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 3Dimensional 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 Nglycosidic 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 (LowDensity 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 (BDNA)
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 (ADNA)
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 (ZDNA)
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
a. DNA crystallized under high-salt conditions
b. DNA crystallized in the presence of alcohol
3. The Z-form of DNA can be present under physiological
conditions when the DNA has long stretches of alternating
guanine and cytosine
4. The Z-DNA is a left handed double helix, and turns in a
counter-clockwise fashion when viewed down its axis
5. The left-handedness of the helix occurs due to alternating syn
and anti conformations of the n-glycosidic bond in consecutive GC nucleotides
6. The backbone of the Z-DNA has a zig-zag appearance
7. The Z-DNA grooves have the following characteristics
a. The major groove is shallow, almost to the point of being
non-existent
b. The minor groove is deep and narrow
III. DNA Denaturation and Renaturation
A. DNA Denaturation and DNA Renaturation: Introduction
1. As we talked about previously, the two strands are held
together by hydrogen bonds
a. Hydrogen bonds are considered weak non-covalent
forces
b. Allows for the two strands to come apart really easily
2. This ability of DNA to denature and renature is important for
two biological processes
a. Replication (in vivo)
b. Gene expression-transcription (in vivo)
3. Nucleic acid hybridization is important for a number of
experimental techniques in Molecular Biology
4. If the DNA is heated just above physiologic temperature (near
100 C) or subjected to high pH, the DNA denatures (the two
strands separate)
5. If the solution containing the DNA is slowly cooled, the DNA
can renature (The two complementary strands can re-form
regular double helices)
6. The process of adding heat to denature the DNA only affects
the hydrogen bonds (weak bonds) that allow base pairing to
occur
7. The phosphodiester bonds are covalent linkages which are
much stronger than hydrogen bonds and are unaffected by
temperature
8. Enzymatic activity is needed to break phosphodiester linkages
a. DNases
b. Restriction Endonucleases
B. DNA Denaturation and DNA Renaturation: Denaturation Kinetics
1. Important insights into the properties of the double helix
were obtained through classic experiments carried out in the
1950s on denaturation kinetics
2. In order to follow DNA denaturation, ultraviolet light at a
wavelength (λ) of 260 nm is used
3. DNA maximally absorbs light at a wavelength of 260 nm due to
the nitrogenous bases
4. Single stranded DNA absorbs UV light at λ= 260 nm more
efficiently than double stranded DNA
a. Base stacking of double-stranded DNA quenches the
ability of the DNA to absorb UV light
b. Native double-stranded DNA will absorb about 40% less
UV light as compared to the same amount of single
stranded DNA
5. To study denaturatation kinetics, a solution of double stranded
DNA is subjected to heat
6. As the solution is heated, the optical density (absorbance) at
260 nm markedly increases in a phenomenon known as
hyperchromicity
7. If the OD of DNA is plotted as a function of temperature, the
increase in absorbance occurs in a narrow range
8. The midpoint of the transition from double stranded to single
stranded DNA is known as the melting temperature or Tm
9. The melting temperature denotes the point at which 50% of
DNA in solution is single-stranded
10. Melting temperature is dependent on the composition of base
pairs in a DNA molecule
a. G:C base pairs contain 3 H bonds
b. A-T base pairs contain 2 H bonds
11. The more G-C base pairs, the higher the melting temperature
12. The less A-T base pairs, the lower the melting temperature
13. Tm = 3(G-C base pairs) + 2 (A-T base pairs)
IV. RNA Structure
A. RNA Structure: Introduction
1. RNA, although single stranded, can also base pair and form
significant secondary structure
2. Instead of base pairing with a second strand, a single RNA
strand can base pair with itself
3. “The structure of RNA is breathtakingly intricate and graceful”
-Harry Noller (2005)
4. There are many RNA can adopt significant number of
structures that are important for biological function
a. tRNA (translation)
b. rRNA (ribosomal RNA)
c. snRNA (splicing)
d. snoRNA (rRNA processing)
e. Ribozymes (enzymatic function)
f. mRNA (gene regulation)
5. RNA is considered a single-stranded molecule
6. The significant secondary structural motifs are stabilized by
base pairing
a. Conventional base pairing (Watson-Crick)
b. Unconventional base pairing
