Download Laboratory Manual for General Biology I (BSC 1010C)

Laboratory Manual for General Biology I
(BSC 1010C)
Lake-Sumter Community College
Science Department
Leesburg
Table of Contents
Note to Students ........................................................................................................................................... 3
Exercise 1 - Measurements and Lab Techniques .......................................................................................... 4
Exercise 2 - Functional Groups, Organic Molecules, Buffers, and Dilutions ............................................... 13
Exercise 3 - Qualitative Analysis of Biological Molecules ........................................................................... 23
Exercise 4 - The Microscope ....................................................................................................................... 31
Exercise 5 - Cell Structure and Membrane Function .................................................................................. 46
Exercise 6 - Enzyme Activity ........................................................................................................................ 56
Exercise 7 - Respiration ............................................................................................................................... 63
Exercise 8 - Photosynthesis ......................................................................................................................... 67
Exercise 9 - Cell Division .............................................................................................................................. 73
Exercise 10 - DNA Fingerprinting ................................................................................................................ 81
Exercise 11 - Genetics ................................................................................................................................. 96
A significant portion of this lab manual is used with the kind permission of the Science Department at
Seminole State College, Sanford, Florida.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
2
Note to Students
Students should read and study the exercises before coming to the laboratory and should supply
themselves with the necessary materials including the text book, lecture notes, laboratory manual,
calculators, pens, and pencils. All students are required to wear appropriate clothing to lab as outlined
by the lab instructor as well as follow all safety precautions during laboratory exercises.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
3
Exercise 1 - Measurements and Lab Techniques
Introduction
In scientific experiments, observation and accurate measurements are essential. The investigations in
this exercise will familiarize you with some of the methodologies and equipment in use in biology
laboratories.
Your objective is to learn to correctly select and use equipment to obtain accurate results, while
avoiding damage to the equipment or yourself.
Materials
Equipment
meter sticks
metric rulers
blocks of various sizes
irregularly shaped objects (fossils, rocks, bones, etc.)
500 ml graduated cylinders
triple beam balances
Part A: The Metric System
Scientific measurements are expressed in the units of the metric system or its modern day successor,
the International System of Units (SI). We will use this system exclusively throughout this course.
The metric system was invented by the French vicar Gabriel Moutin in 1670 and officially adopted as the
standard for weights and measures in France in 1795. Since then it has spread throughout much of the
rest of the world. Although the United States traditionally uses the English system, its use has become
more common in recent years. You may have even noticed canned goods and drinks in grocery stores
are given in metric as well as English units.
Just like in the English system, the metric system has three categories of units. For distance, it is meter,
for volume, liter, and for mass, gram. The metric system makes use of prefixes to change the value of
the unit in multiples of 10 (Table 1.1)
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
4
Exercise 1 – Measurements and Lab Techniques
Table 1.1. Metric System Units
Exponential multiplier
Length
Volume
Mass
103
kilometer (km)
kiloliter (kl)
kilogram (kg)
102
hectometer (hm)
hectoliter (hl)
hectogram (hg)
101
decameter (dam)
decaliter (dal)
decagram (dag)
100 = 1
meter (m)
liter (l)
gram (g)
-1
10
decimeter (dm)
deciliter (dl)
decigram (dg)
10-2
centimeter (cm)
centiliter (cl)
centigram (cg)
-3
10
millimeter (mm)
milliliter (ml)
milligram (mg)
10-4
These units have no prefixes
10-5
10-6
micron (µ)
microliter (µl)
microgram (µg)
-7
10
These units have no prefixes
10-8
10-9
nanometer (nm)
nanoliter (nl)
nanogram (ng)
Use this mnemonic device to remember the order of the prefixes:
kids have dropped over dead converting many blank blank metric blank blank numbers
Conversion between related units is accomplished by moving the decimal point the appropriate number
of places left or right (Fig. 1.1).
Fig. 1.1 Metric Unit Conversion Staircase
kilo (k)
hecto (h)
deca (dam)
m, l, g
deci (d)
centi (c)
milli (m)
micron (µ)
nano (n)
Move “up” the staircase to larger units, “down” to smaller ones. As example, to convert 37.35
decimeters (dm) to millimeters (mm), move the decimal point 2 places to the right (3735).
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
5
Exercise 1 – Measurements and Lab Techniques
Fill in the basic metric unit for each measurement in Table 1.2
Table 1.2 Basic Metric Units
Measurement
Length
Volume
Mass
Basic Metric Unit
Carry out the metric conversions in Table 1.3.
Table 1.3 Practice Metric Conversions
550 ml
3.7 g
20 km
78.4 cm
212 µl
67.5 dam
500 µm
__________ l
__________ mg
__________ m
__________ mm
__________ ml
__________ µm
__________ mm
Part B: Length Measurements
Length measurements are made with a metric ruler. When using a linear device, you should extend
your answer at least to the finest divisions on the device. For example, if you have a meter stick with
markings to the millimeter, you could measure your height to the nearest millimeter (e.g., 1754
millimeters or 1.754 meters). The size of objects falling between marked divisions may be interpolated.
Interpolation is an estimation how the distance an object extends between the smallest marks on the
device.
Part B1: Metric Height
Procedure
1. Obtain a meter stick
2. Find a partner and stand them with their back against a wall or door frame
3. Make a small mark at the level of the top of their head
4. Measure this height in centimeters making the most accurate measurement you can with the
meter stick
5. Repeat the procedure with yourself and record your height here __________ cm
Part B2: Calculating Surface Area to Volume Ratios (SA : Vol)
w
h
l
Procedure
1. Use the dimensions given in table 1.4 for various block sizes, calculate total surface area and
volume and enter in Tables 1.5 and 1.6
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
6
Exercise 1 – Measurements and Lab Techniques
Table 1.4 Block Dimensions
l (cm)
w (cm)
h (cm)
Small
Medium
Large
Calculating Surface Area:
Surface area of a rectangular block = 2 (l x w) + 2 (l x h) + 2 (w x h).
Use the data in Table 1.4 to fill in Table 1.5.
Table 1.5 Surface Area Calculations
Small
Medium
calculations
calculations
SA __________ cm2
SA = __________ cm2
Large
calculations
SA = __________ cm2
Calculating Volume:
Volume of a rectangular block = l x w x h
Fill in Table 1.6
Table 1.6 Volume Calculations
Small
calculations
Vol __________ cm3
Medium
calculations
Vol __________ cm3
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
Large
calculations
Vol __________ cm3
7
Exercise 1 – Measurements and Lab Techniques
Calculating Surface Area : Volume (SA : Vol)
Divide the surface area (cm2) by the volume (cm3) recording your answer in Table 1.7
Table 1.7 Surface Area, Volume, and SA: Vol
Surface area (cm2)
Small
Medium
Large
Volume (cm3)
SA : Vol
Use the data from Table 1.7 to construct a bar plot in Fig. 1.2.
Fig. 1.2 Relationship Between SA : Vol and Block Size
The plot just constructed provides a visual illustration of the changes in SA : Vol with blocks of different
volumes.
Describe the kind of relationship you see:
This SA : Vol ratio is very important in biology and helps to explain why cells have typically not grown
larger than microscope size. The SA : Vol affects the movement of materials in and out of cells. Very
small cells have high ratios and can usually supply most all the cell’s transportation requirements
through diffusion. But, as you noticed in this procedure an object’s ratio decreases relatively quickly as
it grows in size. This larger size means less surface area is available per unit of volume. The result is as
cells grow larger, diffusion is not longer sufficient to meet all the cells needs. Cells must either divide to
maintain that larger ratio or develop elaborate internal transport mechanisms. These topics will be
discussed further in later sections of this course.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
8
Exercise 1 – Measurements and Lab Techniques
Part C: Measuring Volume of Irregular Shaped Solids
Calculation of the volume of regularly shaped objects like rectangular blocks or spheres is
straightforward. However, how can we obtain the volume of something like a piece of bone, or rock, or
a fossil? Their irregular shapes preclude the use of any formula. However, two important facts are
useful to remember
o A submerged object will displace an amount of water equal to its volume
o 1 ml = 1 cm3
Procedure
1. Obtain a 500 ml graduated cylinder
2. Fill cylinder to about the midway mark with tap water
3. Note the level of water in the cylinder in ml
Reading a graduated cylinder
Graduated cylinders are marked off in volume units
Larger units are indicated (e.g., 10 ml, 20 ml, 50 ml, etc.)
Smaller units are not marked but are indicated
You must pay attention to these smaller, unmarked units to get an accurate
reading for volume
Due to capillary attraction, a liquid in a graduated cylinder will not form a flat surface.
Instead, it curves up the sides forming a dip or meniscus. By convention, we always
read the volume of the liquid from the bottom of the meniscus (Fig. 1.3)
Fig. 1.3 Graduated cylinder readings (record you readings in the blanks)
___ ml
___ ml
___ ml
4. Being careful not to splash out any of the water in the cylinder, submerge the irregularly shaped
object. Make sure it is completely underwater. Objects that float should be held underwater
5. Make note of the level of water in the graduated cylinder again
6. Subtract the initial volume of water from this final reading (express your answer in cm3)
7. Record your data in Table 1.8
Table 1.8 Water Displacement Data
Irregularly shaped object
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
Volume (cm3)
9
Exercise 1 – Measurements and Lab Techniques
Part D: Measuring Mass and Density
Procedure
1. Use a triple-beam balance to determine the mass (in grams) of the objects listed in Table 1.9
2. Calculate the volume of these objects using the methods described previously
3. Calculate density of each object
Density = mass (g) / volume (ml or cm3)
4. Record your answers in Table 1.9
Table 1.9 Mass, Volume, and Density of Various Objects
Mass (g)
Volume
(cm3 or ml)
Density
(g / cm3 or ml)
irregularly shaped object _______________________
small block
medium block
The density of water is 1 g /ml or cm3.
In comparing the densities of the objects in Table 1.9 to the density of water,
Which objects float?
The densities of these objects are __________ than that of water.
Which objects sink?
The densities of these objects are __________ than that of water.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
10
Exercise 1 – Measurements and Lab Techniques
Practice Problems
1. Calculate the surface area and volume of a rectangular solid measuring 8.6 cm in length, 2.4 cm
in width, and 3.8 cm in height (use appropriate units). The mass of this block is 121.6 g. What is
its density and will it sink or float in water?
2. Calculate the surface area and volume of a rectangular solid measuring 43 mm in length, 12 mm
in width, and 19 mm in height (report your answer in cm2 and cm3). The mass of this block is
8.5 g. What is its density and will it sink or float in water?
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
11
Exercise 1 – Measurements and Lab Techniques
3. Initial volume of water in a graduated cylinder is 0.26 l. Completely submersing an irregularly
shaped object into the water raises the water level to 512 ml. What is the volume of the object
(express your answer in cm3)? The mass of this object is 60 g. What is its density and will it sink
or float in water?
4. A principle of ecology known as Bergmann’s rule states an organism of a given species will be
larger in colder latitudes than those in warmer ones. For example, grey squirrels (Sciurus
carolinensis) in Florida are significantly smaller than their counterparts in New York. Using what
you have learned about changes in surface area with volume and its implications for membrane
transfer, provide a scientifically reasonable explanation for this observation.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
12
Exercise 2 - Functional Groups, Organic Molecules, Buffers, and Dilutions
Introduction
An overwhelming majority of the elements listed on the periodic table are naturally occurring. A much
smaller proportion of those are found in living systems in anything other than trace amounts. Six of
those elements are most abundant (CHNOPS):
Carbon (C)
Oxygen (O)
Hydrogen (H)
Phosphorus (P)
Nitrogen (N)
Sulfur (S)
Other elements of biological significance include sodium, potassium, calcium, magnesium, iron, and
chlorine. Atoms of these elements combine through bonding in a variety of ways to form molecules.
This exercise will examine some of the basic combinations of atoms that form molecules. Basic
principles of pH and buffers, as well as dilutions will also be covered.
Materials
Equipment
spectrophotometers
molecular model kits
cuvettes
cuvette racks
Kimwipes
Test tubes and racks
10 ml pipettes
pipette pumps
50 ml beakers
marking pencils
Reagents and Solutions
Bogen’s Universal Indicator
1M NaOH
1M HCl
pH 4 buffered solution
pH 4 unbuffered solution
colored dye stock solution, 100%
distilled water
unknown dye solutions
Part A: Functional Groups and Biologically Important Molecules
Most biological molecules are held together by covalent bonds. Covalent bonds result in relatively
stable molecules that do not dissociate in aqueous (water) environments. These stable molecules can
serve as monomers (building blocks or subunits) for the synthesis of larger dimers (2 monomers) or
polymers (chains of many monomers).
Biological molecules are classified according to their functional groups. Functional groups are clusters
of atoms bonded to carbon backbones and are most commonly involved in chemical reactions. They
impart particular characteristics to larger molecules to which they are attached. For example, any
molecule with a carboxyl group behaves as an organic acid like fatty acids or amino acids. Those with a
hydroxyl group are considered alcohols (e.g. glycerol). Carbohydrates contain a carbonyl group (either
an aldehyde if it’s at the end of the molecule or a ketone if not) along with a number of hydroxyl groups.
Table 2.1 illustrates some of the more biologically important functional groups. In this table, each line
represents one covalent bond. Single and double bonds can exist. Each functional group bonds to a
carbon backbone, often symbolized by the letter “R” (e.g. R-OH would be a molecule containing a
hydroxyl functional group). Each functional group must have at least one covalent bond available for
attachment to this carbon backbone.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
13
Exercise 2 –Functional Groups, Organic Molecules, Buffers, and Dilutions
Table 2.1 Biologically Important Functional Groups
Carbonyl
Aldehyde
__
__ C __ H
__
__
__
O
O
__ C __
Phosphate
H
H
O
__
__ S __ H
__
__
__ O __ P __ OH
__ N __ H
__ C __ OH
Sulfhydryl
__
Amine
__
O
__ OH
Carboxyl
Ketone
__
Hydroxyl
O
Procedure
1. Fill in Table 2.2 using the periodic chart in your text.
Table 2.2 Elements Represented in Molecular Model Kits
Element
Atomic
Symbol
Atomic
Number
# of Valence
Electrons
# of e-s
needed to fill
valence shell
Carbon
Hydrogen
Nitrogen
Oxygen
Phosphorus
2. Obtain a molecular model kit
3. Examine the colored balls to determine the number of holes in each. Each ball represents an
atom of a particular element. The holes represent the valence (bonding capacity) of the atom.
Using the information in Table 2.2, you should be able to determine which elemental atom is
represented by each ball
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
14
Exercise 2 –Functional Groups, Organic Molecules, Buffers, and Dilutions
4. Use the molecular kit to construct models of each of the functional groups in Table 2.1. Use the
appropriate colored ball to represent each atom. The grey “sticks” are bonds. Use the longer
“sticks” to bend to create double bonds. When building functional groups, you will always have
one free end of a “stick” that represents the attachment point of the functional group to the
carbon backbone (“R”). Pay attention to the content and shape of each functional group
Circle and label the functional groups within these biologically important molecules in Fig. 2.1.
Fig. 2.1 Some Biologically Important Organic Molecules
H
C
H
__ __
OH
H
C
OH
__ __
__ __
OH
C
H
H
OH
H
OH
__ __
__ __
__ __
__ __
__ __
glucose ring
(hydroxyl)
O
H
OH
__ __
OH
H
__ __
H
C
C
________
H
H
H
__
H
OH
____
__
H
O
______
__
____
__
OH
H
H
__
OH
OH
C
__
C
H
__
H
C
__
OH
OH
____
C
C
____
H
______
H
OH
C
__
H
__
__
H
HO
H
__ __
H
__
C
O
__ __
__ __
__
HO
__
__ __
OH
__
C
OH
__
__
__
H
C
fructose chain
(hydroxyl, ketone)
__
O
__
__
C
__
H
glucose chain
(hydroxyl, aldehyde)
__ __
H
OH
H
fructose ring
(hydroxyl)
H
__
H
__ __
C
OH
H
C
OH
OH
__ __ __
__
__
glycine
(amine, carboxyl)
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
__
H
__ __
__
__ __
C
__
C
OH
C
__
N
O
__
H
__
H
__
H
H
H
glycerol
(hydroxyl)
15
OH
Exercise 2 –Functional Groups, Organic Molecules, Buffers, and Dilutions
Part B: Buffers
The pH of blood and other body fluids is relatively insensitive to the addition of acids or bases. This is
due to the presence of buffers in living systems which help to maintain homeostasis by maintaining
normal pH levels. The pH of a solution can be determined in a variety of ways, including the use of pH
meters, litmus paper, and chemical reagents. In this exercise, we will use the chemical reagent Bogen’s
Universal Indicator to determine pH of specific solutions.
