Answers · Excretion and the kidneys

Inside a nephron — answer key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. NQ1 · Practice

Options as numbered on the pupil sheet: 1 — capsule; 2 — collecting duct; 3 — capillary; 4 — glomerulus; 5 — tubule.

Model answer. A, the knot of capillaries, is the glomerulus. B, the cup round it, is the capsule. C, the long narrow tube that coils and loops, is the tubule. D, the wide tube that C joins, is the collecting duct. E, beside the tubule, is a capillary.

A full answer includes

I matched A to glomerulus.
I matched B to capsule.
I matched C to tubule.
I matched D to collecting duct.
I matched E to capillary.
I did not match A to capsule: the capsule is the cup round the knot.

Also accept

A glomerulus; B capsule; C tubule; D collecting duct; E capillary

Watch for

“A is the capsule.” — A is the knot of capillaries, the glomerulus. The capsule (B) is the cup round it.
“E is the tubule.” — E is a blood capillary beside the tubule. The tubule is C.

2. NQ2 · Practice

Model answer. Water, glucose, urea and ions are small enough to pass through the filter, so all four are in the filtrate. Proteins and red blood cells are too large: they stay in the blood.

A full answer includes

I chose water.
I chose glucose.
I chose urea.
I chose ions.
I did not choose protein: its molecules are too large to pass the filter.

Also accept

water; glucose; urea; ions

Watch for

“Protein.” — Protein molecules are too large to pass through the filter.
“Red blood cells.” — A red blood cell is far larger than the gaps in the filter, so it stays in the blood.

3. NQ3 · Practice

Model answer. The high pressure of the blood in the glomerulus is what pushes water and small molecules through the gaps into the capsule. At a lower pressure less liquid is pushed through, so less filtrate is made, and less urine. The gaps are the same size, so the filtrate still has water, glucose, urea and ions in it, and still no protein.

A full answer includes

I chose: less filtrate is made, because less liquid is pushed through.
I did not choose "the same filtrate": the gap size decides what can pass, but the pressure pushes it through.

Also accept

less filtrate is made, because less liquid is pushed through the filter

Watch for

“More filtrate is made, because the blood flows through more slowly.” — Filtration is driven by pressure, not by time. Lower pressure pushes less liquid through the gaps.
“The same filtrate is made, because the gaps are the same size.” — The gap size decides which molecules can pass. The pressure decides how much liquid is pushed through.
“Filtrate with no glucose is made, because glucose needs more push.” — The pressure pushes every small molecule alike. The filtrate still has glucose in it; there is just less filtrate.

4. NQ4 · Practice

Model answer. All of the glucose, most of the water (about 99%) and some of the ions are reabsorbed into the blood. Far less of the urea is taken back: much of it stays in the tubule and leaves in the urine.

A full answer includes

I chose: all of the glucose is reabsorbed.
I chose: most of the water is reabsorbed.
I chose: some of the ions are reabsorbed.
I did not choose: all of the urea is reabsorbed.

Also accept

all glucose; most water; some ions

Watch for

“All of the urea is reabsorbed.” — Urea is waste: much of it stays in the tubule and leaves in the urine.
“None of the water is reabsorbed.” — Most of the water is reabsorbed. Otherwise the body would lose far more water than it drinks.

5. NQ5 · Practice

Model answer. Reabsorbed substances pass out of the tubule into the blood in the capillaries that wrap round it. The blood then flows to the renal vein.

A full answer includes

I chose: into the blood in the capillaries round the tubule.
I did not choose the collecting duct: it carries what is not reabsorbed.

Also accept

into the blood in the capillaries round the tubule

Watch for

“Into the urine that flows down the collecting duct.” — The collecting duct carries urine, the part that is not reabsorbed.
“Into the bladder, where they are stored for later.” — The bladder stores urine, which is what is not reabsorbed.
“Back into the capsule, to be filtered once again.” — Reabsorbed substances go into the blood, not back into the nephron.

6. NQ6 · Practice

Options as numbered on the pupil sheet: 1 — It flows along the tubule.; 2 — It flows down the collecting duct.; 3 — It drains into the ureter.; 4 — It passes into the capsule.; 5 — It is in the blood in the glomerulus..

