The molecule of heredity: nucleotides, base pairs and the double helix
By the end of this lesson you will be able to
- Label the chain nucleus → chromosome → DNA → gene → nucleotide on a diagram and state what each level is.
- Describe a nucleotide as a phosphate, a deoxyribose sugar and one of four bases, and state the four bases by name and letter.
- Outline the double helix: two sugar–phosphate backbones held together by complementary base pairs, A with T and G with C.
- Explain why the order of nucleotides is the information, and why it differs between species and between individuals of one species.
- Calculate base percentages from complementary pairing and the stretched length of a DNA molecule from its number of base pairs.
You met this before
- Every living thing is made of cells; in animal and plant cells the nucleus controls the cell and holds its genetic material (Year 1).
- Before a cell divides by mitosis it copies all its DNA, then the two identical sets separate so that each daughter cell has a complete copy (Year 2).
- Human body cells have 46 chromosomes in 23 pairs; a chromosome is a very long DNA molecule packed with protein (Year 2).
- Variation within a species is partly inherited (genes) and partly environmental; inherited variation is the raw material of natural selection (Year 3).
- Scale: a typical animal cell is 10–30 μm across, a nucleus about 6 μm; 1 μm = 10⁻⁶ m, 1 nm = 10⁻⁹ m (Year 1).
A cell with 46 chromosomes copies its DNA and divides by mitosis. How many chromosomes does each daughter cell have, and why must the copying happen first?
Reveal
46 each. Mitosis separates two identical copies of every chromosome into the two daughter cells, so the copy must exist before the split — otherwise each daughter would get only half the instructions.
Revise it in full: Year 1 · Cell structure · Lesson 2 — Inside the cell: animal and plant cells, and how to draw them · Year 2 · Cell functions · Lesson 2 — Cells dividing: copy, split, and why they stay small · Year 3 · Classification, variation and natural selection · Lesson 3 — Variation: why no two mopane leaves are the same
The puzzle
Take one of your cells, open its nucleus — a sphere about 6 μm across, far too small to see without a microscope — and pull out the DNA inside, straightening it as you go.
You would be holding a thread about 2 metres long. It is made of only four kinds of unit, repeated 6 billion times, and the whole instruction set for building and running you is written in the order of those four units.
A cell in a banana, a bacterium and a blue whale use exactly the same four units and the same pairing rule. How can four letters be enough — and how can a 2 m thread fit in a 6 μm ball?
Two strangers chosen at random from anywhere on Earth compare their DNA sequences letter by letter. Predict roughly what fraction of the letters will be the same: about half, about 90 %, or more than 99 %? Then predict how a human sequence compares with a chimpanzee’s.
Compare your answer
More than 99 % — about 99.9 % of the 3 billion letters are identical between any two people, so the differences that make you unique are about 1 letter in 1 000 (still some 3 million letters). Human and chimpanzee sequences match at roughly 99 % of aligned letters. The lesson explains why small differences in order can matter so much.
The science — structured notes
1.1 The zoom chain: nucleus, chromosome, DNA, gene, nucleotide
The examination’s favourite first question on this unit is a diagram that zooms in six times, and asks you to drag five labels onto it. Learn the chain as a chain, because each level is made of the next:
| Level | What it is | Size |
|---|---|---|
| Cell | The unit of life; in animals and plants it contains a nucleus | 10–30 μm |
| Nucleus | The organelle that contains the DNA, wrapped in its own double membrane | about 6 μm |
| Chromosome | One very long DNA molecule wound round proteins; visible as an X-shape only when coiled up for cell division. Humans: 46 (23 pairs) | 2–10 μm when condensed |
| DNA | Deoxyribonucleic acid — the double-helix molecule itself | 2 nm wide, centimetres long when stretched |
| Gene | A section of DNA that codes for one protein (or, more precisely, one polypeptide); humans have about 20 000 | typically 1 000 – 100 000 base pairs |
| Nucleotide | The repeating unit of a DNA strand: phosphate + deoxyribose sugar + one base | about 0.34 nm from one to the next along the strand |
Two of the levels are the ones students confuse. A chromosome is not a different substance from DNA: it is one DNA molecule, so long that it has to be coiled around proteins, and it only looks like an X when it has been copied and condensed for cell division.
A gene is not a thing you could pick out with tweezers: it is a length of the DNA molecule — a stretch of the sequence — that carries the instructions for one protein.
The November 2023 markscheme gave one mark for “a section of DNA” and a second for “that codes for a protein”. The May 2021 markscheme allowed the labels DNA and Gene to be swapped on its diagram, because the boxed section of a helix could be read either way; every other label had to be exact.
