Nucleic acid structures

Question 1

What are the 3 major helical structures of DNA

Answer 1

A-DNA, B-DNA and Z-DNA

Question 2

Describe B-DNA

Answer 2

B-DNA is the biologically predominant form of DNA. It consists of a right-handed double helix whose 2 antiparallel sugar-phosphate backbones wrap around each other. The bases in the core of the helix form complementary Watson-Crick base pairs. It has two deep groves between its sugar-phosphate chains: a narrow minor groove and a wide major groove. The Watson-Crick base pairs in either orientation are structurally interchangeable(AT,TA or GC,CG) can replace each other in the double helix without altering the sugar backbone. Any other combination of base pairs will distort the double helix.

Question 3

List the following structural features of B-DNA: helical sense, diameter, base pairs per helical turn, helical twist per base pair, helical pitch(rise per turn), helix rise per base pair, base tilt normal to the helix axis, major groove, minor groove, sugar pucker and glycosidic bond

Answer 3

Right-handed, 20A, 10, 36 degrees, 34A, 3.4A, 6 degrees, wide and deep, narrow and deep, C2’-endo and anti.

Question 4

Describe A-DNA

Answer 4

A-DNA forms a wider and flatter right-handed helix than B-DNA. An important feature of A-DNA is that the planes of its base pairs are tilted 20 degrees with respect to the axis of the helix. It has a narrow and deep major groove and a wide and shallow minor groove. Experimentally, when the relative humidity is reduced to 75%, B-DNA undergoes a reversible conformational change to A-DNA.

Question 5

List the following structural features of A-DNA: helical sense, diameter, base pairs per helical turn, helical twist per base pair, helical pitch(rise per turn), helix rise per base pair, base tilt normal to the helix axis, major groove, minor groove, sugar pucker and glycosidic bond

Answer 5

Right-handed, 26A, 11.6, 31 degrees, 34A, 2.9A, 20 degrees, Narrow and deep, wide and shallow, C3’-endo and anti.

Question 6

List the 3 biological contexts that A-DNA has been observed in

Answer 6

> At the cleavage center of topoisomerase II
At the active site of DNA polymerase
In certain gram-positive bacteria that have undergone sporulation(the formation of spores, a dormant cell type that is the result of environmental stress): such spores contain a high proportion of small acid-soluble spore proteins. Some SASPs induce B-DNA to assume the A form, at least in vitro. DNA in spores are resistant to UV-induced damage- this occurs because the B to A conformational change inhibits the UV-induced covalent cross-linking of pyrimidine bases, by increasing the distance between successive pyrimidines.

Question 7

Describe Z-DNA

Answer 7

Over 25 years after the discovery of the Watson-Crick structure, a left-handed double helix was revealed. The structure of the hexanucleotide d(CGCATGCG) was found to form the left-handed double helix, this helix is called Z-DNA. The features of Z-DNA include: 12 bases per turn, a pitch of 44A and a narrow and deep minor groove and no discernible(distinguishable) major groove. The base pairs in Z-DNA are flipped 180 degrees relative to the base pairs of B-DNA. This results in the repeating unit of Z-DNA being a dinucleotide, d(XpYp), rather than a single nucleotide like the other DNA forms. The line joining the successive phosphorus atoms on a polynucleotide strand of Z-DNA therefore follows a zigzag path around the helix- this is where the name Z-DNA comes from- rather than a smooth curve that is seen in A and B-DNA.

Question 8

List the following structural features of Z-DNA: helical sense, diameter, base pairs per helical turn, helical twist per base pair, helical pitch(rise per turn), helix rise per base pair, base tilt normal to the helix axis, major groove, minor groove, sugar pucker and glycosidic bond

Answer 8

Left-handed, 18A, 12(6 dimers), 9 degrees for pyrimidine-purine steps and 51 degrees for purine-pyrimidine steps, 44A, 7.4A per dimer, 7 degrees, flat, narrow and deep, C2’-endo for pyrimidines and C3’-endo for purines, anti for pyrimidines and syn for purines

Question 9

Describe the formation of Z-DNA

Answer 9

Structural studies have shown that complementary polynucleotides with alternating purines and pyrimidines take up the Z-DNA conformation at high salt concentrations. A high salt [ ] stabilizes Z-DNA relative to B-DNA by reducing the otherwise increased electrostatic repulsions between the closest approaching phosphate groups on opposite stands. A common biological modification is the methylation of cytosine residues at C5 which also promotes Z-DNA formation since a hydrophobic methyl group in this position is less exposed to solvent in Z-DNA than it is in B-DNA.

