Edexcel iGCSE Biology 4BI1 - Paper 2B -Excretion- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)(i) B (carbon dioxide and water)
A is not the answer as lungs do not excrete urea
C is not the answer as lungs do not excrete urea
D is not the answer as lungs do not excrete urea
(a)(ii) An explanation that makes reference to three of the following points:
- volume increases / inhalation occurs / air drawn in (1)
- diaphragm / intercostal muscles contract (1)
- diaphragm moves down / flattens (1)
- ribcage expands (1)
- pressure decreases (inside thorax / lungs) (1)
Accept: internal intercostal muscles relax; ribs move up / move out; thorax / chest expands; pressure higher outside
(b) An explanation that makes reference to three of the following points:
- glucose in urine (1)
- glucose released by ultrafiltration (into filtrate) (1)
- glucose not reabsorbed / too much glucose (in filtrate) to reabsorb (1)
- in the proximal convoluted tubule / PCT / first convoluted tubule (1)
- by active transport (1)
Accept: glucose not absorbed into blood; some glucose not reabsorbed
Reject: if active transport pumping glucose into filtrate
▶️ Answer/Explanation
(a)(i) D (proximal convoluted tubule)
A is not correct as U is not the Bowman’s capsule
B is not correct as U is not the collecting duct
C is not correct as U is not the loop of Henle
(a)(ii) D (ultrafiltration)
A is not correct as it is not ADH production
B is not correct as it is not selective reabsorption
C is not correct as it is not transpiration
(a)(iii) B (Q)
A is not correct as P is the distal convoluted tubule
C is not correct as S is the glomerulus
D is not correct as T is the Bowman’s capsule
(b)(i) Calculation:
Concentration of glucose in blood plasma = 0.1 g/100 cm³
5 dm³ = 5000 cm³
Mass of glucose = (0.1 g/100 cm³) × 5000 cm³ = 5 grams
(b)(ii) An explanation that makes reference to two of the following:
• protein molecules too large / large mass / too heavy (1)
• cannot pass out of glomerulus / into Bowman’s capsule / into nephron / into tubules / through basement membrane (1)
• so stay in blood / not in filtrate (1)
(b)(iii) An explanation that makes reference to two of the following:
• glucose passes out of glomerulus / into Bowman’s capsule / into nephron / into tubules / through basement membrane (1)
• (then) reabsorbed / back into blood (1)
• by active transport (1)
• in proximal convoluted tubule / PCT (1)
▶️ Answer/Explanation
(a) An explanation that makes reference to two of the following:
- Exercise increases body temperature / eq (1)
- (More) sweat / eq (1)
- To cool down / lose heat / eq (1)
- By evaporation (of sweat) (1)
(b)(i) An explanation that makes reference to three of the following:
- Blood concentration increases / less water in blood / eq (1)
- (Detected by) hypothalamus / osmoreceptors / pituitary gland (1)
- Releases ADH (1)
- Collecting duct becomes more permeable / eq (1)
- Increased water reabsorption (so less urine produced) (1)
(b)(ii) An explanation that makes reference to two of the following:
- With (pure water) blood becomes more dilute / less concentrated / has high water concentration / eq (1)
- Less / no ADH released when drinking water / eq (1)
- Less water reabsorbed (by kidney) when water drunk / eq (1)
- Salt / sugar absorbed into blood (in intestine) (from isotonic drink) / eq (1)
Accept converse arguments for the isotonic drink (e.g., more ADH released, more water reabsorbed).
