. The sum of all amino acids should be 1. The sum of all codons is 1. However, S
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Question
. The sum of all amino acids should be 1. The sum of all codons is 1. However, STOP is not a valid amino acid. Yet we should still be able to calculate the probability of any amino acid from the nt frequencies, by proper normalization. In this case, we normalize by the total probability that leads to codons. So P(amino acid) = Pall codons for the amino acid) / P(all codons that are valid amino acids). Assume P(A) = 0.3, P = 0.3, P(C)=0.2, and P(G)= 0.2 (4pts) What is the P(all codons that are valid amino acids)? Using proper normalization, what is the probability of the following amino acids? (3pts) lle (I) (3pts) Trp (W)Explanation / Answer
Here we can calculate using the following premise,
we have total 4 nucleotides which can produce = 43=64 codons out of which 3 stops codon and those are ACT, ATT and ATC (UGA UAA UAG in mRNA)
the probability of all possible codon= 64/64 =1
P(all codon that are valid amino acid) + P(stop codon) = 1
therefore, P(all codon that are valid amino acid) =1- P(stop codon) = 1 - [P(ACT) + P(ATT) + P(ATC)]
Now P(ACT)= P(A) x P(C).x P(T) =0.3x0.2x0.3= 0.018
P(ATT)= P(A) x P(T) x P(T) = 0.3x03x0.3= 0.027
P(ATC)=P(A) x P(T) x P(C) = 0.3x0.3x0.2=0.018
Therefore, p(all codon that are valid amino acid)=1 - (0.018+0.027+0.018)= 0.937
codon for leucin are AAT, AAC, GAA, GAG, GAT, GAC (UUA, UUG, CUU, CUC, CUA, CUG in mRNA)
therefore, p(leucin) = sum of probability of individual codon
=P(AAT)+ P(AAC)+ P(GAA)+P(GAG)+P(GAT)+P(GAC)
= P(A)P(A)P(T)+P(A)P(A)P(T)+P(G)P(A)P(T)+P(G)P(A)P(G)+P(G)P(A)P(T)+P(G)P(A)P(C)
=(0.3x0.3x0.3)+(0.3x0.3x0.3)+(0.2x0.3x0.3)+(0.2x0.3x0.2)+(0.2x0.3x0.3)+(0.2x0.3x0.2)
=0.027+0.027+0.018+0.012+0.018+0.012 = 0.114
codon for tryptophane is ACC ( UGG in mRNA)
therefore, p(Tryptophan)= P(A)P(C)P(C) = 0.3x0.2x0.2=0.012
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