AQA A-level Biology: Nucleic Acids, DNA Replication and ATP (7402)

This guide follows the AQA AS and A-level Biology specification (7401 and 7402, first teaching 2015, exams from June 2016). It covers sections 3.1.5.1 (structure of DNA and RNA), 3.1.5.2 (DNA replication) and 3.1.6 (ATP). Section 3.1 is part of both the AS and the A-level, so everything here applies to either. Nothing in these three sections is marked as A-level only.

Use it with the revision notes and the practice questions. The wider topic is in Biological Molecules. Course page: AQA A-level Biology. Printable list: checklist. To find your weak areas first, try the free diagnostics.

What you must be able to do

Section	Outcome
3.1.5.1	State what DNA and RNA do; describe the nucleotide components of each; explain how a phosphodiester bond forms; describe the double helix and base pairing; describe RNA; appreciate why DNA’s simplicity caused doubt; use incomplete base data to find other base frequencies
3.1.5.2	Describe semi-conservative replication, including the roles of DNA helicase and DNA polymerase; evaluate the work that validated the Watson-Crick model
3.1.6	Describe ATP; describe hydrolysis by ATP hydrolase; explain coupling and phosphorylation; describe resynthesis by ATP synthase

3.1.5.1 Structure of DNA and RNA

Jobs of the two molecules

In all living cells, DNA holds genetic information. RNA transfers genetic information from DNA to the ribosomes. Ribosomes are themselves formed from RNA and proteins.

Nucleotides

DNA and RNA are both polymers of nucleotides. Every nucleotide has three parts: a pentose (five-carbon sugar), a nitrogen-containing organic base and a phosphate group.

	DNA nucleotide	RNA nucleotide
Pentose	deoxyribose	ribose
Phosphate	one phosphate group	one phosphate group
Bases	adenine, cytosine, guanine, thymine	adenine, cytosine, guanine, uracil

The only base difference is thymine in DNA and uracil in RNA. The sugar differs too: deoxyribose in DNA, ribose in RNA.

The phosphodiester bond

A condensation reaction between two nucleotides forms a phosphodiester bond. It joins the phosphate group of one nucleotide to the pentose of the next, and a molecule of water is released. Repeating this builds a sugar-phosphate backbone with the bases projecting from it.

Worked example 1. A single polynucleotide chain has 60 nucleotides. How many phosphodiester bonds and how many water molecules are released in making it?

• Each condensation joins two nucleotides, so a chain of n nucleotides has n − 1 bonds.
• 60 − 1 = 59 phosphodiester bonds, and 59 water molecules released.

Worked example 2. A DNA fragment is 500 base pairs long. How many phosphodiester bonds does it contain?

• Two strands of 500 nucleotides each: 1000 nucleotides in total.
• Each strand has 500 − 1 = 499 bonds.
• Two strands: 2 × 499 = 998 phosphodiester bonds. The hydrogen bonds between the strands are not phosphodiester bonds and are not counted.

The DNA double helix

A DNA molecule is a double helix: two polynucleotide chains wound together and held by hydrogen bonds between specific complementary base pairs.

• Adenine pairs with thymine (A with T).
• Cytosine pairs with guanine (C with G).

The pairing is specific, so the sequence of one strand fixes the sequence of the other. Write the complement of 5 bases to check you can do it: ATGCC pairs with TACGG.

An RNA molecule is a relatively short polynucleotide chain. It is normally a single strand. It contains uracil where DNA contains thymine.

Why DNA’s simplicity caused doubt

Proteins are built from many different amino acids, so they look varied enough to carry a large amount of information. DNA, by contrast, has only four different bases. The specification asks you to appreciate that this relative simplicity led many scientists to doubt that DNA carried the genetic code. The doubt was answered by the structure: a long polymer with four bases can carry a great deal of information in the order of its bases, just as an alphabet of four letters can write long messages. When you answer, make both halves of the point: why it looked too simple, and why order and length overcome that.

Base-frequency calculations (maths skill 0.3)

In double-stranded DNA, A = T and C = G, so you can find every base from one value. The four percentages add to 100.

Worked example 3. A double-stranded DNA sample has 21% cytosine. Find the percentage of each base.

• Guanine pairs with cytosine, so G = 21%.
• C + G = 42%, leaving 100 − 42 = 58% for A and T together.
• A = T, so each is 58 ÷ 2 = 29%.

Worked example 4 (incomplete data on one strand). One strand of a DNA molecule contains 28% A, 16% T, 33% G and 23% C. Find the percentage of each base in (a) the other strand, (b) the whole molecule.

(a) Each base in the second strand pairs with the base opposite it in the first:

Base	Strand 1	Strand 2
A	28%	16% (pairs with T of strand 1)
T	16%	28% (pairs with A of strand 1)
G	33%	23% (pairs with C of strand 1)
C	23%	33% (pairs with G of strand 1)

(b) Average the two strands. A = (28 + 16) ÷ 2 = 22%, so T = 22%. G = (33 + 23) ÷ 2 = 28%, so C = 28%. Check: 22 + 22 + 28 + 28 = 100.

Worked example 5 (counting bases). A double-stranded DNA fragment of 4000 base pairs has 28% guanine. How many adenine bases does it have?

• Total bases = 4000 × 2 = 8000.
• G = C = 28% each, so A + T = 100 − 56 = 44%, and A = 22%.
• A = 0.22 × 8000 = 1760 adenine bases.

The common slip is to use 4000 as the total. A base pair is two bases.

3.1.5.2 DNA replication

Semi-conservative replication ensures genetic continuity between generations of cells. “Semi-conservative” means each new DNA molecule keeps one original strand and gains one new strand.