▶️ Answer/Explanation
(a) D (homeostasis)
A is not correct as it is not absorption
B is not correct as it is not diffusion
C is not correct as it is not egestion
(b)(i) Calculation method:
Insensible loss = gas exchange + sweating = \(0.4 + 0.5 = 0.9\) litres
Total water loss = urine + gas exchange + sweating + faeces = \(1.5 + 0.4 + 0.5 + 0.1 = 2.5\) litres
Percentage insensible loss = \(\frac{0.9}{2.5} \times 100 = 36\%\)
(b)(ii) Calculation method:
Water input from food for 70 kg person = 0.5 litres
Water input per kg = \(\frac{0.5}{70} = 0.007142857\) litres/kg
For 110 kg person: \(0.007142857 \times 110 = 0.7857\) litres ≈ \(0.79\) litres (accept \(0.77–0.80\))
(b)(iii) An explanation that makes reference to four of the following points:
1. more water lost / less water in body /eq (1)
2. water / liquid lost in vomiting / faeces / diarrhoea / eq (1)
3. (more) sweating / eq (1)
4. concentration of blood increases / water potential lowered / eq (1)
5. (more) ADH produced / secreted / eq (1)
6. by hypothalamus / pituitary (1)
7. permeability of collecting duct increases / eq (1)
8. more water reabsorbed / less water lost in urine / more concentrated urine / less urine / eq (1)
(c) An answer that makes reference to four of the following points:
In winter:
1. body mass lower / eq (1)
2. as fewer plants available / less food / hibernating / eq (1)
3. less sunlight / lower temp / less photosynthesis / eq (1)
4. (much) more water intake / eq (1)
5. as more rainfall/ eq (1)
6. less concentrated urine in winter / more urine / eq (1)
7. (so) more water lost in urine / eq (1)
8. Ref to data for urine 3× less concentrated in winter / water input 4 × higher in winter / eq (1)
In summer: converse of above points applies.
▶️ Answer/Explanation
(a)
1. Urea
2. Water (or Salts/Ions, e.g., Sodium, Potassium)
Explanation: The kidneys are vital excretory organs. They filter the blood to remove toxic nitrogenous waste, primarily urea, which is produced from the breakdown of excess proteins in the liver. They also play a crucial role in osmoregulation by removing excess water and mineral ions (like sodium and potassium) to maintain the water and salt balance in the body. Other waste products include creatinine and uric acid.
(b)
1. It is a permanent/long-term solution.
2. It allows for a better quality of life/normal diet/more independence.
Explanation: Unlike dialysis, which is a temporary and ongoing treatment requiring frequent sessions (often 3 times a week for several hours), a successful kidney transplant is a permanent fix. It restores near-normal kidney function, freeing the patient from the grueling schedule of dialysis appointments. Furthermore, transplant recipients have far fewer dietary restrictions (e.g., on fluid, potassium, and phosphate intake) compared to those on dialysis, leading to a significantly better and more independent quality of life.
(c)
The immune system responds to disease through a coordinated attack involving white blood cells. Phagocytes (a type of white blood cell) engulf and digest pathogens in a process called phagocytosis. Lymphocytes (another type of white blood cell) respond to specific antigens on pathogens by producing antibodies. These antibodies help to destroy the pathogens. Furthermore, memory cells are created, which provide long-term immunity by allowing a faster and stronger response if the same pathogen is encountered again.
Explanation: The immune system’s primary function is to defend the body against foreign invaders (pathogens). This defense is multi-layered. First, phagocytes act as the general infantry, patrolling the body and engulfing any foreign cells they find. Simultaneously, lymphocytes act as the special forces; they are highly specific. Each lymphocyte recognizes one unique antigen. When they find their match, they multiply and produce vast numbers of antibodies tailored to neutralize that specific pathogen. After the infection is cleared, some of these lymphocytes remain as “memory cells,” ensuring the body is prepared for a future attack by the same pathogen, which is the basis of vaccination.
(d)
Close relatives are genetically similar. This means they have similar proteins and antigens on their cells. Therefore, the recipient’s immune system is more likely to recognize the donated kidney as “self” rather than as foreign tissue, resulting in a lower immune response and a reduced chance of rejection.
Explanation: The immune system identifies cells based on their surface antigens. If these antigens are too different from the recipient’s own, the immune system will launch an attack, leading to organ rejection. Since close relatives share a significant amount of their genetic material, the antigens on the cells of a kidney from a close relative will be very similar to the recipient’s own antigens. This similarity tricks the immune system into accepting the new kidney, minimizing the aggressive immune response that requires heavy suppression by drugs.
