MS-1 FUND 1: 10:10 – 11:00
Tuesday, September 2, 2014
Dr. Peter J. Detloff
Transcriber: Logan Riley
Editor: Kelsey Real
Page 1 of 4
DNA Structure, Replication, etc.
Abbreviations: H-bond = hydrogen bond; bp = base pair
Introductory Comments: This is the first hour of the lecture on DNA Metabolism. The title of the powerpoint presentation
for this lecture is called “DNA Metabolism”. Dr. Detloff begins by clearing up some student questions from his previous
lectures. Caffeine has the ability to block adenosine receptors, not adenine. He also said to understand the
representations of the structures of DNA; the main one he focused on was only drawing the base pair letter symbols. You
do not need to know the phosphate backbone diagram.
I.
What Sorts of Secondary Structures Can Double-Stranded DNA Molecules Adopt (slide 4, 4:50)
a. The stability of the DNA double helix is due to:
i. Hydrogen bonding between base pairs in two anti-parallel complimentary strands.
ii. Electrostatic interactions – helix is formed to get negatively charged phosphate groups as far away as
possible (facing the outer portion of the helix).
iii. Stacking of base pairs (slide 5, 6:10) – canonical A:T and G:C base pairs have to interact in 3D space. The
benzene rings are hydrophobic (excluding water) and the most thermodynamically favorable position is
achieved by stacking base pairs on top of each other.
b. The “canonical” base pairs (slide 6-7, 7:10): A:T and G:C base pairs have nearly identical overall dimensions
i. The major groove and minor groove are determined by unusual angles formed by the bases and occur
between the two C1 carbons.
ii. The major groove is large enough to accommodate an alpha helix from a protein; allows the alpha helix to
hydrogen bond with some base pairs while excluding others. 8:50
iii. Regulatory proteins (transcription factors) can recognize the pattern of bases and H-bonding possibilities in
the major groove 9:50
iv. Ex: Restriction enzymes recognize the structures of the bases and their side groups and that’s how they
know where to cut at specific sequences
1. They recognize specifically: amine group, methyl group, lack of amine group on adenine, keto group,
etc.
2. Recognizes sequence without melting the DNA by seeing features different between A:T and G:C
base pairs in major groove
3. Shows that base pairing is not the only recognition feature of DNA but can also use
patterns/sequences in major groove 11:00
v. A and T share two H-bonds
vi. G and C share three H-bonds
vii. G:C-rich regions of DNA are much more stable
viii. Polar atoms in the sugar-phosphate backbone also form H-bonds
c. Major and minor grooves (slide 8-9, 11:30)
i. Every time there is a major groove, there is a minor groove right next to it (slide 6, 6:56)
d. Comparison of A, B, Z Types of DNA (slide 10, 12:00)
i. Just know that there are different kinds of tertiary structure of DNA. Do not need to memorize specific pitch
angles, base pairs, etc. 12:15
ii. B DNA is the standard structure that we THINK is the main one.
iii. Basic summary:
Type
Structure
Pitch
A
L/R
handed
Right
Short and
broad
2.3 Å; 11 bp/turn
Base
configuration
Anti
B
Right
3.32 Å; 10 bp/turn
Anti
Z
Left
Longer and
thinner
Longest and
thinnest
3.8 Å; 12 bp/turn
G in syn config
C in anti config
Notables
Slightly dehydrated and
more stretched out
version of B-DNA
This the form normally
seen in the cell, standard.
Very structurally different
from A and B. Only found
in GC rich regions. Rare
iv. Handedness: Same as putting a screw into a wall, the threads being a helix: turn it clockwise to get a
right-handed screw; Counter-clockwise for a left-handed screw.(13:55)
e. Change in Topological Relationships of Base Pairs from B- to Z-DNA (slide 11, 15:06)
i. <Student Question> cannot hear
1. <Dr. Detloff answer>: RNA can be in the A form, and for example, the B-form can be
heterochromatin or euchromatin. There is not a big association between type of DNA (A,B, Z) and
hetero/eu-chromatin.
MS-1 FUND 1: 10:10 – 11:00
Tuesday, September 2, 2014
Dr. Peter J. Detloff
DNA Structure, Replication, etc.
