(Unit II) Chapter 6: The Structure of DNA, and Its Importance For Holding Genetic Formation

I. Introduction

A. Introduction to DNA Structure: The Importance of DNA Structure

1. As we learned in Unit I, DNA is the Prime Genetic Molecule

a. Experiments By Hershey and Chase b. Experiments by Avery Macleod and McCarty

2. This molecule must contain an incredible amount of information

3. Must contain information for proper development of an organism

a. Must allow the proper structures to form at the appropriate timeb. Must allow for appropriate growth at the appropriate time

4. Must contain the information for proper cellular function

a. Our DNA encodes proteins involved in respirationb. Our DNA encodes proteins that are important in sending and receiving signals between cells

5. All the appropriate information is present to allow for reproduction

a. Cellular reproduction (asexual)b. Organismal reproduction (sexual or asexual)

6. As Human Molecular Biologists we have a keen interest in the information in DNA as mistakes (Mutations) in DNA sequence lead to Genetic Disease

B. Introduction to DNA Structure: How It Holds The Information of Heredity

1. What is so special about DNA that gives it the ability to hold important genetic information?

2. The ability of DNA to hold all of this information lies in both its chemistry and 3-Dimensional structure

3. Several biochemists, including Friedrich Meisher were able to determine that DNA was composed of several components

a. Carbonb. Phosphorousc. Nitrogend. Hydrogen

4. Watson and Crick (1952) discovered that the 3-Dimensional structure of DNA is a double helix

a. Understood how the different atoms found in DNA are covalently linked togetherb. Understood how those linkages are viewed in 3-Dimensions

5. The ability for DNA to hold genetic information lies in its structure

a. Compared to proteins, its structure seems quite simpleb. DNA has the ability to hold large amounts of stored information

6. From the 3-Dimensional structure of DNA came several implications

a. All genes would have roughly the same 3-Dimensional structure as most DNA in the cell takes the same 3-Dimensional shapeb. Differences between genes will come from the order and number of nucleotide building blocks

C. Introduction to DNA Structure: Differences in Structure Between DNA Molecules

1. Since Watson and Crick’s discovery, the basic principles of DNA chemistry and structure have not changed

2. However, each DNA molecule may not have the exact same structure

3. Some molecules are circular with no free ends, whereas other DNA molecules are linear and can have at least 2 free ends

4. Depending on how the DNA is crystallized, the double helix can take on any one of three forms

D. Introduction to DNA Structure: DNA is Not The Only Nucleic Acid

1. As a rule, nucleic acids are composed of carbon, nitrogen, phosphorous and hydrogen

2. DNA is not the only important nucleic acid found in nature

3. RNA is another type of nucleic acid also found in nature

a. For some viruses, this is their prime genetic moleculeb. In eukaryotic cells, RNA functions in expression of genesc. In eukaryotic cells, RNA functions as a structural molecule for important intracellular components in the cell (ribosomes)

II. Types of Nucleic Acids: DNA and RNA

A. There are some fundamental similarities as well as differences between DNA and RNA

B. Below are some similarities

1. The basic units to make each are very similar2. Each is polymerized in a very similar way

C. Below are some differences

1. DNA is double stranded, whereas RNA is single stranded2. The basic building blocks (units) of RNA and DNA are slightly different (sugar and nitrogenous bases)3. RNA is more chemically reactive4. RNA can take more elaborate shapes

III. Building a DNA Molecule

A. Building the DNA Molecule: The Chemical Structure of Nucleic Acids

1. Nucleic Acids, whether DNA or RNA are polymers

2. DNA and RNA are composed of repeating units (monomers) called nucleotides

3. Each nucleotide consists of three basic components

a. Phosphate groupb. A five carbon sugar (deoxyribose)c. A nitrogenous base

4. 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

B. 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 phosphorus 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 unequallyb. Oxygen atoms are slightly negativec. Phosphorous is slightly positive

5. At physiological pH, the phosphate group is a proton donor

6. The linking bonds that are formed from the phosphates are esters

a. They have the property of being extremely stableb. These bonds are easily broken by enzymatic hydrolysis (by adding water)

7. The fact that phosphate bonds are stable, yet easily broken is an important quality

a. Allows for polymerization of nucleotidesb. Allows for synthesis of DNA (or RNA) chains

C. Building the DNA Molecule: Nucleotide Structure and The Pentose Sugars

1. Each nucleotide will contain a pentose (5 carbon) sugar, whether it be a nucleotide that gets incorporated into DNA or RNA