Question 10

Is there a biological function of Z-DNA

Answer 10

It has been proposed that the reversible conversion of specific segments of B-DNA to Z-DNA under appropriate circumstances acts as a kind of switch in regulating genetic expression- transcription can induce Z-DNA formation. It was discovered that several proteins specifically bind Z-DNA- the existence of these proteins prove that Z-DNA does exist in vitro.

Question 11

Describe the conformation of double helical RNA

Answer 11

Double helical RNA is unable to assume a B-DNA like conformation because of steric clashes involving its 2’-OH groups. Rather it assumes a conformation resembling A-DNA called A-RNA or RNA-11. Many RNAs, eg. tRNA and rRNA contain complementary sequences that form double helical stems. Also, certain viral genomes.

Question 12

Describe RNA-DNA hybrids

Answer 12

Hybrid double helices, which contain one strand of RNA and one strand DNA, are also predicted to have a A-RNA like conformation and B-DNA like conformation. This structure is biologically significant: short segments of RNA-DNA hybrid helices occur in the transcription of DNA and RNA templates. They also occur in the initiation of DNA replication by short lengths of RNA. The RNA component of the helix is a substrate for RNase H, which specifically hydrolyzes the RNA strands of RNA-DNA hybrid helices in vivo.

Question 13

What determines the conformation of a nucleotide unit

Answer 13

The conformation of a nucleotide unit is specified by the six torsion angles of the sugar-phosphate backbone. The torsion angles describe the orientation of the base about the glycosidic bond. The torsion angles of the sugar-phosphate backbone are greatly restricted. The angles are strain free and the double helices are conformationally relaxed.

Question 14

Explain how the rotation of a base about its glycosidic bond is restricted

Answer 14

The rotation of a base about its glycosidic bond is greatly hindered. Purine residues have two sterically permissible orientations relative to the sugar known as syn and anti conformations. For pyrimidines, only the anti conformation is easily formed because the sugar residue sterically interferes with the pyrimidines C2 substituent. In most double helical nucleic acids, all bases are in the anti conformation, the exception is in Z-DNA. Z-DNA has pyrimidines that are anti and purines that are syn.

Question 15

How does the ribose ring affect the conformation of the sugar phosphate backbone

Answer 15

The ribose ring has a certain amount of flexibility that affects the conformation of the sugar phosphate backbone. The ring substituents are eclipsed when the ring is planar. To relieve the resultant crowding that even occurs between hydrogen atoms, the ring puckers- which means that it becomes slightly non-planar to reorient the ring substituents. If the out-of-plane atom is displaced to the same side of the ring as atom C5’, it has exo conformation. In most nucleotide and nucleoside structures, the out-of-plane atom is either C2’ or C3’. C2’-endo is the most frequently occurring ribose pucker with C3’-endo and C3’-exo also being common.

Question 16

Why is the ribose pucker conformationally important

Answer 16

The ribose pucker is conformationally important because it governs the relative orientations of the phosphate substituents to each residue. B-DNA has a sugar pucker with C2’-endo, A-DNA has a sugar pucker with C3’-endo and Z-DNA has pyrimidines with C2’-endo and purines with C3’-endo.

Question 17

What is base pairing

Answer 17

Base pairing is the ‘glue’ that holds the double-stranded nucleic acids together. When monomeric adenine and thymine derivatives are cocrystallized, the AT base pairs that form have adenine N7 as the hydrogen bonding acceptor(Hoogsteen geometry) rather than N1(Watson-crick geometry). This suggests that hoogsteen geometry is more stable for AT base pairs in comparison to watson-crick geometry. But, steric preferences and other environmental influences make Watson-crick geometry the preferred mode of base pairing in double helices. Monomeric GC base pairs always cocrystallize with Watson-crick geometry because of their triple hydrogen bonded structures.