A diagram shows a cell with a dark sphere inside it; an arrow leads to an X-shaped structure; a further arrow leads to a twisted ladder with a box drawn round part of it; a final arrow leads to a single unit with three parts. Label the dark sphere, the X, the twisted ladder, the boxed part and the three-part unit.
Reveal answer
Dark sphere — nucleus. X — chromosome. Twisted ladder — DNA. Boxed part — gene. Three-part unit — nucleotide. The examiner’s note: a blank box scores nothing, so always place every label; if you are unsure between two, remember the unit is smaller than the section, and the section is smaller than the molecule.
1.2 The nucleotide: three parts, four kinds
A DNA strand is a polymer — a chain of repeating units — and the unit is the nucleotide. Every nucleotide has three parts joined together: a phosphate group (a phosphorus atom with four oxygens, carrying a negative charge), a five-carbon sugar called deoxyribose (the “D” in DNA), and a nitrogen-containing base.
The phosphate of one nucleotide bonds to the sugar of the next, so the chain has a sugar–phosphate backbone — sugar, phosphate, sugar, phosphate — with a base sticking out sideways from every sugar.
The phosphate and sugar are the same in every nucleotide. Only the base differs, and there are exactly four:
| Base | Letter | Shape | Pairs with | Hydrogen bonds in the pair |
|---|---|---|---|---|
| Adenine | A | Larger, two-ring (purine) | Thymine (T) | 2 |
| Thymine | T | Smaller, one-ring (pyrimidine) | Adenine (A) | 2 |
| Guanine | G | Larger, two-ring (purine) | Cytosine (C) | 3 |
| Cytosine | C | Smaller, one-ring (pyrimidine) | Guanine (G) | 3 |
The reason the pairing is so strict is shape and chemistry. A and G are the larger, two-ring bases; T and C are the smaller, one-ring bases. A large base opposite a small one makes a pair of constant width, which is why the helix is exactly 2 nm across everywhere.
And each base has hydrogen-bonding groups in positions that fit only its partner: adenine and thymine line up two hydrogen bonds, guanine and cytosine three. Hydrogen bonds are individually weak — about a twentieth of a covalent bond — which is exactly what the cell needs: strong enough in their millions to hold the strands together, weak enough to be unzipped by an enzyme when the DNA must be copied or read.
A student writes: “DNA is made of four nucleotides — adenine, thymine, guanine and cytosine — held together in a single helix. Adenine pairs with guanine because they are the same size.” Find four errors.
Reveal answer
(1) Adenine and the others are bases; a nucleotide is base + sugar + phosphate. (There are four nucleotides, but they are named by their bases.) (2) DNA is a double helix — two strands.
(3) A pairs with T, and G with C. (4) Pairing puts a large base with a small one, not two of the same size: A and G are both large and would be too wide to fit between the backbones.
1.3 Two strands, one rule: the double helix
A single strand of nucleotides is like one side of a ladder with half-rungs sticking out. DNA is two such strands lying side by side, running in opposite directions, with every base on one strand hydrogen-bonded to its complementary base on the other.
The completed ladder is then twisted into a right-handed spiral: the double helix. Its dimensions matter, because they are how the structure was discovered (Lesson 2) and because the examination expects numbers: the helix is 2 nm wide; one complete turn is 3.4 nm long and contains 10 base pairs, so neighbouring base pairs are 0.34 nm apart along the axis.
Because every A faces a T and every G faces a C, the two strands are complementary: if you know one you can write the other. The strand ATGCCGTAAG must lie against TACGGCATTC. Two consequences follow.
First, in any DNA molecule the amount of A equals the amount of T and the amount of G equals the amount of C — Chargaff’s rules, measured in 1950 before anyone knew why (Lesson 2).
Second, the molecule contains its own copying instructions: unzip the strands, and each can act as a template on which a new complementary strand is built, giving two identical double helices.
That copying is what you watched happening before mitosis in Year 2, and it is why the structure was recognised at once as “the secret of life”.
A sample of DNA from a fish is found to contain 22 % adenine. Calculate the percentage of each of the other three bases.
A = T. Thymine = 22 %.
A + T = 22 % + 22 %
A + T = 44 %
G + C = 100 % − 44 % = 56 %
G = C, so each is half of 56 %:
guanine = 56 % ÷ 2 = 28 %
cytosine = 28 %
Check: 22 + 22 + 28 + 28 = 100
A second check: in any double-stranded DNA, A + G and T + C are each 50 % (here 22 + 28 = 50).
