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Molecular biology · Lesson 7

Read codons and assemble an amino-acid chain

The ribosome reads an RNA frame. Charged transfer RNAs connect codons to amino acids, and release factors respond to stop codons.

By the end: Predict a short peptide, recognize termination, and distinguish a fragment from a complete coding sequence.

Establish the reading frame

A codon is three consecutive RNA bases read 5′ to 3′ in a selected frame. The standard code has 64 codons. Sixty-one specify one of twenty common amino acids; UAA, UAG, and UGA signal termination. Several codons can specify the same amino acid. This redundancy is called degeneracy, not ambiguity about the amino acid for a fixed ordinary codon.

Initiation establishes the frame in a molecular context. AUG usually supplies methionine at initiation, but finding AUG anywhere is not sufficient to identify a functioning gene. Our examples explicitly start at the first base. The fragment reader also allows offsets of one or two bases to show the effect of grouping; it does not predict initiation.

A frame groups bases before the code assigns residues

The number of possible codons follows from independent choices at three positions. Each position can contain A, C, G, or U. Two positions would give four times four, or sixteen combinations, too few to assign a different combination to each of twenty amino acids. Three positions give sixteen times four, or sixty-four combinations. This arithmetic explains the capacity of a triplet code. It does not derive the actual assignments. Chemistry alone does not tell us that GCU means alanine rather than some other amino acid. Those assignments require evidence about translation.

Count codons separately from the residues they specify. In the standard code, subtracting three stop codons from sixty-four leaves sixty-one residue-specifying codons. Several of those sixty-one have the same output. GCU and GCC both specify alanine, for example, while UUU and UUC both specify phenylalanine. Consequently, a peptide does not uniquely determine the RNA that encoded it. Even a two-residue fragment Ala–Phe has several possible RNA sequences. Reading RNA to infer a peptide is a defined mapping under a stated code. Reconstructing one unique RNA from that peptide is not.

Sixteen two-base prefixes each allow four endings

UU prefixUUUUUCUUAUUGUC prefixUCUUCCUCAUCGUA prefixUAUUACUAAUAGUG prefixUGUUGCUGAUGGCU prefixCUUCUCCUACUGCC prefixCCUCCCCCACCGCA prefixCAUCACCAACAGCG prefixCGUCGCCGACGGAU prefixAUUAUCAUAAUGAC prefixACUACCACAACGAA prefixAAUAACAAAAAGAG prefixAGUAGCAGAAGGGU prefixGUUGUCGUAGUGGC prefixGCUGCCGCAGCGGA prefixGAUGACGAAGAGGG prefixGGUGGCGGAGGG
Every printed triplet is a distinct standard-code input. The sixteen groups contain four triplets each, for sixty-four in total. This counts possible codons without deriving their amino-acid assignments.

Now consider an RNA written as AUGGCUUUUUAA. The letters do not contain spaces. Under the stipulated starting position, bases one through three form AUG, bases four through six form GCU, bases seven through nine form UUU, and bases ten through twelve form UAA. The three-letter amino-acid names are abbreviations, not additional codons. Met means methionine, Ala means alanine, and Phe means phenylalanine. The result contains three residues even though the RNA contains twelve nucleotides. One triplet signals release instead of contributing a fourth residue to the chain.

The same twelve bases can be grouped differently without changing a single nucleotide. Starting at base two gives UGG, CUU, and UUU, followed by AA that cannot complete a triplet. Starting at base three gives GGC, UUU, and UUA, followed by a final A that cannot complete a triplet. These are arithmetic readings of offsets one and two. They are not claims that a ribosome would initiate at those positions. Notice that a stop visible in one frame need not be a stop in another. Its three bases must occupy one complete codon in the selected grouping.

Initiation is therefore a prerequisite for interpreting a natural message. A mature RNA can contain an untranslated region before the coding sequence, so its first nucleotide need not be the first nucleotide translated. Initiation machinery selects a start in a biological context. The demonstration removes that problem by stipulating initiation at base one. This lets us examine elongation and termination without pretending to identify a real gene. Likewise, a fragment may contain an internal AUG that specifies methionine during elongation. That occurrence does not automatically start a second protein within the same translation event.

