Investigations in Biology and Chemistry II – Lab work
Name: _____________________________________
Point Value: 68
Date Due: ______________
Points for Completion: Name, Date, Title, Purpose, Safety Precautions, Methods/Procedure, Data/Results (3), Neatness,
(total possible pts for completion: 18)
Lab Rubric Item: Understanding the Purpose of the Experiment #2—Assessed and Graded at “Exceeds Standard” point
value of 10
Lab Rubric Item: Understanding the Purpose of the Experiment #3—Assessed and Graded at “Exceeds Standard” point
value of 20
Lab Rubric Item: Understanding the Design of the Experiment #2B—Assessed and Graded at “Exceeds Standard” point
value of 20
(Total possible rubric points: 50)
Lab #29: Extraction of DNA from Sheep Thymus
Pre-Lab Activities
1. Survey the headings and subheadings to familiarize yourself with this lab.
2. Read the Introduction up to but NOT including the section entitled Helix Structure:
Connecting the Bases, completing the given tasks along the way. Now go back and reread it, this time highlighting main ideas. Don’t highlight everything you read—you are
looking for key points. If you highlight when reading the first time, you likely will
highlight more than the basic cues you need when going back to review it later.
3. Re-read these sections of the Introduction a third time, writing any remaining questions
you have about the concepts in the margins. Be sure to ask these questions in class.
4. Set up your lab notebook for Lab #29.
5. Interactive Lecture: What is an esterification reaction and what role does it play in the
structure of DNA?
6. Add esterification reactions to your SMaRT: Reaction Types.
7. As a class, write out the net reaction for the formation of a nucleotide.
8. Finish reading the Introduction, completing the given tasks along the way. Now go
back and re-read it, this time highlighting main ideas.
9. Re-read these sections of the Introduction a third time, writing any remaining questions
you have about the concepts in the margins. Be sure to ask these questions in class.
10. Interactive Lecture: What is the relationship between dissolving and dissociation?
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11. Create a series of sketches in the boxes below that illustrate the process of solid NaCl
dissolving in water to form a sodium chloride solution.
12. Each lab group will be responsible for preparing 100mL each of NaCl and Dawn™
Detergent solutions described in the Materials List. Each pair in the lab group will write
the protocol for preparing one of the solutions and will be responsible for making that
solution during Procedure Step 1. Determine which team members are preparing which
solution, and then write a procedure. Be sure to include exact masses and volumes in
your lab notebooks. This will be handed in and assessed as the Methods/Procedure
portion of the Completion Points for this lab.
13. Read the Materials and Methods section, completing the given tasks along the way.
Now go back and re-read it, this time highlighting main ideas. Don’t highlight everything
you read—you are looking for key points. If you highlight when reading the first time,
you likely will highlight more than the basic cues you need when going back to review it
later.
14. Re-read these sections of Materials and Methods a third time, writing any remaining
conceptual questions you have in the margins. Be sure to ask these questions in class.
15. Read through the Procedure and circle any new or unfamiliar materials. Access the
MSDS for appropriate circled materials. Flinn Scientific Homepage:
http://www.flinnsci.com/homepage/cfindex.html : Click on Safety on the menu to the
left—select MSDS from the menu on the next page—Click on the letter of the material
you are researching from the alphabet provided. Record any important information, such
as health hazards or emergency procedures in the space below:
16. Transfer any important safety information and/or safety symbols into the “Safety
Precautions” section.
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17. Interactive Lecture: What is polarity and how can this property be used to aid in the
DNA extraction?
18. As a lab group:
a. Use the molecular model kits to build 5-6 molecules each of ethanol and water.
b. Using the provided stickers (red for positive, green for negative), indicate the
location of the poles on your molecules.
c. Use your models to simulate the formation of an ethanol/water solution.
19. Based on your notes from Lab #6, participate in Lab Technique Discussion. Record any
additional notes as necessary.
