AQA A-level Biology: DNA, Genes, Chromosomes and Protein Synthesis (7402) – Revision Notes

These notes condense sections 3.4.1 (DNA, genes and chromosomes) and 3.4.2 (DNA and protein synthesis) of the AQA specification for AS and A-level Biology (codes 7401 and 7402), first teaching 2015. Both sections are AS content and also part of the A-level course. The study guide gives the full explanations and longer worked examples; the practice set has exam-style questions.

Also useful: the subject hub, the checklist to print, the diagnostic tests and the Biological Molecules notes.

3.4.1 Key definitions

Term	Meaning
Gene	A base sequence of DNA that codes for the amino acid sequence of a polypeptide, or for a functional RNA (including ribosomal RNA and tRNAs)
Locus	The fixed position of a gene on a particular DNA molecule
Triplet	Three DNA bases that code for a specific amino acid
Chromosome	In eukaryotes, one DNA molecule plus its associated histone proteins
Exon	Part of a gene that codes for amino acids
Intron	Non-coding sequence within a gene, between exons
Genome	The complete set of genes in a cell
Proteome	The full range of proteins a cell is able to produce
Codon	Three mRNA bases that code for an amino acid or a stop signal
Anticodon	Three bases on a tRNA, complementary to a codon

DNA compared

	Prokaryote	Eukaryote nucleus	Mitochondrion or chloroplast
Size	Short	Very long	Short
Shape	Circular	Linear	Circular
Histones	No	Yes	No

Eukaryotic nuclear DNA also contains non-coding multiple repeats between genes, and introns inside genes.

The genetic code

• Universal: same triplet, same amino acid, in all organisms.
• Non-overlapping: each base belongs to one triplet only.
• Degenerate: most amino acids have more than one triplet. There are 4 x 4 x 4 = 64 triplets and 20 amino acids.
• You do not have to recall codon tables. They are supplied.

mRNA and tRNA

	mRNA	tRNA
Strand	Single, linear	Single, folded into a clover-leaf, held by hydrogen bonds
Carries	Codons	Anticodon and one specific amino acid
Sugar and base	Ribose, uracil	Ribose, uracil

Method boxes

Transcription in steps

1. RNA polymerase binds at the gene; strands separate; one acts as template.
2. Free RNA nucleotides pair with the template bases (A to U, T to A, G to C, C to G).
3. RNA polymerase joins the nucleotides by condensation.
4. Prokaryote: product is mRNA. Eukaryote: product is pre-mRNA, then introns are removed and exons joined (splicing) to give mRNA.

Translation in steps

1. mRNA attaches to a ribosome.
2. tRNA with the matching anticodon brings its amino acid to each codon in turn; two sit on the ribosome at a time.
3. Peptide bonds form between adjacent amino acids; ATP supplies energy.
4. The ribosome moves one codon at a time; a stop codon ends the process.

Codon problem in steps

1. Write the mRNA as the complement of the template, with U in place of T.
2. Split into triplets from the start codon.
3. Look up each codon in the table you are given.
4. Stop at a stop codon. Do not count it as an amino acid.
5. For anticodons, complement the codon again.

Worked reminders

Length of a coding region. A polypeptide of n amino acids needs at least 3n + 3 nucleotides from start to stop codon inclusive. For 150 amino acids: 3 x 150 + 3 = 453.

After splicing. mRNA length = pre-mRNA length minus total intron length. A pre-mRNA of 1,860 nucleotides with introns of 240 and 395 gives 1,860 - 635 = 1,225.

Coding percentage. Total exon length divided by genome size, times 100. Eukaryotes give lower values than prokaryotes because of introns and repeats.

Reading experiments. Say what the label or inhibitor targets, describe the trend with numbers, then conclude, and note a limitation such as the label entering all RNA types.

Organelle DNA in one line

Mitochondria and chloroplasts each contain their own DNA. Like prokaryotic DNA it is short, circular and not associated with protein, so it differs from the long, linear, histone-bound DNA in the nucleus of the same cell.

Gene expression compared

Stage	Prokaryote	Eukaryote
Transcription product	mRNA directly	Pre-mRNA
Splicing	None: transcription gives mRNA directly	Introns removed, exons joined
Where	Cytoplasm	Transcription and splicing in the nucleus
Translation	At ribosomes in the cytoplasm	At ribosomes in the cytoplasm

In eukaryotes the mRNA leaves the nucleus through a nuclear pore before translation.