One strand of a short piece of DNA reads GGATCCTTAG.
(a) Write the complementary strand.
(b) State the number of hydrogen bonds holding the two strands together over these ten base pairs.
(c) The double-stranded piece is 34 % G. State the percentage of thymine.
Reveal answer
(a) CCTAGGAATC.
(b) G–C pairs: 5 (G, G, C, C, G in the given strand) × 3 = 15; A–T pairs: 5 × 2 = 10; total 25 hydrogen bonds.
(c) G = C = 34 %
G + C = 68 %
A + T = 100 % − 68 % = 32 %
T = 32 % ÷ 2 = 16 %
When DNA is heated slowly, its two strands separate. Explain why stretches rich in G–C pairs need a higher temperature to separate than stretches of the same length rich in A–T pairs.
Reveal answer
G–C pairs are held by three hydrogen bonds and A–T pairs by two, so a G–C-rich stretch has more bonds to break per base pair and needs more energy — a higher temperature — before thermal motion pulls its strands apart.
(The full picture involves stacking forces between neighbouring pairs too, but the hydrogen-bond argument is the one the examination expects. Biologists once assumed that bacteria living in hot springs must therefore have G–C-rich genomes; when many were sequenced the link turned out to be weak — they protect their DNA in other ways.)
1.4 The order is the information
If every DNA molecule is built from the same four nucleotides by the same rule, where is the difference between a mouse and a mopane tree? In the order. A gene is a particular sequence of bases — say ATGGTGCACCTGACT… — and, as Lesson 3 shows, that order is read three letters at a time to specify the order of amino acids in a protein.
Change the order and you change the protein. Four letters seem few, but the number of possible sequences grows explosively: a stretch of just 10 bases can be written 4¹⁰ = 1 048 576 different ways, and a gene of 1 000 bases in 4¹⁰⁰⁰ ways — a number with about 600 digits.
There is no shortage of room for every species that has ever lived.
So the IB’s two statements say the same thing at two scales. Between species, the sequences differ enough to build different bodies: about 1 % of aligned letters between humans and chimpanzees, far more between humans and bananas, though whole genes are recognisably shared even there — the machinery of the cell is ancient and common to all life.
Within a species, the sequences differ slightly from one individual to the next: any two people match at about 99.9 % of their 3 billion letters, leaving roughly 3 million positions where they differ.
Those positions are what make one person’s eyes, blood group and disease risks differ from another’s, and they are what DNA profiling (Lesson 4) reads. Identical twins are the exception: they began as one cell and share essentially the whole sequence.
| Organism | Base pairs (haploid set) | Approximate number of genes | Note |
|---|---|---|---|
| Bacterium E. coli | 4.6 × 10⁶ | 4 300 | One circular chromosome |
| Fruit fly | 1.4 × 10⁸ | 14 000 | 4 pairs of chromosomes |
| Human | 3.1 × 10⁹ | 20 000 | 23 pairs; only about 1.5 % of the sequence codes for protein |
| Wheat | 1.6 × 10¹⁰ | 100 000+ | Six sets of chromosomes (hexaploid) |
| Plant Paris japonica | 1.5 × 10¹¹ | not known | About 50 times the human genome — size is not complexity |
“Humans and chimpanzees have almost the same DNA, so the differences between them must be caused by the environment, not by genes.”
Reveal answer
False. “Almost the same” means about 99 % of 3 billion letters, which leaves some 30–40 million differences — more than enough to change many proteins and, above all, when and where genes are switched on. Small differences in order can have large effects because the sequence is read as instructions, not weighed as material.
1.5 How much DNA? Numbers you can calculate
The human genome — one set of 23 chromosomes — is about 3.1 × 10⁹ base pairs. A body cell has two sets, so about 6.2 × 10⁹ base pairs.
With 0.34 nm per base pair, the stretched length of all the DNA in one nucleus is:
length = number of base pairs × spacing
length = 6.2 × 10⁹ × 0.34 × 10⁻⁹ m
length = 2.1 m
That 2 m is coiled around proteins, the coils are coiled, and the result fits in a nucleus 6 μm across: a packing ratio of hundreds of thousands.
The longest human chromosome, chromosome 1, is about 2.5 × 10⁸ base pairs — 8.5 cm of DNA in a structure a few micrometres long at cell division.
An E. coli cell is a rod about 2 μm long. Its single circular chromosome contains 4.6 × 10⁶ base pairs. (1) Calculate the length of the chromosome if it were stretched out. (2) State how many times longer than the cell this is. (3) Calculate how many complete turns the double helix makes.
