Genetics & Evolution

Protein Synthesis

5th Year · 6th Year (Leaving Cert)

  • By the end of this lesson students will be able to describe the process of transcription, including its location and key components.
  • By the end of this lesson students will be able to explain the process of translation, identifying the roles of mRNA, tRNA, and ribosomes.
  • By the end of this lesson students will be able to outline the roles of DNA, mRNA, tRNA, and ribosomes in the overall process of protein synthesis.
  • By the end of this lesson students will be able to define and apply the concepts of codons, anticodons, and the genetic code (HL).
  • By the end of this lesson students will be able to relate the sequence of bases in DNA to the sequence of amino acids in a protein.

Key concepts

Transcription: DNA → mRNA

Transcription is the process where the genetic information from a DNA template strand is copied into a complementary messenger RNA (mRNA) molecule. This process occurs in the nucleus of eukaryotic cells. An enzyme called RNA polymerase binds to a specific region on the DNA (promoter) and unwinds a section of the double helix. RNA polymerase then synthesises an mRNA molecule by adding complementary RNA nucleotides to the DNA template strand. The base pairing rules are: DNA Adenine (A) pairs with mRNA Uracil (U); DNA Thymine (T) pairs with mRNA Adenine (A); DNA Guanine (G) pairs with mRNA Cytosine (C); and DNA Cytosine (C) pairs with mRNA Guanine (G). The mRNA molecule elongates until RNA polymerase reaches a terminator sequence. The newly formed mRNA molecule detaches from the DNA, and the DNA double helix reforms. In eukaryotes, the initial mRNA transcript (pre-mRNA) undergoes processing (splicing, capping, polyadenylation) before it leaves the nucleus and moves to the cytoplasm.

mRNA (messenger RNA)

Messenger RNA (mRNA) is a single-stranded RNA molecule that carries the genetic information from DNA in the nucleus to the ribosomes in the cytoplasm. It acts as a template for protein synthesis. mRNA contains uracil (U) instead of thymine (T).

Translation: mRNA → protein

Translation is the process where the genetic information encoded in an mRNA molecule is used to synthesise a specific protein (polypeptide chain). This process occurs in the cytoplasm, on ribosomes. The mRNA molecule attaches to a ribosome, which then moves along the mRNA, reading the codons (sequences of three bases). Transfer RNA (tRNA) molecules, each carrying a specific amino acid, have an anticodon that is complementary to an mRNA codon. When a tRNA's anticodon pairs with an mRNA codon, the amino acid it carries is added to the growing polypeptide chain. The first codon is typically AUG (start codon), which codes for methionine. Peptide bonds form between adjacent amino acids, linking them together. The ribosome continues to move along the mRNA, adding amino acids until it reaches a stop codon (UAA, UAG, UGA). Stop codons do not code for an amino acid; instead, they signal the termination of translation. The completed polypeptide chain is then released from the ribosome.

tRNA (transfer RNA)

Transfer RNA (tRNA) is a small RNA molecule that plays a crucial role in translation. Each tRNA molecule carries a specific amino acid to the ribosome. It has a unique three-base sequence called an anticodon, which is complementary to a specific mRNA codon, ensuring the correct amino acid is delivered.

Ribosomes

Ribosomes are complex cellular structures, composed of ribosomal RNA (rRNA) and protein, that serve as the sites of protein synthesis (translation). They consist of two subunits (large and small) and facilitate the binding of mRNA and tRNA molecules, allowing for the formation of peptide bonds between amino acids.

Codons & the Genetic Code (HL)

A codon is a sequence of three consecutive bases on an mRNA molecule that specifies a particular amino acid or a stop signal during protein synthesis. There are 64 possible codons (4^3). An anticodon is a sequence of three bases on a tRNA molecule that is complementary to a specific mRNA codon. This pairing ensures the correct amino acid is delivered to the ribosome. The genetic code is the set of rules by which information encoded in genetic material (DNA or RNA sequences) is translated into proteins (amino acid sequences) by living cells. Key characteristics of the genetic code include: - Degenerate: Most amino acids are specified by more than one codon (e.g., both UCU and UCC code for Serine). This redundancy offers some protection against point mutations. - Universal: The genetic code is largely the same across almost all organisms, from bacteria to humans, suggesting a common evolutionary origin. - Non-overlapping: Each base in the mRNA sequence is part of only one codon. Codons are read sequentially, one after another, without skipping bases. - Triplet: Each codon consists of three bases. - Start Codon: AUG (codes for Methionine) initiates translation. - Stop Codons: UAA, UAG, UGA signal the termination of translation and do not code for an amino acid.

