Membrane Transport

Glucose diffuses out until its concentration is the same on both sides. Molecules keep moving at random, so they still cross both ways, but equal numbers cross each way, so there is no net change: a dynamic equilibrium.

• Almost the whole of the glucose has left the bag and ended up in the much larger beaker.: Diffusion evens out concentrations; it does not move every molecule to one side.
• Correct: Equal concentrations, with molecules still crossing both ways.: This describes dynamic equilibrium: equal concentrations, crossings still happening at equal rates both ways.
• Glucose concentration is about equal inside and out, and the glucose molecules have stopped crossing the tubing.: Molecules keep moving and crossing at equilibrium; only the net movement stops.
• Glucose has moved back into the bag, where it is now more concentrated.: Passive movement cannot build up a higher concentration on one side; that would need energy.

Glucose moved from the bag (high) to the beaker (none), and iodine from the beaker (high) to the bag (none). Each went down its gradient, driven by its own motion, which is passive transport.

• Correct: Passive transport: each moved down its own gradient.: Each substance moved from high to low concentration with no energy supplied, which is passive transport.
• Active transport, because the two substances moved in opposite directions across the same tubing at the same time.: Opposite directions just mean the two gradients pointed opposite ways; each move was still downhill.
• Bulk transport, because the bag took in iodine in vesicles.: There are no vesicles in tubing; the molecules passed through pores one by one.
• Active transport, because the movement continued for 30 minutes.: Passive movement continues for as long as a gradient exists; time alone does not make it active.

The untreated cells reach a ratio of 9.6 by 60 minutes; the cyanide-treated cells rise to 1.0 by 30 minutes and stay there.

• In both batches the inside concentration rises well above the outside concentration within the hour.: The cyanide batch never goes above 1.0, so only the untreated batch accumulates the amino acid.
• Correct: Untreated cells reach about ten times the outside level; cyanide-treated cells level off at equal.: This matches both curves.
• With cyanide the cells take up the amino acid faster, then stop at a ratio of 1.0.: At every time point the cyanide batch is below the untreated batch, so its uptake is slower.
• Without cyanide the ratio levels off at 1.0 after about 30 minutes.: The ratio of 1.0 is where the cyanide batch levels off; the untreated batch keeps rising past 9.

A ratio above 1.0 means the amino acid is moving from low to high concentration, which takes energy. When cyanide stops ATP production, the cells can still take it in passively, but only until inside equals outside.

• Cyanide damages the bilayer of the cells, so the amino acid leaks back out as fast as it comes in.: If the bilayer were leaky, the ratio would not climb neatly to 1.0 and hold; the pattern fits passive movement to equal concentrations.
• Correct: Untreated cells use ATP to move it against its gradient; without ATP it moves only until both sides are equal.: This links the ratio above 1.0 to active transport and ATP, and the level at 1.0 to passive transport.
• Cyanide makes the outside concentration of the amino acid rise, so the ratio stays at or below 1.0.: Cyanide stops ATP production in the cells; it does not change the amino acid outside.
• The untreated cells make new amino acid of this kind inside, which raises the concentration measured inside.: The measured amino acid is the labeled one from the medium, so cells making their own would not raise its value.

The untreated range runs from 1.8 to 2.4 and the cyanide range from 0.5 to 0.7. Error bars of ±2 SE that do not overlap suggest the difference is statistically significant, not just sampling noise.

• Correct: No overlap, so the difference is likely real.: Non-overlapping ±2 SE bars are evidence of a real difference.
• The error bars overlap, so the batches are the same at 10 minutes.: The ranges (1.8 to 2.4 and 0.5 to 0.7) do not overlap at all.
• The error bars do not overlap, so there is certainly no difference.: Non-overlapping bars point to a difference, not the absence of one.
• Nothing can be concluded, because error bars matter just for the last time point of a time course.: Error bars can be compared at any time point where both means have them.

Binding to a receptor needs no work from the cell, so it happens in the cold. Taking the bound LDL in needs the membrane to fold inward and pinch off a vesicle, which stops at 4 °C. So LDL piles up on the surface.

• At 4 °C the LDL receptors are destroyed by the cold, so LDL sticks loosely to the bilayer instead of to receptors.: Healthy cells bound 120 ng at 4 °C against 6 ng for receptor-lacking cells, so the receptors are clearly working.
• Correct: LDL binds its receptors, but in the cold no vesicles form to take it in.: This separates binding (still works) from endocytosis (stops in the cold).
• At 4 °C LDL diffuses straight through the bilayer and out the other side.: LDL particles are far too large to diffuse through a bilayer.
• At 4 °C the cells use active transport to push LDL back out.: There is no evidence of LDL being pushed out; it is bound but not taken in.