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I was asked a question, "Considering degenerate and non ambiguous nature of genetic code, Why is that certain mutations don't disturb the protein synthesis leading to synthesis of functional proteins where as certain mutations cause synthesis of non functional proteins or diseased proteins?"

My argument was that according to Wobble hypothesis, the tRNA carries anticodons that have A,U,G,C,I where I is inosine (sugar phosphate with hypoxanthine) that can bind with A,C,U of the codon and also G binding with U and vice versa is possible. Then, I gave the example of serine -AGU- whose anticodon is -UCI- for instance and after mutation say, -AGU- becomes -AGC-, it still codes for serine and -UCI- stills agrees to bind with -AGC- giving us serine as before and hence no disturbance is brought up leading to synthesis of functional protein. But suppose -AGU- got mutated to -CGU- , then the anticodon -UCI- no longer can bind with the codon and hence another tRNA molecule with anticodon of that of arginine gets attached and hence leading to interruption in the amino acid sequence which ultimately leads to non functional protein.

But my professor disagrees with my argument and says I am not clear with the concept. Can anyone please help with where I am going wrong ?

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Nonambiguity refers to the fact that codon X will always code for the same amino acid.

Degeneracy refers to the fact that an amino acid can be coded for by many codons.

I rephrase the question to get to the gist of things:

Why do certain mutations cause no difference for protein synthesis, while others make a big difference?

Certain mutations can change the codon but still code for the same amino acid, so the resulting peptide sequence is unaltered (synthesis of a functionally identical protein). This is due to degeneracy, explained above. Certain other mutations change the codon in a way that produces a completely different amino acid, and these result in the synthesis of a different, distinct peptide that may be dysfunctional. This due to nonambiguity of the code. Simple!

It's best to refer to our translation guide, the codon table. If you know your codon table, you might have noticed that within a codon frame, changing the 3rd nucleotide often has no effect on the translated amino acid (see outer ring, below). Changing the 1st and 2nd nucleotide produces a more drastic effect in the translation of the codon.

That means that the position of a nucleotide mutation within the reading frame is key to the effect of the mutation on protein synthesis.

Codon table

Where I am going wrong?

You are complicating the picture needlessly. You fail to explain that the resulting amino acid depends on its codon, and that the effect of changing the codon in different places (position 1, 2 or 3) has different outcomes. You also have to think about different kinds of mutations; we are discussing substitutions, but you must also consider what deletions and insertions of nucleotides do to the translation of the peptide! You may also get bonus points for discussing the start and stop codons... they are special cases.

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  • $\begingroup$ OK, I completely agree that I haven't been specific to the point considering the degeneracy and non-ambiguity, but I need some clarification. Does the point about wobble hypothesis explain the concept of synthesis of functional and non-functional proteins ? $\endgroup$ – Suraj S Oct 1 at 16:20
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    $\begingroup$ @SurajS — The wobble hypothesis is irrelevant to the question you have been asked. That answers how the degenerate genetic code operates. If the answer were through 61 different tRNAs it was make no difference to the effect of the degeneracy on the likelihood of a mutation having a detrimental effect. The genetic code diagram in this answer may look pretty, but I think you would do better looking at the standard tabular presentation found in text books and Wikipedia. This makes it easier to see how mutations would affect both the amino acid encoded, and the “type” of amino acid encoded. $\endgroup$ – David Oct 1 at 20:05
  • $\begingroup$ @David Oh, now I get it. Thanks a lot $\endgroup$ – Suraj S Oct 3 at 11:47

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