DNA and RNA Structure: Study Guide (Biology 6.1)

The molecule that is passed on

Every time a cell divides, and every time an organism reproduces, a full set of instructions must be passed on. In all cells, from bacteria to oak trees to humans, those instructions are stored in DNA. The information is the order of the bases along the molecule, as you saw in Topic 1.6.

There is one important exception. Some viruses carry RNA as their genetic material, not DNA. Influenza viruses and HIV are well-known examples. So the accurate statement is: hereditary information is stored in DNA, and in some cases (certain viruses) in RNA.

This topic asks two questions:

1. How is the DNA organised inside different kinds of cells?
2. What features of DNA make it a good molecule for storing and passing on information?

Organising the DNA: chromosomes in prokaryotes and eukaryotes

A chromosome is a long DNA molecule together with the proteins that help to pack it. The way chromosomes are arranged differs between the two main kinds of cell.

Prokaryotes (bacteria and archaea):

• Typically have one circular chromosome. The two ends of the DNA are joined, so the molecule is a closed loop with no free ends.
• The chromosome lies in the cytoplasm, in a region called the nucleoid. There is no nuclear envelope around it.
• The loop is far longer than the cell. It is packed by twisting on itself (supercoiling) and by binding to proteins.

Eukaryotes (protists, fungi, plants and animals):

• Typically have several linear chromosomes, each a separate DNA molecule with two free ends. A human body cell, for example, has 46.
• The chromosomes are inside the nucleus.
• The DNA is condensed by wrapping around proteins called histones. Roughly 150 base pairs of DNA wind around a core of eight histone proteins to make one nucleosome. Short stretches of “linker” DNA join one nucleosome to the next, like beads on a string. Other proteins then fold this string into thicker fibres.

Why do histones bind DNA so well? The phosphate groups in the DNA backbone carry negative charges. Histones contain many amino acids with positively charged side chains. Opposite charges attract, so DNA wraps tightly around them.

The word typically matters. A few bacteria have linear chromosomes, and mitochondria and chloroplasts have their own small circular DNA. Exam questions will use the typical patterns unless they give you other information.

Plasmids

Plasmids are small, circular DNA molecules that are separate from the chromosome (extra-chromosomal). Key points:

• They are very common in bacteria, but some eukaryotes, such as yeasts, also carry them.
• A plasmid is copied independently of the chromosome, and a cell can contain many copies.
• Plasmids usually carry a small number of genes that are useful in some conditions but not essential, such as genes for antibiotic resistance.
• Plasmids can pass from one bacterial cell to another, even between different species. This spreads genes sideways through a population, not only from parent to offspring.

Because they are small and easy to move between cells, plasmids are also important tools in biotechnology, which you will meet in Topic 6.8.

What makes DNA a good hereditary molecule?

A molecule that stores and passes on information must be stable, must hold a large amount of information, and must be copied accurately. DNA’s structure meets all three needs.

• Stable backbone. The sugar–phosphate backbone is held by strong covalent bonds, so the base sequence is not easily broken.
• Information in the base sequence. The order of four bases along a very long molecule can encode a huge amount of information.
• Complementary pairing. Each strand predicts the other. If the strands separate, each one can act as a template for building a new partner. You will see how in Topic 6.2.

Purines and pyrimidines

The bases fall into two groups by shape:

Group	Ring structure	Bases
Purines	double ring (two fused rings)	adenine (A), guanine (G)
Pyrimidines	single ring	cytosine (C), thymine (T, DNA only), uracil (U, RNA only)

Two common memory aids: “Pure As Gold” for the purines (A, G), and “CUT the py” for the pyrimidines (C, U, T).

Every base pair contains one purine and one pyrimidine:

• A pairs with T in DNA (and A with U where RNA pairs).
• G pairs with C.

Why this pairing matters.

