During Pichia pastoris yeast cultivation kLa was defined using a dynamic method. The air supply was shut off momentarily and the dissolved oxygen concentration was allowed to decrease $($however$,$ not below the critical limit$).$ After a while," aeration was continued, with the dissolved oxygen DO rising as follows:

At $6$ seconds DO$=50\%,$

At $10$ seconds DO$=56\mathrm{.}1\%,$

At $25$ seconds DO$=63\mathrm{.}0\%$

At $40$ seconds DO$=64\mathrm{.}7\%$

of saturated amount. The equilibrium concentration of dissolved oxygen $68\mathrm{.}4\%$ was reached after the oxygen supply was equalized $($and was same also before the air supply was shut off$).$ Calculate kLa using the first $(6$s$)$ and last value $(40$s$)$ of the measurement, as was done in the calculation exercises. Give the answers to three decimal places and calculate the numerical answer in $1/$s$.$

Question $1$: What is kLa calculated from the measured endpoints?

Question $2$: Your task is now to calculate kLa using all measurement points from the slope of the graph $($the graph should be forced through $0\mathrm{.}0$ origin, which here corresponds to the moment t$1=6$ seconds$).$ Note! when calculating the line to be fitted, the starting point in the evaluation of all the different intervals is always t$1=6$s $($t$2$ and the measured values change$).$ Textbook pages $417-419$ for help. Give the answers to three decimal places and calculate the numerical answer in $1/$s$.$What is kLa calculated from all measurement points?