20. As a group, decide which data you will be gathering during this lab, and construct
appropriate data tables in your lab notebooks.
21. You will be revisiting three rubric items in this lab. Update and then use your Lab Rubric
Summary Sheet to find all the labs where you worked on Understanding the Purpose of
the Experiment #2, Understanding the Purpose of the Experiment #3 and Understanding
the Design of the Experiment #2B. You are revisiting Understanding the Purpose of the
Experiment #2 and mastering the remaining two items. This means the point values for
these two items will be 20 points each. Using the Peer Review Sheet provided, write out
your PIP for these three rubric items.
22. Review the entire lab sheet. As you review, circle concepts and/or activities that relate
directly to Lab Rubric Items. This will help you “Achieve/Exceed Standard” on your lab
write-up.
Introduction
In Lab #6 we conducted this same DNA extraction technique. So why do it again? For one, the
practice of science involves the repeatability of techniques. No one would accept data from an
experiment in which only one trial was completed, or in which a technique was demonstrated to
work only once. For example, scientists who first cloned mammals were unable to do it
consistently. This lack of repeatability was a source of scientific skepticism, and even now it can
take many attempts before a successful clone is produced. The precision of a protocol can be
improved with new technologies, an altered sequence of steps, or the use of different
concentrations of solutions or reagents. We are taking the opportunity to look back at our notes
from Lab #6 and to see if there is information that can be applied to make the protocol more
effective or to simply run more smoothly. Scientists run protocols over and over again to ensure
the accuracy of results as well as the precision of the protocol itself. In a classroom setting we
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often do not have time to revisit protocols, but this is the exception. See if your lab team can
increase the DNA yield (the amount of DNA actually extracted) this second time around.
What are some other reasons for us to revisit this Procedure at this point in the course? Record
your thoughts in the space provided.
This Introduction assumes a working knowledge of the basic structure of DNA. In order to see
how one can chemically extract DNA from a cell, we will look at the chemical structures of
DNA and the cell itself. You may wish to pull together all of your notebook materials on DNA,
especially Lab #6, and put them together in your Bio-Chem II notebook for easy reference.
Nucleotide Structure
Watson and Crick described the basic structure of cellular DNA in their 1953 Letter to Nature,
which we read in Bio-Chem I. While Watson and Crick are credited for articulating the structure
of this biomolecule, the work of many contributed to this final model. As previously discussed,
one of the biggest debates was whether or not the material of heredity was a protein or a nucleic
acid. The work of Avery, Macleod and McCarty (who collaborated at The Rockefeller
University in New York City) can be viewed at the following url:
http://www.nature.com/nature/DNA50/macLeodmccarty.pdf. Their work, along with the work
of Hershey and Chase, heavily supported the idea that nucleic acids are the material for heredity.
As with most great work in science, Watson and Crick have many predecessors and colleagues to
thank for their own success in formulating what is still the accepted molecular model of DNA.
Watson and Crick were aware that DNA consisted of three basic subunits. In the space below
write down the names of these three subunits.
Now name the 4 nitrogenous bases in DNA:
Part of their work required Watson and Crick to determine how these subunits fit together to
form what we now call nucleotides. In Bio-Chem I we learned that the three parts of a nucleotide
are bonded together. Now we can see the type of bonds that are formed. All biotechnology and
genetic engineering techniques depend on the type of bonds present in nucleic acids.
Nucleotide Synthesis: An example of esterification reactions
As we discuss nucleotide synthesis, we will consider an example nucleotide that consists of
deoxyribose, phosphate, and the base adenine. The first reaction in the formation of the
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nucleotide occurs when the adenine molecule, C5N5H5, covalently bonds to deoxyribose,
C5O4H10, through a dehydration reaction.
Complete the following tasks, a-e, based on Figure 1.
a. How do the molecular formulas differ from the Lewis structures?
b. What product confirms that the reaction is a dehydration reaction?
c. Deoxyribose is a 5-carbon sugar. Which reactant formula represents deoxyribose?
d. Complete the following table for elements in the equation AND state whether or the
not the equation in Figure 1 obeys the Law of Conservation of Matter.