1. Constant width. A purine–pyrimidine pair always contains three rings in total. So every rung of the ladder is the same length, and the double helix has the same width all along its length (about 2 nm). A purine–purine pair would bulge; a pyrimidine–pyrimidine pair would pinch.
2. Specific partners. Within the purine–pyrimidine rule, the hydrogen-bonding groups only line up for A with T (or U) and for G with C. So the sequence of one strand fixes the sequence of the other.
3. Conserved through evolution. The same pairing rules are found in every known cellular organism, and in RNA as well as DNA. A feature shared by all known life is strong evidence that it was inherited from a common ancestor.

A useful consequence: in any double-stranded molecule, the number of purines equals the number of pyrimidines. So purines make up 50% of the bases, whatever the species. The ratio of A + T to G + C, by contrast, varies from species to species, because different species have different sequences.

Worked example 1: testing the purine–pyrimidine rule with data

Question. A student is given base composition data for double-stranded DNA from three fictional organisms and for the genetic material of a fictional virus.

Source	A / %	T / %	U / %	G / %	C / %
Soil bacterium	17.6	17.9	0	32.4	32.1
Moss	29.1	28.8	0	21.2	20.9
Freshwater fish	31.0	31.3	0	18.9	18.8
Virus V	30.6	0	17.9	28.3	23.2

(a) For the three organisms, calculate the percentage of purines and of pyrimidines, and the ratio (A + T) ÷ (G + C). (b) Explain what the results show about base pairing across species. (c) What can you conclude about the genetic material of virus V?

(a) Purines = A + G; pyrimidines = C + T.

1. Soil bacterium: purines = 17.6 + 32.4 = 50.0%; pyrimidines = 32.1 + 17.9 = 50.0%. (A + T) ÷ (G + C) = 35.5 ÷ 64.5 = 0.55.
2. Moss: purines = 29.1 + 21.2 = 50.3%; pyrimidines = 20.9 + 28.8 = 49.7%. Ratio = 57.9 ÷ 42.1 = 1.38.
3. Fish: purines = 31.0 + 18.9 = 49.9%; pyrimidines = 18.8 + 31.3 = 50.1%. Ratio = 62.3 ÷ 37.7 = 1.65.

Check. Each row adds up to 100.0%.

(b) The (A + T) ÷ (G + C) ratio differs widely, from 0.55 to 1.65, so the three species have very different sequences. Yet in all three, purines are within 0.3 percentage points of 50%, so purines and pyrimidines are equal within measurement error. This fits the model that every base pair is a purine plus a pyrimidine. The same rule holds in a bacterium, a plant and an animal, which supports the idea that pairing is conserved through evolution.

(c) Virus V contains uracil and no thymine, so its genetic material is RNA. Purines = 30.6 + 28.3 = 58.9%, pyrimidines = 17.9 + 23.2 = 41.1%. These are not equal, and A (30.6%) does not equal U (17.9%). So most bases are not paired: the genome is most likely single-stranded RNA.

Interpretation. The 50 : 50 rule is a property of paired bases, not of DNA as such. A purine fraction far from 50% is evidence of a single-stranded molecule.

Worked example 2: how big is the packing problem?

Question. Use the model value 0.34 nm of length per base pair.

(a) A fictional bacterium has one circular chromosome of 4.2 × 10⁶ base pairs and three copies of a 5 000 base-pair plasmid. The cell is 2.0 µm long. How many DNA molecules does it contain, how many free DNA ends, and how many times longer than the cell is its chromosome? (b) A fictional eukaryote has body cells with 12 chromosomes containing 1.2 × 10⁹ base pairs in total. The nucleus is 8 µm across. Find the number of free ends, the total DNA length, and the number of nucleosomes, if each nucleosome plus its linker uses about 200 base pairs.

