In simple terms
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Protein synthesis
Cambridge 9700 Paper 2 — Protein synthesis (6.2). A-Level Notes diagram-backed lesson with premium structure and live visuals.
- 1
Describe the sequential processes of transcription and translation in protein synthesis.
- 2
Explain the specific roles of mRNA, tRNA, and ribosomes (containing rRNA) in accurate protein production.
- 3
Discuss the characteristics of the genetic code and the differences in gene expression between prokaryotes and eukaryotes.
- 4
Outline the process of post-transcriptional modification in eukaryotes, including the removal of introns and addition of a 5' cap and poly-A tail.
What this topic covers
The official Cambridge syllabus points this lesson works through.
- 6.2.1
State that a polypeptide is coded for by a gene and that a gene is a sequence of nucleotides that forms part of a DNA molecule
- 6.2.2
Describe the principle of the universal genetic code in which different triplets of DNA bases either code for specific amino acids or correspond to start and stop codons
- 6.2.3
Describe how the information in DNA is used during transcription and translation to construct polypeptides, including the roles of: • RNA polymerase • messenger RNA (mRNA) • codons • transfer RNA (tRNA) • anticodons • ribosomes
- 6.2.4
State that the strand of a DNA molecule that is used in transcription is called the transcribed or template strand and that the other strand is called the non-transcribed strand
- 6.2.5
Explain that, in eukaryotes, the RNA molecule formed following transcription (primary transcript) is modified by the removal of non-coding sequences (introns) and the joining together of coding sequences (exons) to form mRNA
- 6.2.6
State that a gene mutation is a change in the sequence of base pairs in a DNA molecule that may result in an altered polypeptide
- 6.2.7
Explain that a gene mutation is a result of substitution or deletion or insertion of nucleotides in DNA and outline how each of these types of mutation may affect the polypeptide produced
Explore the concept
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Full topic notes
Formal explanation with the rigour you need for the exam.
The Central Dogma of Molecular Biology
The flow of genetic information within a biological system is described by the central dogma: DNA → RNA → Protein. This sequence ensures that the genetic information encoded in DNA is accurately expressed as proteins. There are two primary stages in this process: transcription and translation.
The Genetic Code
The genetic code is the set of rules by which information encoded in genetic material (DNA or mRNA sequences) is translated into proteins. The code is read in triplets of nucleotide bases called codons.
Key characteristics of the genetic code:
- Triplet Code: Three consecutive bases on mRNA form a codon, which specifies one amino acid.
- Degenerate (or Redundant): Most amino acids are coded for by more than one codon. For example, Leucine (Leu) is coded by six different codons. This provides a buffer against some point mutations, as a change in the third base of a codon may not change the amino acid being coded for.
- Universal: The same codons specify the same amino acids in almost all organisms, from bacteria to humans. This universality is strong evidence for a common evolutionary ancestor and is crucial for genetic engineering.
- Non-overlapping: The code is read sequentially, one codon at a time, without any bases being shared between consecutive codons.
- Start and Stop Signals: There are specific codons that signal the start and end of translation. AUG is the most common start codon (and also codes for the amino acid Methionine). UAA, UAG, and UGA are stop codons; they do not code for an amino acid but signal for the termination of the polypeptide chain.
Stage 1: Transcription
Transcription is the process where a specific gene's DNA sequence is copied into a complementary messenger RNA (mRNA) molecule. This occurs in the nucleus of eukaryotic cells and the cytoplasm of prokaryotic cells.
- Initiation: The enzyme RNA polymerase binds to a specific region on the DNA known as the promoter. This signals the DNA double helix to unwind and separate, exposing the nucleotide bases of the gene.
- Elongation: RNA polymerase moves along the template strand (also known as the antisense strand) of the DNA in a 3' to 5' direction. It reads the DNA bases and synthesises a complementary mRNA strand in the 5' to 3' direction. RNA nucleotides (containing ribose sugar, phosphate, and bases A, U, C, G) pair with the exposed DNA bases: A with U, T with A, C with G, G with C.
- Termination: When RNA polymerase reaches a terminator sequence on the DNA, it detaches. The newly formed RNA molecule is released, and the DNA double helix reforms.
Post-Transcriptional Modification in Eukaryotes
In eukaryotic cells, the initial RNA transcript, called pre-mRNA, must be processed before it can be translated. This processing occurs in the nucleus.
- Splicing: Eukaryotic genes contain non-coding regions called introns and coding regions called exons. A complex of proteins and RNA called a spliceosome removes the introns and joins the exons together to form a continuous coding sequence.
- 5' Capping: A modified guanine nucleotide is added to the 5' end of the pre-mRNA. This 5' cap protects the mRNA from degradation and helps it bind to the ribosome for translation.
- Polyadenylation (Poly-A Tail): A long chain of adenine nucleotides (the poly-A tail) is added to the 3' end. This tail increases the stability of the mRNA and facilitates its export from the nucleus to the cytoplasm.
The resulting mature mRNA is then ready for translation. Prokaryotic mRNA does not contain introns and does not undergo these modifications; transcription and translation can occur simultaneously.
Stage 2: Translation
Translation is the process where the genetic information in mature mRNA is decoded to synthesise a specific polypeptide chain. This occurs on ribosomes in the cytoplasm.
