Show students two short scenarios side by side: (1) a marathon runner who drinks massive amounts of plain water and develops hyponatremia — brain cells swell and can herniate; (2) freshwater lettuce leaves that go limp in salty dressing at a restaurant. Ask: 'Same molecule — water — is doing damage in both cases. What's driving it, and why do the two cells respond differently?' Take 30 seconds of Turn-and-Talk, then collect 2-3 predictions on the board without confirming. Tell students that by the end of class they will be able to predict and CALCULATE which way water moves in each case. Targets SP 1 Concept Explanation.

1. 8m

   Osmosis and Tonicity — a Comparative Property

   Content

   Osmosis is the net diffusion of water across a selectively permeable membrane. The critical rule students must internalize: water moves from HIGHER water potential to LOWER water potential, which is the same as saying water moves toward the region of HIGHER solute concentration. Tonicity describes a solution relative to the cell inside it. A hypertonic solution has more non-penetrating solute than the cell, so water leaves the cell (animal cell shrivels/crenates; plant cell plasmolyzes). A hypotonic solution has less solute than the cell, so water enters (animal cell lyses; plant cell becomes turgid, protected by the rigid cell wall). An isotonic solution matches the cell's solute concentration — no net movement. Key point: tonicity is COMPARATIVE. A 0.3 M sucrose solution is hypertonic to a red blood cell (~0.15 M solute equivalent) but hypotonic to a seawater-adapted cell (~1 M). Same solution, different label, depending on the cell.

   Delivery

   Anchor everything to the phrase 'water follows solute.' Walk students through the three-panel visual (animal + plant cells in hypertonic, isotonic, hypotonic) — point out that the plant cell in hypotonic goes turgid, not lysed, because the cell wall pushes back. Pre-empt the two big misconceptions: (1) 'water moves toward higher water concentration' — reframe as 'higher Ψ to lower Ψ'; (2) 'tonicity is absolute' — ask them: 'Can 0.3 M sucrose be hypertonic AND hypotonic?' Yes — it depends on the cell. Cold call: 'A red blood cell in distilled water — what happens and why?' Expected: lyses, because distilled water has Ψ = 0, cell Ψ is negative, water rushes in.

2. 10m

   Water Potential: Ψ = Ψp + Ψs

   Content

   Water potential (Ψ) is the free energy of water. Pure water at atmospheric pressure and open to the air has Ψ = 0 bars. Adding solute or applying tension makes Ψ more negative; applying pressure makes it more positive. The governing equation is Ψ = Ψp + Ψs, where Ψp is pressure potential (0 in an open beaker, positive inside a turgid plant cell) and Ψs is solute potential (always ≤ 0 for a solution). Solute potential is calculated as Ψs = −iCRT, where i is the ionization constant (1 for sucrose, ~2 for NaCl), C is molar concentration, R = 0.0831 L·bar/(mol·K), and T is temperature in Kelvin (°C + 273). Worked example: What is Ψ of a 0.15 M sucrose solution in an open beaker at 25 °C? i = 1, C = 0.15, R = 0.0831, T = 298. Ψs = −(1)(0.15)(0.0831)(298) = −3.72 bars. Ψp = 0 (open beaker). So Ψ = 0 + (−3.72) = −3.72 bars. If a plant cell with Ψ = −5.0 bars sits in this solution, water moves from the solution (−3.72) INTO the cell (−5.0), because water flows from higher Ψ to lower Ψ.

   Delivery

   Write the equation on the board and unpack each variable slowly. Emphasize the sign convention — Ψs is negative and adding solute makes it MORE negative (this is the misconception target). Do the worked example fully with students, having them plug in each value at their desks. Ask 'which way does water move?' and force them to compare the two Ψ values numerically, NOT by tonicity words. Then flip the scenario: what if the cell is at Ψ = −2.0 bars? Water would leave the cell. Show that the tonicity label falls out of the numerical comparison — you don't need to name it if you can calculate it.

