Posted by Anonymous on .
For an AST determination, 0.2 mL of serum was used. The change in absorbance (ΔA) was measured at 340 nm with a 3 mL reaction mixture. The initial reading was 0.54 A, the reading after 5 minutes was 0.32 A. Report the AST value in IU.
Equation to use when time is other than 1 min.;
IU/L = ΔA/min x Vt/6.22 x Vs x 1000 x 1/T
Vt is the total volume and Vs is the volume of serum
My math;
= 0.22/5 min x 3.2 / 6.22 x 0.2 x 1000 x 1/5
= 22.6 IU
Do I divide the change in absorbance by the time of 5 min. or is min. just listed as is per the equation? I'm a little confused now on that. And for total volume, would you add together the volume of serum used and the 3mL reaction mixture or would it just be the volume of the reaction mixture to take into account?

Chemistry/lab math 
DrBob222,
The equation is delta A/minute and since your measurement is for 5 min I think you divide by 5 to get per min. As for the 3.2 = Vt, I don't think that is right. The problems states "3 mL reaction mixture" which means the total volume is 3 mL and not 3.2.