(e)(i)
Difference = 39.17% (or a value between 39.15% and 39.2%)
Explanation and Calculation:
First, calculate the percentage change for living donations:
- Change = 1040 – 429 = 611
- Percentage Change = (611 / 1040) × 100 = 58.75% decrease
Next, calculate the percentage change for deceased donations:
- Change = 2283 – 1836 = 447
- Percentage Change = (447 / 2283) × 100 = 19.58% decrease
Finally, find the difference between these two percentage decreases:
- Difference = |58.75% – 19.58%| = 39.17%
This means the decrease in living donations was 39.17 percentage points greater than the decrease in deceased donations from 2020 to 2021.
(e)(ii)
There are more deceased donors available because two kidneys can be taken from each deceased person, and they do not need their kidneys anymore. In contrast, a living person can only donate one kidney and may be reluctant to undergo major surgery.
Explanation: The pool of potential deceased donors is inherently larger. When a person dies, both of their healthy kidneys can be donated. Furthermore, since the donor has passed away, there is no physical risk or long-term health consideration for them. Conversely, a living donor must undergo a significant surgical procedure, which carries inherent risks and a recovery period. The donor must also be healthy enough to live a full life with only one kidney. This combination of medical suitability and personal willingness significantly limits the number of living donors compared to the potential number of organs available from deceased donors.
(f)
A suitable kidney is one that is a close tissue match (and blood type match) to the recipient, is healthy, and is functioning properly, which reduces the risk of rejection.
Explanation: “Suitable” refers to compatibility. The most important factor is a match in tissue type (HLA antigens), which is more likely found in relatives or individuals of the same ethnic background. A matching blood type (e.g., A, B, O) is also essential. Beyond compatibility, the kidney itself must be healthy and free from disease to ensure it will function correctly once transplanted into the recipient.
(g)
People of South Asian, African, and Afro-Caribbean origin are more susceptible to kidney disease, so they more frequently need transplants. However, there are fewer donors from these communities. Furthermore, a better genetic match (and therefore less chance of rejection) is often found within the same ethnic group.
Explanation: There is a higher prevalence of conditions like hypertension and diabetes in these communities, which are leading causes of kidney failure. This creates a higher demand for transplants within these groups. Due to genetic similarities, the best tissue matches for a patient are most likely to come from a donor of the same ethnic background. Therefore, to meet the specific and high demand within these communities and to ensure the best possible outcomes (i.e., lower rejection rates), there is a particular need for donors from South Asian, African, and Afro-Caribbean ethnic origins.
▶️ Answer/Explanation
(a) B (bacteria, fungi, and protoctists)
Explanation: Pathogens are disease-causing microorganisms. While viruses like rabies are one type, other major groups also contain pathogenic species. Bacteria include pathogens like those causing tuberculosis and cholera. Fungi include pathogens responsible for athlete’s foot and ringworm. Protoctists (protists) include pathogenic organisms like Plasmodium which causes malaria and Entamoeba which causes dysentery. Therefore, all three groups – bacteria, fungi, and protoctists – include pathogens.
(b)(i)
Explanation: The rabies vaccine contains weakened or inactivated forms of the rabies virus or its antigens. When this vaccine is administered to dogs, it stimulates their immune system without causing the actual disease. The immune system recognizes these viral antigens as foreign invaders and produces specific antibodies against them. Specialized white blood cells called lymphocytes are activated during this process. Some of these lymphocytes develop into memory cells that remain in the body long-term. If the vaccinated dog is later exposed to the actual rabies virus, these memory cells recognize the pathogen immediately and trigger a rapid, strong immune response. This secondary response produces antibodies much faster and in greater quantities, effectively neutralizing the virus before it can establish an infection and cause disease.
