DNA replication, transcription and translation: from the double helix to a protein

DNA is two antiparallel strands of nucleotides, A paired with T and C with G, wound into a double helix. Replication copies it before a cell divides, each new molecule keeping one old strand (semiconservative). Transcription copies one gene's template strand into messenger RNA, with U in place of T. Translation reads that mRNA three bases at a time, each codon calling one amino acid, and builds a protein on a ribosome. Below, one twelve-base gene traced from the double helix to a peptide, the counts at each step, and the strand that students transcribe by mistake.

DNA structure, in the five facts a test asks

1. A nucleotide is three parts: a phosphate, a five-carbon sugar (deoxyribose), and one of four bases, A, T, C or G.
2. The backbone is sugar–phosphate, and it has a direction. The sugar's carbons are numbered, and a strand runs from a 5′ end (a free phosphate on carbon 5) to a 3′ end (a free hydroxyl on carbon 3). Write 5′ and 3′ on every strand you draw.
3. The two strands are antiparallel: one runs 5′ → 3′ left to right, the other 3′ → 5′.
4. Base pairing is complementary: A with T, held by two hydrogen bonds; C with G, held by three. The bases are inside, the backbones outside.
5. The shape is a double helix, worked out by Watson and Crick in 1953 from Franklin's and Wilkins's X-ray data.

Chargaff's rule follows from fact 4 and is the first computed line of the page. In any double-stranded DNA, the amount of A equals the amount of T and the amount of C equals the amount of G. So a sample that is 30 % A is 30 % T; that leaves 100 − 60 = 40 % for G and C together, split equally, 20 % G and 20 % C. The human haploid genome, one set of chromosomes, is about three billion base pairs long.

DNA replication: semiconservative replication, 5′ to 3′, two different strands

Before a cell divides it copies every chromosome, and the copying is semiconservative: the two strands separate and each is used as a template for a new partner, so each daughter molecule is one old strand and one new strand. Meselson and Stahl (1958) showed this by following heavy and light DNA through rounds of replication, ruling out the alternative where the original stayed whole. The copying is this page's work; the counting of what it does to a chromosome, two sister chromatids where there was one, is on the mitosis and meiosis page.

The six words the course asks for, in order:

• Helicase unwinds the double helix at an origin of replication, separating the strands into a fork. Single-strand binding proteins hold the strands apart.
• Primase lays down a short RNA primer, because the next enzyme cannot start a strand from nothing.
• DNA polymerase adds nucleotides to the 3′ end of the primer, matching each to the template. It can only add to a 3′ end, so every new strand is built 5′ → 3′, and the template is read 3′ → 5′.
• Because of that rule the two strands are built differently. The leading strand runs toward the fork and is made continuously. The lagging strand runs away from the fork and is made in short pieces, Okazaki fragments, each with its own primer.
• The primers are replaced with DNA, and ligase seals the fragments into one strand.
• DNA polymerase proofreads as it goes, removing a wrong base and replacing it.

The one count: after replication there are two molecules where there was one, and each is half old and half new.

Transcription: the template strand, and only the template strand

Transcription makes an RNA copy of one gene. RNA polymerase binds a promoter sequence before the gene, unwinds the DNA there, and builds mRNA 5′ → 3′ by pairing RNA nucleotides with the template strand, which it reads 3′ → 5′. RNA has U where DNA has T, so U pairs with the template's A. No primer is needed.

The two strands have names. The template strand is the one the polymerase reads. The coding strand is the other one, and its sequence matches the mRNA letter for letter, with T where the mRNA has U. That match is your check.

In eukaryotes the transcript is processed before it leaves the nucleus: a 5′ cap is added at the front, a poly-A tail at the back, and the introns are cut out and the exons spliced together.

The error to name before the example: transcribing the coding strand. If you pair against the wrong strand you get the complement of the right mRNA, and every codon after that is wrong.

Worked trace: a twelve-base gene from template to peptide (translation biology, the genetic code)

This gene is ours, written for the page; it is not a real gene. Write 5′ and 3′ on every line.

Line 1 template DNA 3′- TAC AAA CGT ATT -5′ Line 2 mRNA 5′- AUG UUU GCA UAA -3′ Line 3 coding DNA 5′- ATG TTT GCA TAA -3′

Line by line:

• Line 1, the template, written as the polymerase reads it, 3′ → 5′.
• Line 2, the mRNA, antiparallel and complementary, with U for T: T → A, A → U, C → G, G → C. TAC → AUG; AAA → UUU; CGT → GCA; ATT → UAA.
• Line 3, the coding strand, for the check: it is the mRNA with T for U, 5′-ATG TTT GCA TAA-3′, and it is the complement of line 1, as it must be.
• Line 4, the codons, read 5′ → 3′ from the start codon: AUG = Met (start), UUU = Phe, GCA = Ala, UAA = stop.
• Line 5, the peptide: Met–Phe–Ala. Three amino acids from twelve bases, because the fourth codon is a stop and codes for nothing.
• Line 6, the counts: 12 base pairs, 4 codons, 3 amino acids, 3 − 1 = 2 peptide bonds.