Bogen’s Universal Indicator changes color at specific pH end points:
Pink = pH 4
Yellow = pH 6
Green = pH 7
Blue = pH 9
Violet > pH 9
In order to determine the effect of buffers on pH, we will attempt to raise the pH of an unbuffered acid
solution by adding small amounts of a base. For comparison, we will repeat this procedure with a
buffered acid solution. Once both solutions are basic, we will attempt to return them to the original pH
by adding small amounts of acid.
Procedure
1. Obtain two 50 ml beakers and label them A and B
2. Pipette 10 ml of an unbuffered pH 4 solution into beaker A
3. Pipette 10 ml of a buffered pH 4 solution into beaker B
4. Add 3 drops of Bogen’s Universal Indicator to each beaker
5. Note the color. __________ Is this color expected? __________
6. Slowly add 1M sodium hydroxide (NaOH) one drop at a time to beaker A, swirling the beaker
between each drop. Do until you detect a permanent color change to violet
7. Record the number of drop required to change the color to violet in Table 2.3
8. Repeat the last two steps with beaker B
The test you just performed illustrated the effect of a buffer when you attempted to increase the pH
(make it more basic). Did the buffered solution require more or less (circle one) drops to change the
pH? Do you suppose buffers would resist pH changes in either direction? __________
Continue the procedure from above
9. Slowly add 1M hydrochloric acid (HCl) one drop at a time to beaker A, swirling the beaker
between each drop. Do until you detect a permanent color change to pink
10. Record the number of drops required to change the color to pink in Table 2.3
11. Repeat the last two steps with beaker B
Table 2.3 The Effect of Buffer on pH Change
Beaker
Contents
A
unbuffered, pH 4 solution
B
buffered, pH 4 solution
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
# drops to violet
# drops back to pink
16
Exercise 2 –Functional Groups, Organic Molecules, Buffers, and Dilutions
Part C: Dilutions
Part C1: Basic Dilutions
During scientific experiments, it is often necessary to dilute the solution provided (the stock solution).
For example, such a dilution might be made to reduce chemical concentrations so the rate and intensity
of reactions can be controlled.
A stock (100%) dye solution and distilled water will be used in this lab.
How would you go about preparing 10 ml each of 75%, 25%, and 10% solution from an available stock
solution of 100%?
The algebraic equation C1V1 = C2V2 provides our tool to answer this question, where
C1 = concentration (%) of stock solution
V1 = volume (ml) or stock required to prepare the solution (you typically are
solving for this variable)
C2 = concentration (%) of dilution you wish to prepare
V2 = volume (ml) of dilution you wish to prepare
Procedure
1. Use the algebraic equation to determine volumes of 100% stock (ml) and distilled water (ml)
required to create 10 ml each of 0%, 10%, 25% and 75% dilution. Record your answers in Table
2.4.
Table 2.4 Volumes Needed to Prepare Dilutions
Concentrations – C2
10%
25%
75%
Volume of stock solution (ml) - V1
Volume of water (ml)
Total volume of dilution (ml) - V2
2. Obtain 3 test tubes and a test tube rack
3. Prepare the three dilutions from Table 2.4 by pipetting the correct amount of stock in the test
tube first and then diluting the stock with the correct amount of distilled water. There should
be the same amount of liquid in each test tube when you are finished
4. Obtain 5 cuvettes on a cuvette rack
5. Transfer distilled water (0% dye solution) to the first cuvette up to about ¾ full. Distilled water
is used as a blank solution to calibrate the spectrophotometer
6. One at a time and in order of increasing concentration, transfer enough of the other 4 solutions
so that each cuvette is approximately ¾ full
7. Set the spectrophotometer to a wavelength of 450nm
8. Read the % light transmittance for each dye solution you prepared and record your results in
Table 2.5
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
17
Exercise 2 –Functional Groups, Organic Molecules, Buffers, and Dilutions
Table 2.5 % Light Transmittance Associated with Various Concentrations of Dye
Dye Solution
% Concentration of Dye
% Light Transmittance
1
0 (DH2O only)
100
2
10
3
25
4
75
5
100 (stock)
Unknown A, B, C, D (circle yours)
What relationship exists between concentration of dye and % light transmittance?
Part C2: The Standard Curve
Procedure
1. Plot the 0%, 10%, 25%, 75%, and 100% data from Table 2.5 on Fig. 2.2
2. Attempt to draw a “best fit” line through the scatter of data points. Do not simply connect the
dots. Make your line pass through the “average” spread of the dots. This line represents a
standard curve and illustrates the relationship between percent concentration of a dye solution
and percentage of light transmitted. Use this standard curve to complete Part C3
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
18
Exercise 2 –Functional Groups, Organic Molecules, Buffers, and Dilutions
% Light Transmittance
Fig. 2.2 Standard Curve Relating Dye Concentration to % Light Transmittance
Dye Concentration (%)
Describe the kind of relationship you see:
Part C3: Determination of Unknown Dye Concentration
Procedure
1. Select a cuvette of unknown dye concentration (letters A-D) from the samples available
2. Record the letter of your unknown in Table 2.5
3. Use the calibrated spectrophotometer to read the % transmittance of your unknown dye
concentration solution. Record in Table 2.5
4. Determine the concentration of your unknown by finding the value of % transmittance on the Yaxis of Fig. 2.2 and drawing a perpendicular line down from that point to where it crosses the Xaxis. That intersection point is the percent dye concentration of your unknown. Record that in
Table 2.5
5. Return your unknown cuvette to your instructor and tell them your result
6. Rinse out the rest of the cuvettes and place them on the cuvette rack. Do not scrub them with
a test tube brush as it will scratch and render them useless
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
19
Exercise 2 –Functional Groups, Organic Molecules, Buffers, and Dilutions
Practice Problems and Review Questions
1. Given a stock solution of 2.0% dextrose, how would you prepare 10 ml of each of the following
solutions?
a. 0.1% dextrose solution
b. 1.0% dextrose solution
c. 0.5% dextrose solution
2. Given a stock solution of 5.0% sodium chloride (NaCl), how would you prepare 20 ml of each of
the following solutions?
a. 2.0% sodium chloride solution
b. 0.5% sodium chloride solution
c. 3.0% sodium chloride solution
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
20
Exercise 2 –Functional Groups, Organic Molecules, Buffers, and Dilutions
3. Given a stock solution of 10% dextrose, how would you prepare 5 ml of a 0.9% dextrose
solution?
4. Given a stock solution of 0.9% dextrose, how would you prepare 5 ml of a 0.5% dextrose
solution?
5. Given a stock solution of 0.5% dextrose, how would you prepare 5 ml of a 0.004% dextrose
solution?
6. How would you prepare 25 ml of a 15% dye solution beginning with a 20% stock dye solution?
7. How would you prepare 9 liters of a 50% dye solution beginning with a 60% stock dye solution?
Express your answer in ml.
8. How would you prepare 600 ml of a 20% starch solution beginning with a 50% stock starch
solution? Express your answer in liters.
9. You have 10 ml of a 60% stock dye solution. What is the maximum amount of a 12% dye
solution you could prepare?
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
21
Exercise 2 –Functional Groups, Organic Molecules, Buffers, and Dilutions
10. How would you go about preparing the 12% dye solution in question 9?
11. What are buffers and why are they biologically important?
12. List the functional groups present in each of these molecules
glucose
fructose
glycine
glycerol
13. List some possible polymers that can be formed from each of these monomers
glucose
fructose
glycine
glycerol
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
22
Exercise 3 - Qualitative Analysis of Biological Molecules
Introduction
Macromolecules are large molecules formed from aggregates of smaller ones.
Biological
macromolecules are typically classified as carbohydrates, lipids, proteins, and nucleic acids. It is possible
to identify macromolecules and monomers by using chemical indicators.
Reagents used as chemical indicators express their results either qualitatively or quantitatively by
determining the presence or relative amount of a substance in a solution. The example in Table 3.1
should help you understand the basic difference between qualitative and quantitative analyses.
The reagents used in this exercise provide qualitative results. Each reagent exhibits a visible color
change in the presence of a specific substance; however, it does not provide an amount (quantitative)
result. A qualitative test will also be used to track the step-by-step hydrolysis of the polymer starch, a
polysaccharide, into its glucose (monosaccharide) monomers.
Table 3.1 A Case Study Illustrating the Difference Between Qualitative and Quantitative Analyses
Case Study
You are given a beaker containing 100 ml of an aqueous solution
A
B
Question
Are proteins present in this How many mg of protein are
solution?
dissolved in this 100 ml solution?
Would smelling, tasting, or Changing the solution’s color
touching the solution help indicated proteins are present,
determining if it has proteins or but it does not detect exactly
not? (not a good idea in lab)
how much protein is present.
Thinking
The best thing to do is add a An analytical test giving the
protein indicator. If the solution answer in numbers, not just by
changes color, then proteins are presence or absence, needs to
present.
be done.
A qualitative analysis must be A quantitative analysis must be
Response
performed.
performed.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
23
Exercise 3 –Qualitative Analysis of Biological Molecules
Materials
Equipment
test tubes and racks
10 ml pipettes
pipette pumps
10 ml graduated cylinders
marking pencils (Sharpie)
filter paper disks
Petri dish
water baths at 95C
Reagents and Solutions
1% dextrose (glucose)
6% starch (amylose)
1 M NaOH
apple juice
chicken broth
egg white
whole milk
vegetable oil
distilled water
Benedict’s
IKI
Biuret
Sudan IV
Part A: Detection of Carbohydrates
Carbohydrates are molecules consisting of one (monosaccharide), two (disaccharide), or many
(polysaccharide) simple sugars. Examples of carbohydrates include glucose, sucrose, glycogen, maltose,
and starch (amylose).
In this exercise, you will experiment with two carbohydrate reagents:
Benedict’s reagent – usually light blue in color, forms a yellow-green, orange, or red precipitate when
boiled in the presence of reducing sugars such as simple sugars (e.g. glucose)
Iodine-Potassium Iodide (IKI) – amber colored, forms a dark purple or black precipitate in the presence
of starch.
Read the information on the following pages (Parts A1, A2, and A3) and fill in the first three columns of
Table 3.2 before performing the experiments.
Part A1: Detection of Simple Sugars
Procedure
1. Obtain a test tube rack and six test tubes per group
2. Label the test tubes 1 through 6. #1 and #2 will be used in this part
3. Use a 10 ml pipette to transfer 1 ml of the dextrose (glucose) solution to test tube #1
4. Use a different (why?) pipette to transfer 1 ml of the starch solution (swirl to mix before
transferring) to test #2
5. Use a 10 ml graduated cylinder to measure and transfer 1 ml of Benedict’s reagent to each test
tube. Swirl to mix
6. Note the color of each solution
7. Gently heat the contents of each test tube in a 95C water bath for two minutes
8. Observe and record any color change in Table 3.2
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
24
Exercise 3 –Qualitative Analysis of Biological Molecules
Part A2: Detection of Starch
Procedure
1. Use a pipette to transfer 1 ml of dextrose solution to test tube #3
2. Use a different pipette to transfer 1 ml of starch solution (swirl to mix before transferring) to
test tube #4
3. Add one drop of IKI reagent to each test tube and swirl gently
4. Observe and record any color change in Table 3.2
Part A3: Identification of a Carbohydrate Unknown
If you were given an unknown solution and had to perform both the simple sugar (Part A1) and the
starch (Part A2) tests in the same test tube, which test would you perform first? The following
experiment will help to answer this question.
Procedure
1. Use a pipette to transfer 1 ml of dextrose to both test tubes #5 and #6
2. Use a different pipette to transfer 1 ml of starch to both test tubes #5 and #6
3. In test tube #5, perform the Benedict’s test first
4. Make note of any color changes
5. After the Benedict’s test perform the IKI test in test tube #5
6. In test tube #6, perform the IKI test first
7. Make note of any color changes
8. After the IKI test perform the Benedict’s test in test tube #6
9. Make note of any color changes
10. Record your observation in Table 3.2
11. From the results of test tubes #5 and #6, determine which test you should run first if you were
limited to using just one test tube and had to test for both simple sugars and starch. Only one
of these two test tubes will allow you to see the results of both tests correctly
Which test would you perform first and why?
12. Obtain a simple sugar / starch unknown (labeled A, B, C, and D) and test it using the proper
sequence of Benedict’s and IKI reagent
13. Record the letter of your unknown and any color changes in Table 3.2
What (water, glucose, starch, or both) was in your unknown?
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
25
Exercise 3 –Qualitative Analysis of Biological Molecules
Table 3.2 Qualitative Analysis of Simple Sugars, Starch, and a Carbohydrate Unknown
Test
Test Solution
Reagent
Hypothesis
Tube
Results
1
2
3
4
5
Benedict’s 1st
IKI 2nd
6
IKI 1st
Benedict’s 2nd
Unknown
(_____)
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
26
Exercise 3 –Qualitative Analysis of Biological Molecules
Part B: Detection of Lipids
A lipid is a non-polar (hydrophobic) organic molecule which is insoluble in water. One type of lipid are
fats, also called triglycerides or triacylglycerols. A fat molecule is composed on one glycerol and three
fatty (palmitic) acid molecules. Sudan IV-lipid complex will produce an orange spot on filter paper to
which lipid has been added.
Procedure
A
1. Obtain a blank filter paper disk
C
2. Mark the disk with a pencil following the pattern as shown in this figure
A – apple juice
W
C – chicken broth
O
E
E – egg white
M – whole milk
O – vegetable oil
M
W – distilled water (control)
3. Make a hypothesis as to which of the above substances you would expect to contain lipids
4. Record this hypothesis in Table 3.5
5. Transfer a small drop of each substance to the appropriate circle on the filter paper
6. Allow the filter paper to dry
7. Once dry, soak the filter paper for 3 minutes in a petri dish containing Sudan IV reagent. Leave
the dish on the counter where it was originally to avoid spillage
8. Remove the filter paper disk with forceps and gently rinse with distilled water over the sink for
one minute
9. Hold the filter paper over something white for contrast and observe the results
10. Examine the color for the six spots and indicate whether the substances contained lipid using
the by indicating “-“ for negative (no color change; no lipid) and “+” for positive (color change;
lipid)
11. Record your results in Table 3.5
12. Compare your results to your hypothesis
Table 3.5 Sudan IV Test for Lipids
Substance Tested
Apple juice
Chicken broth
Egg white (albumin)
Whole milk
Vegetable oil
Distilled water
Hypothesis
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
Result
27
Exercise 3 –Qualitative Analysis of Biological Molecules
Part C: Detection of Proteins
Proteins are polymers of amino acids in which the carboxyl functional group of one amino acid forms a
peptide bond with the amine functional group of another amino acid.
H
C
__
__
N
C
OH
__
__ __ __
__
+
__
OH
__ __ __
C
O
__
R
H
__
C
H
__
N
O
__
__
H
__
R
H
__
H
H
H
H2O
C
peptide bond
OH N
H
C
__ __ __
__
__
__
__ __ __ __
__
H
C
OH
__
C
O
__
R
H
__
H
O
__
N
__
H
__
R
H
__
H
H
Biuret reagent, which is pale blue, contains copper sulfate (CuSO4). The Biuret reaction is based on the
complex formation of cupric ions with proteins. In this reaction, copper sulfate is added to a protein
solution in strong alkaline solution. A purplish-violet color is produced, resulting from the complex
formation between the cupric ions and the peptide bond.