Transcriber: Logan Riley
Editor: Kelsey Real
Page 2 of 4
Abbreviations: H-bond = hydrogen bond; bp = base pair
ii. Chart shows how B-DNA can have modifications in the middle of a strand to have Z-DNA form. When the
sequence is C-G rich.
iii. Switching from B-DNA to Z-DNA alters major/minor grooves, and could change gene expression as a result
iv. Will alter supercoiling
II. Denaturation of DNA (slide 12, 16:30)
a. The basic model of DNA is the Watson and Crick strands in anti-parallel, complementary form
b. Denaturation is when those strands break apart and can float apart
c. The base stacking (specifically the resonance properties) limits the amount of UV light that can be
absorbed ; when you heat up DNA, the stacking is gone and the absorbance increases
d. Hyperchromic shift: when UV absorbance by bases increases by 30-40% after DNA is denatured
1. This reflects the unwinding of the DNA double helix
2. Stacked base pairs in native DNA absorb less light and denatured bases absorb more light
3. Easy to tell if you have single or double stranded DNA, and how much
ii. Bases can be allowed to re-find complementary base (re-annealing). (18:53)
1. In repairing those damage, some bases can be mis-paired and become mutations
e. Low ionic strength also favors denaturation
i. When there are positive charges in solution, they help stabilize the negative charges on the phosphate
backbone and reduce repulsion.
ii. In pure water, there would be much less positive ions to do this and the DNA would fly apart (19:56)
iii. Slide 13 (20:05) shows different genomes and their hyperchromic shifts compared to the percentage of GC bp.
1. Denaturation occurs at higher temperatures in G-C rich DNA because there are three hydrogen
bonds between G-C. This is stronger than the two hydrogen bonds found between A-T
2. The amount of G is always equal to C, but can be different than amount of A and T.
III. What happens when the 2 strands re-nature (slide 14, 21:20):
i. GGG anywhere in the genome can react with CCC somewhere else other than the correct place on the
other strand
ii. The nucleation step is slow (slide 15, 22:08)
1. This allows breaking and reforming of hydrogen bonds until correct sequence between strands is
found (rate limiting step).
iii. The zippering step is very fast
iv. If you snap cool the DNA, the base pairs are going to go together wrong. (22:30)
1. Small correct structures will go together but the cooling won’t allow incorrect structures to break and
reform
v. When you lower the temperature slowly, the bases will naturally find its complement and reform the helix
over time
IV. Tertiary structures of DNA that involve supercoiling (slide 16, 24:01)
a. Mathematics of Topology
b. Supercoils
i. Long, thin strand that twists around (think phone cord in the old days)
ii. When the DNA crosses over itself - that is a supercoil.
iii. In order to have supercoiling, the two ends have to be anchored (26:40)
1. If one of the anchors lets go, the supercoil is relaxed and topology is gone.
iv. Different ways of twists; part of the DNA backbone
1. (a) Solenoidal twist in a circular molecule of DNA
MS-1 FUND 1: 10:10 – 11:00
Tuesday, September 2, 2014
Dr. Peter J. Detloff
DNA Structure, Replication, etc.
Transcriber: Logan Riley
Editor: Kelsey Real
Page 3 of 4
Abbreviations: H-bond = hydrogen bond; bp = base pair
a. Comes from the word solenoid which is the wrapping of wire around
electric source/magnet/bolt
b. Similarly, DNA wraps around substrate (like how it wraps around
histones to condense form)
2. (b) Plectonemic supercoiling because you can see crossovers with 4 writhes
a. Equivalent to having nucleosomal-type structure
3. (c) Structures have to be either circular (self-anchoring) or anchored to
another substrate
a. Ex: In a chromosome there are anchor points that DNA can’t twist
around. Anything in anchor point (granted phosphate backbone intact
with no nicks) can have supercoiling.
b. Backbone must be intact for supercoiling to be possible and there are
enzymes that ensure that the backbone is intact.
v. Linking number: characteristic of a what a supercoil piece of DNA (slide 17, 28:05)
1. Linking number does not change granted there are no breaks in
phosphodiester bonds or removal of anchors
2. Linking number = twists + writhes (L = T + W)
a. Twists (T): number of twists of double helix (approx. 10 bp/turn) (slide
20, 30:50)
b. Writhes (W): number of supercoils that it has
c. Twists can be interconverted with writhes, because they add up to the
same number, BUT they are not the same.
c. Gyrases are a type of topoisomerase that can introduce or remove supercoils 
i. Eukaryotes and bacteria need a certain level of supercoiling to function, once that density is changed, the
cell will die – this is why gyrase can be a good target for antibiotics. Humans have Type II topoisomerase
that is different from the gyrase found in bacteria. Hence why therapeutically, antibiotics attacking the
gyrase will affect only the bacteria.
