MCB 52: READING AND LECTURE NOTES
[09.04] L ECTURE 1: THE STRUCTURE OF DNA
Goal: To understand elements of DNA stability and information content by applying scientific reasoning
skills to the knowledge you’ve gained from previous courses.
1. Explain how the following contribute to the stability of DNA: hydrogen bonding, entropy,
ionic strength, and base composition.
2. Describe how the geometry of the base pairs leads to the major and minor grooves of the
DNA double helix.
3. Explain how and why sequence information is read in the major groove.
HOW IMPORTANT IS THE ΔG FROM HBOND FORMATION FOR THE STABILITY OF DNA?
>> The formation of Hbonds between the bases has a favorable ΔG. However, base pairing involves
breaking an equal number of Hbonds between the bases and water. Thus, the ΔG from Hbond formation
is not very important for the stability of DNA.
>> Ordered water molecules that satisfy hydrogen bonding on bases of separated polynucleotide strands
have low entropy, which is unfavorable. Releasing water from the bases by joining strands and creating
the double helix results in disordered water molecules, increasing entropy in the system.
>> Hbonding is important for specificity in base pairing. For example, if A and C were to be bonded,
water molecules would be unable to be replaced with Hbonding due to unfavorable angles, resulting in
an unstable pair.
The bases are hydrophobic and are found in the center of the DNA duplex, where their flat
surfaces “stack” on top of each other.
This is advantages because 1) unfavorable contact with water is minimized, 2) favorable ππ
interactions are created, in which electrons are shared between porbitals of adjacent base pairs.
Stacking of GC base pairs with their neighbors contributes more to DNA stability than AT base
HOW DO PROTEINS ACCESS SEQUENCE INFORMATION IN DNA?
The major and minor grooves are created by the edges of the stacked base pairs.
The grooves are unequal in width because the bonds linking the bases to the sugars project at an
angle of 120°. If the bonds linking the bases to the sugars projected at an angle of 180°, then the
two grooves would be identical in width. MCB 52: READING AND LECTURE NOTES
HOW DO PROTEINS KNOW WHERE TO BIND IN DNA?
>> The answer lies in the grooves, particularly the major groove, where proteins make contact with
functional groups on the edges of the base pairs. The edges of the base pairs are rich in chemical
>> There are typically 1020 interactions between a given DNA binding protein and the DNA, and those
interactions can involve hydrogen bonds, ionic bonds, and hydrophobic interactions.
CHAPTERS 1 – 3 (REVIEW), CHAPTER 4 (THE STRUCTURE OF DNA): PP. 7792
DNA Is Composed of Polynucleotide Chains
The nucleotide consists of phosphate joined to a sugar, known as 2’deoxyribose, to which a base is
attached. The sugar and the base alone are called a nucleoside. Adding phosphates makes a nucleotide.
Nucleotides are joined to each other in polynucleotide chains through the 3’OH of 2’deoxyribose of one
nucleotide and the phosphate attached to the 5’OH of another nucleotide, forming a phosphodiester
Each Base Has its Preferred Tautomeric Form
The purines are adenine and guanine, and the pyrimidines are cytosine and guanine. Each of the two
bases exists in two alternative tautomeric states, which are in equilibrium with each other.
The Two Strands of the Double Helix are Would around Each Other in an Antiparallel Orientation
The Two Chains of the Double Helix Have Complementary Sequences
The Double Helix is Satisfied by Base Pairing and Base Stacking
Hydrogen Bonding is Important for the Specificity of Base Pairing
Bases Can Flip Out from the Double Helix
DNA is Usually a RightHanded Double Helix
The Double Helix has Major and Minor Grooves
The Major Groove is Rich in Chemical Information
The Double Helix Exists in Multiple Conformations
DNA Can Sometimes Form a LeftHanded Helix
DNA Strands Can Separate (Denature) and Reassociate MCB 52: READING AND LECTURE NOTES
When the temperature of a solution of DNA is raised to near the boiling point of water, the optical density
(called absorbance) at 260 nm markedly increases, a phenomenon known as hyperchromicity.
Melting temperature of DNA is a characteristic of each DNA that is largely determine by the G:C content
of the DNA and the ionic strength of the solution. The higher the percent of G:C base pairs in the DNA
(and hence the lower the content of A:T base pairs), the higher is the melting point. Likewise, the higher
the salt concentration of the solution, the greater is the temperature at which the DNA denatures. MCB 52: READING AND LECTURE NOTES
[09.06] L ECTURE 2: XR AY D IFFRACTION AND THE STRUCTURE OF DNA
Goal: To understand how Xrays revealed the DNA double helix
1. Identify and explain the five main features of Photo 51 and what they tell us about DNA
2. Predict the diffraction pattern that would result from alterations to DNA structure.
3. Analyze a different diffraction pattern and predict features of the structures that would
generate this pattern.
4. Explain why we need to use xrays to “see” DNA.
WHAT WAS KNOWN IN 1953? WHAT WAS NOT KNOWN?
Known: DNA carried genetic information, chemical composition of DNA, DNA formed large
Unknown: How was the genetic information organized? How did the chemical structure allow for
THE DNA DOUBLE HELIX
Small repeat (3.4 Å basepair repeat), 34 Å helical repeat (period)
10 Å helix radius, 10 base pairs per period
Xpattern >> helix
Helical period >> distance between spot layer lines is 1/P
Radius >> calculated with crossing angle and P
Two molecules (offset) >> missing 4 layer line
Smaller repeat >> density at 10/P
RELATIVE SCALE OF LENGTHS
WHY USE XRAYS TO STUDY THE STRUCTURES OF MOLECULES?
Anything smaller than visible light requires Xrays to measure
OPTICAL IMAGING VS. XRAY SCATTERING MCB 52: READING AND LECTURE NOTES
Xrays hit object of interest, scatters, use film to collect, use computer to recreate structure
DIFFRACTION FROM ORIENTED FIBERS OF DNA
DNA all aligned along helical axis, not necessarily evenly spaced (not a crystal structure)
DIFFRACTION FROM ORIENTED FIBERS OF DNA
CONSTRUCTIVE AND DESTRUCTIVE INTERFERENCE
XRAY SCATTERING: CONSTRUCTIVE INTERFERENCE
Scatters at the same angle theta as the Xray hits the atom
Note that the second wave travels an extra distance compared to first one
>> inphase waves >> dark spots on the film
XRAY SCATTERING: DESTRUCTIVE INTERFERENCE
With a different distance, no longer an integral number of the wavelength
>> out of phase waves >> light spots
OPTICAL SIMULATION OF XRAY DIFFRACTION BY DNA
The diffraction from a horizontal row of lines produces a vertical a series of spots.
