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?

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.

Explanation:

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radioactive decay occurs in unstable atomic nuclei – that is, ones that don't have enough binding energy to hold

the nucleus together due to an excess of either protons or neutrons.

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

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