Biology · Year 11 · Starting
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Predict and explain osmosis, show cell responses and calculate evidence from mass changes.
Learning objectives:
A tissue sample gains mass in one solution and loses it in another. What moved, why did it move, and what do the measurements actually establish?
solute, solvent, concentration, selectively permeable: Dissolved material; dissolving liquid; amount per volume; permits some substances but not others.
Cup A has six solute symbols in two equal volume units. Cup B has four in one unit. Which is more concentrated?
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B is more concentrated: 4 per unit exceeds 3 per unit. More total symbols in A does not reverse the ratio.
A has 8 symbols in 2 units: 8 ÷ 2 = __ per unit. B has 3 in 1 unit: __ per unit. Compare.
A: 8 ÷ 2 = 4. B: 3 ÷ 1 = 3. A is more concentrated because 4 > 3 per equal volume.
Compare 9 symbols in 3 units with 4 in 1 unit. Explain.
9 ÷ 3 = 3 per unit; 4 ÷ 1 = 4 per unit. The second is more concentrated.
initial, final, signed change, percent: Before; after; a difference retaining gain or loss; per hundred.
What percentage is a change of 2 g compared with an initial 10 g?
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2 g compared with 10 g is 20%. Check that 20% of 10 g is 0.20 × 10 = 2 g.
A change of 3 g from an initial 10 g gives 3 ÷ 10 = __; multiply by 100 to get __%.
3 ÷ 10 = 0.3. Then 0.3 × 100 = 30%. Check 0.30 × 10 = 3 g.
A sample loses 1 g from an initial 5 g. Express the signed percentage change.
Change = −1 g. Fraction = −1 ÷ 5 = −0.2. Percentage = −0.2 × 100 = −20%. Check 5 × 0.8 = 4 g, a loss of 1 g.
This Starting lesson teaches initial osmosis direction, plant and animal responses, percentage mass change and cautious interpretation of tissue data. It does not teach a complete membrane-transport course, water-potential equations or experimental proof from one measurement.
Osmosis is net water movement across a selectively permeable membrane from higher to lower water potential. Water molecules move in both directions. Net means the difference between the two opposing transfers; it is not every molecule travelling one way.
For the first models, compare the same non-penetrating solute at equal temperature and equal initial pressure. Then the more dilute side has higher water potential, so initial net water moves toward the more concentrated side. If pressure differs or solute crosses, do not apply this simplified rule without revising the model.
Choose a method from the question: predicting movement → identify permeability and compare water potential; predicting cell response → also identify the wall; calculating data → final minus initial, divide by initial, multiply by 100; interpreting results → read sign or zero and evaluate limits.
A cell membrane and a plant cell wall are different. The membrane controls passage. A plant wall can resist expansion: water entry raises turgor pressure. Animal cells lack that rigid wall. Water loss reduces plant-cell turgor; enough loss can separate the membrane from the wall, called plasmolysis.
In this non-penetrating-solute model, hypotonic means the outside is more dilute relative to the cell and tends to cause water entry; hypertonic means more concentrated and tends to cause water loss; isotonic means no sustained volume change from the effective solutes. These terms compare two sides, not a solution by itself. Use the pressure conditions when discussing a plant cell.
Percentage mass change = (final mass − initial mass) ÷ initial mass × 100. The initial mass must be nonzero. Keep units until the ratio is formed; matching mass units cancel. A plus sign means net gain; a minus sign means net loss. A zero means no detected net mass change, not still water molecules.
Tissue mass is evidence, not a direct count of water crossing. Use consistent blotting, matching samples, controlled exposure and repeated measurements. Damage, surface solution or solute exchange can weaken an osmosis explanation. These examples use invented classroom data; they are not measurements from a real experiment.
Read each question by naming the object, membrane, moving substance, initial conditions and required conclusion. Write the comparison before the arrow. For data, write the formula before substituting. Check the sign, units and whether the conclusion answers the original question.
A flexible model bag contains 0.6 mol/L sucrose and is in 0.2 mol/L sucrose. The membrane passes water but not sucrose. Temperature and initial pressure are equal. Predict the initial net water movement and bag size change.
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Initial net water movement is outside → inside; the bag gains water and initially swells. Sucrose is not the substance crossing in this model. Both directions of water movement remain possible.
A flexible model bag contains 0.2 mol/L sucrose and is in 0.6 mol/L sucrose. The membrane passes water but not sucrose. Temperature and initial pressure are equal. Predict the initial net water movement and bag size change.
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Initial net water movement is inside → outside; the bag loses water and initially shrinks. Sucrose is not the substance crossing in this model. Both directions of water movement remain possible.
A animal cell initially has 0.4 mol/L non-penetrating solute inside and 0.1 mol/L outside. Use the equal initial pressure model to predict the initial water movement and explain the cell response.
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Initial net water movement is outside → cell. The animal cell swells and may burst if water entry is excessive. The outcome depends on the cell structure and continued conditions.
A plant cell initially has 0.4 mol/L non-penetrating solute inside and 0.1 mol/L outside. Use the equal initial pressure model to predict the initial water movement and explain the cell response.
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Initial net water movement is outside → cell. Water entry makes the plant cell turgid as pressure builds against the wall. The outcome depends on the cell structure and continued conditions.
A plant cell initially has 0.2 mol/L non-penetrating solute inside and 0.8 mol/L outside. Use the equal initial pressure model to predict the initial water movement and explain the cell response.
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Initial net water movement is cell → outside. Water loss reduces turgor; sufficient loss can pull the membrane away from the wall (plasmolysis). The outcome depends on the cell structure and continued conditions.
A blotted tissue sample changes from 4.0 g to 4.6 g in solution. Calculate percentage mass change and interpret its sign. Assume mass change mainly reflects water movement.
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(4.6 − 4.0) ÷ 4.0 × 100 = 15%. Reconstruction gives 4.6 g, confirming the calculation. The sign indicates a net water gain under the stated assumptions.
A blotted tissue sample changes from 5.0 g to 4.4 g in solution. Calculate percentage mass change and interpret its sign. Assume mass change mainly reflects water movement.
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(4.4 − 5.0) ÷ 5.0 × 100 = -12%. Reconstruction gives 4.4 g, confirming the calculation. The sign indicates a net water loss under the stated assumptions.
Illustrative tissue data: at 0.1, 0.2, 0.3 and 0.4 mol/L external sucrose, mean mass changes are +8%, +4%, 0% and −4%. What does the zero suggest?
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About 0.3 mol/L gave no mean net mass change. Under the stated controls this supports approximate water-potential balance. Water can still cross in both directions.
Illustrative results show +6% at 0.2 mol/L and −2% at 0.4 mol/L. Neither tested concentration gives 0%. What can be concluded?
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The estimated no-net-change condition lies between 0.2 and 0.4 mol/L under the stated trend. Repeat and test intermediate concentrations; no exact concentration is established.
Complete together: initial 6.0 g, final 6.3 g. Show the change, reference and percentage.
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6.3 − 6.0 = +0.3 g; 0.3 ÷ 6.0 = 0.05; 0.05 × 100 = +5%. Check 6.0 × 1.05 = 6.3 g.
Complete the explanation: outside 0.1 mol/L, inside 0.5 mol/L. Water crosses; solute does not. Equal initial pressure. The lower-solute side is __, so initial net water moves __. Explain why.
Identify water as the moving substance.
Compare the two values under equal pressure.
Name the higher-water-potential side and connect the net arrow to a gain or loss.
Compare 0.1 with 0.5: outside has lower solute concentration.
At equal initial pressure its water potential is higher.
Water crosses both ways, but the initial net movement is outside → inside.
The bag initially gains water.
Complete the working for a sample changing from 8.0 g to 7.2 g: change = __; fraction = change ÷ __; percentage = __. Explain the sign.
Use final minus initial.
Subtract 8.0 from 7.2 and keep the negative sign.
Divide the signed change by 8.0, then express that fraction per hundred.
Change = 7.2 − 8.0 = −0.8 g.
Divide by initial mass: −0.8 g ÷ 8.0 g = −0.1.
Multiply by 100: −0.1 × 100 = −10%.
Check 8.0 × 0.90 = 7.2 g.
The negative sign indicates net mass loss, consistent with net water loss under the assumptions.