There is a reciprocal relationship between the spacings in real space and spacings in diffraction
space (if real space is smaller, space between dots increases)
If you represent the DNA double helix as two waves, each waves have a “zig” and “zag”
component… so only a helix could produce the X pattern.
Also, since molecules are orthogonal to diffraction pattern, the diffraction pattern will reveal the
crossing angle of the actual molecule.
Helical repeat (P, period) >> distance between layer lines of spots is 1/P
Knowing helical period and crossing angle, can solve for radius using trigonometry.
Missing 4 layer line out of 10 >> two molecules that are offset (3/8)
Density increases as you get to the 10 layer line, which is from the smaller basepair repeat
within the larger helical repeats (remember that diffraction produces reciprocal pattern), dark spot
is from relatively high quantity of base pairs compared to a period
CHAPTER 4 (THE STRUCTURE OF DNA): PP. 8889 MCB 52: READING AND LECTURE NOTES
Central “Maltese” cross, composed of broad spots (breadth of spots reflecting disorder in
the fiber), with the spots evenly spaced along horizontal “layer” lines.
Counting up and down from the center of the cross, spots at the fourth layer line are
The Maltese cross and the intensely dark regions at the top and bottom of the image
create a series of four diamondshaped areas.
Principle underlying Xray diffraction: when waves pass through a periodic array,
interference occurs between the waves if the wavelength of the waves is similar to the repeat
distance of the array. If oscillations of the waves are aligned, the waves reinforce each other
(“constructive interference”), but if the troughs and peaks are aligned, there is destructive
A beam of waves passing through a horizontal set of lines should generate a vertical row
of spots. If the horizontal lines are tilted, this produces an array perpendicular to the tilt of the
lines. If waves are passing through sets of tilted lines linked in zigzag fashion, this produces an X
Helices are out of register by 3/8ths of a helical repeat, which creates additional
destructive interference cancelling out the fourth layer line.
Pattern of four diamonds will show that periodicity of the double helix is 10 times the
atomic periodicity, from 10 layer lines from the center of the cross to the north and south poles.
Thus, periodicity of the double helix (3.4 nm) is 10 times the atomic periodicity, corresponding to
10 repeating units at a spacing of .34 nm. Because there is one base per sugarphosphate unit, the
B form of DNA consists of 10 base pairs per helical turn. MCB 52: READING AND LECTURE NOTES
[09.09] L ECTURE 3: METHODS OF M OLECULAR B IOLOGY
Goal: Learn how each of the following can be used in the molecular biology lab—PCR, gel
electrophoresis, DNA cloning & endonuclease restriction enzymes, Southern/northern/western blotting
For each technique, you should be able to answer the following:
In what situations is this technique used?
What are the key reagents and steps involved?
What information is gained from using this method?
AΒ AND ALZHEIMER’S DISEASE
Alzheimer’s disease is characterized by the accumulation of βamyloid peptide (Aβ) in the brain,
along with a hyperphosphorylated form of another protein called tau
Aβ is derived by sequential cleavage of amyloid precursor protein (APP) by cellular proteases.
The precise physiological function of APP is not understood.
APP IS A TRANSMEMBRANE PROTEIN THAT IS CLEAVED BY SECRETASE ENZYMES
>> The secretase enzymes can cleave APP at multiple sites. Your goal is to express Aβ42 in cultured
>> To do this, you plan to: 1) PCR amplify the DNA sequence that codes for this protein, 2) subclone the
PCR product into a plasmid that is suitable for protein expression
SETTING UP THE PCR
If primers have low G/C content: since G/C stacks more favorably with other base pairs (also 3 Hbonds,
not as important), forms stronger bonds >> lower the annealing temperature
If primers have high G/C content: >> raise the annealing temperature, if you lowered, you would get
mismatches that lead to imperfect product
Why is it just the annealing temperature that matches? Denaturing temperature is kind of general, all
DNA > singlestranded. Extension temperature is specific to polymerase being used, 72 degrees standard
for Taq polymerase.
If product is large (>1 Kb): change the extension time, because more time needed to add dNTPs to create
1. Negatively charged DNA molecules migrate to the positive electrode. MCB 52: READING AND LECTURE NOTES
2. DNA molecules are flexible rods which occupy an “effective volume.”
3. Pores in the gel matrix sieve the DNAs according to their effective volume. Large molecules
penetrate the matrix with more difficulty and therefore move more slowly.
4. DNA can then be visualized using EtBr, which intercalates between the base pairs of the DNA.
AGARAOSE VS. POLYACRYLAMIDE
>> Agarose has a large pore size – used for large nucleic acids: 10,000 bp to 500 bp
>> Polyacrylamide has a small pore size – used for small nucleic acids: 10 bp to 300 bp
1. Isolate cellular mRNAs
2. “reverse transcribe” mRNA into DNA using reverse transcriptase (RT)
3. Remove the RNA
4. Made dsDNA: prime with random hexamers, add dNTPs + DNA polymerase
5. The resulting cDNAs can be amplified by PCR, cloned, etc.
Recognize and cleave DNA at specific nucleotide sequences
Most widely known is the restriction enzyme EcoRI from E. coli, which cleaves at the sequence
5’ – GAATTC – 3’
Will cleave any piece of DNA that contains this sequence regardless of the source
Recognizes palindromes, typically 48 nt
Restriction enzymes can make “blunt” ends or “sticky” ends
Sticky ends can base pair with each other. If we add DNA ligase, they can be covalently joined.
We can join any two DNAs from any organism if they are both cut with the same enzyme to give
compatible sticky ends.
WHAT IF PCR PRODUCT DOESN’T CONTAIN RESTRICTION SITES COMPATIBLE WITH
THOSE PRESENT IN THE POLYLINKER?
Include restriction sites on the 5’ ends of the primers. Then, digest the PCR product and vector
with the same restriction enzymes and ligate the digested PCR product into the vector. MCB 52: READING AND LECTURE NOTES
Remember to include start and stop codons!
