Problems: Module 1

Write your answer first, then open the solution. Assignment key items show the points the grader awards; chart problems use the course figures; variants change the numbers or flip the premise; check-yourself items answer the end-of-module review questions.

Assignment key¶

Exercise 1 (Assignment key · 14 pts)

Frame the brain’s dehydration-sensing neurons and vasopressin as a negative feedback loop. Give the regulated variable, how “too high” is sensed, whether the signal travels to an integrating center, the efferent pathway, the effector, and how the effector corrects the problem.

Solution to Exercise 1

Regulated variable (2): concentration of extracellular fluid. “Dehydration” is half credit, because it is the idea but not a variable. (slide 37)

Sensing (2): too-concentrated extracellular fluid draws water out of the cell, which makes it shrivel. The shriveling is how the cell senses the stimulus.

Integration (4): the message does not travel (2 pts). Locally, the physical shriveling pulls open mechanically gated ion channels. Stimulus strength sets the degree of shriveling, and whether enough channels open is the integration and error signal (2 pts).

Efferent pathway (2): release of vasopressin, a chemical messenger circulating through the body.

Effector (2): the kidneys.

Correction (2): vasopressin tells the kidneys to stop getting rid of extra water, preventing further water loss. The key says not to go deeper, since kidneys have not been covered.

Exercise 2 (Assignment key · 10 pts)

A torn vessel exposes collagen; platelets adhere, trace thrombin activates them, and activated platelets make more thrombin. Identify the initial stimulus, the local response, how it amplifies, and what must happen for it to stop.

Solution to Exercise 2

Stimulus (3): platelets first contacting the exposed collagen. Vaguer answers like “a ruptured vessel” get partial credit.

Response (3): adhered platelets are activated by existing thrombin and start producing more thrombin.

Amplification (3): more thrombin activates more platelets, which produce more thrombin, which activates more platelets, and so on.

Cutoff (1): remove either the input (platelet contact with exposed collagen) or the amplifier (thrombin). Either answer is accepted. (slide 7)

Exercise 3 (Assignment key · 15 pts)

You are packaging paclitaxel (nonpolar) and insulin (polar) in phospholipid nanocarriers. Label a phospholipid, explain why bilayers form, state which molecules cross a membrane unaided and why, and design a carrier for each drug.

Solution to Exercise 3

Phospholipid (3): phosphate head, hydrophilic. Fatty acid tails, hydrophobic. (slide 14)

Bilayer (2): tails face inward toward each other, heads face the watery environment on both sides. It forms because each end points toward the environment it has highest affinity for.

Permeability (4): hydrophobic molecules cross easily because they share the nonpolar, fat-soluble character of the membrane core. Hydrophilic molecules cannot, because they have the opposite character to that core. (slides 24, 26)

Paclitaxel (3): a sphere of a single layer of phospholipids, heads out and tails in, with the drug surrounded by hydrophobic tails. The key calls this a micelle and accepts either the term or the description.

Insulin (3): a bilayer envelope, so heads face both outward toward the blood and inward toward the drug, surrounding a hydrophilic payload with hydrophilic heads. The key’s term is liposome or bilayer vesicle.

Exercise 4 (Assignment key · 12 pts)

Give normal extracellular and intracellular concentrations and the gradient direction for Na⁺, K⁺, Cl⁻ and HCO₃⁻.

Solution to Exercise 4

Ion	Outside (mEq/L)	Inside (mEq/L)	Diffusion
Na⁺	135–145	10–15	Into cell
K⁺	3.5–5	120–140	Out of cell
Cl⁻	95–105	4–10	Into cell
HCO₃⁻	22–28	10–12	Into cell

One point per blank. Any value inside the range scores. (slide 23)

Exercise 5 (Assignment key · 20 pts)

Classify five transporters. Alpha moves Na⁺ in and H⁺ out. Gamma lets K⁺ out. Beta moves Na⁺ and HCO₃⁻ in together. Delta moves Na⁺, K⁺ and Cl⁻ in together. Epsilon moves Na⁺ out and K⁺ in using ATP.

