Translation and the Genetic Code: Read the Supplied Frame

MCAT content library / Biology & biochemistry / Translation and the Genetic Code: Read the Supplied Frame

Use a supplied codon table to keep strand direction, reading frame, ribosome steps, and protein processing consistent.

About 15 minutes, including the worked examples and checks.

From a gene to a functional result · Lesson 3 of 5 · Continue in this path

Useful foundations

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Set the reading frame before naming amino acids
Distinguish the message, adaptor, and catalyst
Follow one ribosome through a productive cycle
Worked application: compare sequence, length, and output rate
Separate the encoded chain from its processed form
Check your reasoning
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Set the reading frame before naming amino acids

Translation uses the sequence of an mRNA to build an ordered polypeptide. The ribosome reads the message 5′ to 3′ in successive three-nucleotide codons, while the protein grows from its amino terminus toward its carboxyl terminus. AAMC includes the triplet code, codon–anticodon relationships, initiation, termination, ribosomes, and protein processing within its gene-expression content.¹²

A reading frame specifies which bases form each triplet. Starting one nucleotide later changes the grouping, even if the written RNA sequence is otherwise unchanged. For these examples, use the standard nuclear genetic code and the stated initiation site. An AUG can specify the starting methionine, but an internal in-frame AUG usually adds another methionine rather than restarting an already translating ribosome. A stop codon is a termination signal, not an extra amino acid.³²

The code is degenerate: more than one codon can specify the same amino acid. That does not make a codon ambiguous, nor does it make every third-base substitution harmless. Use the supplied table to compare the actual codons before deciding whether the encoded residue changes.³

Distinguish the message, adaptor, and catalyst

mRNA supplies the codons. tRNA connects a codon to an amino acid: its anticodon pairs with the mRNA, and its 3′ end carries the amino acid. Aminoacyl-tRNA synthetases establish that connection by charging appropriate tRNAs using ATP. The ribosome is made of RNA and protein; its rRNA contributes to the structure and the catalytic center that forms peptide bonds.²⁴

Codon and anticodon pairing is antiparallel. For example, an RNA codon written 5′-AUG-3′ pairs with an anticodon aligned 3′-UAC-5′. The same anticodon written in its own 5′-to-3′ direction is 5′-CAU-3′. An anticodon question can therefore be a direction question before it is an amino-acid question.

Wobble permits some nonstandard pairing at the third codon position, paired with the first position at the anticodon’s 5′ end. This can allow one tRNA to recognize more than one codon for the same amino acid. It is a constrained recognition mechanism; it does not mean any base can substitute at any position or that the ribosome randomly chooses among different amino acids.⁵

Follow one ribosome through a productive cycle

During initiation, factors, the small ribosomal subunit, mRNA, and an initiator tRNA assemble at the selected start. The large subunit joins, and the initiator tRNA occupies the P site. In eukaryotic cap-dependent initiation, the sequence around an AUG affects its recognition, so the first written AUG is not an adequate universal rule for choosing the start.²

During elongation, an appropriate charged tRNA enters the A site. The growing peptide is transferred from the P-site tRNA to the amino acid on the A-site tRNA. Translocation advances the message by one codon and moves the peptidyl-tRNA into the P site; the uncharged tRNA leaves through the E site. The activated tRNA linkage and factor-driven steps make this different from joining two free amino acids by simply removing water.⁴

When an in-frame stop enters the A site, a release factor promotes release of the completed peptide from its tRNA. Thus initiation determines entry into the process, elongation extends the chain, and termination releases it. Blocking one stage need not produce the same immediate pattern as blocking another.²

Worked application: compare sequence, length, and output rate

The following original example supplies a short mature RNA and the necessary codon assignments. Bars divide untranslated flanking sequence from the translated region; they are not nucleotides. Initiation occurs at the indicated AUG, and translation terminates at the first in-frame stop:

5′-CCACC | AUG GCU UUC GAA UGA | CCU-3′

mRNA codon, 5′ to 3′	Meaning
AUG	Met; indicated initiation site
GCU, GCC	Ala
UUC, UUU	Phe
GAA, GAG	Glu
UAA, UGA	Stop
CCU	Pro, if read in a translated frame

Codon assignments used in the worked example

Reading from the selected start gives N–Met–Ala–Phe–Glu–C. There are four amino-acid residues, although five codons are read through termination. The downstream CCU does not add proline because the ribosome has already stopped. The five upstream bases likewise do not become amino acids in this specified reading frame.

