6. Nucleotides and nucleic acids

Question 86. What compounds are nucleic acid monomers?

Answer. Nucleic acids (poly-nucleotides) are biological polymers which monomer parts are nucleotides.

Nucleotide is not an individual compound, it comprises nucleoside and phosphoric acid residue.

Nucleoside is not individual compound also, it is formed at interaction between carbohydrate (ribose or deoxyribose) and hetero-cyclic base.

Question 87. What is carbohydrate form in the structure of nucleic acids?

A nswer. At DNA structure there is ribose at cyclic furanose form - -D-ribofuranose, RNA molecules contain 2-deoxyribose at furanose form – 2-deoxy--D-ribofuranose.

Question 88. What hetero-cyclic bases could be at nucleic acid structure?

Answer. Nucleic bases are nitrogen-containing hetero-cyclic compound derivatives – purine (adenine and guanine) and pyrimidine: thymine (only DNA), uracil (only RNA) and cytosine:

Question 89. What is the difference between DNA and RNA structure?

Answer. DNA and RNA differ from each other in the type of carbohydrate component (at DNA structure there is -D-ribo-furanose and in the structure of RNA there is 2-deoxy--D-ribo-furanose) and in content of pyridine bases (at RNA structure there is uracil and cytosine, and at DNA structure there is thymine and cytosine). In addition DNA exists in the form of double helix which contains two separate molecules. RNA molecules are shorter and mainly they are one-stranded.

Question 90. Define bond type between separate compounds and fragments in nucleic acid structure.

A nswer. Nucleic acids are long monomer chains – nucleotides. Nucleotide is a complex compound comprising nucleoside and attached phosphoric acid residue. Nucleoside, in its turn, is product of reaction between carbohydrate (ribose or deoxyribose) and hetero-cyclic base. Hetero-cyclic base –NН–group participates at reaction with carbohydrate hemi-acetal hydroxyl. As a result N-glycoside bond is formed:

As the form of carbohydrate at nucleic acids is -D-furanose then the bond is called -N-glycoside bond. This type of bond is formed by both pyrimidine and purine hetero-cyclic bases.

After that alcohol hydroxyl group at carbohydrate nucleoside at 5’ or 3’ position should react with phosphoric acid according to esterification reaction. Ester is formed, thus, this bond can be called ester bond. As ester is formed at reaction with phosphoric acid then there is one more name of this bond – phospho-ester bond.

When nucleotides are attached to each other there is one more esterification reaction between phosphoric acid residue at nucleotide and alcohol hydroxyl group at carbohydrate at 5’ or 3’ position. It is possible to say that nucleotides are attached to each other by 3’,5’-ester bond or phosphor-di-ester bond.

At secondary structure formation (double helix) in DNA between complimentary hetero-cyclic bases there are hydrogen bonds providing molecule helix structure existence.

Question 91. What is nucleic acid primary structure?

Answer. Nucleic acid primary structure is nucleotide content and nucleotide succession, i.e. it is alternation order of nucleotide elements at nucleic acid molecule. Nucleotide content is stated studying nucleic acid hydrolytic degradation products.

RNA is easily hydrolyzed in soft conditions at alkaline medium to nucleotides which, in their turn, are able to eliminate phosphoric acid residue at alkaline medium with nucleoside formation. At acid medium nucleosides are hydrolyzed to hetero-cyclic bases and carbohydrates.

DNA chemical analysis is not practically used due to complication by additional processes. More preferable procedure is enzymatic hydrolysis under the action of nucleases.

Question 92. What is DNA secondary structure?

Answer. DNA secondary structure is poly-nucleotide chain space organization. Its model was suggested in 1953 by G. Watson and P. Crick. According to this model, DNA has the form of right-wind spiral formed by two poly-nucleotide chains which are twisted in relation to each other and around common axis. DNA model is represented at figure 6. You can see that distance between rounds (spiral step) is 3,4 nm. This distance has 10 nucleotide residues, one nucleotide size is 0,34 nm; bi-spiral molecule diameter is 1,8 nm.

DNA double-stranded molecule configuration is changed according to water quantity and solution ionic strength. It is proved that there are at least 6 DNA forms, which are called А-, В-, С-, D-, Е- and Z-forms. Configuration of two of them in the simplest form is presented at figure 6. At A-form there is some basic pairs displacement from molecule centre to periphery which is reflected in the size (one round length is 2,8 nm, there are 11 mono-nucleotides instead of 10; distance between nucleotides is also changed etc.).

Figure 6. Double-stranded DNA molecule scheme:

а – according to Watson and Crick (“с” is deoxy-ribose residue, “р” is phosphoric acid residue); б – DNA A-form; в – DNA B-form.

If A- and B-forms are right-wind spirals then Z-form (zigzag-like) has left-wind configuration, in which phosphor-di-ester base is located as zigzag along molecule axis. It is supposed that at A-form DNA plays the role of matrix at transcription process (RNA synthesis on DNA molecule) and B-form is used as matrix at replication process (DNA synthesis on DNA molecule).

Nucleic base residues are directed inside the spiral. To provide more stability to this structure hydrogen bonds are formed between nitrous bases. Their number should be maximal. It is reached by certain correspondence between residues at one strand in relation to residues in the other strand: thymine groups at one strand should be located in front of adenine groups at other strand (two hydrogen bonds are formed between them), cytosine groups should be located in front of guanine groups (three hydrogen bonds are formed between them). These bases make complimentary base pairs. Hydrogen bonds are formed between amino group of one base and carbonyl group at another base, between amide and imine nitrogen atoms.

Question 93. What is DNA tertiary structure?

A nswer. At cell DNA forms super-helices which provide its packing compact character. DNA with 4 cm length is located at chromosome which size is up to 5 nm. DNA length is decreased up to 100 thousand times. Eukaryotic DNA tertiary structure is formed due to interaction with nuclei protein and at certain stage of cell cycle it acquires chromosome form (figure 7).

Figure 7. Scheme of DNA tertiary structure formation.

Question 94. What is RNA secondary structure?

A nswer. In opposition to DNA, ribonucleic acid molecule is made of one poly-nucleotide chain. Separate parts of RNA chain are formed by spiral loops due to hydrogen bonds between complimentary nitrous bases adenine-uracil and guanine-cytosine. RNA chain parts at such spiral structures are anti-parallel and they are not always complimentary, there are un-paired nucleotide residues and one-stranded loops which do not correspond to double helix. Spiral parts are typical of all RNA types.

Figure 8. Scheme of RNA secondary structure formation: а – scheme of bond formation between complimentary bases; b – transport RNA scheme.

Transport RNA comprises four spiral parts and three (sometimes four) one-stranded loops. To draw such a structure on plane special “clover-leaf” figure is used.

Question 95. What is RNA tertiary structure?

Answer. One-stranded RNA molecules are characterized by compact and ordered tertiary structure which appears due to interaction between spiral elements of the secondary structure. Thus, additional hydrogen bonds formation is possible between nucleotide residues located at distance from each other or bonds between OH-groups at ribose residues and bases. RNA tertiary structure is stabilized by di-valent metal ions, for example, Mg²⁺, which form bonds not only with phosphate groups but also with bases.

Question 96. What base is complimentary to thymine? Write down this complimentary pair structure and mark hydrogen bonds.

A nswer. Base complimentary to thymine is adenine. They form two hydrogen bonds.

Question 97. What base is complimentary to cytosine? Write down this complimentary pair structure and mark hydrogen bonds.

Answer. Base complimentary to cytosine is adenine. They form three hydrogen bonds.

Q uestion 98. What biopolymer fragment is at figure 9? What structure is represented (primary, secondary)? Are there any mistakes at this fragment? What is this biopolymer role at human organism? Answer. To define what biopolymer is at the figure we should analyze its structure. At the figure we could see nucleic acid fragment comprising three nucleotides. Carbohydrate component here is ribose as at 2’ position there is hydroxyl group.

Figure 9. Biopolymer scheme.

Thus, at the figure we can see RNA fragment. The structure is primary (nucleotide succession at nucleic acid chain). Nucleotide at the top has hetero-cyclic base thymine and not uracil (there is –СН₃ group). Thus, in the figure at RNA fragment there is a mistake. Its role is to transmit hereditary genetic information and to take part at protein synthesis.

Question 99. What biopolymer fragment is at figure 10? What structure is represented (primary, secondary)? Are there any mistakes at this fragment? What is this biopolymer role at human organism? Answer. To define what biopolymer is represented on the figure we should analyze its structure. At the figure we can see nucleic acid fragment comprising three nucleotides.

Figure 10. Biopolymer scheme.

Carbohydrate component here is deoxyribose as at 2’position there is no hydroxyl group. Thus, at the figure there is DNA fragment. This is primary structure (nucleotide succession at nucleic acid chain). Nucleotide at the top has hetero-cyclic base which is thymine (there is –СН₃ group). Thus, in DNA fragment there are no mistakes. Its role in the organism is to store and transmit genetic information.

Question 100. What RNA types are there?

Answer. There are several RNA types.

Ribosomal RNA molecules (rRNA) are in the structure of ribosomes being their structural basis. Their molecule completeness is necessary for protein biosynthesis on ribosomes.

Transport RNA (tRNA) biological role is addition of activated amino acid residues and their transport on ribosomes, i.e. to the place of poly-peptide chain synthesis.

Information or matrix RNA (iRNA, mRNA) plays the role of matrix (platelet) at protein biosynthesis during translation process (reading nucleotide code and its translation into amino acid succession at protein poly-peptide chains).

The problem to differentiate the following types of RNA is being debated: low-molecular (small) nucleic, antisense and viral.

Small nucleic RNA (snRNA) is RNA class which is located at eukaryotic cell nucleus. They take part at such important processes as splicing (intron removing from immature mRNA), regulating transcription factors or RNA-polymerase and keeping complete telomers.

Antisense RNA is one-stranded RNA which are complimentary to mRNA. Antisense RNA are introduced into cells to inhibit complimentary mRNA translation due to fact that antisense RNA make pair with mRNA-target and physically prevent translation complex formation. This effect is stechiometric.

Viral RNA is structural element of viral or phague ribonucleoproteins, it has all information necessary for virus cell proliferation at host-organism cells.