Solution to Exercise 5

Alpha: exchanger. Na⁺ moves down its gradient, H⁺ up. “Secondary active transport” is justified because the Na⁺ gradient it runs on was created by the Na⁺-K⁺ pump, which is active transport. (slide 28)

Gamma: channel. No energy input, only dissipation of an existing gradient. The outward arrow matches K⁺'s outward gradient, so it is moving down it. (slide 27)

Beta: cotransporter. Na⁺ and HCO₃⁻ both move down their gradients.

Delta: cotransporter. Na⁺ down, K⁺ up, Cl⁻ down. This is the NKCC.

Why the lecture’s “just follow Na⁺” works (4): all gradients actually sum as one vector, but the Na⁺ gradient is so large it dominates the sum. So if the Na⁺ gradient weakens, the net driving force weakens and transport of the other ions falls, even though direction does not change.

Epsilon: pump, both ions moving against their gradients. It needs ATP because neither solute moves down its gradient, so there is no existing gradient energy to exploit. (slide 29)

Exercise 6 (Assignment key · 17 pts)

Normal saline is 9 g NaCl in 1 kg water. Find its molality and osmolality and classify it. Then make 0.45% saline, find its osmolality, and predict what happens to a red blood cell placed in it.

Solution to Exercise 6

\frac{9\ \text{g}}{58.45\ \text{g/mol}} = 0.154\ \text{mol} \;\rightarrow\; 0.154\ \text{mol/kg} = 154\ \text{mmol/kg}

(1)

154\ \text{mmol/kg} \times 2\ \text{particles} = 308\ \text{mOsm/kg}

(2)

Normal range is 280–310 mOsm/kg, so 308 is iso-osmotic. (slides 33–34)

0.45% is half as concentrated, so 4.5 g NaCl in 1 kg water. The rationale is worth 2 of the 3 points.

\frac{4.5\ \text{g}}{58.45\ \text{g/mol}} = 0.077\ \text{mol/kg} \;\rightarrow\; \times 2 = 154\ \text{mOsm/kg}

(3)

Red cell (4): water moves into the cell, because osmosis moves water from the less concentrated to the more concentrated compartment. The cell swells and, left unchecked, bursts. Since structure dictates function, it cannot deliver oxygen properly. (slides 34–37)

Exercise 7 (Assignment key · 6 pts)

Pure water leaks into blood in a bypass circuit, suddenly dropping intravascular osmolality. Why is this life-threatening, and why can’t red cells fix it by moving solute out?

Solution to Exercise 7

Threat (3): the new osmotic gradient between cell interior and plasma drives water into red cells, triggering widespread hemolysis. Losing that many red cells removes oxygen-carrying capacity, and organs start shutting down.

Why they can’t compensate (3): the cell’s major solutes, hemoglobin and the cations Na⁺, K⁺ and Ca²⁺, cannot pass through the membrane, so the cell cannot shrink the gradient by exporting them.

Chart analysis¶

Exercise 8 (Chart analysis · 6 pts)

Diffusion simulation data.

Trial	Solute	Temp	Gradient	Time to equilibrium
1	Medium	40°C	5%	262 s
2	Small	40°C	5%	79 s
3	Small	30°C	5%	111 s
4	Small	40°C	2%	61 s

Calculate each rate, then state which factor each comparison isolates and what it shows.

Solution to Exercise 8

\text{rate} = \frac{\Delta[\text{solute}]}{\text{time}}, \quad \Delta = \text{half the starting gradient}

(4)

Trial 1: 2.5/262 = 0.0095 %/s. Trial 2: 2.5/79 = 0.0316. Trial 3: 2.5/111 = 0.0225. Trial 4: 1.0/61 = 0.0164.

1 vs 2 isolates solute size. The small solute diffuses about 3.3× faster because molecular weight sits in the denominator of the rate expression.

2 vs 3 isolates temperature. Cooling slows molecular movement, so the rate drops by about 29%.

2 vs 4 isolates gradient. The smaller gradient reached equilibrium sooner because less change was needed, but the rate fell by about half because gradient magnitude drives the rate. (Lecture 1D)

Variant¶

Exercise 9 (Variant · 8 pts)

Infuse 2 L of pure water into a person with 14 L extracellular fluid and 28 L intracellular fluid, both at 290 mOsm/kg. Find the final osmolality, the final volume of each compartment, and the volume that shifts.