(b)(ii)
Explanation: The graph shows a clear correlation between the vaccination of domestic dogs and the decline in human rabies cases. Before mass vaccination began in 1947, both dog and human rabies cases were relatively high. After vaccination programs were implemented, we observe a continuous and dramatic decline in rabies cases in both dogs and humans throughout the 1950s and beyond. This strong correlation suggests that most human rabies cases were originating from infected domestic dogs rather than wild animals. The data shows that as dog vaccination reduced the reservoir of infection in the canine population, human cases consequently decreased. However, the graph also indicates that human rabies cases didn’t disappear completely but plateaued at low levels, suggesting that some transmission still occurs from wild animals or unvaccinated dogs. This demonstrates that while vaccinating domestic dogs significantly reduces human rabies risk, it doesn’t completely eliminate it due to other potential sources of infection.
(b)(iii)
Explanation: When the RNA vaccine is injected, cells take up the RNA molecules that code for parts of the rabies virus protein coat. Inside the cell, these RNA molecules move to the ribosomes, which are the cellular structures responsible for protein synthesis. The process of translation then occurs, where the genetic code on the RNA is read and converted into a sequence of amino acids. Transfer RNA (tRNA) molecules with specific anticodons bind to complementary codons on the mRNA, each bringing with it a specific amino acid. As the ribosome moves along the mRNA strand, it facilitates the formation of peptide bonds between adjacent amino acids, creating a growing polypeptide chain. This chain folds into the three-dimensional structure of the viral protein. These viral proteins then act as antigens, stimulating the immune system to produce antibodies against rabies without exposure to the actual virus, thus providing protection against future infection.
▶️ Answer/Explanation
(a) Answer: 9.5 × 108
Explanation: To calculate the number of people with chronic kidney disease, we multiply the world population by 12% (or 0.12).
Calculation: 7,900,000,000 × 0.12 = 948,000,000
In standard form, this is 9.48 × 108, which rounds to 9.5 × 108 when expressed to two significant figures as appropriate for the percentage given (12%).
(b) Answer: Carbon dioxide / CO2 or Water (vapour) / H2O
Explanation: The lungs are responsible for excreting carbon dioxide, which is a waste product of cellular respiration. Water vapor is also excreted through the lungs during exhalation, especially in humid environments.
(c) Answer: The dialyser is designed in two key ways to increase urea removal:
1. The temperature is maintained at 40°C, which is slightly higher than normal body temperature. This increases the kinetic energy of urea molecules, making them move faster and diffuse more rapidly across the partially permeable membrane.
2. The dialysis fluid contains no urea, creating a steep concentration gradient between the blood and the dialysis fluid. This maximizes the rate of diffusion of urea from the blood into the dialysis fluid, following the principle of moving from an area of high concentration to an area of low concentration.
Additional design features include the long, coiled tubing which provides a large surface area for diffusion, and the thin partially permeable membrane which shortens the diffusion pathway.
(d)(i) Answer: A. Bowman’s capsule
Explanation: Ultrafiltration occurs in the Bowman’s capsule of the nephron, where high pressure forces small molecules like water, glucose, urea, and salts out of the blood and into the nephron tubule, while larger molecules like proteins and blood cells remain in the blood.
(d)(ii) Answer: Glucose is reabsorbed from the filtrate in the nephron through selective reabsorption in the proximal convoluted tubule. This process involves active transport, which requires energy (ATP) to move glucose molecules against their concentration gradient from the tubule back into the blood capillaries. Specialized carrier proteins in the cells lining the tubule facilitate this transport.
(d)(iii) Answer: A (renal artery / ureter)
Explanation: The renal artery brings oxygenated blood into the kidney for filtration, while the ureter is the tube that transports urine from the kidney to the bladder. The urethra is not correct as it transports urine from the bladder out of the body, not from the kidney to the bladder.
(e) Answer: In a dehydrated patient:
1. Osmoreceptors in the hypothalamus detect the increased solute concentration (decreased water potential) in the blood.
2. The hypothalamus stimulates the pituitary gland to release more antidiuretic hormone (ADH).
3. ADH travels through the bloodstream to the kidneys (or bioreactor nephron cells).
4. ADH makes the walls of the collecting duct (or bioreactor membrane) more permeable to water.
5. More water is reabsorbed from the filtrate back into the blood, resulting in a smaller volume of more concentrated urine.