Procedure
1. Obtain a test tube and rack and six clean test tubes per group
2. Mark the test tubes with the same symbols used in the lipid experiment (Part C)
3. Make a hypothesis as to which of the above substances you would expect to contain proteins
4. Record this hypothesis in Table 3.6
5. Transfer 1 ml (approximately 20 drops) of the appropriate solution to properly marked test tube
6. Dispense 1 ml of 1M NaOH into each test tube
7. Swirl gently to mix
8. Add 0.5 ml of 1% Biuret reagent to each test tube
9. Swirl gently to mix
10. Look for any instant change in color from blue to violet. This is the positive test for proteins
11. Record your results in Table 3.6 using the same symbols (- and +) as described in Part C
12. Compare your results to your hypothesis
Table 3.6 Biuret Test for Proteins
Substance Tested
Apple juice
Chicken broth
Egg white (albumin)
Whole milk
Vegetable oil
Distilled water
Hypothesis
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
Result
28
Exercise 3 –Qualitative Analysis of Biological Molecules
Practice Problems and Review Questions
1. Explain the difference between a qualitative and quantitative analysis test.
2. What substance is used as a control in the
a. Sudan IV test?
b. Biuret test?
3. Complete the following table concerning the reagents used in detecting these test substances.
Test Substance
Reagent
Test Procedure
Color of Positive
Result
Color of Negative
Result
Starch
Sugar
Lipid
Protein
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
29
Exercise 3 –Qualitative Analysis of Biological Molecules
4. In which order must the sugar and starch test be run? Why?
5. What are the differences among polysaccharides, oligosaccharides, disaccharides, and
monosaccharides?
6. What are the two primary components of a triglyceride?
7. What are the monomers that make up proteins?
8. List and briefly describe the four levels of protein structure.
9. How do proteins of foods differ from those of the organism consuming them?
10. Name a molecule of living systems other than protein which contains nitrogen.
11. What is hydrolysis?
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
30
Exercise 4 - The Microscope
Introduction
The microscope is an essential tool in modern biology. It allows us to view structural details of organs,
tissue, and cells not visible to the naked eye.
This laboratory exercise is designed to demonstrate some of the potential uses of various types of light
microscopes and to help you become familiar with proper microscopic techniques.
Materials
Equipment
compound microscope
dissecting microscope
microscope slides
coverslips
droppers
lens paper
forceps
toothpicks
Biological Specimens
Allium (onion)
pond water
Reagents
IKI
methylene blue
Detain (or Protoslo)
Prepared Slides
newspaper print
colored threads
Paramecium
Part A: Care and Use of the Compound Microscope
 ALWAYS CARRY THE MICROSCOPE UPRIGHT WITH TWO HANDS, ONE ON THE BASE, THE
OTHER ON THE ARM
 MAKE SURE YOUR WORKBENCH IS FREE OF CLUTTER BEFORE YOU PLACE THE MICROSCOPE
ON THE BENCH
 DO NOT DRAG OR SHOVE THE MICROSCOPE ACROSS THE LAB BENCH – ALWAYS LIFT TO MOVE
OR TURN IT
The steps on the next few pages represent the correct procedure for viewing a specimen under a
compound microscope. Your instructor will demonstrate the proper use of the microscope as well as
describe its features. Refer to Fig. 4.1 to familiarize yourself with the parts of the microscope as you
study each step in the procedure.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
31
Exercise 4 –The Microscope
Fig. 4.1 The Compound Light Microscope
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
32
Exercise 4 –The Microscope
Viewing a Specimen with a Compound Light Microscope
Procedure
1. Clean the slide and coverslip by rubbing them gently with lens paper
2. Use the coarse focus adjustment knob to maximize the working distance (the distance between
the stage and the objective lens)
3. Rotate the revolving nosepiece into position with the scanning power (4x) objective lens in the
viewing position
4. Center the slide holder of the mechanical stage on the microscope stage
5. Place the slide between the stage clip and push it all the way back to the bar
6. Plug in the microscope and turn on the light switch
7. Using the mechanical stage drive knobs, center the coverslip and specimen over the stage
aperture
8. While carefully watching the slide on the stage, use the coarse focus adjustment knob to move
the specimen towards the scanning objective lens until it stops. The stage will come close to
the lens but will not touch it
9. Adjust the interpupillary distance until you see a single circle while looking through the
microscope with both eyes open. This circle of light is called the field of view
10. While looking through the ocular lenses, turn the coarse and fine focus adjustment knobs of
the microscope until you see something you believe is the specimen. Stop. Move the slide back
forth using the mechanical stage drive knobs. The item you thought was specimen should
likewise be moving back and forth
11. Cover of close the eye that is not looking through the ocular containing the diopter ring.
Viewing with only that eye focus using the coarse and fine focus adjustment knobs. Adjust the
light using the iris diaphragm adjustment lever and/or the light adjustment. Then close your
other eye adjusting the diopter ring on that ocular lens to bring the object into focus
12. Adjust the condenser to the highest position
13. Using the mechanical stage drive knobs, center the specimen of choice in the viewing area
14. These microscopes are parfocal (if one lens is in focus, all other lenses are, at least, close to
focus). In order to change to the next highest magnification, simply rotate the nosepiece to the
low power (10x) objective lens
15. These microscopes are also parcentral (if an object is in the center of the field of view for one
lens, it will be, at least, close to the center of the field of view at other lenses)
16. Using the mechanical stage drive knobs, re-center the specimen in the viewing area
17. With the low power (10x) objective, use the coarse and fine focus adjustment knobs to focus
the view of the specimen and the iris diaphragm adjustment lever to increase the light intensity
on the specimen
18. Re-center the specimen in the field of view. Rotate the nosepiece to the high power (40x)
objective lens. Use the FINE FOCUS ADJUSTMENT KNOB ONLY to focus and the iris diaphragm
adjustment lever to increase the light intensity on the specimen. If needed, use the light
adjustment to provide additional light
19. When removing the slide, rotate the nosepiece so the scanning power (4x) objective is in the
viewing position, then use the coarse focus adjustment knob to maximize the working distance
20. After you have completed the laboratory activity, turn the light switch off. Clean all microscope
lenses (objective and ocular) with lens cleaner and lens paper
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
33
Exercise 4 –The Microscope
21. Prepare the microscope for storage using the checklist below. Be sure
a. The scanning power (4x) objective is in the viewing position
b. The mechanical stage has been positioned so the stage arm is flush with the right side
of the stage
c. The cord is wrapped securely around the microscope arm
d. The stage has been adjusted all the way down
e. The condenser has been adjusted all the way up
f. The light adjustment is turned all the way down and the light is turned off
Part B: Magnification
There is a set of three objective lenses on your microscope. The magnification (or power) of each
objective lenses is engraved on the side of the objective. The ocular lens is also normally engraved with
its magnification (typically 10x).
To determine the total magnification of a specimen, use the following formula:
Total Magnification = Ocular Magnification x Objective Magnification
Procedure
1. Use Table 4.1 to record the magnification values for each objective lens and the ocular lens on
the microscope
2. Calculate total magnification (using the formula above) for each objective lens and record n
Table 4.1
Table 4.1 Total Magnification of Microscope
Magnification
Objective Lens Name
Objective Lens
Ocular Lens
Total
Scanning
Low Power
High Power
Part C: Working Distance and Diameter of the Field of View
Part C1: Working Distance
Working distance is the distance between the stage and objective lens (Fig 4.2). Because objective
lenses vary in lengths, the working distance will change as you switch from one objective lens to the
next.
In a microscope, as magnification increases, working distance ______________________.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
34
Exercise 4 –The Microscope
Fig. 4.2 Working Distances with Various Objective Lenses
Part C2: Diameter of Field of View
The approximate size of a specimen can be estimated if the diameter of the field of view (DFV) is
known. In parfocal microscopes, if we know the magnification and DFV for one objective lens, we can
calculate the DFV for a second objective on the same parfocal microscope using the following formula:
M1 x DFV1 = M2 x DFV2
where M1 and DFV1 = magnification and diameter of the field of view, respectively, of objective 1, M2
and DFV2 = magnification and diameter of field of view, respectively, of objective 2.
As magnification increases, the diameter of the field of view ______________________ (Fig. 4.3).
Fig. 4.3 Diameter of the Field of View (DFV) with Various Objective Lenses
4x
10x
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
40x
35
Exercise 4 –The Microscope
Fill in Table 4.2 for your microscope using the values given for the scanning objective and the above
formula.
Table 4.2 Diameter of Field of View (DFV) for the Compound Microscope
Objective Lens
Magnification
DFV (mm)
Scanning
4
4
DFV (µ)
__________
Low Power
High Power
10
__________
__________
__________
__________
40
Part C3: Depth of Focus
The depth of focus for a particular objective refers to the power of the objective to produce an in-focus
image from objects that are slightly different distances away from the objective lens. As magnification
power increases, the depth of focus decreases.
When viewing specimens under a microscope, it is beneficial to keep in mind that as magnification
power increases the microscope’s field of view becomes smaller, thinner, and darker (Table 4.3).
Table 4.3 Changes in a Microscope’s Field of View as a Function of Magnification Power
Scanning
Low Power
Diameter of Field of View
(DFV)
Gets Smaller
Depth of Focus
Gets Thinner
Light
Gets Darker
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
High Power
36
Exercise 4 –The Microscope
Part D: Newsprint (dry mount)
Procedure
1. Obtain a prepared slide of newspaper print
2. View the newsprint under the microscope using the scanning power (4x) objective
Move the slide slowly to the right as you view the image in the field of view. In which direction do the
letters appear to move?
Move the slide slowly away from you as you view the image in the field of view. In which direction do
the letters appear to move?
Part E: Depth of Focus
Procedure
1. Obtain a prepared slide of colored threads. The threads have been arranged to intersect at a
single point
2. Focus on the intersection of the three threads first with the scanning power (4x) objective lens
and then the low power (10x) objective lens
3. Very slowly rotate the fine focus adjustment knob while looking at the intersection of the
threads
Which thread is on bottom? __________ In the middle? __________ On top? __________
Part F: Viewing specimens
Specimens are often mounted in water (or other liquids) on a glass slide and then covered with a small
thin glass or plastic coverslip to prepare for microscopic viewing. These wet mounts are unstained and
sometimes difficult to see. Replacement staining can add color and contrast enhancing the detail of the
specimen.
It is important to be able to estimate the sizes of different specimens under the microscope.
Already knowing the diameter of the field of view for a particular objective (Table 4.2), we can utilize the
following formula to estimate size:
FV
si e of cell
of cells across FV
At which magnification do you think you are able to get the most accurate estimate of cell number and
thus the most accurate estimation of cell size? __________. Why?
Part F1: Paramecium
Procedure
1. Obtain a prepared slide of the single-celled protozoan, Paramecium
2. Use the correct focusing technique to find the Paramecium at high power
3. Estimate the # of Paramecium cells required to fill across the DFV end-to-end
4. Use the formula to calculate Paramecium length in microns
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
37
Exercise 4 –The Microscope
5. Estimate the # of Paramecium cells required to fill across the DFV side-by-side
6. Use the formula to calculate Paramecium width in microns
Paramecium cells arranged end-to-end
Paramecium Length __________ (in microns)
Paramecium cells arranged side-to-side
Paramecium Width __________ (in microns)
Part F2: Allium (onion) epidermis (wet mount)
Procedure
1. Prepare a wet mount of Allium (onion) epidermis
2. Place one or two drops of water on a clean slide
3. Peel the epidermis (thin skin) off the inside of a piece of sliced onion using forceps
4. Place the epidermis carefully in the water on the slide
5. Place a coverslip over the epidermis
6. Observe the cells under the microscope and sketch what you see
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
38
Exercise 4 –The Microscope
7. Stain the onion cells with IKI using the replacement staining technique
a. Place a few drops of IKI on the slide against one edge of the coverslip
b. Place the smooth edge of a single layer of paper towel up against the opposite edge of the
coverslip. The paper towel will pull the water out from underneath the coverslip. In turn,
the water as it exits will drag the IKI stain underneath the coverslip
c. Continue this process, adding more IKI if necessary, until the stain covers the area under the
coverslip
d. Examine under the microscope
8. Observed the cells under the microscope again and sketch what you see
9. Can you see more or less detail after staining compared to the unstained cells? _____________
10. Estimate the length and width of an onion cell (in microns)
Onion Cell Length __________ (in microns)
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
Onion Cell Width __________ (in microns)
39
Exercise 4 –The Microscope
Part F3: Cheek Cells (wet mount)
Procedure
1. Place one or two drops of water on a clean slide
2. Obtain a clean toothpick and collect cheek cells by gently scraping the inside of your cheek
3. Swirl the tip of the toothpick in the water on the slide (immediately discard your toothpick)
4. Stain your cheek cells with methylene blue stain
5. Place a coverslip over your cheek cells
6. Observe and sketch the stained cheek cells. Identify the nucleus, cytoplasm, and cell membrane
7. How do these cells differ from onion cells?
8. Estimate the diameter of one of your cheek cells (in microns)
Cheek Cell Diameter __________ (in microns)
Part G: Pond Water
Although staining cells makes it easier to see their detail, most staining techniques also kill any live
specimens. Thus, looking at microorganisms can be a challenge. Living microorganisms are also difficult
to see clearly because many of them are motile and must be chased around the slide while you are
focusing.
Procedure
1. Place a drop of pond water on a clean microscope slide. Try to obtain a sample that is near any
floating debris and organisms tend to congregate there. Be careful not to shake the jar
2. Add a coverslip
3. Examine under the microscope
4. Try to keep motile microorganism in focus by following them around as they move on the slide.
If they move too quickly, carefully lift up the coverslip and add a drop of Detain (or Protoslo)
5. Draw a few of the critters you see in space provided
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
40
Exercise 4 –The Microscope
Part H: The Dissecting Microscope
It is possible to have too much magnification when viewing some specimens. For example, how would
you use your compound light microscope to view an entire earthworm? Larger specimens may require
lower magnification. For this, biologists use dissecting microscopes (Fig. 4.4). Fill in Table 4.4 and notice
the diameter of the field of view in these microscopes is substantially larger than that in compound light
microscope. Your instructor will describe the use and features of this microscope.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
41
Exercise 4 –The Microscope
Fig. 4.4 The Dissecting Microscope
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
42
Exercise 4 –The Microscope
Table 4.4 Diameter of the Field of View for the Dissecting Microscope
Objective Lens
Magnification
DFV (mm)
DFV (µ)
Lowest Power
2
10
10,000
Highest Power
4
5
__________
Procedure
1. Obtain a dissecting microscope using two hands to carry it
2. Identify the parts as per Fig. 4.4 and their functions
3. Observe the various objects made available in lab using the dissecting microscope
Using the information in Table 4.4, complete this sentence for the dissecting microscope
As magnification increases, DFV __________
Have you seen this relationship before? __________ Where? __________________________
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
43
Exercise 4 –The Microscope
Practice Problems and Review Questions
1. What is the total magnification of an object if the ocular lens magnification is 20x and the
objective lens magnification is 45x?
2. Which objective lens is in place if the object you are viewing is magnified 1000x assuming an
ocular lens magnification of 10x?
3. What is the diameter of the field of view (DFV) of a 1000x objective lens if the DFV of a 400x
objective lens is 500 µ? Express your answer in mm.
4. What is the DFV of a 40x objective lens if the DFV of a 10x objective lens is 3 mm? Express your
answer in µ.
5. When viewing an organism using the 40x objective lens from question 4, you estimate 6
organisms could fit across the DFV if they were laid end-to-end and 20 could fit is stacked sideby-side. What is the length and width of this organism (in microns)?
6. What is the DFV of a 25x objective lens if the DFV of a 100x objective lens is 1.5 mm?
7. Using the 100x objective lens from question 6, you estimate 12 organisms could fit across the
DFV if they were laid end-to-end and 30 could fit is stacked side-by-side. What is the length and
width of this organism (in microns)?
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
44
Exercise 4 –The Microscope
8. What is the magnification of an objective lens with a DFV or 0.8 mm if the DFV of a 100x
objective lens is 2 mm?
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
45
Exercise 5 - Cell Structure and Membrane Function
Introduction
The cell is the lowest level of biological organization performing all activities of life. Therefore, it is the
fundamental unit of structure in living things. As such, the characteristics of cells are of monumental
concern to the understanding of biology. The structure of cellular components reflects adaptation to
accomplish those functions necessary for life. The collective functions of individual cells allow for the
activity and behavior of the entire organism of which those cells are a part.
In this laboratory exercise, you will use a compound light microscope to examine cells and observe
cellular activity. You will also conduct experiments illustrating some of the basic mechanisms of cellular
transport.
