This question is from a past exam paper for an introductory bioinformatics module. I'm a computer scientist doing biology for the first time.

"A short bacterial gene has been sequenced, giving the following DNA sequence. Write out the 6 possible reading frames for this sequence and indicate which is the most likely protein translation of this sequence. Explain your reasoning for picking the given translation, and write all the translations in single letter amino acid code form.

5' - ttattcatccgccagcgccatgcgcgccat - 3'  "

I think I understand the 6 possible reading frames: three from the 5' end starting tta, tat and att; and three from the 3' end starting tac, acc and ccg.

I could also write out the translations for a given reading frame with a codon usage table. It's the most probable reading frame part I don't get. I thought perhaps I was looking for the longest open reading frame. I can only see one start codon (atg). What is the correct way to recognize the most probable reading frame?

I can't find this problem exactly on Biology Stack Exchange. I did find:

Help reading chromatogram 

It made me wonder if I'm getting confused between "reading frame" and "open reading frame", this question only asks or the former.


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    $\begingroup$ This is a terribly artificial question, but assuming that the sequence represents the whole reading frame, what do you need to consider as well as start codons? I like your suggestion of codon usage, but it's not as sophisticated as that. $\endgroup$ – David Apr 19 '20 at 12:06
  • $\begingroup$ Thanks for getting back to me so quickly. I'm still not sure but I had a few ideas. $\endgroup$ – azure_reflection Apr 20 '20 at 8:13
  • $\begingroup$ 1) Length of the frame, long enough to code something useful - but that would still be length between a start and stop codon. 2) Presence of expressed sequence tags - though I wouldn't know how to tell you if some sequence contained an EST and I suspect they are longer than 30 bases. 3) The presence of more or less "likely" amino acids eg, does the sequence have more of the most frequent amino acids (which I think are serine and leucine) - but I don't think we'd be expected to know this. $\endgroup$ – azure_reflection Apr 20 '20 at 8:14
  • $\begingroup$ 4) The presence of common subsequences that might indicate, say, common secondary structures - again, this is not something we've talked about in our course. 5) Sequence reading errors in the lab. Look for where there could be start and stop codons if you just changed one (or two) bases. For example, in the reading frame starting "att" from the 5' end, you could change that first triple to "atg" to make it a start codon, then perhaps the last "cca" should have been a tta stop codon. 6) Promoter sites? $\endgroup$ – azure_reflection Apr 20 '20 at 8:15
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    $\begingroup$ The 3 reading frames from the 3' end use the "other" strand of dna, that is ATG and two others. Experts from mathematical and computational sciences contribute greatly to bioinformatics after a bit of education. Good luck! $\endgroup$ – Polypipe Wrangler Apr 21 '20 at 12:05

This is what we classify as a homework question, but as it satisfies the criterion of the poster demonstrating an attempt to answer it, I provide the following suggestion of an answer.

I assume that as it appeared in an introductory bioinformatics module the exam question is just testing reading frames (obviously) and the punctuation of the genetic code. The meaning is not entirely clear, but as one reading frame would start with the initiation codon ATG/AUG (reverse complement of: cat - 3′)† and end with the termination codon TAA/UAA (reverse complement of: 5′ - tta) that will presumably produce “the most likely protein translation”.

This is reading frame F4 in the output from EMBOSS Sixpack, below, in which termination codons are indicated by an asterisk.

      L  F  I  R  Q  R  H  A  R  H                                   F1
       Y  S  S  A  S  A  M  R  A  X                                  F2
        I  H  P  P  A  P  C  A  P  X                                 F3
    1 ttattcatccgccagcgccatgcgcgccat 30
    1 aataagtaggcggtcgcggtacgcgcggta 30
       X  N  M  R  W  R  W  A  R  W                                  F6
      X  I  *  G  G  A  G  H  A  G                                   F5
        *  E  D  A  L  A  M  R  A  M                                 F4

The conceptual translation, reading N to C, is MARMALADE, which is obviously meant to be humorous and suggests that it is indeed the intended answer.

Open Reading Frames

The poster asks for clarification of the difference between reading frame and open reading frame. There is a Wikipedia entry for open reading frame but I provide an explanation of my own to relate it to the example.

There are always six reading-frames for the conceptual translation of a piece of double-stranded DNA, as shown in the example.

I would define an open reading frame as one that is not excluded from being translated by the punctuation of the genetic code. It has the theoretical potential to be translated considering only the punctuation of the code, although it may not actual be translated. It can start either with the first AUG after a termination codon (even though it cannot be certain that this is the actual AUG used) or the start of a sequenced fragment of DNA (with the assumption that an AUG is possible preceding the 5′ end of the fragment). It can end either with a termination codon or the end of the sequenced fragment (with the assumption that a termination codon will lie 5′ to the end of the fragment).

By these criteria, reading frames F1, F2, F3 and F6, above are completely open (even though the internal methionine could theoretically be an initiation codon), F5 contains the open reading frame GAHGAGG, and F4 is a complete open reading frame, perhaps excluding the termination codon (depending on your semantic definition of the precise end of an open reading frame).

† Reverse Complement

If we take a section of DNA written in the 5′-to-3′ direction — according to standard convention — and use the Watson–Crick base-pairing equivalences (A=T, G=C) to generate the complementary strand, this will be in the 3′-to-5′ direction. For the strand in the question,

5′ - ttattcatccgccagcgccatgcgcgccat - 3′

the complementary strand is:

3′ - aataagtaggcggtcgcggtacgcgcggta - 5′

As shown above.

For ease of translation manually — and for any computer program that manipulates sequences — one needs to reverse this to the 5′-to-3′ direction:

5′ - atggcgcgcatggcgctggcggatgaataa - 3′

This is the reverse complement. Now the starts of the three reverse reading frames are easy to read as:

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    $\begingroup$ Could be helpful (especially for someone without a solid biology background) to expand on the reverse complement subject. Based on the question it seems that this was most likely the part that was giving the OP trouble. $\endgroup$ – Astrolamb Apr 21 '20 at 11:39
  • $\begingroup$ @David Well I feel silly, but also grateful for the excellent answer. I had tunnel vision but it's obvious now. Out of interest, what made you describe the question initially as "terribly artificial"? $\endgroup$ – azure_reflection Apr 21 '20 at 17:04
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    $\begingroup$ @azure_reflection — Most proteins are much longer than nine amino acids, and programs predicting genes usually have a cut-off of perhaps 30. (Someone else might be able to give you precise size distributions.) There are exceptions (in biology there are always exceptions). In eukaryotes some small peptides are generated by translation of small orfs that precede the predominant AUG. But, from experience, I sympathize with lecturers setting problem-type exam questions. $\endgroup$ – David Apr 21 '20 at 18:43
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    $\begingroup$ @Astrolamb — done. $\endgroup$ – David Apr 25 '20 at 11:50

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