(1) length = number of base pairs × spacing
length = 4.6 × 10⁶ × 0.34 × 10⁻⁹ m
length = 1.56 × 10⁻³ m
length = 1.6 mm (2 s.f.)
(2) 1.56 × 10⁻³ m ÷ 2 × 10⁻⁶ m ≈ 780 times the length of the cell — which is why even a bacterium must fold its DNA tightly.
(3) One turn per 10 base pairs: 4.6 × 10⁶ ÷ 10 = 4.6 × 10⁵ turns (460 000).
Units. Nanometres to metres is a common slip: 0.34 nm = 0.34 × 10⁻⁹ m = 3.4 × 10⁻¹⁰ m. An answer of “1.6 × 10⁶ nm” is correct but should be converted to a sensible unit, mm, for the comparison.
A human body contains roughly 3 × 10¹³ cells, but most of them are red blood cells, which have no nucleus; the cells with nuclei number roughly 1 × 10¹³. Estimate the total length of DNA in a person, and compare it with the distance from Earth to the Sun, 1.5 × 10¹¹ m.
total length = length per cell × number of cells
total length ≈ 2.1 m × 1 × 10¹³
total length ≈ 2 × 10¹³ m
ratio = 2 × 10¹³ m ÷ 1.5 × 10¹¹ m
ratio ≈ 140 times the Earth–Sun distance.
(An estimate, not a measurement: cell counts are uncertain by tens of per cent, so quote 1 or 2 significant figures.)
A gene is 1 800 base pairs long. Calculate its length in nanometres and in micrometres, and the number of turns of the helix it occupies.
Reveal answer
length = 1 800 × 0.34 nm = 612 nm
612 nm = 0.612 μm ≈ 0.61 μm
turns = 1 800 ÷ 10 = 180
(A gene, then, is about a tenth of the width of a nucleus — and there are 20 000 of them, plus far more non-coding DNA, in the same space.)
Suppose a mutation changed a G–C pair to an A–T pair somewhere in a gene. Would the DNA molecule change in width, length, or neither? What might change?
Reveal answer
Neither. Every pair is one large base plus one small one, so the width stays 2 nm, and the number of base pairs is unchanged, so the length stays the same. What changes is the information: the sequence now reads differently at that position, and (Lesson 3) the protein built from it may have a different amino acid there. The molecule is physically almost identical; the message may not be.
A chromosome from a lily contains 2.0 × 10⁹ base pairs. Its DNA is wrapped around protein “spools” so that each 200 base pairs occupies a bead 11 nm across, and the string of beads is then coiled into a fibre that shortens it a further 6 times. Estimate the length of the fibre and compare it with the stretched length of the naked DNA.
Reveal answer
naked DNA = 2.0 × 10⁹ × 0.34 × 10⁻⁹ m = 0.68 m
number of beads = 2.0 × 10⁹ ÷ 200 = 1.0 × 10⁷
string of beads = 1.0 × 10⁷ × 11 × 10⁻⁹ m = 0.11 m
coiled fibre = 0.11 m ÷ 6 ≈ 0.018 m, about 2 cm
The fibre is 0.68 ÷ 0.018 ≈ 37 times shorter than the naked DNA — and still centimetres long, so further levels of coiling are needed to reach the few micrometres of a condensed chromosome. Each level of packing multiplies the last.
Try it — getting DNA out of a strawberry
You can see DNA without a microscope. Mash fruit with salt water and washing-up liquid, filter, and layer cold ethanol on top: white threads gather at the boundary and can be wound on a rod.
The detergent dissolves the membranes (Year 1: membranes are made of phospholipids — grease, in effect), the salt lets the negatively charged strands clump, and DNA is insoluble in cold ethanol.
The lab lets you change the volume of detergent and weigh the precipitate.
Suppose your mean masses were: 2 cm³ → 25 mg; 4 → 37 mg; 6 → 44 mg; 8 → 48 mg; 12 → 52 mg; 16 → 50 mg; 20 → 47 mg.
(a) Describe the relationship, quoting values.
(b) State the volume of detergent you would recommend and justify it.
(c) Explain, at the level of molecules, why the yield stops rising.
Reveal answer
(a) As the volume rises from 2 to 12 cm³ the mass rises from 25 to 52 mg, steeply at first (25 to 44 mg between 2 and 6 cm³) then slowly (48 to 52 mg between 8 and 12 cm³); above 12 cm³ it falls slightly, to 47 mg at 20 cm³.
