at the bottom of the first column. Click on the Paste Function button on

the tool bar, marked fx. Click Statistical under the function category and specify Average

for the function name. OK. Click OK again in the next dialog box. The average milkweed

bug growth for the bugs in the first treatment should appear in your spreadsheet.

To find the standard deviation, click on cell B26. Click on the Paste Function button on the

tool bar, marked fx. Click on Statistical under function category as before, and specify

Stdev for the function name. OK. Click OK again in the next dialog box.

Now compute the mean and standard deviation for the growth rates of bugs in group two at

the bottom of the second column. Enter all estimate in Table 2 of your lab book.

Hypothesis Testing

Hypothesis testing begins by asking a question. In the data involving diet and cancer, for

example, the question might have been ―Is head size influenced by height?‖ What is an

appropriate question involving growth rates of the two groups of bugs recorded in Table 2?

Question:

The next step is to formulate two hypotheses or possible outcomes related to the question.

The first hypothesis is termed the null hypothesis, and in this instance describes the

outcome if there is NO effect of treatment. What is an appropriate null hypothesis in this

instance?

Null Hypothesis:

The other hypothesis, often termed the alternative hypothesis or prediction, is the outcome

if there is an effect of treatment. State your alternative hypothesis (prediction) for this

experiment. The way you decide among these two possible outcomes is to either accept or

reject the null hypothesis. If you reject the null hypothesis, you are really deciding that there

is an effect of treatment.

Prediction:

A statistical test gives you the necessary information to either accept or reject the null

hypothesis. There many types of statistical tests, appropriate for answering different kinds

of questions. In this instance you will use a two-sample t-test, which evaluates whether the

mean values of observations in two groups of data differ significantly.

Now here is the hardest part. A scientists rarely knows with absolute certainty, if the null

hypothesis is rejected, whether he or she is right!


41 As an example, have another look at Figure 3. The null hypothesis for the data is ―The

two means are equal‖.

It sure LOOKS like the two treatments have different means. But suppose the study were

repeated, with a different set of observations? How sure are you?

Being wrong can be quite embarrassing, so biologists typically limit the likelihood of

incorrectly rejecting the null hypothesis. And each statistical test provides a statement of

significance that tells the scientist what the probability of being wrong actually is.

By convention, most biologists are willing to reject a null hypothesis only if the chance of

being wrong (called ―p‖ or the statistical significance of a test) is less than ―1 in 20‖ (5%),

or if p<0.05. Of course, the smaller the actual ―p‖ value (much less than 0.05), the more

confident you are that your decision to reject the null hypothesis is the correct one.

Why not instead only reject the null hypothesis if your chance of being wrong were, let’s

say, less than one chance in a thousand (p<0.001)?

The answer is ―because then the chances of correctly rejecting the null hypothesis when

there really is a difference become vanishingly small; in other words, real differences

between treatments, or real relationships between variables, would only rarely be uncovered,

and discoveries of potential significance to science would be rare.

Now it’s time to either accept or reject your null hypothesis regarding the bug data. Click

on cell E17 to the right of the block of data. Then click on the Paste Function button on the

tool bar, marked fx. Under the function category click Statistical and t-test for function

name. Click OK.

The t-test dialog box should appear. Now you need to indicate what data you wish to test. If

your cursor isn't already there, click inside the Array 1 box. Now indicate the data in

treatment 1 by dragging your cursor over the data in the second column, containing the

growth rates of the first treatment.

Your data range will automatically appear in the Array 1 box. Now click in the Array 2 box

and drag your cursor over the growth rate data for treatment 2. For the number of tails enter

2, and for the type of test enter 2 (don’t worry about why). Now click OK. Excel will

automatically calculate a (―2-tailed 2-sample‖) t-statistic and enter the resulting statistical

significance, or p value, in the cell.

What is the chance that you would be wrong if you reject your null hypothesis?

____________________________

Based on the t-test, should you decide to accept or reject the null hypothesis?

____________________________

What is your conclusion regarding the results of the experiment?


42

Lab exercise 7: Fermentation and

Metabolism

Lipids, carbohydrates, nucleic acids, and proteins are necessary components to living

things yet are very difficult to synthesize in even the best-equipped laboratories. The

chemical reactions are hard to start and often produce unwanted side reactions and by-

products.

It may seem strange that simple cells can produce macromolecules that highly trained

chemists cannot. This is because even the simplest cells have specific tools that are

synthesized to facilitate difficult and diverse chemical reactions. These tools are catalytic

enzymes.

Enzymes are usually proteins (although some RNA molecules also have catalytic

properties). These enzymes have a complex three-dimensional shape consisting of one or

more polypeptide chains folded to form an active site.

The molecules that fit into the enzyme’s active site are called the substrates. When the

substrate binds to the enzyme, it is affected by the specific functional groups of the amino

acids of the active site and undergoes a chemical reaction to form a product.

The enzyme is not changed by the reaction and is capable of immediately binding to

another substrate molecule.

E + S → E-S → E + P

The conformation of an enzyme is crucial to its activity. Thus, anything that alters the

enzyme structure such as changes in pH, temperature, ions, detergents, or inhibitors can

influence the enzyme’s activity.

Cells depend on the formation (anabolic reactions) of complex macromolecules, such as

DNA, to survive. These types of chemical reactions require chemical energy and complex

pathways catalyzed by enzymes. Adenosine triphosphate (ATP) is the energy-rich

compound most often used by cells to synthesize polymers from monomers.

The major source of ATP for many cells is the oxidation of glucose:

C6H12O6 + 6O2 → 6CO2 + 6H2O

(aerobic respiration)

The oxidation takes place in two major stages. The initial process is called glycolysis and

results in the splitting of the glucose molecule into two 3-carbon molecules of pyruvic acid

(pyruvate). Glycolysis can proceed in the presence or absence of oxygen.

The energy released from this pathway is sufficient to produce a net gain of 2 ATP molecules.

During anaerobic (oxygen absent) conditions, some organisms metabolize pyruvic acid to

other compounds, such as lactic acid (lactate fermentation) or ethanol and CO2 (alcoholic


43 fermentation) in order to regenerate cytoplasmic NAD

+. For example, in the absence of

oxygen, our muscles produce lactic acid but yeast produce alcohol and carbon dioxide (see

Figure 1 on the next page).

C6H12O6 → 2CO2 + 2C2H5OH

(alcoholic fermentation)

When oxygen is available, aerobic respiration proceeds through the Krebs cycle and electron transport.

These reactions will yield 36-38 ATP per molecule of glucose.

http://www.accessexcellence.com/AB/GG/ana_Pyruvate.html


44 Fermentation pathways. (A) When inadequate oxygen is present, for example, in a muscle cell

undergoing vigorous contraction, the pyruvate produced by glycolysis is converted to lactate as

shown. This reaction restores the NAD+ consumed in step 6 of glycolysis, but the whole pathway

yields much less energy overall than complete oxidation.

(B) In some organisms that can grow anaerobically, such as yeasts, pyruvate is converted via

acetaldehyde into carbon dioxide and ethanol. Again, this pathway regenerates NAD+ from NADH,

as required to enable glycolysis to continue.

In this laboratory, you will perform experiments to observe the processes of fermentation and

metabolism.

Exercise 1: Production of carbon dioxide gas by fermentation

Yeasts are simple unicellular fungi that are classified as facultative anaerobes- they can live in

either anaerobic or aerobic environments.

Under anaerobic conditions, yeasts carry out alcoholic fermentation to produce alcohol and carbon

dioxide.

A capillary tube with a culture of fermenting yeast will be used to measure the displacement of the

liquid by gas.

Different groups will investigate the effects of different variables on the rate of gas production by

yeast. Variables to be tested include 1) different substrates and 2) the addition of a metabolic

inhibitor.

Procedure:

1) Each student will fill a plastic disposable tube to the bottom etch mark with baker’s yeast.

Fill to this mark

2) Add 10 drops of buffer and mix for 3 minutes until the mixture becomes a thick slurry.

3) Add 5 drops of the assigned carbohydrate solution (see table on the next page) or water and mix for

an additional 2 minutes.

4) Using a capillary tube open at both ends (1.5-1.8 X 100mm), mark the

tube with a Sharpie marker at the midpoint of the tube. Fill the tube to this mark

by capillary action by placing the capillary tube into the test tube and holding the

tubes so that they are nearly horizontal. Be careful to prevent air from entering

the capillary tube, as it should be a continuous column of the yeast solution.

5) After filling, the open end of the half- filled capillary tube should be

closed with your index finger to prevent loss of solution. The solution end of the

capillary tube is then inserted into the critoseal and gently pushed to the bottom of

the container with a slight twisting motion. The sealed tube is then placed in one

of the numbered receptacles on either end of tray with the open end downwards

(critoseal end is up). The capillary tube should look like this:


45

6) Place the tray containing the tube under an incandescent lamp to increase the temperature of the

reaction.

7) The mark on the tube will be your 0 mm mark at T = 0. Measure the distance that the fluid level

moves downward every 2 minutes until the solution gets to the end of the capillary tube. Remove the

capillary tube from the tray and place the capillary tube in a paper towel to catch the solution as it moves

out of the bottom of tube.

8) If your solution does not move within 12 minutes, stop your reaction.

Data for exercise 1

TIME

(MIN)

Water Gluc Fruct Galacto Sucr Lact lactose +

lactaid

Glucose +

Na bisulfite

0

2

4

6

8

10

12

mm

gas/mi

n

9) Share your results with everyone at your table and fill in the table above.

