jonrkarr
commented
6 years ago Great question.
This is a lumped law. Classically, RNA degradation is thought of as first order in RNA. Because the copy numbers are small, I think it should be first order in RNA. Then logistic in the RNA degradation machinery. This differs from Michaelis-Menten. This is fine. The biochemistry of the reaction is not the same as Michaelis-Menten chemistry. Thus, there is no reason to expect the rate law would be the same.
The basal rate of degradation is fine. Keep in mind that with the basal rate, kcat will not be log(2)/t{1/2}. Rather, the sum of the rates of all mechanisms of RNA degradation should be consistent with the observed RNA half life.
On Sat, Aug 25, 2018, 2:15 PM artgoldberg notifications@github.com wrote:
the RNA degradation rate law is generated by rl.equation = wc_lang.RateLawEquation( expression='{0}[c] (((k_cat {1}) / (k_m + {1})) + {2})'.format(rna.id, deg_rnase.id(), '0.1'))
the result has the form "rna_tu_1_x[c] (((k_cat prot_gene_1_98[c]) / (k_m + prot_gene_1_98[c])) + 0.1)", for various values of '_x'. i suspect that the RNA should be treated as the substrate, and the RNAse (=prot_gene_1_98[c]) as the enzyme, which would be achieved by swapping them, but I'm not an expert in this area. the protein degradation is similar.
Arthur
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the RNA degradation rate law is generated by rl.equation = wc_lang.RateLawEquation( expression='{0}[c] (((k_cat {1}) / (k_m + {1})) + {2})'.format(rna.id, deg_rnase.id(), '0.1'))
the result has the form "rna_tu_1_x[c] (((k_cat prot_gene_1_98[c]) / (k_m + prot_gene_1_98[c])) + 0.1)", for various values of '_x'. i suspect that the RNA should be treated as the substrate, and the RNAse (=prot_gene_1_98[c]) as the enzyme, which would be achieved by swapping them, but I'm not an expert in this area. the protein degradation is similar.
Arthur