Key Components:
- Ribosomes: Composed of ribosomal RNA (rRNA) and proteins, they consist of a large and a small subunit. They provide a scaffold for mRNA and tRNA and catalyse peptide bond formation. They have three binding sites: the A (aminoacyl) site for incoming tRNA, the P (peptidyl) site holding the growing polypeptide chain, and the E (exit) site where uncharged tRNA leaves.
- Transfer RNA (tRNA): Small RNA molecules that act as adaptors. Each tRNA has a specific anticodon (a three-nucleotide sequence) that is complementary to an mRNA codon. At its 3' end, it carries the specific amino acid corresponding to that codon. The attachment of the correct amino acid to its tRNA is catalysed by an enzyme called aminoacyl-tRNA synthetase.
The Process:
- Initiation: The small ribosomal subunit binds to the mRNA at the 5' cap and scans for the start codon (AUG). The initiator tRNA (carrying methionine) with the anticodon UAC binds to the start codon in the P-site. The large ribosomal subunit then joins to form the complete translation initiation complex.
- Elongation: This is a cyclical process:
- Codon Recognition: A new tRNA molecule, carrying its specific amino acid, enters the A-site, its anticodon base-pairing with the mRNA codon.
- Peptide Bond Formation: The ribosome catalyses the formation of a peptide bond between the amino acid in the P-site and the new amino acid in the A-site. The growing polypeptide is transferred to the tRNA in the A-site.
- Translocation: The ribosome moves one codon along the mRNA in the 5' to 3' direction. The tRNA from the P-site moves to the E-site to be released, and the tRNA from the A-site (now carrying the polypeptide) moves to the P-site. The A-site is now vacant for the next tRNA.
- Termination: Elongation continues until a stop codon (UAA, UAG, or UGA) enters the A-site. A protein called a release factor binds to the stop codon. This causes the polypeptide chain to be released from the tRNA in the P-site, and the entire ribosomal complex disassembles. The newly synthesised polypeptide then folds into its specific three-dimensional structure to become a functional protein.
Worked examples
See the formulas applied — reveal one step at a time, like the exam.
A segment of a non-template strand of DNA has the sequence 5'-ATG CTC GAT TTA-3'.
- Deduce the sequence of the mRNA molecule synthesised from the template strand of this DNA segment.
- Using the genetic code table (assume standard table where AUG is Methionine), identify the amino acid sequence of the polypeptide produced from this mRNA.
Standard Genetic Code Table Excerpt:
| Codon | Amino Acid | Codon | Amino Acid |
|---|---|---|---|
| AUG | Met | UUA | Leu |
| --- | --- | --- | --- |
| CUC | Leu | GAU | Asp |
| CGA | Arg | UAA | Stop |
| UAG | Stop | UGA | Stop |
- 1
Identify the template strand: The given non-template (sense) strand is 5'-ATG CTC GAT TTA-3'. The template (antisense) strand runs in the opposite direction (3' to 5') and is complementary.
A specific enzyme in humans has a single polypeptide chain consisting of 250 amino acids. The average relative molecular mass of an amino acid is 110.
- Calculate the minimum length (in number of bases) of the coding region of the mature mRNA molecule required to synthesise this enzyme.
- Calculate the relative molecular mass of the final enzyme polypeptide. (The relative molecular mass of water is 18).
- 1
Bases for amino acids: Each amino acid is coded by a 3-base codon.
How it all connects
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Glossary
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Quick check
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Revision flashcards
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What is transcription?
The process in the nucleus (eukaryotes) where a gene's DNA sequence is copied by RNA polymerase to create a complementary messenger RNA (mRNA) molecule.
Key takeaways
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- ✓
Describe the sequential processes of transcription and translation in protein synthesis.
- ✓
Explain the specific roles of mRNA, tRNA, and ribosomes (containing rRNA) in accurate protein production.
- ✓
Discuss the characteristics of the genetic code and the differences in gene expression between prokaryotes and eukaryotes.
- ✓
Outline the process of post-transcriptional modification in eukaryotes, including the removal of introns and addition of a 5' cap and poly-A tail.
Practice — then mark it
The whole point: a real Cambridge question, marked mark-by-mark.
9700/22 · Q3(e)
Carrot virus Y is a pathogen of carrot plants. The virus, which belongs to a group known as Potyvirus, replicates its viral nucleic acid and proteins within host carrot cells. The general structure of potyviruses is shown in Fig. 3.2. The synthesis of viral proteins in host carrot cells only involves the process of translation. The process of transcription does not occur. Suggest why translation occurs in host carrot cells during the synthesis of viral proteins, but transcription does not occur.
9700/42 · Q3(c)(ii)
Another method being investigated to treat LCA10 is to use a gene editing tool known as the CRISPR/Cas9 system. The CRISPR/Cas9 system uses a short length of RNA called guide RNA. Guide RNA is complementary to the target DNA and is linked to a nuclease enzyme called Cas9. Cas9 breaks phosphodiester bonds in DNA. The cell repair mechanisms repair the cut in DNA after the modification has taken place. • A vector delivers Cas9 and two specific guide RNAs to the photoreceptor cells. • They act on the section of DNA which contains the mutation. • Exon X is no longer added to the mRNA. Explain how this method used to treat LCA10 is an example of gene editing.
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