Worked reminder: a short codon problem

Given: GUC = Val, AAG = Lys, CGA = Arg, UAG = stop.

Template strand: CAG TTC GCT ATC.

1. mRNA: GUC AAG CGA UAG.
2. Amino acids: Val - Lys - Arg, then stop.
3. Anticodon for Lys: complement of AAG = UUC.

Remember that the table only lists what the question needs. If a codon you derive is not in the table, recheck your complement before assuming the table is incomplete.

Worked reminder: percentage coding

A prokaryote genome has 2.4 million base pairs and 2,100 genes with an average coding length of 1,050 base pairs.

Coding DNA = 2,100 x 1,050 = 2,205,000 base pairs. As a percentage of the genome: 2,205,000 / 2,400,000 x 100 = 91.9% (1 d.p.).

For comparison, a eukaryote with 90 million base pairs and 11,000 genes of average coding length 1,500 base pairs has 16,500,000 coding base pairs, or 18.3%. The difference reflects introns, repeats and other non-coding DNA between genes.

Reading experiment data

Work in this order:

1. Identify what was measured, for example radioactivity in protein or in RNA.
2. Identify the variable that was changed, for example an inhibitor of RNA polymerase, or the removal of ATP.
3. Quote figures for the change, using a percentage, a ratio or a difference.
4. Link to the process: if protein synthesis falls more slowly than RNA synthesis, mRNA already made is still being translated.
5. State the conclusion, then one limitation or one extra control that would strengthen it.

Typical roles to recognise: no mRNA means no polypeptide; no ATP means a much lower rate; ribonuclease removes mRNA; ribosomes are required for translation; DNA alone added to ribosomes does not give a polypeptide.

Must-know distinctions

• Codon (mRNA) versus anticodon (tRNA) versus triplet (DNA).
• Genome (genes present) versus proteome (proteins made). Cells of one organism share a genome but not a proteome.
• Gene versus locus: the gene is the sequence, the locus is its position.
• Pre-mRNA (before splicing, eukaryotes only) versus mRNA.
• Exon (coding) versus intron (non-coding, within a gene) versus repeat (non-coding, between genes).

Quick self-test

1. State two differences between prokaryotic DNA and eukaryotic nuclear DNA.
2. Describe the DNA found in a chloroplast.
3. Define a locus.
4. A template strand has the triplet TGC. Give the mRNA codon and the tRNA anticodon.
5. A polypeptide has 98 amino acids. What is the minimum number of nucleotides from start codon to stop codon inclusive?
6. Why can a change to one base in a DNA triplet sometimes leave the amino acid unchanged?
7. A pre-mRNA has 2,430 nucleotides and its introns total 1,086. How long is the mRNA?
8. Name the enzyme that joins RNA nucleotides during transcription.
9. A cell-free extract with ribosomes, tRNA, amino acids and ATP makes a polypeptide when mRNA is added but not when the same amount of DNA is added. What does this suggest?
10. Why do a liver cell and a muscle cell of one person have different proteomes?

Answers

1. Prokaryotic DNA is short and circular and not associated with proteins; eukaryotic nuclear DNA is very long and linear and associated with histones.
2. Short and circular, not associated with protein.
3. The fixed position of a gene on a particular DNA molecule.
4. Codon ACG; anticodon UGC.
5. 3 x 98 + 3 = 297.
6. The code is degenerate: several triplets code for the same amino acid.
7. 2,430 - 1,086 = 1,344 nucleotides.
8. RNA polymerase.
9. Ribosomes use the mRNA, not DNA directly, to build the polypeptide, so DNA must first be transcribed.
10. Different genes are expressed in each cell type, though the genome is the same.

Where marks are usually lost

• Writing “it is circular” for prokaryotic DNA without also saying it is not associated with proteins or is short.
• Putting thymine into mRNA, or A to T pairing during transcription.
• Mixing up codon and anticodon, or giving the anticodon when the codon was asked for.
• Counting the stop codon as an amino acid in length calculations.
• Saying introns are cut from DNA, not from pre-mRNA.
• Saying tRNA “carries amino acids” without stating one specific amino acid per type of tRNA.
• Giving only a description in an “explain” question, with no reason linked to the code or the structure.
• Drawing a conclusion from experimental data that goes further than the data support, for example claiming the label shows mRNA specifically.

Official syllabus

Source document: the AQA specification for AS and A-level Biology (codes 7401 and 7402), first teaching 2015, version 1.5, published by AQA. Sections 3.4.1 and 3.4.2 only.