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A new life form discovered on a distant planet has a genetic code consisting of five unique nucleotides and only one stop codon. If each codon has four bases,what is the maximum number of unique amino acids this life form can use ?

I tried this and got 624 (Using permutation method 5×5×5×5 = 625). But the answer given was 3124. Please help.

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    $\begingroup$ when you say you tried, what was your logic. Can you add that in your question. $\endgroup$
    – WYSIWYG
    Nov 30, 2014 at 7:01
  • $\begingroup$ Using permutation method 5*5*5*5 = 625 amino acid codons of which one is stop codon. $\endgroup$
    – user10367
    Nov 30, 2014 at 7:04
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    $\begingroup$ Plus. It is not that by increasing information content in DNA you can increase the number of amino acids that are put in use. In fact with the current genetic code 63 amino acids can be encoded but there are only 22. Feel free to give me the address of the person who posed this question to you; a cruise missile would be on its way :P BTW 624 is right and 3124 is incorrect. $\endgroup$
    – WYSIWYG
    Nov 30, 2014 at 7:11
  • $\begingroup$ i thought there were only twenty, what are the other two? $\endgroup$ Dec 3, 2014 at 5:14
  • $\begingroup$ @CognisMantis its 23 now. 20 standard and 3 non-standard en.wikipedia.org/wiki/Amino_acid $\endgroup$ Dec 3, 2014 at 6:53

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I also got 624. Maybe the answer is incorrect. My processing is:

Each codon has four bases, the species contain five unique nucleotides, so for the first position in the codon, there are 5 possible results, also, for the second position in the codon, there are 5 possible results, it is same in third and forth position,

so the result should be 5*5*5*5-1=624

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  • $\begingroup$ Hi, the question I posted above is from a competitive exam I appeared. Answer key says 3124 is the correct answer. Help please. I dont know which is correct. But everytime I end up with 624. I could not clear the exam because of 2 marks that is contributed by this question. I really need to know how to solve it. Please help. $\endgroup$
    – user10367
    Feb 8, 2015 at 17:25

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