Question

Now, let’s assume that the distribution of codons is not uniformly random. We also assume that only 5 codons {C1,C2,C3,C4,C5} appear in a dataset with a set of 10 of observations of codon {X1=C2, X2=C1, X3=C1, X4=C4, X5=C1, X6=C3, X7=C5, X8 =C2, X9=C4, X10=C3}. Compute the probability distribution of the five codons. If C2 is a start codon and C3 is a stop codon, with the distribution you just computed, what is the probability that a start codon (C2) is followed by 20 successive triplets of nucleotides none of which is a stop codon (C3)?

Answer 1

First, we will need to calculate probability of each codon on the basis of given data. Then we will need to calculate probability of getting start codon followed by 20 successive triplets (codons) which are non-stop codons

On the basis of 10 observations, calculate frequency of each codon in data set

Frequency of Codon 1 = 3/10 = 0.3 is probability for Codon 1

Frequency of Codon 2 = 2/10 = 0.2 is probability for Codon 2

Frequency of Codon 3 = 2/10 = 0.2 is probability for Codon 3

Frequency of Codon 4 = 2/10 = 0.2 is probability for Codon 4

Frequency of Codon 5 = 1/10 = 0.1 is probability for Codon 5

Note: p (codon 1) + p (codon 2) + p (codon 3) + p (codon 4) + p (codon 5) = 0.3 + 0.2 + 0.2 + 0.2 + 0.1 = 1

Genetic code is universal, triplet AUG is a start codon which also codes for amino acid methionine.

Probability of getting non-stop codon = 1- p(codon3) = 1- 0.2 = 0.8

Probability of getting start codon is = p(codon 2) = 0.2

Probability of 1^st non-stop codon is = 0.8

Probability of 1^st and 2^nd non-stop codon is = 0.8 X 0.8 = (0.8)²

Note: addition of each successive non-stop codon will be independent of earlier triplet (codon)

So, probability of getting 20 successive non-stop codons will be

probability = (0.8)²⁰ = 0.0115

Probability of getting start codon followed by 20 successive non-stop codons will be = (0.2) X (0.8)²⁰ = (0.2) X (0.0115) =0.0023

Answer: Probability that a start codon (C2) is followed by 20 successive triplets of nucleotides none of which is a stop codon (C3) will be 0.0023

Did this help? If yes, do give positive rating for this answer. Thank you !