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Reactant
Element
Carbon
Total # of
Atoms
9
Product
Element
Carbon
Total # of
Atoms
9
Balanced?
√
e. Defend your response to part d, observance of the Law of Conservation of Matter,
using information from the table.
The next step is joining phosphoric acid to the synthetic product of that first reaction (water is a
waste product), through another type of covalent reaction known as an esterification reaction.
This type of reaction results in the formation of an ester, which is a category of organic
compounds. An esterification reaction is basically a dehydration reaction that results in the
formation of an ester. See the second reaction in the formation of this nucleotide in Figure 2.
Complete the following tasks based on the assumption that the equation in Figure 2 obeys the
Law of Conservation of Matter.
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a.
If 4 moles of water were produced, how many moles of phosphoric acid reacted?
b. Based on your answer to part a, how many moles of hydrogen would be needed for
the necessary amount of phosphoric acid to produce 4 moles of water?
c.
If we were to combine the reactions from Figures 1 & 2 to form one net reaction,
how many moles of water are produced during the synthesis of a nucleotide?
Esterification reactions produce esters. In general, an ester has a central atom that is doublebonded to a first oxygen atom, and single-bonded to a second oxygen atom. If we look at
Reference Table R: Organic Functional Groups we will see the functional group of esters,
RCOOR’. The R’s stand for radical, a common symbol used to represent any group of atoms.
According to the NYS PS Chemistry standards, the functional group “… imparts distinctive
physical and chemical properties to organic compounds” and is represented by the highlighted
COO. In the case of nucleic acids, we are dealing with a phosphate atom as the central atom
instead of a carbon atom. Go back to Figure 2 and highlight the functional group of a phosphate
ester. (Remember that even a single hydrogen atom can be an R group.)
Esterification reactions generally link molecules covalently and form a larger macromolecule.
The link is usually a single atom that bonds to both R groups, allowing the subunits to maintain
many of their individual chemical and physical properties while creating a macromolecule with
unique chemical and physical properties.
Helix Structure: Connecting the Bases
Knowing how a nucleotide is formed is only part of the story. Watson and Crick also needed to
determine what shape the nucleotides assumed when they fit together. They used Chargaff’s
1951 work with nitrogenous base ratios to help them (Journal of Comparative Physiology, Vol.
38, p. 41). Chargaff had determined that the amount of adenine bases was equal to the amount of
thymine bases, and the amount of guanine was equal to the amount of cytosine. Watson and
Crick then figured out that the ratios must be due to complementary pairing between nitrogenous
bases: A (on one strand) pairs with T (on the other strand), and G pairs with C. Adenine and
guanine are part of a class of compounds known as purines, and thymine and cytosine are known
as pyrimidines. In the DNA molecule, purines pair with pyrimidines (See Figure 3).
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Therefore, once each of the nucleotides is formed, the complementary pairing can occur between
nucleotides (See Figure 4).
The theme of unity and diversity can be seen even at the level of DNA monomers. Even though
the four bases are found in the DNA of all living things, the sequence of bases within the double
helix is different in each genome. Chemically speaking, the DNA molecule is generally the
same, but the specific sequence of the DNA varies in each individual.
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Both purines and pyrimidines are hydrophobic and relatively insoluble in water at the nearneutral pH of a cell. These molecules are essentially non-polar as a whole. However, they still
have internal partial charges symbolized as δ+/-. Partial charges are created when electrons are
unevenly shared between two bonded atoms. While this can exist within an overall non-polar
molecule, these partial charges can go so far as to create oppositely charged poles of a molecule.
Such molecules are polar (See Figure 5a). Hydrogen bonds form between hydrogen atoms that
have a partial positive charge (δ+) and oxygen, nitrogen or fluorine atoms—all of which are
highly electronegative and will most likely carry a partial negative charge, δ- (See Figure 5b).