Model answer. The urea molecule is filtered out of the blood in the glomerulus into the capsule. It flows along the tubule, which does not take it back, into the collecting duct, and from there it drains into the ureter.

A full answer includes

I put the glomerulus first.
I put the capsule second.
I put the tubule third.
I put the collecting duct fourth.
I put the ureter last.
I did not put the collecting duct before the tubule.

Also accept

glomerulus; capsule; tubule; collecting duct; ureter

Watch for

“It is reabsorbed in the tubule.” — Much of the urea stays in the tubule: it is waste, and this molecule goes on to the ureter.
“The collecting duct comes before the tubule.” — Many tubules join one collecting duct, so the tubule comes first.

7. NQ7 · Practice

Model answer. The urine is 65 times as concentrated as the plasma, so the plasma concentration = 26 ÷ 65 = 0.4 g per dm³. The urea becomes so much more concentrated because most of the water is reabsorbed along the tubule, while far less of the urea is taken back.

Checked automatically

Concentration of urea in the plasma: 0.40 g per dm³ (2 significant figures)
- Common wrong answer 1690 g per dm³: That multiplies. The plasma is less concentrated than the urine, so divide by 65.
- Common wrong answer 2.5 g per dm³: That divides 65 by 26. Divide the urine concentration by 65.

A full answer includes

Method: I divided the urine concentration by 65: 26 ÷ 65.
Answer: I gave 0.4 g per dm³.
I did not multiply 26 by 65.

Also accept

0.4 g per dm³

Watch for

“1690 g per dm³.” — That is 26 × 65. The plasma is the less concentrated liquid, so divide: 26 ÷ 65 = 0.4.
“2.5 g per dm³.” — That is 65 ÷ 26. Divide the urine concentration by 65: 26 ÷ 65 = 0.4.

8. NQ8 · Practice

Model answer. The water not reabsorbed is 100 − 99.2 = 0.8% of the filtrate. Volume of urine = 160 × 0.8 ÷ 100 = 1.28 dm³ = 1.3 dm³ to 1 decimal place. (Or: water reabsorbed = 160 × 99.2 ÷ 100 = 158.72 dm³, and 160 − 158.72 = 1.28 dm³.)

Checked automatically

Volume of urine: 1.3 dm³ (2 significant figures)
- Common wrong answer 158.72 dm³: That is the water reabsorbed. Take it from 160 dm³.
- Common wrong answer 12.8 dm³: That is 8% of the filtrate. Only 0.8% is left in the urine.

A full answer includes

Method: I found the percentage left in the urine: 100 − 99.2 = 0.8%.
Method: I found 0.8% of the filtrate: 160 × 0.8 ÷ 100 = 1.28 dm³.
Answer: I gave 1.3 dm³.
I did not give 158.7 dm³: that is the water reabsorbed.

Also accept

1.3 dm³

Watch for

“158.7 dm³.” — That is the water reabsorbed. The urine is what is left: 160 − 158.72 = 1.28 dm³, which is 1.3 dm³.
“12.8 dm³.” — That uses 8% instead of 0.8%. 100 − 99.2 = 0.8, so the urine is 0.8% of 160 dm³: 1.3 dm³.

9. NQ9 · Practice

Model answer. (a) Ions. 1150 g of ions were filtered but only 14 g were in the urine, so most of the ions were reabsorbed and some were not. Glucose is all reabsorbed (150 g filtered, none in the urine), and protein is not filtered at all. (b) Ions reabsorbed = 1150 − 14 = 1136 g.

A full answer includes

(a) I identified ions.
(a) I quoted 14 g of ions in the urine out of 1150 g filtered.
(a) I gave the evidence that some ions are left in the urine, but far less than was filtered.
(b) Method: I took the mass in the urine from the mass filtered: 1150 − 14.
(b) Answer: I gave 1136 g.
I did not identify glucose: none of it is left in the urine, so all of it is reabsorbed.