Questions to consider:

1) Does the rate of fermentation depend on the carbohydrate fermented? _________ Are there differences

between monosaccharides and disaccharides? ___________

2) Which carbohydrate generates the greatest gas production by the yeast? ______ Which carbohydrate

produces the least amount of gas? ______________

3) What effect does Na bisulfite have on CO2 production? ____________

Sodium bisulfite reacts with acetaldehyde to prevent CO2 formation before it can be reduced to ethanol

and, therefore, inhibits fermentation.

Graph the results from the glucose, fructose, galactose, sucrose, and lactose experiment on the

graph paper and determine the rate of gas production (mm gas/ min).

Exercise 2: Examining Oxidation-Reduction Reactions in living cells

When cells metabolize food for the production of energy, they oxidize energy-rich (electron-

rich) substances, resulting in a loss of electrons.


46 During catabolism, a substrate (such as a carbohydrate) is oxidized initially without the

participation of oxygen. Electrons (H atoms in biological reactions) released by enzymatic

reactions are accepted by coenzymes (like NAD+ and FAD) that become reduced.

The reduced coenzymes (such as NADH + H+ or FADH2) are oxidized by the electron

transport chain.

The final electron transport carrier is oxygen during aerobic respiration.

Part 1: Reduction of methylene blue by yeast

In the absence of oxygen, electrons may be diverted from the electron transport chain by an

oxidation-reduction mediator, such as methylene blue.

Methylene blue is normally blue in the oxidized state but becomes colorless when it is

reduced.

The methylene blue enters the cell, becomes reduced, and leaves the cell in the reduced

state. If the cells are metabolically active, the colorless condition will remain

indefinitely.

The rate of decolorization depends on the metabolic activity of the cells and the presence of

oxygen.

The equations for the reaction are:

O2 + MBred (colorless) → MBox (blue)

MBox → MBred + R

where MBred is the reduced (colorless) form of methylene blue, MBox is the oxidized (blue) form of

methylene blue, and RH represents glucose.

Procedure:

1) Label and set up six test tubes with the following reagents:

Note: (ADD 1 g YEAST TO 5 ML BUFFER) = ONE PER TABLE. FROM THIS MIXTURE, YOU

WILL ADD THE YEAST DROPS INDICATED IN THIS LAST COLUMN

TUBE BUFFER DYE WATER GLUCOSE NA

BISULFITE

YEAST

1 2 ml 1 drop 15 drops

2 2 ml 11 drops 5 drops

3 2 ml 1 drop 10 drops 5 drops

4 2 ml 1 drop 5 drops 5 drops 5 drops


(Boiled)**



47 Cover the tubes with parafilm and incubate at room temperature for 30 min.

2) Record your results in the table below. The intensity of color can be graded 0 for no color, +, ++, +++,

++++ indicate increasing intensities of color.

**The boiled yeast will be provided by your instructor.

TUBE COLOR AT T= 0 COLOR AT T = 30 MIN EXPLANATION OF RESULTS

1

2

3

4

5

6

4) Which tubes were used as controls?

5) Compare the results in tubes 3 & 4? Is there any difference? If so, can you account for the difference?

What effect did boiling the yeast have on the reaction? Explain.

What effect did adding inhibitor have on the reaction? Explain.


48

Part 2: Staining yeast with methylene blue

Living cells will reduce methylene blue, resulting in the colorless form. What happens

when cells are damaged or cease to metabolize fuels?

Prepare a smear (lightly rub the end of the pipette on the slide) of living and boiled yeast

from tubes 4 and 5. Draw your results in the space below.

Pay particular attention to the color of the yeast.

Does methylene blue get into the cells? How do you know?

Is there a different staining between unboiled and boiled yeast? If so, why?

References

http://www.uwrf.edu/biotech/workshop/activity/act1/act1.htm#labfy

http://www.oxy.edu/departments/tops/Yeast/yeasthome.htm

http://www.biology.arizona.edu/biochemistry/problem_sets/metabolism/metabolism.html

http://www.phys.ksu.edu/gene/a1.html

http://tidepool.st.usm.edu/crswr/yeastfermmov.html

"Fermentation, Respiration, and Enzyme Specificity: A Simple Device and Key Experiments with Yeast," by L. Reinking, J.

Reinking, and K. Miller, The American Biology Teacher, Vol. 56, March 1994, pp. 164-168.


49

Lab exercise 8: Mitosis, Meiosis, and Life

Cycles

One of the basic principles of the cell theory is that all cells arise from pre-existing cells. Cells arise

from others by karyokinesis (nuclear division) and cytokinesis (cytoplasmic division). There are

two types of nuclear division: mitosis and meiosis.

During mitosis, two daughter nuclei are formed that are genetically identical to the original cell:

(2n → 2n, diploid to diploid).

In diploid organisms, chromosomes occur as matched pairs (homologous chromosomes), one

paternal and one maternal in origin.

Meiosis, however, reduces the number of chromosomes to half the number present in the

original cell (2n->n, diploid to haploid).

Let us examine the phases the cell cycle when the cell is dividing by mitosis:

I) Interphase= preparing for cell division

* G1 (first gap) is a period of cell growth and doubling of organelles.

* S is the period when new DNA is synthesized.

* G2 (second gap) is when preparations are made for cell division, i. e., synthesis of

spindle proteins and ATP used during division. In animal cells, a pair of centrioles divides to form two

pairs of centrioles.

Notable features: DNA is uncoiled and distinct chromosomes are not visible. The chromosomes

appear as a diffuse, darkly stained material called chromatin.

During the S phase, DNA is replicated and each chromosome is composed of two helices of DNA

called sister chromatids.

The chromatids are joined at a central region called the centromere, which contains a kinetochore-

a place where the spindle fibers can bind.

A chromosome composed of two chromatids is called a dyad.

II) Mitosis = separation of the replicated chromosomes (4 phases)

* Prophase: chromatids become visible, the nuclear membrane and nucleoli disappear, centrioles

migrate to opposite sides of the nucleus, and spindle fibers attach to the kinetochore.

* Metaphase: the chromosomes line up in the middle of spindle apparatus (the metaphase plate).

* Anaphase: the sister chromatids separate into monads and move toward opposite poles. The

monads are called daughter chromosomes.

* Telophase: sister chromatids have completely separated and begin to uncoil, the spindle apparatus

disappears, the nuclear membrane and nucleoli reappear.

III) Cytokinesis = the cellular material is divided between the two daughter cells. Daughter cells are

usually smaller following cytokinesis compared to other cells in interphase.


50 Exercise 1: Mitosis in an eukaryote

Part A: (Allium) root tip Kingdom = Plantae, Phylum = Anthophyta

Use this slide to locate the different stages of the cell cycle in plants. Do you find more cells in various

stages of mitosis at the root cap or at the apical meristem? ________ What do you suppose is the function

of those two regions? ________________________

Focus on the apical meristem. Make drawings and label the nuclear changes that occur during the 5 major

stages in the cell cycle (IPMAT).

Mitosis Jumble: Can you determine the correct sequence of events for mitosis? a) Spindle apparatus disappears and nuclei reform. Nucleoli reappear.

b) Chromosomes line up at the middle of the spindle apparatus.

c) Centrioles replicate and separate.

d) DNA replicates.

e) Cytoplasmic division.

f) Cell grows in size and organelles double.

g) Sister chromatids separate and move toward opposite poles.

h) Spindle apparatus forms.

What is the correct order for mitosis? ___________________________________

Part C: How long does it take for the onion root tip cells to complete the cell cycle?

When studying populations of dividing cells, biologists often need to know the length of the cell

cycle. If a cell biologist wished to know the effect of an environmental factor such as CO2 level on

the cell cycle, it would be necessary to know the length of the cycle under normal and altered

conditions. The technique described below only works for randomly dividing populations of cells.

For our determination of cell cycle, we will use the onion root tip slide. We will assume that the

cells have been treated for 30 min with colchicine, a chemical that stops nuclear division in

metaphase by preventing microtubule polymerization.

In addition, we will assume that the fraction of the cells in metaphase in an untreated population is

0.01 (1%).


51 To calculate the cell cycle,

fraction of cells in metaphase after colchicine

- fraction of cells in metaphase under normal conditions (0.01)

= fraction of cells in metaphase in T = 30 min

Let’s say that 0.21 of the colchicine-treated cells are in metaphase (This is a high estimate for these

kind of cells, used for simplicity). It means that 0.20 (0.21-0.01) of the cells reached metaphase in

30 min. Thus, 30 min represents 0.20 (20% or 1/5) of the cell cycle:

Cell cycle = T/F

where T = time of the colchicine treatment and = fraction of cells reaching metaphase.

For the example above, the cell cycle would be 30 min / 0.2 = 150 minutes

Work with a partner where one of you serves as the observer while the other will record the data.

The observer scans up and down the vertical rows of cells visible within the field at one time,

calling off judgments on the stages of the cell cycle encountered. The recorder marks down the data

in the table below.

Six such field counts should be made (two slides, as there are 3 root tips per slide), with the

observer and recorder changing roles after the third field count.

It is suggested that only the three most central rows be counted in each field. Using the 40 x

objective, you should see something like this (keep in mind that your field of view will be a circle):

In this example, there are about 144

cells, of which 4 are in metaphase

(4/144 = 0.028).