Change	Predicted encoded product	Reason
GCU → GCC	Met–Ala–Phe–Glu	Both codons specify Ala
UUC → UAA	Met–Ala	Termination now follows the second residue
Weaker recognition of the same start	Met–Ala–Phe–Glu whenever that start is used	Initiation frequency can change without changing the frame

Three distinct changes to the same illustrative message

In a reconstituted assay with equal intact RNA amounts, no RNA decay, and unchanged elongation and termination, suppose weakening the start context reduces completed chains from 100 to 25 per minute. Output becomes 25/100 = 0.25 of control, a 75% decrease. It does not become a one-residue peptide. The measured rate and the sequence of each successfully completed product answer different questions.

Likewise, a synonymous change establishes an unchanged encoded residue at that position, not identical behavior of the RNA in every cell. A premature stop predicts an earlier termination point if the RNA is translated; whether that RNA persists and how much protein accumulates require additional evidence. The explicit assay conditions above keep those separate effects out of the arithmetic.

Separate the encoded chain from its processed form

A newly made chain can fold, undergo proteolytic processing, or receive chemical modifications. For example, removing an N-terminal segment changes the length of the remaining chain; phosphorylation changes a side chain without replacing the codon that encoded that residue. Chemical modifications can influence activity, localization, or stability, so a protein’s abundance and behavior are not specified by translation alone.⁴⁶

For a new problem, identify the requested product: the codon sequence, the newly translated chain, a processed protein, or the amount produced. Then apply only the stated operations in order. Translate from the supplied start, stop without adding a “stop residue,” remove any specified cleaved segment, and keep a chemical modification separate from residue counting. This sequence of decisions also tells you which missing measurement would be needed before claiming a functional effect.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Check 1

For the original RNA 5′-AA | AUG UUU GGC UAA CCU-3′, initiation occurs at the indicated AUG. Use AUG = Met, UUU = Phe, GGC = Gly, UAA = stop, and CCU = Pro. Translation uses this frame without readthrough or post-translational cleavage. Which product is made, written from N terminus to C terminus?

A. Met–Phe

Review this choice. GGC is an in-frame glycine codon before the stop. Termination occurs after that glycine has been added.

B. Met–Phe–Gly

Correct. Read AUG, UUU and GGC to add three residues. UAA terminates translation, so CCU downstream does not add proline.

C. Met–Phe–Gly–Pro

Review this choice. This skips the supplied termination signal and translates a downstream codon despite the explicit absence of readthrough.

D. Phe–Gly–Met

Review this choice. This changes the specified starting position and residue order. The initiator methionine is made first in the supplied frame.

The principle to carry forward

Starting position, direction and the first in-frame stop determine which supplied codons contribute residues.

Check 2

At time zero, a treatment prevents all new translation initiation on an mRNA but does not affect ribosomes already elongating. Existing ribosomes can finish and release their chains; mRNA amount remains constant, and protein degradation is negligible during observation. What happens to the rate of production of newly completed chains?

A. It becomes zero immediately, even while existing ribosomes finish.

Review this choice. Finishing an already initiated chain still produces a newly completed protein. An initiation block does not prevent the supplied remaining elongation and termination.

B. It continues briefly, then falls to zero as the existing ribosomes finish.

Correct. Previously initiated ribosomes can complete their work, but no new ones replace them. Once those ribosomes finish, no further chains are completed.

C. It increases continuously as ribosomes are released from the message.

Review this choice. Released ribosomes cannot initiate again under the supplied treatment, so their release cannot sustain increasing production.

D. It remains constant indefinitely because the RNA is still present.

Review this choice. A stable template is not sufficient when new initiation is completely blocked. Existing ribosomes provide only a finite period of continued output.

The principle to carry forward

Predict the immediate effect at the inhibited step. Existing intermediates can continue after entry into a process is blocked.

Check 3

A coding segment is 159 nucleotides long from the start of its AUG through the end of its only in-frame stop codon. Every codon before that stop adds one residue. After translation, exactly eight N-terminal residues are removed, and one remaining serine is phosphorylated; no other cleavage occurs. How many amino-acid residues remain in the processed chain?

A. 44

Correct. 159/3 gives 53 codons including one stop, so translation produces 52 residues. Removing eight leaves 44. Adding a phosphate does not add an amino-acid residue.

B. 45

Review this choice. This counts the stop codon as a residue: 53 − 8 = 45. The stop is a signal for release, not an encoded amino acid.

C. 52

Review this choice. This correctly excludes the stop but does not subtract the eight residues removed during processing.

D. 53

Review this choice. This counts every codon as an amino acid and also omits the specified proteolytic processing.

The principle to carry forward

Count translated residues first, then apply the stated processing. A chemical group is not another residue.