This negative feedback mechanism helps conserve water in the body and restore normal blood concentration.
▶️ Answer/Explanation
(a)(i) B (nephron)
Explanation: Structure P represents the functional unit of the kidney, which is the nephron.
(a)(ii) B (blood)
Explanation: Tube S carries blood to or from the kidney, specifically it is likely the renal artery or vein.
(a)(iii) C (ureter)
Explanation: Tube Q is the ureter, which transports urine from the kidney to the bladder.
(b)(i) An explanation that makes reference to:
• Patient W: Presence of protein indicates failure of ultrafiltration in the glomerulus/Bowman’s capsule.
• Patient X: Presence of glucose indicates failure of selective reabsorption in the proximal convoluted tubule.
• Patient Y: High water content indicates reduced water reabsorption in the collecting duct, possibly due to low ADH.
(b)(ii) A description that includes:
• Use Benedict’s reagent
• Heat in a water bath
• Observe colour change (green → red indicates glucose)
• Alternatively, use a glucose test strip and compare colour to a chart.
▶️ Answer/Explanation
(a) A
Explanation: The proximal convoluted tubule is the first part of the tubule system in the nephron, located right after the Bowman’s capsule. In a standard nephron diagram, it’s often the first coiled section labelled. Option B is the distal convoluted tubule (later section), C is the collecting duct (where urine is collected), and D is the loop of Henle (the U-shaped part that dips into the medulla).
(b)(i) 0.0314 mm²
Explanation: The radius is 100 μm. First, convert this to millimetres: 100 μm = 100/1000 mm = 0.1 mm. Now, use the area formula: Area = π × (radius)² = 3.14 × (0.1 mm)² = 3.14 × 0.01 mm² = 0.0314 mm².
(b)(ii)
Explanation: Blood vessel X is the afferent arteriole (wider lumen), and blood vessel Y is the efferent arteriole (narrower lumen). This size difference is crucial for creating high hydrostatic pressure within the glomerulus. The wider afferent arteriole allows a large volume of blood to flow into the glomerulus. The narrower efferent arteriole restricts the outflow of this blood. This “bottleneck” effect significantly increases the blood pressure inside the glomerular capillaries. This high pressure is the driving force for ultrafiltration, where water, ions, glucose, and urea are forced out of the blood and into the Bowman’s capsule to form the glomerular filtrate.
(c)
Explanation: The standard test for protein in urine is the Biuret test. Pour a small sample of urine into a test tube. Add an equal volume of sodium hydroxide (NaOH) solution and mix. Then, add a few drops of very dilute copper(II) sulfate (CuSO₄) solution. Do not shake. If protein is present, the solution will turn a lilac or purple colour. Alternatively, a urine test strip (dipstick) can be used. The strip is dipped into the urine sample, and after the specified time, the colour change on the protein pad is compared to a chart on the bottle.
▶️ Answer/Explanation
(a)(i) C (S)
Explanation: The Bowman’s capsule (S) is the cup-like sac at the beginning of the nephron that surrounds the glomerulus and receives the filtrate.
(a)(ii) B (Q)
Explanation: The loop of Henle (Q) is a U-shaped tubule that descends into and ascends from the medulla of the kidney. It is crucial for creating a concentration gradient for water reabsorption.
(a)(iii) A (P)
Explanation: ADH (Antidiuretic Hormone) affects the collecting duct (P). ADH increases the permeability of the collecting duct walls to water, allowing more water to be reabsorbed back into the blood, producing more concentrated urine.
(b)(i) An explanation that makes reference to three of the following points:
• Glucose passes from the blood in the glomerulus (R) into the Bowman’s capsule / renal capsule (S) during ultrafiltration. (1)
• (All) glucose is (then) reabsorbed / absorbed back into the blood / eq. (1)
• This reabsorption occurs in the proximal convoluted tubule / PCT (T). (1)
• It is reabsorbed by active transport (which requires energy). (1)
Explanation: During ultrafiltration, small molecules like glucose enter the nephron. The body cannot afford to lose this valuable energy source, so 100% of filtered glucose is normally reclaimed from the filtrate in the proximal convoluted tubule via active transport against its concentration gradient.
