Question

A genome has a GC content of 40%. Assuming the sequence is IID, what is the probability that three random consecutive nucleotides form a stop codon?
Edit: IID means independently identically distributed.

Answer 1

Answer:

The probability of three codons will be 0.063 and the nonstop codons are ATT, ATC, ACT .

Explanation:  

Given:

The genome content = 40%  

To find:

Probability of three random consecutive nucleotide to form a stop codon.

Solution:

The Percentage of AT+ percentage  of GC = 100.

The Percentage of AT = 100-40 = 60.             [GC is given 40]

The Percentage of T = 30

= $\frac{30}{100}$

= $\frac{3}{10}$

The Percentage of A = 30

 =$\frac{30}{100}$

  =$\frac{3}{10}$

The percentage  of G = 20

= $\frac{20}{100}$

=$\frac{2}{10}$

The percentage  of C = 20;

= $\frac{20}{100}$

=$\frac{2}{10}$

The nonstop codons are ATT, ATC, ACT

So, The probability of ATT

=  $\frac{3}{10}\times \frac{3}{10}\times \frac{3}{10}$

= $\frac{27}{1000}$

= 0.027

So, The probability of ATC

= $\frac{3}{10}\times \frac{3}{10}\times\frac{2}{10}$

=$\frac{18}{1000}$

= 0.018

So, The probability of ACT

=  $\frac{3}{10}\times\frac{2}{10}\times \frac{3}{10}$

= $\frac{18}{1000}$

= 0.018

So, the answer is 0.027+0.018+0.018 = 0.063