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A plant cell gains water. Complete the chain: water entry → larger cell contents → pressure against __ → increased __. Explain why this differs from an animal cell.
Separate cell wall from membrane.
Trace what expanding contents press against.
Name the pressure effect and compare the missing animal-cell structure.
Water enters the cell contents and vacuole.
They expand against the cell wall, increasing turgor pressure.
The wall resists expansion and can oppose further net entry.
An animal cell has no rigid wall and can swell or lyse.
Water can still cross both ways in a turgid cell.
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Put the steps for finding a no-net-change condition in order. Then apply them to +5% at 0.1 mol/L, 0% at 0.2 mol/L and −5% at 0.3 mol/L.
Start with the measured response, not the largest concentration.
Locate the zero percentage row.
Read its concentration and distinguish balanced crossing from absent movement.
Find the 0% response first, then read its concentration: about 0.2 mol/L.
No net mass change means gains and losses balance overall.
It does not mean water molecules stop crossing.
Check repeated samples and measurement precision before claiming an exact balance.
A plant cell has lower water potential than the surrounding solution. Use the hints to explain its initial response.
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Net water initially enters the plant cell. The contents expand against the wall and turgor increases. Pressure may eventually oppose further net entry.
A bag contains 0.3 mol/L non-penetrating solute; outside is 0.7 mol/L. Predict initial net water movement at equal pressure and explain the volume change.
Mark which substance can cross.
Compare 0.3 and 0.7 under equal initial pressure.
Draw the net arrow from higher water potential and connect it to volume.
Water can cross; the named solute cannot.
Inside has lower solute concentration (0.3 < 0.7 mol/L), hence higher initial water potential at equal pressure.
Initial net water movement is inside → outside.
The flexible bag loses water and initially shrinks.
This does not calculate its final volume.
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A sample changes from 2.5 g to 3.0 g. Calculate percentage mass change and verify it backwards.
Use the initial mass as reference.
Calculate final minus initial: 3.0 − 2.5.
Divide the change by 2.5 and convert the fraction to a percentage.
Change = 3.0 − 2.5 = +0.5 g.
Fraction = 0.5 g ÷ 2.5 g = 0.2.
Percentage = 0.2 × 100 = +20%.
Reverse check: 2.5 × (1 + 20/100) = 2.5 × 1.2 = 3.0 g.
The positive sign indicates net gain.
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At water-potential balance across a water-permeable membrane, which statement is correct? Explain why.
Separate an individual movement from a net total.
Draw two opposing arrows through a water-permeable membrane.
Explain what must balance, then check whether pressure was specified.
The statement confuses net movement with individual movement.
If water can cross, its molecules continue moving in both directions.
At water-potential balance the opposing transfers balance overall, so there is no net water gain.
Equal solute concentration alone is insufficient if pressure differs.
Mean mass changes are +3% at 0.25 mol/L and −4% at 0.45 mol/L. Give a defensible estimate range and a useful next step.
Interpret the positive and negative signs.
Place zero between a gain and a loss.
State only the supported interval and choose measurements inside it.
The sign changes from gain to loss, so a no-net-change condition is expected between 0.25 and 0.45 mol/L under the same continuous trend.
The data do not prove linearity or an exact midpoint.
Test repeated samples at several concentrations within that interval to narrow the estimate.
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Outside 0.8 mol/L; inside 0.4 mol/L. A membrane passes water but not this solute. Equal initial pressure. Explain initial net movement and what could limit further change.
Name the moving substance and conditions.
Compare the two solute concentrations.
Trace the resulting water loss and consider what changes as water moves.
Inside has lower solute concentration and higher initial water potential under equal pressure.
Initial net water movement is inside → outside.
The bag loses water.
Concentrations change as water moves; any developing pressure difference also affects water potential.
Thus the initial comparison does not determine a final volume.
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Sample A changes from 3.0 g to 3.3 g; sample B changes from 6.0 g to 6.3 g. Which has the larger proportional increase? Show both calculations.
Equal gram gains need not be equal proportions.
Find the signed change for each sample.
Divide each change by its own initial mass before comparing.
A: 3.3 − 3.0 = +0.3 g; 0.3 ÷ 3.0 = 0.1; 0.1 × 100 = 10%.
