Unit 1: Chapter 6 Lecture Notes
I. Introduction
A. Introduction to DNA Structure: The Importance of DNA Structure
1. DNA, since it carries all the information for a given organism, must be a molecule must contain an incredible amount of information
2. All organisms have their own genomic DNA
3. Contains information for proper development of an organism
a. Allows the proper structures to form at the appropriate time
b. Allows appropriate growth at the appropriate time
4. Contains the information for proper cellular function
a. DNA encodes the information to produce proteins involved in respiration
b. DNA encodes the information to produce proteins that are important in sending and receiving signals between cells
5. All the appropriate information is also passed on to subsequent generations
a. Cellular reproduction (asexual)
b. Organismal reproduction (sexual or asexual)
B. Introduction to DNA Structure: How It Holds The Information of Heredity
1. The ability of DNA to hold all of this information lies in both its chemistry and 3-Dimensional structure
2. DNA contains only five different types of atoms
a. Carbon
b. Phosphorous
c. Nitrogen
d. Hydrogen
e. Oxygen
3. When Watson and Crick (1952) discovered that the 3-Dimensional structure of DNA
a. Found that the molecule takes the shape double helix
b. More importantly understood how the different atoms found in DNA are covalently linked together and how these linkages are viewed in 3-dimensions
4. Watson and Crick saw that DNA was a polymer made of repeating building blocks known as nucleotides
a. Genes differ in the number of nucleotides
b. Genes differ in the sequence of nucleotides
II. Building a DNA Molecule
A. Building the DNA Molecule: Introduction to the Chemical Structure of Deoxyribonucleic Acid
1. Each nucleotide consists of three basic components
a. Phosphate group
b. A five carbon sugar (deoxyribose)
c. A nitrogenous base
2. The phosphate group and the deoxyribose are part of the DNA backbone, whereas the nitrogenous bases are located towards the interior of the DNA molecule
3. More specifically, it is the sequence and number of these nitrogenous bases (which are part of nucleotides) that give each gene its own identity
B. Building the DNA Molecule: Nucleotide Structure and The Pentose Sugars
1. To start, each nucleotide will contain a central pentose (5 carbon) sugar
2. The sugar that is used in DNA is deoxyribose
3. Within the ring, there are four carbon atoms (labeled 1’, 2’, 3’ etc) joined by an oxygen atom
4. The fifth carbon (the 5’ carbon) projects upward from the ring
5. To build the nucleotide, we are going to attach other chemically reactive groups to specific carbons in the pentose sugar
C. Building the DNA Molecule: The Nitrogenous Base Component
1. The presence of the nitrogenous bases in nucleic acids was discovered by Friedrich Miecher after he started to determine the chemistry of his nuclein
2. They are called nitrogenous bases due to the fact that they are have a high nitrogen content
3. They are considered a base due to the fact that they have the properties of a base (proton acceptors)
4. By and large, the structure of DNA the nitrogenous bases are non-polar, which is important for DNA structure
a. The bases are hydrophobic
b. The bases are located towards the interior of a molecule of DNA
5. There are four common nitrogenous bases found in DNA
a. Adenine
b. Guanine
c. Cytosine
d. Thymine
6. Adenine and guanine are known as purines and have a double ring
7. Cytosine, Thymine are known as pyrimidines and have a single ring
8. In nature, each nitrogenous base can take one of two conformations
9. Each base exists in two tautomeric states in equilibrium with each other
a. Tautomers are isomers that readily interconvert at equilibrium
b. Tautomerization results in the migration of a proton and a resulting shift from single to double bond, or vice versa
10. For the nitrogenous bases, there are two conformations
a. Conventional form
b. Tautomeric state
11. For all of the nitrogenous bases, the equilibrium strongly favors the conventional form
D. Building the DNA Molecule: Nucleotide Structure and The Phosphate Group
1. The chemistry of the phosphate group is important in allowing DNA to be a polymer (i.e. the phosphate group is important in linking nucleotides together)
2. The phosphate group consists of a phosphorous and four oxygen atoms
3. The phosphorous is located centrally in the phosphate group, and each of the four oxygen atoms are bound to the phosphorous
4. The bonds between the phosphorous and each oxygen atom is unequal
