Probability and ribonucleotides

[Yann]

There's question in Hardt's Essential Genetics that bugs me, because I'm not sure if the answer is very simple or if I just don't get the question right. If you have a sequence of 4 ribonucleotides (with equal frequency), what is the probability to have a start codon/stop codon and what would be the average distance between stop codons in a DNA sequence...

The probability of a start codon is 1/64, it's 3/64 for stop codons (of course it depends on the organism but in this case it must be the "convenctional" start/stop codons). A codon is made of 3, not 4 nucleotides. Within a sequence of 4 codons, there could be 2 codons (XXXY or YXXX), so it seems the probability would be 2/64 and 6/64, but I'm not sure if it's that simple...

For the average distance it's easier; 64/3 codons or 64 nucleotides.

 

[farful]

The probability of codon frequency is dependent on the GC content. To make life easier, suppose GC content is 50%. Then, the probability of a stop codon is indeed 3/64. Asking for the probability of a start codon is a dumb question and is dependent on the locations of the stop codon. Even the naive approach to gene finding first choose the longest regions without stop codons, then look for the start codon.

A codon is made of 3, not 4 nucleotides. Within a sequence of 4 codons, there could be 2 codons (XXXY or YXXX)

That made no sense. Within a sequence of 4 codons, there are four codons - not two. If you meant that within a sequence of 4 nucleotides, there are two codons, that's not true either - there are four. Two in the 5' to 3' direction and two in the 3' to 5' direction.

As for average distance, assuming a GC content of 50%, I agree with your answer.