Diffusion

Passive Transport and Diffusion

• Diffusion: The passive net movement of molecules down their concentration gradient, moving from an area of higher concentration to an area of lower concentration driven by random thermal motion.

• Selective Permeability: Cellular membranes consist of a phospholipid bilayer that is selectively permeable.

  • Small, nonpolar molecules (such as $O_2$ and $CO_2$) cross freely via simple diffusion.

  • Large, polar, or charged species (such as $Na^+$ or glucose) require channel or carrier proteins via facilitated diffusion.

Osmosis and Solution Tonicity

• Osmosis: The net diffusion of free water molecules across a selectively permeable membrane from a region of higher water potential (dilute solute concentration) to a region of lower water potential (concentrated solute concentration).

• Toxicity and Solution Types:

  1. Hypertonic Solution

  • Has a higher solute concentration than the cytoplasm.

  • Net movement of water is out of the cell.

  • Plant cells become plasmolyzed (cytoplasm pulls away from cell wall).

  • Animal cells become crenated (shriveled).

  1. Hypotonic Solution

  • Has a lower solute concentration than the cytoplasm.

  • Net movement of water is into the cell.

  • Plant cells develop turgor pressure against the cell wall and become turgid.

  • Animal cells lack a rigid cell wall and may undergo lysis (bursting).

  1. Isotonic Solution

  • Has an equal solute concentration relative to the cytoplasm.

  • No net water movement occurs (dynamic equilibrium).

Water Potential Calculations

• Water Potential Equation:   $\Psi = \Psi_P + \Psi_S$

  • Water flows spontaneously from an area of higher water potential to an area of lower water potential.

  • In an open container, the pressure potential ($\Psi_P$) is $0\,bars$.

• Solute Potential Equation:   $\Psi_S = -iCRT$

  • $i$: Ionization constant (e.g., $1.0$ for sucrose or glucose, $2.0$ for $NaCl$).

  • $C$: Molar concentration ($M$).

  • $R$: Pressure constant ($0.0831\,L\,bar\,mol^{-1}\,K^{-1}$).

  • $T$: Temperature in Kelvin ($K = ^\circ C + 273$).

AP Biology Practice Questions: Diffusion, Osmosis, and Water Potential Question 1: A plant cell with a solute potential ($\Psi_S$) of $-7.5\,bars$ is placed in an open beaker containing a solution with a solute potential ($\Psi_S$) of $-4.0\,bars$. The cell reaches dynamic equilibrium with the solution without changing volume. What is the pressure potential ($\Psi_P$) of the plant cell at equilibrium? Answer 1: In an open beaker, the pressure potential of the solution is $0\,bars$, so $\Psi_{\text{solution}} = 0 + (-4.0) = -4.0\,bars$. At equilibrium, the water potential of the cell equals that of the solution ($\Psi_{\text{cell}} = -4.0\,bars$). Using $\Psi = \Psi_P + \Psi_S$: $-4.0\,bars = \Psi_P + (-7.5\,bars)$, which yields $\Psi_P = 3.5\,bars$. Question 2: Calculate the solute potential ($\Psi_S$) of a $0.2\,M$ sucrose solution at $21^\circ C$ in an open container ($R = 0.0831\,L\,bar\,mol^{-1}\,K^{-1}$). Answer 2: First, convert temperature to Kelvin: $T = 21 + 273 = 294\,K$. For sucrose, the ionization constant $i = 1.0$. Using $\Psi_S = -iCRT$: $\Psi_S = -(1.0)(0.2\,M)(0.0831\,L\,bar\,mol^{-1}\,K^{-1})(294\,K) = -4.89\,bars$. Question 3: Compare what happens to an animal cell and a plant cell when placed in a hypotonic solution. Answer 3: In a hypotonic solution, water enters the cells via osmosis down the water potential gradient. An animal cell will swell and undergo lysis (burst) because it lacks a cell wall. In contrast, a plant cell will take in water until turgor pressure against its rigid cell wall prevents further net entry, causing the cell to become turgid without bursting.