a. They share electrons unequally
b. Oxygen atoms are slightly negative
c. Phosphate is slightly positive
5. At physiological pH, the phosphate group is a proton donor
a. Phosphate group is polar
b. Phosphate group has a slight negative
6. Ester bonds link the phosphate to the rest of the nucleotide
a. They have the property of being extremely stable
b. These bonds are easily broken by enzymatic hydrolysis (by adding water)
7. The chemistry of the phosphate group also allows for linking of nucleotides together
8. Phosphate bonds are stable, yet easily broken
a. Allows for polymerization of nucleotides
b. Allows for synthesis of DNA (or RNA) chains
E. Building the DNA Molecule: Constructing a Nucleotide-The Basic Building Block of DNA
1. To build a nucleotide one must start with a nucleoside
2. A nucleoside consists of only a pentose sugar and a nitrogenous base
3. The nitrogenous base is bound to the 1’ carbon through an N-glycosidic linkage, which is formed through a condensation reaction
4. There are proper naming conventions for each type of nucleoside
a. Deoxyguanosine (if containing guanine and dexoyribose)
b. Deoxycytidine (if containing cytosine and deoxyribose)
c. Deoxyadenosine (if containing adenine and deoxyribose)
d. Deoxythymidine (if containing thymine and deoxyribose)
5. In a nucleotide the phosphate group is bound to the 5’ carbon
6. Like the nitrogenous base and the phosphate group are added to the pentose via condensation reactions with water as the byproduct
7. A nucleotide can have one, two or three phosphates bound to the 5’ carbon
a. The phosphate that is bound to the 5’ carbon is known as the α phosphate
b. The second phosphate from the 5’ carbon is the β phosphate
c. The third phosphate from the 5’ carbon is the γ phosphate
F. Building the DNA Molecule: Naming the Nucleotides
1. The name of a nucleotide comes uses as a root the name of the nucleoside followed by the number of phosphates the nucleotide contains
a. A nucleotide containing deoxyribose, adenosine and one phosphate is deoxyadenosine monophosphate
b. A nucleotide containing deoxyribose guanosine and two phosphates is deoxyguanosine diphosphate
c. A nucleotide containing deoxyribose, thymidine and three phosphates is deoxythymidine triphosphates
G. Building a DNA Molecule: A Strand of DNA Is Composed of Chains of Polynucleotides
1. A single polymer of DNA is considered a strand
2. To create a DNA strand, a polymer must be formed of repeating nucleotides
3. A strand of DNA is only formed in the 5’ 3’ direction and never in the 3’ 5’ direction
4. In forming a strand of DNA, the nucleotides will only be added onto the 3’ end of a growing DNA strand
5. In order to join two nucleotides together, a condensation reaction must occur between the free 3’OH group of the final nucleotide in a growing strand and the 5’ PO4 group in the nucleotide to be added
a. A phosphodiester bond is formed between the two nucleotides
b. A byproduct of the reaction is one molecule of water
H. Building the DNA Molecule: DNA Base Pairing
1. DNA is a double stranded molecule and therefore, two strands must be able to interact with each other
2. The results of two very important experiments were important for showing how the two strands of a DNA molecule interact
a. Erwin Chargaff’s biochemical experiments (first)
b. Watson and Crick’s X-ray diffraction studies (second)
3. Chargaff wanted to determine the relative concentration of each nitrogenous base within a molecule of DNA
4. In 1940, Chargaff developed a paper chromatography method to analyze the amount of each nitrogenous base present in a molecule of DNA
5. Chargaff observed several important relationships among the molar concentrations of the different bases
6. In 1940 Chargaff proposed three important rules with regards to the nitrogenous base composition of DNA, which became known as Chargaff’s rules
7. Chargaff rules are as follows
a. [A] = [T]
b. [G] = [C]
c. [A] + [G] = [T] + [C] or the concentration of purines is equal to the concentration of pyrimidines
8. Chargaff also found that the base composition, as defined by the percentage of G and C (G+C content) for DNA is the basically the same for organisms of the same species, and different for organisms of different species
9. The G + C content can vary from 22 – 73% depending on the organism
10. Watson and Crick built off Chargaff’s work
11. Watson and Crick isolated and crystallized DNA then subjected it to X-ray diffraction analysis to determine the structure of the DNA
12. Their results show that the secondary structure of DNA was a double helix
