Structure of Nucleic Acids

AS · 12 min

DNA stores the instructions for building every protein in an organism, and RNA carries those instructions to the ribosomes. Both are nucleic acids: long chains of nucleotides. This note builds DNA and RNA from their nucleotides, explains the double helix with its antiparallel strands and complementary base pairing, and covers ATP, the phosphorylated nucleotide that is the cell's energy currency. The structure of DNA explains how it can be copied exactly and how it codes for proteins, so this note underpins the rest of the unit. Expect labelling, comparison tables, base-pairing calculations and short "explain" questions on hydrogen bonding.

Nucleotides

Definition

A nucleotide is made of three components joined by condensation reactions:

a pentose (five-carbon) sugar: deoxyribose in DNA, ribose in RNA;
a phosphate group;
a nitrogen-containing (nitrogenous) base.

The phosphate group is attached to carbon 5 of the sugar, and the base to carbon 1. The carbon atoms of the sugar are numbered 1′ to 5′ ("one prime" to "five prime") to distinguish them from atoms in the base. Deoxyribose differs from ribose only in having one fewer oxygen atom: it has –H instead of –OH on carbon 2′ (hence "deoxy").

You do not need to learn structural formulae of nucleotides, but you should be able to recognise and draw a nucleotide as a simple diagram: a circle (phosphate) joined to a pentagon (pentose), joined to a rectangle (base).

The five bases

There are five bases in nucleic acids, in two structural groups:

Key result

Group	Structure	Bases	Found in
Purines	double ring (two fused rings of carbon and nitrogen atoms); larger	adenine (A), guanine (G)	DNA and RNA
Pyrimidines	single ring; smaller	cytosine (C), thymine (T), uracil (U)	C in DNA and RNA; T only in DNA; U only in RNA

Two memory aids: "pure As Gold" for the purines (A and G), and "CUT the py" for the pyrimidines (C, U and T).

ATP: a phosphorylated nucleotide

ATP (adenosine triphosphate) is a nucleotide with three phosphate groups. It consists of:

the base adenine;
the sugar ribose;
three phosphate groups in a chain.

Adenine joined to ribose is called adenosine; adding one, two or three phosphates gives AMP, ADP or ATP. When the bond to the last phosphate is hydrolysed, ATP becomes ADP and inorganic phosphate (Pi), releasing about $30\ \text{kJ mol}^{-1}$ of energy that can be used by the cell, for example for active transport, muscle contraction and synthesising molecules.

\text{ATP} + \text{H}_2\text{O} \longrightarrow \text{ADP} + \text{P}_\text{i} \quad (\text{energy released})

ATP is resynthesised from ADP and Pi during respiration. The role of ATP in respiration is an A Level topic; at AS you need its structure and that it is a phosphorylated nucleotide.

Polynucleotides and the phosphodiester bond

Nucleotides join to form a long chain called a polynucleotide. The phosphate group of one nucleotide forms a covalent bond with the sugar (carbon 3′) of the next nucleotide in a condensation reaction, releasing water. The linkage sugar–phosphate–sugar is called a phosphodiester bond.

The result is a sugar–phosphate backbone with the bases sticking out to one side. The sequence of bases along the chain is the information the molecule carries.

Every polynucleotide has two different ends:

the 5′ end, with a free phosphate group attached to carbon 5′ of the sugar;
the 3′ end, with a free –OH group on carbon 3′ of the sugar.

So a strand has a direction, written $5' \to 3'$ . New nucleotides are always added to the 3′ end.

The structure of DNA

DNA (deoxyribonucleic acid) consists of two polynucleotide strands wound around each other to form a double helix.

Key result

Features of the DNA double helix

Two polynucleotide strands, each with a sugar–phosphate backbone on the outside and bases pointing inwards.
The two strands are antiparallel: they run in opposite directions, one $5' \to 3'$ and the other $3' \to 5'$ .
The strands are held together by hydrogen bonds between complementary bases.
Complementary base pairing: adenine always pairs with thymine (A–T) and cytosine always pairs with guanine (C–G).
A–T pairs are held by two hydrogen bonds; C–G pairs by three hydrogen bonds.
Each base pair consists of one purine and one pyrimidine, so all base pairs are the same width, and the double helix has a constant diameter (2 nm).
The two strands twist into a double helix, with about 10 base pairs per complete turn ( $3.4\ \text{nm}$ ), so adjacent base pairs are $0.34\ \text{nm}$ apart.

