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A-Level · Biology

AB11 Biology: Homeostasis and Response

AQA 7402 · Calculator allowed · about 180 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Organisms coordinate responses to changes in their internal and external environment using two main communication systems: the nervous system and the endocrine (hormonal) system.

(a)State four structural or functional differences between nervous communication and hormonal (endocrine) communication in the body.(4)

(b)Explain why a target cell only responds to a particular hormone, even though that hormone is carried throughout the whole body in the blood.(2)

(c)Using body temperature as an example, explain how negative feedback control differs from positive feedback control.(3)

(Total for Question 1 is 9 marks)

Homeostasis relies on receptors, coordination centres and effectors working together within negative feedback loops to keep the internal environment of the body relatively stable.

(a)State what is meant by the term 'homeostasis'.(1)

(b)Describe the general roles of a receptor, a coordination centre and an effector within a negative feedback loop.(3)

(c)Explain why it is important for mammals to keep their core body temperature and blood pH within narrow limits.(3)

(d)State one homeostatic variable, other than core temperature, blood glucose concentration or blood pH, that is controlled in the human body.(1)

(Total for Question 2 is 8 marks)

During childbirth, the hormone oxytocin is involved in a positive feedback mechanism that increases the strength and frequency of uterine contractions.

(a)Using the example of childbirth, explain how positive feedback operates, referring to the hormone oxytocin.(3)

(b)Suggest why positive feedback mechanisms, such as this one, are relatively rare in the control of body systems, whereas negative feedback mechanisms are common.(3)

(Total for Question 3 is 6 marks)

Blood glucose concentration is monitored and controlled by cells within the islets of Langerhans in the pancreas.

(a)Describe how a rise in blood glucose concentration is detected and leads to the release of insulin.(3)

(b)Explain how insulin lowers blood glucose concentration once released into the blood, referring to liver and muscle cells.(5)

(c)Using the term 'second messenger', explain how insulin binding to a receptor on the outside of a liver cell membrane can bring about changes inside the cell without insulin itself entering the cell.(3)

(Total for Question 4 is 11 marks)

When blood glucose concentration falls, α cells in the islets of Langerhans secrete the hormone glucagon, which acts on liver cells via a second messenger system.

(a)Describe the second messenger model of glucagon action in a liver cell, referring to adenylate cyclase and cyclic AMP (cAMP).(5)

(b)State the name of the process by which the liver produces glucose from non-carbohydrate sources such as amino acids or glycerol, and explain when this process becomes important.(2)

(c)The response described in (a) is described as an 'amplification' mechanism. Explain why this second messenger system amplifies the original hormonal signal.(3)

(Total for Question 5 is 10 marks)

A glucose tolerance test (GTT) was carried out on a patient. The patient drank a glucose solution at time 0 minutes, and blood glucose concentration was measured at 30-minute intervals over 3 hours. The results are shown in Table 1.

Table 1: Blood glucose concentration during the glucose tolerance test

Time / min	Blood glucose concentration / mmol dm-3
0	6.5
30	9.2
60	9.8
90	8.6
120	7.4
150	6.8
180	6.5

The normal fasting blood glucose range is 3.5-5.5 mmol dm-3.

(a)Using Table 1, calculate the mean rate of decrease in blood glucose concentration between 60 and 120 minutes. Give your answer in mmol dm-3 per minute, to 3 significant figures.(3)

(b)The patient's blood glucose concentration remained above the normal fasting range throughout the test, and took longer than 2 hours to fall back towards the starting value. Suggest an explanation for this observation, referring to insulin and receptor sensitivity.(3)

(c)Suggest why blood glucose concentration was measured over a period of 3 hours, rather than for only 30 minutes after the glucose drink.(2)

(Total for Question 6 is 8 marks)

Compare the underlying causes of type 1 and type 2 diabetes mellitus, and discuss how the treatments used for each type relate to these causes.

(Total for Question 7 is 6 marks)

The nephron is the functional unit of the kidney. Blood entering the kidney is filtered at high pressure in the Bowman's capsule, in a process called ultrafiltration.

(a)Describe how the structures of the glomerulus and Bowman's capsule bring about ultrafiltration.(5)

(b)State two substances that would be present in the glomerular filtrate but not (normally) in the urine, and explain why this is the case.(3)

(c)Explain why plasma proteins are not normally found in the glomerular filtrate.(2)

(Total for Question 8 is 10 marks)

The loop of Henle is the region of the nephron that runs deep into the medulla of the kidney and back out again, and is responsible for creating the conditions needed to produce concentrated urine.

(a)Describe how the proximal convoluted tubule (PCT) reabsorbs glucose and sodium ions from the glomerular filtrate back into the blood.(5)

(b)Explain how the loop of Henle, acting as a countercurrent multiplier, creates a region of low water potential in the tissue fluid of the medulla.(5)

(c)State why animals adapted to live in very dry (arid) environments, such as desert rodents, tend to have proportionally longer loops of Henle than animals living in wet habitats.(2)

(Total for Question 9 is 12 marks)

Antidiuretic hormone (ADH) controls the permeability of the collecting duct to water, allowing the kidney to regulate the water potential of the blood.