Not linear and not proportional: doubling the volume from 4 to 8 cm³ raises the mass by only 30 %, from 37 to 48 mg.
(b) About 8–12 cm³: 8 cm³ already gives 92 % of the maximum and 12 cm³ the maximum itself; more is wasted and slightly harmful. (The November 2022 paper’s “lowest concentration that gives the maximum yield” is the logic to copy.)
(c) Each detergent molecule can dissolve a limited amount of membrane. At low volumes the detergent is used up before every cell and nucleus is broken open, so more detergent releases more DNA; once there is enough to dissolve all the membranes, all the DNA is already free and additional detergent has nothing left to act on.
The extra volume dilutes the extract and makes foam that traps threads, which is why the mass drifts down.
Common mistakes to avoid
- Calling the bases “nucleotides” — a nucleotide is phosphate + sugar + base; A, T, G and C are the bases. There are four nucleotides because there are four bases.
- “A pairs with G” — the pairs are A–T and G–C; large with small, 2 bonds and 3 bonds.
- Gene and chromosome muddled — a chromosome is a whole DNA molecule (plus protein); a gene is a section of it.
- Applying Chargaff’s rule to one strand — A = T only for the double-stranded molecule.
- Forgetting the nanometre conversion — 0.34 nm = 3.4 × 10⁻¹⁰ m; keep units on every line.
- “Bigger genome = more complex organism” — wheat and Paris japonica say otherwise; most DNA is not genes.
- “Humans are 99.9 % identical, so DNA differences do not matter” — 0.1 % of 3 billion is 3 million letters.
- Blank boxes on the label question — every label must be placed; a blank scores 0.
Study smart: how this is examined
Study routine. Today: draw the zoom chain from memory with a definition beside each level, then draw one nucleotide and an A–T and a G–C pair with their hydrogen bonds.
In two days: do the base-percentage EXAMPLE and the E. coli length EXAMPLE with the book closed, then write the complementary strand of a random 12-letter sequence. In a week: run the extraction lab as a full investigation, take the quiz, and explain to someone why 0.1 % of 3 billion is a lot.
How we found out — an acid from pus, and a rule nobody could explain
DNA was discovered before anyone knew what it was for. In 1869 Friedrich Miescher, a young Swiss physician working in a laboratory in Tübingen, was trying to work out the chemistry of white blood cells, which he collected from the pus on discarded surgical bandages.
From their nuclei he extracted a substance unlike any protein: rich in phosphorus, acidic, and resistant to the protein-digesting enzymes he tried. He called it “nuclein”. Its parts were worked out over the next fifty years — the phosphate, the sugar and the four bases were all identified by 1929 — but the conclusion drawn from them was wrong.
The biochemist Phoebus Levene proposed that the four nucleotides simply repeated in a fixed order, ATGC ATGC ATGC; a molecule so monotonous could not carry information, so the genes, everyone assumed, must be proteins, which came in twenty different building blocks.
Two results overturned that. In 1944 Oswald Avery, Colin MacLeod and Maclyn McCarty in New York showed that a harmless strain of pneumonia bacteria could be transformed into a deadly one by DNA extracted from dead deadly bacteria — and by nothing else: destroy the protein and transformation still worked; destroy the DNA and it stopped.
And between 1949 and 1951 Erwin Chargaff in New York used the new technique of paper chromatography to measure the four bases in DNA from different species.
He found that the proportions differed from species to species — so Levene’s fixed order was wrong and the sequence could carry information — and that in every species the amount of A equalled the amount of T and G equalled C. Chargaff could not explain his own rule.
Two years later the double helix explained it in a sentence, which is where Lesson 2 begins.
Explain why Levene’s repeating-tetranucleotide idea made DNA seem an impossible carrier of information, and how Chargaff’s results disposed of it.
Reveal answer
If the bases always ran ATGC ATGC…, every DNA molecule would be the same message, and a molecule that cannot vary cannot store instructions — so heredity was assigned to proteins, whose twenty amino acids could vary.
Chargaff showed the base proportions differ between species (for example roughly 30 % A in humans, higher in some bacteria, lower in others), which is impossible if the order is fixed at one of each in four; the order therefore varies, and varying order is exactly what a code needs.
His A = T and G = C rule, in the same data, turned out to be the fingerprint of complementary pairing.
Where it is used now
- Cattle and disease testing in Botswana. Veterinary laboratories test for foot-and-mouth disease and other infections by detecting the pathogen’s DNA or RNA sequence — a method that works because every species has its own sequence. The same principle identifies the strain of a virus in an outbreak.
