Question

. Given that 3 of the 64 possible codons are stop codons, what is the chance of having a stop codon at any given position, assuming that the sequence is random?

Answer 1

Answer:

$1 - (\frac{61}{64})^n$

Explanation:

Given

There are three stop codon

TAA, TAG, TGA.

Out of 64 available codons, 3 are stop codon thus the remaining are non-stop codon.

So the probability of choosing a non stop codon is

$\frac{61}{64}$

Let us suppose the number of trials are "x" then the probability of choosing a non stop codon is  

$(\frac{61}{64})^x$

Probability of choosing a stop codon is equal to

$1 -$ probability of choosing a non stop codon

$1 - (\frac{61}{64})^n$