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?

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.

To calculate the AST value in IU, you need to use the given equation:

IU/L = ΔA/min x Vt/6.22 x Vs x 1000 x 1/T

Let's break down and substitute the values into the equation:

ΔA/min = (0.54 A - 0.32 A) / 5 min = 0.044 A/min

Vt = 3 mL (volume of reaction mixture)
Vs = 0.2 mL (volume of serum)

Substituting these values into the equation:

IU/L = 0.044 A/min x 3 mL / 6.22 x 0.2 mL x 1000 x 1/5

Now, simplify the equation:

IU/L = 0.044/5 x 3/6.22 x 1000

Calculating this expression will give you the AST value in IU.

To calculate the AST value in IU, you correctly used the equation for when the time is other than 1 minute. The change in absorbance is divided by the time in minutes, which in this case is 5 minutes.

Regarding the total volume (Vt), it should include both the volume of serum used (Vs) and the volume of the reaction mixture (3 mL). So, in this case, Vt would be 3.2 mL (0.2 mL + 3 mL).

Now let's put all the values into the equation:

IU/L = ΔA/min x Vt/6.22 x Vs x 1000 x 1/T

IU/L = 0.22/5 min x 3.2 mL / 6.22 x 0.2 mL x 1000 x 1/5

Simplifying the equation:

IU/L = 0.044 x 3.2 / 6.22 x 0.2 x 1000

Calculating further:

IU/L = 0.044 x 16 / 12.44

IU/L = 0.0561

Therefore, the AST value is approximately 0.0561 IU/L.

It is important to note that these calculations are based on the given equation and assumptions. Always make sure to verify if the equation and method are appropriate for the specific laboratory assay being performed.