Lecture 2 Nucleic Acids

Question 1

Cooperative bonding

Answer 1

The formation of base pairing trough hydrogen bonding is cooperative -> As soon as asingle formation of hydrogen bonding has accured and a base pair has come together it will be easier for the next bond to be formed and the rest of the bases to pair as well.The first one increases the probability of formation of the next base pair.

Question 2

stabilization provided by the hydrogen bonds in the double helix

Answer 2

the difference in stability between the hydrogen bonds that the bases form in water and the hyrdrogen bonds between bases in the double helix.
Before the helix is formed there is interaction with water molecules because you are in a watery environment. Because of that the effective contribution of hydrogen bonds to the stability of DNA is less than the intrinsic strenght of the hydrogen bonds, because of the competition with water.

Question 3

Pi-pi stacking

Answer 3

Non-covalent attraction between aromatic rings.
Happens between base pairs in the DNA helix and provides the dominant contribution to stabalizing the double-helix.

Question 4

anti conformation of nucleotide

Answer 4

Conformation with lowest energy and therefore the most frequently observed.
H1 atom of sugar and C8 atom of th base are in a trans formation (C8 on the left of the glycosidic bond and H1 on the right)

Question 5

Syn conformations of nucleotide

Answer 5

H1 atom of the sugar and the C8 atom of the base are in a cis conformation (both on the same side of the glycosidic bond)

Question 6

Sugar puckering (what is it?)

Answer 6

C2 and C3 of the sugar can move to be more energetically favorable.
When the out-of-plane atom is located on the same side of the plane as C5, the formation is referred to as endo (inside). When it is located on the opposite side of C5 it is called exo (outside).
C2’ exo -> C2 on oposite side of C5
C2’ endo -> C2 on the same side of C5
C3’ exo -> C3 on oposite side of C5
C3’ endo -> C3 on same side of C5

Question 7

Sugar puckering in DNA

Answer 7

• Endo sugar puckers are more common than exo.
• In DNA sugar pucker can be eather C2’ or C3’ endo -> C2’ endo in B-form and C3’ endo in A-form.

Question 8

Sugar puckering in RNA

Answer 8

C2’ endo is not possible in RNA, because of the steric hindreance between the OH group on C2 and the phosphate group on C3, so only C3 can be observed.
The distance is only 1,9 A, which is very close. This is not a problem in DNA because there is a H atom there but in RNA there is a OH and this will bring a clash between the OH and the phosphate in the C2 endo confomation.

Question 9

B-form of DNA

Answer 9

• Most common form of DNA
• C2’ endo sugar pucker
• The complementary chains are parallel, but run in opposite directions, and together they twist into a right-handed double helix. The helix forms two grooves of unequal size.

Question 10

Major groove

Answer 10

Wider and more accessible groove that allows regulatory proteins or other molecules to gain access to the nucleotide functional groups on the edges of the groove.

Question 11

Minor groove

Answer 11

The narrower an less accessible groove that allows much more limited access to the functional groups lying within the groove.

Question 12

Why interaction with major groove instead of minor groove?

Answer 12

-Recognition sites are placed at the edges of base pairs in ds-DNA, which are exposed witin the major and minor groove. They are recognized by transcription factors.
-The pattern of recognition elements on the sites are unique for each 4 base pairs. But in the minor groove the pattern for C-G in indistinuishable from G-C (same for A-T/T-A)
-So the major groove of DNA allows each of the 4 kind of base pairs to be distinguished from each other, whereas interactions in the minor groove don’t

Question 13

B-form and A-form in RNA

Answer 13

RNA can’t form the B-form because it can’t form the C2’ endo sugar pucker.