B: 6.3 − 6.0 = +0.3 g; 0.3 ÷ 6.0 = 0.05; 0.05 × 100 = 5%.
A has the larger proportional increase although both gain 0.3 g.
Reverse checks: 3.0 × 1.1 = 3.3; 6.0 × 1.05 = 6.3.
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A plant cell in concentrated external solution loses water. Explain the sequence from the membrane boundary to plasmolysis, including a limit on the prediction.
Follow water, not solute, across the membrane.
Connect water loss to the volume of cell contents.
Compare the flexible membrane with the supporting wall and state a limit.
Water crosses the cell membrane outward overall when outside has lower water potential.
The cell contents and vacuole lose volume; turgor decreases.
With sufficient loss, the cell membrane pulls away from the wall: plasmolysis.
The wall itself does not shrink in the same way.
The extent and timing are not determined without further information.
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A group forgets to blot tissue before the final mass measurement. Explain the likely bias, its effect on calculated percentage change and how to improve the comparison.
Ask what the balance weighs besides tissue.
Extra surface liquid increases the final measurement.
Follow that increase through final minus initial, then propose a consistent correction.
Surface solution adds to the final measured mass without necessarily entering cells.
Final minus initial becomes more positive or less negative, so the calculated percentage change is biased upwards.
Blot all samples using the same gentle procedure before weighing, and keep starting dimensions, solution exposure, temperature and weighing method consistent.
Repeat samples to assess variation.
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Return on another day. Try these fresh questions before revealing a hint or answer. Explain what changed in your approach.
A bag has 0.2 mol/L solute inside and pure water outside. Water crosses; solute does not. Equal initial pressure. Predict initial movement.
Identify the substance that crosses.
Compare zero added solute with 0.2 mol/L.
Use higher to lower water potential, then describe the bag.
Pure water has zero added solute, lower than 0.2 mol/L.
At equal pressure it has higher water potential.
Initial net water moves outside → inside; the bag gains water.
Solute remains inside and water still moves individually both ways.
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A bag has pure water inside and 0.5 mol/L non-penetrating solute outside. Predict initial movement at equal pressure.
Mark the two sides.
Inside has less non-penetrating solute.
Follow water from higher water potential and connect to volume.
Inside has zero added solute and higher water potential at equal pressure.
Initial net water moves inside → outside, so the bag loses water.
The non-penetrating solute cannot enter to produce this change.
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Both sides have 0.3 mol/L of the same non-penetrating solute at equal pressure. Describe both individual and net water movement.
There are two different meanings of movement.
Draw arrows in both directions through a permeable membrane.
State what balances and what continues.
The water potentials are equal in the model.
Water can cross in both directions, with no net transfer when the opposing movements balance.
Equal means no net change, not motionless molecules.
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A diagram shows four solute symbols in 1 volume unit on the left and six in 3 units on the right. Which side is more concentrated?
Concentration compares quantity with volume.
Divide each solute count by its own volume.
Compare the ratios rather than total counts.
Left: 4 ÷ 1 = 4 symbols per volume unit.
Right: 6 ÷ 3 = 2 symbols per volume unit.
Compare 4 with 2: the left is more concentrated despite fewer total symbols.
With blocked solute and equal initial pressure, net water would initially move right → left.
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An animal cell begins in a more dilute external solution with non-penetrating solute and equal initial pressure. Explain why “it must instantly burst” is too strong.
First establish initial water direction.
Relate the direction to cell volume and lack of wall.
Separate a possible later result from an inevitable immediate result.
The more dilute outside has higher initial water potential, so net water initially enters.
The cell may swell because it lacks a rigid wall.
Bursting depends on the amount and duration of entry and membrane properties, so initial direction alone cannot prove instantaneous lysis.
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A plant cell is placed in a more dilute solution. Explain how its wall affects the response without claiming the wall blocks all water.
Distinguish membrane and wall.
Track expansion of the contents against the wall.
Explain pressure rather than inventing an impermeable wall.
Water crosses the cell membrane into the contents overall.
The contents expand against the cell wall; turgor pressure rises.
The supporting wall resists expansion, and the resulting pressure can oppose further net entry.
The wall is not an impermeable water barrier.
