How does the structure of DNA store information and direct the synthesis of proteins?
The structure of DNA and RNA, semi-conservative DNA replication, the genetic code, and the processes of transcription and translation that make proteins.
An Edexcel A-Level Biology B answer on DNA and RNA structure, semi-conservative replication, the genetic code, and transcription and translation.
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What this dot point is asking
Edexcel wants you to describe the structure of DNA and RNA, explain semi-conservative replication, describe the features of the genetic code, and explain how transcription and translation build a polypeptide from a gene. You should be able to apply base-pairing rules and read off codon sequences.
The structure of DNA and RNA
A nucleotide has three parts: a pentose sugar, a phosphate group and a nitrogenous base. In DNA the sugar is deoxyribose; in RNA it is ribose. Nucleotides join by phosphodiester bonds (condensation reactions) between the phosphate of one and the sugar of the next, forming a sugar-phosphate backbone.
DNA is a double helix of two antiparallel strands. One runs to and the other to , and they are joined by hydrogen bonds between complementary bases on the inside.
There are three types of RNA: messenger RNA (mRNA) carries the code from nucleus to ribosome; transfer RNA (tRNA) carries amino acids and has an anticodon; ribosomal RNA (rRNA) is part of the ribosome.
Semi-conservative replication
The double helix unwinds and the hydrogen bonds break (catalysed by DNA helicase), so the strands separate. Free activated nucleotides pair with the exposed bases by complementary base pairing. DNA polymerase joins adjacent nucleotides with phosphodiester bonds, working only in the to direction. Each daughter molecule has one old and one new strand. The Meselson and Stahl experiment, using heavy nitrogen and density-gradient centrifugation, gave the classic evidence by showing an intermediate-density band after one generation.
The genetic code
The genetic code is a triplet code: each sequence of three bases (a codon) codes for one amino acid. It is degenerate (most amino acids have more than one codon, which limits the effect of some mutations), non-overlapping (each base is read once, in one reading frame) and almost universal (the same codons mean the same amino acids in nearly all organisms, which is why gene technology can move genes between species).
Transcription and translation
- Transcription. RNA polymerase binds the gene, the DNA unwinds, and a complementary pre-mRNA copy of the template strand is made using free RNA nucleotides (uracil pairs with adenine). In eukaryotes, non-coding introns are spliced out and the exons are joined before the mature mRNA leaves the nucleus through a nuclear pore.
- Translation. The mRNA attaches to a ribosome. tRNA molecules, each with an anticodon and a specific amino acid, pair their anticodon with the mRNA codon. Adjacent amino acids are joined by peptide bonds, the ribosome moves along codon by codon, and a polypeptide is built until a stop codon is reached.
Examples in context
Example 1. Sickle-cell from a single substitution. A single base substitution in the haemoglobin gene changes one codon so that glutamic acid is replaced by valine. This shows how the triplet code links a tiny DNA change to an altered protein, and why a degenerate code sometimes (but not here) buffers mutations. It connects directly to gene mutation in cystic fibrosis.
Example 2. Insulin made by bacteria. Because the genetic code is near-universal, the human insulin gene transcribed and translated in genetically modified bacteria produces functional human insulin. This underpins the gene-technology dot point and explains why moving a gene between species works at all.
Try this
Q1. Explain why DNA replication is described as semi-conservative. [2 marks]
- Cue. Each new molecule has one original template strand and one new strand made from free nucleotides.
Q2. Describe the role of tRNA in translation. [2 marks]
- Cue. tRNA has an anticodon that pairs with the mRNA codon and carries the specific amino acid to the ribosome.
Q3. The template strand reads . Deduce the mRNA sequence. [2 marks]
- Cue. (complementary bases, with U replacing T).
Exam-style practice questions
Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
Edexcel 20186 marksDescribe the process of translation, from the binding of mRNA to a ribosome to the release of a completed polypeptide.Show worked answer →
Markers reward an ordered, mechanistic account.
The mRNA binds to a ribosome so that the first two codons lie in the ribosome. A tRNA with an anticodon complementary to the start codon binds, carrying its specific amino acid. A second tRNA binds the next codon, bringing a second amino acid. A peptide bond forms between the two amino acids, catalysed by the ribosome. The ribosome moves along the mRNA by one codon (translocation); the first tRNA leaves and is reused. The process repeats, codon by codon, extending the polypeptide. When a stop codon is reached, no tRNA binds, and the completed polypeptide is released.
Award marks for: mRNA binds ribosome (1), tRNA anticodon pairs with codon (1), tRNA carries specific amino acid (1), peptide bond forms (1), ribosome moves along mRNA (1), stops at stop codon and polypeptide released (1).
Edexcel 20224 marksA length of DNA template strand has the base sequence . Deduce the mRNA codon sequence transcribed from it and State how many amino acids this codes for.Show worked answer →
Markers reward correct complementary base pairing and the triplet rule.
Transcription copies the template strand using complementary bases, with uracil replacing thymine. Pairing with , with , with and with , the template gives the mRNA .
There are three codons, so this codes for three amino acids (here methionine, the start, then proline and asparagine).
Award: correct mRNA sequence (2, lose one for any error), uses uracil not thymine (1), states three amino acids (1).
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Sources & how we know this
- Pearson Edexcel A-Level Biology B (9BN0) specification — Pearson Edexcel (2015)