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Hydrogen bonds are a special class of bonds within a group known as dipole-dipole forces.
These types of bonds (forces) form between polar molecules, or dipoles. A polar molecule is
often referred to as a dipole because it has two poles of charge within the molecule. A dipole is a
molecule that, when covalently bonded, has an uneven sharing of charge between its atoms.
Where have you heard the word parts di- and pole before? Give an example using each:
The hydrogen bonds between complementary bases are intermolecular forces since they link
separate molecules. In the case of a DNA helix, a purine bonds to a pyrimidine (Hint: take out
The Double Helix Colorplate #82 to see these bonds articulated). The hydrogen bond is often
indicated using dotted or dashed lines as opposed to the solid lines used to indicate covalent
bonds in Lewis Structures.
When a covalently bonded molecule, one that is supposed to share
electrons equally, has a charge, this is due to a shift in electron
density toward one of the elements in the molecule. The arrow
above the Lewis Structure indicates the direction of the charge shift.
So, in this hydrogen fluoride molecule, the hydrogen atom is more
positive than the fluorine atom. At any given time, there are more
electrons orbiting around the fluorine atom than around the hydrogen
atom; the electrons are not being evenly shared. In this depiction, the
thicker cloud around fluorine represents the greater electron density
that creates the partial negative charge. Consequently, this bond is
not purely covalent. This bond type is referred to as a polar covalent
bond.
A chemical property that can help us distinguish purely covalent bonds from polar covalent
bonds is electronegativity. We used electronegativity difference to distinguish between ionic and
covalent bonds in The Quest for Energy I. The difference in electronegativity is quantifiable.
Remember that an ionic bond results when the difference in electronegativity between bonding
atoms is 2.0 or greater, whereas covalent bonds result when the difference in values in less than
2.0. Using Reference Table S, we can determine bond types of simple molecules based on this
simple rule. For example, we can determine the bond type between Na and Cl. Na has a value
of 0.9 and Cl has a value of 3.2. The difference is 2.3; this is an ionic bond. The greater the
difference in electronegativity, the more uneven is the electron sharing. Based on your
knowledge of the property of electronegativity, propose a reason for the previous statement.
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When attempting to identify a polar covalent
bond, we look for differences approaching 2.0.
So let us look again at hydrogen fluoride, HF.
Using Reference Table S, we see that the value
of H is 2.1 and the value of F is 4.0. The
difference is 4.0 – 2.1 = 1.9. While this is a
covalent bond, the value is very close to the
ionic “threshold.” It is thus identified as polar
covalent bond. If we had a container of pure
HF, the molecules would arrange themselves so
that the H of one molecule was close to the F of
another molecule.
Helix Structure: The DNA Backbone
Watson and Crick determined that phosphates
and the sugars comprise the outer backbone of
the DNA molecule. We have already looked at
how the nucleotides are formed via dehydration
and esterification reactions. We will now explore
how successive phosphates and sugars are
bonded to one another, thereby connecting all of
the nucleotides into the backbone of the double helix, or the “rails” of its spiral staircase.
(Remember: the DNA backbone actually runs along the sides of the macromolecule, whereas a
human backbone runs up the middle.)
The phosphates are linked to each other through phosphodiester linkages. This means that each
phosphate group functions as an ester for each sugar to which it is attached. Once again, refer to
The Double Helix Colorplate #82. At the bottom of the diagram called Structural Formula, we
see the helix broken into its subunits. You can see the phosphate group attached to a sugar
above it and to a sugar below it. Sketch this combination to create Figure 7.
Figure 7: Sugar-Phosphate Backbone
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The single phosphate acts as the central atom for two esters. Circle the P atom in Figure 7. Now
using two different colored hi-liters, indicate the two esters of phosphorus.
Why is DNA an acid?