Therefore, the fraction of cells in

metaphase is will be given by the

following relationship (as above):

o Fraction of cells in

metaphase after colchicine

(0.028) - fraction of cells in

metaphase under normal

conditions (0.010) = 0.018

Assuming a T = 30 minutes, the length of

the cell cycle would be:

30/0.018 = 1,667 minutes (~ 27 hours or

about 1 day and 3 hours).


52 Using your own data, determine the length of the cell cycle using the formula on the previous page.

Complete the table below.

Field Of

View

Total Number Of

Cells

Number Of

Cells In

Metaphase

Fraction of Cells

In Metaphase

Length Of The Cell

Cycle

1

2

3

4

5

6

Exercise 2: Meiosis and life cycles

Meiosis is a form of nuclear division in which the newly formed nuclei contain one half the number

of sets of chromosomes as were present in the original nucleus. Meiosis turns diploid nuclei (nuclei

with two sets of chromosomes, 2n) into haploid nuclei (nuclei with only one set of chromosomes;

one chromosome from each and every "pair" in the original).

Any organism that reproduces sexually must have a stage in the life cycle where meiosis occurs to

ensure that the total number of chromosomes stays constant over generations. The fusion of

haploid (n) nuclei following fertilization produces a zygote (2n).

IMPORTANT NUCLEAR EVENTS DURING MEIOSIS

Meiosis consists of two nuclear divisions (Meiosis I and Meiosis II) and results in the

production of 4 daughter nuclei. The interphase stage of the cell cycle immediately

preceding meiosis I is similar to interphase in mitosis (see first page of this lab).

MEIOSIS I

PROPHASE I:

Uncondensed homologous dyad chromosomes are brought together to form uncondensed

tetrads. Chromosomes condense, the nuclear envelope disappears, and tetrads become

attached to the spindle (a tetrad is composed of two homologous dyad chromosomes).

Homologous chromosomes synapse (get close together) and may break and rejoin. This

process is called crossing over.

This is the first major difference between mitosis and meiosis.

METAPHASE I:

Condensed tetrads are moved to the equator of the spindle.

ANAPHASE I:

The tetrads are separated as homologous dyad chromosomes move toward opposite poles. This

is a second major difference between meiosis and mitosis.

In mitosis, each homologous chromosome independently separates into monads.


53 TELOPHASE I:

Condensed dyads are at the poles of the spindle and the spindle subsidesor disappears. By this time

the sets of chromosomes have been halved.

Each nucleus is haploid, but chromosomes are still in the dyad condition. The condensed dyads

may go to the uncondensed state (resulting in a stage called interkinesis.)

Interkinesis is a third major difference between mitosis and meiosis. During this time the DNA

may again become functional.

In some cells, the dyad chromosomes may remain in the condensed state and proceed directly to

Prophase II.

MEIOSIS II

METAPHASE II:

The dyad chromosomes are moved to the equator of the spindle.

ANAPHASE II:

Each dyad chromosome is split into two monads, which move to opposite poles of the spindle.

TELOPHASE II:

The condensed monad chromosomes become uncondensed monad chromosomes and the nuclear

envelope reforms. Meiosis is now complete.

In summary: The first meiotic division separates sets of chromosomes, resulting in haploid nuclei,

but the chromosomes are still dyads. The second meiotic division reduces dyad chromosomes to

monads.

THE ROLE OF MEIOSIS IN LIFE CYCLES

The objective of this exercise is to illustrate the role of meiosis in the reproductive life cycle. There are

three basic life cycle patterns which can be described by the role that meiosis plays. They are:

A) Gametic- meiosis is involved in providing nuclei for gametes. This pattern is typical of animals and

some protista.

B) Zygotic - the zygote nucleus goes through meiosis before the life cycle continues. This pattern is very

common among the protista (algae) and the fungi.

C) Sporic - meiosis occurs in 2n sporogenous cell nuclei, providing nuclei for spores. This pattern is found

in the plants and some protista.

The Gametic life cycle = e.g. Homo sapiens (Sorry, no lab experiments for this one)


54

Spirogyra is a green filamentous alga commonly found in ponds and water filled ditches. In

the vegetative state, the cells contain one or more spiral chloroplasts and a haploid nucleus

suspended in the cell vacuole by cytoplasmic bridges.

At the time of sexual reproduction, two filaments come to lie parallel to each other.

Projections from the cells begin to grow toward each other. These eventually join, forming a

conjugation tube.

The vegetative cells then go through a morphological change and the protoplast (cell

membrane and the cellular contents) functions as a gamete. The protoplast in one cell moves

by amoeboid movement through the conjugation tube and fuses with the gamete in the other

cell, forming a zygote.

The zygote produces a thick wall around itself and becomes a zygospore. The parent filaments

disintegrate, releasing the zygospore into the environment.

The zygospore is a resting stage and allows the species to withstand environmental conditions that

are unfavorable to the vegetative form. When environmental conditions are favorable, the 2n

zygospore nucleus goes through meiosis producing four haploid nuclei.

Three of these nuclei disintegrate, leaving a cell with one haploid nucleus. This cell then escapes

from the zygospore wall and produces the typical vegetative filament by mitosis.

Part A: Observing the life cycle of Spirogyra

Using prepared slides of Spirogyra, observe the organism in its vegetative form and locate a

chloroplast and nucleus. Observe the sexual stages and identify - the vegetative form, conjugation

tube, gamete, and zygote. Label the diagram below.


55 The Sporic Life Cycle = e.g. fern

The ferns will be used to illustrate the sporic life cycle. In this life cycle there are two generations

produced, a sporophyte generation (nuclei diploid) that produce spores, and a gametophyte

generation (nuclei haploid) that produce gametes.

Part B: Observing the life cycle of fern

The sporophyte generation is composed of roots, stem, and leaves. The most obvious structures are

the leaves or fronds. Observe the frond of the fern on demonstration and note the brownish clusters

on the underside of the leaflets. These clusters are called sori (singular = sorus).

Each sorus is a collection of sporangia in which meiosis and cytokinesis has occurred to produce

spores. Obtain a slide of Fern leaflet and locate the sporangia and spores. Draw a picture of the

fern leaflet in the space below.

After the spores geminate, they produce a tiny, green flat heart-shaped, plant body called a

prothallium. This is the gametophyte generation. The prothallium grows on the surface of the soil.

The sex organs are borne on the lower or ventral surface of the prothallium. Study the mature

prothallia on demonstration and identify the sex organs, the antheridia (male) and the archegonia

(female).

The antheridia are usually formed at the point of the heart-shaped prothallium and appear as small

bumps. The archegonia are usually formed at the cleft of the heart-shaped prothallium and appear as

small finger- like projections. Sperm are released at a time when there is a layer of water on the soil

to allow the sperm to move to an egg in an archegonium. The zygote is retained within the

archegonium and during the early stages of embryonic development. The prothallium provides

nourishment for the developing sporophyte. Study the prothallium with attached sporophyte on

demonstration.


56 Know the structures of the fern on the diagram belowgote, sporophyte, sori, gamete, spore,

gametophyte

References:

http://www.umanitoba.ca/faculties/science/biological_sciences/lab3/biolab3_5.html#Mitosis

http://www.umanitoba.ca/faculties/science/biological_sciences/lab5/biolab5_2.html#Meiosis

http://www.biology.arizona.edu/cell_bio/activities/cell_cycle/cell_cycle.html

http://biog-101-104.bio.cornell.edu/BioG101_104/tutorials/cell_division.html


57

Lab exercise 9: Mendelian Genetics

Gregor Mendel discovered that ―hereditable factors‖ come in pairs. We now call these factors

genes.

In some cases a gene is a piece of DNA in a chromosome that codes for a specific protein that

will become a distinct trait, such as the height of a plant, the color of a bird, or part of a

chemical pathway.

More often, however, a gene codes for a protein that is a part of a complex set of instructions

that may include other genes.

While diploid cells have two copies of each gene, the copies may not be identical. For

example, each of our cells has two genes that control hair color, but one gene may code for

light hair and one may code for dark hair.

The different versions of a gene are called alleles. The different alleles of a gene both produce

a protein. However, these proteins are not expressed equally based on the appearance of an

individual (phenotype).

For each trait, one allele is usually dominant and the other is recessive. A dominant allele is

evident in the phenotype whenever it is present, even if only one copy is present. A recessive

allele must be present in two copies to be apparent in the phenotype.

Mendel studied pea plants in order to understand hereditable factors. He discovered that when

purebred (homozygous, RR = genotype) plants that produce round seeds are crossed with

purebred (homozygous, rr = genotype) plants with wrinkled seeds, the first generation (F1)

of plants all have round seeds.

Thus, the round seed trait is dominant over the wrinkled seed trait (recessive). The offspring

from this mating are called hybrids (heterozygous, Rr = genotype). Mendel found that

crossing the hybrid offspring led to 75% of the offspring with the round seeds and 25% of the

offspring with wrinkled seeds (3:1 ratio). These experiments led Mendel to postulate that each

individual carries two factors for each trait.

Exercise 1: Why Do Mendel's Peas Wrinkle?

This lab is designed to demonstrate the connections between genotype and phenotype. The

sequence of events that influence the production of round and wrinkled seeds is shown on the

next page.

This complex process involves the synthesis of different forms of an enzyme that lead to the

production of starch, accumulation of sugar (sucrose) with the recessive form of the enzyme,

differences in the osmotic potential, and water loss with desiccation.

Remember that starch is a polysaccharide made up of many glucose monomers bonded

together. Specific enzymes are needed to link the glucose onto the growing starch molecule.