(a)Explain the role of ADH in regulating blood water potential when a person becomes dehydrated, referring to osmoreceptors, the pituitary gland and aquaporins.(5)

(b)Predict and explain what would happen to ADH secretion and urine concentration if a person drank a large volume of water.(3)

(c)The kidneys produce approximately 180 dm3 of glomerular filtrate per day. A healthy person excretes 1.5 dm3 of urine per day. Calculate the percentage of the filtrate that is reabsorbed back into the blood. Give your answer to 3 significant figures.(2)

(d)A patient with a rare condition affecting ADH receptors in the collecting duct excretes an unusually large volume of dilute urine, even when dehydrated. Suggest and explain the likely effect of this condition on the volume of filtrate reabsorbed, compared with the healthy value calculated in (c).(2)

(Total for Question 10 is 12 marks)

Required practical. A group of students dissected a fresh sheep kidney (obtained from a butcher) to observe its internal structure, then photographed a longitudinal section next to a mm scale for later measurement and analysis.

(a)State two safety precautions that should be taken when dissecting a fresh mammalian kidney.(2)

(b)Explain why the kidney should be cut longitudinally (rather than transversely) to see the cortex, medulla and pelvis clearly in a single section.(2)

(c)Identify the three main regions visible in a longitudinal section of the kidney, and state one function associated with each.(6)

(d)The photograph of the dissected kidney was printed at a magnification of x2.5. On the photograph, the width of the cortex measured 20 mm. Calculate the actual width of the cortex, in mm.(2)

(e)The whole kidney's actual length was measured directly with a ruler as 90 mm. On the same (x2.5) photograph, this length measured 220 mm. Calculate the percentage error between the expected photograph length (based on the actual length and the stated magnification) and the measured photograph length. Give your answer to 3 significant figures.(3)

(f)Suggest one reason why the measured photograph length might differ from the expected length, and suggest one improvement to the method that could reduce this source of error.(3)

(g)The students noted that individual glomeruli could not be seen with the naked eye during the dissection. Suggest an appropriate technique, in addition to gross dissection, that could be used to observe the detailed structure of a single glomerulus and Bowman's capsule, and explain why gross dissection alone is insufficient for this purpose.(3)

(h)Explain why the kidney tissue used for dissection should be as fresh as possible, and state one appropriate storage method that could be used if immediate dissection is not possible.(3)

(i)Suggest why measuring the same structure (e.g. cortex width) on three different kidneys, rather than on just one, would improve the reliability of any conclusions drawn about typical kidney dimensions.(2)

(Total for Question 11 is 26 marks)

The hypothalamus contains thermoreceptors that monitor the temperature of blood flowing through the brain and coordinates responses that keep core body temperature close to 37 degrees C.

(a)Explain the role of the hypothalamus as a coordination centre in thermoregulation.(3)

(b)Explain how vasodilation of arterioles supplying skin capillaries, and increased sweat gland activity, help to lower core body temperature when it rises above normal.(4)

(c)Explain how shivering and vasoconstriction help to raise core body temperature when it falls below normal.(2)

(d)Explain why thermoregulation is described as operating by negative feedback.(2)

(e)A student used a hand-held infrared thermometer to measure forehead skin temperature in ten volunteers, and concluded that forehead skin temperature is a reliable way to estimate core body temperature. Evaluate this conclusion.(2)

(Total for Question 12 is 13 marks)

Small mammals have a much larger surface area to volume (SA:V) ratio than large mammals, which affects the rate of heat loss and the thermoregulatory demands placed on each animal.

(a)Using the formulae surface area = 4 x π x r² and volume = 4/3 x π x r³, calculate the surface area to volume (SA:V) ratio of a sphere of radius 1 cm. Give your answer in cm^-1, to 3 significant figures.(2)

(b)Calculate the SA:V ratio of a sphere of radius 5 cm, to 3 significant figures, and state how the SA:V ratio changes as the radius (size) of a sphere increases.(2)

(c)Using your answers to (a) and (b), explain why small mammals (modelled by the sphere of radius 1 cm) generally lose heat more rapidly, relative to their body mass, than large mammals (modelled by the sphere of radius 5 cm). Suggest one adaptation that helps small mammals to reduce this heat loss.(3)

(d)A small mammal has a mass of 25 g and a large mammal has a mass of 4000 g. Both generate heat at a rate of 12 J per gram of body mass per hour. Calculate the total rate of heat generation, in J per hour, for each mammal, and comment on which animal must lose heat proportionally faster to avoid overheating, using your SA:V ratio findings above.(3)

(Total for Question 13 is 10 marks)