We now know all the chemicals that are referred to in the term “deoxyribonucleic.” Deoxyribrefers to the deoxyribose sugar; nucleic refers to the nucleotides. However, we have not yet
addressed why DNA is an acid. To do this we will again study the phosphate groups that make
up part of the DNA backbone.
In The Double Helix Colorplate #82 you will notice that there is a negatively-charged oxygen
associated with the phosphate ester. This oxygen accounts for DNA’s acidic character. Consider
phosphoric acid, H3PO4. Each single-bonded oxygen atom in the acid molecule has a hydrogen
atom attached. Through the esterification reactions that form the DNA nucleotide, two hydrogen
atoms are lost to form water molecules (See Figure 2). However, a third hydrogen remains
attached to an oxygen. This hydrogen is what makes DNA an acid.
In solution, this hydrogen will dissociate from phosphoric acid to form an H+ or hydrogen ion.
This dissociation behavior leaves behind the O atom that we see in the structural formula on the
colorplate (See Figure 8). As we saw in Lab #27, the dissociation of an Arrhenius acid produces
H+ ions. The high concentration of these ions creates the low pH associated with acids. As the H+
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ions dissociate from the phosphate ester, they enter into the nuclear solution containing DNA,
thereby lowering the pH to acidic ranges. Therefore, DNA is identified as an acid because its
chemical structure causes it to behave like an acid.
Returning to DNA, we can summarize its solubility as follows. At cellular pH, which is close to
neutral, the nitrogenous bases are hydrophobic. However, they occupy the interior space of the
molecule and do not interact much with the polar cytosol. The sugar-phosphate backbone of the
DNA is quite negative due to the negative oxygen atom in each phosphodiester bond. This
backbone makes up the exterior of the DNA molecule and interacts closely with its surroundings.
Earlier you drew water molecules surround Na+ and Cl- ions. In the same way, the partial
charges in water (cytosol is mostly water) will surround the opposite charges in DNA’s backbone
and cause it to dissolve.
In summary, Watson and Crick proposed their model of DNA as a molecule with hydrophobic,
complementary bases that are hydrogen-bonded across the helix and with a hydrophilic backbone
composed of phosphoric acid and deoxyribose covalently along backbone of the helix.
Materials and Methods
DNA Extraction Based on Chemical Properties
We are revisiting the lab protocol from Lab #6 so that we can compare our understanding of the
science behind this extraction from Bio-Chem I to Bio-Chem II. In this lab you will be revisiting
the extraction of DNA from sheep thymus. We performed several extractions in Bio-Chem I as
well as in ED II. State the purpose of an extraction:
The ability to extract DNA from organisms is essential to researchers. This extraction is very
simple, but it may cause the genetic information in the DNA to be lost. More advanced
techniques have been developed, but they are based in part on the simple methods utilized in this
extraction. You will see how amazingly easy it is to access DNA! Once DNA is extracted, it
can be amplified for study using a technique called PCR, which we will discuss later in this
Strand.
The protocol for the extraction of DNA from sheep thymus depends entirely on the chemical
properties of the thymus cells and the DNA molecule. Studies of the DNA molecule and the cell
membrane have led to the development and refinement of these protocols.
First: Preparing the Sheep Thymus
Sheep thymus can be ordered in some restaurants and in butcher stores by the name
“sweetbreads.” There is quite a bit of tissue in this gland, and, in order for us to conduct
efficiently an extraction, we need to obtain a manageable sample size and prepare it for the steps
that follow. For this reason, we cut a small sample off of the larger gland. We then grind the
sample in a NaCl solution using a mortar and pestle. This ensures that when we add additional
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solutions, they can get to the components of the cell we are most interested in, the cell membrane
and the nucleus. Once we have physically prepared the sample we can begin with the chemical
treatment.