1. 20m

   Elodea Under the Microscope: Turgor vs. PlasmolysisLab

   Targets SP 2 Visual Representations, SP 4 Representing and Describing Data, and SP 6 Argumentation. Students prepare two wet mounts of Elodea and observe the SAME leaf as its bathing solution changes. Pairs rotate through microscopes at ×400. Teacher setup: Pre-fill dropper bottles. Have paper towels at every station. Warn students that 20% NaCl will damage the plant — this is expected and observable. Student handout: Elodea Osmosis Lab — Student Handout Part 1 — Wet mount in distilled water 1. Place one Elodea leaf on a slide in a drop of distilled water; add coverslip. 2. Observe at ×400. Locate chloroplasts along the edges of a single cell. 3. Sketch ONE cell in the box below. Label: cell wall, chloroplasts, central vacuole region. Sketch box (distilled water, ×400): [ draw here ] Part 2 — Perfuse with 20% NaCl 1. Place a small drop of 20% NaCl at ONE edge of the coverslip. 2. Touch a paper towel wedge to the OPPOSITE edge to wick the salt solution across the leaf. 3. Wait 2-3 minutes. Re-observe the SAME cell. 4. Sketch the same cell now. Sketch box (20% NaCl, ×400): [ draw here ] Part 3 — Analysis (answer in complete sentences) a. Describe what happened to the chloroplasts and the protoplast between Part 1 and Part 2. ______ b. In which direction did net water movement occur in Part 2? Justify using the words water potential, Ψs, and Ψp. ______ c. The Elodea cell's Ψ is approximately −6 bars. The 20% NaCl solution has Ψ ≈ −160 bars. Explain, using these numbers, why water moved the way it did. ______ d. An animal cell placed in 20% NaCl would shrivel (crenate) rather than plasmolyze. Explain the structural reason for the difference. ______ e. Argument (SP 6): A classmate claims 'The 20% NaCl killed the cell by pushing salt INTO it.' Write a 2-3 sentence rebuttal using evidence from your observation. ______ Safety: Do not taste any solution. Wash hands after handling specimens. Circulate and check: (1) students identify the SAME cell before and after, (2) sketches show chloroplasts pulled inward in Part 2 (plasmolysis), (3) analysis uses Ψ values, not just tonicity words.

   Materials

   • Compound microscope (one per pair)
   • Fresh Elodea (or red onion epidermis) sprigs
   • Microscope slides and coverslips
   • Distilled water in dropper bottles
   • 20% NaCl (or 1.0 M sucrose) solution in dropper bottles
   • Paper towel wedges (for wicking)
   • Student handout (below)
   • Colored pencils

   Example outputs

   • Part 3b sample: 'Water moved OUT of the Elodea cell. The 20% NaCl solution has a much more negative Ψs than the cell interior, so its overall Ψ is much lower than the cell's Ψ. Water moves from higher Ψ (inside the cell, ~−6 bars) to lower Ψ (the salt solution, ~−160 bars), causing the protoplast to shrink from the wall.'
   • Part 3d sample: 'Elodea cells have a rigid cellulose cell wall that holds the wall's shape even when the protoplast shrinks (plasmolysis). Animal cells lack a cell wall, so when they lose water in a hypertonic solution, the whole cell — plasma membrane and all — shrinks, producing a crenated shriveled appearance.'
   • Part 3e sample rebuttal: 'The NaCl did not enter the cell — the plasma membrane is selectively permeable and Na⁺ and Cl⁻ do not freely cross. What we observed is WATER leaving the cell down its water-potential gradient, causing the protoplast to pull away from the wall. Solute did not push in; water pulled out.'
2. 15m

   Water-Potential Problem Set (AP-Style Calculation)

   Targets SP 5 Statistical Tests and Data Analysis and SP 1 Concept Explanation. Students work in pairs on 3 short calculation/reasoning problems that mirror AP FRQ style. After 10 minutes, review problem 3 as a class. Student problem set: Problem 1 — Calculate Ψ A 0.4 M sucrose solution sits in an open beaker at 22 °C. - i = ______ (sucrose) - C = ______ M - R = 0.0831 L·bar/(mol·K) - T = ______ K - Ψp = ______ bars - Ψs = −iCRT = ______ bars - Ψ of the solution = ______ bars Problem 2 — Predict water movement A plant cell has an internal Ψ = −6.5 bars. It is placed in the 0.4 M sucrose solution from Problem 1. a. Compare the two Ψ values. Which is higher (less negative)? ______ b. Predict the direction of net water movement (into or out of the cell). Justify. ______ c. Is the sucrose solution hypertonic, hypotonic, or isotonic to this cell? Explain how you know from the numbers. ______ Problem 3 — AP-style data analysis (potato cores) A student soaked potato cores in sucrose solutions of increasing molarity and plotted percent mass change vs. molarity. The best-fit line crosses the x-axis at 0.30 M sucrose. The experiment was conducted at 27 °C in open beakers. a. What does the x-intercept physically represent? ______ b. Calculate the Ψ of the 0.30 M sucrose solution at 27 °C. ______ bars c. Therefore, what is the Ψ of the potato tissue? Justify. ______ bars d. If the same experiment were repeated at 37 °C using the same molarities, would the x-intercept shift, stay the same, or is more information needed? Explain in one sentence. ______ Circulate and check: students convert °C to K correctly, keep the negative sign on Ψs, and recognize that at the x-intercept Ψ(cell) = Ψ(solution).