(b)(ii) A description that makes reference to two of the following points:
• Add Benedict’s solution to the urine sample (and heat). (1)
• A positive result is indicated by a colour change to green / yellow / orange / brick-red. (1)
Alternative: Use a test strip (e.g., Clinistix) which changes colour (e.g., to brown) in the presence of glucose. (1 each)
Explanation: Benedict’s test is a standard biochemical test for reducing sugars like glucose. Heating with Benedict’s reagent causes a reduction reaction, producing a coloured precipitate of copper(I) oxide.
(c) A description that makes reference to two of the following points:
• Less urine is produced / lower volume. (1)
• The urine becomes more concentrated / contains less water / appears darker in colour. (1)
• (It may contain) a higher concentration of urea / other solutes. (1)
Explanation: Dehydration lowers the water potential of the blood. This is detected by osmoreceptors, leading to increased secretion of ADH. ADH causes more water to be reabsorbed from the collecting duct back into the blood, conserving water. This results in a smaller volume of more concentrated, darker yellow urine.
▶️ Answer/Explanation
(a)
Explanation: The difference in glucose concentration between area X and area Y occurs due to selective reabsorption in the nephron. Area X represents the filtrate in the Bowman’s capsule, which contains glucose at the same concentration as blood plasma. As the filtrate moves through the proximal convoluted tubule (area Y), all glucose is actively reabsorbed back into the blood. This process requires energy (ATP) and specific carrier proteins to transport glucose against its concentration gradient. Since glucose is a valuable nutrient that the body needs to conserve, it is completely removed from the filtrate in the proximal convoluted tubule, resulting in zero glucose concentration in area Y.
(b)(i) 3.9
Explanation: To calculate how many times more concentrated urea is in area Z compared to area Y, we divide the concentration in Z by the concentration in Y: 1700 ÷ 440 = 3.8636… Rounded to two significant figures, this gives 3.9. This means urea is approximately 3.9 times more concentrated in area Z than in area Y.
(b)(ii)
Explanation: The difference in urea concentration between area Y and area Z occurs primarily due to water reabsorption. As the filtrate moves through the distal convoluted tubule and collecting duct (area Z), water is reabsorbed back into the blood by osmosis. This process is regulated by the hormone ADH (antidiuretic hormone). While urea itself is not actively reabsorbed in significant amounts, the removal of water from the filtrate causes the remaining substances, including urea, to become more concentrated. Therefore, the urea concentration increases significantly from area Y to area Z as water is removed from the filtrate.
(c)(i)
Explanation: To test for protein in urine, a common method is to use urine test strips (dipsticks) that contain a chemical indicator. The strip is dipped into the urine sample for a specified time, then removed. The test pad on the strip will change color if protein is present – typically turning green or blue depending on the concentration. Alternatively, a chemical test can be performed using reagents like Biuret solution, which would change from blue to purple in the presence of protein. The intensity of the color change corresponds to the amount of protein present in the sample.
(c)(ii)
Explanation: High blood pressure during pregnancy can cause protein to appear in urine because the increased pressure can damage the filtration barrier in the glomerulus. Normally, the glomerular filter prevents large molecules like proteins from passing through into the filtrate. However, when blood pressure is excessively high, it creates increased mechanical stress on the glomerular capillaries and basement membrane. This can cause the filtration pores to enlarge or the membrane to become more permeable, allowing proteins (which are large molecules) to be forced through into the filtrate. Once in the filtrate, proteins are too large to be reabsorbed effectively by the nephron, so they remain and are excreted in the urine.
▶️ Answer/Explanation
(a)(i) D S
Explanation: The structure labelled S is the glomerulus, which is a network of capillaries where ultrafiltration occurs. Blood pressure forces water, ions, and small molecules out of the blood and into the Bowman’s capsule, forming the filtrate.
(a)(ii) D U
Explanation: The structure labelled U is the proximal convoluted tubule (PCT). This is where the majority of glucose reabsorption takes place through active transport, returning this valuable nutrient to the bloodstream.