Select for transformants on medium containing ampicillin.
SOUTHERN BLOTS ARE USED TO TEST FOR THE PRESENCE OF SPECIFIC DNA SEQUENCES
Overview: cleave target DNA with restriction enzyme(s), separate on a gel, denature separated
DNA, transfer to a membrane
Incubate the membrane (blot) with a radioactive probe that will bind to its complement.
Expose blot to a piece of Xray film and a band will appear wherever the probe bound.
Why an alkali solution? Denatures the DNA, makes it singlestranded, makes it singlestranded,
facilitates the probe binding in the next step
Probe: from plasmid itself or a piece of the copied DNA
DO THE TRANSFECTED CELLS EXPRESS AΒ PEPTIDE?
• Western blot: uses an antibody probe to detect a target protein molecule within a population of
• Western blotting starts with electrophoresis of denatured proteins (by SDS).
• Transfer the separated proteins to a membrane, using electroblotting
• Incubate the membrane with antibodies that bind to the protein of interest
• Detect where the antibody bound to the blot (use one of several detection techniques), not usually
CHAPTER 7 (TECHNIQUES OF MOLECULAR BIOLOGY): PP. 147158 + ONLINE SUPP.
NUCLEIC ACIDS: BASIC METHODS MCB 52: READING AND LECTURE NOTES
[09.11] LECTURE 4: DNA TOPOLOGY
Goal: To understand how cellular DNA is topologically constrained and explain why that matters for
many DNA “transactions”.
1. Explain how the mathematical concept of linking number and two related geometric properties
(twist and writhe) relate to the topology of DNA molecules.
2. Describe the physiological consequences of maintaining DNA in a supercoiled state.
3. Predict how the topology of DNA molecules affects their mobility in an electrophoretic field and
diagram how that was used to experimentally determine the helical periodicity of DNA in
4. Illustrate how enzymes called “topoisomerases” change the topology of DNA and how drugs that
target topoisomerases are important therapeutic topics.
5. Explain how intercalating agents (such as ethidium bromide) affect the topology of a DNA
I. LINKING NUMBER
>> Many DNA molecules are topologically constrained, covalently closed circles
>> If you try to separate…create knots and tangles
Linking number is always an integer. To change Lk, must break at least one ring.
Linking number of covalently closed circular DNAs (cccDNAs)
By convention, righthanded DNA has a positive linking number.
The linking number is the number of times one strand passes over the other strand, which is
flattened into the plane.
is the average linking number for a population of DNA molecules.
II. TWIST AND WRITHE
The topological property of linking number is comprised of two geometric components called “twist” and
Twist: the number of helical turns of one strand about the other.
Writhe: when the double helix crosses over itself in threedimensional space
Lk= Tw +Wr MCB 52: READING AND LECTURE NOTES
Tw and Wr are interconvertible.
Two types of writhe:
Interwound (or plectonemic)
Toroidal (or spiral), such as around a DNA binding protein
III. NEGATIVE SUPERCOILING
LK : PREFERRED TWIST
Under physiological conditions, Bform DNA has a preferred twist of 10.5 bp/turn
The preferred twist of a DNA molecule N base pairs in length under physiological conditions is:
Ex. 10,000 bp >> Lk must be an integer, so Lk does not equal Lk
Occurs when Lk does not equal preferred twist
ΔLk = Lk – Lk
Negative and positive supercoiling
SUPERCOILED DNA CAN BE DISTORTED IN TWO WAYS:
Overall shape of the ring may be distorted (writhe)
Helical twist may be distorted
Identical DNA molecules that differ only in their linking number are called topoisomers.
IV. GEL ELECTROPHORESIS OF TOPOISOMERS
A highly supercoiled cccDNA will have a smaller volume and thus travel faster in a gel, while a circular,
totally relaxed cccDNA will take the longest time to travel through the gel. Note that linear DNA will
travel faster than a circular DNA because it can weave through the pores of the gel more effectively.
The fact that DNA topoisomers can be separated from each other by electrophoresis was the basis for the
experiment that showed that DNA has a helical periodicity of 10.5 bp per turn of the helix in solution.
SUPERCOILING AND WRITHE
NEGATIVELY supercoiled DNA has: RIGHThanded interwound writhe & LEFThanded
toroidal writhe (most organisms have lefthanded toroidal writhe around histone proteins >
negatively supercoiled DNA) MCB 52: READING AND LECTURE NOTES
POSITIVELY supercoiled DNA has: LEFThanded interwound writhe & RIGHThanded toroidal
DNA is about 6% underwound
*Organisms that live in hightemperature environments, in order to prevent their DNA from denaturing,
will maintain DNA in positive supercoils; DNA is tightly wound together.
>> Topoisomerases are essential for DNA replication, untangling DNA during mitosis and many other
>> Topoisomerases are the targets of important medicines. For example, quinolones, such as
ciprofloxacin, are antibiotics that inhibit a special bacterial topoisomerase called DNA gyrase. Other
topoisomerase called DNA gyrase. Other topoisomerase are used in anticancer therapy.
TYPE I TOPOISOMERASE: DOES NOT REQUIRE DNA, INCREASES LK BY 1
Work by making transient singlestranded breaks in the DNA
No ATP required
TYPE II TOPOISOMERASES: CHANGE LK IN STEPS OF 2
Work by making transient doublestranded breaks in the DNA
Requires energy from the hydrolysis of ATP
Can catenate, decatenate and unknot ds circular DNAs
DNA GYRASE: A SPECIAL PROKARYOTIC TYPE II TOPOISOMERASE
DNA gyrase introduces negative supercoils into DNA and introduces negative supercoils into
bacterial chromosomal DNA.
Quinolones (such as Ciprofoxacin) are antibiotics that target DNA gyrase
TYPE IIA TOPOISOMERASE INHIBITION BY A NEW CLASS OF ANTIBACTERIAL AGENTS
Binds one molecule of topoisomerase
No doublestranded break
INTERCALATION OF ETHIDIUM INTO DNA DECREASES THE HELICAL TWIST
Ethidium is a planar molecule that intercalates (“slips in”) between the base pairs of DNA. MCB 52: READING AND LECTURE NOTES
Intercalation of 1 molecule of ehtidium causes the DNA to unwind by 26°, decreasing the twist of
Since ethidium does not break either of the strands of DNA, if ethidium is added to a cccDNA,
that decrease in twist must be accompanied by an increase in writhe.
Thus, adding a small amount of ethidium to a negatively supercoiled cccDNA will cause it to
If you add enough ethidium to a negatively supercoiled cccDNA, it will completely relax.
If you add lots of ethidium, it will become positively supercoiled.
CHAPTER 4 (THE STRUCTURE OF DNA): PP. 93103 MCB 52: READING AND LECTURE NOTES
[09.13] LECTURE 5: N UCLEOSOMES AND C HROMATIN
Goal: To understand how DNA is packaged in the nucleus
1. Define the following terms: chromosome, 30nm fiber, nucleosome, histone.
2. Describe the composition and structure of the nucleosome, including:
a. The stoichiometry of the histone core proteins,
b. The significance of the positively charged amino acids that make up the histone proteins
c. How histone H1 affects the arrangement of nucleosomes on the DNA
d. How the DNA wraps around the nucleosome and its significance
3. Explain how nucleases can be used to reveal how nucleosomes interact with DNA
DNA IN EUKARYOTIC CHROMOSOMES IS COMPACTED IN CHROMATIN
Chromatin refers to DNA and all the proteins associated with it in chromosomes.
Most of the proteins in chromatin are small proteins called histones, which package DNA into
beadlike structures called nucleosomes.
Nucleosomes are packed into 30 nm fibers, which are assembled into loops on a nuclear scaffold.
Nucleosome = protein core and DNA
147 bp of DNA is wrapped ~1.65 times around the core
“Linker DNA” separates adjacent nucleosomes
HOW IS DNA WOULD AROUND THE NUCLEOSOME?
Negative supercoil, allows eukaryotic cells to maintain DNA in negative supercoils
INTERPRETING A GEL: NUCLEASE + NUCLEOSOMES
Middle concentration is core + linker, while first lane also shows increments of 150 bp which further
suggests that core + linker = 150 bp.
Lane 1: fragments are 450 bp, 300 bp, 150 bp MCB 52: READING AND LECTURE NOTES
Lane 2: every core + linker fragment separated: 150 bp
Lane 3: linker DNA completely digested, resulting in 100 bp being the length of the histone DNA
NUCLEOSOMES ARE MADE OF HISTONE PROTEINS
Histones are small protein
The proteins are rich in lysine and arginine
Hence they are called “basic” proteins
THE NUCLEOSOME CORE IS AN OCTAMER OF HISTONE PROTEINS
Two molecules each of H2A, H2B, H3, and H4
Nterminal tails of the histones protrude from the core and are accessible to modifying enzymes
H1 binds linker DNA
Histonefold DNA = conserved region of alphahelices
SUMMARY OF MAIN POINTS
DNA is wrapped around the nucleosome in a lefthanded toroid.
The nucleosome is an octamer composed of two copies each of the histones H2A, H2B, H3, and
Histones have tails that form grooves for wrapping of the DNA strands
Histone H1 compacts nucleosomes into 30nm fibers.
Chromosomes consist of loops of 30nm fibers.
CHAPTER 8 (GENOME STRUCTURE, CHROMATIN, THE NUCLEOSOME): PP. 220255 MCB 52: READING AND LECTURE NOTES
[09.16] L ECTURE 6: DNA R EPLICATION
Goal: To understand how the genetic material is duplicated and assess how duplication is achieved with
1. Explain how we know that DNA replication is semiconservative.
2. Explain how the driving force for DNA synthesis is pyrophosphate hydrolysis.
3. Identify the DNA components of the replication fork and predict the necessary features of each
for successful replication.
4. Describe continuous and discontinuous synthesis at the replication fork.
5. Compare and contrast the steps of DNA synthesis, DNA sequencing, and PCR.
THE MESELSONSTAHL EXPERIMENT (“THE MOST BEAUTIFUL EXPERIMENT IN BIOLOGY”)
Semiconservative replication (conservative, semiconservative, dispersive)
Grew E. coli in medium containing N for many generations. Next, shifted bacteria to medium
containing N for various times. Extracted DNA and subjected it to ultracentrifugation in cesium
chloride, separates DNAs of light, heavy, and hybrid densities.
1 Generation 2 Generation (only in L, N)
Conservative L, H L, H (more of L)
SemiConservative I L, I
Dispersive I I* (ratio slightly shifted)
CHEMISTRY OF DNA SYNTHESIS
DNA is synthesized from 2’deoxynucleoside triphosphates (dNTPs)
DNA synthesis requires a primer, and more specifically the 3’OH
The incoming nucleotide is specified by base pairing
Synthesis occurs in a 5’ to 3’ direction with the release of pyrophosphate
The driving force for DNA synthesis is hydrolysis of pyrophosphate
SUBSTRATES FOR REPLICATION MCB 52: READING AND LECTURE NOTES
Base (A, G, C, or T) attached to 1’ of deoxyribose sugar
αphosphate attached to 5’ of deoxyribose sugar
SN2 attack of the 3’OH to αphosphate, produces pyrophosphate which is broken down into two
phosphates by pyrophosphatase
DNA SYNTHESIS IS AN ENERGETICALLY COUPLED REACTION
Reaction one: NMP )+XTP → (XMP ) +P−P , ΔG = small, modestly favorable
Reaction two: P−P→2P i , ΔG = 7.3 kcal/mol
Combined reaction: ΔG > filter reaction through positively charged membrane (DNA sticks and nucleotides
Measure label associated with filter to determine amount of new DNA synthesis
Over time, more radioactivity associated with the filter
Equilibrium density gradient centrifugation: separate by density (MeselsonStahl)
Sedimentation gradient centrifugation: separate by size (Okazaki’s experiment) >> such as using
sucrose inside the tube, large strands will move more successfully
HOW DID THIS EXPERIMENT HELP OKAZAKI DEMONSTRATE THE EXISTENCE OF SHORT,
DISCONTINUOUSLY SYNTHESIZED DNA STRANDS?
Okazaki grew E. coli in a medium containing radioactive precursors for a brief period of time
called a “pulse”. He immediately isolated the DNA, denatured it, and determined the sizes by
On the left, at very short times of labeling (short pulses) very short pieces of DNA are found.