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In a plant cell losing water, which changes first in the explanation: reduced cell contents or a wall that disappears? Repair the false option.
Identify what loses water.
Compare flexible contents with the rigid supporting wall.
Describe the membrane position after sufficient loss.
Water loss reduces the volume of cell contents and turgor.
With sufficient loss, the membrane may detach from the wall.
The cell wall does not need to disappear for plasmolysis; it remains a supporting boundary.
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A blotted sample changes from 10.0 g to 11.5 g. Calculate percentage mass change and reverse-check it.
Use final minus initial.
Divide the signed change by the initial 10.0 g.
Multiply the fraction by one hundred and reconstruct the final mass.
Change = 11.5 − 10.0 = +1.5 g.
Fraction = 1.5 ÷ 10.0 = 0.15.
Percentage = 0.15 × 100 = +15%.
Reverse check: 10.0 × 1.15 = 11.5 g.
The positive sign means mass gain.
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A blotted sample changes from 4.0 g to 3.2 g. Calculate percentage mass change and interpret it.
Keep the order final minus initial.
The change is negative; do not discard its sign.
Divide by 4.0 and convert to a percentage.
Change = 3.2 − 4.0 = −0.8 g.
Fraction = −0.8 ÷ 4.0 = −0.2.
Percentage = −0.2 × 100 = −20%.
Check: 4.0 × 0.8 = 3.2 g.
The sample lost mass, consistent with net water loss if other mass changes are negligible.
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A learner calculates a 6.0 g to 6.6 g change as 0.6 ÷ 6.6 × 100. Identify and correct the denominator error.
A change is compared with where it started.
Label 6.0 g as initial and 6.6 g as final.
Divide 0.6 by the initial value before multiplying by one hundred.
The change is 6.6 − 6.0 = +0.6 g.
The reference is the starting mass 6.0 g, not final mass 6.6 g.
Correct fraction: 0.6 ÷ 6.0 = 0.1.
Percentage: 0.1 × 100 = +10%.
Check: 6.0 × 1.1 = 6.6 g.
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Samples A and B each gain 0.4 g. Their initial masses are 2.0 g and 8.0 g. Compare proportional gains.
The same gain can be a different fraction of the start.
Form one ratio for each starting mass.
Convert both ratios before comparing.
A: 0.4 ÷ 2.0 = 0.2; 0.2 × 100 = 20%.
B: 0.4 ÷ 8.0 = 0.05; 0.05 × 100 = 5%.
A has the greater proportional gain.
Check final masses: 2.0 × 1.2 = 2.4 g; 8.0 × 1.05 = 8.4 g.
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A tissue sample keeps a measured mass of 5.0 g. Calculate percentage change and explain what it does not prove.
Apply the same formula even when the values match.
Subtract first, then divide by the initial mass.
Distinguish no detected net change from no microscopic movement.
Change = 5.0 − 5.0 = 0 g.
Fraction = 0 ÷ 5.0 = 0.
Percentage = 0 × 100 = 0%.
There was no detected net mass change.
This does not prove water stopped moving, exact balance in every cell, or absence of measurement error.
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Illustrative means are +7% at 0.1 mol/L, 0% at 0.25 mol/L and −6% at 0.4 mol/L. Identify the observed no-net-change condition and qualify the claim.
Read response and concentration as a pair.
Choose the zero response, not the largest concentration.
State it as an estimate with the relevant measurement limits.
Locate 0%, then read the matched concentration: 0.25 mol/L.
Under the protocol this is an estimate of a no-net-mass-change condition.
Repeats and measurement precision are needed; it is not proof of identical internal solute concentration in every cell.
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Illustrative means are +2% at 0.3 mol/L and −3% at 0.5 mol/L. Can you state an exact no-net-change concentration?
Check whether either response is zero.
A gain and loss bracket a crossing under the stated trend.
Give the supported interval and identify further evidence needed.
No.
The sign change supports a value between 0.3 and 0.5 mol/L if the response is continuous and the protocol is comparable.
Two measurements do not establish a linear relationship.
Test intermediate concentrations and repeat samples; do not invent an exact midpoint.
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Two repeats at the same concentration give +1% and −1%. Their mean is 0%. Explain why this is weaker than proof that each sample was unchanged.