Why complementary base pairing matters

Only A–T and C–G fit. A purine paired with a purine would be too wide; a pyrimidine with a pyrimidine too narrow. And the hydrogen-bonding groups line up only for A with T and C with G.
Each strand determines the other. If you know the base sequence of one strand, you know the other. This allows DNA to be copied exactly in replication, with each strand acting as a template, and allows a gene to be transcribed into mRNA.
Stability. Although each hydrogen bond is weak, there are millions of them, so the double helix is very stable. Yet the strands can be separated by breaking hydrogen bonds (by enzymes, without breaking the covalent backbone) for replication and transcription.
Protection. The bases, which carry the information, are inside the helix, shielded by the backbone.
C–G content. Because C–G pairs have three hydrogen bonds and A–T pairs two, DNA with a higher proportion of C–G pairs needs more energy (a higher temperature) to separate its strands.

Chargaff's rules

In any double-stranded DNA molecule, because A always pairs with T and C with G:

the amount of A equals the amount of T, and the amount of C equals the amount of G;
so the amount of purines (A + G) equals the amount of pyrimidines (T + C).

The ratio of (A + T) to (C + G) is not fixed: it varies between species.

Base percentages (routine)

A sample of double-stranded DNA contains 22% adenine. Calculate the percentages of thymine, guanine and cytosine.

Solution

A pairs with T, so thymine = 22%.

A + T $= 44\%$ , so C + G $= 100 - 44 = 56\%$ .

C pairs with G, so guanine = cytosine = 56 / 2 = 28%.

Writing the complementary strand (routine)

One strand of a DNA molecule has the base sequence

5′ ATGCCGTAAGCTTGCA 3′

(a) Write the sequence of the complementary strand, labelling its ends. (b) Calculate the number of hydrogen bonds holding this length of double helix together.

Solution

(a) Each base pairs with its complement and the strands are antiparallel:

5′ ATGCCGTAAGCTTGCA 3′ 3′ TACGGCATTCGAACGT 5′

(Read from left to right, the complementary strand runs 3′ to 5′. Written in the conventional 5′ to 3′ direction it is 5′ TGCAAGCTTACGGCAT 3′.)

(b) Count the pairs in the original strand: A and T together $= 8$ (A–T pairs); G and C together $= 8$ (C–G pairs).

Hydrogen bonds $= (8 \times 2) + (8 \times 3) = 16 + 24 = 40$ .

The structure of RNA

RNA (ribonucleic acid) is also a polynucleotide, but it differs from DNA in several ways.

Key result

Structure of messenger RNA (mRNA)

A single polynucleotide strand (not a double helix).
The pentose sugar is ribose.
The bases are A, U, C and G: uracil replaces thymine.
Nucleotides are joined by phosphodiester bonds, forming a sugar–phosphate backbone with exposed bases.
It is a linear molecule, much shorter than DNA: a copy of one gene (or a few genes), not a whole chromosome.
Its base sequence is complementary to the template strand of the gene from which it was transcribed; every three bases form a codon.
It is made in the nucleus by transcription, and leaves through nuclear pores to the ribosomes, where it is translated. It is broken down after use.

There are two other types of RNA, both involved in protein synthesis:

Transfer RNA (tRNA): a single strand folded into a "clover-leaf" shape held by hydrogen bonds between complementary bases within the strand. It carries a specific amino acid at one end and has an anticodon of three bases at the other.
Ribosomal RNA (rRNA): combines with proteins to form ribosomes.