However, with longer times, the pieces of DNA get increasingly longer (120 sec).
He tried the same experiment with a mutant virus that was defective in DNA ligase, so the labeled
pieces of DNA remained short, even after long times of radiolabeling. MCB 52: READING AND LECTURE NOTES
DNA SEQUENCING: TAKING ADVANTAGE OF THE CHEMISTRY IN DNA SYNTHESIS
DNA sequencing requires a primer.
DNA sequencing uses 2’deoxynucleoside triphosphates (dNTPs) plus dNTP analogs.
Dideoxychain termination method for DNA sequencing
CHAPTER 9 (THE REPLICATION OF DNA): PP. 257276
THE CHEMISTRY OF DNA SYNTHESIS
DNA Synthesis Requires Deoxynucleoside Triphosphates and a Primer: Template Junction
The four dNTPs: three phosphoryl groups attached via the 5’OH of the 2’deoxyribose
Primer:template junction: template provides the ssDNA that directs the addition of each
complementary deoxynucleotide, primer is complementary to, but shorter than, the template
Template must have exposed 3’OH adjacent to the singlestrand region of the template
DNA is Synthesized by Extending the 3’ End of the Primer MCB 52: READING AND LECTURE NOTES
Phosphodiester bond formed in an S 2Nreaction in which 3’OH at the end of the primer strand
attacks the αphosphoryl group of the incoming dNTP. Leaving group for the reaction is
The dNTP that pairs with the template strand is highly favored for addition to the primer strand.
Hydrolysis of Pyrophosphate is the Driving Force for DNA Synthesis
Free energy for addition of a nucleotide to a growing polynucleotide change is rather small.
Additional free energy is provided by the rapid hydrolysis of the pyrophosphate into two
phosphate groups by an enzyme known as pyrophosphatase.
Coupled process is a highly favorable reaction with a ΔG of 7 kcal/mol and an equilibrium
constant of ~10 , DNA synthesis is effectively irreversible.
THE MECHANISM OF DNA POLYMERASE
DNA Polymerases Use a Single Active Site to Catalyze DNA Synthesis
DNA polymerase uses a single active site to catalyze addition of any of the four dNTPs, through
exploiting nearly identical geometry of the A:T and G:C base pairs.
Does not detect exact nucleotide that enters the active site, instead, monitors whether the
incoming nucleotide can form an A:T or G:C base pair
Only when a correct base pair is formed are the 3’OH of the primer and αphosphate of the
incoming nucleoside triphosphate in the optimum position for catalysis to occur.
Incorrect base pairing leads to dramatically lower rates of nucleotide addition as a result of
catalytically unfavorable alignment of substrates, example of kinetic proofreading
Kinetic proofreading: enzyme favors catalysis using one of several possible substrates by
increasing rate of bond formation only in presence of correct substrate.
DNA polymerases are able to distinguish between rNTPs and dNTPs through steric exclusion of
rNTPs from the DNA polymerase active site. Nucleotidebinding pocket cannot accommodate a
2’OH on the incoming nucleotide, space is already occupied by two amino acids that make van
der Waals contact with the sugar ring. If these amino acids are changed to ones with smaller side
chains, DNA polymerase is no longer effective at distinguishing between dNTPs and rNTPs.
DNA Polymerases Resemble a Hand That Grips the Primer: Template Junction
DNA substrate sits in a large cleft that resembles a partially closed right hand: thumb, fingers,
Palm domain: β sheet, contains primary elements of catalytic site, binds two divalent metal ions
(Mg or Zn ) that later chemical environment around correctly basepaired dNTP and the 3’OH MCB 52: READING AND LECTURE NOTES
of the primer. Monitors base pairing of most recently added nucleotides, making extensive H
bond contacts with base pairs in the minor groove, with contacts only forming if recently added
nucleotides are correct. Slowed catalysis and reduced affinity for mismatched DNA >> release of
primer strand and signals for proofreading nuclease to remove the mismatched DNA.
1. One metal ion reduces affinity of 3’OH for its hydrogen, generating a 3’O primed for
nucleophilic attack of the αphosphate on the incoming dNTP.
2. Second metal ion coordinates negative charges of the other two phosphates of the dNTP,
stabilizes the pyrophosphate produced by joining primer and incoming nucleotide.
Fingers: residues within fingers bind to incoming dNTP. Once a correct base pair is formed
between dNTP and template, finger domain encloses the dNTP. Stimulates catalysis by moving
incoming nucleotide into close contact with catalytic metal ions. Also associates with the
template region, creating 90 ° turn of the phosphodiester backbone between first and second bases
of the template, which exposes the first template bases after the primer and makes it clear which
template base should pair with the next nucleotide.
Thumb: interacts with DNA that has been most recently synthesized, 1) maintains correct position
of thumb and active site, 2) helps maintain strong association between DNA polymerase and
Ordered series of events:
1. Incoming nucleotide basepairs with next available template base
2. Fingers of polymerase enclose basepaired dNTP, placing critical catalytic metal ions in
position to catalyze formation of next phosphodiester bond
3. Attachment of basepaired nucleotide to the primer leads to reopening of the fingers and
movement of primer:template junction by one base pair
DNA Polymerases are Processive Enzymes
Processivity: characteristic of enzymes that operate on polymeric substrates.
Degree of processivity: average number of nucleotides added each time the enzyme binds a
primer: template junction
Each DNA polymerase has characteristic processivity ranging from only a few nucleotides to
more than 50,000 bases added per binding event.
Initial binding of polymerase to primer:template junction is rate limiting, ~1 sec, once bound,
adds nucleotides very fast (in millisecond range).
Processivity facilitated by sliding of DNA polymerases along DNA template. Once bound, DNA
polymerase interacts tightly with much of doublestranded portion of DNA, including
electrostatic interactions between phosphate backbone and thumb and interactions between the MCB 52: READING AND LECTURE NOTES
minor groove of the DNA and the palm domain. Sequenceindependent nature of these
interactions permits easy movement of the DNA even after it binds to polymerase.
Each time a nucleotide is added to the primer strand, DNA partially releases from the polymerase
(Hbonds with minor groove broken, but electrostatic interactions with thumb maintained.) DNA
then rapidly rebinds to the polymerase in a position that is shifted by 1 bp using the same
Exonucleases Proofread Newly Synthesized DNA
One major limit to DNA polymerase accuracy is occasional (1 in 10 times) flickering of bases
into “wrong” tautomeric form. Alternate forms of bases permit incorrect base pairs to be correctly
positioned for catalysis. When the nucleotide returns to “correct” state, incorporated nucleotide is
mismatched and must be removed.