1. How gyrases work: see figure on right (slide 18, 31:30)
a. Start with DNA loop (every DNA loop has certain linking number)
b. Add DNA gyrase and ATP
c. Break phosphodiester bonds on one of the strands, let it cross over to the other strand to
form a loop (conformational change)
d. Gryase re-ligates and causes formation of two negative supercoils  this requires energy
i. Negative supercoils can unwind a portion of the DNA molecule and make the single
strands easier to access
e. Every catalytic cycle causes two negative supercoils (bottom image on the figure = after
1 cycle) (slide 19, 33:40)
2. Why this is important: gyrases relieve tension in DNA coil by changing the linking number:
(a-c) follow the figure on the slide (slide 20 34:20)
a. (a) Shows a relaxed state circular DNA with no supercoils. You’re given that it has 400 bp.
i. Assume that the DNA is in the normal B form, you can calculate the linking number
knowing that you have10 bp/turn
ii. T, W = 0 because there are no supercoils
iii. L = 400 bases/10 bp/turn = 40
b. (b) Now add gyrase to add negative supercoiling
i. 2 catalytic cycles of gyrase = 2 ATPs consumed
ii. From part 1, we know that 1 cycle = 2 supercoils. So for 2 cycles, that means 4
supercoils.
iii. T = 0 because everything is base-paired, but this makes W = - 4 (crossover points in
the neg. supercoil)
iv. The linking # is changing b/c you broke phosphodiester bonds. Therefore, since L = T +
W, change in L = 40 + (-4) and the new linking number is now 36
v. Since L is now 36 instead of 40, the change in L is 4. Since you know that there are
10bp/turn, this is topologically and biologically equivalent to removing/melting 40 bp
(4x10bp; reverse of item a. iii above)
vi. Negative supercoiling is equivalent to opening DNA up (i.e. destroying BP) (37:25)
1. DNA opening is important for when the cell needs to replicate or
transcription via RNA Polymerase.
MS-1 FUND 1: 10:10 – 11:00
Tuesday, September 2, 2014
Dr. Peter J. Detloff
DNA Structure, Replication, etc.
Transcriber: Logan Riley
Editor: Kelsey Real
Page 4 of 4
Abbreviations: H-bond = hydrogen bond; bp = base pair
vii. All ambient temperature organisms (bacteria, yeast, humans, etc.) have a negative
superhelical density to their chromosomes because opening needs to be favored.
1. Some organisms, bacteria living in boiling water, do have positive
supercoiling, and it makes it more difficult to unwind the DNA.
c. Twist of -2 can be reoriented to writhe of -2, either be plectonemic or solenoidal. (slide 21,
39:00)
<Break occurred early, lecture resumes at 49:10>
V. DNA Replication
a. Basis of Molecular Biology
i. DNA, as seen in its structure, has 2 strands that are complements of each other, allowing for an enzyme to
come in, separate the two strands, and make copies of each original strand so that now 4 strands
ii. Thus DNA is semiconservative – new DNA made is composed of one old strand and one new strand.
Every new piece created was based off an original template.
1. Illustrated by figure 28.1
iii. DNA is also bidirectional
b. How is DNA replicated
i. Polymerase adds to the 3` free hydroxyl end of the DNA
ii. Therefore for most polymerases to work, you need a primer, that is a small piece of RNA or DNA that
contains a 3` end and that attaches to the DNA you want to replicate.
iii. Nucleotide triphosphate comes in to attach and two of its phosphates come off as pyrophosphate.
1. Remember that the bonds between the phosphate are very high energy. This make the addition of a
nucleotide to the DNA even more favorable.
2. Why us a triphosphate bond?  If you had a diphosphate bond, then would have inorganic
phosphate coming off in the reaction and it would build up and stop the reaction (because of Le
Chatlier’s principle wherein the buildup of the products causes the forward reaction to slow down).
3.
iv.
VI.
V
No student questions
<END OF LECTURE 39:20>