IB Biology D2.3: Water potential worked solutions

A. Osmosis needs a membrane that lets water cross but holds back the solute, and a difference in solute concentration. It is passive, so ATP (B), a carrier protein (C) and a cell wall (D) are not required.

B. Distilled water is hypotonic to the cell contents, so water enters by osmosis. With no wall to resist, the cell swells and may burst. A describes a hypertonic bath, and C and D describe cells with a wall.

B. Plasmolysis means water left the cells, so solution X was hypertonic to the sap. A and C would give turgid or unchanged cells, and D confuses solute movement with the water movement that causes plasmolysis.

A. Water is lost, so the pressure of the protoplast on the wall falls; the cells become flaccid and the leaf loses support. Walls do not dissolve (B), membranes do not burst (C), and plasmolysis is caused by water leaving, not entering (D).

A. Fertiliser makes the soil solution more concentrated, which reduces or reverses the concentration difference between soil and root sap, so water cannot enter by osmosis. B and C contradict this, and D has the cell wall do something it does not.

C. An isotonic bath has the same solute concentration as the cells, so there is no net water movement and the cells keep their volume. Hypotonic baths cause swelling and hypertonic baths cause shrinkage. Isotonic does not mean that no water crosses.

B. \psi_w=\psi_s+\psi_p=-650+250=-400 kPa. A adds the magnitudes, C changes the sign and D is both mistakes.

A. Water entering dilutes the sap, so \psi_s becomes less negative, and the protoplast presses on the wall, so \psi_p rises. B reverses the first, C the second and D both.

B. Water moves from higher to lower water potential, and -300 is higher than -450, so from Y to X. A has the direction reversed, C needs equal values and D cannot be right because the total, not one component, sets the direction.

A. As water enters, the solution is diluted and the difference in concentration falls, so the net movement slows. The sucrose does not leave (B), the membrane keeps its properties (C) and the water is not used up (D).

(a) The pond water has a lower solute concentration than the cytoplasm, so water enters by osmosis (1). There is no cell wall to resist the swelling, so the vacuole collects the excess water and expels it, which keeps the volume constant (1). (b) The medium now has a solute concentration nearer to that of the cytoplasm, so the concentration difference is smaller (1) and water enters more slowly, so the vacuole fills and empties less often (1).

(a) (4.6-5.0)/5.0\times100=-8\,\% (1). (b) The tissue gained mass at 0.0 and 0.1, so those solutions were hypotonic (1). (c) The change is +6\,\% at 0.1 and -2\,\% at 0.2; the zero crossing is 0.1+0.1\times6/8=0.175, so 0.18 mol dm^{-3} (2; one mark for a range of 0.1 to 0.2 only). (d) At 0.4 the solution is hypertonic to the cells, so water leaves by osmosis and the mass falls (1).

(a) Independent variable: concentration of the sodium chloride solution, in mol dm^{-3} (1); dependent variable: percentage change in strip length (1). (b) Two controls, each with its effect, such as the same cross-sectional area, because surface area changes how fast water moves, and the same time in solution, because the extent of osmosis depends on time (2; one mark each). (c) The initial lengths are not identical, so percentage change allows strips to be compared (1).

Any six of: osmosis is the net movement of water across a partially permeable membrane from lower to higher solute concentration (1); distilled water is hypotonic to the cells, so water enters (1); an animal cell has no wall, so it swells and may burst (1); a plant cell becomes turgid, and the wall stops it bursting (1); the salt solution is hypertonic, so water leaves (1); an animal cell shrinks and its outline becomes notched: crenation (1); in a plant cell the membrane pulls away from the wall: plasmolysis (1); in an isotonic solution there is no net change (1).

(a) Fluid R, because its concentration matches the cells, so there is no net water movement (1). (b) Fluid S at 0.60 is hypertonic to the cells, so water leaves by osmosis (1) and the cells shrink and become crenated (1). (c) Fluid P has no solute, so it is hypotonic to the cells; water would enter and the cells would swell and burst, and being clean does not change this (1).

(a) \mathrm{SE}=2.24/\sqrt5=2.24/2.236\approx1.0 (1); SE at 0.3: 2.0/\sqrt5=0.9 (1). (b) The ranges are 3.0 to 5.0 and -3.9 to -2.1, which do not overlap, so the two means probably differ (1) and the tissue gains mass at 0.2 and loses mass at 0.3, so the isotonic point is between them (1). (c) A systematic error such as the same unequal blotting for every disc is not reduced by repeats (1).

(a) \psi_w=\psi_s+\psi_p: P =-800+400=-400; Q =-600+100=-500; R =-900+300=-600 kPa (3). (b) Water moves from higher to lower water potential, so from P to Q and from Q to R (1). (c) P has the more negative solute potential, so its sap is the more concentrated, yet it has the higher water potential because its pressure potential is much higher (1); water moves by total water potential, not by solute potential alone (1).

(a) A flaccid cell has \psi_p=0, so \psi_w=\psi_s=-500 kPa (1). (b) The bath has the higher water potential (-150 against -500), so water enters the cell (1). (c) At equilibrium the cell's \psi_w=-150 kPa; ignoring the small change in \psi_s, \psi_p=\psi_w-\psi_s=-150-(-500)=+350 kPa (2). (d) The sap is diluted by the incoming water, so the solute potential becomes less negative (1).

(a) The mean change decreases as the salt concentration rises (1), from +14.0\,\% at 0.0 to -9.0\,\% at 1.2, changing from a gain to a loss between 0.6 and 0.9 (1). (b) Zero crossing: 0.6+0.3\times2/6=0.70\,\% NaCl (2). (c) Pure water is hypotonic, so water enters by osmosis (1); the cell wall resists the swelling and the cells become turgid, so the mass rises but the cells do not burst (1). (d) The ranges 7.1–8.9 and 1.2–2.8 do not overlap, so the difference between the 0.3 and 0.6 results is probably real (1) and it supports a gradual change rather than a one-off effect; five discs per concentration is a small sample, so the claim is cautious (1).

(a) Root hair: -500+150=-350 kPa; cortex: -600+100=-500 kPa (2). (b) Water moves from higher to lower potential: -100>-350>-500>-800, so soil to root hair to cortex to xylem (2). (c) The salty soil solution has no pressure in an open soil, so \psi_w=\psi_s=-450 kPa (1), lower than the root hair at -350 kPa, so water moves out of the root hair into the soil (1). (d) The cells lose water, so their pressure potential falls towards zero; they lose turgor, the leaves lose support and the plant wilts although the soil is wet (2).

(a) Rate 0–5 min: (11.2-10.0)/5=0.24 g per minute (1); 20–25 min: (12.9-12.8)/5=0.02 g per minute (1). (b) The sucrose is hypertonic to the distilled water, so water enters the bag by osmosis (1); entry dilutes the solution, so the concentration difference falls and the rate falls (1); the mass levels off as the difference approaches zero, because sucrose cannot leave (1). (c) A more concentrated solution gives a steeper gradient, so a higher initial rate (1) and a larger final mass because more water enters before the concentrations match (1).

(a) Fresh water at 0.02 is more dilute than the sap at 0.35 (1), so it is hypotonic to the root cells and water enters by osmosis (1). (b) Brackish water at 0.60 is more concentrated than the sap (1), so water leaves the root cells by osmosis (1); the cells lose turgor pressure, become flaccid, and the non-woody parts lose their support, so the plant wilts (1). (c) Changing back to fresh water restores the concentration difference so water enters again and turgor returns (1), provided the cells have not been damaged.

(a) Cohesion, hydrogen bonds between water molecules, lets the water column be pulled as a chain (1); adhesion to the xylem walls helps hold the column up (1). (b) Water lost from the mesophyll cells makes their sap more concentrated (1), so water moves into them from the xylem by osmosis (1). (c) Water leaves the guard cells and they lose turgor, becoming flaccid (1), so the stomatal pore closes and less water is lost (1).

(a) Water is polar and the interior of the bilayer is hydrophobic, so few water molecules cross through the lipids (1); aquaporins provide a channel, so more water crosses by facilitated diffusion (1). (b) The treated cells swell more slowly (1) because water crosses mainly through the lipid and not through channels (1). (c) The direction of net movement is set by the difference in solute concentration, which is unchanged (1), so water still enters the treated cells and they would still swell towards bursting (1).

(a) More concentrated plasma and tissue fluid are hypertonic to the cells, so water leaves cells by osmosis and they shrink (1). (b) ADH from the pituitary gland (1) makes the collecting duct walls more permeable to water, so more water is reabsorbed by osmosis from the filtrate and a small volume of concentrated urine is made (1). (c) Isotonic saline has the same solute concentration as the cells, so there is no net water movement into or out of the blood cells (1); pure water is hypotonic and could make red blood cells burst (1).