Proofreading exonuclease : degrade DNA starting from 3’ DNA end (exonuclease = only degrade
from DNA end; endonuclease = cut within a DNA strand)
Proofreading occurs without release of DNA from the polymerase. When mismatched base pair
present, primer:template junction destabilized, creating several base pairs of unpaired DNA.
Exonuclease active site has high affinity for singlestranded 3’ ends, so newly unpaired 3’ end
moves to the exonuclease active site. Removal of mismatched base allows primer:template
junction to reform and rebind the polymerase active site.
THE REPLICATION FORK
Both Strands of DNA are Synthesized Together at the Replication Fork
Replication fork: junction between newly separated template strands and the unreplicated duplex
Leading strand: newly synthesized DNA strand, moving towards overall direction of DNA
replication towards the replication fork
Lagging strand: direction of replication is away from the replication fork, strands of DNA
synthesized in discontinuous fashion, formation of Okazaki fragments which are later covalently
The Initiation of a New Strand of DNA Requires an RNA Primer
Primase: specialized RNA polymerase dedicated to making short RNA primers on an ssDNA
Unlike RNA polymerases involved in mRNA, rRNA, and tRNA synthesis, primase does not
require an extended DNA sequence to initate RNA synthesis.
Primase activity dramatically increases when associated with DNA helicase, which unwinds
DNA at the replication fork. MCB 52: READING AND LECTURE NOTES
Requirement for ssDNA template and DNA helicase association ensures that primase is only
active at replication site.
RNA Primers Must Be Removed to Complete DNA Replication
To replace RNA primers with DNA, RNase H recognizes and removes most of each RNA primer,
specifically degrades RNA that is basepaired with DNA, removing all of the RNA primer except
the one directly linked to the DNA end, because RNase H can only cleave bonds between two
Final ribonucleotide removed by a 5’ exonuclease.
DNA polymerase can then fill the gap until every nucleotide is basepaired, with the “nick” in the
DNA being repaired by DNA ligase, which uses highenergy cofactors (ATP) to create a
DNA Helicases Unwind the Double Helix in Advance of the Replication Fork
DNA helicases catalyze separation of the two strands of duplex DNA, bind two and move
directionally along ssDNA using energy of nucleoside triphosphate (usually ATP) binding and
hydrolysis to displace any DNA strand that is annealed to the bound ssDNA.
Typically, DNA helicases that act as replication forks are hexameric proteins that assume the
shape of a ring, complexes encircle one of the two single strands at the replication fork adjacent
to the singlestranded:doublestranded junction.
DNA helicases act processively, unwinding multiple base pairs of DNA each time they bind.
High processivity is due to the ring shape >> release of the helicase from the DNA requires the
opening of the ring (which is a rare event). Alternatively, helicase can dissociate when end of the
DNA strand is reached.
How does helicase bind to DNA in the first place? Obvious problem in circular chromosomes,
same problem in linear because helicase usually loaded onto internal site. >> Specialized
mechanisms that open DNA helicase rings and place it around DNA before reforming the ring.
Each DNA helicase moves in a defined direction, referred to as polarity of DNA helicase. Can be
a polarity of either 5’ > 3’ or 3’ to 5’, as direction is defined according to the strand of DNA
bound/encircled, rather than the strand that is displaced. So if the helicase is functioning on the
laggingstrand template, the polarity is 5’ > 3’.
Movement along DNA requires input of chemical energy, energy provided by ATP hydrolysis.
DNA Helicase Pulls SingleStranded DNA through a Central Protein Pore
SingleStranded DNABinding Proteins Stabilize ssDNA before Replication
Topoisomerases Remove Supercoils Produced by DNA Unwinding at the Replication Fork MCB 52: READING AND LECTURE NOTES
Replication Fork Enzymes Extend the Range of DNA Polymerase Substrates MCB 52: READING AND LECTURE NOTES
[09.18] LECTURE 7: THE REPLICATION MACHINE
Goal: To understand how the genetic material is duplicated and assess how duplication is achieved with
1. Explain the roles of the protein components of the replication fork.
2. Predict the effect on replication when one of these components is mutated or removed.
3. Explain the benefits and drawbacks of proofreading activity in a polymerase.
4. Explain why not all polymerases have the same processivity.
5. Explain why lagging strand synthesis needs a DNA primase.
THE REPLICATION MACHINE
DNA is synthesized by DNA polymerase: fast, proofreads, processive
Leading strand synthesis requires: DNA helicase, singlestranded DNA binding protein
Lagging strand synthesis requires: DNA primase to produce RNA primers
Leading and lagging strand synthesis are linked: the trombone model
DNA POLYMERASE IS FAST
DNA polymerase synthesizes DNA at rate of 800 nt/sec
But chromosomes are large: chromosome of E. coli is 4.6 x 10 base pairs
Since there are two replication forks, the overall rate is 1,600 nucleotides per second.
Thus it takes ~40 minutes to replicate the E. coli chromosome.
Eukaryotic chromosomes are much larger; the human genome contains 6 x 10 base pairs. Hence,
replication in higher cells initiates at multiple origins.
DNA POLYMERASE PROOFREADS ITS OWN WORK
DNA polymerase has a builtin 3’ to 5’ exonuclease that serves as a proofreader for detecting and
removing misincorporated nucleotides.
Clicker question: Given the enzyme and substrates used in DNA replication, why does DNA
synthesis proceed in the 5’ to 3’ direction? The reason is 3’ to 5’ synthesis is: MCB 52: READING AND LECTURE NOTES
× 5’ to 3’ synthesis chemically impossible: no, if 3’ to 5’ synthesis, the incomingOH
can still attack the phosphate groups of the chain.
× Unfavorable energy of reaction: actually, energy of the reaction is the same
because there is still formation of pyrophosphate
× Interferes with base pairing: no, still occurs as normal
× Proofreading is the issue ! If DNA synthesis proceeded in a 3’ to 5’ direction and
the nucleotide at the growing end were removed by a “proofreading”
exonuclease, the resulting growing end would be left with only an α phosphate
and no pyrophosphate leaving group, DNA synthesis would abort.