Read individual values before their average.
Add the signed values and divide by two.
Describe what the opposite signs reveal about the samples.
Add the results: +1 + (−1) = 0.
Divide by two repeats: 0 ÷ 2 = 0% mean.
One sample gained mass and the other lost mass; they were not individually unchanged.
Variation and measurement error must be considered before interpreting the mean as exact balance.
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A group blots only the initial tissue, leaving visible solution on the final sample. Explain the effect on its calculated change.
Identify extra material counted in the final weighing.
Trace a larger final value through subtraction.
Explain the direction of bias and how to standardise the method.
The final weighing includes surface solution.
This raises final mass, so final minus initial is more positive or less negative.
Dividing by the same positive initial mass preserves that upward bias in percentage change.
Blot initial and final samples consistently without damaging them.
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The membrane is freely permeable to both water and the stated solute. Explain why the simple non-penetrating-solute prediction needs revision.
Check which assumption changed.
Both substances can now change their distributions.
Explain why a model that traps solute cannot simply be reused.
The earlier model assumed solute stayed on its own side.
If solute crosses too, both water movement and solute diffusion can alter the concentrations.
The initial solute comparison alone is insufficient to predict sustained tonicity or the final cell volume.
State permeability and pressure information before choosing a model.
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A turgid plant cell has more dissolved solute inside than outside but no net water entry. Does this automatically disprove osmosis?
Recall the equal-pressure condition.
A turgid cell has pressure against its wall.
Explain how pressure can oppose the solute effect without stopping individual movement.
No.
The simple lower-solute-to-higher-solute rule assumed equal pressure.
Pressure inside a turgid cell can raise its water potential and balance the solute effect.
At balance there is no net entry even though water still crosses both ways.
Pressure must be included before drawing a conclusion.
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A gardener sees wilted leaves after soil becomes very concentrated in dissolved salts. Explain a possible water-movement mechanism and why this observation alone is not a complete diagnosis.
Relate outside solution to root-cell water potential.
Connect reduced water supply with turgor in plant tissues.
Distinguish a plausible mechanism from proof of the sole cause.
Concentrated soil solution can have lower water potential than root-cell contents.
Water uptake may be reduced or net water loss may occur, reducing cell turgor and contributing to wilting.
The observation alone does not measure soil water potential or rule out root damage, heat or other causes; further evidence is needed.
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A report says: “The tissue lost 12%, so solute must have moved out.” Replace it with a reasoned interpretation and state the necessary assumptions.
A percentage describes total measured mass, not molecule identity.
Use the non-penetrating-solute assumption from the model.
Give the supported water interpretation and state what the measurement cannot establish alone.
A −12% mass change means final mass was 88% of initial mass: 100 − 12 = 88.
If samples were blotted consistently, membranes remained functional and solute movement was negligible, the loss is consistent with net water leaving by osmosis.
Mass data alone do not identify the moving molecule without those assumptions.
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Sample X changes from 4.0 g to 4.4 g; Y changes from 8.0 g to 8.4 g. A report says their equal gram gains prove equal osmotic responses. Evaluate the claim and identify two further controls.
Optional extension task. It does not change lesson access.
Outside 0.2 mol/L, inside 0.9 mol/L blocked solute, equal initial pressure. Explain the initial water direction.
Outside has higher water potential under these conditions; net water initially enters. Individual crossings still occur both ways.
Initial mass 12.0 g, final mass 10.8 g: calculate signed percentage change.
10.8 − 12.0 = −1.2 g; −1.2 ÷ 12.0 = −0.1; ×100 = −10%; 12.0 × 0.9 = 10.8 g.
How does a plant wall change the consequence of net water entry?
The expanding contents press against the supporting wall, raising turgor pressure that can oppose further entry; the wall does not stop all water passage.
A mean is zero but repeats are +2%, 0% and −2%. What should the conclusion include?
The mean is (2 + 0 − 2) ÷ 3 = 0%; individual samples vary. Report no mean net change under the protocol, not identical unchanged samples or stopped molecular movement.
Viewing examples, using help and independent correct working are different kinds of evidence. Completing these pages does not automatically mark this skill as mastered.