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. 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?

Respuesta :

Answer:

[tex]1 - (\frac{61}{64})^n[/tex]

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

[tex]\frac{61}{64}[/tex]

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

[tex](\frac{61}{64})^x\\[/tex]

Probability of choosing a stop codon is equal to

[tex]1 -[/tex] probability of choosing a non stop codon

[tex]1 - (\frac{61}{64})^n[/tex]

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