PROOFREADING IS IMPORTANT FOR REPLICATION
Bases can exist in two tautomeric forms (ex. rare “flickering” of a C into a tautomer)
The rare C tautomer can base pair with A instead of G during replication, leading to
misincorporation of an A.
Tautomerization sets a theoretical limit on the accuracy of nucleotide incorporation.
At a frequency of about 10 to 10 , adenosine gets incorporated instead of guanine.
Yet the rate of spontaneous mutation is about 10 per round of replication.
DNA POLYMERASE IS PROCESSIVE
DNA polymerase binds multiple nucleotides each time it binds.
A sliding clamp tethers DNA polymerase to the DNA template.
A clamp loader loads the sliding clamp on DNA, also able to recognize primer:template junction,
will be able to correctly load where the DNA polymerase needs to be.
A PRIMER EXTENSION ASSAY TO MEASURE PROCESSIVITY
Load: DNA polymerase (with a primer), processivity clamp, clamp loader, ATP
With clamp: full synthesis of the template
Without clamp: low concentrations of polymerase, keeps falling off > unreliable synthesis and
production of many short pieces, high concentrations > polymerase can keep attaching and
produce longer stretches
Sliding clamp greatly increases the processivity of DNA polymerase.
ELECTROPHORETIC MOBILITY SHIFT ASSAY (EMSA) IS OFTEN USED TO STUDY PROTEIN
DNA INTERACTIONS MCB 52: READING AND LECTURE NOTES
Take a labeled DNA, add varying concentrations of protein of interest.
Electrophorese the protein: DNA complexes under nondenaturing (sometimes called “native”)
EMSA with sliding clamp: even at high concentrations, small concentrations of DNAprotein
complex detected… because the ring can just slide off linear DNA.
HOW TO STUDY SLIDING CLAMP ON DNA?
Use clamp loader + nicked circular DNA + clamp
Add ATP, gelfiltration chromatography (another way to separate by size)
× Elution column filled with porous gel beads
× Molecules larger than the pores in the beads will come out first, because they
have to travel on the outside of the beads.
× Molecules smaller than the pores in the beads will enter the beads, take more
time to come out.
Sliding clamp associated with the DNA is elutionshifted to the left compared with the
free clamp (and in the sample with clamp + DNA, some clamp didn’t associate with
If you cut with a restriction enzyme that cuts once, you will get one peak of free clamp.
LEADING STRAND SYNTHESIS
The enzyme DNA helicase unwinds duplex DNA downstream of the replication fork in an
Singlestranded binding protein (SSB) holds the separated strands apart so that they
DNA polymerase with its sliding clamp continuously elongates the daughter strand.
LAGGING STRAND SYNTHESIS
The enzyme DNA primase generates an RNA primer on the lagging strand.
A molecule of DNA polymerase (also, with a sliding clamp) elongates the primer to
create an Okazaki fragment.
RNase H removes the RNA primer.
DNA polymerase fills in the gap. MCB 52: READING AND LECTURE NOTES
DNA ligase joins the Okazaki fragment to previously synthesized DNA.
CHAPTER 9 (THE REPLICATION OF DNA): PP. 277311 MCB 52: READING AND LECTURE NOTES
[09.20] L ECTURE 8: STRUCTURE AND MECHANISM OF DNA R EPLICATION
Goal: To understand how the genetic material is duplicated and assess how duplication is achieved with
1. Analyze data from and propose experiments to probe the function of the components of the
2. Predict the effect on replication when a topoisomerase is mutated or deleted.
3. Describe the basic elements for replication initiation and predict the consequence of mutations in
one of the components in replication initiation.
4. Explain the end replication problem and describe the requirement for replicating DNA ends.
COMBINING THE CHEMISTRY AND THE POLYMERASE
Role of the Ohelix: in the closed conformation, tyrosine interaction with the base, can
perform base stacking to stabilize the catalytic site.
*additional tyrosine located on the end of the Ohelix that brings the next base near the
catalytic site (in the closed conformation) and holds the next based to be copied away while in the
MULTIPLE POLYMERASES INVOLVED IN DNA REPLICATION
Major polymerases: Pol I, Pol III core + Pol III holoenzyme, Pol δ, Pol ε
Pol I: RNA primer removal DNA repair
Pol III core + Pol III holoenzyme: chromosome replication
Pol δ: laggingstrand synthesis; nucleotide and base excision repair
Pol ε: leadingstrand synthesis; nucleotide and base excision repair
REPLICATION CHANGES THE SUPERCOILING STATE OF THE DNA AND THEREFORE
Replication generates positive supercoils ahead of the replication fork
Topoisomerase II required to break the DNA, polymerase can pass DNA through the break >>
ORIGINS OF REPLICATION
Replication of chromosomes commences at specific sites called origins. MCB 52: READING AND LECTURE NOTES
Initiator proteins bind to specific DNA elements to activate replication.
Replication proceeds bidirectionally from origins with two replication forks that move away
from each other.
Replication of circular chromosomes yields interlocked daughter chromosomes and replication of
linear chromosomes yields tangled chromosomes.
Initiator proteins bind to specific DNA elements to activate replication.
*In circular DNA, topoisomerase II required to cut the two linked daughter molecules
Eukaryotic chromosomes contain multiple origins: initiation is very tightly controlled so that each
origin is activated once and only once per cell cycle.
It takes ~40 minutes to replicate the E. coli chromosome. In the lab, E. coli can divide every 20
minutes. How is this possible?
Replicating chromosomes are segregated > segments of chromosome that are replicated twice in a
circle that is otherwise replicated once.
REPLICATING CHROMOSOME ENDS
The lagging strand cannot be completely synthesized because on the end, there is no room for an
additional Okazaki fragment being formed.
Information loss during replication >> cells die
Telomere repeats help protect DNA of essential genes.
Telomerase: builds additional telomeric repeats, which solves the end replication problem. The
telomerase has an RNA component that can pair with the end of the template, creating a primer:
template junction that allows for DNA synthesis. The telomerase then translocates, happening in
successive rounds to keep adding repetitive sequences onto the end of the DNA.
