Question

Consider a ribosome translating a codon to an amino acid. Compared to a perfect pairing, imagine that a codon-anticodon pairing with one incorrect base pair is 10 kJ•mol-1 higher in energy, two incorrect base pairs is 15 kJ•mol-1 higher, and three incorrect base pair is 20 kJ•mol-1 higher. What percent of decoding events occur without mismatches at 300 K?

Answer 1

Answer: There are 9 ways to make a single base mismatch, 27 ways to make a two base pair mismatch and, again, 27 ways to make a three base pair mismatch (and only 1 way to do it correctly). Q = e − U i k b T i ∑ = e − U zeromismatches k b T + 9 × e − U onemismatche k b T + 27 × e − U twomismatches k b T + 27 × e − U threemismatches k b T = e 0 k b T + 9 × e − 10 kJ / mol k b T + 27 × e − 20 kJ / mol k b T 27 × e − 30 kJ / mol k b T = 1 + 9 × (0.018) + 27 × (0.00033) + 27 × (0.000006) = 1 + (0.163) + (0.0089) + (0.00016) =1.17 p 0 = e − U 0 k b T Q = e − 0 k b T 1.17 = 1 1.17 = 0.853 => 85.3% Therefore, approximately 85.3% of all amino acids are translated faithfully. The redundancy built into the genetic code and other energetic considerations raise this number considerably.

Step-by-step explanation:

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