Solution to Exercise 9

Assumptions: 1 kg water = 1 L. Solute cannot cross the membrane. Water moves until both compartments match.

\text{ECF solute} = 14\ \text{L} \times 290\ \text{mOsm/L} = 4{,}060\ \text{mOsm} \qquad \text{ICF solute} = 28\ \text{L} \times 290\ \text{mOsm/L} = 8{,}120\ \text{mOsm}

(5)

\text{final osmolality} = \frac{12{,}180\ \text{mOsm}}{44\ \text{L}} = 276.8\ \text{mOsm/kg}

(6)

\text{ECF}_{\text{final}} = \frac{4{,}060\ \text{mOsm}}{276.8\ \text{mOsm/L}} = 14.67\ \text{L} \qquad \text{ICF}_{\text{final}} = \frac{8{,}120\ \text{mOsm}}{276.8\ \text{mOsm/L}} = 29.33\ \text{L}

(7)

Right after infusion the ECF is 16 L, so 1.33 L shifts into the cells. Check: 14.67 + 29.33 = 44 L. (slide 34)

Exercise 10 (Variant · 4 pts)

Calculate the osmotic pressure of normal saline at body temperature, given an osmotic coefficient of 0.93.

Solution to Exercise 10

\pi = n \cdot C \cdot R \cdot T \cdot \phi = 2 \times 0.154\ \tfrac{\text{mol}}{\text{kg}} \times 62.36\ \tfrac{\text{L}\cdot\text{mmHg}}{\text{K}\cdot\text{mol}} \times 310\ \text{K} \times 0.93

(8)

\pi \approx 5{,}540\ \text{mmHg}

(9)

State the assumption that 1 kg water = 1 L so the units reduce to mmHg. (slide 35)

Exercise 11 (Variant · 4 pts)

In the diffusion simulation, halving the gradient from 5% to 2% shortened time to equilibrium but lowered the rate. Explain both.

Solution to Exercise 11

Time fell, from 79 s to 61 s, because equilibrium sat at 1% rather than 2.5%, so there was less concentration change to accomplish. The rate, change divided by time, fell from 0.0316 to 0.0164 %/s, because gradient magnitude sits in the numerator of the diffusion expression and directly drives the rate.

Exercise 12 (Variant · 6 pts)

Make 0.6% NaCl (6 g in 1 kg water). Find its molality and osmolality, classify it, and predict what a red blood cell does in it.

Solution to Exercise 12

\frac{6\ \text{g}}{58.45\ \text{g/mol}} = 0.103\ \text{mol/kg} = 103\ \text{mmol/kg}

(10)

103\ \text{mmol/kg} \times 2\ \text{particles} = 205\ \text{mOsm/kg}

(11)

Below 280, so hypo-osmotic. Water moves into the red cell, which swells and may hemolyze, so it cannot deliver oxygen properly. (slides 33–37)

Exercise 13 (Variant · 5 pts)

A 3% NaCl solution (30 g in 1 kg water). Osmolality, classification, and the red cell response?

Solution to Exercise 13

\frac{30\ \text{g}}{58.45\ \text{g/mol}} = 0.513\ \text{mol/kg} \;\rightarrow\; \times 2 = 1{,}027\ \text{mOsm/kg}

(12)

Hyper-osmotic. Water leaves the cell and it shrivels: crenation.

Exercise 14 (Variant · 4 pts)

A 5% glucose solution (50 g in 1 kg water, molecular weight 180.18). Osmolality and classification?

Solution to Exercise 14

\frac{50\ \text{g}}{180.18\ \text{g/mol}} = 0.2775\ \text{mol/kg} \;\rightarrow\; \times 1\ \text{particle} = 277.5\ \text{mOsm/kg}

(13)

Glucose is covalent and does not dissociate, so n = 1. At 277.5 it sits just below the 280–310 range: slightly hypo-osmotic. The trap is multiplying by 2.

Exercise 15 (Variant · 6 pts)

Infuse 1 L of normal saline (308 mOsm/kg) into a person with 14 L ECF and 28 L ICF at 290 mOsm/kg. Where does the fluid end up?

Solution to Exercise 15

Assumptions: 1 kg = 1 L; NaCl stays in the ECF; water equilibrates.