   Materials

   • Student problem set (below)
   • Calculator

   Example outputs

   • Problem 1: i = 1; C = 0.4; T = 295 K; Ψp = 0; Ψs = −(1)(0.4)(0.0831)(295) = −9.81 bars; Ψ = −9.81 bars.
   • Problem 2: The cell's Ψ (−6.5) is higher (less negative) than the solution's Ψ (−9.81). Water moves OUT of the cell (from higher Ψ to lower Ψ). The solution is hypertonic to the cell because its Ψ is more negative — the cell loses water.
   • Problem 3: (a) At 0.30 M there is no net mass change, so Ψ(potato) = Ψ(solution). (b) Ψs = −(1)(0.30)(0.0831)(300) = −7.48 bars; Ψ = −7.48 bars. (c) Ψ(potato) ≈ −7.48 bars, because at the x-intercept the two water potentials are equal. (d) The x-intercept would shift slightly toward lower molarity — at higher T, a given molarity produces a more negative Ψs, so a less concentrated solution would now match the potato's Ψ (assuming Ψ of the potato itself doesn't change much).

1. A red blood cell (Ψ ≈ −7 bars) is placed in a 0.5 M NaCl solution in an open beaker at 25 °C. (a) Calculate the Ψ of the NaCl solution (use i = 2). (b) Predict what happens to the cell and justify using both Ψ values. (SP 5, SP 1)

   calculation(a) Ψs = −iCRT = −(2)(0.5)(0.0831)(298) = −24.76 bars. Ψp = 0. Ψ = −24.76 bars. (b) The solution's Ψ (−24.76 bars) is much lower than the cell's Ψ (−7 bars). Water moves from higher Ψ (inside the cell) to lower Ψ (the NaCl solution), so water exits the cell. The RBC crenates (shrivels) because it lacks a cell wall to resist the volume loss.

2. A student claims: 'Adding more solute to a solution makes its water potential more positive because there is more stuff dissolved.' Identify and correct the error in one or two sentences. (SP 6)

   short answerThe claim is incorrect. Adding solute makes Ψs MORE NEGATIVE (since Ψs = −iCRT increases in magnitude with C), which lowers the overall Ψ = Ψp + Ψs. More solute → lower (more negative) water potential, not more positive.

3. On a potato-core percent-mass-change vs. sucrose molarity graph, the best-fit line crosses zero at 0.25 M. What does that x-intercept represent about the relationship between the potato tissue and the solution? (SP 4)

   short answerAt 0.25 M sucrose, there is no net mass change, meaning no net water movement into or out of the potato cores. This is the point where the water potential of the potato tissue equals the water potential of the sucrose solution (Ψ cell = Ψ solution), so osmosis reaches equilibrium at that molarity.

4. A 0.3 M sucrose solution is placed around Cell A (internal solute ~0.1 M) and around Cell B (internal solute ~0.5 M). Which of the following is correct? A) The solution is hypertonic to both cells. B) The solution is hypotonic to both cells. C) The solution is hypertonic to Cell A and hypotonic to Cell B. D) The solution is hypotonic to Cell A and hypertonic to Cell B. (SP 1)

   multiple choiceC. Tonicity is comparative. Relative to Cell A (0.1 M), the 0.3 M solution has more solute — hypertonic, and water leaves Cell A. Relative to Cell B (0.5 M), the 0.3 M solution has less solute — hypotonic, and water enters Cell B. Same solution, different tonicity labels — proving tonicity is a relative property, not an absolute one.

• ☐Compound microscopes (one per student pair)
• ☐Fresh Elodea sprigs (or red onion epidermis as backup)
• ☐Microscope slides and coverslips
• ☐Distilled water in dropper bottles
• ☐20% NaCl solution (or 1.0 M sucrose) in dropper bottles
• ☐Paper towel wedges
• ☐Colored pencils
• ☐Calculators
• ☐Printed Elodea lab handout (one per student)
• ☐Printed water-potential problem set (one per student)
• ☐Printed formative assessment / exit ticket