One RNA has three possible triplet groupings

Offset 0, begin at base 15′AUGGCUUUUUAA3′Offset 1, begin at base 25′AUGGCUUUUUAA3′Offset 2, begin at base 35′AUGGCUUUUUAA3′
Each row repeats the same twelve bases, always 5′ to 3′. Underlines mark complete codons from the stated offset. Neither offset one nor offset two reaches a stop; their final AA or A remains incomplete. These groupings do not predict biological initiation.

A useful contrast is between changing a frame and changing a codon within a fixed frame. Replacing GCU with GCC changes one nucleotide while preserving the alanine assignment. Moving the starting position by one base changes how every subsequent triplet is collected, even though the RNA itself is unchanged. Neither operation alone predicts cellular function. The first establishes a coding equivalence under this code; the second establishes a different reading of the supplied fragment. Keep the chosen frame visible whenever comparing peptide predictions, because the code cannot repair an incorrectly grouped input.

Connect recognition to chemistry

A transfer RNA has an anticodon that pairs with a codon and an attached amino acid. Aminoacyl-tRNA synthetases help load the appropriate amino acid onto each tRNA. Correct codon pairing cannot by itself verify the amino acid attached at the other end. Ribosomal RNA catalyzes peptide-bond formation. Energy-coupled steps support charging, selection, and movement.

The chain grows from its amino end to its carboxyl end while the ribosome advances along the RNA. At a stop codon, a release factor promotes release rather than adding a special stop amino acid. The animation groups these events into one teaching step per codon; it is not a timing model or a detailed depiction of ribosomal sites.

Charging and chain transfer solve different recognition problems

A ribosome brings together molecules with different recognition roles. The messenger RNA supplies a sequence of codons. A transfer RNA, abbreviated tRNA, has a region that pairs with a codon and a separate end that carries an amino acid. The codon and anticodon pair antiparallel, as other complementary nucleic-acid strands do. For a simple aligned example, codon 5′-GCU-3′ pairs with anticodon 3′-CGA-5′. If that anticodon is reported conventionally as a sequence running 5′ to 3′, it is AGC. Changing the reporting direction requires reversing the letters as well as the end labels.

Attaching an amino acid to a tRNA is called charging. An aminoacyl-tRNA synthetase is an enzyme that recognizes an amino acid and suitable tRNAs, then uses an ATP-dependent reaction to attach the amino acid at the tRNA's acceptor end. The name describes its product, an amino-acid-bearing tRNA. This creates a chemical connection between amino-acid identity and RNA recognition before the tRNA enters elongation. Some synthetases also edit unsuitable products. This selection is distinct from pairing an anticodon with a codon, so fidelity requires more than correct complementarity at the ribosome.

The anticodon and attached amino acid occupy different regions

mRNA5′3′GCCGUA3′5′tRNA, schematicAlaAttached cargo
GCU and CGA are aligned antiparallel. The anticodon written 5′ to 3′ would be AGC. The tRNA outline is schematic; it links a recognition region to an amino-acid attachment site rather than showing atomic structure.

An incorrectly charged tRNA exposes the distinction. Suppose an alanine-reading tRNA carries a different amino acid. Its anticodon can still match the RNA positions that normally call for alanine. The codon has not changed, but the attached molecule has. Codon recognition therefore cannot establish the identity of the cargo by itself. A sequence reader reports the expected product of the ordinary translation machinery, not a direct chemical inspection of every newly formed chain. This is a limit on what the code table models, rather than evidence that ordinary codon assignments are ambiguous.

During elongation, the ribosome holds a tRNA carrying the existing chain and accepts a charged tRNA for the next codon. A peptide bond joins the incoming amino acid to the chain. The chain transfers from the old tRNA onto the incoming amino acid's tRNA. The first tRNA is now empty. Ribosomal movement then places the longer chain in position for another round. The conventional names P site and A site refer to the chain-bearing and incoming-tRNA positions in this simplified account. Ribosomal RNA supplies the catalytic center for peptide-bond formation.