- Wildlife forensics. Sequences from confiscated ivory, rhino horn or bushmeat are matched to species and even to populations, tracing where an animal was killed; southern African laboratories run such tests for anti-poaching prosecutions.
- Sequencing crops. The genomes of sorghum, cowpea and maize — staple crops across Africa — have been sequenced, and breeders use the sequence differences between varieties to select for drought tolerance without waiting for a harvest.
- Medicine. Newborn screening, cancer diagnosis and matching organ donors all depend on reading sequence differences between individuals; the 99.9 % that is shared is what makes the 0.1 % readable.
- Data storage. Because DNA stores information densely and lasts thousands of years when dry, research groups have encoded books, images and software as synthetic DNA sequences — still slow and costly, but a genuine industry in the making.
Science now — and the scientist you need to be
What is happening in 2026
Reading the order of the four bases has gone from a decade-long project to a routine. The Human Genome Project, declared complete in 2003, cost billions and still left gaps; in 2022 a consortium published the first truly complete human genome, gap-free from end to end, using long-read sequencers that read tens of thousands of bases in one pass.
A human genome can now be sequenced for a few hundred dollars in a day, and national programmes in the UK, the USA, Qatar, Nigeria and elsewhere are sequencing hundreds of thousands of people to link sequence differences to health.
The 2023 “pangenome” — a reference built from dozens of people of different ancestries rather than one — was a correction to a real bias: most sequenced genomes are of European ancestry, so variants common in African populations, the most genetically diverse on Earth, are under-studied; African genome projects are working to close that gap.
Meanwhile the four-letter alphabet itself has been extended in the laboratory: bacteria have been engineered to copy DNA containing extra, synthetic base pairs, an experiment in whether four letters were a necessity or an accident of history.
And synthetic DNA is being tested as an archive medium, with the first commercial DNA data-storage systems now being demonstrated.
The mindshift
What you need to be able to do
Label the zoom chain with confidence, describe the nucleotide and the pairing rule with the numbers of hydrogen bonds, and handle the arithmetic — base percentages and lengths in standard form with units carried through. Then hold the idea that examiners keep rewarding: structure and function belong together. The double helix is beautiful, but it is beautiful because complementary pairing is what lets it be copied and read.
See an example
Threads appeared within a minute of adding the cold spirit; they wound onto a cocktail stick as a glistening blob. The assistant said the precipitate was “DNA”; when pushed, it agreed that the strawberry’s pectin and some protein precipitate with it and that a pure sample would need further steps — a good reminder that a demonstration is not a measurement.
Key terms
| DNA (deoxyribonucleic acid) | The molecule that carries the inherited instructions of a cell; a double helix of two nucleotide strands. |
| Nucleotide | The repeating unit of DNA: a phosphate group, a deoxyribose sugar and one of four bases. |
| Base | The variable part of a nucleotide: adenine (A), thymine (T), guanine (G) or cytosine (C). |
| Sugar–phosphate backbone | The covalently bonded chain of alternating sugars and phosphates that forms each strand. |
| Complementary base pairing | A always pairs with T (two hydrogen bonds) and G with C (three), holding the two strands together. |
| Double helix | The twisted-ladder shape of DNA: two strands wound round each other, 2 nm wide, 10 base pairs per 3.4 nm turn. |
| Chromosome | One very long DNA molecule packaged with protein; visible as an X-shape when condensed for division. |
| Gene | A section of DNA whose base sequence codes for one protein. |
| Genome | The complete DNA sequence of an organism — about 3.1 × 10⁹ base pairs in one human set. |
| Chargaff’s rules | In double-stranded DNA the amount of A equals T and G equals C; base proportions differ between species. |
Think like a scientist
Criterion B A school has: strawberries, kiwi fruit and onions; table salt; washing-up liquid; ethanol kept in a freezer; a balance reading to 1 mg; measuring cylinders (10 cm³ and 100 cm³); a thermometer; coffee filters, beakers, glass rods, stopwatches and zip-lock bags. A student wants to find out how the concentration of salt in the extraction solution affects the amount of DNA that can be spooled from strawberries.