2. The sugar that is used in DNA is deoxyribose, whereas ribose, is used in another type of nucleic acid called RNA

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

D. Building the DNA Molecule: Nucleotide Structure and The Pentose Sugars

1. Ribose and Deoxyribose differ in structure only by the presence or absence of a 2’ hydroxyl group

a. For RNA, the 2’ carbon has a hydroxyl group bound to itb. 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 structuresb. DNA is more stable than RNA; RNA is more prone to degradation

E. 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, when part of the structure of DNA the nitrogenous bases are non-polar, which is important for DNA structure

a. The bases are hydrophobicb. The bases are located towards the interior of a molecule of DNA

5. There are five different types of nitrogenous bases in nucleic acids

a. Adenineb. Guanine c. Cytosine d. Thyminee. Uracil

6. Adenine and guanine are known as purines and have a double ring

7. Cytosine, Thymine and Uracil are known as pyrimidines and have a single ring

8. Both DNA and RNA contain Adenine, Guanine, Cytosine

9. Instead of Thymine as DNA has, RNA has Uracil

10. In nature, each nitrogenous base can take one of two conformations

11. Each base exists in two tautomeric states in equilibrium with each other

a. Tautomers are isomers that readily interconvert at equilibriumb. Tautomerization results in the migration of a proton and a resulting shift from single to double bond, or vice versa

12. For the nitrogenous bases, there are two conformations

a. Conventional formb. Tautomeric state

13. For all of the nitrogenous bases, the equilibrium strongly favors the conventional form

F. 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

a. Guanosine (if containing guanine)b. Cytosine (if containing cytidine)c. Adenosine (if containing adenine)d. Thymidine (if containing thymine)e. Uridine (if containing uracil

3. A nucleoside becomes a nucleotide once at least one phosphate group is bound to the 5’ carbon)

a. If one phosphate is bound then it is a nucleotide monophosphate (NMP)b. If two phosphates are bound then the nucleotide is a nucleotide diphosphatec. If three phosphates are bound then the nucleotide is a nucleotide triphosphate

4. In a nucleotide, the nitrogenous base is bound to the 1’ carbon of the pentose sugar

5. In a nucleotide the phosphate group is bound to the 5’ carbon

6. Both 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 phosphateαb. The second phosphate from the 5’ carbon is the βphosphatec. The third phosphate from the 5’ carbon is the γphosphate

H. Building the DNA Molecule: Naming the Nucleotides

1. To name a nucleoside, we use the name of the appropriate nitrogenous base in the nucleoside and add a sine, unless the nitrogenous base is thymine or uracil, then a dine is added

a. Guanosine (if containing guanine)b. Cytosine (if containing cytidine)c. Adenosine (if containing adenine)d. Thymidine (if containing thymine)e. Uridine (if containing uracil)

2. 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. If one phosphate is bound then it is a nucleotide monophosphate (NMP)b. If two phosphates are bound then the nucleotide is a nucleotide diphosphatec. If three phophates are bound then the nucleotide is a nucleotide triphosphate

J. Building a DNA Molecule: A Strand of DNA Is Composed of Chains of Polynucleotides

1. To create a DNA strand, a polymer must be formed of repeating nucleotides

2. A strand of DNA is only formed in the 5’ 3’ direction and never in the 3’ 5’ direction

3. In forming a strand of DNA, the nucleotides will only be added onto the 3’ end of a growing DNA strand

4. 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 nucleotidesb. A byproduct of the reaction is one molecule of water

K. 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. How the two strands interact were determined through 2 sets of experiments

a. Erwin Chargaff’s biochemical experimentsb. Watson and Crick’s X-ray diffraction studies

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 two 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 number of purines is equal to the number 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 and determined that the secondary structure of DNA was a double helix

11. In the double helix, the two DNA strands interacted through base pairing

a. Adenine pairs with thymine (2 H bonds)b. Guanine pairs with cytosine (3 H bonds)

12. To allow for proper base pairing, each strand in a DNA molecule must be anti-parallel

a. Allows the nitrogenous bases to align properly for efficient base pairingb. The free 5’ ends of each strand are on opposite sides of the moleculec. The free 3’ ends of each strand are also on opposite sides from each other

13. Base pairing is advantageous to the DNA chemistry

a. Excludes water from the interior of the DNA moleculeb. Creates entropy which allows for stabilization of the double helix

L. Building the DNA Molecule: DNA Base Pairing

1. Each strand of a DNA molecule has complementary sequence

a. Due base pairing between the two strandsb. 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 separatelyb. Only the sequence of the nitrogenous bases is written outc. The sequence of each strand is ALWAYS written in the 5’ 3’ directiond. 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’

III. DNA Secondary Structure

A. 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 informationb. 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 geneb. Sequence of bases varies for each gene

