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Q.
What would be $\Psi_{p}$ of cell sap in cell-$A$ and cell-$B $ respectively?
Transport in Plants
Solution:
Cell-$A$
$\Psi_{ w }=-900 \,kPa$
$\Psi_{ s }=-1200\, kPa$
$\Psi_{ w }=\Psi_{ s }+\Psi_{ p }$
$ \Psi_{ p } =\Psi_{ w }-\Psi_{ s } $
$=-900-(-1200) $
$=-900+1200 $
$ \Psi_{ p } =+300\, kPa$
Cell-$B$
$\Psi_{ w }=-600\, kPa $
$\Psi_{ s }=-1000 \,kPa $
$\Psi_{ w }=\Psi_{ s }+\Psi_{ p }$
$ \therefore \Psi_{ p } =\Psi_{ w }-\Psi_{ s } $
$=-600-(-1000) $
$=-600+1000) $
$ \Psi_{ p } =400 \,kPa $
