Chapter 4

Question 0

Phosphodiester linkage yeilds__ defined by the asymmetry of the nucs and how they’re joined

Answer 0

Polarity

Question 1

Two ___ molecule are removed from between the base and the OH on the 1’ carbon of deoxyribose.

This forms a __ bond bt ___ and ___

Answer 1

Water

Glycosidic bond bt the sugar and the phosphoric acid.

Question 2

DNA chains have a free __ or __ at one end and a free __ or ___ at the other end.

Answer 2

5’ phosphate or 5’ OH at one end and a free 3’ phosphate or 3’ OH at the other.

Question 3

N atoms rarely assume the __ tautomeric form, which has no 2nd H on the ___. But instead has an H on the ___.

The common form is the __ tautomer.

Answer 3

Imino. No second H on the 4’ NH group, instead has an H on the 3’ N

Amino tautomer

Question 4

Guanine rRely takes on its __tautomeric form! which places a H on the top __ rather than the adjacent __. Guanine common form is the __ tautomer.

Answer 4

Enol. O rather than N. keto tautomer.

Question 5

The capacity to form an alt. tautomer is a frequent source of error during

Answer 5

DNA replication and synthesis

Question 6

H bonds bt G and C form bt:

Answer 6

1. N1 of G & N3 of C
2. Carboxyl at C6 on G & exocyclic N2 @ carbon #4 of C
3. exocyclic NH2 @ carbon #2 on G and carboxyl at carbon #2 on C

*these aren’t ordered yet, must edit

Question 7

H bonds bt A&T:

Watson crick base pairing requires what?

Answer 7

1. Exocyclic amino group at carbon 6 of A with the carboxyl at carbon 4 of T
2. N1 of A with N3 of T

The preferred tautomeric states

Question 8

The disorder caused by the breaking of H2O molecules during nucleotide synthesis increases __ which stabilizes the __

Answer 8

Entropy, which stabilizes the double helix

Question 9

____contribute to the double helix stability via van der Waals interactions

Answer 9

Electron cloud configurations

Question 10

Surfaces of the bases are __ and stacking them in the middle lowers their interaction with water, this lowers the helix’s __ ___.

Answer 10

hydrophobic, free energy.

The hydrophobic effect

Question 11

Hydrogen bonds are important for ________ and thermodynamic stability. There is no ___when they are broken or made during nucleotide synthesis

Answer 11

complementary base pair specificity, change in free energy

Question 12

___ is when the base is flipped out, protruding from the helix. The base sits in the catalytic cavity of the methylation enzymes. Aka base flipping

Answer 12

Extra helical configuration

Question 13

Enzymes involved in the removal of damaged bases and homologous recombination scan DNA for homologies and lesions via __ __.

Answer 13

Base flipping

Question 14

Each base pair is displaced 36° from the previous base pair. One stack of 10 base pairs gets around one helical turn. The grooves are caused by the 120° angle and the 240° angles between ___. If the sugars pointed away from each other, in a straight line, it would be 180° and no grooves would form

Answer 14

glycosidic bonds

Question 15

Edges of each base pair are exposed in the grooves. ___signifies an AT pair and a ___ signifies a GC pair.

The protruding edges are important because they stick out and _____

the minor groove of A:T is ___ & the minor group of G:C is ___.
The minor groove is useful for proteins recognizing correct B form DNA, but not useful for seq specificity.

Answer 15

ADAM, AADH

allow proteins to recognize the DNA sequence without opening that helix.

AHA, ADA

Question 16

B form DNA is seen with high humidity it is the average structure for DNA and it has 10 base pairs per turn. It has a wide major groove and a narrow minor groove. A form DNA is seen under low humidity it contains ___ base pairs per turn and has a major groove that is ______and has a _____minor groove.

Answer 16

11, deep and narrow, broad, shallow

Question 17

A propeller twist is seen when the bases in a pair ____

Answer 17

aren’t on the same plane

Question 18

Left-handed helix (Z DNA) occurs when the ____. There is syn confirmation at the ___ residues but it stays anti at the ___ residues. This causes the sugar to change it’s pucker

Answer 18

glycosidic bond of a 1’ position on the sugar is in syn confirmation (right-handed DNA has this bond in anti-confirmation.)

syn confirmation at the purine residues, but stays anti at the pyrimidine residues.

Question 19

Left-handed helix occurs at high concentrations of positively charged ions, which ____

Answer 19

shield the negatively charged phosphate groups

Question 20

____ occurs in base pairs when they are heated, this causes more UV absorbance from the nucleotides.

Answer 20

Hyperchromicity

Question 21

When denaturing DNA, melting point is impacted by ___ & ____. Melting point is higher when the percentage of GC is ___. And, the higher the concentration of salt the ___ the melting temperature will be.

Answer 21

GC content and ionic strength of the solution

Higher, higher

Question 22

What will occur when you put DNA in high ionic concentrations?

