One particular codon codes only for one amino acid, but an amino acid can be coded for by several different codons. Now according to the genetic code, the codon UUU codes for the amino acid phenylalanine and UUA codes for leucine. But, according to the Wobble Hypothesis, the base on the third position of the codon and that on the anticodon need not be complementary (which helps explain why there are very few types of tRNA molecules, inspite of there being 61 codons). If this hypothesis is true, then we could have a phenylalanine placed in a position which was meant to be for leucine, and vice versa (since the codons coding for them differ only in their third base). The same holds true for pairs like aspartic acid & glutamic acid and serine & arginine. So how does translation of a particular mRNA molecule result in the right polypeptide sequence?


2 Answers 2


Wobble pairing is just a phenomenon and not a hard and fast rule. There are some justifications for why it should exist and that is why it is still called a hypothesis. And this statement is not true:"the base on the third position of the codon and that on the anticodon need not be complementary". The anticodon residue corresponding to the third residue of codon can be a promiscuous base which can pair with two or many different bases. The tRNA for Phenylalanine has an anticodon - GAA which can pair with both UUU and UUC but not UUA.

So the statement of wobble hypothesis is that the first base of the anticodon (often is a modified/atypical nucleobase) can show promiscuity of binding.


You are correct in saying that Crick, in his Wobble Hypothesis, proposed that “the base on the third position of the codon and that on the anticodon need not be complementary”, but the “need not be” in your statement is a paraphrase of the “some” in Crick’s original statement:

“It is suggested that while the standard base pairs may be used rather strictly in the first two positions of the triplet, there may be some wobble in the pairing of the third base.”

If you read that paper — or consult the Wikipedia entry under Wobble — you will become aware that Crick is using the word “some” to indicate that:

(i) The wobble proposed is specific for certain base pairs.

(ii) That such wobble base pairs will only be found in cases where they do not violate the genetic code.

The Wobble Hypothesis — as stated above — has been unequivocally shown to be correct. The specific Wobble Rules that Crick proposed to satisfy point (i) were based on an examination of the chemistry of the bases, and have been shown to be partially correct:

Wobble Bases: Prediction and Reality Wobble Rules: Crick’s origin predictions compared with observed 5'-anticodon bases and their base-pairing with codons.

Thus, the prediction that the 5'-tRNA anticodon bases, G and I could wobble (and C could not) have been borne out. Crick was aware of the paucity of A at this position in anticodons, and both it and U are normally found in chemically modified forms, the base-pairing of which he did not attempt to predict (he was unaware of most of them) and which is different in different cases. The point to remember is that there is a scientific rationale for this in terms of the three-dimensional structure of the anticodon in the tRNA (which holds the first two bases in position by base stacking) and the proximity of the potentially hydrogen-bonding groups in the various bases.

Point (ii) is that Nature will only use Wobble where the genetic code allows. G pairing with C or U always works, whereas I pairing with A, C or G will work with amino acids encoded by all four bases in a block (e.g. Leu, Val, Ser), but not where there are blocks of two (e.g. Tyr, His, Asn).

One final point that emphasizes the chemical basis of all this. Mammalian mitochondria have a different set of wobble rules on account of their peculiarly truncated tRNAs.

  • $\begingroup$ I appreciate that this is an old question that has already been answered (it came to my attention on a 'related' listing. However as I had some material on the topic I thought it would be useful to flesh the previous answer out a little. $\endgroup$
    – David
    May 31, 2016 at 20:18

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