Follow a two-residue chain through one addition. Before the reaction, Met–Ala is attached through its carboxyl end to one tRNA, while another tRNA carries Phe. After peptide-bond formation, Met–Ala–Phe is attached to the second tRNA. The Met amino end remains the N end throughout. The newest residue contributes the growing carboxyl end, called the C end. This establishes N-to-C chain growth. At the same time, codon progression runs toward the RNA's 3′ end. These are directions on different polymers and should not be described as though amino acids had nucleotide 5′ and 3′ ends.

Peptide formation transfers the chain onto the incoming tRNA

Before peptide-bond formationMetAlaPheN endC endtRNA 1tRNA 2After transferMetAlaPheN endC endtRNA 1 emptytRNA 2 carries chain
The upper state has a two-residue chain and a separate Phe-bearing tRNA. The lower state has a three-residue chain attached to the second tRNA. Connecting lines denote covalent attachments schematically, not molecular lengths or reaction trajectories.

Energy use also spans distinct steps. ATP-dependent charging prepares an amino acid attached through an energy-rich bond. Other energy-coupled steps help select tRNAs and move the ribosome. It would be misleading to draw peptide formation as simply joining free amino acids because their codons touched. The RNA sequence orders recognition events, while activated substrates and molecular machinery make synthesis possible. The teaching animation compresses these operations into an addition event. Its discrete steps preserve residue order, but they cannot reveal which chemical operation consumed energy or how long any operation took.

Work through the mechanism

Translation in discrete codon events

1AUGNot read yet
2GCUNot read yet
3UUUNot read yet
4UAANot read yet
N end
No chain yet

Event 0 of 4. No codon processed.

Step 0 / 4
Complete selected codons and result

RNA read 5′ to 3′. Offset 0.

Base 1AUGMet
Base 4GCUAla
Base 7UUUPhe
Base 10UAAStop

Peptide: Met · Ala · Phe

An in-frame stop was reached.

Standard RNA code and amino-acid names

Read each RNA codon 5′ to 3′ in the established frame. This table excludes alternative codes and context-dependent recoding.

AbbreviationNameRNA codons
AlaAlanineGCU, GCC, GCA, GCG
ArgArginineCGU, CGC, CGA, CGG, AGA, AGG
AsnAsparagineAAU, AAC
AspAspartateGAU, GAC
CysCysteineUGU, UGC
GlnGlutamineCAA, CAG
GluGlutamateGAA, GAG
GlyGlycineGGU, GGC, GGA, GGG
HisHistidineCAU, CAC
IleIsoleucineAUU, AUC, AUA
LeuLeucineUUA, UUG, CUU, CUC, CUA, CUG
LysLysineAAA, AAG
MetMethionineAUG
PhePhenylalanineUUU, UUC
ProProlineCCU, CCC, CCA, CCG
SerSerineUCU, UCC, UCA, UCG, AGU, AGC
ThrThreonineACU, ACC, ACA, ACG
TrpTryptophanUGG
TyrTyrosineUAU, UAC
ValValineGUU, GUC, GUA, GUG
StopTermination, no amino acidUAA, UAG, UGA
Changing the message resets playback. Reset retains the message. The clock is not biological time. Initiation is stipulated at base one; charging, proofreading, ribosomal sites, and physical translocation are omitted.
RNA codonEventChain after event
AUGAdd MetMet
GCUAdd AlaMet–Ala
UUUAdd PheMet–Ala–Phe
UAAReleaseMet–Ala–Phe
All four events remain available without playback.

Read the playback as a trace of a bounded calculation

Use the three-residue message to separate event count from chain length. Before the first event, no codon has been processed and the displayed chain is empty. After the first addition, the chain contains Met. After the second, it contains Met–Ala. After the third, it contains Met–Ala–Phe. The fourth event recognizes UAA and releases that same three-residue chain. Thus four processed codons give three residues. The difference is mechanistic: a release factor acts at the stop instead of a charged tRNA supplying another amino acid for incorporation.

The release reaction breaks the attachment between the completed chain and its tRNA. It does not remove the RNA bases that supplied the stop signal. Nor does it add an amino acid named Stop. The RNA and ribosomal components can participate in later translation events after appropriate recycling. A diagram that places a stop symbol at the end of a peptide is therefore using punctuation, not showing part of its primary sequence. Read the released chain from N end to C end, with the last incorporated amino acid still present after termination.

