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Flashcards in Chapter 4 Deck (48)
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0
Q

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

A

Polarity

1
Q

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

This forms a __ bond bt ___ and ___

A

Water

Glycosidic bond bt the sugar and the phosphoric acid.

2
Q

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

A

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

3
Q

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.

A

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

Amino tautomer

4
Q

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

A

Enol. O rather than N. keto tautomer.

5
Q

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

A

DNA replication and synthesis

6
Q

H bonds bt G and C form bt:

A
  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

7
Q

H bonds bt A&T:

Watson crick base pairing requires what?

A
  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

8
Q

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

A

Entropy, which stabilizes the double helix

9
Q

____contribute to the double helix stability via van der Waals interactions

A

Electron cloud configurations

10
Q

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

A

hydrophobic, free energy.

The hydrophobic effect

11
Q

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

A

complementary base pair specificity, change in free energy

12
Q

___ 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

A

Extra helical configuration

13
Q

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

A

Base flipping

14
Q

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

A

glycosidic bonds

15
Q

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.

A

ADAM, AADH

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

AHA, ADA

16
Q

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.

A

11, deep and narrow, broad, shallow

17
Q

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

A

aren’t on the same plane

18
Q

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

A

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.

19
Q

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

A

shield the negatively charged phosphate groups

20
Q

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

A

Hyperchromicity

21
Q

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.

A

GC content and ionic strength of the solution

Higher, higher

22
Q

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

A

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.

23
Q

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

A

small replicating genetic elements.

sticky ends

24
Q

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)

A

it doesn’t have free linear ends.

the chains can twist about each other

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

25
Q

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 _____.

A

permanently breaking sugar-phosphate bonds of the backbone

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

26
Q

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

A

torsionally stressed

Interwound/plectonemic

27
Q

__/___ 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

A

Torrid/spiral

spiral and Interwound writhes

28
Q

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° = ___

A

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).

29
Q

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

A

change in Lk.

torsionally strained AKA supercoiled.

30
Q

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

Superhelical density (sigma) =

A

Lk < Lk° and the change in Lk < 0

Lk > Lk° and change in Lk is > 0

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

31
Q

Average sigma = ___.

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

A

-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.

32
Q

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

A

denaturing at high temps.

33
Q

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.

A

torid/spiral writhe, torid/spiral writhe

negative supercoils!

34
Q

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

Type I topoisomerase makes ____.

A

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

35
Q

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 ___.

A

makes negative supercoils

Circular DNA

separate into individual daughter cells

36
Q

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

A

a nick or a gap

37
Q

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

A

Type II

38
Q

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.

A

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.

39
Q

Topoisomerase Enzyme Bridge. Type I: (seven steps)

How is type II topoisomerase different?

A
  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.

40
Q

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.

A

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.

41
Q

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

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

A

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

Increased

42
Q

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.

A

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.

43
Q

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

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

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

A

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)

44
Q

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:___

A

(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

45
Q

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 ___.

A

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

46
Q

Advantages of pseudoknot formation in ribozymes and riboswitches:

A

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

47
Q

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°?
A

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.

  1. 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.