The R allele produces a starch branching enzyme that, as the name implies, adds branches of

glucose units to the growing starch molecule

The starch branching enzyme also influences other enzymes (coded by other genes) and their

ability to synthesize starch. The r allele codes for a mutant form of starch branching enzyme,

leading to an inefficient production of starch.


58

You will prepare an extract of both round and wrinkled pea cells to test their:

1) Capacity to form large starch granules whose shape is determined by the type of starch formed, and

2) Difference in osmotic potential that results from sugar accumulation.

Figure 1: Mechanism responsible for round and wrinkled pea shape

Procedure:

1) Take a sample of 5 dried peas. Determine the mass of 5 dried round peas and 5 dried wrinkled

peas. Record your results in the table below.


59

2) Take a sample of 5 presoaked peas. Determine the mass of 5 soaked round peas and 5 soaked

wrinkled peas. Record your results in the table below.

PEA TYPE DRY MASS (G) WET MASS (G) CHANGE IN MASS (G)

round

wrinkled

Which peas gained the most weight? ___________________________

3) Take 2 presoaked peas of each type from the side counter. Grind the round peas with a small

amount of water using a mortar and pestle. Try pushing hard with the pestle while twisting against the

mortar until no large pieces remain. Repeat this procedure for the wrinkled peas. MAKE SURE YOU

THOROUGHLY WASH THE MORTAR AND PESTLE WITH SEVERAL RINSES OF TAP

WATER between each variety.

4) Using a pipet, make a wet mount slide for each variety of peas and cover it with a coverslip.

Observe the extract from both the round and wrinkled peas using the 40x objective.

5) Stain the slide with Lugol’s solution by touching a drop of Lugol’s to one side of the coverslip and

a piece of a paper napkin to the other side of the coverslip. Observe with the high power objective.

6) Compare the types of starch granules. Some grains are compound (see fig. 2 below), i.e., an

aggregation of individual granules and some are simple with just one grain.

Figure 2: Diagrammatic representation of compound starch grains found in wrinkled peas (simple

grains not shown).


60

7) Draw a picture of your observations below.

ROUND

WRINKLED

Question 1: Did you measure any differences in the osmotic potential of the round and wrinkled peas?

Explain.

Question 2: Are there differences in the starch grains of wrinkled and round peas? Explain.

Exercise 2: Human Genetics: what will your children be like?

The traits described by classical Mendelian genetics demonstrate simple dominant-recessive

characteristics because the trait has only 2 alleles. Polygenic traits, such as skin color or height,

depend on several genes and their interaction with each other and with the environment. Traits that

have only a few discreet phenotypic states are called meristic traits.

Several meristic traits have been identified in humans that illustrate Mendelian genetics quite well. A

few are listed below:

1) Acondroplastic dwarfism: The dominant allele causes dwarfed arms and legs; normal proportions

are recessive.

2) Attached earlobes: Free earlobes are dominant.

3) Bent little finger: Bent little finger is dominant, a straight little finger is recessive.

4) Pigmented irises: Lack of pigment in the iris results in blue eye color, which is recessive (p).

Any other eye color (brown, green, hazel, violet, etc.) is due to the deposition of pigment in the iris

and the expression of the dominant allele of the gene (P).

5) Curly hair: Curly hair is dominant over straight hair. Wavy hair is heterozygous.

6) Dimples: Dimples in your cheeks or in your chin are dominant.


61

7) Freckles: Freckles on your face are dominant.

8) Interlacing fingers: Relax and casually fold your hands together so that your fingers interlace. If

you left thumb is on top of your right thumb, you have the dominant allele. If your right thumb is on

top, you are homozygous recessive.

9) Hitchhiker’s thumb: Place your thumb in a hitchhikers position. If the tip of your thumb curves

downward, you have the dominant allele. A straight thumb is the recessive allele.

10) PTC tasting: Touch some test paper impregnated with phenylthiocarbamide to your tongue. If

you detect a bitter taste, you have a dominant allele.

11) Tongue rolling: The ability to curl your tongue in a U-shape indicates the dominant allele.

12) Widow’s peak: A pointed hairline is determined by the dominant allele.

Determine your own phenotype for each trait. Can you determine the genotype?

Your instructor will tabulate the data for the whole class for these traits. Fill in the chart below

with the results. Are the dominant phenotypes always more common than the recessive

phenotypes?

______________

DOMINANT TRAIT PRESENT ABSENT

Acondroplastic dwarfism

Free earlobes

Bent little finger

Pigmented iris

Curly hair

Dimples

Freckles

Interlacing fingers (L/R)

Hitchhiker’s thumb

PTC tasting

Tongue rolling

Widow’s peak

Not all common traits are a matter of simple dominance. Blood types, for example, exhibit a

property called codominance. As you probably know, there are 4 major blood groups, A, B,

AB and O. These four types are influenced by multiple alleles of a single gene.

Blood type describes a particular glycolipid on the surface of red blood cells. The allele IA

produces the type-A glycolipid and the allele IB produces a type-B glycolipid. Both of these

alleles are dominant over ―i‖, which produces neither glycolipid A nor B. The presence of the

two dominant alleles is what makes blood type an example of codominance. Thus, six possible

genotypes will give the four phenotypes.

IA I

A or I

Ai → Type A I

B I

B or I

Bi → Type B

IA I

B → Type AB i i → Type O


62

Several traits in humans are located on the X or Y-chromosomes. These are called sex-linked

traits. Genes for sex-linked traits are usually carried on the X chromosome and are absent on

the Y chromosome. As a result, males only have a single copy of any gene located on the X

chromosome and they will express whichever allele is present, even if it is normally a recessive

allele.

The gene that controls color vision is located on the X chromosome and has two alleles:

normal (XB) and colorblind (X

b). The normal condition is dominant over the colorblind

condition. Females that have normal color vision, but are heterozygous for colorblindness (XB

Xb) are often referred to as carriers.

Males that have inherited the gene for color-blindness (XbY) are color blind. Can females be

colorblind? ___________________

Why?_____________________________________________________________________________

__________________________________________________________________________________

What will your children be like?

For this activity, we will observe patterns of inheritance for three general types of traits: an

autosomal trait controlled by two alleles (hair texture), an autosomal trait controlled by

multiple alleles (blood type), and a sex-linked trait (color-blindness).

To gain experience with genetics, you will work with a partner to determine the combination of

traits that would be possible in your offspring. Here is how it works:

Step 1: Get a partner: Men will draw a card from the bag marked males and women will draw a card

from the bag marked females. If there isn’t an equal number of men and women in the class, some

people will play the role of the opposite sex.

Step 2: Determine your traits. The three traits listed on the card are hair type, blood type, and color-

blindness. Write them in the space below:

Step 3: Determine your genotype. Your hair is curly (HH), wavy (Hh), or straight (hh). Your blood

type is A, B, AB or O. If your blood type is A or B, you can assume that the genotype is IAi or I

Bi,

respectively. You are a normal-vision (XBY) or color blind male (X

bY). Alternatively, you are a

normal vision (XB X

B ), carrier (X

B X

b), or color-blind female (X

bX

b). Write them in the space below.


63

Step 4: Determine all the possible gametes that you could contribute to your offspring. Remember,

each gamete will only get one allele from each pair of homologous chromosomes. Write them in the

space below.

Step 5: How many possible gametes could your mate have? Write them in the space below.

Step 6: Look at the various combinations of your gametes to answer the following questions.

1. Will any of your children be colorblind? If so, what are the chances that your sons will be color

blind? Your daughters?

2. Will any of your children have curly hair? Wavy hair? Straight hair? What are the ratios of the

different hair types?

3. What are the possible blood types of your children?


64

4. Could you possibly produce a son who was colorblind, had wavy hair, and A type blood?

5. Can you predict the phenotype of your first child based on the most likely combination of these

three traits?

References

http://esg-www.mit.edu:8001/esgbio/mg/mgdir.html

http://biology.clc.uc.edu/courses/bio105/geneprob.htm

http://www.blc.arizona.edu/courses/181gh/rick/genetics1/mendel.html

http://www.biology.arizona.edu/mendelian_genetics/mendelian_genetics.html

http://www.furman.edu/~lthompso/pealab.htm

http://www3.ncbi.nlm.nih.gov/Omim/


65

Lab exercise 10: Population Growth of

Duckweed

In this laboratory you will 1) determine whether the growth of individual duckweed

plants is density dependent (involving intraspecific competition), 2) experimentally

manipulate nutrients and/or light to determine their effects on growth, and 3)

mathematically simulate the effects of varying growth rate and carrying capacity on

duckweed growth using a computer model.

This is a two-week lab. During the first week you will decide on an experimental

design and set up your experiment. During the second week you will analyze the

growth response of the plants.

Duckweeds (genus Lemna) are free-floating aquatic plants that

undergo continuous growth. Individual plants are stemless and

have 1-4 leaves. Roots from the leaves hang down into the water

and absorb nutrients needed for growth. Leaves typically

reproduce asexually by producing new leaves which, when large

enough and equipped with their own roots, break off from the

parent to form a new plant.

Week 1:

You first wish to determine whether the growth of the duckweed plants is density

dependent. To do this you will measure the ―per individual growth rate‖ (PIG for

short). The PIG can be approximated for relatively short time periods using the

equation

PIG = ln(Nwk2) – ln(Nwk1) (1)

where ln(N) is the natural log of the population size. PIG can be computed using most calculators, but

in this lab you will use Excel to do your calculations.

