Assuming there were 20 different amino acids and less than 40 types of different tRNAs found in this alien organism. How many amino acids would be found in the translational product of a 600-nucleotide mRNA?

I am very confused over how I should solve this question.

If we assume 1 codon = 1 nucleotide,

20 amino acids require 20 different codons = 20 different nucleotides < 40 different tRNAs.

If we assume 1 codon = 2 nucleotides,

20 amino acids require 20 different codons = 5 x 5 combination of nucleotides? (considering A, T, C, G, U)?

5 x 5 = 25 < 40 different tRNAs.

The correct answer is to consider 1 codon = 2 nucleotides.

My question,

Since both 1 codon = 1 nucleotide and 1 codon = 2 nucleotides fit the requirement of this question, why is 1 codon = 2 nucleotides the correct answer?

  • $\begingroup$ You could have more than 1 tRNA for each amino acid, as well as stop codons. Both of these features are found in Earth's genetic code. $\endgroup$
    – user137
    Commented Nov 18, 2014 at 14:50
  • $\begingroup$ Wait, are we supposed to assume that this alien uses both T and U in it's RNA? To solve this we need to know how many types of nucleotide it uses. $\endgroup$
    – user137
    Commented Nov 18, 2014 at 14:56
  • $\begingroup$ @user137 thanks for your reply, that is the exact question, no additional information given. I am assuming the organism uses all 5 different nucleotides. $\endgroup$
    – Kenny
    Commented Nov 18, 2014 at 15:06
  • $\begingroup$ The problem is I don't know how you could use both U and T and tell them apart because they both bind to A. Furthermore we can't assume an alien biochemistry would be at all similar to Earth biochemistry and their "RNA" could use entirely different bases. $\endgroup$
    – user137
    Commented Nov 18, 2014 at 15:16
  • $\begingroup$ If this is the original question accurately and in full it is meaningless and unanswerable. If, as your answer about codon length implies, it is a garbled version of the question, please present it carefully and word for word. As it stands, If the question means how many amino acids in the peptide produced you can't tell because it doesn't tell you where the start and stop codons are. If it means how many different amino acids there are, you cannot tell because you do not know what the sequence is. In either case the number of tRNAs can have no bearing on the answer. $\endgroup$
    – David
    Commented Nov 28, 2017 at 21:32

1 Answer 1


I think I have this figured out. Had to do some math. With our 4 nucleotide system, the number of possible codons is 43 = 64. That is number of nucleotides to the codon length power. The question asks us to determine codon length while giving us the number of amino acids and an upper limit on tRNAs. Using the same formula,

 #Nucleotides(Codon-Length) = #Codons,
we must find a combination of nucleotide number and codon length that gives between 20 and 40 codons.

                               Codon Length
Nucleotides    1         2        3        4        5
     2         2         4        8        16       32
     3         3         9        27       81       243
     4         4         16       64       256      1024
     5         5         25       125      625      3125
     6         6         36       216      1296     7776
     7         7         49       343      2401     16807

So using this table we see that a few different combinations of Nucleotide number and codon length will fit in that range. A 2-nucleotide system could work if each codon was 5 bases long. A 3-nucleotide system could work with 3 base long codons. A 4-base system cannot work, it goes from less than 20 to more than 40. A 5-base system can work with 2 base codons. A 6-base system can also work with 2-base codons. 7 or more bases fails because even a 2-base codon will make more than 40 codons.

A codon length of 1 nucleotide would require at least 20 nucleotides, so it would not be a good answer.

5 is probably the correct answer because it works with the shortest codon length, though I don't know how you'd get base pairing with an odd number of bases. Perhaps the 5th base pairs with itself. A 6 base system can also work with the same codon length and wouldn't need a self pairing base. However that table shows several combinations could work, so this question might have been to see how you defend your answer.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .