AS & A Level Biology (9700)
Homeostasis Banker
51 questions · about 6% of the marks we indexed · examined in 9 of 9 sessions.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Kidney & osmoregulation · Core · Paper 4 style · 5 marksEach human kidney contains about one million nephrons.
- (a) Put these parts of a nephron in the order in which filtrate passes through them: distal convoluted tubule, Bowman's capsule, collecting duct, proximal convoluted tubule, loop of Henle. [2]
- (b) State the region of the kidney (cortex or medulla) in which the glomeruli are found and the region in which most of each loop of Henle is found. [2]
- (c) Name the blood vessel that brings blood to a glomerulus and the vessel that carries blood away from it. [1]
Show answer
- (a) Bowman's capsule → proximal convoluted tubule → loop of Henle → distal convoluted tubule → collecting duct
B2 all correct; B1 if one pair is reversed.
- (b) Glomeruli: cortex; loops of Henle: medulla.
B1 each.
- (c) Afferent arteriole in; efferent arteriole out.
B1 both needed.
Ali plays cricket for three hours in Lahore on a hot afternoon. He sweats a lot and eats salty pakoras, but drinks very little.
- (a) State the effect on the water potential of Ali's blood. [1]
- (b) Describe the changes in ADH secretion and in Ali's urine that follow. [3]
- (c) Ali then drinks a large glass of water. State how the volume of his urine will change over the next hour, and why. [2]
Show answer
- (a) It decreases (becomes more negative).
B1
- (b) More ADH is released from the posterior pituitary; the collecting ducts become more permeable to water, so more water is reabsorbed into the blood; a small volume of concentrated (dark) urine is produced.
B1 more ADH. B1 more water reabsorbed in collecting duct / DCT. B1 small volume, more concentrated urine.
- (c) The volume will increase (dilute urine); blood water potential rises, so less ADH is released and less water is reabsorbed in the collecting ducts.
B1 more urine / dilute. B1 less ADH / collecting duct less permeable.
- (a) Describe how ultrafiltration takes place in the glomerulus and Bowman's capsule. [4]
- (b) Explain why the filtrate in Bowman's capsule has a lower concentration of protein than blood plasma but the same concentration of glucose. [2]
Show answer
- (a) The afferent arteriole is wider than the efferent arteriole, so blood in the glomerulus is at high hydrostatic pressure; water and small solutes are forced out through gaps (fenestrations) in the capillary endothelium; the basement membrane acts as the filter, holding back large proteins (relative molecular mass above about 69 000) and cells; the fluid passes between the podocytes' foot processes (filtration slits) into the capsule space.
1 mark each for any four of: afferent wider than efferent; high hydrostatic pressure; pores / fenestrations in endothelium; basement membrane is the filter; large proteins and blood cells retained; podocytes with filtration slits; filtrate contains water, glucose, urea, ions, amino acids.
- (b) Proteins are too large to pass through the basement membrane, so they stay in the blood; glucose molecules are small and pass freely, so their concentration is the same as in plasma.
B1 proteins too large / retained by basement membrane. B1 glucose small, filtered freely.
- (a) Describe how glucose is reabsorbed from the filtrate in the proximal convoluted tubule. [4]
- (b) Explain how the structure of the proximal convoluted tubule cells is adapted for reabsorption. [2]
Show answer
- (a) Na⁺/K⁺ pumps in the basal membrane of the tubule cells actively pump Na⁺ out into the tissue fluid/blood, using ATP; this lowers the Na⁺ concentration inside the cell; Na⁺ diffuses into the cell from the lumen through cotransporter proteins that carry glucose with it (indirect active transport / cotransport); glucose then moves out of the cell into the blood by facilitated diffusion through the basal membrane.
1 mark each for any four of: Na⁺/K⁺ pump on basal membrane; uses ATP / active; low Na⁺ concentration in cell; Na⁺ moves in by facilitated diffusion with glucose (cotransporter / symporter); glucose against its concentration gradient; glucose leaves by facilitated diffusion into capillaries.
- (b) Microvilli on the luminal surface give a large surface area for transport proteins; many mitochondria supply ATP for active transport.
B1 microvilli → large surface area. B1 many mitochondria → ATP for active transport. (Also accept: many cotransporter proteins / tight junctions.)
- (a) Describe how a fall in the water potential of the blood is detected and how ADH is released. [2]
- (b) Explain how ADH increases the permeability of the collecting duct to water. [4]
- (c) Explain why this is an example of negative feedback. [2]
Show answer
- (a) Osmoreceptors in the hypothalamus detect the fall in water potential (they lose water and shrink); neurosecretory cells (in the hypothalamus) send impulses so ADH is released from the posterior pituitary gland into the blood.
B1 osmoreceptors in hypothalamus. B1 ADH released from posterior pituitary.
- (b) ADH binds to receptors on the cell surface membrane of collecting duct cells; this activates adenylyl cyclase, producing cAMP as a second messenger; cAMP activates a protein kinase (cascade) that phosphorylates proteins; vesicles containing aquaporins move to and fuse with the luminal membrane; more aquaporins in the membrane allow more water to move out by osmosis.
1 mark each for any four of: ADH binds to receptor (on membrane of collecting duct cells); adenylyl cyclase activated / cAMP produced (second messenger); protein kinase activated / phosphorylation; vesicles with aquaporins; vesicles fuse with luminal (cell surface) membrane; more water channels → more water reabsorbed by osmosis.
- (c) The change (low water potential) triggers responses that reverse it – more water is reabsorbed so blood water potential rises back to the set point; as it returns to normal, less ADH is released.
B1 response reverses the change / returns to set point. B1 less ADH once water potential is restored.
In a healthy adult the kidneys produce glomerular filtrate at a rate of 125 cm³ per minute.
- (a) Calculate the volume of filtrate produced per day in dm³. [1]
- (b) The person produces 1.5 dm³ of urine per day. Calculate the percentage of the filtrate that is reabsorbed. Give your answer to 1 decimal place. [2]
- (c) Suggest why a person with kidney failure who is treated by dialysis must limit how much water and salt they consume. [2]
Show answer
- (a) 180 dm³
125 × 60 × 24 = 180 000 cm³ = 180 dm³
- (b) 99.2 %
M1 (180 − 1.5) = 178.5 dm³ reabsorbed. A1 178.5 ÷ 180 × 100 = 99.2 %.
- (c) Their kidneys cannot remove excess water and salt (cannot form urine normally), so these build up in the blood between dialysis sessions, raising blood volume / pressure and causing oedema.
B1 kidneys cannot excrete excess water / salt. B1 build-up → high blood pressure / oedema / fluid overload.
In diabetes insipidus the posterior pituitary gland releases little or no ADH. A patient with this condition used a nasal spray of desmopressin, a drug that acts on the collecting duct in the same way as ADH. Her rate of urine production was measured.Time after spray / h: 0, 1, 2, 4, 8, 12Urine production / cm³ min⁻¹: 10.0, 6.5, 2.0, 1.2, 1.5, 4.0
- (a) Before the spray, the patient produced urine at 10.0 cm³ min⁻¹. Calculate the volume of urine she would produce in one day at this rate, in dm³. [1]
- (b) Explain why a person who makes no ADH produces a large volume of dilute urine. [3]
- (c) Suggest why the rate of urine production rises again between 8 h and 12 h after the spray. [2]
Show answer
- (a) 14.4 dm³
B1 10.0 × 60 × 24 = 14 400 cm³ = 14.4 dm³
- (b) Without ADH, few aquaporins are inserted into the luminal membranes of the collecting duct cells (vesicles containing aquaporins stay in the cytoplasm); so the collecting duct walls have low permeability to water; little water moves out by osmosis into the medulla (even though its water potential is low), so a large volume of dilute urine passes out.
1 mark each for any three of: no ADH binding to receptors on collecting duct cells; few aquaporins in the luminal membrane; collecting duct (and DCT) walls less permeable to water; little water reabsorbed by osmosis into the medulla / blood; large volume of dilute urine.
- (c) The desmopressin is gradually broken down / removed from the blood; so fewer aquaporins remain in the collecting duct membranes, less water is reabsorbed and urine output increases – another dose is needed.
B1 drug broken down / excreted / concentration falls. B1 aquaporins removed from membrane / less water reabsorbed.
Concentrations of some substances in a healthy person (g dm⁻³):Substance — blood plasma, filtrate in Bowman's capsule, urineProtein — 80.0, 0.05, 0Glucose — 1.0, 1.0, 0Urea — 0.30, 0.30, 20.0
- (a) Explain the concentrations of protein shown in the table. [2]
- (b) Explain why glucose is present in the filtrate but absent from the urine. [2]
- (c) Calculate how many times more concentrated urea is in the urine than in the plasma. Give your answer to 3 significant figures. [1]
- (d) Explain why the urea concentration in urine is much higher than in the filtrate, although urea is not actively secreted into the tubule in large amounts. [2]
Show answer
- (a) Proteins are too large to pass through the basement membrane in ultrafiltration, so almost none enters the filtrate, and therefore none appears in the urine.
B1 too large to be filtered / held back by basement membrane. B1 so (almost) absent from filtrate and urine.
- (b) Glucose is small and is filtered; it is all reabsorbed (by cotransport with Na⁺) in the proximal convoluted tubule.
B1 filtered because small. B1 all reabsorbed in the PCT.
- (c) 66.7 times
20.0 ÷ 0.30 = 66.7
- (d) About 99 % of the water in the filtrate is reabsorbed (by osmosis in the PCT, loop of Henle and collecting duct); only some urea is reabsorbed, so the same mass of urea is in a much smaller volume.
B1 most water reabsorbed. B1 urea not reabsorbed to the same extent, so concentrated.
- (a) Explain how the loop of Henle creates a low water potential in the tissue fluid of the medulla. [4]
- (b) Explain how the low water potential of the medulla allows concentrated urine to be produced. [2]
- (c) Desert gerbils in the Thar desert produce urine far more concentrated than human urine. Suggest how the structure of their kidneys differs from human kidneys. [1]
Show answer
- (a) Cells of the ascending limb actively transport Na⁺ and Cl⁻ out of the filtrate into the tissue fluid of the medulla; the ascending limb is impermeable to water, so water cannot follow; this lowers the water potential of the medulla; the descending limb is permeable to water, so water leaves it by osmosis and the filtrate becomes more concentrated as it goes down; the countercurrent arrangement builds up the highest solute concentration (lowest water potential) at the tip of the loop / deep in the medulla.
1 mark each for any four of: Na⁺ and Cl⁻ pumped out of ascending limb (active transport); ascending limb impermeable to water; low water potential / high solute concentration in medulla tissue fluid; descending limb permeable to water, water leaves by osmosis; filtrate most concentrated at the bottom of the loop; countercurrent multiplier – opposite flow maintains gradient along the whole loop.
- (b) Filtrate in the collecting duct passes through the medulla; water moves out of the collecting duct by osmosis into the medulla tissue fluid (when the walls are permeable, with ADH), so the urine becomes concentrated.
B1 collecting duct passes through medulla. B1 water leaves by osmosis down water potential gradient (when ADH present).
- (c) They have relatively longer loops of Henle (thicker medulla), so a steeper water potential gradient builds up and more water can be reabsorbed.
B1 longer loops of Henle / thicker medulla.
In an untreated diabetic patient, the glomerular filtration rate is 0.125 dm³ min⁻¹. The kidney tubules of this patient can reabsorb glucose at a maximum rate of 1.25 mmol min⁻¹. Assume the glucose concentration in the filtrate equals that in the plasma.
- (a) Calculate the plasma glucose concentration above which glucose will start to appear in this patient's urine. [1]
- (b) Calculate the rate at which glucose is lost in the urine when the plasma glucose concentration is 15 mmol dm⁻³. [2]
- (c) Explain why there is a maximum rate at which glucose can be reabsorbed. [2]
- (d) Suggest why a person with glucose in the urine often feels thirsty and produces large volumes of urine. [2]
Show answer
- (a) 10 mmol dm⁻³
1.25 ÷ 0.125 = 10 mmol dm⁻³
- (b) 0.625 mmol min⁻¹
M1 filtered = 0.125 × 15 = 1.875 mmol min⁻¹. A1 1.875 − 1.25 = 0.625 mmol min⁻¹.
- (c) Glucose is reabsorbed by a limited number of cotransporter proteins in the PCT cell membranes; when all are working at their maximum rate (saturated), extra glucose cannot be reabsorbed.
B1 limited number of carrier / cotransporter proteins. B1 all occupied / saturated.
- (d) Glucose in the filtrate lowers its water potential, so less water is reabsorbed by osmosis from the tubules; more water is lost in urine, lowering blood water potential, which stimulates thirst.
B1 glucose in filtrate lowers water potential / less water reabsorbed by osmosis. B1 large urine volume → dehydration → thirst.
- (a) Describe how a stoma opens in the light. [4]
- (b) Explain why stomata opening in the light is an advantage to the plant. [1]
Show answer
- (a) Proton pumps in the guard cell membrane use ATP to pump H⁺ out of the guard cells; this creates an electrochemical gradient so K⁺ ions diffuse in through channel proteins; the water potential of the guard cells falls; water enters by osmosis and the guard cells become turgid; the thicker, less elastic inner walls (cellulose microfibrils arranged radially) make the cells bend apart, opening the pore.
1 mark each for any four of: H⁺ pumped out (active, ATP); K⁺ enters through channels; water potential of guard cells lowered; water enters by osmosis; guard cells become turgid; uneven wall thickening / microfibril arrangement causes cells to curve and pore opens.
- (b) Allows carbon dioxide to diffuse into the leaf for photosynthesis when light is available.
B1
Wheat growing in rain-fed fields in Potohar often experiences drought. When the soil dries, the roots and leaves produce more abscisic acid (ABA).
- (a) Describe how abscisic acid (ABA) causes stomata to close. [4]
- (b) Explain why stomatal closure during drought helps the plant survive but reduces its growth. [2]
Show answer
- (a) ABA binds to receptors on the guard cell surface membrane; this inhibits the proton pumps (H⁺ no longer pumped out); Ca²⁺ ions enter the cytoplasm and act as a second messenger; Ca²⁺ opens channels so K⁺ (and anions) leave the guard cells; water potential of guard cells rises, water leaves by osmosis; guard cells become flaccid and the stoma closes.
1 mark each for any four of: ABA binds receptors on guard cells; proton pumps inhibited; Ca²⁺ enters cytoplasm / second messenger; K⁺ channels open, K⁺ leave; water potential of guard cells increases; water leaves by osmosis; guard cells flaccid → stoma closes.
- (b) Closing stomata reduces water loss by transpiration so the plant does not wilt / die; but it also reduces carbon dioxide uptake, so photosynthesis and growth slow down.
B1 less transpiration / water conserved. B1 less CO₂ uptake → less photosynthesis.
The mean width of stomatal pores on the leaves of a cotton plant in a field near Bahawalpur was measured during a hot, dry summer day.Time: 00:00, 04:00, 08:00, 10:00, 14:00, 17:00, 20:00Mean pore width / µm: 0.5, 0.5, 6.0, 9.0, 5.5, 4.0, 1.0
- (a) Describe the pattern shown by the data. [2]
- (b) Suggest why the stomata partly closed at 14:00 even though light intensity was high. [2]
- (c) Explain why closing stomata at night does not reduce the rate of photosynthesis. [1]
Show answer
- (a) Stomata are almost closed at night (0.5 µm at 00:00 and 04:00); they open after dawn to a maximum of 9.0 µm at 10:00, partly close at 14:00 (5.5 µm) and close again by evening (1.0 µm at 20:00).
B1 closed at night, open in morning, with data. B1 partial closure at midday / early afternoon.
- (b) At 14:00 temperature is highest and humidity lowest, so transpiration is very fast; the plant loses water faster than roots can absorb it; ABA is produced / guard cells lose turgor, so stomata partly close to reduce water loss.
B1 high temperature / low humidity → high transpiration / water stress. B1 ABA / loss of turgor causes partial closure to conserve water.
- (c) There is no light at night so photosynthesis cannot take place anyway; closing saves water.
B1
A student painted clear nail varnish on the lower surface of a leaf, peeled it off and viewed the impression with a microscope. The diameter of the field of view at high power was 0.40 mm. The mean number of stomata in ten fields of view was 23.
- (a) Calculate the area of the field of view in mm². Give your answer to 3 significant figures. [1]
- (b) Calculate the stomatal density of the lower epidermis in stomata per mm². Give your answer to the nearest whole number. [1]
- (c) Leaves treated with ABA solution were examined and many stomata were found to be closed. Suggest why ABA treatment does not change the stomatal density. [1]
- (d) Suggest why the student counted stomata in ten fields of view rather than one. [1]
Show answer
- (a) 0.126 mm²
π × 0.20² = 0.126 mm²
- (b) 183 stomata mm⁻²
23 ÷ 0.1257 = 183
- (c) Stomatal density is the number of stomata per unit area, which is set during leaf development; ABA only changes whether existing stomata are open or closed.
B1
- (d) Stomata are not evenly spread, so a mean of several fields gives a more representative / reliable value.
B1
Scientists compared wild-type plants of a species with a mutant strain. Mean stomatal pore widths (µm):Well watered: wild type 8.2, mutant 8.4After 3 days without water: wild type 1.9, mutant 7.6 (mutant plants wilted)
- (a) Compare the responses of the wild type and the mutant to drought. [2]
- (b) When ABA solution was sprayed on the leaves of well-watered plants, wild-type stomata closed to 2.0 µm but mutant stomata stayed at 8.1 µm. Use all the data to suggest what is wrong in the mutant, and explain your reasoning. [3]
- (c) Suggest why the mutant wilts during drought. [1]
Show answer
- (a) Both have similar wide pores when well watered (8.2 and 8.4 µm); after drought the wild type stomata close to 1.9 µm, but the mutant's stay open (7.6 µm) and the mutant wilts.
B1 similar when watered. B1 wild type closes in drought, mutant does not, with data.
- (b) The mutant can probably make ABA but cannot respond to it: it is likely to have a faulty ABA receptor (or a faulty part of the signalling pathway in the guard cells); if it simply could not make ABA, spraying ABA would have closed its stomata; without the response, K⁺ is not lost from guard cells so they stay turgid.
B1 mutant does not respond to ABA / receptor or signalling defect. B1 reasoning: externally applied ABA had no effect, so not a lack of ABA synthesis. B1 guard cells stay turgid / K⁺ not lost / Ca²⁺ signal absent.
- (c) Stomata stay open, so water continues to be lost by transpiration faster than it can be taken up.
B1
After a meal of rice and daal, the blood glucose concentration of a healthy person rises.
- (a) Name the cells in the pancreas that secrete insulin. [1]
- (b) Describe how insulin causes the blood glucose concentration to decrease. [4]
- (c) Explain why only certain cells respond to insulin. [1]
Show answer
- (a) β (beta) cells of the islets of Langerhans.
B1
- (b) Insulin binds to receptors on target cells (liver, muscle, fat cells); in muscle and fat cells, vesicles containing GLUT4 glucose transporters fuse with the cell surface membrane, so more glucose enters by facilitated diffusion; glucose is converted to glycogen (glycogenesis) in liver and muscle by activating enzymes such as glycogen synthase; the rate of respiration of glucose increases; glucose is converted to fat in adipose tissue.
1 mark each for any four of: binds to receptor on target cell membrane; more GLUT4 (glucose transporter) proteins in membrane; increased uptake of glucose by facilitated diffusion; glycogenesis / glucose → glycogen (enzymes activated); increased respiration of glucose; glucose → fat; (liver) inhibits glycogenolysis / gluconeogenesis.
- (c) Only target cells have insulin receptors with a shape complementary to insulin on their cell surface membranes.
B1
Glucagon is secreted when blood glucose concentration falls below the set point, for example during the daytime fast in Ramadan.
- (a) Describe how glucagon binding to a liver cell leads to the production of cAMP. [3]
- (b) Explain how cAMP leads to an increase in blood glucose concentration. [3]
- (c) Explain the advantage of using a cascade of enzymes in this pathway. [2]
Show answer
- (a) Glucagon binds to a specific receptor on the liver cell surface membrane; this changes the shape of the receptor and activates a G protein; the G protein activates the enzyme adenylyl cyclase, which converts ATP to cAMP (second messenger).
B1 binds receptor on cell surface membrane. B1 G protein activated. B1 adenylyl cyclase converts ATP → cAMP.
- (b) cAMP activates protein kinase A; this starts an enzyme cascade in which enzymes are activated by phosphorylation; glycogen phosphorylase is activated, so glycogen is broken down to glucose (glycogenolysis); (gluconeogenesis also increases); glucose leaves the cell by facilitated diffusion into the blood.
1 mark each for any three of: protein kinase (A) activated; enzyme cascade / phosphorylation; glycogen phosphorylase activated; glycogenolysis; gluconeogenesis; glucose diffuses out into the blood.
- (c) Each activated enzyme activates many molecules of the next enzyme, so the signal is amplified; binding of a few glucagon molecules leads to release of a very large number of glucose molecules quickly.
B1 amplification. B1 small amount of hormone → large / rapid response.
Two people fasted overnight and then drank a solution containing 75 g of glucose. Their blood glucose concentrations (mmol dm⁻³) were measured.Time / min: 0, 30, 60, 90, 120, 150Person A: 4.8, 7.6, 6.2, 5.3, 4.9, 4.7Person B: 8.5, 13.9, 16.2, 15.4, 14.1, 12.8
- (a) Compare the results for the two people. [3]
- (b) Person B has type 1 diabetes. Explain the results for person B. [3]
Show answer
- (a) Person B starts higher (8.5 v 4.8 mmol dm⁻³); B's concentration rises higher (peak 16.2 v 7.6) and peaks later (60 min v 30 min); A returns to its starting value by 120 min, whereas B is still well above its starting value (12.8) at 150 min.
1 mark each for any three comparisons with data.
- (b) The β cells do not produce (enough) insulin; so after the glucose drink, liver, muscle and fat cells take up little glucose and little is converted to glycogen; glucose remains in the blood for a long time; even the fasting level is high because glycogen breakdown/gluconeogenesis are not reduced by insulin.
1 mark each for any three of: little / no insulin secreted; less uptake of glucose by cells (fewer GLUT4 in membranes); less glycogenesis; glucose stays high for a long time; high fasting level as glucagon effects unopposed.
- (a) Glucagon causes liver cells to release glucose, but muscle cells do not respond to glucagon. Suggest why. [1]
- (b) Glucagon is a protein and does not enter the liver cell. Explain why a second messenger is needed. [2]
- (c) A drug inhibits the enzyme that breaks down cAMP. Predict and explain the effect of this drug on the blood glucose concentration of a person who has been fasting. [3]
Show answer
- (a) Muscle cells do not have glucagon receptors on their cell surface membranes.
B1
- (b) Glucagon is a large, water-soluble (hydrophilic) molecule, so it cannot cross the phospholipid bilayer; binding to a surface receptor must be converted into a signal inside the cell, which cAMP carries to the enzymes in the cytoplasm.
B1 cannot cross membrane / too large / hydrophilic. B1 second messenger relays the signal inside the cell.
- (c) cAMP stays in liver cells for longer; protein kinase A and the enzyme cascade stay active; more glycogen is broken down to glucose (and more gluconeogenesis), so blood glucose concentration would rise above normal / stay high.
B1 cAMP remains / accumulates. B1 cascade / glycogen phosphorylase stays active. B1 more glucose released → blood glucose higher.
A student ate a large plate of biryani. Her blood was tested every 30 minutes.Time / min: 0, 30, 60, 90, 120, 180Glucose / mmol dm⁻³: 5.0, 8.2, 7.4, 6.1, 5.3, 4.8Insulin / arbitrary units: 8, 60, 60, 35, 18, 10Glucagon / arbitrary units: 90, 62, 50, 60, 74, 85
- (a) Describe the changes in insulin and glucagon concentrations after the meal. [2]
- (b) Explain how these hormone changes return the blood glucose concentration to normal. [3]
- (c) Suggest why the blood glucose concentration falls slightly below the starting value at 180 min. [1]
Show answer
- (a) Insulin rises from 8 to a peak of 60 at 30–60 min, then falls back to 10 by 180 min; glucagon falls from 90 to a minimum of 50 at 60 min and rises back to 85 by 180 min.
B1 insulin rises then falls, with data. B1 glucagon falls then rises, with data.
- (b) The rise in blood glucose is detected by α and β cells in the islets of the pancreas; more insulin and less glucagon increase glucose uptake by cells and glycogenesis, and reduce glycogenolysis/gluconeogenesis; as glucose falls back to the set point, insulin secretion falls and glucagon secretion rises again (negative feedback).
1 mark each for any three of: pancreatic α / β cells detect glucose; insulin increases uptake / glycogenesis; less glucagon reduces glycogenolysis / gluconeogenesis; changes stop as glucose returns to normal; negative feedback.
- (c) Insulin is still slightly raised / glucagon not yet fully restored, so the correction overshoots (the hormones take time to be removed from the blood).
B1
Urine test strips for glucose contain two immobilised enzymes, glucose oxidase and peroxidase, and a colourless chromogen. The colour of the strip is compared with a chart.
- (a) Explain how the enzymes on the strip produce a colour that depends on the glucose concentration. [3]
- (b) Explain why testing urine with a strip is less useful for managing diabetes than measuring blood glucose. [2]
Show answer
- (a) Glucose oxidase catalyses the oxidation of glucose (with oxygen) to gluconic acid (gluconolactone) and hydrogen peroxide; peroxidase catalyses the reaction of hydrogen peroxide with a colourless chromogen, forming a coloured compound; more glucose → more hydrogen peroxide → more intense colour.
B1 glucose oxidase: glucose → gluconic acid + H₂O₂. B1 peroxidase: H₂O₂ + chromogen → coloured product. B1 colour intensity proportional to glucose concentration.
- (b) Glucose only appears in urine when the blood concentration exceeds the renal threshold, so normal or low (hypoglycaemic) blood glucose cannot be detected; urine reflects glucose over the hours since last urination, not the current blood level.
B1 only detects high blood glucose (above renal threshold) / cannot detect low. B1 not a current reading / delayed / colour chart only semi-quantitative.
A patient in Karachi with type 1 diabetes uses a biosensor meter. A drop of blood is placed on a disposable strip in the meter, which displays the glucose concentration within seconds.
- (a) Explain how a biosensor uses glucose oxidase to produce a reading of blood glucose concentration. [3]
- (b) Give two advantages of the biosensor over a urine test strip read with a colour chart. [2]
- (c) The same meter is used for both readings. Suggest why the patient should check the meter with a control solution of known concentration from time to time. [1]
Show answer
- (a) Glucose oxidase is immobilised on the electrode/strip; glucose in the blood drop is oxidised by the enzyme, producing hydrogen peroxide (and transferring electrons); the transducer/electrode detects this, producing a small electric current proportional to glucose concentration; the meter converts the current into a digital concentration reading.
1 mark each for any three of: immobilised glucose oxidase; specific reaction with glucose; H₂O₂ / electrons produced; electrode / transducer produces current; current proportional to concentration; digital display.
- (b) Gives a precise, quantitative reading (not a colour judgement); measures the current blood glucose concentration, including low values; quick, small sample; results can be stored.
1 mark each for any two.
- (c) To make sure the meter (and enzyme on the strips) still gives accurate readings – enzymes on old strips can lose activity / calibration.
B1
- (a) Define homeostasis. [1]
- (b) Explain the roles of receptors and effectors in a homeostatic control system. [2]
- (c) Explain what is meant by negative feedback. [2]
Show answer
- (a) Keeping the internal environment (e.g. body temperature, blood glucose, blood water potential) within narrow limits / close to a set point.
B1
- (b) Receptors detect a change (stimulus) away from the set point and send information to a coordination centre; effectors (muscles or glands) carry out the response that corrects the change.
B1 receptor detects change / stimulus. B1 effector brings about corrective response.
- (c) A change from the set point causes a response that reverses the change, bringing the factor back towards the set point; once back to normal, the corrective response stops.
B1 response opposes / reverses the change. B1 returns to set point, then corrective action reduced.
A hiker in Murree is caught outdoors in freezing weather in January. Her core body temperature begins to fall.
- (a) Describe how the hypothalamus detects that the body is losing heat. [1]
- (b) Explain three responses that reduce heat loss or increase heat production in the cold. [3]
- (c) Explain why the responses of the nervous system and endocrine system work together in thermoregulation. [2]
Show answer
- (a) Thermoreceptors in the hypothalamus detect a fall in blood temperature (and it receives impulses from thermoreceptors in the skin).
B1
- (b) Vasoconstriction of arterioles supplying skin capillaries, so less blood flows near the surface and less heat is lost by radiation; shivering – rapid muscle contractions release heat from respiration; increased metabolic rate (thyroxine / adrenaline) releases more heat; hair erector muscles contract, trapping a layer of air (less important in humans); sweating reduced.
1 mark each for any three responses with explanation.
- (c) Nervous responses (vasoconstriction, shivering) are rapid and short-term; hormonal responses (thyroxine raising metabolic rate) are slower but longer-lasting; together they correct both sudden and prolonged cold.
B1 nervous fast / short-lived. B1 endocrine slower / longer-lasting.
A hospital laboratory technician in Islamabad is training students. She gives them five samples of artificial urine with unknown glucose concentrations.
- (a) Describe a method the technician could use to estimate the glucose concentration of each artificial urine sample, using Benedict's solution and a colorimeter. [6]
- (b) Explain why a calibration curve is needed rather than just comparing the colours by eye. [2]
Show answer
- (a) Make a dilution series of known glucose concentrations (e.g. 0, 2, 4, 6, 8, 10 mmol dm⁻³) from a stock solution; add the same volume of each standard and unknown to the same volume of Benedict's solution (in excess); heat in a boiling water bath for the same time (e.g. 5 min); cool / filter or centrifuge to remove the precipitate; measure the absorbance (or % transmission) of the remaining blue solution in a colorimeter with a red filter, zeroed with a blank; plot a calibration curve of absorbance against glucose concentration and read off the concentrations of the unknowns; repeat three times and take means.
1 mark each for any six of: standards of known concentration (at least five) made by dilution; same volumes of sample and Benedict's; Benedict's in excess; same heating time and temperature (boiling water bath); remove precipitate / let it settle; colorimeter with suitable (red) filter, zeroed with blank; calibration curve; read unknowns from curve; repeats and means; safety (eye protection, hot water).
- (b) Judging colour by eye is subjective and only semi-quantitative; a calibration curve converts colorimeter readings into actual concentrations, giving quantitative and more accurate results.
B1 colour by eye subjective / semi-quantitative. B1 calibration curve gives quantitative values.
A glucose oxidase–peroxidase reagent produces a colour whose absorbance increases with glucose concentration. Absorbance of standards:Glucose / mmol dm⁻³: 2.0, 4.0, 6.0, 8.0, 10.0Absorbance: 0.12, 0.25, 0.37, 0.49, 0.61
- (a) Calculate the volumes of 10.0 mmol dm⁻³ glucose stock and distilled water needed to make 10 cm³ of each standard of 2.0, 4.0, 6.0 and 8.0 mmol dm⁻³. [2]
- (b) The diluted urine sample had an absorbance of 0.31. Use linear interpolation between the standards to find the glucose concentration in the diluted sample. [2]
- (c) The urine had been diluted 1 in 5 before testing. Calculate the glucose concentration of the original urine. [1]
Show answer
- (a) 2.0: 2 cm³ stock + 8 cm³ water; 4.0: 4 + 6; 6.0: 6 + 4; 8.0: 8 + 2.
B1 stock volumes. B1 water volumes.
- (b) 5.0 mmol dm⁻³
M1 between 4.0 (0.25) and 6.0 (0.37). A1 4.0 + (0.06 ÷ 0.12) × 2.0 = 5.0.
- (c) 25 mmol dm⁻³
5.0 × 5 = 25
Students used glucose test strips and a colour chart to estimate the glucose concentration in artificial urine samples.
- (a) Identify two sources of error when glucose test strips are read by comparing their colour with a printed chart. [2]
- (b) Suggest how each error could be reduced. [2]
- (c) Students repeat each test three times and get identical colour readings each time. Explain why this does not show that the readings are accurate. [1]
Show answer
- (a) Colour matching is subjective / differs between people; the chart has only a few colour blocks, so the concentration is only known within a range; strip read too early or too late (colour keeps developing); lighting affects perceived colour; strips stored badly lose enzyme activity.
1 mark each for any two.
- (b) Use a reflectance meter / colorimeter or biosensor to read the strip; read at the exact time stated using a timer; same observer in standard lighting; use a chart with more intermediate colours; check strips with a control solution.
1 mark each for improvements matched to errors.
- (c) Identical readings show the method is repeatable (precise), but all readings could share the same systematic error, e.g. old strips or a misjudged colour, so they may not be close to the true value.
B1 precision / repeatability is not accuracy – systematic error possible.
A student plans to investigate the effect of different concentrations of abscisic acid (ABA) on stomatal closure using strips of lower epidermis peeled from Commelina leaves.
- (a) State the independent and dependent variables in this investigation. [2]
- (b) Explain why the epidermal strips are first kept in light in a solution containing potassium chloride before ABA is added. [2]
- (c) State three variables that must be standardised and how one of them could be controlled. [3]
- (d) Describe a suitable control. [1]
Show answer
- (a) Independent: concentration of ABA. Dependent: stomatal pore width (or percentage of stomata open).
B1 each.
- (b) So that all the stomata are open at the start (K⁺ available for uptake and light stimulates opening); then any closure can be attributed to the ABA.
B1 stomata open at start / standard starting condition. B1 so change is due to ABA.
- (c) Temperature (water bath / temperature-controlled room); light intensity (same lamp at the same distance); time in ABA solution; pH (buffer); leaves of the same age / same plant / same position; size of strip; volume of solution.
B1 B1 two variables. B1 a valid method for controlling one of them. (Max 3: three variables, or two variables and one method.)
- (d) Strips placed in the same solution without ABA (0 mmol dm⁻³ ABA, same volume, same time).
B1
A student will measure stomatal pore width in epidermal strips placed in ABA solutions of 0, 0.01, 0.1, 1.0 and 10 µmol dm⁻³.
- (a) Write a hypothesis for the effect of ABA concentration on stomatal pore width. [1]
- (b) The student will use Spearman's rank correlation. Write a suitable null hypothesis. [1]
- (c) Predict the results for strips placed in a solution of ABA at a very high concentration, and explain your prediction. [2]
Show answer
- (a) As ABA concentration increases, mean stomatal pore width decreases.
B1 directional relationship.
- (b) There is no correlation between ABA concentration and stomatal pore width.
B1 must say no correlation (not 'no difference').
- (c) Almost all stomata closed (pore width near zero); at high ABA concentration all receptors on guard cells are occupied, so maximum K⁺ loss and loss of turgor in every guard cell – further increase in ABA has no further effect.
B1 stomata fully / nearly closed (levels off). B1 receptors saturated / maximum response.
A volunteer drank 1.0 dm³ of water on one day and 1.0 dm³ of 0.9 % sodium chloride solution (about the same water potential as blood plasma) on another day. Urine was collected every 30 minutes.Time period / min: 0–30, 30–60, 60–90, 90–120, 120–150, 150–180Urine after water / cm³: 40, 160, 310, 220, 90, 50Urine after salt solution / cm³: 40, 50, 60, 55, 45, 40
- (a) Describe and compare the effects of the two drinks on urine volume. [2]
- (b) Explain the difference between the effects of the two drinks. [3]
- (c) State two limitations of this investigation that make it hard to draw a firm conclusion. [2]
Show answer
- (a) After water, urine volume rose sharply to a peak of 310 cm³ in the 60–90 min period and then fell back to 50 cm³; after salt solution, urine volume stayed low and almost constant (40–60 cm³).
B1 water: large increase then decrease, with data. B1 salt solution: little change, with data.
- (b) Pure water raises the water potential of the blood, so less ADH is released and less water is reabsorbed in the collecting ducts, giving a large volume of dilute urine; the 0.9 % salt solution has about the same water potential as blood plasma, so the blood water potential hardly changes and ADH secretion stays the same; the extra volume is removed slowly.
B1 water raises blood water potential → less ADH. B1 less water reabsorbed → more urine. B1 isotonic salt solution: water potential unchanged so ADH unchanged.
- (c) Only one volunteer (no replicates); the two drinks were tested on different days so other factors (food, sweating, temperature, activity) may differ; no statistical analysis possible; urine volume only collected every 30 min.
1 mark each for any two.