(a) Write a research question for the investigation. [2]
(b) Formulate a hypothesis with a testable prediction and scientific reasoning introduced by “because”, reaching the level of charges on the DNA molecule. [3]
(c) Identify the independent and dependent variables, each with its unit and the instrument used to set or measure it, and explain why “amount of DNA” would not be accepted as the dependent variable. [3]
(d) Identify three control variables with the value at which each is held and why it must be held. [3]
(e) Write a numbered method another student could follow, with five salt concentrations, three trials each, what is recorded and what is calculated, and one safety point specific to this experiment. [5]
(f) Explain one systematic error in what the balance measures and how a real laboratory avoids it. [2]
Model answer
(a) How does the concentration of sodium chloride in the extraction solution, from 0 to 8 g per 100 cm³, affect the mass of DNA precipitate spooled from 50 g of strawberry? [Both variables named; a measurable question with a range.]
(b) If the salt concentration is increased from 0 to 8 g per 100 cm³, then the mass of precipitate will increase, steeply at first and then levelling off, because each nucleotide carries a negatively charged phosphate group, so DNA strands repel one another and stay dispersed in water.
Sodium ions (Na⁺) surround the phosphates and cancel the repulsion, letting the strands clump together and come out of solution when ethanol is added — once there are enough ions to neutralise all the phosphates, extra salt makes no further difference.
(c) IV: mass of NaCl dissolved in 100 cm³ of water (g per 100 cm³), weighed on the balance and dissolved in water measured in the 100 cm³ cylinder. DV: mass of spooled precipitate in mg, blotted 30 s and weighed on the 1 mg balance.
“Amount of DNA” is not a quantity any instrument reads — it has no unit and could mean volume, mass or number of molecules; the dependent variable must be the measured reading, from which anything else is calculated.
(d) Volume of detergent — 10 cm³ per extraction by 10 cm³ cylinder, because detergent volume changes how many cells are broken open. Mass and source of fruit — 50 g from the same punnet on the balance, because different fruit have different DNA content per gram (strawberries are octoploid).
Ethanol temperature — used straight from the freezer (below 0 °C), checked with the thermometer, because DNA is more soluble in warm ethanol and less would precipitate. (Also: 40 cm³ of ethanol; 5 min standing time by stopwatch; the same filter type and drain time.)
(e) 1. Make five salt solutions: 0, 2, 4, 6 and 8 g NaCl weighed on the balance, each dissolved in 100 cm³ of water.
2. Weigh 50 g of strawberries into a zip-lock bag and mash for 2 min.
3. Add 100 cm³ of the salt solution and 10 cm³ of detergent (10 cm³ cylinder); mix gently for 1 min.
4. Filter through a coffee filter into a beaker for 2 min.
5. Pour 40 cm³ of ethanol straight from the freezer slowly down the side of the beaker; start the stopwatch and wait 5 min.
6. Spool the threads on a glass rod, blot on filter paper for 30 s, weigh to 1 mg and record in a table with units in the headings.
7. Repeat steps 2–6 twice more at each concentration (15 extractions) with washed glassware.
The results table to draw (fill in the empty cells during the experiment):
| NaCl concentration / g per 100 cm³ | mass of DNA, trial 1 / mg | mass of DNA, trial 2 / mg | mass of DNA, trial 3 / mg | mean mass of DNA / mg |
|---|---|---|---|---|
| 0 | ||||
| 2 | ||||
| 4 | ||||
| 6 | ||||
| 8 |
8. Calculate the mean mass at each concentration to the nearest mg, plot mean mass against salt concentration with a curve of best fit, and describe the relationship quoting values.
Safety specific to the experiment: ethanol is flammable and its vapour irritates the eyes — no flames in the room, keep the bottle capped, and wear eye protection when pouring.
(f) The balance weighs everything on the rod: DNA, but also pectin and protein that precipitate with it, and any ethanol not blotted off. Every mass is therefore too high by a similar amount — a systematic error that shifts all points upward without changing the shape of the curve. A laboratory measures DNA by the absorbance of the dissolved sample at 260 nm, which pectin does not absorb, or purifies the DNA before weighing.
Probe deeper — exam-style questions
Attempt every question before opening the answer. Marks in brackets show what a full answer is worth.
Describe the structure of a nucleotide and state how nucleotides are joined to make a strand.
Reveal answer
A phosphate group, a deoxyribose sugar and one of four bases (A, T, G, C) bonded together [1]; the phosphate of one nucleotide bonds to the sugar of the next, forming a sugar–phosphate backbone with the bases projecting from it [1].
One strand of DNA reads TTGACCGATA. Write the complementary strand and state the total number of hydrogen bonds between the two strands.
Reveal answer
AACTGGCTAT [1]; G–C pairs 4 × 3 = 12, A–T pairs 6 × 2 = 12, total 24 [1].
Outline the structure of DNA as a double helix, including how the two strands are held together and why the molecule has a constant width.
