If there's 4 nucleotides, what is the probability to get a start codon ?
If I understand correctly; a codon = 3 nucleotides, so there's 2 possible combinations with 4 nucleotides (AAAX and XAAA), of the 64 possible codons, there's only 1 start codon. So 2 possible combinations * 1/64 = 1/32. But I'm not sure about this.
Your understanding is almost correct, but let me explain it in a bit more detail to clarify.
Given that there are 4 nucleotides (A, C, G, and T), each codon is made up of 3 nucleotides. So, the number of possible codon combinations is calculated as 4^3 = 64.
However, not all 64 codons are unique, as some codons code for the same amino acid. Among these 64 codons, there is indeed only one specific start codon, which is AUG (where A refers to adenine, U refers to uracil, and G refers to guanine).
So, your calculation of 2 possible combinations out of 64 codons is slightly incorrect because it assumes that the only combinations with 4 nucleotides are AAAX and XAAA. In reality, there are more possibilities, such as AXAX, AXAA, etc. But since we are looking for the probability of getting the start codon, the number of possible combinations is not relevant.
Therefore, the probability of randomly selecting the start codon (AUG) out of all possible codons is indeed 1/64. So, in your case, the probability would be 1/64.
Remember that this calculation assumes random selection, but in actual genetic sequences, there are other factors involved, such as specific regions where start codons are more likely to occur.