(a)

1. DNA molecules: 1 chromosome + 3 plasmids = 4.
2. Free ends: all four are circles, so 0.
3. Chromosome length = 4.2 × 10⁶ × 0.34 nm = 1.428 × 10⁶ nm = 1.43 mm.
4. Cell length = 2.0 µm = 2 000 nm. Ratio = 1.428 × 10⁶ ÷ 2 000 ≈ 710 times longer than the cell.
5. (Extra) The plasmids hold 3 × 5 000 = 15 000 bp, only about 0.36% of the cell’s DNA.

(b)

1. Free ends: 12 linear chromosomes × 2 ends = 24.
2. Length = 1.2 × 10⁹ × 0.34 nm = 4.08 × 10⁸ nm = 0.408 m, about 41 cm of DNA.
3. Compared with the nucleus: 0.408 m ÷ (8 × 10⁻⁶ m) ≈ 51 000 times longer.
4. Nucleosomes ≈ 1.2 × 10⁹ ÷ 200 = 6.0 × 10⁶, which needs 8 × 6.0 × 10⁶ = 4.8 × 10⁷ histone proteins.

Interpretation. Both cells must compact their DNA enormously. The bacterium does it mainly by supercoiling its loop. The eukaryote, with far more DNA in a small nucleus, relies on histones. The count of free ends also matters later: linear chromosomes have ends that need special handling when they are copied, while circles do not.

Worked example 3: which molecule carries the information?

Question. Researchers extract the genetic material of a fictional plant virus. They split the extract into four equal portions. Three are treated with an enzyme that destroys one type of molecule; one gets buffer only. Each portion is then rubbed onto identical leaves, and the number of infection spots is counted after 5 days.

Treatment	What it destroys	Infection spots
Buffer only	nothing	240
Protease	proteins	236
DNase	DNA	231
RNase	RNA	0

(a) Express each result as a percentage of the buffer-only result. (b) What is the hereditary material of this virus? Justify your claim. (c) Why is the buffer-only portion needed?

(a) Protease: 236 ÷ 240 × 100 = 98.3%. DNase: 231 ÷ 240 × 100 ≈ 96%. RNase: 0 ÷ 240 × 100 = 0%.

(b) Claim: the hereditary material is RNA. Evidence: destroying protein or DNA left infection almost unchanged (96–98% of the control), but destroying RNA prevented all infection. Reasoning: the molecule that carries the instructions for making new viruses must be intact for infection to occur. Only the loss of RNA stopped infection.

(c) The buffer-only portion is the control. It shows how much infection the untreated extract causes, so each treatment can be compared with it. It also shows that the handling and incubation alone do not stop infection.

Limits. Small differences such as 231 against 240 are likely to be normal variation; repeated trials would show this. The test also assumes each enzyme is pure and destroys only its own target.

Common misconceptions

• “All genetic material is DNA.” All cells use DNA, but some viruses use RNA.
• “Prokaryotes have no chromosome because they have no nucleus.” They do have a chromosome: typically one circular one, lying in the nucleoid region.
• “Plasmids are only found in bacteria.” They are most common in bacteria, but some eukaryotes, such as yeasts, carry them.
• “Plasmids are part of the chromosome.” They are separate (extra-chromosomal) molecules that are copied on their own.
• “Histones are a kind of DNA.” Histones are proteins. DNA wraps around them because the negative backbone is attracted to their positive charges.
• “Purines are single-ringed because the word is shorter.” It is the other way round: purines (A, G) are double-ringed; pyrimidines (C, T, U) are single-ringed.
• “A pairs with G because both are purines.” Like pairs with unlike: a purine always pairs with a pyrimidine.
• “Purines are always 50% of a nucleic acid.” Only in double-stranded molecules, where every base is paired.

Where this leads

Complementary base pairing, together with the antiparallel strands you met in Topic 1.6, is exactly what lets DNA be copied. Continue with Topic 6.2, DNA Replication. Before you move on, try the practice questions, then use the revision notes and the checklist. The previous topic is Topic 5.5, Environmental Effects on Phenotype.