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Plant Physiology - Transport in Plants (Diffusion, Osmosis, Transpiration, Phloem Transport)

Grade 11ICSEBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Diffusion is the passive movement of individual molecules from a region of higher concentration to a region of lower concentration, occurring until equilibrium is reached. It is influenced by the concentration gradient, temperature, and pressure.

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Osmosis is a special type of diffusion involving the movement of solvent molecules (usually H2OH_2O) through a semi-permeable membrane from a region of higher solvent chemical potential to lower chemical potential.

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Water Potential (Ψw\Psi_w) is the potential energy of water per unit volume relative to pure water. Pure water has the highest potential, defined as 00 MPaMPa. Adding solutes decreases Ψw\Psi_w, making it negative.

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Plasmolysis occurs when a plant cell is placed in a hypertonic solution, causing water to move out and the protoplast to shrink away from the cell wall. The cell becomes flaccid.

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Transpiration is the loss of water in the form of water vapor from the aerial parts of the plant, primarily through stomata. It creates a 'Transpiration Pull' that facilitates the ascent of xylem sap.

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The Cohesion-Tension Theory explains water movement in xylem through the cohesive force between H2OH_2O molecules, adhesive forces between water and xylem walls, and surface tension.

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Phloem Transport involves the translocation of organic solutes (mainly sucrose) from 'Source' (leaves) to 'Sink' (roots/storage organs) via the Mass Flow Hypothesis or Pressure Flow Hypothesis.

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Active Transport involves the movement of molecules against a concentration gradient using energy in the form of ATPATP and specific membrane proteins.

📐Formulae

Ψw=Ψs+Ψp\Psi_w = \Psi_s + \Psi_p

DPD=OP−TPDPD = OP - TP

Rate of Diffusion∝1Density\text{Rate of Diffusion} \propto \frac{1}{\sqrt{\text{Density}}}

Ψw=Water Potential,Ψs=Solute Potential,Ψp=Pressure Potential\Psi_w = \text{Water Potential}, \Psi_s = \text{Solute Potential}, \Psi_p = \text{Pressure Potential}

💡Examples

Problem 1:

Calculate the water potential (Ψw\Psi_w) of a plant cell if its solute potential (Ψs\Psi_s) is −15-15 barsbars and its pressure potential (Ψp\Psi_p) is 55 barsbars.

Solution:

Ψw=Ψs+Ψp\Psi_w = \Psi_s + \Psi_p Ψw=−15+5=−10 bars\Psi_w = -15 + 5 = -10\text{ bars}

Explanation:

The water potential is the algebraic sum of the solute potential and the pressure potential. Since solute potential is always negative in a solution, it reduces the overall water potential.

Problem 2:

A cell AA with OP=10OP = 10 atmatm and TP=5TP = 5 atmatm is in contact with cell BB having OP=15OP = 15 atmatm and TP=12TP = 12 atmatm. In which direction will water flow?

Solution:

DPDA=OPA−TPA=10−5=5 atmDPD_A = OP_A - TP_A = 10 - 5 = 5\text{ atm} DPDB=OPB−TPB=15−12=3 atmDPD_B = OP_B - TP_B = 15 - 12 = 3\text{ atm} Water flows from BB to AA.

Explanation:

Water always moves from a region of lower DPDDPD (Diffusion Pressure Deficit) to a region of higher DPDDPD. Since Cell BB has a lower DPDDPD (33 atmatm) compared to Cell AA (55 atmatm), water moves from BB to AA.

Problem 3:

Explain the state of a plant cell placed in a 10%10\% NaClNaCl solution.

Solution:

The cell will undergo plasmolysis. H2OH_2O will move out via exosmosis.

Explanation:

A 10%10\% NaClNaCl solution is hypertonic relative to the cell sap. This creates a water potential gradient where Ψw(outside)<Ψw(inside)\Psi_{w(\text{outside})} < \Psi_{w(\text{inside})}, causing water to leave the cell and the plasma membrane to retract.