For example, if a duckweed population increased from 10 leaves to 30 leaves in a week,

the PIG = ln(30) – ln(10) = 3.4 – 2.3 = 1.1 individuals/individual/wk. If the

population is growing, the PIG will be positive; if it is declining the PIG will be

negative.

A convenient way to describe the population growth rate is to calculate the number of

new leaves produced using the equation

Fig. 1. Four plants of Lemna

minor, each with three leaves


66

PGR = Nwk2 – Nwk1 (2)

If the PIG remains approximately the same regardless of the initial number of leaves,

then growth is presumed to be density independent (with no limiting resources and no

competition among individual duckweed plants).

If, however, the PIG declines when initial densities are greater, this implies that growth

is density dependent, and indicates the presence of intraspecific competition for a

limiting resource.

In Experiment 1, you will manipulate the number of plants (2 vs. 10 vs. 25 plants) and

see whether changes in starting density affect the per-individual growth.

Discuss with your group an appropriate question and null hypothesis for this experiment:

State your Question:

What is you Null hypothesis regarding the effect of density on PIG?:

Now obtain 6 glass vials, and fill each with 15 mL of growth medium. Count the

number of leaves on 2 healthy duckweed plants, then add the plants to vial #1.

Record the number of leaves in column 3 of Table 1. Do the same for vial #2. Place 10

plants in each of vials #3 and #4 (again count the number of leaves for each before

adding them to the vials and record in Table 1). Place 25 plants in vials #5 and #6,

again after recording the total number of leaves in each.


67

Table 1. Growth rates of duckweed (your data).

Vial #

Initial no.

plants

Init. no.

leaves

(Nwk1)

ln(N1) Final no.

leaves

(Nwk2)

ln(N2) PIG

ln(N2)-ln(N1)

No. new

leaves

1 2

2 2

3 6

4 6

5 10

6 10

7** 2

8 2

9 2

10 2

**Vials 7 to 10 will have your experimental variable.

In Experiment 2, you can experimentally determine whether particular resources

control the growth of the duckweed. Vials #7-10 are reserved for your use, and

additional materials needed to test for possible growth limitation by light intensity

and/or quality, and by nutrient supply, are available in the lab as shown in Table 2.

Table 2. Supplies available for Experiment 2.

Potentially Limiting Resource Supplies Available

Amount of Light Window screening

Quality of light Colored cellophane

Nutrients Nutrient stock

Discuss with your group how you could test whether a particular resource limits

growth. The vials with 2 plants should be used as ―controls‖, with which your

―treatments‖ will be compared. For example, you could ask whether light limits growth

by comparing the PIG in your treatment vials with that of vials #1-2. Or you could see

whether adding extra nutrients stimulated growth compared to vials #1-2. Vials #7-10

should consist of two treatments, each with two replicates.

Suppose you want to know whether either of the two treatments causes the PIG to differ

from that of your controls. What is an appropriate null hypothesis appropriate to that

question?

Null Hypothesis:


68

Now state the experimental design you will use to address the question.

Experimental Design:

For Experiment 2, record the total number of leaves in each of vials #7-10 as before in Table 1.

Label the entire row of 10 vials with your initials and lab section, and place the vials under one

of the fluorescent lights on the side benches. Next week you will again count the total leaves

in each vial, entering them in Table 1 as before.

Week 2:

Your ability to create, and understand, graphic displays of data is a critical skill in biology.

The following three graphs describe relationships among population size, population growth

rate (PGR) and per-individual growth rate (PIG). The dashed line describes the expected

relationships for exponential growth. In this lab, you are testing to see whether duckweed

plants compete for growth-limiting resources, and display logistic growth to carrying capacity.

As preparation, draw a solid line corresponding to the prediction for logistic growth in Figure

2b-c below.

Pour out the contents of each vial into a Petri dish and again count the total number of leaves

present in each.

Fig. 2a-c. Expected relationships among population size, population growth rate

and per-individual growth rate for exponential growth (dashed lines), and logistic

growth (solid lines). Add the solid lines to graphs b and c.


69

Enter the final leaf counts from week 2 in column 5 of Table 1 (your initial number of leaves

from last week should already be entered in column 3. Don’t worry about the calculations for

the other columns yet. They’ll be computed using Excel. When done, please clean all vials,

shake off most of the liquid and place them upside down in the cardboard tray.

Once the lab bench is dry, get out your laptop and download the file ―Logistic Growth Model‖

to your desktop (making sure the ―desktop‖ is in fact your screen and not a subdirectory of the

hard drive). Now open the file.

The Excel workbook has two worksheets, one for calculations and one with a simulation of

expected duckweed growth. Examine the worksheet with ―calculations‖. You should see a

copy of Table 1.

Enter the initial and final numbers of leaves in the Excel table as you did in your lab manual.

Note that Excel automatically calculates the per-individual growth rate (PIG) and the increase

in number of leaves.

In the first experiment, you want to measure the degree to which the PIG declines as

population size (N) increases. To do this you need to know something about simple linear

regression.

Simple linear regression (called a ―trendline‖ in Excel) describes the degree to which the value

of one variable ―depends‖ on the value of another variable. For example, suppose you

suspected that how you perform on BIO 110 exams is at least partly related to the amount of

time spent studying. A mathematical model of this relationship might take the form:

EXAM SCORE = b(STUDY TIME) + a

In that simple model, EXAM SCORE and STUDY TIME are the variables, and are

related to each other algebraically by two constants, a and b. STUDY TIME is called

the ―independent variable‖ and can be used to predict the ―dependent variable‖

EXAM SCORE.

Stated differently, you want to see how much exam score depends on the amount of

time spent studying. A possible relationship between STUDY TIME and EXAM

SCORE is shown in Figure 3 below:

Notice that Exam Score was

positively related to study time.

The actual relationship is

shown by an algebraic equation

(model) of the form Y = b(X) +

a, in this case ES = 3.3(ST) +

66.6, where 66.6 (=a) is the

score predicted if you didn’t

study at all (0 hours) (DON’T

TRY THIS!) and 3.3 (=b) is the

number of additional points

above 66.6 that might be

expected for each hour spent

studying.


70

As stated above, both numbers (66.6 and 3.3) are considered constants, and EXAM

SCORE and STUDY TIME are dependent and independent variables, respectively. To

test your understanding of simple linear regression, answer the following questions:

If a student increased his/her study time by 3 hour, how many additional points might he/she

expect on the exam? ____________

If the student studied for 6 hours, what is the expected grade? _________________

In the regression equation you’ll create for Experiment 1 the per-individual growth rate

(PIG) is the dependent variable and the initial number of leaves (N) is the

independent variable. You wish to determine whether PIG declines with increasing

N.

In the Excel worksheet with Table 1, highlight cells D5:D10 (the initial number of

leaves in vials #1-6). Then, while holding down the Ctrl key, also highlight cells

H5:H10 (the estimates of PIG for those vials). Now pull down under Insert to Chart,

specify XY (Scatter), and choose the subtype without lines (upper left). Next. In

Chart Wizard Step 3 enter Axis titles for both the X axis (= N) and Y-axis (= PIG).

Leave the Chart Title blank. Next. In Chart Wizard Step 4, for chart location, specify

the current spreadsheet (Sheet 1). Single click on the resulting graph to highlight it,

then move it so you can see both it and the datasheet.

To customize your graph, make sure the whole chart area is highlighted, click once to

highlight the legend at the right, then delete it. Now double-click on the Y-axis. In the

Format Axis Window which should appear, specify Scale, then remove the check

from the checkbox for automatic Minimum and set the minimum for the Y-axis at 0 (at

a value of 0 there is no growth). OK. Then (with the chart still highlighted) pull down

under Chart to Add Trendline.

In the Trendline Window, specify Linear. OK. This adds a regression line to the

chart. If the regression line doesn’t extend all the way to the X axis, double-click on the

regression line to open the Format Trendline Window, and click on Options. Click

the checkbox for Display Equation on Chart. Then, under Forecast, extend the line

by enough units so that it intersects the X axis (how many units are required will

depend on your data; overshooting is OK).

Rescale the X-axis if necessary as you did the Y-axis so that the line is nicely centered

in the graph. Also estimate the number of leaves where the regression line meets the X

axis. This is the carrying capacity (=K) (the population size at which the per-

individual growth rate has declined to zero)!

Sketch the points and regression line in Figure 4 below. In the equation above the

figure, remember that in this instance PIG= ___N + ___ is the same as y = b(x) + a.

Insert the constants for the slope (b) and intercept (a) in the equation.

You can now calculate the carrying capacity based on this equation. What should be

the value for PIG at carrying capacity? _____ Enter this value into the equation, then

solve for N. Enter your estimate of K next to the equation at the top of Figure 4.

Regression: PIG = ____(N) + ____ K: ______


71

Fig. 4. Relationship between per-individual growth and population size in duckweed

(Experiment 1)

You now need to decide whether to accept or reject your null hypothesis. If the

regression line shows a negative slope (the value for b in the regression equation that

has been added to your chart is negative) this means that the per-individual growth rate

is reduced by intraspecific competition for some limiting resource (in effect, that the

PIG is density-dependent).

Decision:

Just what resource is limiting is investigated in Experiment 2. Begin by entering the

PIG estimates for vials #1-2 and the treatment vials as 3 separate columns in your

spreadsheet (see Fig. 5 below as an example).

The first column should have the heading ―control‖, with the estimates for PIG directly

beneath it. In the two adjacent columns insert titles for your two treatments, and place

the PIG values underneath them.

