I understand that AUG is the "start codon", and, because of this, most proteins begin with methionine as their first amino acid.

However, this ORF problem on Rosalind.info states that "AUGCUGAC... can be translated as ...AUGCUG..., ...UGCUGA..., and ...GCUGAC". That is, something (the Ribosome? The GTP cap mentioned here? Something else?) must cause the translation to be shifted by one or two nucleotides in the sequence.

My first question is : What mechanism(s) cause(s) this shift in translation to another Open Reading Frame (ORF)?

Specifically, I want to know how the start codon, AUG, can have one ("A") or two ("AU") of its bases ignored in the Ribosome during translation.

My second question is : How often does this happen? That is, what is the percentage of (known) proteins that start with methionine vs. the percentage that start with something else due to this "ignoring" of the bases of the start codon?

I am interested in the answers for both prokaryotes and eukaryotes.

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    $\begingroup$ This is standard material you will find in any modern textbook of biochemistry or molecular biology. Aspects of it have also been discussed here, but I advise you to start with NCBI Bookshelf — a textbook generally provides a better explained and illustrated account of an elementary topic. Here are links in Berg et al. to accounts of prokaryotic and eukaryotic initiation of translation. $\endgroup$ – David Jun 16 '20 at 17:07
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    $\begingroup$ I now suspect that you are being mislead by the use of the word "can". In problems of this type "can be translated" refers to the pure informatics conceptual possibility, without any reference to biological reality, except, in this case, that of the genetic code. It is a question of considering the six possible reading frames of any stretch of DNA, and deciding which have the theoretical potential to encode a protein — see my answer to this question. The biology depends on transcription as well as the translation mechanism referenced above. $\endgroup$ – David Jun 17 '20 at 10:10
  • $\begingroup$ David the part that confuses me is how the start codon, AUG, can have one or two of its bases ignored (skipped over). I would expect "AUG" to always get translated. I don't understand how the "A" or "AU" can be skipped, as the start codon has a specific function in the Ribosome. $\endgroup$ – Greg Thatcher Jun 17 '20 at 14:14
  • $\begingroup$ @GregThatcher I do not understand what you mean by "A or AU can be skipped". 1) Do you mean at translation initiation? E.g which start codon to choose if "AUGAUG"? 2) Or do you ask what happens when the ribosome encounters a "AUG" in a different frame, e.g. "TAU"G? I feel like Joshua Morris great answer below should answer the second option. $\endgroup$ – dtadres Jun 17 '20 at 15:35
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    $\begingroup$ @GregThatcher I hope I now understand your confusion. DNA is long. X = "AUGCUGAC" will always have other sequences around it. If there is another 'AUG' before the X sequence it can be read as either "'...'AUG'CUG'...,...'UGC'UGA'... or ...'GCU'GAC'" depending on the reading frame as explained in Joshua's answer. IF translation initiation happens at the 'AUG' of the X, there is only one possibility: AUG, nothing will be truncated. $\endgroup$ – dtadres Jun 17 '20 at 20:32

Tl Dr: In the rosalind example they are showing the 3 reading frames that stem from the definition of a reading frame(non-overlapping triplets), not an example of open reading frames. Open reading frames are non-overlapping triplets between a start codon and a stop codon.

Its important to step back and understand what is meant by reading frame first. It has been determined experimentally that translation occurs in non-overlapping triplets. That is to say that if the ribosome starts reading at position n it will translate n, n+1, and n+2 into an amino acid, then it will start at the next non-overlapping position which is n+3. For example, in the sequence: ABCABCABCABC if you start at A it will read ABC ABC ABC.... But if you start at B you will have an equally valid reading frame by the definition of a non-overlapping triplet and it will read BCA BCA BCA. If you start at C it will read CAB CAB CAB. What would happen if I start at the next A? ABC ABC ABC... See how you are back in the first reading frame? No matter how long a given sequence it will only have 3 reading frames.

Why does that matter if they all start at an AUG? You don't know which AUG translation starts at. Not all AUGs are actually used as the first codon in translation, in fact, most are either internal to a translated protein or not translated at all. An open reading frame is any sequence starting from an AUG to a stop codon. An open reading frame can be within another open reading frame. E.g: AUGCTAUGGCGAUGUUAGAUGA. All of the valid open reading frames are:


               AUG UUA GAU GAC UAA
                      AUG ACU AAC UAA
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    $\begingroup$ What do you mean by "triplet pair"? It is a term that I have never encountered and does not come up in a Google search (except in its chemical meaning) even with DNA appended to the search term. It seems to me that it is incorrect. $\endgroup$ – David Jun 17 '20 at 9:59
  • $\begingroup$ Not sure why i put that. Edited. $\endgroup$ – Joshua Morris Jun 17 '20 at 15:11
  • $\begingroup$ A shifting of the reading frame downstream of a given point can occur with either an insertion or a deletion of a nucleotide. I encourage you to pick a point after aug in my hypothetical sequence and add one nucleotide (A U G C) and see what happens when you group them into codons, then add one more and then another after that. A change in the reading frame can cause the ribosome to encounter a very early stop codon and produce a non-functional protein. A great early experiment by (Brenner et. all, 1961) carried out this experiment inducing point mutations in viruses. $\endgroup$ – Joshua Morris Jan 4 at 23:10
  • $\begingroup$ ehu.eus/biomoleculas/proteinas/pdf_doc/genetic_code.pdf $\endgroup$ – Joshua Morris Jan 4 at 23:10
  • $\begingroup$ Basically one insertion/deletion shifts the reading frame and at some point downstream of that a stop codon occurs far too early. An additional insertion/deletion (++ or --) it shifts reading frame to the third one and remains disabled. If you add one more (+++ or ---) the reading frame is back to the original one and functionality reemerges. $\endgroup$ – Joshua Morris Jan 4 at 23:18

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