The common wrong line: a student copies the template with U for T instead of complementing it, writes 3′-UAC AAA CGU AUU-5′, and reads it as codons: UAC Tyr, AAA Lys, CGU Arg, AUU Ile. Four amino acids, no start codon and no stop. That last part is the tell. A trace that begins without AUG or never reaches a stop almost always has the strands swapped or the complementing skipped.

The three amino acids are the start of a protein's primary structure; every level of folding above it begins with this sequence.

Translation: the ribosome's three steps

Translation happens on a ribosome, in the cytoplasm (or on the rough endoplasmic reticulum), and uses transfer RNA, tRNA, as the adaptor: each tRNA carries one amino acid and has a three-base anticodon that pairs with a codon.

• Initiation. The small ribosomal subunit binds the mRNA and finds the start codon, AUG. The first tRNA, carrying methionine, pairs with it, and the large subunit joins.
• Elongation. A tRNA whose anticodon matches the next codon enters the ribosome's A site. A peptide bond forms between its amino acid and the growing chain, the ribosome shifts along one codon, and the emptied tRNA leaves. Repeat.
• Termination. A stop codon enters the A site. No tRNA matches it; a release factor binds instead, and the finished chain is released.

The anticodon for AUG is 3′-UAC-5′. Say it once and you will see why an anticodon looks like the template DNA with U: both are complementary to the mRNA.

The genetic code's four properties, one line each: it is a triplet code, three bases per codon; it is redundant, 64 codons for 20 amino acids and 3 stops, 61 of them coding, so most amino acids have more than one codon; it is unambiguous, each codon meaning one thing; and it is nearly universal across living things.

The two things students swap

The template strand and the coding strand. The mRNA is built on the template and matches the coding strand. A question that hands you "the strand" without saying which needs reading twice: try it as the template and see whether the mRNA starts with AUG. If it says "the coding strand" or "the sense strand", convert it to mRNA by swapping T for U and nothing else.

Transcription and translation themselves. Transcription makes RNA from DNA, in the nucleus. Translation makes protein from RNA, on a ribosome. The words are close; the products are not.

A third, on direction: 5′ → 3′ is the direction of synthesis for both DNA and RNA, which is why the template is always read 3′ → 5′. Writing the primes on every strand is what keeps this straight under pressure.

Check: a sibling to trace cold

Template strand: 3′-TAC GGA CTT ACT-5′. Write 5′ and 3′ on every line before any base. Give the mRNA, the coding strand, the codons and the peptide, then open the page and mark it; Karpicke and Blunt (2011) found that this kind of recall and check beats re-studying the worked one, and the honest-marking rules apply line by line.

Answer. mRNA, complementary and antiparallel with U for T: TAC → AUG; GGA → CCU; CTT → GAA; ACT → UGA, so 5′-AUG CCU GAA UGA-3′. Coding strand: 5′-ATG CCT GAA TGA-3′. Codons: AUG Met (start), CCU Pro, GAA Glu, UGA stop. Peptide: Met–Pro–Glu, three amino acids, two peptide bonds, from twelve bases and four codons.

If your mRNA began with UAC, you copied instead of complementing. If it read Met–Pro–Glu–stop as four amino acids, the stop codes for nothing.

Where this sits in the course

These are ideas 14 to 17 of the 27 in the SBI4U course map: The structure of DNA, DNA replication, Transcription, and Translation and the genetic code, the opening of strand D. Before them idea 13, The Calvin cycle, closes metabolism, which the photosynthesis page traces. After them come idea 18, Mutations and their consequences, which is this page's trace with one base changed, and idea 19, The control of gene expression; later in the strand, Punnett squares follow the gene once it has become an allele. How to study biology has the order for the unit: the vocabulary (the six enzyme names, the two strand names) first, then the trace.

DNA to protein on one screen

• Five facts: nucleotide = phosphate + deoxyribose + base; 5′ and 3′ ends; antiparallel; A–T (two hydrogen bonds) and C–G (three); a double helix.
• Replication's six words: helicase, primase, DNA polymerase, leading and lagging strands, Okazaki fragments, ligase; every new strand built 5′ → 3′. Semiconservative: one old strand, one new.
• The template rule: mRNA is built on the template (read 3′ → 5′) and matches the coding strand with U for T.
• The trace in three lines: 3′-TAC AAA CGT ATT-5′ → 5′-AUG UUU GCA UAA-3′ → Met–Phe–Ala; 12 bases, 4 codons, 3 amino acids, 2 peptide bonds.
• The code: 64 codons, 61 coding, 3 stops; AUG starts; triplet, redundant, unambiguous, nearly universal.
• The two swaps: template against coding strand; transcription (DNA to RNA) against translation (RNA to protein).