CHAPTER 9 (THE REPLICATION OF DNA): PP. 277311 MCB 52: READING AND LECTURE NOTES
[09.23] L ECTURE 9: DNA DAMAGE , DNA REPAIR , AND MUTAGENESIS
Goal: to understand how cells repair and tolerate different types of DNA damage
1. List the sources of DNA damage and the repair pathway that is associated with each type of
2. Explain why cells need mismatch repair and describe a mechanism for how the proteins know
which base is correct.
3. Describe, based on the chemistry of the bases, why evolution has selected for T and not U, and
for C and not 5methyl C in DNA.
4. Compare and contrast DNA repair and DNA damage tolerance.
5. Design experiments to measure survival after DNA damage, test whether or not a chemical is a
mutagen, and quantitate mutagenesis.
6. Predict the phenotype for survival and mutagenesis if a gene is deleted or mutated in a DNA
repair or tolerance pathway.
HOW DOES DNA ACQUIRE MUTATIONS?
1. Not all replication errors are caught by proofreading.
2. Chemicals damage DNA.
3. Water damages DNA. (DNA is bathed in 55 molar water!)
4. Ultraviolet light damages DNA.
5. Xrays break DNA.
SYSTEMS THAT ENSURE GENETIC INTEGRITY
1. Mismatch Repair corrects replication errors that escape proofreading.
2. BaseExcision Repair mends single damaged bases.
3. Nucleotide Excision Repair mends damage involving more than one base.
4. DoubleStrand Break Repair fixes breaks in DNA.
5. Translesion Synthesis bypasses or “tolerates” damage.
OCCURRENCE OF MISMATCHES
If a mismatch is not corrected, it can lead to a mutation after the second round of replication. MCB 52: READING AND LECTURE NOTES
BASIC STEPS FOR REPAIR
3. DNA synthesis
4. Ligation MCB 52: READING AND LECTURE NOTES
MISMATCH REPAIR (IN E. COLI)
Mismatch repair catches and repairs mismatches.
Mismatch repair is carried out by three proteins: MutS, MutL and MutH.
MutS scans the chromosome for mismatches; recognizes the mismatch and kinks the DNA.
MutS bound to a mismatch recruits MutL and MutH, which nicks the DNA.
UvrD helicase, an exonuclease, DNA polymerase III, and DNA ligase completes the repair.
WHICH STRAND CONTAINS THE CORRECT BASE?
Dam methylase (DNAadenine methylase) recognizes and methylates the A in 5’–GATC3’ in E.
coli, but only after a delay.
Delay in methylation leads to the formation of hemimethylated DNA (one DNA strand has
methyl group, one strand does not).
The nick occurs on the unmethylated strand, which is the newly synthesized strand.
Thus, mismatch repair has to occur quickly, right after replication, because once both strands are
replicated, it’s not possible to distinguish between the two strands.
Nick could happen upstream or downstream.
A different nuclease will be involved in each case (name not important)
Must be able to work at some distance, because a 5’GATC3’ isn’t necessarily right next to a
HEREDITARY NONPOLYPOSIS COLORECTAL CANCER
Genetic defect in one of the genes encoding the human homologs of MutS or MutL.
Increased number of mismatches increases subsequent number of mutations from those
mismatches that are not repaired.
Mutations in genes controlling growth and division – increased risk of colorectal cancer.
CELLULAR DEFENSES AGAINST DNA DAMAGE
Two possible approaches to DNA damage:
1. DNA repair MCB 52: READING AND LECTURE NOTES
2. DNA damage tolerance >> translesion synthesis
SOURCES OF DNA DAMAGE
Exogenous: benzopyrene (byproducts of smoking tobacco – intercalates between base pair and
distorts the DNA helix), chemical (alkylation), UV radiation
Endogenous: chemical (alkylation and oxidation), hydrolytic (deamination and depurination) MCB 52: READING AND LECTURE NOTES
TYPES OF DNA DAMAGE
Example: oxidation of a G produces 8oxoG, one of the most stable mutant versions
Example: hydrolysis of cytosine > repair proteins recognize uracil in DNA
Repair proteins do not recognize deaminated 5methyl C in DNA
BASE EXCISION REPAIR
Damage generates unnatural bases.
Unnatural bases are recognized and removed by special repair systems known as DNA
Next, the gap is filled in by DNA polymerase and ligase, restoring the original nucleotide.
1) glycosylase cleaves off the purine, 2) AP endo cleaves to make 3’OH for Pol I and
exonuclease removes the sugar
UVINDUCED THYMINE DIMERS
Formation of cyclobutane ring > thymine dimer
Changes the structure of the helix
Cisplatin > treatment of cancer
NUCEOTIDE EXCISION REPAIR
Nucleotide excision repair is mediated by UvrA, UvrB, and UvrC and other proteins, which:
× Scan the DNA for lesions
× Unwind the DNA over the lesion
× Nick the DNA 12 nucleotides apart on both sides of the lesion
× Replace the oligonucleotide with newly synthesized DNA (UvrD removes the
strand of DNA with the lesion)
HOW DO WE TEST IF A SPECIFIC GENE/PROTEIN IS INVOLVED IN DNA REPAIR OR
Plate equal number of cells onto each plate.
Expose some plates with cells to UV light. MCB 52: READING AND LECTURE NOTES
Incubate, count, and calculate.
If wildtype: should be able to repair damage, plate exposed to UV will have similar number of
cells growing as plate not exposed to UV
If UVsensitive mutant: when exposed to UV, only very few cells grow
TRANSLESION SYNTHESIS (TLS): MODEL 1, POLYMERASE SWITCHING
Processive DNA replication >> encounters DNA damage
Replication stall and polymerase switching to a translesion polymerase which can synthesize
across the damage (highly errorprone)
Return of the replicative polymerase
TRANSLESION SYNTHESIS: MODEL 2, GAPFILLING TLS
No switching of polymerase, but the replicative polymerase just bypasses the lesion
Gap is filled by translesion polymerases
TLS DNA polymerase: more errorprone, lacks proofreading, less processive (do not want them
synthesizing DNA for the rest of the strand)
TLS DNA polymerase lacks the Ohelix, creating a large pocket; lacks the level of regulation in
Can be lesionspecific: one TLS polymerase specific for thymine dimers, different TLS
polymerase better suited for other lesions
AMES TESTASSAY FOR MUTAGENS
Use Salmonella bacteria culture requiri