Materials
Equipment
compound light microscope
microscope slides
coverslips
test tubes and racks
beakers
droppers
dialysis tubing
dental floss or string
scissors
triple beam balances
95C water bath
Biological Specimens
Elodea
Reagents and Solutions
Benedict’s
IKI
10.0% saline solution
concentrated glucose
concentrated starch
Part A: Cellular Transport
Cellular transport mechanisms are typically divided into two categories: Passive Transport and Active
Transport. The basic differences between them is summarized in Table 5.1
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
46
Exercise 5 –Cell Structure and Membrane Function
Table 5.1 Basic Differences Between Passive and Active Transport
Passive Transport
Involves the movement of water
or solute through a semipermeable membrane down their
Types and Direction of
concentration gradient (i.e., from
Transported Substances
regions of higher concentration of
water or solutes to regions of
lower concentration).
Cellular Energy
Active Transport
Involves the movement of
solutes through a semipermeable membrane against
their concentration gradient
(i.e., from regions of lower
concentration of solutes to
regions
of
higher
concentration).
Does NOT require cellular energy Requires cellular energy in the
in the form of ATP.
form of ATP.
Passive
transports
systems Requires membrane transport
include two types of diffusion and proteins.
osmosis:
Simple diffusion – membrane
transport proteins not required
Membrane Transport Proteins
Facilitated diffusion – membrane
transport proteins required
Osmosis – specific to the passive
transport of water from an area
of higher water concentration to
an area of lower water
concentration (lower to higher
concentration of solutes). Water
moves through protein channels
known as aquaporins.
Solutions are often described using the terms hypotonic, hypertonic, and isotonic. Tonicity is a
comparative term related to the concentration of solutes in a solution. It may be defined as the ability
of a solution to cause a cell to gain or lose water. Hypotonic solutions contain less solute by % (i.e.,
more water) when compared to hypertonic solutions, which contain more solutes by % (i.e., less water).
With a hypotonic solution that is separated from a hypertonic one by a selectively permeable membrane
that allows water molecules to pass through but not solutes, the net movement of water molecules will
be from a region of high water concentration (i.e., low solute – hypotonic) to a region of lower water
concentration (i.e., high solute – hypertonic). Isotonic solutions are equal to one another in solute
concentration; therefore, a concentration gradient does not exist and water moves in equal rate back
and forth across the membrane.
This exercise will explore some of these basic principles of cellular transport.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
47
Exercise 5 –Cell Structure and Membrane Function
Part A1: Passive Transport in a Model Cell
Procedure
1. Obtain a piece of dialysis tubing
2. Working quickly so the dialysis tubing won’t dry out, fold one end and tie off with floss or string
3. Open the other end of the tubing by sliding your fingers back and forth across the top
4. Place 10 ml of concentrated glucose and 10 ml of concentrated starch into the bag
5. Squeeze out the excess air from the bag before folding its other end and tying off
6. Rinse the bag gently under running water at the sink and blot dry with a paper towel. Make
sure the bag is not leaking
7. Weigh the bag to the nearest 0.1 g and record as initial mass of the bag in Table 5.2
8. Fill a beaker with distilled water
9. Add just enough IKI to the beaker water to turn it light yellow
10. Place the dialysis bag in the beaker. The bag should be fully submerged
11. Let your beaker sit no less than 30 minutes
This model cell system consists of four different molecules which could possibly move through the small
holes in the dialysis bag. What are they?
1. _______________ 2. _______________ 3. _______________ 4. _______________
Based on the molecular size of these four molecules, develop a hypothesis to describe which molecules
will move into the bag, which will move out and why. Record your hypothesis in Table 5.3
12. After your bag has soaked for the appropriate amount of time (no less than 30 minutes),
remove it from the beaker and gently blot dry with a paper towel
What color is the solution in the bag? _______________
What color is the solution in the beaker? _______________
13. Weigh the bag again to the nearest 0.1 g and record in Table 5.2
14. Calculate the change in the mass of the bag by subtracting the initial mass from the final mass of
the bag and record in Table 5.2
15. Calculate the % mass change of the bag using this formula and record in Table 5.2
% mass change of bag
final bag ass - initial bag
initial bag ass
Table 5.2 Mass and Time of Dialysis Bag Experiment
Mass
Final
__________ g
Initial
__________ g
Change in Mass of Bag
__________ g
% Mass Change of Bag
__________ %
ass
Time (hr : min)
_____ : _____
_____ : _____
16. Pour about 1 ml of the contents of the bag into a test tube
17. Test the bag contents with Benedict’s reagent
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
48
Exercise 5 –Cell Structure and Membrane Function
18. Pour about 1 ml of the contents of the beaker into a test tube
19. Test the beaker contents with Benedict’s
20. Fill in Table 5.3
Table 5.3 Results of Dialysis Bag Experiment
Net Movement of Molecules Across the Dialysis Bag (In / Out / None)
Molecular
Final Results
Component of
Based on
the Dialysis
Hypothesis
Explanation
Observations and
Bag System
Testing
Part A2: Osmosis in Elodea
Elodea is a common aquatic plant related to Hydrilla. It has leaves of only two layers of thickness.
In this exercise, the thin leaves of Elodea will be useful in exploring some of the principles of osmosis.
As seen under the compound microscope, the movement of cytoplasm with the Elodea leaf cells along
the perimeter of the cell called cyclosis or cytoplasmic streaming will be observed.
Procedure
1. Using forceps, remove one leaf from an Elodea plant
2. Prepare a wet mount of the leaf using distilled water
3. Observe the leaf at high power under the microscope
4. Identify the parts of an Elodea leaf (Fig. 5.1)
Where are the chloroplasts located? _______________
Do you see cyclosis (cytoplasmic streaming)? _______________
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
49
Exercise 5 –Cell Structure and Membrane Function
5. Draw your Elodea cell and label the visible parts
6. Using the replacement staining technique, replace the distilled water under your coverslip with
the saline solution
7. After 5-10 minutes, observe the cells again and make note of any changes that have occurred
8. Draw the cell again
Where are the chloroplasts located now?
What cellular structure (not visible previously) has receded from the cell wall?
What happened to the volume of the central vacuole to cause this change?
In what type (hypotonic, hypertonic, isotonic) of environment is the Elodea cell in?
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
50
Exercise 5 –Cell Structure and Membrane Function
Fig. 5.1 Elodea Cell
Part A3: Osmoregulation in Protists
Some single-celled organisms live in a fresh water environment that is hypotonic to their cellular fluid
which means they are continually taking on water through osmosis. They stay alive because they
possess abilities to regulate internal water pressure using contractile vacuoles. These contractile
vacuoles remove excess water from the cell. Contractile vacuoles typically appear a “fluid-filled
bubbles” in the cytoplasm that slowly get large and then suddenly disappear.
Procedure
1. Using the web, find and view pictures and video clips of contractile vacuole function in
Paramecium. Your instructor may also make some clips available online or have you view them
in class
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
51
Exercise 5 –Cell Structure and Membrane Function
2. Draw a Paramecium and label the contractile vacuole
What is its function?
Part B: Structure and Motility in Protists
Most groups of protists are capable of movement. This motility is made possible by one of three types
of structures. Organisms like Amoeba move by means of pseudopodia (“false foot”) which are
extensions of the cytoplasm. Paramecium and similar organisms move using cilia, fine hair-like
structures covering the cell membrane. Organisms typically have many, many cilia. Other protists, such
as Euglena, move using flagella, which are whipped back and forth. Organisms usually have one or just
a few flagella. Finally, some protists lack the ability to move at all.
Procedure
1. Using the web, find and view pictures and video clips of protist structure and movement. Your
instructor may also make some clips available online or have you view them in class
2. Using online resources and the text book, draw and label the following parts for Amoeba,
Paramecium, and Euglena
cell membrane
cytoplasm
pseudopod (Amoeba)
cilia (Parmecium)
flagella (Euglena)
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
contractile vacuole
nucleus
chloroplast (Euglena)
food vacuole
52
Exercise 5 –Cell Structure and Membrane Function
Amoeba
Paramecium
Euglena
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
53
Exercise 5 –Cell Structure and Membrane Function
Practice Problems and Review Questions
1. If the initial mass of a dialysis bag was 8.2 g and final mass was 10.9 g, what is the % mass
change of the bag?
2. If the initial mass of a dialysis bag was 10.6 g and final mass was 11.1 g, what is the % mass
change of the bag?
3. If the initial mass of a dialysis bag was 9.9 g and final mass was 8.8 g, what is the % mass change
of the bag?
4. A pre-weighed dialysis bas which contained a solution of 10% glucose was placed in a beaker
containing a solution of 20% glucose. After one hour, the bag was weighed again. Calculate the
% mass change of this dialysis bag from the following information:
Mass of bag before experiment: 15.3 g
Mass of bag after experiment: 12.7 g
5. Was the beaker solution in question 4 hypertonic, hypotonic, or isotonic to the dialysis bag
contents?
6. What are the major differences between the following pairs of cells?
prokaryotic and eukaryotic
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
54
Exercise 5 –Cell Structure and Membrane Function
plant and animal
protists and generalized animal cells
7. How was the dialysis bag in your experiment an example of a semi-permeable membrane?
8. Define these terms:
hypertonic
hypotonic
isotonic
9. Complete the following sentence: When two aqueous solutions are separated by a semipermeable membrane, the net water movement is always from a ________tonic to a
________tonic solution.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
55
Exercise 6 - Enzyme Activity
Introduction
Enzymes are biological catalysts that regulate the rate of chemical reactions. Their 3-dimensional
conformation and therefore their function can be affected by several variables.
In this laboratory exercise, you will manipulate various factors that affect an enzyme’s activity. The
enzyme is catalase, which is found in most all living organisms. Catalase decomposes hydrogen
peroxide (H2O2), a toxic compound, into water and oxygen:
2H2O2 + catalase  2H2O + O2
The amount of oxygen created is directly proportional to the rate of the enzymatic reaction. Therefore,
measuring the amount of oxygen produced provides a measure of the speed at which the reaction is
proceeding.
The effect of external factors such as substrate concentration, temperature, and pH will be examined.
Materials
Equipment
test tubes and racks
metric rulers
graduated cylinders
droppers
marking pens (Sharpies)
thermometers
ice water bath
warm (40C) water bath
hot (95C) water bath
Reagents and Solutions
catalase (sheep’s blood)
3% hydrogen peroxide (H2O2)
pH 3 buffered H2O2 solution
pH 5 buffered H2O2 solution
pH 7 buffered H2O2 solution
pH 9 buffered H2O2 solution
pH 11 buffered H2O2 solution
Part A: Enzyme Activity as a Function of Substrate Concentration
An enzyme requires a substrate which it converts into product.
Drawing from what you have learned about enzymes so far, develop a hypothesis regarding the effect of
substrate (H2O2) concentration on enzymatic reaction rate.
Hypothesis: As substrate concentration increases, reaction rate will _______________.
Procedure
1. Obtain six test tubes and a test tube rack per group
2. Add 3 drops of blood and add to each of the six tubes
3. Using a dropper, add 3 drops of H2O2 to test tube #1
4. After timing for 60 seconds, mark the maximum height of the bubble column with a Sharpie
5. Doing each tube one at a time, repeat steps 3 and 4 to the remaining five test tubes increasing
by three drops the amount of H2O2 in each test tube (i.e., test tube #2 receives 6 drops of H2O2,
test tube #3 receives 9 drops of H2O2, etc.)
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
56
Exercise 6 –Enzyme Activity
6. Upon completion, return to each test tube and measure in mm the distance from the bottom of
the test tube to the height of the mark you made
7. Fill in Table 6.1 and plot the results as a bar chart in Fig. 6.1
Table 6.1 Reaction Rates for Catalase at Various Substrate Concentrations
Test Tube
Blood (# drops)
H2O2 (# drops)
Height of Bubble Column (mm)
1
3
3
2
3
6
3
3
9
4
3
12
5
3
15
6
3
18
Height of Bubble Column (mm)
Fig. 6.1 Reaction Rate of Catalase as a Function of Substrate Concentration
# of Drops of Substrate (H2O2)
Make a general statement regarding the effect of substrate concentration on enzymatic reaction rate:
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
57
Exercise 6 –Enzyme Activity
Part B: Enzyme Activity as a Function of Temperature
Temperature is a measure of the speed at which molecules are moving. As temperature increases, the
molecular movement speed does so as well. Increasing temperature increases the probability and rate
at which enzyme and substrate come together, thereby increasing the reaction rate. However, enzymes
are subject to denaturation at excess temperatures. A denatured enzyme’s active site conformation is
changed not allowing the substrate to bind. The result is that at these temperatures, the reaction will
decrease. An enzyme’s optimum temperature is that point which has the greatest reaction rate but
does not denature the enzyme.
Drawing from what you have learned about enzymes so far, develop a hypothesis regarding the effect of
temperature on enzymatic reaction rate.
Hypothesis: As temperature increases, reaction rate will _______________ and then _______________.
Procedure
1. Obtain eight test tubes and a test tube rack per group
2. Using a pipette, measure 1 ml of H2O2 into four of the eight test tubes
3. Using a dropper, add 3 drops of blood to each of the other four test tubes
4. Place one test tube each of H2O2 and catalase into each of the water baths (make note of which
tubes are yours for later retrieval). Leave the remaining pair of test tubes in the rack
5. Allow all tubes to acclimate for 15 minutes
6. After 15 minutes, proceeding one pair of tubes at a time, pour the H2O2 into the catalase
7. After timing for 60 seconds, mark the maximum height of the bubble column with a Sharpie
8. Upon completion, return to each test tube and measure in mm the distance from the bottom of
the test tube to the height of the mark you made
9. Fill in Table 6.2 and plot the results as a bar chart in Fig. 6.2
Table 6.2 Reaction Rates for Catalase at Various Temperatures
Temperature (C)
Cold
Room
Warm
Hot
Catalase (# drops)
H2O2 (ml)
3
3
3
3
1
1
1
1
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
Height of Bubble
Column (mm)
58
Exercise 6 –Enzyme Activity
Height of Bubble Column (mm)
Fig. 6.2 Reaction Rate of Catalase as a Function of Temperature
Temperature (C)
Make a general statement regarding the effect of temperature on enzymatic reaction rate:
What is the optimum temperature for catalase? __________
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
59
Exercise 6 –Enzyme Activity
Part C: Enzyme Activity as a Function of pH
Another variable that can affect enzyme conformation and therefore activity levels is pH. Like with
temperature, enzymes also have an optimum pH.
Drawing from what you have learned about enzymes so far, develop a hypothesis regarding the effect of
pH on enzymatic reaction rate.
Hypothesis: As pH moves away from optimum, reaction rate will _______________.
Procedure
1. Obtain five test tubes and a test tube rack per group
2. Using a pipette, add 1 ml of buffered (3, 5, 7, 9, 11) H2O2 to each of the five tubes
3. Using a dropper, add 3 drops of blood to test tube #1
4. After timing for 60 seconds, mark the maximum height of the bubble column with a Sharpie
5. Doing each tube one at a time, repeat steps 3-5 above to the remaining four test tubes
6. Upon completion, return to each test tube and measure in mm the distance from the bottom of
the test tube to the height of the mark you made
7. Fill in Table 6.3 and plot the results as a bar chart in Fig. 6.3
Table 6.3 Reaction Rates for Catalase at Various Temperatures
Test Tube
pH
Catalase (# drops)
H2O2 (ml)
1
2
3
4
5
3
5
7
9
11
3
3
3
3
3
1
1
1
1
1
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
Height of Bubble
Column (mm)
60
Exercise 6 –Enzyme Activity
Height of Bubble Column (mm)
Fig. 6.3 Reaction Rate of Catalase as a Function of pH
pH
Make
a
general
statement
regarding
the
effect
of
pH
on
enzymatic
reaction
rate:
What is the optimum pH for catalase? __________
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
61
Exercise 6 –Enzyme Activity
Practice Problems and Review Questions
1. What is meant by an organic catalyst?
2. List and describe the affect of the three major factors that cause changes in rate of enzymatic
activity.
3. Why did certain temperatures and pH exhibit little or no activity at all?
4. During the reaction, you may have noticed a slight bit of heat given off. Explain the source of
this heat.
5. What is the general equation for all enzymatic reactions?
6. Fill in these blanks
The rate of enzymatic reaction is _______________ (directly / inversely) proportional to
substrate concentration.
At optimum, enzymatic reaction rate is __________ (greatest / least).
