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

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

=
`(30)/(100)`

=
`(3)/(10)`

The Percentage of A = 30

=
`(30)/(100)`

=
`(3)/(10)`

The percentage of G = 20

=
`(20)/(100)`

=
`(2)/(10)`

The percentage of C = 20;

=
`(20)/(100)`

=
`(2)/(10)`

The nonstop codons are ATT, ATC, ACT

So, The probability of ATT

=
`(3)/(10)* (3)/(10)* (3)/(10)`

=
`(27)/(1000)`

= 0.027

So, The probability of ATC

=
`(3)/(10)* (3)/(10)*(2)/(10)`

=
`(18)/(1000)`

= 0.018

So, The probability of ACT

=
`(3)/(10)*(2)/(10)* (3)/(10)`

=
`(18)/(1000)`

= 0.018

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