Second: Dissolving the Membrane
If we are to extract this precious molecule of life from a cell, it is important to know about its
location. Think of yourself as a cellular bank robber—how would you penetrate the safe to reach
the cash? DNA is contained within the nucleus of a eukaryotic cell. It is too big to pass through
the nuclear membrane into the cytoplasm so we must break through the membrane ourselves. In
Bio-Chem I you were introduced to the structure of cell membranes. Locate your copy of The
Fluid Mosaic Model Colorplate #36 from Bio-Chem I.
When we originally discussed the membrane, we focused on the basic structure of the
phospholipid that makes up the membrane. A phosphate group is the polar head and two fatty
acids make up the non-polar tails. Connecting these two components is glycerol, a chain of three
carboxyl groups. In phospholipids, the phosphate group swings around and sits at the head,
leaving the two remaining fatty acids to hang below as the tail. This gives the phospholipids the
look of an old-fashioned clothespin. In Figure 9, fill in the correct Lewis structures for only the
phosphate and the carboxyls.
Polarity is important when determining solubility. There is a saying in science: “like dissolves
like.” “Like” refers to the polarity of a substance. Polar dissolves polar; non-polar dissolves nonpolar. However, a polar substance cannot dissolve a non-polar substance.
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When cell membranes form, the intracellular and extracellular environments repel the non-polar
tails; these “escape” by facing each other to form the interior of the membrane (See The Fluid
Mosaic Model Colorplate #36). The polar heads are attached to the water in both the intra- and
extracellular environments, so they face outwards from the tails. This is the basis for the cell’s
semi-permeable membrane.
As we read in the Introduction, polar molecules are generally hydrophilic, meaning they will
dissolve in water. This makes sense because water is also a polar molecule. When a polar solvent
comes in contact with a polar solute, there is an attraction between the opposite poles of the
solvent and solute particles. These attractions allow for a homogenous distribution of solute
particles in the solvent, producing the homogenous mixture known as a solution (See Figure
10a). By the same reasoning, non-polar molecules are generally hydrophobic. When a non-polar
solute such as oil is added to a polar solvent such as water, the oil molecules are repelled by the
water molecules and form visible oil droplets in the water. The resulting uneven distribution of
matter forms a heterogeneous mixture such as a suspension or a colloid (See Figure 10b). Oil
(hydrophobic) and water (hydrophilic) are said to be immiscible, or unable to mix. These general
rules apply when mixing two liquids and when mixing a solid and a liquid.
Water is known as the universal solvent because not only do most polar molecules dissolve in it,
but so do most ionic compounds. The number of polar and ionic substances on the planet is
huge; a great many molecules dissolve in water. If you look at Reference Table F-Solubility
Guidelines, you can see which types of substance can dissolve in water and which are insoluble
in water. There are several listed exceptions. In the space below, state why you think these
exceptions exist.
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In the phospholipid bilayer of a cell membrane, water will be attracted to the polar heads and
start to pass through, but once the polar water molecule comes in contact with the non-polar tails,
it is repelled. Water can only get through the cell membrane through special protein channels
called aquaporins (See Lab #20). If an oily substance comes into contact with the polar heads,
the oil molecules will be repelled and start to form an aggregate of non-polar molecules.
Meanwhile, the non-polar tails will be attracted to
the non-polar substance. If this happens, the nonpolar tails can start to change orientation within the
structure of the cell membrane; this may cause a
breakdown of the cell membrane. It is important to
remember that the membrane “floats” in the
cytosol; it is not rigid like a wall. Scientists have
taken advantage of this property to get things into
and out of cells.
This process is similar to how dish detergents
work. Detergents and soaps have a polar head and
a non-polar tail (See Figure 11). Often times the
dirt on dishes is oil-based, which means it is also
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non-polar. So when we add detergent to sink of dirty, oily dishes, the non-polar oil molecules are
repelled by the water molecules and attracted to the non-polar tails of the soap molecules. The
non-polar tails of the soap molecules encircle the oil molecule to form a micelle that has the
polar heads facing out. The micelle is then able to dissolve in polar water, and it appears that the
oils off the dishes have dissolved into the water, but in fact they are “hiding” inside the micelle
(See Figure 12). Based on how detergents work, suggest the role of soap in our protocol:
Third: Extracting the DNA from the Soap Solution
The negatively-charged oxygen atoms of the
phosphate groups, coupled with the hydroxyl
groups of the sugars, create an overall hydrophilic
DNA backbone. However, these negative charges
are generally neutralized by ionic interactions
with proteins, metal ions, and other molecules
when a DNA molecule is inside a cell.
Extracellular fluids contain a variety of ionic
compounds such as NaCl (think back to Lab #14).
In cellular solutions, the NaCl dissociates into Na+
and Cl- ions. This means that positively charged
sodium ions are moving freely about in the
cellular solution and are able to bond to the
negatively charged oxygen atoms associated with
the phosphate groups of DNA. Overall, this
neutralizes the phosphate groups.
During the course of the protocol, we add
additional NaCl to the sheep thymus mixture. The
additional sodium ions ensure that the DNA molecule will in fact exhibit non-polar
characteristics. However, this leaves the DNA molecules scattered about the soap solution and
therefore invisible to the naked eye. This is where the ethanol plays an important role.
Ethanol is an organic acid. Its formula and structure can be determined using the NYSPS
Chemistry Reference Tables as follows:
a. Using NYSPS Chemistry Reference Table P—Organic Prefixes, determine how many
carbons are in ethanol.
C
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b. Then using NYSPS Chemistry Reference Table Q—Homologous Series of
Hydrocarbons, determine if ethanol has single, double or triple bonds. Do this by
replacing the –ol suffix with an e, to obtain ethane.
c. Again using NYSPS Chemistry Reference Table Q—Homologous Series of
Hydrocarbons, determine how many hydrogen atoms would be present, if the
molecule were ethane instead of ethanol.
C
H
d. Draw a Lewis Structure for ethane in the space provided.
e. Using NYSPS Chemistry Reference Table R—Organic Functional Groups, determine
the functional group for an alcohol. Record in the space provided.
f. To construct the Lewis Structure of ethanol, replace one of the hydrogen atoms in
your Lewis Structure for ethane in Step d with the functional group for alcohols you
determined in Step e. Do this in the space provided for Figure 13.
Figure 13: Structure of Ethanol
Ethanol, like most organic acids, is a polar molecule. The covalent bond between the oxygen
and the hydrogen of the functional group is polar (Figure 14a). Consequently, the entire
molecule is classified as polar covalent. When water and ethanol are mixed, a solution is created
as the partially positive hydrogen in the OH group of the ethanol is attracted to the partially
negative oxygen of the water molecule (Figure 14b).
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In this lab we are using a 95% by mass solution of
ethanol and water. Calculate the number of moles of
ethanol in 100 grams of our ethanol/water solution in
this space.
This solution has an overall polar nature. We are adding
ice-cold ethanol to our mixture of dish detergent, cell
membrane micelles and neutralized DNA. Carefully add
the ethanol down the side of the test tube to create layers
that do not mix (see Figure 15). The lower temperature
of the ethanol limits the amount of interaction between
the polar ethanol solution and the polar particles in the
detergent mixture. Explain, on a molecular level, why
this is true:
The neutralized DNA in the detergent layer
is much less soluble in ethanol than it is in
water. For this reason, we now can see the
DNA at the interface between the water
and the ethanol. It is interesting to note that
we are seeing DNA that has been extracted
from the sheep thymus tissue as a whole.
There is no organization to the DNA clump
and no way of discerning where specific
genes start and other genes end. As a
matter of fact, there also is RNA in the
clump. Therefore, it is difficult to use this
technique to study specific DNA
sequences. However, modifications to this
technique have led to more recent
extraction techniques that isolate specific
parts of DNA.
