AP® Biology review sheet from Aim for Five (aimforfive.com/bio/units/6/6-4)
Unit 6 · Topic 6.4
6.4 Translation
Translation turns the base sequence of mRNA into the amino acid sequence of a protein. Ribosomes read mRNA three bases at a time, tRNAs bring matching amino acids, and the chain grows from the start codon AUG until a stop codon. The near-universal genetic code is evidence that all life shares common ancestry.
Key terms
- codon
- anticodon
- tRNA
- ribosome
- start codon
- reverse transcriptase
Where translation happens
Translation takes place on ribosomes. In both prokaryotes and eukaryotes, free ribosomes sit in the cytoplasm. Eukaryotes also have ribosomes attached to the cytoplasmic side of the rough endoplasmic reticulum (ER); those make proteins that will be sent to the membrane, to other organelles or out of the cell. In prokaryotes, which have no nucleus, ribosomes can start translating an mRNA while it's still being transcribed.
Codons and the genetic code
The mRNA is read in triplets called codons. With four bases there are 4³ = 64 possible codons. Sixty-one of them code for amino acids, and three (UAA, UAG and UGA) are stop codons. AUG is the start codon and also codes for methionine, so most new proteins begin with methionine.
There are only 20 amino acids, so many amino acids are coded by more than one codon. For example, both GAA and GAG code for glutamic acid. This redundancy is why some mutations don't change the protein (6.7).
You don't need to memorize the codon table. On the exam you'll be given one; you only need to know that AUG is the start codon.
The steps of translation
The names of the proteins and factors that run each step are beyond the scope of the exam. Focus on the order of events and what each RNA does.
- Initiation: the small ribosomal subunit binds the mRNA, and the ribosome's rRNA helps line the mRNA up so that reading begins at the start codon, AUG. A tRNA carrying methionine pairs with AUG, and the large subunit joins.
- Elongation: the ribosome reads the next codon. A tRNA whose anticodon is complementary to that codon brings its amino acid. For example, the codon 5′-GGA-3′ pairs with the anticodon 3′-CCU-5′. The ribosome links the new amino acid to the growing chain with a peptide bond, then shifts one codon along the mRNA.
- Termination: when a stop codon enters the ribosome, no tRNA matches it. The finished polypeptide is released and the ribosome comes apart.
One code for almost all life
Bacteria, plants, fungi and animals nearly all use the same codon assignments. That's why a human insulin gene placed into bacteria can be translated into working human insulin. Such a shared, arbitrary code is hard to explain unless all organisms inherited it from a common ancestor, so it's strong evidence for common ancestry.
Retroviruses: information flowing backward
Usually information flows DNA → RNA → protein. Retroviruses, such as HIV, are the special case. Their genome is RNA, and they carry an enzyme called reverse transcriptase that copies the RNA genome into DNA. That viral DNA is inserted into the host cell's genome. The host then transcribes and translates it like its own genes, producing viral RNA genomes and proteins that assemble into new viruses. Because the viral DNA becomes part of the host genome, the infection can't simply be cleared, and drugs that block reverse transcriptase are one way to treat HIV.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Translating an mRNA (find the start first)
Translate this mRNA: 5′-GCAUGUUCGGAUACUAAGC-3′. Codons: AUG = Met (start), UUC = Phe, GGA = Gly, UAC = Tyr, UAA = stop.
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- Step 1: Translation doesn't start at the first base. Scan from the 5′ end for AUG: G-C-AUG… The start codon begins at the third base.
- Step 2: Read in triplets from there: AUG | UUC | GGA | UAC | UAA.
- Step 3: Translate: AUG = Met, UUC = Phe, GGA = Gly, UAC = Tyr, UAA = stop.
- Step 4: The stop codon ends translation and adds no amino acid. The bases before AUG and after the stop codon aren't translated.
Answer: Met–Phe–Gly–Tyr (4 amino acids).
- Example 2
How many nucleotides?
A polypeptide is 150 amino acids long. What is the minimum number of mRNA nucleotides in its coding sequence, including the stop codon?
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- Step 1: Each amino acid needs one codon of 3 nucleotides: 150 × 3 = 450.
- Step 2: The stop codon adds 3 more nucleotides but no amino acid: 450 + 3 = 453.
- Step 3: The real mRNA is longer, because it also has untranslated regions, a cap and a poly-A tail, and the gene in a eukaryote is longer still because of introns.
Answer: 453 nucleotides (450 coding plus a 3-base stop codon).
Common mistakes
- Matching tRNA anticodons directly to DNA. Anticodons pair with mRNA codons.
- Starting translation at the first base of the mRNA. Find the first AUG, then read in triplets.
- Saying the stop codon codes for an amino acid. It signals the end and no amino acid is added.
- Saying reverse transcriptase makes RNA from DNA. It makes DNA from an RNA template.
On the exam
- Expect to use a given codon chart to translate a sequence or to predict how a sequence change alters the protein. Show the reading frame clearly.
- If asked for evidence of common ancestry at the molecular level, the shared genetic code is a strong, specific answer.
Connected topics
Videos
Check yourself
4 questions on 6.4 Translation. Pick an answer to see if you got it, and why.
A tRNA delivers lysine to a ribosome reading the mRNA codon 5′-AAG-3′. What is the anticodon of this tRNA?
When the human insulin gene is inserted into bacteria, the bacteria produce human insulin. Which of the following best explains why this is possible, and what it suggests?
HIV is a retrovirus. A drug that blocks reverse transcriptase would most directly prevent which step of the virus's life cycle?
A polypeptide is 150 amino acids long, counting the methionine at the start. What is the minimum number of nucleotides in its mRNA coding sequence, from the start codon through the stop codon?
0 of 4 answered