Feature	DNA	RNA (mRNA)
Number of strands	two (double helix)	one
Pentose sugar	deoxyribose	ribose
Bases	A, T, C, G	A, U, C, G
Base pairing	A–T and C–G between the two strands	no pairing within mRNA (tRNA and rRNA have some internal pairing)
Length	very long (millions of base pairs per molecule)	short (hundreds to thousands of nucleotides)
Location in eukaryotic cells	nucleus (also mitochondria and chloroplasts)	made in nucleus; moves to cytoplasm
Stability	very stable; lasts the lifetime of the cell	relatively unstable; broken down after a short time
Function	stores genetic information	carries a copy of the genetic code of one gene to the ribosomes

Length and size of DNA (moderate)

Adjacent base pairs in DNA are $0.34\ \text{nm}$ apart.

(a) A gene is 1500 base pairs long. Calculate its length in nm and in µm. (b) A human diploid cell contains about $6.4 \times 10^{9}$ base pairs of DNA. Calculate the total length of DNA in the cell in metres, and comment on the answer given that the nucleus is about $10\ \mu\text{m}$ across.

Solution

(a) $1500 \times 0.34 = 510\ \text{nm} = 0.51\ \mu\text{m}$ .

(b) $6.4 \times 10^{9} \times 0.34 \times 10^{-9}\ \text{m} = 2.2\ \text{m}$ .

About 2 m of DNA fits inside a nucleus only about $10 \times 10^{-6}\ \text{m}$ across, which is 200 000 times smaller. The DNA must be tightly coiled and packed by winding around histone proteins to form chromatin (and, during division, condensed chromosomes).

Exam-hard: interpreting base composition data (6 marks)

The table shows the base composition of nucleic acid from four sources.

Source	A / %	G / %	C / %	T / %	U / %
Human liver	30.3	19.5	19.9	30.3	0
Bacterium	24.7	26.0	25.7	23.6	0
Virus X	25.0	24.0	18.0	33.0	0
Virus Y	29.0	21.0	30.0	0	20.0

(a) Calculate the ratio $\dfrac{\text{A} + \text{G}}{\text{C} + \text{T}}$ for human liver DNA, giving your answer to two decimal places, and explain why it is close to 1. (b) Suggest the type of nucleic acid in virus X and in virus Y, explaining your reasoning. (c) Suggest why the bacterial values do not show exactly A = T and C = G.

Solution

(a) $\dfrac{30.3 + 19.5}{19.9 + 30.3} = \dfrac{49.8}{50.2} = 0.99$ .

In double-stranded DNA every purine (A or G) pairs with a pyrimidine (T or C) by complementary base pairing: A with T, G with C. So the number of purines equals the number of pyrimidines and the ratio is 1.

(b) Virus X: contains thymine and no uracil, so it is DNA; but A (25%) ≠ T (33%) and G (24%) ≠ C (18%), so the bases are not paired: it is single-stranded DNA.

Virus Y: contains uracil and no thymine, so it is RNA; A (29%) ≠ U (20%) and G (21%) ≠ C (30%), so the bases are not paired: it is single-stranded RNA.

(c) Experimental error in measuring the amount of each base (the differences are small, about 1%); the sample may contain some single-stranded DNA or RNA contamination.

Watch out

The base thymine is in DNA only; uracil is in RNA only. Do not write "thiamine" (a vitamin) or "uracil in DNA".
Hydrogen bonds hold the two strands together (between bases). The bonds within each strand (sugar to phosphate) are covalent phosphodiester bonds.
"Antiparallel" means the strands run in opposite directions ( $5' \to 3'$ and $3' \to 5'$ ), not that they are "not parallel".
Purines are double-ringed (A and G); pyrimidines are single-ringed (C, T, U). Learn which is which.
ATP contains ribose, not deoxyribose.
Chargaff's rule is A = T and C = G, not A = G.

Exam tip

"Describe the structure of DNA" (5–6 marks): nucleotides (deoxyribose, phosphate, base); phosphodiester bonds forming sugar–phosphate backbone; two strands, antiparallel; double helix; complementary base pairing A–T, C–G; hydrogen bonds (2 between A and T, 3 between C and G); purine pairs with pyrimidine.
In comparison questions, set out a two-column table or write paired statements with "whereas".
When asked for a complementary sequence, check you have swapped every base correctly, and if RNA is involved, use U instead of T.
Show percentages working step by step: A = T, then (C + G) = 100 − (A + T), then halve.

