Investigations in Biology and Chemistry II – Lab work Point Value: 68

Name: ______Date Due: ______

Lab #29: Extraction of DNA from Sheep Thymus

Pre-Lab Activities

  1. Survey the headings and subheadings to familiarize yourself with this lab.
  1. 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 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.
  1. 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.
  1. Set up your lab notebook for Lab #29.
  1. Interactive Lecture: What is an esterification reaction and what role does it play in the structure of DNA?
  1. Add esterification reactions to your SMaRT: Reaction Types.
  1. As a class, write out the net reaction for the formation of a nucleotide.
  1. Finish reading the Introduction, completing the given tasks along the way. Now go back and re-read it, this time highlighting main ideas.
  1. Re-read these sections of the Introductiona third time, writing any remaining questions you have about the concepts in the margins. Be sure to ask these questions in class.
  1. Interactive Lecture: What is the relationship between dissolving and dissociation?
  1. 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.
  1. 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.
  1. 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.
  1. 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.
  1. Read through the Procedure and circle any new or unfamiliar materials. Access the MSDS for appropriate circled materials. Flinn Scientific Homepage: : 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:
  1. Transfer any important safety information and/or safety symbols into the “Safety Precautions” section.
  2. Interactive Lecture: What is polarity and how can this property be used to aid in the DNA extraction?
  1. As a lab group:
  2. Use the molecular model kits to build 5-6 molecules each of ethanol and water.
  3. Using the provided stickers (red for positive, green for negative), indicate the location of the poles on your molecules.
  4. Use your models to simulate the formation of an ethanol/water solution.
  1. Based on your notes from Lab #6, participate in Lab Technique Discussion. Record any additional notes as necessary.
  1. As a group, decide which data you will be gathering during this lab, and construct appropriate data tables in your lab notebooks.
  1. 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 Understandingthe Purpose ofthe 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.
  1. 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 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:

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 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.

  1. How do the molecular formulas differ from the Lewis structures?
  1. What product confirms that the reaction is a dehydration reaction?
  1. Deoxyribose is a 5-carbon sugar. Which reactant formula represents deoxyribose?
  1. 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.

Reactant Element / Total # of Atoms / Product Element / Total # of Atoms / Balanced?
Carbon / 9 / Carbon / 9 / √
  1. 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.

  1. If 4 moles of water were produced, how many moles of phosphoric acid reacted?
  1. 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?
  1. 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 double-bonded to a first oxygen atom, and single-bonded to a second oxygen atom. If we look at Reference Table R:Organic Functional Groupswe 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).

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.

Both purines and pyrimidines are hydrophobic and relatively insoluble in water at the near-neutral 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).

[CML1]

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 HelixColorplate #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 TheQuest 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.

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.