Question 14

A-form of DNA (and RNA)

Answer 14

• DNA can be in A-form when it is dehydrated, RNA is always in A-form
• In A-form helices, the base pairs are tiltet away from perpendicular an are moved away from the centre of the helix. As a result, the major groove is wider and shallower than in the B-form.
• The narrowing and deepening of the major groove in A-form helices means that it is more difficult for proteins to read out the sequence-specific information at the edges of the bases in A-form helices

Question 15

Z-form of DNA

Answer 15

• The Z-form is adopted preferentially by segments of DNA that have strictly alternating C and G nucleotides. The base pairs in Z-form DNA obey the Watson-Crick rules, so the alternation of C and G must ocur on both strands.
• In the Z-form structure, there are 12 base pairs per turn, the sugars alternate between 2’ endo and 3’ endo puckers, and the G (or A) residues are in the syn conformation, while the C (or T) residues are in the normal anti conformation.
• The alternaton of sugar pucker gives the helix backbone a zigzag appearance -> not possible in RNA because C2 endo is not possible.
• Left helical sense

Question 16

Distortion of DNA by small molecules

Answer 16

B-form of DNA is ideal, but in reality local reagons of DNA deviate considerably from the B-form while still maintaining a double-heliacl structure that preserves normal base pairing.
The structure of DNA can easily be deformed locally by certain small molecules. These molecules bind with high affinity to the DNA and deform the structure away from the ideal B-form.
intercalators (bijv. TOTO)can slip between base pairs of the DNA and stack with bases above and below it. The increased seperation of the base pairs causes local unwinding.
(B/C) Other aromatic molecules bind in the minor groove, interacting with the sugar and the edges of the bases rather than stacking between bases (bijv. distamycin a polyamide-antibiotic). Up to 2 molecules of distamycin can bind side by side in a single region of the minor groove, but the minor groove has to expand considerably in order to accommodate this.

Question 17

Persistence length of DNA

Answer 17

The persistence length of DNA corresponds to the maximum length of a
segment of DNA that behaves as a rather rigid rod.
DNA segments that are shorter than the persistence length behave as
relatively rigid rods.
DNA segments that are longer than the persistence length can be bent whithout needing a lot of energy.

Question 18

The packaging of DNA in chromatin

Answer 18

Nucleosomes compact chromosomal DNA.
DNA is rapped twice around the nucleosome core. This costs energy that is provided by interactions between DNA and histone proteins that the nucleosome is made of.

Question 19

TBP deforming DNA

Answer 19

B-DNA can be distorted when regulatory proteins, such as transcription factors bind to it.
TATA-box binding protein (TBP) binds to regions of DNA with a central TATA sequence, inducing a sharp bend in the double helix and drastically widening the minor groove. The DNA bending recruites RNA-polymerase and the opening of the transcription bubble in the promotor where TBP is bound.
The degree to which the DNA can be bent or distorted varies for different sequences and this sequence specific deframability can, in turn affect the affinity of a protein for a particular DNA sequence.

Question 20

Supercoiled DNA

Answer 20

Supercoiling is a large-scale conformational effect in DNA in which the whole double helxi winds into a sperhelix. This effect is a consequence of the elastic deformation of DNA when the ends of the double helix cannot rotate relative to one another.

Question 21

G-U Wobble base pair

Answer 21

G-U wobble base pairs are the most common noncanonical base pairing
- They are well suited to binding RNA molecules, proteins or other ligands.
- The N7 and O6 groups of guanosine and the O4 of uridine contribute to the negative electrostatic potential in the major groove. his increases the RNA molcule’s ability to bind divalent metal ions.
- The stability of G-U approaches that of Watson-Crick pairs and they are the most stable of mismatched base pairs.

Question 22

Hoogsteen base-pairing

Answer 22

Hoogsteen base pairs involve hydrogen bonds between the Watson-Crick base-pairing edge of one base and the major groove edge of another base. this is possible because within the major groove the exposed edges of the purines and pyrimidines contain several atoms that are available for hydrogen bonding.
Hoogsteen base pairs are formed when a single-stranded piece of DNA or RNA enters the major groove of double-stranded nucleic acid and forms hydrogen bonds with one of the strands in the double helix. The result is the formation of a triple helix. These kind of interactions can be used to detext specific sequences in the DNA, for example, shut down transcription of certain genes.