Termination increases event count without increasing residue count

EventCodon read 5′ to 3′Chain, N to CReleased?
0NoneNo chainNo
1AUGMetNo
2GCUMet–AlaNo
3UUUMet–Ala–PheNo
4UAAMet–Ala–PheYes
These are the complete states of the ordinary preset under initiation at base one. Four codon events yield three residues. Release is a separate state change, not an amino acid added to the chain.

Next compare the earlier-stop preset without changing the stipulated start. Its RNA is AUGUAGUUUUAA. AUG contributes Met, and the next triplet, UAG, terminates the event. The remaining UUUUAA is downstream RNA, not two additional translated codons in this event. A reader that translates every triplet to the end of the string would incorrectly add Phe after termination. The unchanged downstream letters provide a useful control in the comparison: changing the second codon alone changes which later positions this particular translating ribosome reaches before releasing its chain.

Finally choose the no-stop fragment, AUGGCUUUU. Three complete triplets specify Met–Ala–Phe, but the supplied interval does not contain a release signal. The demonstration can report the chain implied so far. It cannot conclude that a complete protein was released. The missing boundary could lie beyond the sampled fragment. If this were the actual end of a message, cellular rescue and quality-control processes would matter, but the model does not simulate them. Ending the display is therefore a limit of the input, not a molecular substitute for a stop codon.

Use Next step to inspect one event at a time, then compare the complete selected codons with the intermediate state. Predict whether the next event should change residue count or only release status. In the ordinary preset, the last transition changes release status while leaving the count at three. In the earlier-stop preset, that transition occurs after one residue. Changing the message resets the trace, so event two in one message need not have the same result as event two in another. Reset within a message retains its sequence and only restarts the trace.

A stop and an input boundary are different endpoints

Three residues then stop5′AUGGCUUUUUAA3′Three residues; release establishedEarlier stop5′AUGUAGUUUUAA3′One residue; downstream UUU UAA not translatedNo stop in fragment5′AUGGCUUUU3′Three residues so far; release not established
Solid blue underlines mark residue-specifying triplets reached in the stipulated frame. Amber marks the first stop; dashed underlines mark downstream sequence not translated in that event. The no-stop fragment has three residues predicted so far, without established release.

The visual trace establishes ordered consequences of a fixed sequence, frame, and code. It does not measure translation speed, tRNA abundance, or the probability of an error. Even if one codon takes one visible step in every preset, real biochemical waiting times need not be equal. The finished peptide display also omits chemical modifications and folding. A valid conclusion from playback is that UAG is reached before the downstream UUU in the selected frame. A claim that the cell must produce a particular amount of active protein would require information absent from this calculation.

Respect code and fragment limits

The standard code is widely shared, but mitochondrial codes and context-dependent recoding provide exceptions. These are not included in the reader. It reports three-letter amino-acid abbreviations such as Met for methionine, Ala for alanine, and Phe for phenylalanine.

If a fragment ends before a stop, the reader reports an incomplete boundary, not a complete protein. One or two trailing bases cannot form a full codon. Bases after the first in-frame stop are not translated in this event. A peptide sequence alone does not establish folding, location, or activity.

A peptide prediction leaves molecular questions unresolved

A translation prediction has several inputs that can fail independently. The RNA must be the relevant processed product, the start must be specified correctly, and the code must fit the biological setting. Splicing can change which bases are adjacent before translation begins. An exon may contribute only untranslated RNA, so counting retained exon bases does not automatically give coding length. The earlier processing example avoids those complications with fixed coding blocks. Including or excluding its three-base GCU block changes the peptide by one residue while leaving the following UUU and UAA grouping intact.

A frame-preserving change and a synonymous change answer different questions. Adding the coding block GCU increases coding length by three and inserts alanine in the stipulated position. Replacing GCU with GCC leaves the length and encoded alanine unchanged. Both preserve downstream grouping, but only the substitution preserves that local peptide sequence. Conversely, deleting a single coding base changes later grouping in the simple model. It does not specify the resulting stop position without the remaining sequence. Length arithmetic can identify a frame problem before the code table identifies the new residues.