Mark scheme · AB11 Biology: Homeostasis and Response

Question 1

(a) B1 nervous communication uses electrical impulses carried along neurones, whereas hormonal communication uses chemical messengers (hormones) carried in the blood (oe)
(a) B1 nervous impulses travel very rapidly (up to around 100 m/s), whereas hormones travel more slowly, limited by the speed of blood flow (oe)
(a) B1 nervous responses act on a very specific, localised effector (connected directly by a neurone), whereas hormones travel throughout the whole body and can act on any cell with the complementary receptor (oe)
(a) B1 nervous responses tend to be rapid but short-lived, whereas hormonal responses tend to take longer to begin but can be longer-lasting (oe). Any four creditable differences.
(a) Answer: Nervous communication is electrical and very fast, acting on a specific, localised effector for a short duration; hormonal communication is chemical, carried in the blood, generally slower to act, but can affect many cells throughout the body for a longer duration.
(b) M1 the hormone can only bind to and affect cells that have a complementary-shaped (specific) receptor for that hormone, on or in the cell
(b) A1 cells that lack this specific receptor are not affected, because the hormone cannot bind to them (oe)
(b) Answer: Only cells with a complementary receptor for that specific hormone can bind it and respond; cells without the receptor are unaffected even though the hormone reaches them in the blood.
(c) B1 in negative feedback, a change away from the normal level (e.g. a rise in body temperature above 37 degrees C) triggers a response (e.g. sweating, vasodilation) that reverses/counteracts the change, returning the level back towards normal
(c) B1 in positive feedback, a change away from the normal level triggers a response that increases/amplifies the original change further, moving the level further away from normal, rather than restoring it
(c) B1 negative feedback is the basis of most homeostatic mechanisms (maintains a stable set point), whereas positive feedback is comparatively rare and usually drives a process towards a definite endpoint, rather than maintaining long term stability (oe)
(c) Answer: Negative feedback reverses a deviation from the normal level (e.g. correcting a rise in body temperature back towards 37 degrees C), whereas positive feedback amplifies the original deviation further away from normal.

Question 2

(a) B1 the maintenance of a stable/constant internal environment, within restricted (narrow) limits, despite changes in the internal or external environment (oe)
(a) Answer: The maintenance of a stable internal environment within narrow limits, despite external or internal changes.
(b) B1 receptor - detects a stimulus/change (deviation) from the normal level and communicates this (as a nerve impulse or hormone) to a coordination centre
(b) B1 coordination centre - receives and processes information from the receptor(s), comparing it with a normal/set level, then sends instructions to (an) effector(s)
(b) B1 effector - a muscle or gland that brings about a response, which counteracts the original change and restores the level back towards normal
(b) Answer: Receptor detects the deviation and informs the coordination centre; the coordination centre processes this and instructs an effector; the effector (a muscle or gland) brings about a response that restores the normal level.
(c) M1 enzymes (and other proteins) have an optimum temperature and pH at which the rate of the reaction they catalyse is greatest
(c) M1 a change in temperature or pH away from the optimum alters the tertiary structure/shape of the enzyme's active site (e.g. by breaking hydrogen or ionic bonds), so substrate molecules can no longer bind (denaturation at extremes)
(c) A1 this reduces the rate of enzyme-catalysed reactions (metabolism), so cell and tissue function becomes less efficient, or could stop altogether/be fatal, if enzymes were denatured or working well below their optimum (oe)
(c) Answer: Enzymes work fastest at their optimum temperature and pH; deviation from these values changes the shape of the active site, reducing the rate of enzyme-catalysed reactions and impairing metabolism, which could be fatal if severe.
(d) B1 e.g. blood water potential (water content of the blood) / osmoregulation, or blood pressure, or blood carbon dioxide concentration (any one valid physiological variable, not temperature, glucose or pH)
(d) Answer: For example, the water potential of the blood (osmoregulation).

Question 3

(a) B1 stretching of the cervix (by the baby's head) stimulates the release of oxytocin from the posterior pituitary gland
(a) B1 oxytocin causes the muscle of the uterus wall to contract more strongly and more frequently, pushing the baby further against the cervix
(a) B1 this increases cervical stretching further, stimulating (even) more oxytocin release, so contractions become progressively stronger, until the process ends with the birth of the baby (removing the original stimulus)
(a) Answer: Cervical stretching stimulates oxytocin release, which strengthens uterine contractions and increases cervical stretching further, so more oxytocin is released; this cycle intensifies until birth occurs and the stimulus is removed.
(b) M1 positive feedback amplifies a change away from the normal state rather than correcting it, so it could not be used to maintain a stable internal environment in the way homeostasis requires
(b) A1 if positive feedback operated continuously (without a natural endpoint), it could rapidly drive a variable to a harmful or fatal extreme, so it is generally only suited to rapid, self-limiting processes with a definite endpoint (such as childbirth), rather than long-term regulation (oe)
(b) B1 a further valid supporting example, e.g. the blood clotting cascade is also positive feedback with a natural endpoint (it switches off once a fibrin mesh has sealed the wound), illustrating why such mechanisms are reserved for rapid, self-limiting events rather than continuous regulation (oe)
(b) Answer: Positive feedback amplifies change rather than restoring stability, so if left unchecked it could drive a variable to a dangerous extreme; it is therefore only suited to rapid processes with a natural endpoint, unlike the continuous stability negative feedback provides.

Question 4

(a) B1 β cells in the islets of Langerhans (in the pancreas) detect a rise in blood glucose concentration; glucose diffuses into the β cells and is respired, raising the ATP concentration in the cell
(a) B1 the raised ATP concentration causes potassium channels in the β cell membrane to close, which depolarises the cell membrane
(a) B1 depolarisation causes voltage-gated calcium channels to open; Ca2+ ions enter the cell, causing insulin-containing vesicles to move to the membrane and release insulin by exocytosis
(a) Answer: Beta cells detect raised blood glucose, respire the glucose to raise ATP, which closes K+ channels and depolarises the membrane; this opens Ca2+ channels, and Ca2+ influx triggers exocytosis of insulin.
(b) B1 insulin binds to specific (glycoprotein) receptors on the cell-surface membrane of liver and muscle cells
(b) B1 this causes more glucose transporter proteins (e.g. GLUT4) to be moved to and inserted into the cell-surface membrane
(b) B1 this increases the rate of facilitated diffusion of glucose into the cells, removing glucose from the blood faster
(b) B1 insulin activates enzymes that convert glucose to glycogen for storage (glycogenesis) in liver and muscle cells
(b) B1 insulin increases the rate at which cells (particularly liver and muscle) respire glucose, further lowering blood glucose concentration
(b) Answer: Insulin binds to receptors on liver/muscle cells, increasing glucose uptake (more GLUT4 transporters), stimulating glycogenesis, and increasing the rate of glucose respiration, all of which lower blood glucose concentration.
(c) M1 insulin (a large, water-soluble/protein hormone) cannot cross the phospholipid cell-surface membrane, so it binds to a specific receptor protein on the outside of the membrane
(c) M1 binding of insulin causes a conformational (shape) change in the receptor, which activates a chain/cascade of enzyme-controlled reactions in the cytoplasm, producing or activating a second messenger
(c) A1 this intracellular second messenger brings about the cell's response (e.g. activation of glycogenesis / increased glucose uptake), without insulin itself needing to enter the cell
(c) Answer: Insulin binds an external receptor, causing a shape change that triggers an internal second messenger cascade, which brings about the cell's response without insulin crossing the membrane.

Question 5

(a) B1 glucagon binds to a specific (glycoprotein) receptor on the cell-surface membrane of the liver cell
(a) B1 this activates the enzyme adenylate cyclase on the inside of the membrane
(a) B1 adenylate cyclase converts ATP into cyclic AMP (cAMP)
(a) B1 cAMP acts as a second messenger, activating a cascade of enzyme-controlled reactions inside the cell (e.g. activating protein kinase enzymes)
(a) B1 this cascade activates the enzyme (glycogen) phosphorylase, which catalyses the breakdown of glycogen to glucose (glycogenolysis)
(a) Answer: Glucagon binds an external receptor, activating adenylate cyclase, which converts ATP to cAMP; cAMP (the second messenger) triggers an enzyme cascade that activates phosphorylase, breaking down glycogen to glucose.
(b) B1 gluconeogenesis
(b) B1 becomes important when blood glucose concentration is low and liver glycogen stores have been (largely) depleted, e.g. during prolonged fasting or exercise (oe)
(b) Answer: Gluconeogenesis; it becomes important once glycogen stores are depleted, for example during prolonged fasting or exercise.
(c) M1 one glucagon-receptor binding event activates only a small number of adenylate cyclase enzyme molecules
(c) M1 each activated enzyme in the cascade can catalyse the conversion of many substrate molecules to product (e.g. one adenylate cyclase produces many cAMP molecules, and each subsequent enzyme in the cascade acts on many further substrate molecules)
(c) A1 so at each successive step of the cascade the number of molecules affected multiplies, meaning a small number of hormone molecules binding at the membrane can bring about a very large response inside the target cell (oe)
(c) Answer: Each enzyme activated in the cascade converts many substrate molecules, and this multiplies at each step, so a single hormone-receptor binding event is amplified into a very large intracellular response.

Question 6

(a) M1 change in blood glucose concentration = 9.8 - 7.4 = 2.4 (mmol dm-3)
(a) M1 rate = 2.4 / (120 - 60) = 2.4 / 60
(a) A1 0.0400 mmol dm-3 per minute (awrt 0.0400, cao, allow 4.00 x 10^-2)
(a) Answer: 0.0400 mmol dm-3 per minute (a decrease).
(b) M1 suggests the patient may have type 2 diabetes
(b) M1 body cells (liver and muscle) may have insulin receptors that have become less sensitive/responsive to insulin (insulin resistance), even though insulin is still secreted (oe)
(b) A1 so glucose uptake into cells (and glycogenesis) occurs more slowly than normal, so blood glucose concentration remains elevated for longer than in a person without diabetes (oe)
(b) Answer: The patient may have type 2 diabetes: their cells show reduced sensitivity to insulin (insulin resistance), so glucose is removed from the blood more slowly than normal, keeping blood glucose raised for longer.
(c) M1 measuring for only 30 minutes would show only the initial rise in blood glucose concentration, and not how effectively the body responds to and removes the glucose afterwards
(c) A1 a longer sampling period allows the rate and extent of the return of blood glucose concentration towards the normal/fasting range to be observed, which is needed to assess how well blood glucose is being regulated (e.g. to help diagnose diabetes) (oe)
(c) Answer: A longer test period is needed to observe how quickly and how completely blood glucose concentration returns to normal, which is essential information for assessing glucose regulation.

Question 7

Level 3 (5-6): A clear, logically structured comparison of the causes of type 1 and type 2 diabetes, with treatment for each type explicitly and accurately linked back to its underlying cause. Appropriate scientific terminology used accurately throughout.
Level 2 (3-4): Some accurate description of the causes of type 1 and/or type 2 diabetes, with some links made between treatment and cause, but the comparison is incomplete or lacks clarity in places.
Level 1 (1-2): Basic or limited statements about type 1 and/or type 2 diabetes causes or treatments, with little or no linkage made between cause and treatment.
Level 0 (0): No relevant content.
Indicative content:
- Type 1 diabetes is an autoimmune condition in which the immune system destroys the insulin-secreting beta cells of the islets of Langerhans, so little or no insulin is produced.
- Type 1 diabetes usually develops in childhood or early adulthood and is not linked to lifestyle or obesity.
- Type 1 diabetes is treated with regular insulin injections, because the body cannot produce its own insulin.
- Insulin cannot be taken as a tablet/orally because it would be digested/broken down by protease enzymes in the gut before it could act, so it must be injected.
- Type 2 diabetes develops when body cells (liver, muscle, fat) become less responsive to insulin (insulin resistance), and/or the pancreas cannot produce enough insulin to compensate for this.
- Type 2 diabetes is strongly associated with lifestyle factors such as obesity, poor diet and lack of exercise, and usually develops later in life.
- Type 2 diabetes is initially managed with lifestyle changes (diet, exercise, weight loss) and/or oral medication that improves insulin sensitivity or secretion, rather than insulin injections, because the patient's own insulin production and response may still be partly functional.
- In more advanced or severe type 2 diabetes, insulin injections may eventually become necessary.
- Both types result in chronically raised blood glucose concentration if poorly controlled, which can damage blood vessels and organs over time.

Question 8

(a) B1 the afferent arteriole (entering the glomerulus) has a larger diameter than the efferent arteriole (leaving it), which creates a high hydrostatic pressure within the glomerular capillaries
(a) B1 this pressure forces small molecules (water, glucose, amino acids, mineral ions, urea) out of the blood, through fenestrations (pores) in the capillary endothelium
(a) B1 and through the basement membrane, which acts as a filter, preventing large molecules such as plasma proteins from passing through (they remain in the blood)
(a) B1 and through filtration slits between the foot-like processes (pedicels) of the podocyte cells forming the wall of the Bowman's capsule
(a) B1 into the capsule space, forming the glomerular filtrate; blood cells are also too large to be filtered and remain in the blood (any 5 for 5 marks)
(a) Answer: High hydrostatic pressure, caused by the wider afferent than efferent arteriole, forces water and small solutes out of the glomerular capillaries through the fenestrated endothelium, basement membrane and podocyte filtration slits into the Bowman's capsule, forming the filtrate; proteins and blood cells are too large to pass through.
(b) B1 glucose and/or amino acids named as present in filtrate but not (normally) in urine
(b) M1 these molecules are small enough to be filtered at the glomerulus, so are present in the glomerular filtrate
(b) A1 but they are completely (selectively) reabsorbed back into the blood in the proximal convoluted tubule (by active transport/co-transport), so none remains to be excreted in the urine of a healthy person
(b) Answer: Glucose and amino acids: both are small enough to be filtered into the glomerular filtrate, but are completely reabsorbed by active transport/co-transport in the proximal convoluted tubule, so none normally remains in the urine.
(c) B1 plasma proteins have a large relative molecular mass and are too large to pass through the basement membrane/fenestrations of the capillary, so they remain in the blood
(c) B1 this retained protein maintains a (relatively) low water potential in the blood plasma compared with the tissue fluid/filtrate, which helps water re-enter the capillaries by osmosis at the venous end (oe, any other valid additional detail)
(c) Answer: Plasma proteins are too large to pass through the basement membrane and capillary fenestrations, so they are retained in the blood plasma.

Question 9

(a) B1 microvilli on the epithelial cells lining the PCT increase the surface area available for reabsorption
(a) B1 Na+ ions are actively transported (using ATP, via Na+/K+ pumps) out of the epithelial cells into the blood, maintaining a low intracellular Na+ concentration
(a) B1 this maintains a concentration gradient for Na+ to move from the tubule lumen into the epithelial cells via co-transporter proteins, carrying glucose (or amino acids) into the cell at the same time (facilitated diffusion/co-transport)
(a) B1 glucose then moves out of the epithelial cell into the blood, down its concentration gradient, by facilitated diffusion (via further carrier proteins)
(a) B1 numerous mitochondria in the epithelial cells provide ATP for the active transport of Na+ ions
(a) Answer: Microvilli increase surface area; active transport of Na+ out of the cells (using ATP) maintains a gradient that drives Na+ (and co-transported glucose) into the cells from the filtrate; glucose then diffuses into the blood; many mitochondria supply the ATP needed.
(b) B1 Na+ and Cl- ions are actively transported out of the ascending limb (which is impermeable to water) into the surrounding medulla tissue fluid
(b) B1 this lowers the water potential of the tissue fluid in the medulla
(b) B1 the descending limb is permeable to water (but not very permeable to ions), so water moves out of the descending limb into the medulla tissue fluid by osmosis, down the water potential gradient, concentrating the filtrate as it flows down the loop
(b) B1 fluid flows in opposite directions in the descending and ascending limbs (countercurrent flow), which maintains/multiplies the water potential gradient along the length of the loop
(b) B1 the deeper into the medulla, the lower the water potential of the tissue fluid becomes, i.e. water potential decreases from the cortex to the tip of the medulla (pyramid)
(b) Answer: Active transport of Na+/Cl- out of the water-impermeable ascending limb lowers the water potential of the medulla tissue fluid; water leaves the permeable descending limb by osmosis; the countercurrent flow in the two limbs maintains and multiplies this gradient, so water potential decreases progressively towards the tip of the medulla.
(c) M1 a longer loop of Henle extends further into the medulla, creating a steeper (more concentrated) water potential gradient in the medulla tissue fluid
(c) A1 this allows more water to be reabsorbed from the collecting duct by osmosis, producing a smaller volume of more concentrated urine, which conserves water - an adaptation to living in a dry environment (oe)
(c) Answer: A longer loop of Henle produces a steeper water potential gradient in the medulla, allowing more water to be reabsorbed from the collecting duct, so a smaller volume of more concentrated urine is produced, conserving water.

Question 10

(a) B1 osmoreceptor cells in the hypothalamus detect a decrease in the water potential of the blood (the blood becomes more concentrated)
(a) B1 this stimulates the posterior pituitary gland to release (more) antidiuretic hormone (ADH) into the blood
(a) B1 ADH binds to (glycoprotein) receptors on the cell-surface membrane of the cells lining the collecting duct (and distal convoluted tubule)
(a) B1 causing more aquaporin (water channel) proteins to be inserted into the cell-surface membrane, increasing the permeability of the tubule wall to water
(a) B1 more water is reabsorbed from the filtrate back into the blood by osmosis, producing a smaller volume of more concentrated urine, and restoring blood water potential towards normal (negative feedback)
(a) Answer: Osmoreceptors detect low blood water potential and stimulate the posterior pituitary to release more ADH; ADH increases aquaporin insertion in the collecting duct, increasing water reabsorption and restoring blood water potential, producing concentrated urine.
(b) M1 drinking a large volume of water increases the water potential of the blood (dilutes the blood)
(b) M1 this is detected by osmoreceptors in the hypothalamus, which reduces (inhibits) the secretion of ADH from the posterior pituitary gland
(b) A1 fewer aquaporins are present in the collecting duct membrane, so less water is reabsorbed, producing a larger volume of more dilute urine (oe)
(b) Answer: ADH secretion decreases because blood water potential rises; fewer aquaporins are inserted in the collecting duct, so less water is reabsorbed, producing a large volume of dilute urine.
(c) M1 volume reabsorbed = 180 - 1.5 = 178.5 (dm3); percentage = (178.5 / 180) x 100
(c) A1 99.2% (awrt 99.2, cao)
(c) Answer: 99.2%
(d) M1 if the collecting duct cells cannot respond to ADH (unresponsive/faulty receptors), aquaporins are not inserted into the membrane even when ADH is released, so the collecting duct remains relatively impermeable to water
(d) A1 (much) less water is reabsorbed than the healthy value (less than 178.5 dm3 per day), so a larger volume of filtrate remains in the tubule and is excreted as dilute urine, regardless of hydration status (oe)
(d) Answer: Far less water would be reabsorbed than the healthy 178.5 dm3 per day, because the collecting duct cannot respond to ADH by inserting aquaporins, so a much larger volume of dilute urine is produced even when dehydrated.

Question 11

(a) B1 wear (disposable) gloves to avoid direct contact with blood/bodily fluids (biological hazard) (oe)
(a) B1 use a sharp scalpel/dissecting instruments carefully, cutting away from the body and on a dissecting board/mat, and wash hands thoroughly afterwards, and/or dispose of animal tissue according to appropriate biological waste procedures (any two distinct valid precautions)
(a) Answer: For example: wear disposable gloves when handling the kidney, and use a sharp scalpel carefully on a dissecting board, washing hands thoroughly afterwards.
(b) B1 a longitudinal section cuts along the length of the kidney, from the outer curved surface through the hilum to the opposite side, exposing all three regions (cortex, medulla, pelvis) in one continuous section
(b) B1 a transverse section (cut across the width) would only show the cortex and medulla as rings/segments, without clearly showing the pelvis and its connection to the ureter (oe)
(b) Answer: A longitudinal cut runs the full length of the kidney through the hilum, showing the cortex, medulla and pelvis together; a transverse cut would not show the pelvis and ureter connection clearly.
(c) B1 cortex identified correctly (outer region)
(c) B1 cortex function - contains the glomeruli and Bowman's capsules, the site of ultrafiltration (oe)
(c) B1 medulla identified correctly (middle region, containing pyramids)
(c) B1 medulla function - contains the loops of Henle and collecting ducts, where water is reabsorbed and urine is concentrated (oe)
(c) B1 (renal) pelvis identified correctly (innermost, funnel-shaped region)
(c) B1 pelvis function - collects/funnels urine from the collecting ducts before it passes into the ureter (oe)
(c) Answer: Cortex (contains glomeruli/Bowman's capsules, site of ultrafiltration); medulla (contains loops of Henle/collecting ducts, concentrates urine); pelvis (collects urine before it enters the ureter).
(d) M1 actual size = image size / magnification = 20 / 2.5
(d) A1 8 mm (cao)
(d) Answer: 8 mm
(e) M1 expected photograph length = actual length x magnification = 90 x 2.5 = 225 (mm)
(e) M1 percentage error = (measured - expected) / expected x 100 = (220 - 225) / 225 x 100
(e) A1 -2.22% (allow 2.22%, awrt 2.22, ecf from a correct expected length)
(e) Answer: -2.22% (the measured photograph length was 2.22% less than expected).
(f) M1 valid reason for the discrepancy, e.g. the kidney or camera may not have been perfectly flat/perpendicular when photographed (introducing parallax error/distortion), or the mm scale on the photograph may have been read imprecisely (oe)
(f) A1 a valid, linked improvement, e.g. photograph the kidney directly from above, on a flat surface, with the camera perpendicular to it, or use digital image analysis software or vernier callipers to measure the photograph more precisely (oe)
(f) B1 further creditable evaluative point, e.g. repeating the photograph/measurement and calculating a mean would reduce the impact of random measurement error (oe)
(f) Answer: For example, the kidney may not have been perfectly perpendicular to the camera (parallax error); photographing it flat and directly from above, and using more precise measuring equipment, would reduce this error.
(g) M1 use light microscopy of a thin, stained section of kidney tissue (or an electron micrograph, for finer detail such as podocyte foot processes/filtration slits)
(g) A1 the glomerulus and Bowman's capsule are microscopic structures, far too small to resolve with the naked eye (or hand lens) used in gross dissection
(g) B1 a light microscope provides sufficient magnification and resolution to see the glomerulus and Bowman's capsule, though an electron microscope would be needed to resolve podocyte foot processes/filtration slits, as these are smaller than the resolution limit of light microscopy (oe)
(g) Answer: Light microscopy (or electron microscopy for finer detail) of a thin, stained kidney section; gross dissection alone cannot resolve microscopic structures like the glomerulus, which are far too small to see with the naked eye.
(h) B1 fresh tissue better preserves the natural size, texture and colour distinction between regions (before autolysis/decomposition alters the tissue), improving the accuracy of identification and any measurements taken
(h) B1 if dissection cannot occur immediately, the kidney should be refrigerated (or frozen) to slow decomposition/autolysis/bacterial action
(h) B1 preservation in a fixative (e.g. formalin) may cause the tissue to shrink or harden, which could introduce error into any linear measurements or magnification calculations taken from the tissue (oe)
(h) Answer: Fresh tissue preserves natural structure and colour, improving accuracy; if delayed, the kidney should be refrigerated or frozen, since fixatives such as formalin can shrink or harden tissue and distort measurements.
(i) M1 using three (or more) kidneys accounts for natural biological variation between individual animals (e.g. differences in size, age or sex), rather than relying on a single, possibly atypical, kidney
(i) A1 calculating a mean value from repeated biological replicates gives a more representative and reliable estimate of the typical dimension, and allows any anomalous results to be identified (oe)
(i) Answer: Using several kidneys accounts for natural variation between individual animals; taking a mean of repeated biological replicates gives a more reliable, representative estimate and helps identify anomalies.

