Replication, transcription and translation as chemical processes

Sequence has a chemical direction

A nucleic-acid strand has chemically different 5′ and 3′ ends. A template written 3′→5′ can be read to make a complementary product written 5′→3′. The new nucleotide is added at the growing product’s 3′ end. Reversing how a sequence is displayed is different from changing its bases.

Replication copies DNA; transcription makes RNA from a template; translation uses an RNA sequence to organize a polypeptide sequence. These descriptions identify different molecular operations. They do not mean that one polymer becomes another by changing letters on a page.

Count the extension and declare how the sequence is read

Take an invented existing primer eighteen nucleotides long and a final strand seventy-five nucleotides long. Only extension is counted, so each complete product needs fifty-seven new nucleotides. Specify incoming A/C/G/T counts 17/13/11/16 per chain and a template-primer inventory 90.0 pmol. These assigned counts describe the incoming sequence; the primer inventory and chemistry remain separate.

Since 1 pmol=0.001 nmol, incoming dATP/dCTP/dGTP/dTTP needs are 1.53/1.17/0.990/1.44 nmol. Their sum is 5.13 nmol incorporated nucleotides and, for this extension account, 5.13 nmol released diphosphate. Incoming triphosphates carry 15.39 nmol phosphorus atoms: 5.13 retained in the new links plus 10.26 in diphosphate. Stipulate available pools 2.00/1.80/1.40/1.70 nmol; all can support the requested complete-chain amount. This inventory bound is not a yield or reaction-rate prediction.

For a separate invented template written 3′-TACCGATTTACT-5′, the complementary RNA product is 5′-AUGGCUAAAUGA-3′. Choose the standard genetic code, table 1, and the reading frame starting at the first AUG. Triplets AUG/GCU/AAA/UGA specify Met/Ala/Lys/stop: three residues and two peptide links. Other codes, frames or processing would need their own account.

For this twelve-nucleotide RNA, specifically stipulate NTP-initiated de-novo synthesis with an uncapped, unprocessed 5′ triphosphate end. Twelve NTPs are incorporated; eleven later extensions release eleven diphosphates. Phosphorus counts close: 36 incoming = 14 retained in RNA + 22 released. The retained inventory is eleven link phosphorus atoms plus three at the initiating end. This is different from adding twelve nucleotides to an existing primer and is not a universal mature-RNA end convention.

Polymer formation uses activated substrates

For the classified DNA-directed DNA polymerase, a deoxyribonucleoside 5′-triphosphate supplies an incoming nucleotide to an existing primer, and diphosphate is released during extension. The primer’s 3′ end and the incoming nucleotide’s phosphate groups belong to a chemical reaction, rather than a condensation of free bases.

DNA-directed RNA polymerase likewise uses nucleoside triphosphates during elongation and can initiate a strand without an existing primer. In a stipulated unprocessed transcript initiated with an NTP, the first nucleotide can retain its 5′ triphosphate. Count initiation separately from later extension; mature RNA may have processed or otherwise modified ends.

Carrier, link and error counts describe different events

Specify the synthetic seven-residue sequence Met–Ser–Gly–Leu–Gly–Asp–Phe and 45.0 pmol complete chains. Assign one ATP-to-AMP activation per charged tRNA and two GTP events per post-initiation addition. Charging uses 7×45=315 pmol ATP and gives 315 pmol each of AMP and diphosphate in this restricted account.

Each chain has six peptide links, so link formation totals 270 pmol. The assigned post-initiation GTP budget is 2×6×45=540 pmol. Two glycine residues occur per chain. If only 70.0 pmol Gly-tRNA equivalents can be supplied and other resources are ample, the complete-chain limit is 35.0 pmol; its charging/link/GTP counts are 245/210/420 pmol. Initiation, termination, repair and other costs are excluded, so this is not a full cellular energy budget.

For another independent teaching model, assign 930 independent incorporation events, each with incorrect-incorporation probability p=0.000700 before later repair or selection. Expected error count is np=0.651; it need not be an integer because it is an average over repeated realizations.

The zero-error probability is (1−p)⁹³⁰=0.521405121, and at-least-one-error probability is 0.478594879. The expected count 0.651 is not that probability. Reducing only p tenfold gives expected count 0.0651 and zero-error probability 0.936971626. These are copying-event quantities under an independence assumption, not organism mutation risk or clinical outcomes.

Translation depends on chemically activated amino acids

An aminoacyl-tRNA synthetase links the appropriate amino acid to a tRNA using ATP activation. The net charging account includes ATP conversion to AMP and diphosphate. The charged tRNA then carries an amino acid through an ester linkage at its terminal region. Ribosomal peptide formation transfers a growing peptide between such carriers.

A stop codon is a termination instruction in the chosen code, rather than an amino acid incorporated into the product. Charging, initiation, elongation, proofreading and termination are different stages. A count of amino acids or peptide links alone is not a complete energy budget for protein synthesis.

An error event is not automatically a lasting mutation

Complementarity, enzyme selectivity, proofreading and repair contribute to copying fidelity. A probability of an initial incorporation error and the probability that it escapes later correction describe different events. State which sequence, opportunity count and conditional stages a model includes.

A remaining mismatch does not itself establish an inherited mutation, altered protein, phenotype or disease. Subsequent copying, repair, location, expression and biological context affect those outcomes. Chemical copying explains how information can be preserved and changed, while evidence is needed for each further conclusion.

Worked examples

Try it yourself

1. Reduce only available dGTP to 0.660 nmol with all other assigned pools and 90.0 pmol templates/primers unchanged. Find maximum complete chains and incorporation.

   Show the answer and reasoning

   60.0 pmol complete chains; A/C/G/T needs 1.02/0.780/0.660/0.960 nmol, total incorporation and diphosphate 3.42 nmol.

   Eleven G units per chain limit the complete-chain amount. Possible partial products are outside this count.

2. Change the second RNA triplet from GCU to GUU under the same code/frame and end chemistry. State residue and event-count changes.

   Show the answer and reasoning

   RNA 5′-AUGGUUAAAUGA-3′; Ala becomes Val. Still three residues/two links, twelve NTPs and eleven diphosphates.

   Changing one specified codon changes its amino-acid assignment without changing length or the stipulated terminal account.

3. Only 70.0 pmol Gly-tRNA equivalents are available for the seven-residue sequence, with other resources ample. Find complete chains, charging, links and assigned GTP.

   Show the answer and reasoning

   35.0 pmol chains; 245 pmol charging ATP; 210 pmol peptide links; 420 pmol assigned GTP.

   Each chain consumes two Gly equivalents, so the complete-chain limit is 70/2.

4. Reduce p tenfold without changing 930 independent copying events. Find expected incorrect count and zero-error probability.

   Show the answer and reasoning

   p=0.0000700; expected count 0.0651; zero-error probability 0.936971626.

   np and (1−p)⁹³⁰ describe different quantities; neither establishes inherited mutation or a clinical outcome.

What to carry forward

• Count an existing primer separately from newly incorporated nucleotides.
• Declare template direction, reading frame and genetic code.
• Initiation and elongation can leave different phosphate inventories.
• Carrier budgets and copying probabilities belong to their stated event models.