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
62
Exercise 7 - Respiration
Introduction
Regardless of species, all living organisms carry on the process of respiration. The equation for
respiration is
C6H12O6 + 6O2  6CO2 + 6H2O + energy
In this process, the energy released from food molecules is coupled into the synthesis of ATP. The
energy in ATP is then used to power metabolic reactions. Respiration can occur with oxygen (aerobic) or
without oxygen (anaerobic) respiration. In anaerobic respiration, only glycolysis occurs while in aerobic
respiration glycolysis is followed by the Kreb’s cycle and a process known as oxidative phosphorylation.
During glycolysis, 2 ATPs for each glucose molecules are produced. This is the total ATP production for
anaerobic respiration. Aerobic respiration increases that output approximately 19 fold (38 ATPs).
In this laboratory exercise, you will examine aspects of respiration by qualitative examination of carbon
dioxide production.
Materials
Equipment
large test tubes and racks
bean seeds
ungerminated
germinated
Erlenmeyer flasks
rubber stoppers
thistle funnels
glass tubing
Reagents and Solutions
phenol red
Part A: Carbon Dioxide Production
Seeds contain stored food material in the form of carbohydrates. When a seed germinates, the
carbohydrate is broken down liberating energy (ATP) needed for growth of the enclosed embryo.
For this procedure, dry bean seeds have been soaking for some time in water to begin the germination
process. Another set of beans was not soaked and is, therefore, not germinating.
Procedure
1. Obtain two respiration flask setups (Fig. 7.1)
2. Place about 30 ml of ungerminated seed into one of the flasks. Repeat this for the other flask
with germinated seeds
3. Fit rubber stoppers securely into the flasks and the thistle funnels
4. Add enough water to each test tube to cover the ends of the glass tubing coming out of the flask
5. Set the flasks aside for approximately 1 hour
6. After this time, add enough phenol red to each test tube to turn the water a medium red color.
Phenol red is a pH indicator that changes color in response to changes in pH. Red indicates a
neutral pH. A yellow color signifies an acid
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
63
Exercise 7 –Respiration
7. Pour water through the thistle funnels into each flask forcing the gas in the flask out through the
glass tubing and into the test tube. Carbon dioxide (CO2) when bubbled through water forms a
mild acid called carbonic acid (H2CO3). Any CO2 given off by the seeds in the flask will interact
with the water in the test tube creating carbonic acid changing the phenol red solution to yellow
8. Record your results in Table 7.1
Fig. 7.1 Respiration Procedure Setup
Table 7.1 Carbon Dioxide (CO2) Production in Bean Seeds
Ungerminated
Germinated
color
CO2 present?
Which set of seeds was respiring?
How do you know?
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
64
Exercise 7 –Respiration
Practice Problems and Review Questions
1. What is the chemical equation for respiration?
2. Define or describe the following:
aerobic
anaerobic
NADH
FADH2
ATP
3. How many ATPS are produced during
glycolysis (total)
glycolysis (net)
Krebs cycle
aerobic respiration (eukaryotes)
anaerobic respiration (net)
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
65
Exercise 7 –Respiration
4. Some wines are marketed as “sparkling” wines. Sparkling wines are considered so due to a
dissolved gas that makes them fizzy. Apply what you know about the process of respiration to
explain the type and origin of this gas.
5. Animals are either homeothermic or poikilothermic (homeo – same, poikilo – varied, thermic –
warm), more commonly known as “warm-blooded” or “cold-blooded”, respectively.
Homeothermic animals include mammals and birds. Examples of poikilothermic species would
include the invertebrates, reptiles, fish, amphibians, etc. Using what you have learned about
respiration and enzymes, explain why poikilotherms move more slowly in the winter than the
summer.
6. Very small mammals such as shrews and many rodent species have very high metabolic rates
compared to larger mammals like ourselves or elephants. Based on what you’ve learned about
temperature’s effect on the rate of enzymatic reactions, as well as the relationship between
surface area and volume, provide an explanation.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
66
Exercise 8 - Photosynthesis
Introduction
The equation for photosynthesis is essentially the reverse of respiration.
6CO2 + 6H2O  C6H12O6 + 6O2
In this process, energy from the sun is used to reduce carbon dioxide (CO2) into glucose (C6H12O6).
Photosynthesis is best understood as a set of two linked sets of chemical reactions. The light reactions
require sunlight and water as inputs and output energy in the form of ATP, NADPH as a reducing agent.
Oxygen is also formed during the light reactions. The outputs of the light reactions then become inputs
for the dark reactions. Dark reactions, as the name suggests, do not require sunlight. Instead, the dark
reactions reduce carbon dioxide into glucose using the ATP and the electrons carried by NADPH from
light reactions.
In this laboratory exercise, you will examine aspects of photosynthesis by spectral examination of the
pigment chlorophyll. A qualitative analysis of carbon dioxide production as evidence of the process of
photosynthesis will also be conducted.
Materials
Equipment
large test tubes and racks
ring stands
test tube clamps
spectrophotometer
kimwipes
cuvettes
light source
razor blades or scalpel
beakers
drinking straws
Biological Specimens
Elodea
Reagents and Solutions
dilute chlorophyll
phenol red
Part A: Absorption Spectrum of Photosynthetic Pigments
The absorption spectrum is the pattern of light absorption of a particular pigment. In this exercise, a
spectrophotometer will be used to determine the absorption spectrum of chlorophyll.
Procedure
1. Fill a spectrophotometer cuvette with a dilute chlorophyll solution
2. Fill a second cuvette with the chromatography solvent to act at a control
3. Prepare the spectrophotometer
4. Measure the light absorbency of the chlorophyll extract for wavelengths 375 nm to 725 nm in 25
nm increments (Table 8.1)
5. Plot light absorbency as the dependent variable (y-axis) and wavelength as the independent
variable (x-axis) in Fig. 8.1
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
67
Exercise 8 –Photosynthesis
Table 8.1 Light Absorbance of Chlorophyll
Wavelength
Wavelength
Absorbance
(nm)
(nm)
375
500
400
525
425
550
450
575
475
600
Absorbance
Wavelength
(nm)
625
650
675
700
725
Absorbance
% Light Absorbance
Fig. 8.1 Absorption Spectrum of Chlorophyll
Wavelength (nm)
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
68
Exercise 8 –Photosynthesis
Part B: The Uptake of Carbon Dioxide During Photosynthesis
During the dark reactions of photosynthesis, plants take up carbon dioxide and use NADPH to reduce it
to carbohydrate (glucose).
Hypothesis: The amount of carbon dioxide in the water surrounding aquatic plant, will
_______________.
Procedure
1. Fill two test tubes with tap water. Add enough phenol red to each test tube to turn the water a
medium red color (phenol red is a pH indicator: yellow is acid, red is neutral)
2. Using a drinking straw, exhale bubbles into both test tubes until the phenol red solution turns
yellow. Carbon dioxide dissolving in water creates a weak acid known as carbonic acid (H2CO3)
3. Obtain a healthy green sprig of Elodea and place into one of the test tubes
4. Place both test tubes side-by-side in a test tube rack in front of a lamp
5. Fill the beaker with tap water and place in between the lamp and the test tubes
6. Periodically, check the color of the solutions until you detect a change from yellow to red (this
generally takes anywhere between 45-60 minutes)
7. Record your results in Table 8.2
Table 8.2 Carbon Dioxide (CO2) Uptake by Elodea
Start Time: ____ : ____
Test Tube
with Elodea
without Elodea
Initial color
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
Final color
69
Exercise 8 –Photosynthesis
Practice Problems and Review Questions
1. What is the chemical equation for photosynthesis?
2. Define or describe the following:
light reactions
dark reactions
NADPH
Rubisco
absorption spectrum
3. What are the three stages of the Calvin Cycle?
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
70
Exercise 8 –Photosynthesis
4. How many peaks were there in the chlorophyll absorption spectrum? At which wavelengths did
they occur? At which wavelengths, was the least amount of light absorbed? In viewing this
data, provide a reasonable hypothesis as to why the leaves of most plants are green.
5. In the Elodea and carbon dioxide experiment, what does the gradual change in color from
yellow to red indicate about the pH of the solution? What does this say about the amount of
carbon dioxide in the solution? Where did the carbon dioxide go?
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
71
Exercise 8 –Photosynthesis
6. Many Florida waterways have extensive aquatic vegetation as well as an abundance of
photosynthetic algae. Would you expect the pH of these waterways to fluctuate through a 24
hour period? If so, how would it change?
7. What was the purpose of the empty tube in the Elodea and carbon dioxide experiment?
8. You may have noticed small bubbles in the solution with the Elodea. What gas would you
expect to be in these bubbles?
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
72
Exercise 9 - Cell Division
Introduction
There are two distinct types of cell division in eukaryotes. In mitosis, genetically identical daughter cells
are created while in meiosis the resultant cells not only differ from each other genetically but only
contain half the original genetic information.
This laboratory exercise will demonstrate the steps the mitosis and meiosis using demonstration and
simulation. Additionally, Allium (onion) cells in various stages of mitosis will be examined under the
microscope.
Materials
Equipment
compound microscopes
two colors of pop beads
magnetic “centromeres”
Prepared Slides
Allium root tip mitosis
whitefish mitosis
Part A: The Cell Cycle
Part A1: Observation of the Cell Cycle in Allium (onion)
The cell cycle is an ordered sequence of events in the life of a eukaryotic cell from its origin at the
division of the parent cell until it itself divides. In this exercise, interphase and the mitotic phases
(prophase, metaphase, anaphase, telophase) of Allium will be viewed under the microscope and/or
through images available on the web.
Procedure
1. Obtain a compound light microscope
2. Obtain a prepared Allium root tip slide
3. Focus your observations on the region of cell division (meristematic region). This area of rapidly
dividing cells is located just behind the root tip (the rounded portion of the root tip which
consists of relatively larger cells). Center this area in the field of view and use the high power
objective to make your final observation
4. Draw and label the stages in Fig. 9.1
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
73
Exercise 9 – Cell Division
Fig. 9.1 Sketches of the Stages of Mitosis in Allium
2 daughter cells
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
74
Exercise 9 – Cell Division
Part A2: Observation of the Cell Cycle in Animal Cells (whitefish)
A good place to observe the stages of mitosis in an animal is in a cross section of a blastula. A blastula is
a stage during embryonic development of animals in which cells are rapidly dividing.
Procedure
1. Obtain a compound light microscope
2. Obtain a prepared whitefish slide
3. Draw and label the stages in Fig. 9.2
How do the cell cycle phases of animal cellular division differ from those in plants?
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
75
Exercise 9 – Cell Division
Fig. 9.2 Sketches of the Stages of Mitosis in Whitefish
2 daughter cells
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
76
Exercise 9 – Cell Division
Part A3: Simulation of Mitosis with Pop Beads
Procedure
1. Use the pop beads provided to assemble 4 chromosomes (2 homologous pairs) consisting of two
sister chromatids each. Construct your chromosomes like this: yellow (paternal) – 2 long, 2
short and red (maternal) – 2 long 2 short (Fig. 9.3)
2. Manipulate the chromosomes constructed to illustrate each of the stages of mitosis
Fig. 9.3 Initial Arrangement of Pop Beads in Mitosis Simulation
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
77
Exercise 9 – Cell Division
Part B: Meiosis
Meiosis is a cellular division in which a diploid (2n) cell divides to produce haploid (n) cells which
become gametes or sex cells. It is necessary to reduce the chromosome number by half in these sex
cells so when fertilization (the union of the gametes) occurs in sexual reproduction the resulting zygote
(offspring) will have the correct diploid number of chromosomes.
Procedure
1. Use the same set of pop beads and initial arrangement (Fig. 9.3) from the previous exercise
2. Manipulate the chromosomes constructed to illustrate each stage of meiosis
3. Disconnect and reconnect pop beads to simulate crossing-over in prophase I thereby creating
genetically non-identical daughter cells (Fig. 9.4)
Fig. 9.4 Crossing-Over Sequence for Meiosis Simulation
synapsis and tetrad
arrangement during
prophase I
chiasma formation
during crossing over
in prophase I
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
recombined
chromatids formed in
anaphase I
78
Exercise 9 – Cell Division
Practice Problems and Review Questions
1. Define or describe the following:
cytokinesis
spindle fiber
synapsis
homologous chromosomes
chromosome vs. chromatid
2. How does metaphase of mitosis differ from metaphase I of meiosis?
3. During mitosis, a cell with 52 chromosomes will produce daughter cells with how many
chromosomes?
4. During metaphase I of meiosis you observe a cell with 84 chromatids. How many chromosomes
will there be in the cells at the completion of meiosis?
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
79
Exercise 9 – Cell Division
5. When, during the life of an organism, would you expect to find the most intensive mitotic
activity?
6. Describe a chiasma. When are they formed? What is the end result of the formation of
chiasma?
7. What is the benefit of the genetic variation created during meiosis as it pertains to
environmental pressures and the process of natural selection?
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
80
Exercise 10 - DNA Fingerprinting
Introduction
Deoxyribonucleic acid (DNA) is a double stranded genetic molecule consisting of many monomers called
nucleotides, hence DNA is a polynucleotide. The two strands of DNA are connected to one another by
hydrogen bonds between the nitrogenous bases of each strand. The DNA base pair sequence and DNA
quantity (base pair total) vary from species to species. There would be less, but still measurable
differences among conspecifics. Indeed, no two organisms of the same species, unless they are clones
or identical twins, share exactly the same base pair sequence. It is the differences in these base pair
sequences that allows for the identification of genetic similarities between DNA from two sources using
a process known as DNA fingerprinting.
This laboratory exercise will investigate the basic concepts in DNA fingerprinting involving techniques
such as polymerase chain reaction (PCR) and the analysis of short tande repeats (STR’s).
Materials
Equipment
scissors
marking pen or pencils
Part A: Polymerase Chain Reaction (PCR)
The most current form of DNA fingerprinting begins with a technique known as polymerase chain
reaction (PCR). The advantage of PCR is that only a tiny amount of DNA is needed and the sample can
be old, stored under less than ideal conditions or even partially degraded. PCR involves the following
steps:
1. The DNA sample is placed in a small test tube with a solution of deoxyribonucleotides,
small pieces of DNA to act as primers, and the enzyme DNA polymerase. The mixture is
then placed in a thermal cycling device, which will raise and lower the temperature of
the tube at precisely timed intervals.
2. Denaturing – occurs to the DNA when the mixture is raised to 94C. The hydrogen
bonds between the nitrogenous bases break down from the heat and the two
complementary strands of DNA separate.
3. Annealing – the primers attach themselves to the long pieces of DNA through
complimentary base pairing (A-T; G-C) when the temperature is lowered to 65C.
4. Extending – DNA polymerase extends the new strands of DNA from the primers as the
temperature is raised to 72C.
5. The process (denaturing, annealing, extending) is repeated.
The PCR process amplifies the original amount of DNA in a very short period of time (~ 2 minutes per
cycle) so investigators will have much more DNA to use for subsequent analyses with little delay.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
81
Exercise 10 – DNA Fingerprinting
Procedure
1. On the following page is a “sample” of DNA and some primers
2. Use scissors to cut out all the pieces and lay them on the table
3. Starting with the two strands next to one another and lined up so the complementary base pairs
are aligned
4. Denaturing - slide the two strands of DNA away from one another
5. Annealing - move two of the primers in between the DNA strands, lining them up to form
complementary base pairs
6. Extending - use a marking pen or pencil to write in the complementary bases to the new DNA
strand until both strands are complete
After one round of PCR, one molecule of DNA consisting of two complementary strands yields _____
molecules of DNA for a total of _____ strands.
How many DNA molecules would exist after 2 PCR cycles? _____
5 cycles?
_________________
10 cycles?
_________________
20 cycles?
_________________
30 cycles?
_________________
Write a formula to calculate the number of DNA molecules which will be created for a given number
of PCR cycles.
Are all the DNA molecules created identical? _____
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
82
Exercise 10 – DNA Fingerprinting
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
83
Exercise 10 – DNA Fingerprinting
Part B: Short Tandem Repeat (STR) Analysis
Once the amplification of the original DNA through PCR has occurred, analysis of the DNA fingerprint
can begin. Although estimates of the differences in DNA between individuals are very small (~ 1/10 of
one percent), the sheer volume of DNA an individual possesses results in about 3 million bases pairs of
unique sequence (i.e., each person differs by about 3 million DNA base pairs). The analysis of short
tandem repeats examines some of this individual variation.