Genetic Unity and Diversity II—Lab #29: Extraction of DNA from Sheep Thymus
©2002-2006 EduChange®
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Materials per Team of 2
Sodium Chloride or Dawn™ Dish Detergent
Balance
Weighing Paper
100mL Volumetric Flask
2 cm x 2 cm piece of sheep thymus
1 mortar and pestle—275-400 mL
10mL graduated cylinder
Scissors or scalpel
2 cm x 2 cm square of cheesecloth (double thickness) or gauze
2 test tubes—20 mL or greater
Test tube rack
1 glass stirring rod
2 pairs of goggles
Per Class
Cooler with Ice
500-mL bottle of 95% Ethanol
2-3 Pasteur pipettes
200 mL of 0.155 M NaCl solution—Student Prepared
200 mL of 10% Dawn Solution—Student Prepared
Procedure
1. Based on the procedure you wrote for Pre-Lab Activity #9, prepare your assigned
solution. Be sure to label the container.
2. Obtain a piece of thymus from the prep area, and place it in the mortar and pestle.
3. Using the scissors or scalpel, mince the thymus.
4. Using your graduated cylinder, obtain 10 mL of NaCl solution.
5. Carefully pour NaCl(aq) into the mortar and grind the mixture using a pestle. DO NOT
BANG the pestle into the mortar. Record observations as to what type of mixture this is
and why you think so.
6. Place one test tube in the rack, and layer the cheesecloth on the top of the test tube.
Gently press the center down to form a small well.
7. Pour the thymus/NaCl mixture through the cheesecloth into the test tube. The
cheesecloth will absorb some of the solution. You may gently massage the cheesecloth
to release some of the absorbed fluid but DO NOT SQUEEZE. Record observations that
describe the nature of the mixture that passes through the cheesecloth, and note what
material remains in the cheesecloth.
Genetic Unity and Diversity II—Lab #29: Extraction of DNA from Sheep Thymus
©2002-2006 EduChange®
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8. Discard the cheesecloth and solid matter as directed by your instructor.
9. Rinse your graduated cylinder with water 2-3 times.
10. Obtain 5 mL of 10% Dawn™ Solution.
11. Carefully add the solution to the test tube containing the strained thymus solution.
Record observations that describe the nature of this latest mixture.
12. Place your filled test tube and your empty test tube next to one another in the test tube
rack.
13. Walk over to the cooler containing the 95% ethanol solution.
14. Using the provided Pasteur pipette, transfer to the empty test tube a volume of 95%
ethanol that is equal to the volume of your thymus solution. You can eyeball this; the
amount does not need to be precise.
15. Return to your lab bench.
16. Tipping the test tube containing the thymus solution at a 45o angle to the lab bench (if
you are using wire test tube racks you can achieve this angle right in the rack), carefully
drip the ice-cold ethanol down the side of the test tube into the thymus solution. Do not
allow yourself to make big splashes. Record observations as to the nature of the resulting
mixture.
17. You should see an interface forming. At this interface, a white material should be
precipitating out of the thymus solution. This white matter is sheep DNA & RNA. From
a genetic unity and diversity standpoint, the chemical similarities between nucleic acids
are such that they both precipitate out by this method! Record observations.
18. Carefully insert your stirring rod. The DNA should be attracted to the stirring rod. Record
observations.
19. Leave your test tube in your rack, and when the teacher gives the sign, view other groups’
DNA. If you do not see “white stuff” right away, try letting it sit for a while and you
should eventually see some. Record observations.
20. Clean up and dispose of materials as directed by your teacher.
21. Wash your hands.
Genetic Unity and Diversity II—Lab #29: Extraction of DNA from Sheep Thymus
©2002-2006 EduChange®
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Post-Lab Activities
1. How did this lab experience compare to the experience you had with Lab #6?
2. Imagine that we changed the sequence of steps in the Procedure. Create a new sequence
and explain how the results might or might not be affected?
3. Lab Debrief
Genetic Unity and Diversity II—Lab #29: Extraction of DNA from Sheep Thymus
©2002-2006 EduChange®
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