Answer 22

The negatively charged phosphate groups of the backbone repel each other. In high ion concentrations, the cations shield the phosphates’ charges and the helix stabilizes.

Question 23

Some bacteria have plasmids which are:

Bacteriophage lambda is a DNA virus of E. coli. It is initially linear and becomes circular after infection due to

Answer 23

small replicating genetic elements.

sticky ends

Question 24

Covalently closed circular DNA is constrained topologically because ___. So the absolute number of times that ____ is restricted.

Linear DNA is somewhat constrained due to its _____. (3 things)

Answer 24

it doesn’t have free linear ends.

the chains can twist about each other

long length, entrainment in chromatin, and interaction with cellular components.

Question 25

Two strands of cccDNA cannot be separated without ____, due to it having no interruptions in this sugar phosphate backbone. So instead, one strand is passed through the other repeatedly. The linking number is how many times one strand _____.

Answer 25

permanently breaking sugar-phosphate bonds of the backbone

must be passed through the other in order to separate the two strands entirely.

Question 26

The linking number is equal to the sum of the twist and the writhe. Writhe is ____ cccDNA in which the long axis which crosses over itself repeatedly.
AKA ___/___writhe = long axis twists around itself

Answer 26

torsionally stressed

Interwound/plectonemic

Question 27

__/___ writhe is when the long axis is wound in a cylindrical manner (like when DNA wraps around a protein).

Linking number is the total number of ___ in cccDNA. Twist and writhe are interconvertible, but the sum of them MUST remain equal to the linking number. Lk = Tw + Wr

Answer 27

Torrid/spiral

spiral and Interwound writhes

Question 28

DNAse I can ___ if used mildly to break, on average, one phosphodiester bond in each DNA molecule. This is known as nicking. If nick is repaired, the cccDNA molecule will ____.

Lk° = ___

Answer 28

relax negative supercoils

be relaxed and will have a Lk approx equal to Lk° with slight supercoiling due to the Lk° not being an integer because Lk° = (#bps)/(10.5).

Question 29

Supercoiling is measured by the ___. Lk - Lk°. if change in Lk is significantly different than zero, the cccDNA is ___ AKA ___.

Answer 29

change in Lk.

torsionally strained AKA supercoiled.

Question 30

If ____ and ___, DNA is negatively supercoiled. If ___ and ___, DNA is positively supercoiled.

Superhelical density (sigma) =

Answer 30

Lk < Lk° and the change in Lk < 0

Lk > Lk° and change in Lk is > 0

Superhelical density (sigma) = (change in Lk)/Lk°

Question 31

Average sigma = ___.

Negatively supercoiled DNA strands open more easily than relaxed strands because ____, making the bases_____.

Answer 31

-0.06

the free energy stored in the Lk = Tw + Wr negative supercoils can be converted into energy used to partially untwist the double helix, making the bases accessible and the strands opened up.

Question 32

Only certain thermophiles have positively supercoiled DNA to keep their DNA from ____. Their strand separation requires more energy.

Answer 32

denaturing at high temps.

Question 33

Nucleosomes are a form of ____ which occurs in a ___ manner.

Writhe in the form of left handed spirals is also known as ____! Thus, nucleosomes give DNA its negatively supercoiled density.

Answer 33

torid/spiral writhe, torid/spiral writhe

negative supercoils!

Question 34

Type II topoisomerase changes the Lk by ____. Type II requires energy from ___.

Type I topoisomerase makes ____.

Answer 34

Making double stranded breaks in the DNA through which they pass a segment of uncut duplex DNA before resealing the DNA.

ATP hydrolysis

single stranded breaks, allowing the uncut strand to pass through the break before resealing

Question 35

Special type II topoisomerase in bacteria which ____, rather than removes them, is DNA gyrase. DNA gyrase can catenate and decatenate ___ DNA. Catenated DNA molecules usually occur after a round of replication and must be decatenated in order to ___.

Answer 35

makes negative supercoils

Circular DNA

separate into individual daughter cells

Question 36

Decatenation uses Type II topoisomerase, but if there is ___, Type I can be used.

Answer 36

a nick or a gap

Question 37

Eukaryotes use type ___ topoisomerase to detangle their linear, catenated DNA molecules during mitosis.

Answer 37

Type II

Question 38

Topoisomerases don’t require any outside proteins or energy (i.e. ATP) because ____. Then, the topoisomerase is covalently joined to one of the ___ via ____. The other end terminates with a _____ held by topoisomerase. This cleavage uses energy from the ___ bond.

Answer 38

Topoisomerases don’t require any outside proteins or energy ie ATP because: they are placing a tyrosine residue in the active site of the topoisomerase, which attacks a phosphodiester bond in the backbone of the target DNA.