Removing a three-base coding block keeps the following frame

Include middle coding block5′AUGGCUUUUUAA3′N–Met–Ala–Phe–C; then releaseSkip middle coding block5′AUGUUUUAA3′N–Met–Phe–C; then release
These are the fixed coding blocks from the processing model, shown after joining. Each nucleotide takes equal horizontal space, so block widths encode three, three, and six bases. Exons in real messages can also include untranslated sequence; these blocks do not model splice recognition.

Several codons for an amino acid do not imply that every nucleotide difference is biologically invisible. Synonymous messages have the same expected peptide under the stated frame, yet their RNA sequences differ. Those differences can affect RNA interactions or expression in contexts not represented by the code table. The inference should remain narrow: the selected substitution does not change the assigned residue. Whether message abundance, translation behavior, or cellular activity changes is a separate question. This avoids both the claim that synonymous always means harmless and the opposite claim that every synonymous change must have an effect.

The standard code is also an explicit assumption rather than a universal dictionary for every compartment. Some mitochondrial systems use different assignments, and special molecular contexts can alter how particular codons are interpreted. The reader does not detect those contexts. Applying its default assignments to an unidentified RNA gives a conditional result, not proof of the biological product. For these invented messages, the standard code and starting position are supplied, so there is no need to speculate about exceptions to finish the calculation. Exceptions matter when deciding whether this model fits a different case.

Sequence evidence, peptide evidence, and functional evidence constrain different stages. A transcript sequence supports a prediction about the residues ordinary translation would assemble. Detecting a matching peptide fragment supports the presence of that segment, but may not show that every full-length molecule has the same boundaries or state. Measuring activity asks whether material performs a specified task under particular conditions. None of these observations automatically replaces the others. Agreement between them strengthens a proposed chain of explanation only when the observations concern compatible products, samples, and conditions rather than unrelated molecular pools.

Two trailing bases cannot supply a fourth residue

Frame begins at base one5′AUGGCUUUUCA3′CA is incomplete; no residue and no termination assignedN endMetAlaPheC end so far
The constructed eleven-base fragment contains three full codons followed by CA. Under the standard code from base one, the peptide predicted so far is Met–Ala–Phe. No stop is reached, and CA must await another base before it can specify a codon.

Consider a new constructed fragment, AUGGCUUUUCA. Its first nine bases form the same three residue-specifying codons as the no-stop preset, while CA remains incomplete. Under initiation at base one, the justified output is Met–Ala–Phe with no in-frame stop in the supplied interval. Do not invent a residue for CA or infer that the fragment encodes a functioning three-residue protein. If additional sequence were supplied, it would complete that partial codon before the next triplet began. This example combines frame bookkeeping with an evidence boundary without requiring a prediction about cellular function.

Exercises

Stop before the end of the message

Read 5′-AUGGAAUAGGCU-3′ from its first base using the standard code.

Hint

GAA specifies glutamate, abbreviated Glu. UAG is a stop.

Worked answer

RNA read 5′ to 3′. Offset 0.

Base 1AUGMet
Base 4GAAGlu
Base 7UAGStop

Peptide: Met · Glu

An in-frame stop was reached. After stop, not translated here: GCU.

Codons read from the first base under the standard code. This fragment does not predict initiation or cellular function.
CodonInterpretation
AUGMet
GAAGlu
UAGStop
GCUDownstream, not translated here
There is no stop residue in the peptide.

The peptide is Met–Glu. UAG terminates this translation event, so the following GCU does not add alanine. The RNA can still contain bases after the stop.

Correct pairing, wrong cargo

A tRNA pairs with a codon normally specifying alanine, but it has been incorrectly loaded with another amino acid. Does base pairing alone guarantee alanine incorporation?

Hint

Which molecule carries the amino acid?

Worked answer
CheckWhat it concerns
Anticodon pairingRNA recognition
tRNA chargingAmino acid attached
Final sequence fidelityBoth processes and quality control
The RNA-to-protein mapping requires molecular machinery.

No. Pairing recognizes the tRNA anticodon. The attached amino acid is its cargo. Charging accuracy and additional quality control are therefore important parts of translation fidelity, separate from codon recognition.

Sources and further reading

Selected textbook passages support the mechanisms below. The prose and diagrams are original, and examples are constructed rather than measured. These references are not a full literature review or independent specialist validation. Optional links are not runtime dependencies.