Question 12

(a) B1 thermoreceptors in the hypothalamus (and in the skin) detect changes in blood/core temperature (and skin temperature)
(a) B1 the hypothalamus compares this information with a genetically determined normal/set core temperature (about 37 degrees C)
(a) B1 the hypothalamus (acting as the coordination centre) sends nerve impulses, via the autonomic nervous system, to effectors to bring about an appropriate corrective response
(a) Answer: Thermoreceptors in the hypothalamus detect deviations from the normal core temperature; the hypothalamus compares this to the set point and sends impulses via the autonomic nervous system to effectors.
(b) B1 arterioles supplying skin capillaries dilate (vasodilation), increasing the flow of warm blood close to the skin surface
(b) B1 this increases the rate of heat loss (by radiation, convection and conduction) from the blood to the surrounding environment
(b) B1 sweat glands increase the secretion of sweat onto the skin surface
(b) B1 heat energy from the body is used to evaporate water in the sweat (latent heat of vaporisation), transferring heat away from the skin/body, producing a cooling effect
(b) Answer: Vasodilation increases blood flow near the skin surface, increasing heat loss by radiation/convection/conduction; increased sweating uses body heat to evaporate water from the skin, cooling the body.
(c) M1 skeletal muscles contract rapidly and repeatedly (shivering); this increased muscle contraction requires increased respiration, which releases heat as a by-product, warming the blood
(c) A1 arterioles supplying skin capillaries constrict (vasoconstriction), reducing blood flow near the skin surface (diverting blood through shunt vessels), reducing heat loss from the skin and helping retain heat in the body core
(c) Answer: Shivering increases muscle contraction and respiration, releasing heat as a by-product; vasoconstriction reduces blood flow to the skin surface, reducing heat loss and conserving core heat.
(d) M1 a deviation from the normal core temperature (37 degrees C) in either direction (a rise or a fall) triggers effector responses that counteract/reverse the change
(d) A1 this returns core temperature back towards the normal set point, which removes the original stimulus and switches off the corrective response (oe)
(d) Answer: A deviation from 37 degrees C in either direction triggers responses that reverse the change, returning temperature to the set point and removing the original stimulus.
(e) M1 skin temperature is strongly influenced by ambient/environmental temperature and by blood flow to the skin (vasodilation/vasoconstriction), so it may not accurately reflect core (deep body) temperature, particularly during a thermoregulatory response
(e) A1 the conclusion is not fully supported without comparing forehead readings to a direct measure of core temperature (e.g. tympanic, oral or rectal) as a reference method, so further evidence would be needed to justify it (oe)
(e) Answer: The conclusion is not well supported: skin temperature is affected by ambient conditions and blood flow, so it may not reliably reflect core temperature without comparison to a validated core-temperature method.

Question 13

(a) M1 surface area = 4 x π x 1² = 12.57 (cm²); volume = 4/3 x π x 1³ = 4.19 (cm³)
(a) A1 SA:V ratio = 12.57 / 4.19 = 3.00 cm^-1 (cao, awrt 3.00; allow the shortcut SA:V = 3/r for full credit)
(a) Answer: 3.00 cm^-1 (a ratio of 3:1).
(b) M1 surface area = 4 x π x 5² = 314 (cm²); volume = 4/3 x π x 5³ = 524 (cm³); ratio = 314 / 524 = 0.600 cm^-1 (cao, awrt 0.600; allow the shortcut SA:V = 3/r)
(b) A1 as the radius (size) of a sphere increases, the SA:V ratio decreases (oe)
(b) Answer: 0.600 cm^-1; the SA:V ratio decreases as the radius (size) of the sphere increases.
(c) M1 the smaller sphere (small mammal) has a much higher SA:V ratio (3.00 cm^-1) than the larger sphere (0.600 cm^-1), meaning it has a proportionally larger surface area, relative to its volume (mass) of heat-generating tissue
(c) A1 so small mammals lose heat to the environment faster, relative to their body mass, than large mammals, and must generate/replace heat at a proportionally higher rate to maintain a constant core temperature
(c) B1 valid suggested adaptation, e.g. a thick fur coat or layer of insulating fat relative to body size, a more compact body shape with a lower SA:V ratio (e.g. shorter limbs/ears), a higher metabolic rate, huddling in groups to reduce effective surface area, or behavioural thermoregulation such as nesting/burrowing (any one, well explained)
(c) Answer: The high SA:V ratio of small mammals means proportionally more heat is lost through the surface than is generated in the volume, so they lose heat faster relative to body mass; adaptations such as thick fur, a compact body shape, or huddling reduce this heat loss.
(d) M1 small mammal: 25 x 12 = 300 J per hour
(d) M1 large mammal: 4000 x 12 = 48 000 J per hour
(d) A1 although the large mammal generates far more heat in total, its much lower SA:V ratio (0.600 cm^-1, compared with 3.00 cm^-1) means it loses heat more slowly relative to the amount it generates; the small mammal, with the higher SA:V ratio, must lose heat proportionally faster relative to its heat production/mass (oe)
(d) Answer: Small mammal = 300 J per hour; large mammal = 48 000 J per hour; the small mammal (higher SA:V ratio) must lose heat proportionally faster relative to its mass/heat production.

Biology: Homeostasis and Response - Worksheets, Questions and Revision