Classical formula for filtered samples
The classical formula can be found here or here
$$Pvol = DIC \cdot \frac{DPMvol}{DPMadded} \cdot 1.05 \cdot 1000$$
and plugging the value for $DPMadded$ (see below):
$$Pvol = DIC \cdot \frac{DPMvol}{SA \cdot V} \cdot 1.05 \cdot 1000$$
where:
$Pvol$ = Primary Productivity per unit volume in $mgC \cdot m^{-3} \cdot d^{-1}$
$DIC$ = Dissolved inorganic carbon in $mgC \cdot L^{-1}$
$DPMvol$ = DPM measured during 24 h for volume $V$ (filtered)
$DPMadded$ = $SA \cdot V$
$SA$ = Standard activity in 1 mL (DPM per mL)
$V$ = Volume of incubation (mL)
$1.05$ is to account for differential uptake between $^{14}$C and $^{12}$C
$1000$ is to go from from $L^{-1}$ to $m^{-3}$ since DIC is in $mgC \cdot L^{-1}$
For cells sorted by flow cytometry
Let us define:
$DPMcells$ = DPM measured during 24 h for $N$ cells
$N$ = number of cells sorted
$C$ = cell concentration per mL
$Pcell$ = Primary productivity per cell ($fgC \cdot cell^{-1} \cdot h^{-1}$)
$Pcell$ can be computed from $Pvol$ by dividing it by the total number of cells in 1 $m^{-3}$:
$$Pcell = Pvol \cdot \frac{1}{C \cdot 10^{6} } \cdot \frac{1}{24} \cdot 10^{12}$$
$10^{12}$ is to convert from milligrams to femtograms
${24}$ is to convert from day to hours
$C \cdot 10^{6}$ is the number of cells per $m^{-3}$
Which simplifies to:
$$Pcell = Pvol \cdot \frac{1}{C} \cdot \frac{1}{24}\cdot 10^{6}$$
We now replace $Pvol$ by the formula provived at the top :
$$Pcell = DIC \cdot \frac{DPMvol}{SA \cdot V} \cdot 1.05 \cdot 1000 \cdot \frac{1}{C} \cdot \frac{1}{24} \cdot 10^{6}$$
Simplifies to :
$$Pcell = DIC \cdot \frac{DPMvol}{SA \cdot V \cdot C} \cdot 1.05 \cdot \frac{1}{24} \cdot 10^{9}$$
$DPMvol$ can be computed from the individual DPM per cell ($DPMcells/N$) as:
$$DPMvol = \frac{DPMcells \cdot C \cdot V }{N}$$
So that:
$$Pcell = DIC \cdot \frac{DPMcells \cdot C \cdot V}{SA \cdot V \cdot C \cdot N} \cdot 1.05 \cdot \frac{1}{24} \cdot 10^{9}$$
which simplifies finally to:
$$Pcell = DIC \cdot \frac{DPMcells}{SA \cdot N \cdot 24} \cdot 1.05 \cdot 10^{9}$$
With contributions from Adriana Lopes dos Santos (ASE, NTU) and Andres Gutierrez-Rodriguez (NIWA)