Question 18

Does Hoogsteen geometry have any biological importance

Answer 18

Yes, they help stabilize the tertiary structures of tRNA.

Question 19

Why don’t we find non-Watson-Crick base pairs in helical environments

Answer 19

The bases of the double helix associate such that any pair position is interchangeable without affecting the structure of the sugar-phosphate backbone. It was believed that this requirement of geometric complementarity of Watson-Crick base pairs is the only reason that other base pairs do not occur in a double helical environment. So, the failure to detect pairs of different bases in a non-helical environment led scientists to used infrared spectrum analyses to determine the values of K(association constant) for the various bases. It was found that the self-association constants were smaller in comparison to the Watson-Crick association constants and the non-Watson-Crick base pairs have constants that were negligible in comparison to the self-association constants. Therefore a second reason as to why non-Watson-Crick base pairs do not occur in double helices is because they have low stability.

Question 20

Discuss the contribution of hydrogen bonding to the stability of the double helix

Answer 20

Hydrogen bonding is required for the specificity of base pairing in DNA that is ultimately responsible for the enormous fidelity required to replicate DNA with almost no error. Yet, hydrogen bonding contributes very little to the stability of the double helix.

Question 21

What is base stacking

Answer 21

Purines and pyrimidines tend to form extended stacks of planar parallel molecules. The bases on these structures are usually partially overlapped.

Question 22

Explain the stacking of nucleic acid bases in aqueous solution

Answer 22

Bases aggregate in aqueous solution, this has been demonstrated by the variation of osmotic pressure with concentration. The van’t Holfman law of osmotic pressure. If the species under investigation is of known molecular mass, but aggregates in solution then we must rewrite the van’t Holfman law. The osmotic coefficient, indicates the solutes degree of association. It varies from 1 to 0( no association to indefinite association). The variation of the osmotic coefficient with the molality of the solute for nucleic acid bases in aqueous solution is consistent with a model in which the bases aggregate in successive steps. This association cannot be a result of hydrogen bonding since N^6N^6-dimethyladenosine, which cannot form interbase hydrogen bonds, has a greater degree of association than adenosine. The aggregation results from the stacking of bases.

Question 23

Explain how hydrophobic forces stabilize nucleic acid structures

Answer 23

Stacking associations in aqueous solutions are stabilized by hydrophobic forces. The hydrophobic interactions that stabilize nucleic acids are different from the ones that stabilize protein structures. They differ in character. When looking at the the reaction: dinucleoside phosphate(unstacked) to dinucleoside phosphate(stacked)- the thermodynamic analysis of the melting curves indicates that the base stacking is enthalpically driven and entropically opposed. Therefore the hydrophobic forces are different in nucleic acids and proteins. Hydrophobic forces in nucleic acids are poorly understood- the fact that the hydrophobic forces that stabilize the nucleic acid structures and the protein structures are different can be seen by the nitrogenous bases of nucleic acids being more polar than the hydrocarbon residues that participate in hydrophobic interactions. However there is no theory that explains the hydrophobic forces that stabilize the nucleic acid structures.

Question 24

Explain the ionic interactions that stabilize the nucleic acid structures

Answer 24

Charged phosphate groups have electrostatic interactions. The melting temperature of duplex DNA increases with cation concentration because these ions bind more tightly to duplex DNA than to single-stranded DNA- this is because of the duplex DNAs higher anionic charge density. An increase in salt concentration therefore shifts the equilibrium towards the duplex form, thus increasing the DNAs Tm. Divalent cations, such as Mg2+, Mn2+ and Co2+ specifically bind to phosphate groups, whereas monovalent cations, such as Li+ and K+ bind nonspecifically to phosphate groups. Therefore divalent cations are more effective shielding agents for nucleic acids than monovalent cations. Eg, Mg2+ is required by enzymes that mediate reactions with nucleic acids or just nucleotides for activity and Mg2+ ions also play an essential role in stabilizing the complex structures assumed by many RNAs, such as tRNAs and rRNAs.