Also accept

(a) ions; (b) 1136 g

Watch for

“(a) Glucose.” — All 150 g of the glucose filtered was reabsorbed: none is left in the urine.
“(a) Protein.” — Protein is not filtered at all (0 g), because its molecules are too large, so it is not reabsorbed either.
“(b) 1164 g.” — That adds the masses. Reabsorbed = filtered − left in the urine.

10. NQ10 · Practice

Model answer. The blue dye appears in the urine. Its molecules are smaller than glucose, so they pass through the filter in the glomerulus into the capsule. The tubules do not take it back, so it stays in the filtrate and leaves in the urine. The red dye's molecules are larger than protein molecules, which are already too large to pass the filter, so the red dye stays in the blood.

A full answer includes

I predicted that the blue dye appears in the urine.
I explained that its small molecules pass through the filter into the capsule.
I explained that it is not reabsorbed, so it stays in the filtrate and leaves in the urine.
I predicted that the red dye stays in the blood.
I explained that its molecules are too large to pass through the filter.
I did not say that both dyes are filtered because both are in the blood.

Also accept

blue dye in urine: small, filtered, not reabsorbed; red dye stays in the blood: too large

Watch for

“Both dyes appear in the urine.” — Only molecules small enough to pass the filter enter the capsule. The red dye's molecules are larger than protein, so they stay in the blood.
“Neither dye appears, because the body keeps everything useful.” — The filter does not pick useful substances: every small molecule passes it. The blue dye is not reabsorbed, so it leaves in the urine.

11. NT1 · Apply

Model answer. Glucose is found in their urine. Glucose passes the filter, so the filtrate has the same glucose concentration as the plasma: about three times as much glucose is filtered as normal. The tubules can reabsorb at most about twice the normal amount, so about the normal amount's worth of glucose is left in the tubule. It cannot be taken back, so it leaves in the urine. (A doctor can test urine for glucose for this reason.)

A full answer includes

I predicted that glucose is found in the urine.
I explained that about three times the normal amount of glucose is filtered.
I explained that the tubules can reabsorb only about twice the normal amount.
I explained that the glucose left in the tubule leaves in the urine.
I did not say that the glucose is not filtered.

Also accept

glucose in urine; three times as much filtered; only about twice can be reabsorbed; the rest leaves in urine

Watch for

“No glucose, because the tubules always reabsorb all of it.” — They reabsorb all of it only up to their limit, about twice the normal amount. Three times the normal amount is filtered here.
“The glucose is not filtered.” — Glucose is always filtered. Here, more is filtered than the tubules can take back.

12. NT2 · Apply

Model answer. Protein normally stays in the blood because its molecules are too large to pass the filter in the glomerulus. If protein appears in urine, the filter must be damaged: its gaps have become large enough to let protein through into the capsule, and the tubule does not take it back.

A full answer includes

I suggested that the filter in the glomerulus is damaged.
I explained that its gaps now let protein molecules through into the capsule.
I did not say that the tubule stopped reabsorbing protein: protein is normally never filtered.

Also accept

the filter is damaged, so protein passes into the filtrate

Watch for

“The tubule stopped reabsorbing protein.” — Protein is normally never in the filtrate. It gets there only if the filter is damaged.

13. NT3 · Apply

Model answer. Y has protein (75 g per dm³), which is too large to be filtered, so Y is plasma. Z has the same glucose (0.9) and urea (0.25) as Y but no protein, so Z is filtrate. X has no glucose (all reabsorbed) and a high urea concentration, 16 g per dm³ (most of the water reabsorbed), so X is urine.

A full answer includes

I identified X as urine.
I explained that X has no glucose.
I explained that X has a far higher urea concentration than Y and Z.
I identified Y as plasma.
I explained that Y is the only one with protein.
I identified Z as filtrate.
I explained that Z has no protein.
I explained that Z has the same glucose concentration as Y (0.9 g per dm³).
I explained that Z has the same urea concentration as Y (0.25 g per dm³).
I did not identify Z as plasma: plasma has protein.

Also accept

X urine; Y plasma; Z filtrate, with reasons

Watch for

“Z is plasma.” — Plasma contains protein; Z has none, so it has passed the filter: it is filtrate.
“X is filtrate.” — Filtrate has the same glucose as plasma. X has none, so all of it was reabsorbed: X is urine.