\text{ECF solute} = 14\ \text{L} \times 290\ \text{mOsm/L} + 308\ \text{mOsm} = 4{,}368\ \text{mOsm} \qquad \text{ICF solute} = 8{,}120\ \text{mOsm}

(14)

\text{final} = \frac{12{,}488\ \text{mOsm}}{43\ \text{L}} = 290.4\ \text{mOsm/kg} \qquad \text{ECF} = \frac{4{,}368\ \text{mOsm}}{290.4\ \text{mOsm/L}} = 15.04\ \text{L}

(15)

Almost all of the liter stays in the ECF. Only about 0.04 L moves out of the cells, because saline is nearly iso-osmotic. Compare pure water, which pulls most of its volume into the cells.

Exercise 16 (Variant · 8 pts)

Map body temperature regulation on a hot day onto every part of the negative feedback loop.

Solution to Exercise 16

Regulated variable: body temperature. Perturbation: hot day. Sensors: thermoreceptors in skin, brain, muscles, liver. Afferent: sensory nerves. Integrating center and set point: preoptic anterior hypothalamus. Error signal out along efferent motor nerves. Effectors: skin blood vessels dilate (convective and radiative heat loss) and sweat glands sweat (evaporative). (slide 6)

Exercise 17 (Variant · 6 pts)

Childbirth as positive feedback: stimulus, response, how it amplifies, and the cutoff.

Solution to Exercise 17

Stimulus: the baby’s head presses on the cervix. Response: cervical stretch releases oxytocin, which drives stronger uterine contractions. Amplification: stronger contractions press the head harder, releasing more oxytocin. Cutoff: delivery removes the pressure. (slide 7)

Exercise 18 (Variant · 8 pts)

Classify each and give its energy source: (a) the Na⁺-Ca²⁺ exchanger, (b) the sodium-glucose transporter, (c) SERCA, (d) an aquaporin.

Solution to Exercise 18

(a) Countertransporter. Na⁺ in down its gradient, Ca²⁺ out up its gradient. Secondary active: the Na⁺ gradient was built by the Na⁺-K⁺ pump.

(b) Cotransporter. Na⁺ in down its gradient drives glucose in.

(c) Pump (ATPase). Ca²⁺ moved up its gradient into the SR using ATP.

(d) Channel for water. Osmosis supplies the driving force. (slides 27–29, 32)

Check yourself¶

Exercise 19 (Check yourself · Check yourself 1)

Draw the negative feedback schematic from memory and label all seven components, then map body temperature regulation onto it.

Solution to Exercise 19

Regulated variable on a seesaw → perturbation → sensor → afferent pathway → integrating center compares with the set point → error signal → efferent pathway → effector. Body temperature: thermoreceptors in skin, brain, muscle, liver → afferent sensory nerves → preoptic anterior hypothalamus (center and set point) → efferent motor nerves → skin vessels dilate and sweat glands sweat. (M1 slides 5–6)

Exercise 20 (Check yourself · Check yourself 2)

Why does positive feedback require an external cutoff switch when negative feedback does not?

Solution to Exercise 20

Negative feedback removes its own stimulus: as the variable returns to normal the error signal shrinks, so the loop shuts itself off. Positive feedback intensifies its stimulus, so nothing inside the loop stops it. Left running it drives the variable dangerously far from normal, which is why an outside event such as delivery of the baby has to end it. (M1 slide 7)

Exercise 21 (Check yourself · Check yourself 3)

Explain, using the phospholipid structure, why the membrane is a good barrier between two watery compartments.

Solution to Exercise 21

Hydrophilic phosphate heads face the watery fluid on both sides and the hydrophobic fatty acid tails face each other, forming a lipid core. Water and anything that dissolves in water cannot cross that core, so the two watery compartments cannot mix. (M1 slide 14)

Exercise 22 (Check yourself · Check yourself 4)

Given a molecule’s structural formula, decide whether it is nonpolar, polar, or ionic, and predict whether it needs a transport protein.