Then pull down under Insert to Chart, and specify Column. Select the subtype

showing a Clustered Bar (upper left option). Next. In Chart Wizard Step 2, click to

indicate the data are in rows, then click once in the data range window and highlight the

data for Experiment 2, including column titles. Next. In Chart Wizard Step 3 enter

appropriate names for the X and Y axes. Next.

Make sure the chart will appear on your spreadsheet, then click on Finish. You can

again delete the legend, and should again position the graph near the data used to create

it.


72

control 1 layer 2 layers

2.3 2.1 1.9

2.5 2 1.7


73 Sketch your chart in Figure 6 below. You will evaluate the null hypothesis of no treatment

effect in Experiment 2 qualitatively (without statistical testing) by comparing the heights of the

bars.

Do the bars for the treatment vials show consistent differences in height from those of the control

vials? What can you conclude from the results? Write your decision in the space below.

Decision:

Fig. 6. Effects of resource manipulation on per-individual growth rate (Experiment 2).

Modelling Population Growth using the Logistic Equation:

When a population is allowed to grow from a small initial number of individuals in an

environment in which resources are limited, it is expected that population size will

eventually reach carrying capacity (K) as shown in Figure 7.

In Experiment 1 you estimated the carrying capacity by extending the regression of PIG

down to the X axis (to the point where PIG = 0 and the population is no longer growing).

The carrying capacity can be thought of as the maximum number of leaves that the vials can

stably sustain. This is likely determined by the number of leaves that can coexist on the

surface of the water.

Treatment

PIG


74

Fig. 7a-b. (a) Logistic growth model of the number of duckweed leaves (N) expected over

time. (b) the effect of duckweed density (N) on the per-individual growth rate (PIG).

You will now create a logistic model of duckweed growth. Switch to the second worksheet

in your Excel file. A sheet with the title Logistic Growth Simulation should now appear,

with three graphs (of Population trends over time in green, trends in per-individual growth

(PIG) in blue, and population growth rate (PGR) in brown.

There is also an area in yellow at the right containing equations that produce the three

graphs. You will examine the impacts of changing the three values in the magenta box on

the three graphs of logistic growth.

Begin by setting the carrying capacity (K) in cell N5 to the value you obtained in

experiment 1. Similarly, enter the initial number of leaves (the average of the number of

leaves in vials #1-2 in week 1 (column 3 in Table 1) in cell N6. Finally insert the value 0.1

for the maximum PIG in N4.

Now examine the three graphs. First, the graph of N vs. t (in green) should show an

approximately S-shaped curve (or, if the growth rate is rapid, a curve whose slope simply

declines with time) beginning at the initial number of leaves and approaching carrying

capacity.

The second plot describes the predicted effect of population size on the per-individual

growth rate (PIG). The third plot describes the population growth rate (estimated as the

number of new leaves produced) in relation to population size. Note that the population

growth rate is expected to be greatest at a value of roughly ½ of the carrying capacity.

Now replace the estimate of MAXPIG in cell N4 of your spreadsheet with the following 5

values: 01, 0.2, 0.4, and 0.6. For each value of MAXPIG, examine the third (brown) graph

to estimate the maximum population growth rate (PGR).

Enter these data in Table 3. Enter your estimates in Table 3.

Table 3. Effect of changing MAXPIG on PGR

MAXPIG Greatest PGR

01

0.2

0.4

0.6

N

Time

K

PIG

N K

MAXPIG

1


75

How does MAXPIG affect the population growth rate?

Similarly, examine the effect of changing the carrying capacity on the maximum population growth

rate. Enter the value 0.4 for MAXPIG in cell N4, and enter the following 5 values into cell N5: 80,

100, 120, 140, 160.

Then estimate the maximum population growth rate in the brown graph as before, entering the

estimates in Table 4.

Table 4. Effects of carrying capacity on maximum population growth rate.

K Maximum PGR

80

100

120

140

160

What is the effect of increasing the carrying capacity on the maximum population growth rate?

Finally, do you think the treatment manipulations you performed in #7-10 affected the MAXPIG or the

carrying capacity (or both)? What evidence can you give to support your answer?


76

Daphnia Morphology and Feeding Ecology

In this portion of the lab you will examine feeding by Daphnia, a common member of the

zooplankton in lakes and ponds in this region. It is a ―filter feeding‖ herbivore that strains

phytoplankton algae from the water around it.

Your observations should address the following question: Is Daphnia “picky” about what it eats?

If so, what kind(s) of algae does it prefer? You’ll examine two kinds of potential algal food first,

then observe the response of Daphnia to the food.

Types of Algae

There are many kinds of phytoplankton algae, differing in their size, shape and nutritional content.

Work with a partner. Begin by looking at two kinds of algae under the microscope.

The first is Chlamydomonas, a small (about 10-20 m) eukaryotic, single-celled, and uses flagella

to swim through the water. Make a wet mount (with a drop of culture medium and a coverslip), and

view it under high power with a compound microscope.

Sketch a Chlamydomonas in the space below. The second is Anabaena. Recall from the lab on The

Microscope, Measurement and Cell that Anabaena is prokaryotic member of the cyanobacteria,

with many very small cells that form long filaments.

Again, make a wet mount, view it under high power and sketch it in the space below.

Based on what you’ve seen so far, which kind of food do you predict will be preferred by Daphnia? Why?

Daphnia Chemotaxis

Daphnia are microcrustaceans, very small but related to shrimp and lobsters.

Like milkweed bugs, they pass through several immature stages called instars, molting in order to

grow. In the culture, you’ll see different size individuals of various instars.

Typically they are all females, and produce eggs that are genetic clones of themselves.

Chlamydomonas Anabaena


77 ―Chemotaxis‖ is movement toward (or away from) a chemical cue. In this part of the lab you

want to determine whether Daphnia can chemically detect patches of preferred food.

Work as a group of 4. Set up your experiment as follows:

1) Place a small amount of water (approximately 1-2 pipetfuls) in the center of a Petri dish.

2) Place suspensions of Chlamydomonas and Anabaena in separate capillary tubes. Seal one end of

each tube.

3) Place two pieces of florist’s putty on the rim on opposites of the Petri dish. Place the open end of

the capillary tubes at the edge of the puddle of the water in the center of the Petri dish. Mark one end of the

Petri dish with a C (for Chlamydomonas) and the other with an A (for Anabaena). Your setup should

resemble the diagram below).

4) Add 4-5 Daphnia to the puddle of water, turn on a light and view under the dissecting scope.

5) Observe the swimming behavior of the Daphnia. Is their swimming random or do they stay near

one of the two capillary tubes?

Daphnia Morphology

Now it’s time to get a close-up view of the morphology and feeding behavior of Daphnia.

Again, work in groups of four. First fill a ―well slide‖ with aged tapwater, and place a small

piece of florest’s putty on one side.

Second, obtain a short (< 1 cm) bristle, to be used as a ―tether‖, and dip on end of it in small

amount of petroleum jelly (too much and your Daphnia will be too gooey to perform for

you!). Keep your tether ready next to the well slide. Third, you now need to snag a

Daphnia.

Use a wide-bore plastic pipet (with the tip snipped off to increase the size of the aperture),

then chase down a healthy looking specimen from the Daphnia culture. Once the Daphnia

is in you pipet, it will typically swim against the current. Carefully evacuate most of the

water until just a few drops (including your Daphnia) remain.

Birds-eye view of Petri dish with capillary tubes containing

Chlamydomonas (C) and Anabaena (A). Daphnia not shown.


78 Then empty the Daphnia temporarily onto a clean microscope slide (no coverslip!), soak

up excess water with a piece of paper towel so the Daphnia can’t swim around, and

immediately place the slide on the stage of a dissecting microscope.

While observing your Daphnia through the microscope and using a forceps, touch the bristle

tip with the petroleum jelly to the back or side of the carapace, being careful to avoid the

swimming antennae and filtering appendages (see figure below).

\ Now insert the other end of the bristle into the putty of the well slide. Adjust it if necessary

so the Daphnia is once more under water and approximately in the center of the well.

Observe the well slide first under the dissecting microscope to make sure your Daphnia is

clearly visible at the end of the bristle, then transfer the well slide to a compound

microscope and observe the Daphnia under your scanning objective.

Begin by identifying the following structures.

1) The ―head‖ region has a single compound eye (note the many ―lenses‖ that together allow the

Daphnia to distinguish the general direction of light, but are not adequate to form clear images. Can the

eye rotate? (Look for any apparent movement of the lenses).

2) Below the head are the two sides of the carapace, which together are a bit like a trenchcoat that

opens toward the front. The carapace encloses the 5 pairs of filtering appendages used to concentrate

algae from the water. Do the appendages beat in sequence or in synchrony?

3) The filtered algal food is moved forward between the paired filtering appendages and becomes

concentrated next to the mandibles. Look for a pair of moving structures just under the head. Note how

they push food into the gut. Does the gut have food in it? (You’ll have a chance to compare feeding on

various types of food in a bit).

4) Food exits the gut at an anus located at the back of a postabdomen. The front part of the

postabdomen has a pair of claws that help keep the feeding area between the legs clean of unwanted algae.

5) At the back of the carapace note the heart, beating rapidly! Daphnia is an ectotherm, and its

heartbeat increases with the temperature of its environment.

Tethered Daphnia in the well of a well slide.


79 6) Note the swimming antennae. Your Daphnia may try to use them at times, unaccustomed to

such a confined space. Are the antennae branched? Can you see setae on the branches, serving to increase

propulsive power? Daphnia swim in a ―hopping‖ motion, alternating movement with feeding.

7) Note the brood chamber just below the heart at the back of the carapace. If your specimen is large

enough and is healthy, you may be able to see some eggs in the brood chamber. Eggs are deposited all at

once in the brood chamber about a half hour after molting, and will be released by the mother (by pulling

forward her postabdomen) shortly before her next molt.

Anatomy of a female Daphnia→→→→

Daphnia Feeding Behavior The group at one end of the table should now add a

single, very small drop of one of Chlamydomonas to the

well slide. The group at the other end of the table should

add a drop of Anabaena. You will barely see the algal

particles, but can observe several aspects of feeding by the

Daphnia. Answer the following questions in the table

below.

1) Do the filtering legs continue to beat regularly and

successfully concentrate the food?

2) Is there evidence of the food in the gut (does it get

increasingly green)?

3) Does the Daphnia appear to be increasingly

rejecting the food with frequent upward movements of the

postabdomen to clean out the area between the filtering

appendages? Make sure you compare notes with the other

group to get the responses for both kinds of algae. When done, return your Daphnia to the culture

chamber.

Feeding behavior of Daphnia on two types of phytoplankton.

Chlamydomonas Anabaena

Filtering?

Food in gut?

Rejection movts.

by the

postabdomen?

Based on your observations, which of the phytoplankton types is likely to be the most suitable food for

Daphnia?


80

Do your observations of food preference based on feeding behavior agree with your observations of

chemotaxis? Which type of algae is preferred? Does this agree with your original prediction at the start of

the experiment?


81 Lab exercise 11: Molecular Genetics Molecular genetics part 1: DNA extraction from wheat germ

Introduction:

DNA (deoxyribonucleic acid) is the genetic material of all living organisms. A common

technique used by molecular biologists is to isolate DNA from cells.

The DNA can be used for finding and cloning specific genes, for constructing chromosome

maps, or for sequencing, to name a few applications.

There are a few basic steps in DNA extraction. The cell must be lysed (broken open) to

release the nucleus. The nucleus (if present) must also be opened to release the DNA. At this

point the DNA must be protected from enzymes (DNAses) that will degrade it. Once the

DNA is released, it must then be separated from other cellular components.

The rationale for each step in the procedure is listed below:

In order for the cell to be lysed, the cell wall (if present) and lipid membranes must

be broken down. Mechanical disruption in a blender and the detergent solution are used for this

purpose. Detergent also functions to denature and unfold proteins, making them more susceptible to

protease cleavage

The meat tenderizer has proteases, enzymes that help digest the contaminating

proteins. Protease is added to destroy nuclear proteins that bind DNA and cytoplasmic enzymes

that breakdown and destroy DNA.

Finally, cold alcohol in a high salt environment is used to precipitate the DNA.

Na+ ions of NaCl bind to the phosphate groups of DNA molecules, neutralizing the electric

charge of the DNA molecules. The addition of NaCl allows the DNA molecules to come together

instead of repelling each other, thus making it easier for DNA to precipitate out of solution when

alcohol is added. In water, DNA is soluble. When it is in alcohol, it uncoils and comes out of

solution. Precipitated DNA molecules appear as long pieces of fluffy, stringy, web-like strands.

DNA Extraction from Wheat Germ

Protocol (for each table to step 7)

For the first 7 steps, it is important to keep the temperature at 60 degrees

1. Add 25 ml distilled water to a beaker and place in water bath at 60o C.

2. Add 0.4 g wheat germ and mix.

3. Add 1.25 ml detergent. Mix and maintain at 50-60o C.

4. Add 0.8 g meat tenderizer

5. Add many drops (approximately 30) of sodium bicarbonate (baking soda) so that the final pH is 8.0.

Use the pH paper to check the pH.

6. Maintain 50-60o C and stir for 10 minutes.

7. Remove from the water bath.

Each person at the table can perform the next steps.


82 8. Add 3 ml of the solution to a test tube and cool to room temperature.

9. ***This is the most important step! Pour 3 ml of ice cold ethanol carefully down the side of the

tube to form a layer

10. Let the mixture sit undisturbed 2-3 minutes until bubbling stops.

11. The DNA will precipitate at the alcohol interface. Shake the tube gently to increase the

precipitation of DNA. Swirl a glass stirring rod at the interface of the two layers to see the precipitated

threads of DNA.

Molecular genetics part 2: DNA Fingerprinting

Technicians working in forensic labs are often asked to do DNA profiling or ―fingerprinting‖ to

analyze evidence in law enforcement cases and other applications. DNA fingerprinting may

involve polymerase chain reaction (PCR) amplification to analyze minute quantities of DNA or

restriction fragment length polymorphism (RFLP) analysis, if large amounts of DNA are recovered.

A step in human RFLP analysis requires a comparison of band patterns produced by cleavage of

DNA samples when separated on an agarose gel.

Two major factors affecting the reliability of DNA fingerprinting technology in forensics are

population genetics and genetic statistics. In humans there are thousands of RFLP loci or DNA

segments that can be selected and used for fingerprinting analysis.

Depending on demographic factors such as ethnicity or geographic isolation, some segments will

show more variation than others. If 90% of a given population has the same frequency in its DNA

fingerprinting pattern for a certain DNA segment, then very little information will be attained.

But if the frequency of a DNA pattern turning up in a population for a particular segment is

extremely low, then this segment can serve as a powerful tool to discriminate between individuals

in that population.

The patterns in this exercise are produced from one sample that represents DNA taken at the crime

scene and five samples obtained from suspects in the case. This laboratory exercise models the

more elaborate technique that is performed on complex human DNA samples and uses 2 important

research tools 1) restriction enzymes and 2) agarose gel electrophoresis.

Restriction Enzymes

A restriction enzyme acts like molecular scissors, making cuts at specific sequences of DNA base

pairs that it recognizes. A restriction enzyme sits on a DNA molecule and slides along the helix

until it recognizes specific sequences of base pairs that signal the enzyme to stop sliding.

The enzyme then cuts or chemically separates the DNA molecule at that site—called a restriction

site.

If a specific restriction site occurs in more than one location on a DNA molecule, a restriction

enzyme will make a cut at each of those sites, resulting in multiple fragments.

Therefore, if a given linear piece of DNA is cut with a restriction enzyme whose specific

recognition code is found at two different locations on the DNA molecule, the result will be three

fragments of different lengths.

If the given piece of DNA is circular and is cut with a restriction enzyme whose specific recognition

code is found at two different locations on the DNA molecule, the result will be two fragments of

different lengths.

The length of each fragment will depend upon the location of restriction sites on the DNA molecule.


83 Agarose Gel Electrophoresis

DNA that has been cut with restriction enzymes can be separated and observed using a process

known as agarose gel electrophoresis. The term electrophoresis means to carry with electricity.

Agarose gel electrophoresis separates DNA fragments by size.

DNA fragments are loaded into an agarose gel slab, which is placed into a chamber filled with a

conductive buffer solution. A direct current is passed between wire electrodes at each end of the

chamber. Since DNA fragments are negatively charged, they will be drawn toward the positive pole

(red electrode = ―run to red‖) when placed in an electric field.

The matrix of the agarose gel acts as a molecular sieve through which smaller DNA fragments can

move more easily than larger ones. Therefore, the rate at which a DNA fragment migrates through

the gel is inversely proportional to its size in base pairs.

Over a period of time, smaller DNA fragments will travel farther than larger ones. Fragments of the

same size stay together and migrate in single bands of DNA. These bands will be seen in the gel

after the DNA is stained.

Using restriction enzymes

Because they cut DNA, restriction enzymes are the "chemical scissors" of the molecular biologist.

When a particular restriction enzyme "recognizes" a particular four - or six –base pair (bp)

recognition sequence on a segment of DNA, it cuts the DNA molecule at that point. The

recognition sequences for two commonly-used enzymes, EcoRI and PstI, are shown below. The

place on the DNA backbones where the DNA is actually cut is shown with a (✄) symbol:

The line through the base pairs

represents the sites where bonds will

break if a restriction

endonuclease recognizes the site

GAATTC.

Using agarose gel electrophoresis

The electrophoresis apparatus creates an electrical field with positive and negative poles at the ends

of the gel. DNA molecules are negatively charged and will migrate to the positive pole.

Smaller fragments move more easily through the gel when the electric field is applied.

DNA is colorless so DNA fragments in the gel cannot be seen during electrophoresis. A loading dye

containing two blue dyes is added to the DNA solution.

The loading dye does not stain the DNA itself but makes it easier to load the gels and monitor the

progress of the DNA electrophoresis. The dye fronts migrate toward the positive end of the gel, just

like the DNA fragments.

The ―faster‖ dye comigrates with DNA fragments of approximately 500 bp, while the ―slower‖ dye

comigrates with DNA fragments approximately 5 kb in size. Staining the DNA pinpoints its

location on the gel. When the gel is immersed in Fast Blast DNA stain, the stain molecules attach to

the DNA trapped in the agarose gel. When the bands are visible, you can compare the DNA

restriction patterns of the different samples of DNA.


84

For fingerprinting analysis, the following information is important to remember:

• Each lane has a different sample of DNA

• Each DNA sample was treated with the same

restriction endonucleases.

With reference to the numbered lanes, analyze the

bands in the gel drawing at right, and then answer the

questions.

