Bacterial cell walls are quite rigid and can withstand a pressure of 100 atm. Ca

Question

Bacterial cell walls are quite rigid and can withstand a pressure of 100 atm.

Calculate the turgor pressure generated across a cell wall at 298K if the bacterial cytoplasm has a concentration of 0.6 M and it is immersed in a hypotonic solution of with a solute concentration 0.1 M. Assume that the cell walls are permeable to water, but not to these solutes (which are sugars, macromolecules,etc..)

Will this turgor pressure be sufficient to rupture the bacterial cell walls? Note: Turgor pressure is the difference in osmotic pressure inside the cell vs outside the cell.

( Hint: R = 8.314 J/ mol K = 0.082 L atm/mol K)

Explanation / Answer

Pturgor = P1-P2

P1 = M1*RT1

P2 = M2*RT2

since T2 = T1 then

(P1-P2) = (M1-M2)*RT

(P1-P2) = (0.6-0.1)*0.082*298

(P1-P2) =12.218 atm

it is not sufficient sicne

dTurgor < 100 atm

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