Humans, like other eukaryotes, contain interruptions in the protein-coding genes, called introns.
Because introns do not contain protein synthesis information, the base sequences in these regions tend
to be repetitive. The same sequence of four, five, or six bases repeats itself over and over again. For
example, intron 3 of the human α fibrinogen (a blood clotting protein) gene contains a sequence of
bases “TTTC”, which repeats:
TTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTC
Because these repetitive sequences are short (4-6 bases) and occur side-by-side (in tandem) they are
termed short tande repeats (STR’s). The objective of DNA fingerprinting is to determine how many
times a sequence of an STR is repeated in a DNA sample.
How many times does the STR “TTTC” repeat itself in the above DNA sequence? _____
For intron 3 in the human α fibrinogen gene, an individual has between a 5 and 20% chance of sharing
the same number of repeats with another individual. Although that seems like a relatively low chance, it
must be remembered that the human population is quite large, so there would still be quite a few
people in this group. Therefore, knowing the number of STR’s for just one particular intron is not
sufficient enough to develop a unique DNA fingerprint. To achieve that, more than one STR must be
examined.
For example, intron 1 of the human tyrosine hydroxylase gene repeats the sequence “AATG”. Like the
human α fibrinogen gene, only about 5-20% of people will repeat the “AATG” sequence the same
number of times. Combining both these STR percentages narrows the field considerably.
As an illustration, assume that 1 out of every 20 people (1/20 or 5%) repeats the intron 3 of the human
α fibrinogen gene STR (“TTTC”) 14 times and that 1 out of every 5 people (1/5 or 20%) repeats the intron
1 of the human tyrosine hydroxylase gene STR (“AATG”) 10 times. The chances that two individuals will
share the same number of repeats for both STR’s would be calculated as such:
Chance of sharing the same number of repeats in both STR’s = chance of sharing the same
number of repeats in the first STR x chance of sharing the same number of repeats in the
second STR
For this example:
Chance of sharing the same number of repeats in both STR’s = 1/20 x 1/20 = 1/400
So, an individual has a 1 in 400 chance (0.25%) of sharing the same number of repeats in both STR’s.
The more STR’s examined, the more that chance decreases until, within statistical certainty, a unique
DNA fingerprint is assembled.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
84
Exercise 10 – DNA Fingerprinting
Since 1997 the Federal Bureau of Investigation (FBI) has set standards for DNA fingerprinting analysis for
forensic and law enforcement purposes. To meet those standards, 13 specific genes areas (loci; singular
locus) are evaluated. These loci are found on autosomes (non-sex chromosomes). A 14th locus is used
to determine the sex of the individual and measures STR’s on the X and Y chromosomes. Requiring a
match of at least 14 loci virtually guarantees a conclusive identification as the odds of two, unrelated
person matching all 14 loci is approximately 1 in 1,000,000,000,000,000 (1 quadrillion).
After PCR, the scientist or technician must find out how many repeats occur at each loci examined.
There are number of ways this is accomplished, but most methods compare the DNA molecules
produced by PCR to DNA fragments of known lengths. DNA fragments are created by cutting the DNA
sample with special molecules known as restriction enzymes.
Part C: Restriction Enzymes
The preparation of a sample of DNA for fingerprinting involves treating the DNA with specific restriction
enzymes (endonucleases). Restriction enzymes cut both strands of DNA at specific locations known as
recognition sites (Table 10.1). The original intent of these enzymes is to provide protection for bacteria
against some invading bacteriophages. Bacteriophages are viruses that attack bacteria. Cutting DNA
with restriction enzymes results in smaller DNA fragments (restriction fragments) of various base pair
lengths.
Table 10.1 Recognition Sites for Various Restriction Enzymes
Restriction Enzyme
BstEII
EcoRI
HindIII
Recognition Site
---GGTNACC-----GGTNACC-----CCANTGG-----CCANTGG---
---GAATTC-----GAATTC-----CTTAAG-----CTTAAG---
---AAGCTT-----AAGCTT-----TTCGAA-----TTCGAA---
Digested DNA
Fragments
---GXXXXGTNACC-----CCANTGXXXXG--(N = any nucleotide)
---GXXXXAATTC-----CTTAAXXXXG---
---AXXXXAGCTT-----TTCGAXXXXA---
A DNA Restriction Map provides the exact locations of the recognition sites for a particular restriction
enzyme on a DNA sample. The map indicates distances from the origin where the enzyme cut the DNA.
Size of each DNA fragment produced by restriction enzyme treatment can be measured in base pair (bp)
units. Figure 10.2 illustrates a map of a DNA molecule of 50,000 bp in length. This DNA molecule has
been treated with three different restriction enzymes.
Procedure
1. Examine Fig. 10.2 and notice that each enzyme (A, B, and C) cuts the DNA sample at different
recognition sites (for example, enzyme A cuts the DNA at 15,010 bp, again at 24,650 bp, and a
third time at 30,003 bp)
2. Calculate the base pair (bp) size for each DNA fragment and write that number above the
fragment
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
85
Exercise 10 – DNA Fingerprinting
Fig. 10.2 Restriction Map of 50,000 Base Pair DNA Sample
uncut DNA
enzyme A
enzyme B
enzyme C
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
86
Exercise 10 – DNA Fingerprinting
Part D: Gel Electrophoresis and Fingerprint Analysis
Electrophoresis means to “carry with an electric current.” The digested DNA fragments created by the
restriction enzymes are loaded into an agarose gel in which an electric current will flow through.
Because DNA is a negatively charged molecule (due to the presence of large numbers of phosphate
groups in its backbone), it will migrate through the gel towards the positive pole of the electrophoresis
chamber. This procedure will separate the digested DNA fragments based upon molecular size. Smaller
fragments will migrate further through the gel than larger ones. The different fragments can then be
stained to make them visible. The resultant pattern of stained DNA “bands” or fragments constitutes a
DNA fingerprint (see Fig. 10.3 as example).
Fig. 10.3 Sample DNA Fingerprint
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
87
Exercise 10 – DNA Fingerprinting
Part D1: Restriction Maps
Procedure
1. Compare the Restriction Map in Fig. 10.2 to the DNA fingerprint in Fig. 10.4. Does the number
of bands or fragment created by each restriction enzyme in the fingerprint coincide with the
number of fragments illustrated in the Restriction Map? __________
2. Label each of the DNA bands in Fig. 10.4 with its actual size in base pairs. Remember, in
electrophoresis fragments are separated upon size, with smaller fragments moving farther from
the wells at the negative end of the chamber than larger ones
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
88
Exercise 10 – DNA Fingerprinting
Fig. 10.4 DNA Fingerprint of 50,000 Base Pair DNA Sample
uncut
DNA
enzyme
A
-
enzyme
B
enzyme
C
+
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
89
Exercise 10 – DNA Fingerprinting
Part D2: Using a DNA Fingerprint to Determine the Number of STR’s
The examination of banding patterns in gel electrophoresis can be used to determine the number of
STR’s in a given DNA sample. Intron 3 of human α fibrinogen and intron 1 of the human tyrosine
hydroxylase genes may result in a banding pattern such as Fig. 10.5.
Fig 10.5 STR Banding Patterns for Two Human Genes
α fibrinogen locus
-
tyrosine hydroxylase locus
size markers
72 bp
68 bp
64 bp
60 bp
56 bp
52 bp
48 bp
44 bp
40 bp
+
Why is there only one band for the human α fibrinogen loci? (recall an organism can be homozygous or
heterozygous for a particular gene)
To determine the nu ber of STR’s for each band, divide the size of the DNA fragment by the number of
bases that repeat (usually 4-6) in that particular STR. For example, a DNA fragment of an STR which
repeats the 5 bases “AGGTA” that is 55 base pairs long would repeats that STR 55/5 or 11 times.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
90
Exercise 10 – DNA Fingerprinting
For each allele in Fig. 10.5, how many times does the STR in intron 3 of the human α fibrinogen gene
repeat?
_____ _____
For each allele in Fig. 10.5, how many times does the STR in intron 1 of the human tyrosine hydroxylase
gene repeat? _____ _____
An example of a DNA fingerprint from 14 different loci is presented in Table 10.2.
Table 10.2 Sample DNA Fingerprint from 14 Loci
Locus
D3S1358
vWA
FGA
Genotype
15, 18
16, 16
19, 24
Locus
Genotype
D13S17
10, 11
D7S820
10, 10
D8S1179
12, 12
D21S11
29, 31
D18S51
12, 13
D5S818
11, 11
THO1
9, 10
TPOX
8, 8
CSF1PO
11, 12
AMEL
X, Y
D16S539
11, 11
Is the person with this DNA fingerprint male or female? __________ {10.9}
Which loci are heterozygous?
Which loci are homozygous?
Procedure
1. Assume the DNA fingerprint in Table 10.2 above comes from a blood sample collected at a crime
scene. Law enforcement authorities currently hold three suspects. A sample of DNA has been
collected from each suspect and their DNA fingerprints for the loci in Table 10.2 have been
determined (Table 10.3)
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
91
Exercise 10 – DNA Fingerprinting
Table 10.3 DNA Fingerprints for Three Suspects
Suspect “A”
Locus
D3S1358
vWA
FGA
Genotype
15, 18
16, 17
20, 24
D8S1179
12, 14
D21S11
29, 31
D18S51
10, 12
D5S818
11, 13
Locus
Genotype
D13S17
12, 12
D7S820
13, 13
D16S539
11, 12
THO1
9, 9
TPOX
7, 8
CSF1PO
12, 12
AMEL
X, Y
Suspect “B”
Locus
Genotype
D3S1358
15, 15
vWA
16, 16
FGA
19, 25
D8S1179
12, 12
D21S11
30, 31
D18S51
10, 11
D5S818
11, 13
Locus
Genotype
D13S17
12, 13
D7S820
9, 10
D16S539
11, 11
THO1
9, 10
TPOX
9, 9
CSF1PO
11, 14
AMEL
X, Y
Suspect “C”
Locus
Genotype
D3S1358
15, 18
vWA
16, 16
FGA
19, 24
D8S1179
12, 12
D21S11
29, 31
D18S51
12, 13
D5S818
11, 11
Locus
Genotype
D13S17
10, 11
D7S820
10, 10
D16S539
11, 11
THO1
9, 10
TPOX
8, 8
CSF1PO
11, 12
AMEL
X, Y
Based on these data, which of these suspects is the most likely culprit? __________
How would the answer change if it was found out Suspect “C” had recently been the recipient of a bone
marrow transplant? (hint: what are the functions of bone marrow?)
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
92
Exercise 10 – DNA Fingerprinting
Practice Problems and Review Questions
1. Define or describe the following:
DNA
STR
PCR
locus (loci)
annealing
2. In DNA, __________ binds to thymine, guanine binds to __________ through __________
bonds.
3. How many loci are required by the FBI to positively identify an individual? Is it reasonably
possible for two, non-related persons to share the same genotypes at all these loci? Why or
why not?
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
93
Exercise 10 – DNA Fingerprinting
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
onon
stain
bloodblood
stain
victim’s
clothes
clothes
victim’s
victim
victim
size
size markers
markers
2
suspect
suspect
1
suspect
suspect
4. The following set of DNA fingerprints is from an actual crime scene. It contains a set of size
markers and banding patterns from blood samples collected from the victim, two suspects, and
blood found on the victim’s clothing. From what is shown here, determine which suspect’s
blood more likely matches the blood found on the victim’s clothes. Justify your decision.
94
Exercise 10 – DNA Fingerprinting
5. DNA fingerprinting can also be used in issues of family relationships. Since each person receives
half their chromosomes from each parent, half of a person’s DNA fingerprint should match the
mother and half should match the father.
Consider the following DNA fingerprints as evidence in a paternity case. Is it reasonably possible
the male in question is the baby’s biological father? __________
How would the answer change if the male in question had an identical twin?
Mother
Locus
Genotype
D3S1358
13, 18
vWA
15, 15
FGA
18, 24
D8S1179
12, 12
D21S11
28, 31
D18S51
13, 13
D5S818
11, 11
Locus
Genotype
D13S17
10, 10
D7S820
11, 11
D16S539
10, 11
THO1
9, 9
TPOX
10, 11
CSF1PO
11, 11
AMEL
X, X
Baby
Locus
Genotype
D3S1358
15, 18
vWA
15, 16
FGA
19, 24
D8S1179
12, 12
D21S11
29, 31
D18S51
12, 13
D5S818
11, 11
Locus
Genotype
D13S17
10, 11
D7S820
10, 11
D16S539
11, 11
THO1
9, 10
TPOX
8, 10
CSF1PO
11, 12
AMEL
X, X
Alleged Father
Locus
D3S1358
Genotype
13, 15
vWA
16, 16
FGA
18, 19
D8S1179
12, 14
D21S11
29, 30
D18S51
12, 12
D5S818
10, 11
D7S820
10, 10
D16S539
11, 11
THO1
10, 10
TPOX
8, 8
CSF1PO
11, 12
AMEL
X, Y
Locus
Genotype
D13S17
11, 12
6. Calculate the number of repeats in the DNA fragments for the STR’s given.
a.
STR: “AAGCTA”
DNA fragment length: 72 bp
# of repeats: _____
b.
STR: “TTAT”
DNA fragment length: 52 bp
# of repeats: _____
c.
STR: “TAGGG”
DNA fragment length: 105 bp
# of repeats: _____
d.
STR: “AAGCT”
DNA fragment length: 60 bp
# of repeats: _____
e.
STR: “GAGGCT”
DNA fragment length: 144 bp
# of repeats: _____
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
95
Exercise 11 - Genetics
Introduction
Many features or characteristics of plants and animals are acquired during its lifetime. For example, a
person may get a tattoo and although the mark is, for all intents and purposes, permanent on their skin
their children will not be born with tattoos. Other characteristics, however, are inheritable and can be
passed from one generation to the next. These inheritable characteristics are encoded in DNA in units
called genes.
Genes may exist as alternative forms known as alleles. An allele of a gene is nothing more than a gene
choice (blue or brown for eye color, unattached or attached for earlobes, widow’s peak or no widow’s
peak, etc.). Some alleles are dominant because, when present, they control the appearance or
phenotype of the organism. Other alleles are recessive and only phenotypically expressed when no
dominant allele for that gene is present.
Diploid organisms receive one set of chromosomes each from the maternal (female) and the paternal
(father) parent. If the alleles for a given gene inherited from the parents are the same, the organism is
considered homozygous for that trait and heterozygous if the alleles are different. The actual
combination of alleles inherited by the offspring for a trait is known as its genotype.
This laboratory exercise will investigate how dominant and recessive alleles affect an organism and
demonstrate how these effects occur in a predictable pattern.
Materials
Equipment
ears of purple : yellow genetic corn
straight pins
PTC taste paper
color vision diagnostic charts
Part A: Monohybrid Cross in Corn (Zea mays)
Corn kernels are actually the fruit of the corn plant, each containing an individual corn embryo. Around
the embryo are a number of structures providing nourishment and protection. Between the endosperm
(where starch is stored) and the pericarp (covering of the kernel) is a layer of cells called the aleurone.
The color of the aleurone is controlled by several genes. One gene produces a purple aleurone, the
other the yellow kernels with which we are most familiar. The purple allele is dominant, the yellow
allele being recessive. The dominant purple allele can be symbolized by “R” while “r” represents the
recessive yellow allele. Therefore, corn kernels with the homozygous genotype “RR” are purple as are
those kernels who are heterozygous (“Rr”). Only kernels homozygous for the yellow allele (“rr”) will
express the recessive phenotype, yellow.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
96
Exercise 11 – Genetics
Procedure
1. Obtain an ear of genetic corn. Notice it contains a mixture of purple and yellow kernels. This
ear of corn represents the F2 generation from the following cross:
Rr
x
Rr
RR, Rr, rr
The parents (“Rr” and “Rr”) are the F1 generation and were obtained from the P generation
cross of a homozygous dominant (“RR”) and recessive (“rr”) individual. The offspring (“RR, ‘Rr”,
“rr”) are then the F2 generation. This cross results in a mixture of phenotypes in the F2
generation. Most of the kernels are purple, a fewer number are yellow. Punnett square
diagrams are used to diagram genetic crosses. Using this mating, fill in the Punnett square in
Table 11.1
Table 11.1 Monohybrid Cross Punnett Square for Kernel Color in Corn (Zea mays)
F1 gametes
R
r
genotype __________
genotype __________
phenotype __________
phenotype __________
genotype __________
genotype __________
phenotype __________
phenotype __________
R
r
Based upon the Punnett square in Table 11.1, about what proportion of the kernels in the ear of
corns should be purple? _____
What proportion should be yellow? _____
2. Place a straight pin at the end of the row being counted to serve as a place holder. Count the
number of purple kernels and yellow kernels in one row only. Enter data in Table 11.2
Table 11.2 Kernel Counts in F2 Generation Corn (Zea mays) – One Row
# of kernels
Number
Proportion*
purple
yellow
total
*divide the number of kernels of each color by the total number of kernels in that row
Is the proportion of purple to yellow kernels close to the predicted 3:1 (75% purple, 25% yellow)
ratio? __________
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
97
Exercise 11 – Genetics
While the numbers might be close (and even if they are not) to the expected phenotypic ratio of
3:1, statistical theory says the bigger the sample size the closer our actual data should fit
theoretical predictions.