Summary

A nucleotide = pentose sugar + phosphate group + nitrogenous base.
DNA nucleotides have deoxyribose and bases A, T, C, G; RNA nucleotides have ribose and bases A, U, C, G.
Purines (A, G) have a double ring; pyrimidines (C, T, U) have a single ring.
ATP = adenine + ribose + three phosphates; a phosphorylated nucleotide; hydrolysis to ADP + Pi releases energy.
Phosphodiester bonds (condensation) join nucleotides into a sugar–phosphate backbone with 5′ and 3′ ends.
DNA: double helix of two antiparallel strands; complementary base pairing A–T (2 hydrogen bonds) and C–G (3 hydrogen bonds); purine pairs with pyrimidine.
mRNA: single strand, ribose, uracil instead of thymine, shorter, complementary to the template strand of a gene.
In double-stranded DNA, A = T and C = G.

Practice questions

Question

Name the three components of a nucleotide.
Name the two purines and state one structural difference between purines and pyrimidines.
Describe the structure of an ATP molecule.
Name the bond that joins nucleotides in a polynucleotide and the type of reaction that forms it.
A DNA sample contains 31% guanine. Calculate the percentage of adenine.
Explain what is meant by the term antiparallel in describing DNA.
Give four differences between the structure of DNA and the structure of mRNA.
A length of DNA has 60 base pairs, of which 24 are C–G pairs. Calculate the total number of hydrogen bonds between the two strands.
Explain why the two strands of DNA can be separated easily by enzymes during replication, yet the molecule is stable. (3 marks)
Two samples of double-stranded DNA were heated, and the temperature at which the strands separated was recorded. Sample P had 35% C–G pairs; sample Q had 60% C–G pairs. (a) Predict which sample separated at the higher temperature, and explain your answer. (b) Calculate the percentage of thymine in each sample. (c) Explain why complementary base pairing is essential for the functions of DNA. (6 marks)

Answers

Pentose sugar; phosphate group; nitrogen-containing (nitrogenous) base.
Adenine and guanine. Purines have a double ring structure; pyrimidines have a single ring (so purines are larger).
A nucleotide containing the base adenine, the pentose sugar ribose and three phosphate groups (a phosphorylated nucleotide).
Phosphodiester bond; condensation reaction.
G = C = 31%, so G + C = 62%; A + T = 38%; A = 19%.
The two polynucleotide strands run in opposite directions: one strand runs from 5′ to 3′ and the complementary strand from 3′ to 5′.
Any four: DNA double-stranded / mRNA single-stranded; DNA has deoxyribose / mRNA ribose; DNA has thymine / mRNA uracil; DNA much longer / mRNA shorter; DNA is a double helix / mRNA linear single strand; DNA has hydrogen-bonded base pairs between strands / mRNA has no base pairing; DNA in nucleus (stable) / mRNA moves to cytoplasm (short-lived).
24 C–G pairs × 3 = 72; 36 A–T pairs × 2 = 72; total = 144 hydrogen bonds.
The strands are held together only by hydrogen bonds between bases, which are individually weak, so enzymes (helicase) can break them without breaking the covalent sugar–phosphate backbone; but there are very many hydrogen bonds along the molecule, so together they make the double helix stable; the bases are inside the helix, protected by the backbone.
(a) Q: it has more C–G pairs, which are held by three hydrogen bonds rather than two (A–T); more hydrogen bonds overall need more energy to break. (b) P: A–T pairs = 65% of pairs, so A + T = 65% of bases, T = 32.5%. Q: A + T = 40%, T = 20%. (c) Each strand acts as a template for a new complementary strand in replication, so DNA is copied exactly / genetic information is passed accurately to daughter cells; the template strand of a gene is used to make complementary mRNA in transcription, so the base sequence (genetic code) is transferred; base pairing between codon and anticodon in translation; pairing holds the strands together (stability) while allowing them to be separated.

Nucleotides

The five bases

ATP: a phosphorylated nucleotide

Polynucleotides and the phosphodiester bond

The structure of DNA

Why complementary base pairing matters

Chargaff's rules

The structure of RNA

Practice questions

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