A marine alga has an internal osmotic pressure of 0.39 MPa. Seawater has an osmo
ID: 218671 • Letter: A
Question
A marine alga has an internal osmotic pressure of 0.39 MPa. Seawater has an osmotic pressure of 0.24 MPa.
a. If the alga is at equilibrium, what is its turgor?
b. If, due to heavy rain, seawater is diluted by 5%, what is the new external osmotic pressure? When the cell comes to equilibrium, what is the new turgor pressure? What is the percentage change in turgor pressure?
c. If, due to evaporation on a shallow reef flat, the original seawater is concentrated by 5%, what is the new external osmotic pressure? When the cell comes to equilibrium, what is the new turgor pressure? What is the percentage change in turgor pressure?
Explanation / Answer
We need to use the water potential equation. (Consider some factors as matric potential and gravitational potential to be negligible). We will have two equations (for inside and outside of the alga)
Internal ?w = ?o + ?t and external: ?w = ?o + ?t
water potential = osmotic potential + turgor potential
At equilibrium there is no further net influx of water by osmosis and water potentials on both sides of the membrane are equal. Then we have:
?o + ?t (I) = ?o + ?t (E)
?t= ?o (E) - ?o (I)
?t = -0.24 MPa. - (-0.39 MPa) (negative because they are potentials)
Turgor pressure: 0.63 MPa
b)( 5% x 0.24MPa)/ 100% = 0.012
0.24 MPa - 0.012 MPa = 0.228 MPa is the new external osmotic pressure.
?t = -0.228 MPa. - (-0.39 MPa) = 0.162 MPa is the new turgor pressure.
(0.162 x 100 %)/ 0.63 = 25,7 % , 100% - 25.7 % = 74.3 % is the percentage change in turgor pressure.
c) If seawater is concentrated by 5 %
( 5% x 0.24MPa)/ 100% = 0.012
0.24 MPa + 0.012 MPa = 0.252 MPa is the new external osmotic pressure.
new turgor pressure?
?t = - 0.252MPa. - (-0.39 MPa) = 0.138 MPa is the new turgor pressure.
(0.138 x 100 %)/ 0.63 = 21.90 %
100% - 21.90 % = 78 % is the percentage change in turgor pressure.
Hope it helps!
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