Solution to Exercise 22

Mostly C–H bonds with few oxygens and even sharing: nonpolar, crosses unaided by simple diffusion. Covalent with uneven sharing (O–H, many oxygens) and regional partial charges: polar, needs a carrier. Lost or gained electrons and a net charge: ionic, needs a channel. Polar and ionic solutes are both hydrophilic and cannot cross the lipid core. (M1 slides 18–20)

Exercise 23 (Check yourself · Check yourself 5)

List the seven factors in the diffusion rate expression and state whether each raises or lowers the rate.

Solution to Exercise 23

Raise the rate: concentration gradient, surface area, temperature, membrane permeability. Lower it: membrane thickness, solute molecular weight, density of the medium. Electrical forces and distance also matter outside the expression. (M1 slide 24)

Exercise 24 (Check yourself · Check yourself 6)

The simulation cut the gradient from 5% to 2% and equilibrium arrived sooner, yet the rate fell. Explain both facts.

Solution to Exercise 24

Equilibrium sits halfway, so a 2% gradient only needs a 1% change against 2.5% for a 5% gradient. Less change takes less time (61 s against 79 s). The rate, change divided by time, still fell (0.0164 against 0.0316 %/s) because gradient magnitude drives the rate. (Lecture 1D)

Exercise 25 (Check yourself · Check yourself 7)

Name the four receptor types and state which one is not membrane-bound and why.

Solution to Exercise 25

Ligand-gated ion channel, G protein-coupled receptor, catalytic receptor, and intracellular receptor. The intracellular receptor is not in the membrane because its ligands are hydrophobic: they cross the membrane on their own and bind inside, then act on gene expression. (M1 slide 16)

Exercise 26 (Check yourself · Check yourself 8)

Why does a cotransporter still count as passive transport when solute B may move up its gradient?

Solution to Exercise 26

No ATP is used. Solute A runs down its own gradient and the energy stored in that gradient pushes the “revolving door,” carrying B along even when B moves up its gradient. The transporter only spends an existing gradient, so it counts as passive (the key calls it secondary active, since a pump built the Na⁺ gradient elsewhere). (M1 slide 28)

Exercise 27 (Check yourself · Check yourself 9)

Calculate the osmolality of a 0.15 molal NaCl solution and of a 0.30 molal glucose solution. Which pulls water harder?

Solution to Exercise 27

0.15\ \text{molal NaCl} \times 2 = 0.30\ \text{Osm/kg} = 300\ \text{mOsm/kg} \qquad 0.30\ \text{molal glucose} \times 1 = 300\ \text{mOsm/kg}

(16)

They are equal, so they pull water equally. NaCl dissociates into two particles; glucose, covalently bonded, stays as one. (M1 slide 33)

Exercise 28 (Check yourself · Check yourself 10)

Infuse 2 L of pure water into a 14 L extracellular compartment at 290 mOsm alongside a 28 L intracellular compartment at 290. Find the new equilibrium osmolality and the volume shift.

Solution to Exercise 28

\text{total solute} = 42\ \text{L} \times 290\ \text{mOsm/L} = 12{,}180\ \text{mOsm} \;\rightarrow\; \frac{12{,}180\ \text{mOsm}}{44\ \text{L}} = 276.8\ \text{mOsm/kg}

(17)

ECF ends at 4,060 ÷ 276.8 = 14.67 L, ICF at 29.33 L. Right after infusion the ECF held 16 L, so 1.33 L moves into the cells. Assume 1 kg water = 1 L and that no solute crosses. (M1 slide 34)

Exercise 29 (Check yourself · Check yourself 11)

A red cell is placed in a 320 mOsm bath. Predict the direction of water movement, the resulting shape, and the name of the process.

Solution to Exercise 29

320 mOsm is more concentrated than the cell (about 290), so water leaves the cell. The cell shrivels like a raisin: crenation. Its changed shape impairs its function. (M1 slide 37)

Exercise 30 (Check yourself · Check yourself 12)

Explain how a neuron in the brain converts a change in extracellular fluid osmolality into an electrical signal.

Solution to Exercise 30

Concentrated extracellular fluid pulls water out of the neuron, which shrivels. The shriveling physically pulls open mechanically gated ion channels. Ions flow, the membrane potential changes, and the neuron fires, signaling vasopressin release and thirst. Dilute fluid swells the cell, the channels close, and the signals stop. (M1 slide 37)