13. In the double helix, the two DNA strands interacted through base pairing (showing a physical reason for Chargaff’s observations
a. Adenine pairs with thymine (2 H bonds)
b. Guanine pairs with cytosine (3 H bonds)
14. To allow for the nitrogenous bases to properly lie and form hydrogen bonds between them, the strands must be in the opposite orientation
a. Opposite orientation = anti-parallel
b. The free 5’ ends of each strand are on opposite sides of the molecule
c. The free 3’ ends of each strand are also on opposite sides from each other
d. Allows the nitrogenous bases to align properly for efficient base pairing
15. Base pairing is advantageous to the DNA chemistry
a. Excludes water from the interior of the DNA molecule
b. Creates entropy which allows for stabilization of the double helix
J. Building the DNA Molecule: The Consequences of DNA Base Pairing
1. Each strand of a DNA molecule has complementary sequence
a. Due to the base pairing between the two strands
b. The two strands do not have the same sequence
2. There are important conventions that need to be followed when writing the sequence of a DNA strand
a. The sequence of each strand is written separately
b. Only the sequence of the nitrogenous bases is written out
c. The sequence of each strand is ALWAYS written in the 5’ 3’ direction
d. A 5’ is written before the 5’ most nitrogenous base and a 3’ is written after the 3’ most nitrogenous base
3. For the DNA molecule on the right the sequence of the two strands are as follows
a. For the strand 5’ 3’ bottom to top (left strand) the sequence is 5’ CAGT 3’
b. For the strand 5’ 3’ top to bottom (right strand) the sequence is 5’ ACTG 3’
K. DNA Secondary Structure: The Structure Confers Stability and Allows The Molecule To Hold Vast Amounts of Information
1. If DNA is to be the prime genetic molecule, then it must have two important characteristics
a. It must hold vast amounts of information
b. The molecule must be extremely stable
2. DNA is able to hold vast amounts information in its sequence of nitrogenous bases
3. Although there are only four nitrogenous bases each gene can still has its own identity
a. The number of bases varies for each gene
b. Sequence of bases varies for each gene
c. The reason why we say “bases” is that a gene is only defined by the base sequence for only one of the strands
4. The stability of the double stranded DNA molecule comes from two important forces
a. Hydrogen bonding between the base pairs
b. Base stacking interactions
5. In reality, the bases in DNA stack together, which results in increased stability
a. The base stacking eliminates water from the interior of the DNA molecule
b. This is slightly different than the popularized Watson and Crick’s model of the DNA structure where the base pairs appear as rungs on a ladder
6. In actuality, the base pairs lie flat upon one another and so instead of looking like “rungs on a ladder” they look like a stack of coins
7. In order to have the base pairs lie flat on one another, each base pair must be slightly twisted with respect to previous base pair
L. DNA Secondary Structure: Important Physical Features of the Double Helix
1. The shape of base pairs results in two extremely important physical features
a. Major Groove
b. Minor Groove
2. The grooves are present because the two bonds that attach a base pair to its deoxyribose sugar rings are not directly opposite (not a true 180 degrees)
3. The major and minor groove form as a result of the angle at which the two sugars protrude from the base pairs
a. 120 degrees for the narrow angle (minor groove formation)
b. 240 degrees for the wide angle (major groove formation)
4. The major groove is about twice as wide (22 A) as the minor groove (12 A)
5. The grooves allow for proteins to bind to the DNA
6. There are two aspects of the major groove that allow proteins to bind in a sequence specific manner
a. The wide geometry of the major groove allows proteins to gain access to the sequence information
b. Each base pair has its own unique combination of hydrogen bond acceptors and donors which line the edge of the major groove
7. DNA binding proteins use these unique patterns of hydrogen bond acceptors and donors for each base pair to find the specific sequence in DNA they should bind
8. Below are the patterns of donors and acceptors for each of the four possible base pairs (A= acceptor D = Donor H=non-polar hydrogen M=methyl group)
a. A-T (ADAM)
b. T-A (MADA)
c. G-C (AADH)
d. C-G (HDAA)
9. By contrast, each base pair does not really have its own unique combination of hydrogen bond acceptors and donors lining the minor groove