(a)(iii) B R
Explanation: The structure labelled R is the loop of Henle. This hairpin-shaped section of the nephron creates a concentration gradient in the kidney medulla, which is essential for water reabsorption and urine concentration.
(b)(i) Substances/chemicals present in solution/dissolved/carried in the liquid part of the blood/plasma.
Explanation: Plasma components refer to the various dissolved substances found in blood plasma, which is the liquid matrix of blood. These include water, electrolytes (like sodium), nutrients (like glucose), waste products (like urea), hormones, and proteins.
(b)(ii) 30.24 g
Explanation: The calculation is based on the percentage reabsorbed. If 44% of urea is reabsorbed, then 56% is excreted. The mass excreted per day is therefore 56% of the mass filtered: \( \frac{56}{100} \times 54 = 30.24 \) g.
(b)(iii) To prevent glucose from being excreted/lost from the body and to maintain blood glucose levels for respiration/energy release.
Explanation: Glucose is a vital energy source for cells. The body carefully regulates blood glucose levels. Reabsorbing glucose in the nephron ensures that this important fuel is not wasted in urine and is kept in the bloodstream to be used for cellular respiration, which releases the energy needed for all bodily functions.
(b)(iv)
Explanation: A high-protein, high-salt meal with low water intake would have several effects. The breakdown of excess protein increases urea production, leading to a higher concentration of urea in the plasma and subsequently more urea being filtered and excreted. The high salt intake increases the sodium concentration in the blood, lowering its water potential. This is detected by osmoreceptors, which signal the pituitary gland to release more Anti-Diuretic Hormone (ADH). ADH makes the walls of the collecting duct more permeable to water. As a result, more water is reabsorbed back into the blood from the filtrate, producing a smaller volume of more concentrated urine to conserve water and excrete the excess salts and urea.
▶️ Answer/Explanation
(a)
A: Ureter
B: Bladder
Explanation: The ureter is the tube that carries urine from the kidney to the bladder. The bladder is the muscular sac that stores urine until it is expelled from the body. These are two key components of the urinary system’s transport and storage pathway.
(b)(i)
Explanation: The concentration of protein drops from 100 arbitrary units in area X to 0 in area Y because proteins are too large to pass through the filtering system of the glomerulus. The glomerulus acts like a sieve, allowing small molecules like water, salts, glucose, and urea to pass into the Bowman’s capsule (area Y), while large protein molecules remain in the blood (area X). This is a crucial filtration mechanism that prevents essential proteins from being lost in urine.
(b)(ii)
Explanation: The concentration of glucose decreases from 50 arbitrary units in area Y to 0 in area Z because glucose is actively reabsorbed back into the blood from the filtrate. This process occurs primarily in the proximal convoluted tubule. The body recognizes glucose as a valuable energy source and uses active transport, which requires energy, to reclaim it from the filtrate and return it to the bloodstream, ensuring no glucose is wasted in the urine under normal conditions.
(c)
Explanation: When the body is dehydrated, the blood becomes more concentrated (has a lower water potential). This is detected by osmoreceptors in the hypothalamus of the brain. In response, the pituitary gland releases more Antidiuretic Hormone (ADH) into the bloodstream. ADH travels to the kidneys and makes the walls of the collecting ducts more permeable to water. As a result, more water is reabsorbed from the filtrate back into the blood by osmosis. This process conserves water for the body, produces a smaller volume of more concentrated urine, and helps to raise the blood water potential back towards normal levels.
▶️ Answer/Explanation
(a) Eat a balanced diet / less lipid / fat / oil / eat fewer foods that contain cholesterol (e.g., eat fewer eggs).
Explanation: To lower the risk of high cholesterol, one should reduce the intake of foods high in saturated fats and cholesterol. This means eating less fatty or oily food and consuming fewer items like eggs, which are known to contain cholesterol. A balanced diet ensures that the body gets necessary nutrients without excess harmful fats.
(b) Osmoregulation is the control of water and salt levels (or water potential) in the body / body fluids / blood / plasma / cells.