Reveal answer
Two strands of nucleotides, each with a sugar–phosphate backbone, twisted round each other in a helix [1]; the strands are held together by hydrogen bonds between complementary bases, A with T (two bonds) and G with C (three) [1]; each pair is one large two-ring base with one small one-ring base, so every rung is the same width and the backbones stay 2 nm apart [1].
A sample of double-stranded DNA is 18 % cytosine. Calculate the percentage of adenine.
Reveal answer
G = C = 18 %
G + C = 36 %
A + T = 100 % − 36 % = 64 % [1]
A = 64 % ÷ 2 = 32 % [1]
Explain what is meant by the statement “the order of nucleotides varies between species and between organisms within a species”, and why this variation matters.
Reveal answer
All species use the same four nucleotides and the same pairing rule; what differs is the sequence in which the bases occur along the molecule [1]. Different species have substantially different sequences (about 1 % between human and chimpanzee, far more between distant species), while individuals of one species differ at a small fraction of positions (about 0.1 % between two people) [1].
The sequence is the information: it determines which proteins are made, so differences in order produce differences in structure and function — and the individual differences are what make DNA profiling possible [1].
A fruit-fly chromosome contains 2.3 × 10⁷ base pairs. Calculate the stretched length of its DNA in millimetres and the number of helical turns, and state how many times longer than a 4 μm condensed chromosome the stretched DNA is.
Reveal answer
length = 2.3 × 10⁷ × 0.34 × 10⁻⁹ m
length = 7.8 × 10⁻³ m = 7.8 mm [1]
turns = 2.3 × 10⁷ ÷ 10 = 2.3 × 10⁶ [1]
ratio = 7.8 × 10⁻³ m ÷ 4 × 10⁻⁶ m ≈ 2 000 times [1]
7. Two students measure the base composition of DNA from a virus. Student A reports 24 % A, 24 % T, 26 % G, 26 % C. Student B, working on a different virus, reports 31 % A, 19 % T, 22 % G, 28 % C. Deduce what each result tells you about the structure of the virus’s DNA, and explain how student B’s virus could still copy its DNA inside a host cell.
Reveal answer
Student A: A = T and G = C, consistent with double-stranded DNA obeying complementary pairing [1]. Student B: A ≠ T and G ≠ C, so the rule is broken — the DNA must be single-stranded, since only a strand paired with its complement is forced to have equal A and T [1]; some viruses (such as the parvoviruses) do carry single-stranded DNA [1].
Inside a host cell the single strand serves as a template: host enzymes build the complementary strand by pairing A with T and G with C, producing a double helix that can then be copied normally — the pairing rule applies to the copy even though the packaged virus breaks it [1].
8. A DNA molecule of 5 000 base pairs is heated slowly while the temperature at which its strands separate is measured. A second molecule, also 5 000 base pairs, separates at a temperature 8 °C higher. Using ideas from this lesson, suggest the difference between the two molecules, explain the reasoning, and predict how the separation temperature of a 500-base-pair piece of the first molecule would compare with the whole molecule.
Reveal answer
The second molecule has a higher proportion of G–C pairs [1], because G–C pairs are held by three hydrogen bonds and A–T pairs by two, so more energy — a higher temperature — is needed to break enough bonds to separate the strands [1].
Length also matters: a longer molecule has more bonds along its length holding it together and separates at a higher temperature, up to a limit [1], so a 500-bp piece of the first molecule would separate at a lower temperature than the whole molecule, unless the piece happens to be much richer in G–C than the average [1].
Summary
- The chain to learn: nucleus → chromosome (one long DNA molecule) → DNA → gene (a section that codes for a protein) → nucleotide (phosphate + deoxyribose + base).
- Four bases — adenine, thymine, guanine, cytosine — so four nucleotides; A pairs with T by two hydrogen bonds, G with C by three, always a large base with a small one.
- DNA is a double helix: two sugar–phosphate backbones, complementary base pairs between them, 2 nm wide, 10 base pairs per 3.4 nm turn. Complementary pairing means A = T and G = C, and lets the molecule be copied.
- The information is the order of the bases. It differs between species (about 1 % human–chimpanzee) and between individuals of a species (about 0.1 % between people — 3 million letters).
- Numbers: 3.1 × 10⁹ base pairs per human set, about 2 m of DNA per cell, 0.34 nm per base pair; genome size is not complexity.
Next time → The race for the structure: what Photo 51 actually showed, what Chargaff’s numbers meant, who did what in 1953 — and why a model is only as good as the evidence that tests it.