B. DNA Secondary Structure: The Structure Confers Stability and Allows The Molecule To Hold Vast Amounts of Information

1. The stability of the DNA molecule comes from two important forces

a. Hydrogen bonding between the base pairsb. Base stacking interactions

2. 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

3. 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

4. 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

C. 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 Grooveb. 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 which is about 120 degrees for the narrow angle and 240 degrees for the wide angle

4. As the bases stack on top of one another, the wide angle between the sugars on one edge generates the major groove

5. As the bases stack on top of one another the narrow angle on one edge generates the minor groove

6. The major groove is about twice as wide (22 A) as the minor groove (12 A)

7. 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

8. 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

9. 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)

10. 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

D. 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 Cornaeb. Tendon Xanthomas

c. Recurrent nonprogressive polyarthritisd. Tenosynovitise. 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 the major groove

a. For the C-G G-C change, (HDAA AADH)b. For the C-G T-A change (HDAA MADA)

E. 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 relevantb. Some of these conformations are not physiologically relevant

3. The three conformations DNA forms are as follows

a. B-DNAb. A-DNAc. 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 Concentrationb. Water Content (humidity)

5. Each conformation will have its own structural properties

F. 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 depthb. 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

G. 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 axisb. There are 11 bases per turn

5. The grooves of A-DNA have the following geometry

a. The major groove is deep and narrowb. The minor groove is shallow and broad

H. 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

a. DNA crystallized under high-salt conditionsb. 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 G-C 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-existentb. The minor groove is deep and narrow

IV. Strand Denaturation and DNA Renaturation

A. Strand 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 forcesb. Allows for the two strands to come apart really easily

2. If the DNA is heated just above physiologic temperature (near 100 C) or subjected to high pH, the DNA denatures (the two strands separate)

3. If the solution containing the DNA is slowly cooled, the DNA can renature (The two complementary strands can re-form regular double helices)

4. This ability of DNA to denature and renature is important for two biological processes

a. Replication (in vivo)b. Gene expression-transcription (in vivo)

5. Besides to DNA strands base pairing, a single strand of DNA can base pair with a single strand of RNA

6. Nucleic acid hybridization is important for a number of experimental techniques in Molecular Biology 7. The process of denaturation and renaturation only affects the hydrogen bonds (weak bonds) that allow base pairing to occur

8. The phosphodiester bonds are covalent linkages which are much stronger than hydrogen bonds and are unaffected by temperature

9. Enzymatic activity is needed to break phosphodiester linkages

a. DNasesb. Restriction Endonucleases

B. Strand 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 ( ) of 260 nm is usedλ

a. DNA maximally absorbs light at a wavelength of 260 nmb. Specifically the nitrogenous bases are necessary for the absorption of light at 260 nm

3. Single stranded DNA absorbs UV light at = 260 nm more λefficiently 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

4. In the denaturatation kinetics experiment, a solution of double stranded DNA is subjected to heat

5. As the solution is heated, the optical density (absorbance) at 260 nm markedly increases in a phenomenon known as hyperchromicity

6. If the OD of DNA is plotted as a function of temperature, the increase in absorbance occurs in a narrow range

7. The midpoint of the transition from double stranded to single stranded DNA is known as the melting temperature or Tm (Point at which 50% of DNA is single stranded)

8. Each molecule of DNA will have its own melting temperature and is largely dependent on G-C content

a. G:C base pairs contain 3 H bondsb. A-T base pairs contain 2 H bonds

9. The higher the G-C content, the higher the melting temperature

10. The higher the A-T content, the lower the melting temperature

11. It is possible to estimate melting temperature without experimentation by using the following formula

12. Tm = 3(G-C base pairs) + 2 (A-T base pairs)

V. 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. The significant secondary structural motifs are stabilized by base pairing

a. Conventional base pairing (Watson-Crick)b. Unconventional base pairing (non-canonical)

B. 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, non-conventional 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)

C. RNA Secondary Structure: Base-Paired RNA Adopts an A-type Helix

1. 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

2. The RNA A-Type Helix cannot adopt a B-conformation due to the 2’OH group

3. 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 pairsb. Base stacking interactions

4. The RNA A-Type Helix has 11 base pairs per turn and two grooves

a. Major Grooveb. Minor Groove

5. The major groove is deep and narrow and is not well suited to protein-RNA interactions

6. 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

7. Although there are ample 2’ OH groups in the minor groove, RNA binding proteins are unable to bind there in a sequence specific manner

8. 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 non-complementary sequence, there are many variations on the stem-loop

9. RNA Secondary Structure: Base-Paired RNA Adopts an A-Type Helix (Fig. 6-30)

D. 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. tRNAb. rRNAc. 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 Motifsb. A-Minor Motifc. Tetra-loop Motifd. Ribose Zipper Motife. Kink-turn motif