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Consider a codon of the form NNK (where N = Adenine, Cytosine, Guanine or Uracil & K = Uracil or Guanine). How many codons are now available? I know if all were available there would be 4^3 = 64 codons. How many are possible now? When I have tried combinations manually I got it to be 32, is this correct?

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    $\begingroup$ 4 * 4 * 2 = 32. $\endgroup$
    – user24284
    May 9 '17 at 11:35
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    $\begingroup$ What is being meant by N and NN of NNK? Does it means the first 2 nucleotides have to be same? $\endgroup$ May 9 '17 at 11:43
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Yes, 32 is correct.

Technically, I have nothing to add what Gerardo Furtado and a tiger haven't already mentioned but a graphical representation of all permutations might help to understand this a bit better.

For the first 2 positions in the codon we have 4 bases to choose from (adenine, guanine, uracil and cytosine). So this can mathematically be represented as 4 x 4 = 16 or visually as:

enter image description here

Now, for the next position (the position of interest) we only have two bases to choose from (uracil and guanine). Leaving us with 4 x 4 x 2 = 32 different permutations, or visually:

enter image description here

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  • $\begingroup$ You are welcome. I am glad it helped. $\endgroup$
    – Johnny
    May 10 '17 at 12:01
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Yes, it is.

Gerardo Furtado has already provided a short answer in the comments, but allow me to explain why.

The 4^3 = 64 if all bases can be chosen freely comes down to 4*4*4=64, because you can choose from 4 options for each of the 3 bases.

If you restrict the possibilities of one of the bases, e.g. there are only 2 options for the third base (as in your example), the formula changes to 4*4*2=32, as you can choose from 4 options each for the first 2 bases and from 2 options for the 3rd base.

Edit: If the 2 'N's from 'NNK' have to be the same, you can choose the first base freely (4 options), the second base is already defined (it has to be the same as the first base, 1 option) and there are 2 options for the 3rd base as described above. Thus, we get 4*1*2=8 possible codons

I hope I could clear this up.

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    $\begingroup$ Good explanation. For completeness, it would be extremely unusual if NNK denoted that the first two bases had to be identical: first off, it’s not the convention; IUPAC ambiguity codes are character wildcards, not variables in the common mathematical sense. And secondly, there’s little biological reason to consider codons where the first two bases are identical. $\endgroup$ May 9 '17 at 16:21
  • $\begingroup$ I absolutely agree. $\endgroup$
    – a tiger
    May 15 '17 at 15:38

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