3. Increase the sample size for this dataset by counting all rows in the ear of corn and fill in Table
11.3
Table 11.3 Kernel Counts in F2 Generation Corn (Zea mays) – All Rows
# of kernels
Number
Proportion*
purple
yellow
total
*divide the number of kernels of each color by the total number of kernels in that row
Did counting more kernels make the numbers come more closely to the expected 3:1 ratio of
purple to yellow color? __________
Part B: Phenylthiocarbimide (PTC) Taste Test
Phenylthiocarbimide is an organic compound which interacts with the taste buds on the tongue and
mouth of some people but not in others. Therefore, some people can taste PTC and are considered
“tasters.” Other people cannot taste PTC and are considered “non-tasters.” Being able to taste is the
dominant condition (“T”), non-tasters the recessive (“t”).
Procedure
1. Obtain a piece of paper treated with a small amount of PTC and place it on the tongue. Tasters
will sense a very strong flavor, non-tasters will taste only “paper”
2. Collect results on the number of tasters and non-tasters from the population in class and use
that data to fill in Table 11.4
Table 11.4 PTC Tasting
Phenotypes
Number
Proportion*
taster
non-taster
total
*divide the number of phenotypes by the total class size
What is the genotype of a taster? __________
What is the genotype of a non-taster? __________
What is the predicted phenotypic ratio for a cross between two heterozygous (Tt) individuals?
__________
Were the resultant proportions close to this ratio? __________
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
98
Exercise 11 – Genetics
If the proportions were not close (and even if they were), provide an explanation as to why they
would be (hint: sample size).
Part C: Multiple Alleles – Human Blood Types
Some genes have more than two alleles. Although each organism can only have two copies of a gene
(why? _________________), within an entire population there may be several alleles. One example of
this is seen with the ABO blood types in human beings. Blood types are produced by multiple alleles.
On the membranes of red blood cells are proteins which can stimulate an immune response. These
proteins are called antigens. The dominant allele “IA” produces the A antigen protein, the dominant
allele “IB” produces the B antigen protein. The recessive allele “i” produces no protein. When a genetic
trait has more than one dominant allele, a situation of codominance may exist. The genetic trait for
human ABO blood typing contains three alleles (IA, IB, i).
Procedure
1. Fill in Table 11.5 with the possible genotypes for the listed phenotypes
Table 11.5 Human ABO Blood Typing Genotypes and Phenotypes
Genotype(s)
_____ _____ or _____ _____
Phenotype
A
_____ _____ or _____ _____
B
_____ _____
AB
_____ _____
O
Part D: X-Linked Characteristics
In humans, all somatic cells (typical body cells) contain 23 pairs of chromosomes. Of these, 22 pairs are
autosomes, the last pair are the sex chromosomes. The sex chromosomes are related to the gender of
the individual and are called X and Y. Women have two X chromosomes, men have an X and a Y.
Haploid sex cells (gametes) produced by women (ova) have only the X chromosome, male gametes
(sperm) have either X or Y. Sex of the offspring is determined by the male.
The X chromosome is large and carries many genes such as those for essential muscle proteins and
retinal pigments. The Y chromosome, on the other hand, is quite small and carries only a few genes,
mostly related to male gender development. A defective gene on the X chromosome will be
phenotypically expressed in a male because there is no other X chromosome to compensate. However,
a woman with the same defective gene will not express it phenotypically if her other X chromosome is
normal. She will be a carrier though in that she has the defective gene but does not express it.
Red-green colorblindness is caused by a mutation in a gene for retinal pigments on the X chromosome.
The defective allele is recessive to the normal one so a woman with one normal X chromosome (X) and
one colorblind carrying chromosome (XC) will have normal color vision because the X chromosome is
dominant for this trait over the XC chromosome. Men, however will be colorblind if they possess the XC
chromosome since there is no other X to be dominate over it and will exhibit red-green colorblindness.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
99
Exercise 11 – Genetics
Procedure
1. Obtain a color vision diagnostic chart
2. Test for colorblindness in yourself and among lab partners
What is your gender? __________ (if you don’t know, excuse yourself to the restroom to find
out)
What is your sex chromosome genotype? __________
Do you exhibit red-green colorblindness? __________
Do any lab partners exhibit colorblindness? __________
If so, what are their sexes and genotypes? ____________________
Part E: Dihybrid Crosses
Dihybrid crosses are used to examine the inheritance patterns in more than one gene (di = two). As an
example, consider fur color and texture in guinea pigs. Among these organisms, black fur color is
dominant (“B”) over white (“b”) and rough fur coat is dominant (“R”) over smooth (“r”).
Procedure
Use a Punnett Square to answer the questions regarding the offspring in the F1 generation from the
P generation crosses indicated
1.
BBRR
x
bbrr
What will the genotype be of all of the offspring from this cross?
What will be the phenotypes of all of the offspring from this cross?
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
100
Exercise 11 – Genetics
2.
Bbrr
x
bbRr
What proportion of the guinea pigs will be
3.
black, rough? __________
black, smooth? __________
white, rough? __________
white, smooth? __________
BbRr
x
bbRr
What proportion of the guinea pigs will be
black, rough? __________
black, smooth? __________
white, rough? __________
white, smooth? __________
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
101
Exercise 11 – Genetics
4.
BbRr
x
BbRr
What proportion of the guinea pigs will be
black, rough? __________
black, smooth? __________
white, rough? __________
white, smooth? __________
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
102
Exercise 11 – Genetics
Practice Problems and Review Questions
1. Define or describe the following:
monohybrid cross
dihybrid cross
testcross
allele
dominant allele
recessive allele
genotype
phenotype
homozygous
heterozygous
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
103
Exercise 11 – Genetics
2. What are the expected phenotypic and genotypic ratios in a monohybrid cross between two
heterozygous individuals?
3. What is the expected phenotypic ratio in a dihybrid cross between two organisms that are
heterozygous for both traits?
4. List all the possible types of gametes that can be produced from an organism with the genotype
AaBb.
5. Humans as well as many other mammals show a genetic condition known as albinism. A
recessive allele interferes with the ability to produce the brown pigment melanin which colors
eyes and hair in addition to protecting the skin from the harmful effects of UV light. People who
are homozygous recessive produce little or no melanin and have very pale eyes, white skin, and
yellow or white hair. Normal pigmentation is produced by a dominant allele (“A”). The albino
allele is recessive (“a”). Use Punnett Squares to calculate genotypic and phenotypic proportions
for the parental generation crosses indicated and denote which F1 genotypes (if any) will be
carriers.
a. AA x aa
b.
Aa x aa
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
104
Exercise 11 – Genetics
c.
Aa x Aa
d.
AA x Aa
e.
aa x aa
6. A woman with normal pigmentation, whose mother was an albino, mates with an albino. What
is the probability their first child will be an albino?
7. Normal parents have an albino child. Give the genotypes for the parents and the child.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
105
Exercise 11 – Genetics
8. A person is a taster as is their mother. The father is a non-taster. What is this person’s
genotype?
9. A non-taster has two taster parents. What is the genotype of this person and their parents?
10. For ABO blood typing in humans, for which phenotypes is the genotype definitive?
11. A person has blood type O. Their mother is B and father is A. What is the genotype of this
person and their parents?
12. A paternity suit involves a child whose blood type is AB. The mother is blood type B, the alleged
father is O. Make a ruling on this case as to whether it is reasonably possible this is the
biological father.
13. Is it possible for a female human to be colorblind? What would her genotype be?
14. Can two normal vision parents produce a colorblind son? Explain with a diagram.
15. Can two normal vision parents produce a colorblind daughter? Explain with a diagram.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
106
Exercise 11 – Genetics
16. Can two colorblind parents, produce a normal vision son? Explain with a diagram.
17. A normal vision woman, whose father was colorblind, mates with a colorblind man. What
proportion of their sons would be colorblind? What proportion of their daughters would be
colorblind and what proportion would be carriers? If one of their normal vision sons mates with
a homozygous, normal vision woman, would it be possible for them to have a colorblind child?
18. A guinea pig that is heterozygous for fur color and texture mates with another guinea pig
heterozygous for both traits. They produce a total of 96 offspring. Use a Punnett Square to
diagram this cross then answer the questions below.
How many of the 96 offspring will phenotypically be
black with rough fur? __________
black with smooth fur? __________
white with rough fur? __________
white with smooth fur? __________
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
107
Answers
Answers to Lab Exercise Practice Problems and Review Questions
Exercise 1
#1
Surface area: 124.9 cm2
Volume: 78.4 cm3
Density: 1.5 g/cm3
Sinks
#2
Surface area: 31.2 cm2
Volume: 9.8 cm3
Density: 0.9 g/cm3
Floats
#3
0.26 L = 260 ml
Volume = 252 ml (cm3)
Density = 0.2 g/cm3
Floats
#4
Squirrels are homeothermic (warm-blooded) mammals. In northern latitudes, one of the more obvious
environmental pressures is the cold during the winter months. For southern latitudes, the opposite is
true. Staying cool in the hot summer months is most important there. So, squirrels in the north should
evolve adaptations that would provide for the retention of heat while their southern counterparts
would find ways to stay cool in the summer. Recall the change in surface area to volume ratio observed
in the block of various sizes. The highest ratio was in the smallest block. This means the smallest block
had more surface area exposed to the environment per unit of volume. When applied to animals this
means the smaller ones expose more of their skin (surface area) to the environment (hot, cold, wet,
etc.) per unit of body mass than larger animals. In homeotherms like mammals, this means smaller
animals lose heat from their body through their skin faster than larger ones. This would be beneficial in
the southern squirrels while being able to hold onto body heat longer would allow northern versions to
survive the cold winters.
Exercise 2
#1
a. Stock: 0.5 ml
b. Stock: 5.0 ml
c. Stock: 2.5 ml
DH2O: 9.5 ml
DH2O: 5.0 ml
DH2O: 7.5 ml
a. Stock: 8.0 ml
b. Stock: 2.0 ml
c. Stock: 12.0 ml
DH2O: 12.0 ml
DH2O: 18.0 ml
DH2O: 8.0 ml
#2
#3
Stock: 04.5 ml
DH2O: 4.55 ml
#4
Stock: 2.8 ml
DH2O: 2.2 ml
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
108
Answers
#5
Stock: 0.04 ml
DH2O: 4.96 ml
#6
Stock: 18.75 ml
DH2O: 6.25 ml
#7
Stock: 7500 ml
DH2O: 1500 ml
#8
Stock: 0.24 L
DH2O: 0.36 L
#9
50 ml
#10
Stock: 10 ml
DH2O: 40 ml
#11
Buffers are substances that absorb H+ when they are in excess which happens when the environment is
too acidic and release H+ when the environment is too basic. They work to stabilize pH which is crucial
for living organisms in order to maintain homeostasis. For example, excessively acidic or basic
conditions in the body may cause enzyme denaturation and a resultant loss of physiological function.
#12
Glucose – aldehyde and hydroxyl (linear), hydroxyl (ring)
Fructose – ketone and hydroxyl (linear), hydroxyl (ring)
Glycine – amine, carboxylic acid
Glycerol - hydroxyl
#13 (examples, not necessarily just these)
Glucose – starch, maltose, glycogen
Fructose – sucrose
Glycine – proteins
Glycerol – triglycerides, phospholipids
Exercise 3
#1
In this lab, qualitative tests are used to simply detect the absence or presence of a substance. There is
no particular indication of amount just a "yes" or "no" to whether the substance exists or not. For
example, a positive test for starch using IKI only tells you starch is present, but not how much starch. A
quantitative test determines the actual amount of something present. If wanted to know how much
starch was present in a solution we would use a test that would return a value (for example, 3 mg of
starch per liter).
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
109
Answers
#2 (a & b)
Since it can reasonably be assumed that distilled water does not contain lipids or proteins it makes a
perfect control. Notice it is distilled water and not tap water. Tap water contains many other things and
would not make a good control. Distilled water is just water, nothing else.
#3
Test Substance
Reagent
Test Procedure
Color of Positive
Result
Color of Negative
Result
Starch
IKI
1 drop
dark blue / black
Amber
Sugar
Benedict’s
1 ml
red, orange,
green, yellow
Blue
Lipid
Sudan IV
Soak 3 mins
pink / red
no color or very
light pink
violet
no color change
1 ml NaOH,
Protein
Biuret
0.5 ml reagent
#4
The IKI test must be performed first. If a Benedict's is done first, the heat will hydrolyze any starch
present and it will go undetected.
#5
They differ in number of monomers. Monosaccharides, like glucose, consist of one monomer. Two
saccharide monomers (for example sucrose which is made up of a glucose and a fructose
monosaccharide), would be a disaccharide. Oligosaccharides are typically three to ten monosaccharides
long. Polysaccharides are typically >10 monosaccharides (for example, starch and glycogen).
#6
Glycerol and palmitic (fatty acids)
#7
Amino acids
#8
Primary - order of the amino acids in the protein. Secondary - a protein's "orientation in space"; evident
as either an α helix (spiral) or ß sheet. Tertiary - protein folding back on itself due to chemical
interactions among the amino acid side (or "R") groups. Quaternary - occurs when two or more
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
110
Answers
polypeptide chains associate (for example, hemoglobin which is made up of four polypeptide chains
bonded together). Proteins also exhibit further complexities of structure and form motifs (for example,
ß-α-ß motif) and domains which are functional units of a protein.
#9
They differ in their particular arrangement of their amino acids. This, of course, affects all other levels
(secondary, tertiary, quaternary, motifs, domains). Organisms ingest proteins, hydrolyze them into
amino acids and then re-assemble those amino acids into organism specific proteins.
#10
DNA, RNA, ATP all contain nitrogen
#11
The word "hydrolysis" literally means "water" ("hydro"), "splitting" ("lysis"). The opposite of this is when
organic molecules are combined to form polymers. This process is called dehydration synthesis because,
in this process, water is lost.
Exercise 4
#1
900x
#2
100x
#3
0.2 mm
#4
750 µ
#5
Length: 125 µ
Width: 37.5 µ
#6
6 mm
#7
Length: 125 µ
Width: 50 µ
#8
250x
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
111
Answers
Exercise 5
#1
33.0%
2
4.7%
#3
-11.1%
#4
-17.0%
#5
Since the bag lost weight, water must have flowed from inside of the bag to the beaker. Water always
flows to the side of the membrane with the greatest solute concentration, which means the solution in
the beaker was hypertonic to the solution inside the bag.
#6
Prokaryotic cells are members of the Domains Archaea and Bacteria do not have a nucleus, are much
less complex than eukaryotic cells, and possess only a single chromosome. Eukaryotic cells (all other
kingdoms) have a nucleus, exhibit much greater complexity of structure and have multiple
chromosomes.
Plant cells have a cell wall, chloroplasts, and typically a large central vacuole typically used to store
water. They also do not have centrioles. Animal cells have no cell wall or chloroplast and, if they have
it, their vacuoles are much smaller. Animal cells have centrioles.
Protists differ from animal cells in that they often possess a cilia or flagella (although some animal cells
have a flagella - sperm), and are the complete organism (that is, protozoans are single celled creatures).
#7
A semi-permeable membrane is one that allows only certain substances to pass through it. Cell
membranes are semi-permeable as they regulate what passes through them letting some things through
while excluding others. The dialysis bag was an example of a semi-permeable membrane because it only
allowed items to pass that could fit through the pores in the tubing.
