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
= [tex]\frac{30}{100}[/tex]
= [tex]\frac{3}{10}[/tex]
The Percentage of A = 30
=[tex]\frac{30}{100}[/tex]
=[tex]\frac{3}{10}[/tex]
The percentage of G = 20
= [tex]\frac{20}{100}[/tex]
=[tex]\frac{2}{10}[/tex]
The percentage of C = 20;
= [tex]\frac{20}{100}[/tex]
=[tex]\frac{2}{10}[/tex]
The nonstop codons are ATT, ATC, ACT
So, The probability of ATT
= [tex]\frac{3}{10}\times \frac{3}{10}\times \frac{3}{10}[/tex]
= [tex]\frac{27}{1000}[/tex]
= 0.027
So, The probability of ATC
= [tex]\frac{3}{10}\times \frac{3}{10}\times\frac{2}{10}[/tex]
=[tex]\frac{18}{1000}[/tex]
= 0.018
So, The probability of ACT
= [tex]\frac{3}{10}\times\frac{2}{10}\times \frac{3}{10}[/tex]
= [tex]\frac{18}{1000}[/tex]
= 0.018
So, the answer is 0.027+0.018+0.018 = 0.063