a. For A-T or T-A base pairs (ADA)
b. For G-C or C-G base pairs (AHA)
c. The minor groove does not support sequence specific binding very well
M. DNA Secondary Structure: Important Physical Features of the Double Helix and Disease
1. Many diseases result from a changes in DNA sequence that abrogate (block) DNA binding
a. The changes in DNA sequence result in a change in the pattern of hydrogen bond acceptors and donors in the major groove
b. The protein that is supposed to bind the DNA in a sequence specific manner can no longer do so because the pattern has changed
2. Familial Hypercholersterolemia (FH) is a genetic disorder caused by changes in DNA sequence in the LDLR gene (Low-Density Lipoprotein Receptor)
3. The LDLR gene encodes a protein that is expressed in the liver and adrenal cortex
4. This protein encoded by the LDLR gene is responsible for removing 66-80% of all LDL from the blood
5. Patients with FH exhibit disease symptoms at birth starting with a cholesterol level above the 95 percentile
6. By the second decade of life, other secondary symptoms arise due to the extremely high cholesterol levels
a. Arcus Cornae
b. Tendon Xanthomas
c. Recurrent nonprogressive polyarthritis
d. Tenosynovitis
e. Artheroschlerosis
7. Without aggressive treatment, patients can die of secondary symptoms by age 30
8. There is no cure for FH, patients require aggressive normalization of LDL levels
a. Dietary management
b. Drug therapy to reduce the amount of free LDL in the blood
9. There are two identified changes in the sequence of the LDL gene that can result in loss of a specific protein called from Sp1 from specifically binding the LDLR gene
a. One is a change from a C-G base pair G-C base pair at a specific position (-139)
b. The other is a change from a C-G base pair T-A base pair at another specific position (-60)
10. A patient needs only one of these changes to lose Sp1 binding, which will lead to FH development
11. Each of these base changes in DNA sequence will change the pattern of hydrogen bond acceptors and donors in the major groove
a. For the C-G G-C change, (HDAA AADH)
b. For the C-G T-A change (HDAA MADA)
N. DNA Secondary Structure: DNA Can Form Multiple Types of Double Helices
1. When Watson and Crick determined the secondary structure of DNA, it was thought to be fairly simple without significant structural variation between DNA molecules
2. As it turns out that is not quite true as DNA can adopt multiple conformations
a. Some of these conformations are physiologically relevant
b. Some of these conformations are not physiologically relevant
3. The three conformations DNA forms are as follows
a. B-DNA
b. A-DNA
c. Z-DNA
4. The DNA conformation present is generally determined by conditions of the solution in which the DNA is present in (or experimentally, the conditions in which crystallized)
a. Salt Concentration
b. Water Content (humidity)
5. Each conformation will have its own structural properties
O. DNA Secondary Structure: DNA Can Adopt A B-Type Double Helix (B-DNA)
1. The B-DNA form is considered the Watson and Crick conformation and is the most predominant conformation in vivo
2. The B-form of DNA is seen when the DNA is present in conditions of high humidity (95%) and relatively low salt
3. The B-DNA forms a right handed double helix (has a right handed twist)
4. The grooves present in B-DNA have the following characteristics
a. In B-DNA, the major groove is wide and of moderate depth
b. In B-DNA the minor groove is also of moderate depth, but is narrower
5. The distance between base pairs is about 0.34 nm
6. For each turn of the helix there will be 10.5 bp/turn at a distance of approximately 3.4 nm
P. DNA Secondary Structure: DNA Can Adopt an A-Type Double Helix (A-DNA)
1. The A-DNA form can be observed if the water content is decreased and the salt concentration is increased during crystallization
2. The A-form has only been observed in vitro and is thus thought to not be physiologically relevant
3. The A-DNA form takes the shape of a right-handed double helix
4. The A-DNA form is more compact and slightly tilted
a. The bases are tilted with respect to the axis
b. There are 11 bases per turn
5. The grooves of A-DNA have the following geometry
a. The major groove is deep and narrow
b. The minor groove is shallow and broad
Q. DNA Secondary Structure: DNA Can Adopt an Z-Type Double Helix (Z-DNA)
1. The Z-form of DNA was discovered by the Alexander Rich Lab in 1979 (MIT)
2. Z-DNA was visualized in the laboratory when the DNA was crystallized under one of two conditions