Key facts to remember

  • 1Protein synthesis is the process by which cells make proteins and involves two main stages: transcription and translation.
  • 2Transcription occurs in the nucleus (in eukaryotes) and involves copying genetic information from DNA into mRNA.
  • 3Translation occurs in the cytoplasm, on ribosomes, and involves converting the mRNA sequence into an amino acid sequence (protein).
  • 4mRNA carries the genetic message from the DNA to the ribosomes.
  • 5tRNA molecules transport specific amino acids to the ribosome, matching them to mRNA codons.
  • 6Ribosomes are the cellular machinery responsible for reading mRNA and synthesising polypeptide chains.
  • 7Codons are triplets of bases on mRNA that specify particular amino acids or stop signals (HL).
  • 8The genetic code is degenerate, universal, and non-overlapping (HL).

Worked examples

Example 1

A segment of a DNA molecule has the following template strand sequence: 3'-TAC GGC CTA GAT-5'. Determine the sequence of the mRNA molecule transcribed from this DNA segment.

IIdentify the DNA template strand: 3'-TAC GGC CTA GAT-5'.
IIRecall the base pairing rules for transcription: DNA A pairs with mRNA U; DNA T pairs with mRNA A; DNA G pairs with mRNA C; DNA C pairs with mRNA G.
IIISynthesise the complementary mRNA strand, remembering that mRNA is synthesised in the 5' to 3' direction:
IVDNA template: 3'-T A C G G C C T A G A T-5'
VmRNA: 5'-A U G C C G G A U C U A-3'

Answer

5'-AUG CCG GAU CUA-3'

Always ensure the mRNA sequence is written in the 5' to 3' direction, as this is how it is read during translation.

Example 2

An mRNA molecule has the following sequence: 5'-AUG CCG GAU CUA UAA-3'. Using the genetic code table (as provided in exams), determine the amino acid sequence of the polypeptide produced. (HL)

IIdentify the mRNA sequence: 5'-AUG CCG GAU CUA UAA-3'.
IIBreak the mRNA sequence into codons (triplets of bases): AUG, CCG, GAU, CUA, UAA.
IIIUse the genetic code table to find the amino acid for each codon:
IVAUG = Methionine (Met) - Start codon
VCCG = Proline (Pro)
VIGAU = Aspartate (Asp)
VIICUA = Leucine (Leu)
VIIIUAA = Stop codon
9Assemble the amino acid sequence.

Answer

Met-Pro-Asp-Leu (or Methionine-Proline-Aspartate-Leucine)

The stop codon does not code for an amino acid; it signals the end of the polypeptide chain. Do not include it in the amino acid sequence.

Example 3

A segment of a DNA molecule has the following coding strand sequence: 5'-ATG CCG GAT CTA TAA-3'. Determine the mRNA sequence transcribed from the template strand and then the amino acid sequence of the polypeptide produced. (HL)

IIdentify the DNA coding strand: 5'-ATG CCG GAT CTA TAA-3'.
IIDetermine the mRNA sequence. The mRNA sequence is complementary to the DNA template strand. Since the coding strand has the same sequence as the mRNA (with T replaced by U), we can directly convert:
IIIDNA coding strand: 5'-A T G C C G G A T C T A T A A-3'
IVmRNA: 5'-A U G C C G G A U C U A U A A-3'
VBreak the mRNA sequence into codons: AUG, CCG, GAU, CUA, UAA.
VIUse the genetic code table to find the amino acid for each codon:
VIIAUG = Methionine (Met)
VIIICCG = Proline (Pro)
9GAU = Aspartate (Asp)
10CUA = Leucine (Leu)
11UAA = Stop codon
12Assemble the amino acid sequence.

Answer

mRNA: 5'-AUG CCG GAU CUA UAA-3'. Polypeptide: Met-Pro-Asp-Leu.

When given the DNA coding strand, the mRNA sequence will be identical to it, except all Thymine (T) bases are replaced by Uracil (U).

Common mistakes

  • Confusing the locations of transcription (nucleus) and translation (cytoplasm/ribosomes).
  • Incorrect base pairing during transcription (e.g., pairing T with A in RNA instead of U with A, or pairing U with T).
  • Not understanding the distinct roles of mRNA, tRNA, and rRNA in protein synthesis.
  • Misreading the genetic code chart, especially for the first, second, and third bases of a codon (HL).
  • Forgetting that translation starts at the AUG codon and ends at a stop codon, not necessarily at the very beginning or end of the mRNA molecule.

Exam tips

  • Practise drawing and labelling diagrams of both transcription and translation to solidify your understanding of the processes and components.
  • Learn the key terms and their precise definitions, such as 'codon', 'anticodon', 'transcription', 'translation', 'RNA polymerase', and 'ribosome'.
  • Understand the flow of genetic information: DNA → mRNA → Protein. This central dogma is fundamental.
  • For Higher Level students, become proficient in using the genetic code table to translate mRNA sequences into amino acid sequences quickly and accurately.

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