The process

1. Unwinding. The double helix unwinds.
2. Strand separation. DNA helicase breaks the hydrogen bonds between complementary bases and so unwinds the DNA. Each strand is now exposed as a template.
3. Base pairing. Free DNA nucleotides are attracted to the exposed bases on each template strand and base-pair with them: A with T, C with G.
4. Joining. DNA polymerase catalyses the condensation reaction that joins adjacent nucleotides, forming phosphodiester bonds in the new strand.
5. Result. Two molecules, each made of one old and one new strand.

Do not say helicase “cuts” the DNA or breaks the backbone. It breaks hydrogen bonds. DNA polymerase makes phosphodiester bonds. It does not make hydrogen bonds between bases: those form by base pairing.

Worked example 6. A bacterial chromosome of 3.0 × 10⁶ base pairs is copied from one starting point, with two replication forks moving in opposite directions. Each fork copies 800 base pairs per second. How long does replication take?

• Combined rate = 2 × 800 = 1600 base pairs per second.
• Time = 3.0 × 10⁶ ÷ 1600 = 1875 s.
• 1875 ÷ 60 = 31 minutes (2 s.f.).

(The rate is data supplied for this example, not a value to learn.)

Evaluating the validation of the Watson-Crick model

The specification wants you to evaluate the work of scientists in validating the Watson-Crick model of DNA replication. The standard evidence is the Meselson-Stahl experiment in bacteria.

Method. Bacteria were grown for many generations in a medium whose nitrogen was the heavy isotope ¹⁵N, so all their DNA contained ¹⁵N. They were then moved to a medium containing the lighter ¹⁴N and left to divide. DNA was extracted at intervals and spun in a caesium chloride density gradient, where DNA settles at a height matching its density: ¹⁵N DNA lower, ¹⁴N DNA higher, mixed DNA between.

What each model predicts (starting with all-heavy DNA, then growing in ¹⁴N):

Generation	Semi-conservative	Conservative	Dispersive
1	one intermediate band	one heavy and one light band	one intermediate band
2	intermediate and light bands in equal amounts	heavy and light bands, light three times heavy	one band, lighter than intermediate

Result. After one generation there was a single intermediate band. After two there were two bands, intermediate and light, in about equal amounts. The first result rules out conservative replication. The second rules out dispersive replication and fits semi-conservative replication.

Worked example 7. After n generations in ¹⁴N, what fraction of the DNA molecules contain some ¹⁵N, and what fraction of all strands are ¹⁵N?

• Molecules after n divisions: 2ⁿ. The original two ¹⁵N strands stay in just 2 molecules, and the rest are all ¹⁴N.
• After 3 generations: 8 molecules, 2 hybrid, 6 light. Hybrid = 2 ÷ 8 = 25%.
• Strands: 16 in total, 2 of them ¹⁵N. That is 2 ÷ 16 = 12.5%.

Evaluating the work.

• Strengths: it distinguishes three clear models using a prediction that can fail; the result is quantitative, from band positions and amounts; a sample at each generation shows a pattern, not one snapshot.
• Limits: it shows how the old and new strands are distributed. It does not show the enzymes involved. It used bacteria, so you would need further work to claim the same in other organisms. Density alone cannot tell you the base sequence was copied accurately.
• Earlier work, such as X-ray diffraction images and the pairing ratios in DNA, supported the structure of the molecule but did not by themselves show how it is copied.

3.1.6 ATP

Structure

A single molecule of adenosine triphosphate (ATP) is a nucleotide derivative. It is made of ribose, adenine and three phosphate groups. Compare it with an RNA nucleotide: same sugar and same kind of base, but three phosphates instead of one.

Hydrolysis

ATP + water → ADP + Pi

ATP hydrolase catalyses the hydrolysis of ATP to adenosine diphosphate (ADP) and an inorganic phosphate group (Pi). Hydrolysis uses water. This releases energy.

• Coupling. The hydrolysis of ATP can be coupled to energy-requiring reactions in cells, so the energy released drives the other reaction.
• Phosphorylation. The Pi released can be used to phosphorylate other compounds, often making them more reactive.

Worked example 8 (coupling, with data supplied). A reaction needs an input of 14.0 kJ per mole. The hydrolysis of ATP releases 30.5 kJ per mole. If the two are coupled, what is the net energy change?

• Energy released − energy needed = 30.5 − 14.0 = 16.5 kJ per mole released overall.
• Coupling lets an unfavourable reaction go ahead. Net: 16.5 kJ per mole released.

Worked example 9 (amount of ATP). The relative molecular mass of ATP is 507. How much energy is released when 2.54 g of ATP is hydrolysed, using 30.5 kJ per mole?

• Moles = 2.54 ÷ 507 = 0.00501 mol.
• Energy = 0.00501 × 30.5 = 0.153 kJ (3 s.f.), or 153 J.

Resynthesis

ATP is resynthesised by the condensation of ADP and Pi, catalysed by ATP synthase. This happens during photosynthesis, or during respiration. Hydrolysis uses water and is catalysed by ATP hydrolase; synthesis releases water and is catalysed by ATP synthase. Learn that pair of enzymes and the opposite direction of each.

Common errors

• Writing “deoxyribose is in RNA” or “uracil is in DNA”.
• Saying helicase breaks phosphodiester bonds. It breaks hydrogen bonds.
• Saying DNA polymerase joins bases. It joins adjacent nucleotides by condensation.
• Counting the bases of a double helix as the number of base pairs.
• Writing “ATP contains energy” without saying that energy is released on hydrolysis.
• Mixing up ATP hydrolase and ATP synthase.

Next steps

Work through the revision notes, then the practice questions.

Official syllabus

AQA, AS and A-level Biology (7401 and 7402), specification for first teaching 2015, sections 3.1.5.1, 3.1.5.2 and 3.1.6.