V. RNA Structure
A. RNA Structure: DNA and RNA Structure Comparison
1. The structure of RNA and DNA are fundamentally quite similar,
with and one significant chemical difference
2. Below are some similarities
a. Each is synthesized from the monomer building blocknucleotides
b. Nucleotides are polymerized in exactly the same way
3. The main difference between DNA and RNA lies with how the
deoxyribonucleotides and ribonucleotides are constructed
a. The basic building blocks (units) of RNA and DNA are
slightly different (sugar and nitrogenous bases)
b. RNA is more chemically reactive
4. RNA is generally found as a single stranded molecule that can
take more elaborate shapes due to the fact that it can base pair
with itself
B. RNA Structure: Ribonucleotide Structure and The Pentose Sugars
1. Ribose and Deoxyribose differ in structure only by the
presence or absence of a 2’ hydroxyl group, which is the
CRITICAL CHEMICAL DIFFERENCE
a. For RNA, the 2’ carbon has a hydroxyl group bound to it
b. For DNA, the 2’ carbon does not have a hydroxyl group
(deoxy) bound, instead it has a hydrogen bound to it
2. The difference in whether there is a 2’ hydroxyl (as in RNA) or
a 2’ hydrogen (as in DNA), gives DNA and RNA different chemical
properties due to the fact that the hydroxyl group is more
reactive than the hydrogen
a. RNA can fold into a greater array of structures
b. Allows RNA to form a whole array of different types of
base pairs
c. DNA is more stable than RNA; RNA is more prone to
degradation
C. RNA Structure: Base Pairing Is Critical For Allowing Secondary
Structure To Form
1. The conventional base pairs found in RNA are as follows
(Watson-Crick Base Pairs):
a. G:C base pair (3 H bonds)
b. A:U base pair (3 H bonds)
2. In order for a single strand of RNA to base pair with itself, nonconventional base pairing is also critical
a. More than 20 types of non-canonical base pairs form
with at least two H bonds
b. A common theme for non-canonical base pairing is that
one of the nitrogenous bases of the pair will need to be
shifted sideways to allow for hydrogen bonds to form
c. The most common non-canonical base pair is the G-U
base pair (will be present in almost all secondary
structure) and base pairs through 2 H bonds
3. Other less common non-canonical base pairs found in RNA
secondary structure are as follows
a. AU reverse Hoogstein (Adenine is shifted sideways in
comparison to the canonical AU base pair)
b. Sheared G-A base pair(2 H bonds)
c. G-A imino (3 H bonds)
D. RNA Secondary Structure: Base-Paired RNA Adopts an A-type Helix
1. RNA readily forms secondary structure in the form of a helix
2. DNA cannot adopt an A-Type Helix under physiological
conditions, however RNA can adopt something similar to the DNA
A-Type double helix under physiologic conditions
3. The RNA A-Type Helix cannot adopt a B-conformation due to
the 2’OH group
4. The A-Type Helix RNA adopts is stabilized by the same forces
as the DNA B-Type Double Helix
a. Hydrogen bonding between the base pairs
b. Base stacking interactions
5. The RNA A-Type Helix has 11 base pairs per turn and two
grooves
a. Major Groove
b. Minor Groove
6. The major groove is deep and narrow and is not well suited to
protein-RNA interactions
7. The minor groove is shallow and wide and is much better
suited to protein-RNA interactions due to the presence of 2’ OH
groups that extend out into the minor groove
8. Although there are ample 2’ OH groups in the minor groove,
RNA binding proteins are unable to bind there in a sequence
specific manner
9. When two complementary stretches of sequence are near each
other, a stem-loop structure may form
a. Not all sequences within the stretches are
complementary (especially at the end)
b. Intervening, non-complementary sequence is looped out
from the double-helical segment as a hairpin, bulge or
simple loop
c. Depending on the amount and location of the noncomplementary sequence, there are many variations on
the stem-loop
E. RNA Structure: Overview of Tertiary Structure
1. Beyond secondary structure, RNA can form higher order
tertiary structure
a. RNA binding proteins can recognize specific portions of
an mRNA due to higher order 3-dimensional structure
b. The higher order 3D structure allows for proper
functions of certain RNA (eg tRNA , rRNA, ribozymes)
2. Tertiary structures can arise from the interaction of multiple
secondary structures making use of significant non-conventional
base pairing
a. tRNA
b. rRNA
c. snRNA
3. In some cases proteins are necessary to allow for the
formation of higher order tertiary structure
4. Below are several common examples of tertiary structure
a. Pseudoknot Motifs
b. A-Minor Motif
c. Tetra-loop Motif
d. Ribose Zipper Motif
e. Kink-turn motif