#8
Hypertonic: environments that contain a higher concentration of solutes when compared to the
environment on the other side of the membrane. Water always flows to the side with the greater
concentration of solutes; therefore cells in hypertonic environments would lose water.
Hypotonic: environments with a lesser concentration of solutes when compared to the environment on
the other side of the membrane. In this instance, cells in these environments would gain water.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
112
Answers
Isotonic: condition of equal solute concentration across the membrane and although water flows back
and forth across the membrane it does so equally. Cells in isotonic conditions exhibit neither a net loss
nor gain of water.
#9
When two aqueous solutions are separated by a semi-permeable membrane, the net water movement
is always from a hypotonic to a hypertonic solution.
Exercise 6
#1
An organic catalyst is an organic molecule (compounds that contain carbon) that “spurs on” or helps to
promote a chemical reaction. Enzymes are examples of organic catalysts. Enzymes enter reactions the
same way they them which is beneficial because enzymes are metabolically "expensive" molecules and
nature would favor them being used over and over again. Of course, like any system they eventually
suffer from entropy and become useless.
#2
Enzymatic reactions are affected by
(1) substrate concentration - increased substrate concentration give the enzyme more to work
on so reaction rate increases
(2) temperature - temperature increases reaction rate until the enzyme has been denatured
(3) pH – an environmental pH different from an enzyme's optimum will cause denaturation and
a subsequent decrease in reaction rate
#3
Low temperatures exhibit low enzymatic activity levels because the molecules (enzyme and substrate)
are moving slowly which decreases the probability they will contact one another. Higher temperatures
and a pH that is substantially different from the enzyme's optimum will cause enzyme denaturation.
Denaturation affects the shape of the enzyme and its active site. A change in the shape of an enzyme's
active site will not allow a substrate to bind and if no substrate can bind to the enzyme then no product
can be produced.
#4
The enzymatic reaction in this lab degrades a more complex molecule (H2O2) into more simple ones (H2O
and O2). This is an example of a catabolic reaction. Catabolic reactions are exergonic and release energy.
#5
E + S  ES complex  E + P
#6
The rate of enzymatic reaction is directly proportional to substrate concentration.
At optimum, enzymatic reaction rate is greatest.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
113
Answers
Exercise 7
#1
C6H12O6 + O2  CO2 + H2O + ATP
#2
Aerobic - aerobic respiration occurs when the final electron acceptor is oxygen. Aerobic respiration is
respiration with oxygen.
Anaerobic - anaerobic respiration is when the final electron acceptor is something other than oxygen.
When this final electron acceptor is an organic molecule (e.g. acetaldehyde) it is a type of anaerobic
respiration known as fermentation.
NADH, FADH2 - nicotinamide adenine dinucleiotide (NADH) and flavin adenine dinucleotide (FADH2) are
electron carriers. Through a series of oxidation and reduction reactions, these molecule shuttle
electrons throughout the respiration processes.
ATP – adenosine triphosphate is the energy currency created in respiration and used by cells for cellular
work. It is created by using the energy released in respiration to add an inorganic phosphate (Pi) to
adenosine diphosphate (ADP).
#3
Glycolysis (total): 4
Glycolysis (net): 2
Krebs cycle: 2
Aerobic respiration (eukaryotes): 36
Anaerobic respiration (net): 2
#4
Alcoholic beverages are produced by yeast cells undergoing respiration in the absence of oxygen. This
type of anaerobic respiration utilizes acetaldehyde as a final electron acceptor (fermentation). During
this process CO2 is also created. It is the accumulation of this carbon dioxide gas that gives these wines
their "sparkle" or fizziness.
#5
There is a direct relationship between the amount of heat in an environment and the speed of chemical
reactions. As temperature increases, the rate of reaction increases. Homeothermic animals are "warm
blooded" and keep their body at a constant temperature regardless of environmental temperatures.
This constant warmth allows chemical reactions to occur at elevated rates in spite of the temperature in
the environment. On the other hand, the metabolic rate of “cold blooded” (poikilothermic) creatures is
dependent upon the temperature of the environment. Since they are unable to generate body heat in
colder temperature their metabolic rates slow down. Often times, poikilothermic animals use
"behavioral thermoregulation" to increase their body temperature and therefore the rate of their
physiological reactions by laying on warm surfaces (e.g. rocks, concrete, etc.). This also explains why
there are no amphibians in the Arctic or Antarctic.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
114
Answers
#6
Smaller mammals lose heat faster than larger mammals due to their increased surface area to volume
ratio. Just like in cells, as homeothermic organisms get smaller, their amount of surface area (essentially
their skin) increases in relation to their volume. Since organisms lose heat to the surrounding
environment through their skin, smaller "warm blooded" animals lose heat faster than larger ones. To
compensate, they have evolved increased metabolic rates.
Exercise 8
#1
CO2 + H20  C6H12O6 + O2 (this reaction requires an input of energy from sunlight)
#2
Light reactions - those occurring on the thylakoid membrane. These reactions strip the electrons and
protons (H+) from water, energize the electrons using sunlight and then use those energy rich electrons
to activate a proton pump in the cytochrome complex. The pumping of protons across the membrane
sets up an unbalance. To reach equilibrium, protons flow back through the membrane passing through
an ATP synthase protein generating ATP to be used in the dark reactions. The electrons originally
stripped from water are then transferred to NADP+ reducing it to NADPH. Like ATP, NADPH is then
transferred to the dark reactions.
Dark reactions - occur in the stroma of chloroplasts. In this reaction, atmospheric CO2 is reduced
through a series of reactions creating glucose. The dark reactions require the ATP as the source of
energy and electrons to complete this task.
NADPH – the electron carrier responsible for shuttling electrons from the light to the dark reactions to
be used in the reduction of CO2 to glucose.
Rubisco - (ribulose bisphosphate carboxylase/oxygenase), the enzyme that catalyzes the reaction which
uses carbon dioxide to form 3-phosphoglycerate (PGA) thereby initializing the Calvin cycle.
Absorption spectrum - describes the range and efficiency of light absorbed by a particular molecule. The
molecule chlorophyll's absorption spectrum shows two peaks. One occurring at the shorter
wavelengths, the other at longer wavelengths. Minimal absorption occurs among the wavelengths in
between (the greens and yellows).
#3
The dark reaction (in order) consist of
(1) carbon fixation - the enzyme rubisco uses RuBP and atmospheric CO2 to form PGA. It is
called "fixation" because it is in this step where CO2 is "captured" or "fixed."
(2) reduction - energy in the form of ATP and electrons from NADPH created in the light
reactions are used to reduce molecules creating glyceraldehyde 3-phosphate (G3P)
which is eventually converted into glucose.
(3) regeneration - ATP is used to create RuBP necessary for the next cycle round.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
115
Answers
#4
There were 2 peaks in the chlorophyll absorption spectrum occurring at the shorter wavelengths (blue,
purples) and longer wavelengths (red, orange). The least amount of absorption occurred in the
intermediate wavelengths (green, yellow). Since the green wavelengths of light are not absorbed by
plants, it would be this color they would reflect.
#5
Phenol red is yellow in the presence of an acid. The blowing of CO2 into the solution at the beginning of
the lab caused the formation of carbonic acid (H2CO3); hence the yellow color. Through time as the
Elodea photosynthesized it used the CO2 in the water. This decreased the amount of CO2 in the solution,
thereby decreasing the amount of carbonic acid, causing the pH to rise and the phenol red solution to
turn red.
#6
Photosynthesis consists of two sets of reactions. The ATP and NADPH created in the light reactions are
used to drive the dark reactions where CO2 is converted into sugar. During daylight, the abundance of
energy from the sun drives these reactions, but at night the absence of sun causes a slow-down to
occur. The result is that during the day, CO2 is being used by the dark reactions at a much more rapid
rate than at night. Because of this, the amount of CO2 would be greater at night than during the day.
Therefore, the pH in these aquatic systems would be more acidic at night due to the build-up of carbonic
acid formed from the CO2 and water. During the day, the rapid uptake of CO2 by the dark reactions
would remove it from the water and cause the pH to rise (i.e., the water would become less acidic.).
#7
The empty tube contained everything the tube with the Elodea did except for the one variable that was
being tested. In this case, that one variable was the Elodea itself. If the solution in the empty tube
changed color just like the tube with the Elodea then it could not be conclusively stated that is was the
Elodea alone that caused the change. A control is an experimental set up that is the same as the
treatment experimental set up but with only one variable different.
#8
Oxygen
Exercise 9
#1
Cytokinesis - process by which cell cytoplasm is divided between two daughter cells. In animals, it occurs
through cleavage furrowing, in plants through cell plate formation.
Spindle fiber - microtubule structures used to separate sister chromatids and/or homologous
chromosomes.
Synapsis - alignment and pairing of homologous chromosomes during prophase I of meiosis. This forms
a tetrad (4) of sister chromatids.
Homologous chromosomes - in diploid cells, these are the pairs of the same type of chromosomes. One
homologous chromosome is inherited from each parent.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
116
Answers
Chromosome vs. chromatid - a chromatid is a duplicated chromosome joined to its copy by a single
centromere.
#2
During metaphase in mitosis, sister chromatids line up on the middle of cell (cell equator). In metaphase
I of meiosis, homologous chromosomes (each made up of a pair of sister chromatids), line up on the cell
equator.
#3
52
#4
21
#5
Mitosis creates diploid cells from diploid cells and is used to replace old cells or to create new ones. The
process of growing in a multicellular organism involves the creation of new cells. Therefore, we would
see the most mitotic activity during the early developmental (fetal) stages of an organism.
#6
A chiasma is a structure that occurs during prophase I of meiosis and is formed when two chromatids
from homologous chromosomes cross over one another. During anaphase I as homologous
chromosomes separate, chiasma break, swapping genetic material. This swapping of genetic
information will eventually serve to form genetically non-identical gametes (sex cells - sperm, egg).
#7
Genetic variation is crucial for the survival of a species. Environmental conditions on Earth have changed
considerably throughout Earth's history. These different conditions exert a "pressure" on species to
survive. Those that survive best, exhibit this by leaving behind more offspring which perpetuates the
parent's genetics into the future. What constitutes which organism survives best cannot be predicted so
variability in a species means a particular set of genetic information, which has arisen simply through
chance, may impart an advantage to an organism allowing it to leave behind more offspring than its
counterparts.
Exercise 10
#1
DNA - deoxyribonucleic acid, genetic molecule found in all organisms. Genetic traits are coded in this
molecule via the sequence of base pairs.
STR - short tandem repeats, repeating sequences of base pairs found in the introns (noncoding portions)
of DNA of an individual. DNA fingerprinting strives to enumerate the number of times these sequences
repeat themselves yielding a genotype for that particular gene locus.
PCR - polymerase chain reaction, the process by which a DNA sample is copied over and over again in
order to create enough DNA to conduct an analysis.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
117
Answers
Locus (pl. loci) - the location of a particular gene on a chromosome.
Annealing - the complementary base-pairing of a DNA primer to a strand of DNA that is being copied.
Annealing is followed by extending when DNA polymerase starts at the primer and begins adding new
bases to the DNA molecule.
#2
In DNA, adenine binds to thymine, guanine binds to cytosine through hydrogen bonds.
#3
The FBI requires a minimum of 13 autosomal loci to confirm an identification. The odds of two people
sharing the same genotype at all 13 loci is, for all intents and purposes, practically impossible. Just using
three loci as an example will demonstrate. If the odds of two people sharing the same genotype for
locus #1 is 1/20, and 1/40 for locus #2, and 1/100 for locus #3, the total probability would be 1/20 x 1/40
x 1/100 = 1/80,000! That means the odds of two people sharing identical genotypes at 3 loci is 1 in 80
thousand. Increasing that up to 13 loci would decrease this probability considerably. Therefore, while it
is technically possible for two people to match at all 13 loci, the odds are definitely against it.
#4
Suspect #2's banding pattern is an identical match to the banding pattern from the blood found on the
victim's clothing. Although one of fragments in the victim's DNA matches a fragment from the blood
stain, the second band from the victim is nowhere near close. Therefore, suspect #2 is the most likely
candidate for a match.
#5
An offspring will receive one chromosome from each parent. Therefore, the genotype at each loci for
the baby would be a combination of the mom and dad's genotype. To determine if this man is the baby's
father, examine each loci independently. One genotype will come from the mother (obviously) . . . the
other should match the father. In this particular case, each fragment length making up the genotype at
each loci has a match in the mother and the father. At 13 loci, this makes the odds that this man is not
the father very unlikely . . . unless he has an identical twin.
#6
a.
b.
c.
d.
e.
12
13
21
12
24
Exercise 11
#1
Monohybrid cross - cross examining only one trait.
Dihybrid cross - cross examining two traits simultaneously.
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
118
Answers
Testcross - cross between an organism that exhibits the phenotypically dominant trait and one that is
phenotypically recessive. The goal is to determine whether the phenotypically dominant organisms is
true-breeding (i.e., homozygous) or not (i.e., heterozygous).
Allele - alternative form of a gene. For example, in Mendel's pea plants the genetic trait of flower color
have two alleles, purple and white.
Dominant allele - allele that is always phenotypically expressed.
Recessive allele - allele that is phenotypically expressed only when the organism is homozygous for
recessive alleles.
Genotype - actual genetic makeup of an organism. Organisms are homozygous dominant, heterozygous,
or homozygous recessive.
Phenotype - physical or physiological expression of a genetic trait.
Homozygous - when an organism contains identical alleles. Homozygous organisms can be homozygous
dominant when both alleles are the dominant ones or homozygous recessive when they are recessive.
Heterozygous - when an organism contains alleles that are different. Heterozygous individuals
phenotypically express the dominant trait. Recessive alleles are not expressed in heterozygous
individuals.
#2
phenotypic: 3:1
genotypic: 1:2:1
#3
9:3:3:1
#4
AB
Ab
aB
ab
#5
a.
A
A
a
Aa
Aa
a
Aa
Aa
no albinism; all heterozygous (carriers)
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
119
Answers
b.
A
a
a
Aa
aa
a
Aa
aa
½ albino; ½ heterozygous (carriers); ½ homozygous recessive
c.
A
a
A
AA
Aa
a
Aa
aa
¼ homozygous dominant; ½ heterozygous (carriers); ¼ albino
d.
A
A
A
AA
AA
a
Aa
Aa
½ homozygous dominant; ½ heterozygous (carriers)
e.
a
a
a
aa
aa
a
aa
aa
all albino
#6
50%
#7
Parents: Aa (carriers)
Child: aa (albino)
#8
Tt
#9
Parents: both Tt
Person: tt
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
120
Answers
#10
AB only
#11
Mother: IBi
Father: IAi
Person: ii
#12
Not possible
#13
Yes; XcXc
#14
Yes, if the mother is a carrier.
XY x XcX
Xc Y
#15
No, not possible.
XY x XcX
XcX, XX
or
XY x XX
XX
#16
No, this is not possible. If all parents are colorblind, there is no chance the son will inherit an X
chromosome that does not have the colorblind trait.
#17
In the first cross, the mother must be heterozygous (a carrier). She has to be this because her father was
colorblind and passed on the one X chromosome he had to her. Her normal X chromosome came from
her mother. She marries a man "just like dear old dad" meaning he is also colorblind. So, she's a carrier
and he's colorblind. Their sons have a 50/50 chance of getting a colorblind X chromosome because the
mother has one colorblind X and one normal X chromosome. For their daughters, half will be colorblind
having inherited the colorblind X chromosome from their father and, unluckily, the same kind of X
chromosome from their mother. Daughters also have a 50% chance of being a carrier. In the second
cross, since neither parent has a colorblind X chromosome, they cannot pass that on to any of their
children, male or female.
#18
BR
BR
BBRR
Br
BBRr
bR
BbRR
br
BbRr
black with rough fur (B_R_): 9/16 * 96 = 54
white with rough fur (bbR_): 3/16 * 96 = 18
Br
bR
BBRr
BbRR
BBrr
BbRr
BbRr
bbRR
Bbrr
bbRr
black with smooth fur (B_rr): 3/16 * 96 = 18
white with smooth fur (bbrr): 1/16 * 96 = 6
Lake-Sumter Community College, Leesburg Laboratory Manual for BSC 1010C
br
BbRr
Bbrr
bbRr
bbrr
121