I have no idea if this would be the right place to post. I am a biologist by trade so mathematics is not my forte.
I am having problems calculating the degradation/synthesis constants and protein half-life in a protein turnover experiment using metabolic labeling. I apologize if something similar has been answered already, I have tried to find a answer but to no avail.
The experiment: I have obtained caluculated ratio's (r) from a temporal protein turnover pulse labeling experiment where heavy isotopes of an amino acid is added during treatment conditions and incorporated into the proteome at various time points. The ratio's are part of the output from the software used to search and quatify peptides. The heavy isotopes are incorporating therefore the ratio's display a positive slope instead of a negative. I tried $\frac{1}{r}$ transformation but when accounting for dilution I have the same problem as detailed below.For data analysis, this is the paper I am working from: https://www.nature.com/articles/nature10098#supplementary-information.
$$ r = \frac{P_H}{P_L} $$
I calculate the rate constant ($K_H$) with the following equation.
$$ K_H=\frac{\sum_{i=0}^m ln(r_{t_i}+1)}{\sum_{i=0}^m t_i^2} $$
Where $m$ is the amount of time points, $t_i$ is each specific time point and $r_{t_i}$ is the the ratio at each time point. I have cross referenced this with the excel function $Linest$ which returns the same value. Up until here I have no problems.
In order to estimate the correct turnover we have to account for dilution of heavy amino acids ($K_{dill}$) occurring as a natural process of cell division and subtract that from the rate constant (${K_H}$). The cell cycle ($t_{cc}$) in hours is pre-determined.
$$ K_{dill} = \frac{Ln(2)}{t_{cc}} $$ $$ K_{dp} = K_H - K_{dill} $$
The problem is when I do this step I end up with a set of both negative and positive values for $K_{dp}$. The negative values result in a negative half-lives $(t_{\frac{1}{2}})$ as calculated in the next equation. The negative values are due to $K_H < K_{dill}$
$$ t_{\frac{1}{2}} = \frac{ln(2)}{K_{dp}} $$
However if I plug in my positive $K_{dp}$ values in the normal half-life equation and solve $Ln(0.5)$ I get the correct answer, therfore I am completely lost.
$$ \frac{N_0}{2} = e^{(-K_{dp}){(t_\frac{1}{2})}} $$
$$ Ln(0.5) = (-K_{dp})(t_{1/2}) $$
This can't be right since half life should not drop below zero. Somewhere I am missing soemthing but I cant figure out where I am going wrong. Alternatively, this is correct and any negative values can be assumed to be in steady state as they should not be allowed to drop below zero in the first place, where we can assume $K_{dp} = 0$.
Any help would be greatly appreciated. Thanks!