Then, the topoisomerase is covalently joined to one of the BROKEN ENDS via phosphotyrosine LINKAGE. The other end terminates with a FREE OH GROUP help by topoisomerase. This cleavage uses energy from the phosphodiester bond.

Question 39

Topoisomerase Enzyme Bridge. Type I: (seven steps)

How is type II topoisomerase different?

Answer 39

1. Binds to duplex DNA and melts the strands
2. One of the DNA strands binds to a cleft in the enzyme topoisomerase that places it near the active site of tyrosine
3. The DNA strand is cleaved to generate a DNA-tyrosine intermediate, while the other end is bound by the enzyme topoisomerase.
4. Topoisomerase goes through a confirmation change to open up a gap
5. The the un-cleaved strand passes through the gap and binds to the DNA binding site located in a hole in tyrosine.
6. A second confirmation change occurs in the enzyme-DNA complex which brings the cleaved strand back together via attack of the OH end on the phospho-tyrosine bond.
7. After rejoining, the enzyme topoisomerase opens up and released the DNA.

Type II is dimeric and uses two subunits to cleave both strands.

Question 40

The more ___ and the higher _____ a cccDNA has, the faster it will run through a gel. The more ___ the cccDNA is, the slower it will run.

Answer 40

The more COMPACTED and the higher THE LINKING NUMBER a cccDNA has, the faster it will run through a gel. The more RELAXED the cccDNA is, the slower it will run.

Question 41

Ethidium intercalates between the bps and causes a ____, which ____.

Lk = Tw + Wr…so if twist lessened, the Wr is ___.

Answer 41

26° unwind, reducing it from 36° per bp to 10° (decreases the twist)

Increased

Question 42

When ethidium is added to a linear DNA or a nicked DNA, it cause ___ to increase. When added to a cccDNA, the ___ doesn’t change but the twist ___ for each molecule of ethidium that binds. Thus, ethidium will relax the DNA and can even relax it enough to cause ____. This will then cause the cccDNA to migrate on a gel more rapidly. But if only enough ethidium is added for the cccDNA to be ___, and not ___, it will migrate more slowly.

Answer 42

helical pitch

The Lk doesn’t change, but the twist decreases by 26° for each molecule of ethidium that binds.

To cause positive supercoils.

But if only enough ethidium is added for the cccDNA to be RELAXED, and not SUPERCOILED, it will migrate more slowly.

Question 43

G:C vs C:G is distinguable using:____

A:T vs G:V is distinguishable using:____

A:T vs T:A is distinguishable using:____

Answer 43

G:C vs C:G is distinguable using major groove
(AADH vs HDAA)

A:T vs G:C is distinguishable using major and minor grooves
(ADAM/MADA vs AADH/HDAA)

A:T vs T:A is distinguishable using major groove
(ADAM vs MADA)

Question 44

4 helical turns of B form DNA will have ___bps.

4 helical turns of B form DNA in solution will have ___bps.

4 helical turns of A form DNA will have ___bps.

4 helical turns of Z form DNA will have __ bps.

Vertical helix length in 4 helical turns:___

Answer 44

(4 turns)(10bps) = 40 bps in B form DNA

(4 turns)(10.5 bps) = 42 bps in solution

(4 turns)(11 bps) = 44 bps in A form DNA

(4 turns)( 12 bps) = 48 bps in Z form DNA

(1.2 nm minor groove + 2.2 nm major groove)(4)
= (3.4nm)(4 turns) = 13.6 nm

Question 45

Z DNA is ___ handed, with majors and minor grooves that are___.

Formation of Z DNA is generally unfavorable, but it can be promoted via ___, ____, or ___.

Answer 45

Left handed, unlike A and B form DNA. The grooves show little difference in width.

Formation of z DNA is generally unfavorable. It can be promoted via alternating purine-pyrimidine sequence, negative DNA supercoiling, or high salt and some cations

Question 46

Advantages of pseudoknot formation in ribozymes and riboswitches:

Answer 46

Pseudoknots add stability to stem loop structures which are important for specific seq interactions with proteins.

Question 47

You treat a 10000 bp cccDNA plasmid with DNase I under conditions such that the DNase I nicks, on average, once per DNA molecule.

1. Name the bond which DNase I breaks
2. How does this change the topological state of the 10000 bp plasmid?
3. You then inactivate the DNase I and treat the DNA with DNA lipase + ATP, what does this do? What does this do in respect to Lk & Lk°?

Answer 47

1, phosphodiester bond (DNase I will nick this bond in the sugar-phosphate backbone)
2. The DNA is originally supercoiled. After DNase I treatment, the DNA is no longer a cccDNA and becomes relaxed.

3. Treatment with DNA lipase reseals the nick, allowing for the formation of cccDNA again.
   The value of Lk° is not an integer (10,000/10.5). Lk is ALWAYS an integer, so they are not equal. Therefore, there must still be some slight supercoiling.