Agenome 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.

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

=

=

The Percentage of A = 30

 =

  =

The percentage  of G = 20

=

=

The percentage  of C = 20;

=

=

The nonstop codons are ATT, ATC, ACT

So, The probability of ATT

=  

=

= 0.027

So, The probability of ATC

=

=

= 0.018

So, The probability of ACT

=  

=

= 0.018

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

c is the answer : )               your welcome

cuz he is really really short, assuming an unorthodox and inherently unlucky configuration of elevator buttons…

explanation: