2.6 Facilitated Diffusion

1. AThey diffuse through a channel protein, from high to low Na⁺ concentration. Correct
   Facilitated diffusion of an ion is passive movement through a membrane protein, here a channel, down the ion's concentration gradient. The charged ions cannot cross the hydrophobic interior of the bilayer in useful amounts, and no ATP is used.
2. BThey slip directly between the phospholipids, as the ions are small enough to fit.
   A student who thinks size alone decides what crosses the bilayer picks this. Na⁺ is small, but its charge keeps it out of the hydrophobic interior of the bilayer; facilitated diffusion means crossing through a protein.
3. CThey are pushed across by a channel protein that uses ATP to move each ion in.
   A student who thinks any movement through a protein uses ATP picks this. Facilitated diffusion is passive: the ions move down their concentration gradient and the cell supplies no metabolic energy.
4. DThey pass through a one-way channel protein that lets ions enter the cell only.
   A student who thinks membrane proteins are one-way doors into the cell picks this. In facilitated diffusion the direction of net movement is set by the concentration gradient, not by the protein.

CED 2.6.A.1 · Read this in Fix

1. AX stops entering, as the transport proteins cannot move any substance without ATP.
   A student who thinks any movement through a protein needs ATP picks this. Transport proteins in facilitated diffusion do not use ATP; only active transport does.
2. BX keeps entering at about the same rate, as moving down its gradient uses no ATP. Correct
   Facilitated diffusion is passive: X moves through transport proteins down its concentration gradient without any input of metabolic energy, so stopping ATP production does not slow it while the gradient remains.
3. CX enters faster, as a cell that is short of ATP needs more X and takes in more.
   A student who thinks cells take in substances according to need picks this. The rate of facilitated diffusion is set by the concentration gradient and the number of transport proteins, not by what the cell needs.
4. DX enters more slowly, as diffusion is partly powered by energy from the cell itself.
   A student who thinks passive transport is partly driven by the cell's energy picks this. Diffusion is driven by random molecular motion and the concentration gradient; the cell's ATP supply does not power it.

CED 2.6.A.2 · Read this in Fix

1. A25%
   A student who divides the change by the final value picks this: 0.24 ÷ 0.96 × 100 = 25%. Percent change uses the starting value, 1.20 µL, as the base.
2. B80%
   A student who writes the final volume as a percent of the starting volume picks this: 0.96 ÷ 1.20 × 100 = 80%. That is how much volume remains; the decrease is 100% − 80% = 20%.
3. C20% Correct
   The volume fell from 1.20 µL to 0.96 µL, a change of 0.24 µL. Dividing by the starting volume gives 0.24 ÷ 1.20 × 100 = 20%.
4. D24%
   A student who reports the absolute change as a percent picks this: 1.20 − 0.96 = 0.24 µL, written as 24%. The change must be divided by the starting volume, 1.20 µL.

Working Eggs with aquaporins: volume at 0 s = 1.20 µL; at 60 s = 0.96 µL. Change = 1.20 µL − 0.96 µL = 0.24 µL. Percent change = 0.24 µL ÷ 1.20 µL × 100 = 20%, a 20% decrease.

CED 2.6.A.3 · Read this in Fix

1. ACells given neither ethanol nor the drug in any form
   A student who thinks a control must receive no treatment at all picks this. These cells differ from the treated cells in both the drug and the ethanol, so a difference could be caused by either one.
2. BCells given twice as much drug in the same volume of ethanol
   A student who treats any comparison group as a control picks this. A higher dose is another experimental group; every group still contains the drug, so the effect of the ethanol alone is never measured.
3. CCells given the same volume of ethanol, with no drug added Correct
   This control differs from the experimental group only in the drug. If K⁺ loss differs between the two groups, the difference can be attributed to the drug, because both received the same ethanol. If the drug-treated cells lose K⁺ much more slowly than the ethanol-only cells, the result supports the hypothesis that K⁺ leaves mainly through channels; charged K⁺ cannot cross the bilayer in useful amounts.
4. DA second batch of cells given the drug in ethanol, as a repeat
   A student who confuses replication with a control picks this. Repeating the treatment shows whether the result is consistent, but both batches receive drug and ethanol, so their effects cannot be separated.

CED 2.6.A.1.ii · Read this in Fix

1. AK⁺ moves out, but the inside becomes positive as it still has more K⁺.
   A student who equates a high concentration of a positive ion with positive charge picks this. The inside started with equal positive and negative charges; losing K⁺ while keeping all its A⁻ leaves it with extra negative charge, whatever K⁺ concentration remains.
2. BK⁺ moves out, which leaves the inside negative relative to the outside. Correct
   K⁺ is more concentrated inside, so when the channels open it diffuses out. Each K⁺ that leaves carries a positive charge, while the A⁻ ions cannot follow, so the inside gains a net negative charge and the outside a net positive charge: the membrane becomes polarized.
3. CK⁺ moves out, but no charge difference forms unless ATP is used.
   A student who thinks only ATP-driven transport can separate charge picks this. Passive movement of K⁺ through channels, without matching movement of negative ions, separates charge across the membrane.
4. DK⁺ moves out until both sides hold equal K⁺, leaving no charge difference.
   A student who treats ions like uncharged molecules picks this. As K⁺ leaves, the inside becomes negative and attracts K⁺ back, so net movement stops before the concentrations are equal, and a charge difference remains.

CED 2.6.A.1.i · Read this in Fix

1. AIt fits through the channel, which is wider than the gaps between phospholipids.
   A student who thinks the bilayer blocks particles only by size picks this. The bilayer has no gaps that sieve particles by size; Na⁺ is kept out of the bilayer by its charge, and the channel offers a hydrophilic path.
2. BIt is pushed past the charged groups lining the pore by energy released from ATP.
   A student who thinks every protein-mediated movement uses ATP picks this. The model shows no ATP; Na⁺ moves passively through the open channel, down its concentration gradient, along the hydrophilic lining.
3. CIt binds to the charged groups, and the protein then flips over to release it inside.
   A student who confuses channels with carrier proteins picks this. A channel is an open pore lined with hydrophilic groups that ions pass along; it does not bind each ion and change shape to carry it across.
4. DIt passes along a pore lined with polar and charged groups, avoiding the nonpolar core. Correct
   The pore of the channel is lined with hydrophilic (polar or charged) R groups, so the charged Na⁺ ion can pass through without entering the hydrophobic interior of the bilayer, which ions cannot cross in useful amounts.

CED 2.6.A.1.ii · Read this in Fix

1. AG uptake levels off because its transport proteins are all occupied at high G. Correct
   G's uptake rises at first and then levels off, while E's rises in proportion to concentration. A plateau is what is expected when a substance crosses through a limited number of transport proteins: once all are in use, a higher concentration cannot increase the rate. E's straight line fits diffusion through the bilayer.
2. BG crosses the bilayer directly, since its uptake rate rises as its concentration rises.
   A student who takes any rise in rate with concentration as a sign of simple diffusion picks this. G's rate rises only at low concentration and then levels off, unlike E's straight line; the plateau points to a limited number of transport proteins.
3. CG uptake levels off because the cells run out of the ATP needed to move G in.
   A student who thinks protein-mediated movement always uses ATP picks this. Nothing in the data involves ATP; facilitated diffusion is passive, and the plateau is explained by saturation of a limited number of transport proteins.
4. DG uptake levels off because the cells have taken in as much G as they need.
   A student who thinks cells take in substances according to need picks this. These are initial rates into cells that started with no G; the rate is limited by the transport proteins, not by the cells' needs.

CED 2.6.A.2 · Read this in Fix

1. AGlucose moves out of the cell, pumped by the proteins using energy from ATP.
   A student who thinks movement through a protein always uses ATP picks this. Glucose is moving down its concentration gradient, which is passive; the transport proteins do not use ATP.
2. BGlucose stays in the cell, as these proteins can carry glucose inward only.
   A student who thinks transport proteins are one-way doors into the cell picks this. In facilitated diffusion the protein allows movement in either direction; net movement follows the gradient, which now points out of the cell.
3. CGlucose keeps moving into the cell, as the liver cell needs it for its energy.
   A student who thinks cells take in what they need picks this. Net movement by facilitated diffusion is set by the concentration gradient, not by need; glucose is now more concentrated inside, so it leaves.
4. DGlucose diffuses out of the cell, down its gradient, through the proteins. Correct
   In facilitated diffusion, the net direction is set by the concentration gradient. With glucose more concentrated inside, the same transport proteins let glucose leave the cell, with no ATP used.

CED 2.6.A.2 · Read this in Fix

1. AWater no longer enters the cells, as the bilayer will not let water through.
   A student who thinks water can cross membranes only through aquaporins picks this. Water is small and uncharged and crosses the bilayer itself in small amounts; aquaporins increase the rate.
2. BWater enters at the same rate, as water molecules slip easily between lipids.
   A student who thinks small size alone lets particles pass freely picks this. Water does cross the bilayer, but slowly; aquaporins are what allow large quantities of water to cross.
3. CWater enters faster, as cells without aquaporins need water and so take in more.
   A student who thinks cells take in what they need picks this. The rate of osmosis depends on the gradient that drives water into the cells and on how permeable the membrane is to water, not on need; losing aquaporins lowers permeability.
4. DWater still enters, but far more slowly, as it now crosses through the bilayer itself. Correct
   Water crosses the phospholipid bilayer slowly by itself; aquaporins allow it to cross rapidly in large quantities. The mutant root cells are in the same soil as the nonmutant cells, so water still moves into them by osmosis, but without working aquaporins it must cross the bilayer itself, at a much lower rate.

CED 2.6.A.3 · Read this in Fix