$\mathrm{k}=\frac{1}{60}\ln \frac{{\mathrm{A}}_0}{{\mathrm{A}}_{60}}$	→	initial absorbance
	→	absorbance at 60 s
${\mathrm{k}}_{\mathrm{total}/\mathrm{ml}}=\mathrm{k}\times \frac{6}{\mathrm{ml\; sample}}$	→	total mls in reaction; both blank and sample
	→	total mls of sample; both blank and sample (example if 20 μl of sample was added before splitting the sample in half, denominator = 0.02 ml)
$\mathrm{k}/\mathrm{mg}=\frac{{\mathrm{k}}_{\mathrm{total}/\mathrm{ml}}}{\mathrm{mg}{\mathrm{ml}}^{­1}}$	→	divide by the protein concentration of the original sample