Question 23

Post-transcriptional modifications of RNA

Answer 23

• RNA is known to be post-transciptionally modified.
• Some nucleotides are eliminatd, some are inserted and the remaning nucleotides at certain positions have bases that are chemically distinct from the normal RNA complement of A, U, C and G.
• Thus to determine the sequence of mature RNA molecules correctly , it is not enough to just know the sequence of the gene encoding it; the RNA must be extracted from its native source and characterized

Question 24

Post-translational modifications of DNA

Answer 24

Typically, cytosines that lie immediately 5’ to a guanosine are methylted. The methylation of cytosine does not affect its ability to form base pairs in the normal way, but the presence of the methylated cytosine can be recognized by various proteins.
In this way methylation of DNA can regulate processes, such as transcription without affecting the underlying genetic instructions. It can also be used to regulate gene expression -> a high level of methylation within a gene can lead to he represion of transcritpion.

Question 25

GNRA loop motif

Answer 25

Hairpin loops are formed when an RNA strand folds sharply back on itself and the resulting ‘stem’ is stablized by complemetary base pairing.
A particulary common RNAmotif is one that stabilizes the sharp turn of the phosphate backbone in a stem-loop structure. the strucure is stabalized by the formation of base-pairing interactions in the stem, but the loop region itself often forms a specific structure that drives the formation of a sharp turn.
GNRA tetraloop is one of these (Guanine, Nucleotide, R=purine, Adenine).

Question 26

RNA secondary structural elements / motifs

Answer 26

RNA can have many different conformations. A primary sequence of nucleotides folds into one or more helicases and intervening ss-regions, all of which finally fold togheter into a specific tertiary three-dimensional shape. The helices and the elements that connect them togehter are called RNA secondary structural elements or motifs, and include helicases, ss-regions, hairpins, bulges, internal loops and junstions.

Question 27

metal-RNA interactions

Answer 27

Diffuse ion interactions
Outer-sphere interactions
Inner-sphere interactions.

Question 28

Diffuse ion interactions

Answer 28

Diffuse ions have water between the hydrated ion and the nucleic acid.

Question 29

Outer-sphere interactions

Answer 29

Outer-sphere ions are those where a water that is coordinated to the metal also makes contact to the nucleic acid.

Question 30

Inner-sphere interactions

Answer 30

Inner-sphere ions make direct contact with the nucleac acid

Question 31

Coaxial stacking

Answer 31

There are 2 types of interactions between helices in RNA. The interacting helicases can form one continious helix, in an arrangement known as coaxial base stacking. The base pare at the end of one helix stacks against the base pair at the beginning of the other helix. This is favored by base-stacking interactions analogous to those that stabilize duplex RNAs or DNAs. These stacking interactions are an important feature of the overall tertiary structure of RNA.

Question 32

Pseudoknot

Answer 32

The other tertiary structural motif formed by two interacting helicases is the pseudoknot, in which nucleotides within the loop of a hairpin form base pairs with a segment at the other end of the stem. The hairpin loop pairs with a complementary squence next to he hairpin stem to form a contiguous coaxially stacked helix

Question 33

A-platform

Answer 33

Another structural motif in RNA is the adenosine platform (A-platform). This consist of two consecutive adenosine residues that are arranged side by side rather than staked on top of each other. the two adenosines form hydrogen bonds to each other to create a pseudo-base pair, resuting in a relatively large, flat surface. The bases in the A-platform stack against base pairs from other elements of the structure to form coaxially stacked double helices.

Question 34

A-minor interactions

Answer 34

interactions with the minor groove instead of the major groove.
Less specific and less strong than in the major strand.

Question 35

Tetraloop

Answer 35

Tetraloops leave a cap for the helical hairpin structures. The turn region of the tertraloop leaves some nucelotide bases unpaired, and these bases can interact with the minor groove of adjacent double helical segments. The docking site for the tetraloop is referred to as the tetraloop receptor, and such interactions help stabilize the tertiary structure of RNA molecules.

Question 36

Ribose zipper motif

Answer 36

• The tertiary structure of RNA can also be stabilized by interactions made by the sugar-phophate backbone.
• The ribose groups of two RNA strand that are running alongside each other in an antiparallel orientation can form a ribose zipp, in which the ribose suger of the two strands are interdigitated.
• Hydrogen bonds between te 2’-OH group of a ribose from one helix and the 2’-OH group and N3 of a puine or the O2 of pyrimidine in the opposite helix create a ‘zipper’ of ribose sugars in the minor grooves of two helices.