1. What would be a logical explanation as to why

there is more than one band of DNA for each of the

samples?

2. Which of the DNA samples have the same number of restriction sites for the restriction endonucleases

used? Write the lane numbers.

3. Which sample has the smallest DNA fragment?


85

4. Assuming a circular piece of DNA (plasmid) was used as starting material, how many restriction sites

were there in lane three?

5. Which DNA samples appear to have been ―cut‖ into the same number and size of fragments?

6. Based on your analysis of the gel, what is your conclusion about the DNA samples in the drawing?


86

Obtain a piece of parafilm and ―spot‖ the various sample/enzyme mixtures according to the figure below:


87

Using an automatic pipetter, load 3 µL of distilled water in each well

of the flash gel (FG) system (at right). Discard the tip.

*Insert cassette into dock

*Plug in electrophoresis power supply

*Pre-run for 15 seconds (200 V)

*Turn off voltage and disconnect cables

*Using a new tip, mix and collect the loading dye with the

Enzyme/sample from Suspect #1 and transfer it to well #1 of the FG

system. Discard the tip. Continue this procedure until you put all the

samples, including the crime scene system and the molecular weight

standards. USE A FRESH TIP FOR EACH SAMPLE!!!

*Plug in and turn on light and voltage

*Watch until desired separation is achieved. Record your observations.


88 Analysis of results

1) Draw the pattern resulting from the gel after staining in the box below:

2) Compare the fragment sizes of the suspects and the crime scene.

Is there a suspect that matches the crime scene?

How sure are you that this is a match?


89

Molecular Genetics part 3: Bioinformatics using Internet Resources

Introduction to Bioinformatics:

In the last couple of decades, there has been a technical revolution in molecular biology, which

has made it possible to get enormous amounts of information about the sequences of amino

acids in proteins and the sequences of bases in nucleic acids.

To construct meaning from the sequence information, the array of computer techniques called

"bioinformatics" has been developed. The components of bioinformatics are sequence

databases and computer programs that analyze the sequences for patterns and similarities.

To understand the basic idea of bioinformatics, one might think of a written language. The text

you are reading consists of a series of letters, words, sentences, and paragraphs. If you did not

know the meanings of the words and the rules of the language, this page would just be a

collection of meaningless symbols. Similarly, the first time scientists saw gene and protein

sequences, they saw a string of symbols with no clear meaning in terms of biological function.

But now, bioinformatics is showing us many things about what sequences mean.

Using bioinformatics, sequences are being used to reveal relationships among different life

forms that we could not find out any other way. Bioinformatics is revealing the rules and

meaning of a language that is new to human beings but in fact is more than a billion years old-

the Language of Life.

The genetic code

The genetic code consists of 64 triplets of nucleotides. These triplets are called codons.With three

exceptions, each codon encodes for one of the 20 amino acids used in the synthesis of proteins. That

produces some redundancy in the code: most of the amino acids being encoded by more than one codon.

One codon, AUG serves two related functions:

it signals the start of translation

it codes for the incorporation of the amino acid methionine (Met) into the growing polypeptide

chain

The genetic code can be expressed as either

RNA codons or DNA codons. RNA codons

occur in messenger RNA (mRNA) and are the

codons that are actually "read" during the

synthesis of polypeptides (the process called

translation). But each mRNA molecule

acquires its sequence of nucleotides by

transcription from the corresponding gene.

Note that in the table at right, the left-hand

column gives the first nucleotide of the codon,

the 4 middle columns give the second

http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/N/Nucleotides.html

http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/A/AminoAcids.html

http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/T/Translation.html

http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/C/Cys_met.gif

http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/T/Transcription.html#mRNA

http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/T/Transcription.html


90 nucleotide, and the last column gives the third nucleotide.

For example, below is a piece of a sequence for the mRNA strand that codes for a piece of an imaginary

short peptide:

AUG AAA AAG UCA AGA UUU GGA CCC

Using the table in the previous page, determine the amino acid sequence of this short peptide:

______________________________________________________

Show your amino acid sequence to your instructor before continuing to the next page.

Bioinformatics can also be used to compare protein sequences from different organisms.

Part 1: Manual Comparison of a few Amino Acid in hemoglobin- refer to Table 1

Table 1: Selected amino acid positions (19 total) in the Hemoglobin of some vertebrates.

1 2 3 4 5 6 7 8 9 10

Primate Human SER THR ALA GLY ASP GLU VAL GLU ASP THR

Primate Chimpanzee SER THR ALA GLY ASP GLU VAL GLU ASP THR

Primate Gorilla SER THR ALA GLY ASP GLU VAL GLU ASP THR

Primate Baboon ASN THR THR GLY ASP GLU VAL ASP ASP SER

Primate Lemur ALA THR SER GLY GLU LYS VAL GLU ASP SER

NonPrimate Dog SER SER GLY GLY ASP GLU ILU ASP ASP THR

NonPrimate Chicken GLN THR GLY GLY ALA GLU ILU ALA ASN SER

NonPrimate Frog ASP SER GLY GLY LYS HIS VAL THR ASN SER

11 12 13 14 15 16 17 18 19

Primate Human PRO GLY GLY ALA ASN ALA THR ARG HIS

Primate Chimpanzee PRO GLY GLY ALA ASN ALA THR ARG HIS

Primate Gorilla PRO GLY GLY ALA ASN ALA THR LYS HIS

Primate Baboon PRO GLY GLY ASN ASN ALA GLN LYS HIS

Primate Lemur PRO GLY SER HIS ASN ALA GLN LYS HIS

NonPrimate Dog PRO SER ASN LYS ASN ALA ALA LYS LYS


91 NonPrimate Chicken PRO THR THR LYS ASN SER GLN ARG ALA

NonPrimate Frog ALA HIS ALA LYS ASN ALA LYS ARG ARG

Fill in Table 2 below based on the data provided in Table 1.

Table 2: Similarities and differences in the amino acid sequences of hemoglobin

ORGANISM NUMBER OF SIMILAR

AMINO ACID POSITIONS

NUMBER OF DIFFERENT

AMINO ACID POSITIONS

Human vs. Chimpanzee

Gorilla

Baboon

Lemur

Dog

Chicken

Frog

Based on your results on Table 2 above , prepare a bar graph (below) showing the number of similar amino

acid positions (compared with humans) for each type of organism.

Similarities in the amino acid sequences ofhemoglobin in selected vertebrates

Hum

an

Chim

panze

e

Goril

la

Bab

oon

Lemur

Dog

Chic

ken

Frog

0

2

4

6

8

10

12

14

16

18

20

Nu

mb

er

of

Ide

nti

ca

lp

os

itio

ns


92 Using hemoglobin as a model protein, which primate is most closely related to humans?

_______________________

Which non-primate is most closely related to humans? _____________________

Imagine having to compare hundreds of nucleotides or amino acids in this way (DOH!!!!). There are

several analysis tools available on the Internet that will help you compare amino acid (or nucleotide)

sequences very quickly.

Get a laptop computer.

Using internet resources, we will compare mitochondrial DNA sequences. Because mitochondrial DNA

mutates more rapidly than nuclear DNA, scientists use it to study the evolution of humans.

Mitochondrial DNA Bioinformatics Inquiry Lab

Go to the Dolan DNA Learning Center: http://www.bioservers.org/bioserver/

After you enter the Bioservers site, enter the Sequence server.

Click on Bioservers

http://www.bioservers.org/bioserver/


93

Click on Manage groups:

From the pull down menu on the top right called “Sequence sources” and briefly look at the

wide range of sequence choices. Choose “Modern Human mt DNA. Click in all of the boxes

to the left of the six choices so that you can see all of different human samples.

Click here


94

The goal of this activity is for you to develop a hypothesis about different types of peoples

or organisms and compare them using computer Internet resources. Here is an example:

This is my hypothesis: The mitochondrial DNA from a Native American is more closely related to

the mitochondrial DNA from an individual from Asia than it is to an African American.

Here is how I will test my hypothesis:

1) I will compare the sequence of a Native American mtDNA to an Asian mtDNA and count the

number of differences in the nucleotide sequence of the mtDNA. I choose ―Bali #1‖ as my Asian

sample and ―Blackfoot #1‖ as my Native American sample.

2) I will compare the sequence of a Native American mtDNA to an African American mtDNA and

count the number of differences in the nucleotide sequence of the mtDNA. I choose ―African

American #1‖ as my sample and ―Blackfoot #1‖ as my Native American sample.

3) Click on ―Compare‖ at the top left.

Pull down menu

Click on all

six boxes


95

I choose to compare

these two sequences

Click on ―Compare‖ after the

sequences are selected.


96

The initial portions of the sequence are shown on the next page. As you scroll down

through the sequence, you will see that there are yellow highlighted areas that correspond to the

nucleotide sequences that are different. I am going to choose to ignore the beginning part of the

sequence where there is no overlap (the first 53 nucleotides) and the end of the sequence (after

nucleotide 444). **The computer program will do this for you if you choose to “trim” the

sequence.

Here are my results:

Number of nucleotide differences between Asian Bali #1 and Native American Blackfoot #1 = 7

I repeat the same process to determine the differences for an African American and a Native American.

Number of nucleotide differences between African American #1 and Native American Blackfoot #1 = 9

Based on these results, I can conclude that my hypothesis is correct (or is it? Do I need to collect more

data?)


97

bio 110 lab manual spring 2011

Documents

fine adjustment

free amino

selectively

paste function

lugols reagent

standard deviation

red blood

fine adjustment