Water Potential in a Root Hair
A root hair cell has a solute potential of −0.60 MPa and a pressure potential of +0.20 MPa. It sits in soil solution with a solute potential of −0.30 MPa and negligible pressure. The cell wall is rigid but slightly elastic. A student measures net water movement over 30 minutes and finds the cell's volume increased by 4%. The soil solution is then diluted with distilled water to a solute potential of −0.10 MPa, and the same cell is observed.
1. Calculate the water potential of the cell and of the soil solution in the first condition, and predict the direction of net water movement.
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Water potential ψ = ψs + ψp. Cell: ψ = (−0.60) + (+0.20) = −0.40 MPa. Soil: ψ = (−0.30) + 0 = −0.30 MPa. Water moves from higher (less negative) to lower ψ, so from soil (−0.30) into the cell (−0.40). This matches the observed 4% volume increase.
2. After dilution, predict how the cell's ψp and ψs change as it approaches equilibrium, and explain why the cell does not simply keep swelling indefinitely.
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Diluted soil ψ = −0.10 MPa, still higher than the cell's −0.40 MPa, so water continues to enter. As water enters, the elastic wall resists, raising ψp. Simultaneously, dilution of the cytoplasm makes ψs less negative (closer to zero). Both changes raise the cell's ψ toward −0.10 MPa. The cell stops swelling when ψcell = ψsoil; at that point ψp has increased enough (and ψs risen) that further influx has no driving gradient. The wall's elasticity, not an active pump, sets the limit.
3. A classmate says the cell is 'pumping water in against the gradient.' Critique this statement.
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The statement misstates the physics. Water moves down its water-potential gradient (from −0.10 to −0.40 MPa), not against it. No ATP-driven water pump is required; the gradient itself drives osmosis. What the cell does expend energy on is solute accumulation (e.g., ions, sugars), which lowers ψs and thereby creates the gradient. The wall provides the opposing pressure that eventually balances influx.