Explanation: Osmoregulation is a vital process where the kidney maintains the balance of water and dissolved salts (like sodium and potassium) in the blood and body fluids. This ensures that cells function properly by preventing them from losing too much water (dehydration) or taking in too much water (swelling), thus maintaining a stable internal environment.
(c) The body produces urea / salt / toxins / water / metabolic waste that need to be excreted to prevent build-up / poisoning, and kidney disease is incurable / has no cure until a transplant.
Explanation: In severe kidney disease, the kidneys fail to remove waste products like urea and excess salts from the blood. These substances accumulate and can become toxic, leading to poisoning or other complications. Since the disease is chronic and incurable (unless a transplant is performed), dialysis must be continued for life to artificially perform the kidney’s excretory function and keep the patient alive.
(d)(i) A partially permeable membrane allows some molecules / substances / water to pass through but stops others / does not let large or charged molecules pass through.
Explanation: A partially permeable membrane, like the peritoneum, acts as a selective barrier. It lets small molecules such as water, ions, and waste products diffuse across it but blocks larger molecules like proteins and blood cells. This selective nature is crucial for processes like dialysis, where only specific substances need to be removed from the blood.
(d)(ii) The dialysis solution contains purified water, glucose, and mineral ions so that these substances do not leave the blood / can return to the blood by diffusion / down a concentration gradient, as cells require water for water balance, ions for water balance / metabolic reactions, and glucose for respiration / energy.
Explanation: The dialysis fluid is carefully formulated to match the natural concentration of useful substances in the blood. This prevents the loss of essential molecules like glucose and mineral ions from the blood into the dialysis fluid by diffusion. If the fluid lacked these, the body would lose vital nutrients. The presence of glucose provides energy, and the ions help maintain osmotic balance, ensuring that necessary substances are retained in the bloodstream.
(e) The dialysis fluid has a lower concentration of / no urea / salts / waste products compared to the blood, so these wastes diffuse from the blood (where they are in high concentration) into the dialysis solution (where they are in low concentration).
Explanation: Waste removal relies on the principle of diffusion. The dialysis fluid is designed with a low (or zero) concentration of waste products like urea and excess salts. Since blood from the patient has a high concentration of these wastes, they naturally move down their concentration gradient from the blood, across the partially permeable peritoneum, and into the dialysis fluid. This cleanses the blood of toxins.
(f)(i) \( 9 \div 24 = 0.375 \); \( 0.375 \times 100 = 37.5\% \) or \( 38\% \).
Explanation: For APD, the total treatment time per night is 9 hours. There are 24 hours in a day, so the fraction of a day used is \( \frac{9}{24} = 0.375 \). To find the percentage of time in a week, we multiply by 100: \( 0.375 \times 100 = 37.5\% \). This means roughly 37.5% of the patient’s week is dedicated to this treatment.
(f)(ii) You can walk around / can be done at home / when travelling / does not require a machine.
Explanation: CAPD offers greater independence and flexibility compared to haemodialysis. Patients are not tied to a machine for hours at a time and can perform exchanges themselves at home or even while traveling. This allows them to maintain a more normal daily routine and lifestyle.
(g)
- Proteins / large molecules cannot leave the glomerulus / cannot go into Bowman’s capsule.
- Reabsorption of glucose / amino acids into the blood by the proximal convoluted tubule.
- Water is reabsorbed from the collecting duct.
Explanation: The kidney has specialized structures to retain essential substances. In the glomerulus, the filter is fine enough to prevent large proteins from leaving the blood and entering the filtrate. In the proximal convoluted tubule, useful substances like glucose and amino acids are actively reabsorbed back into the blood. Finally, in the collecting duct, water is reabsorbed depending on the body’s hydration needs, concentrating the urine and conserving water. This coordinated process ensures that vital nutrients and water are kept in the bloodstream while wastes are excreted.
▶️ Answer/Explanation
(a) A
Explanation: The nephrons are the functional units of the kidney responsible for filtration, reabsorption, and secretion. In the urinary system diagram, part A represents the kidney, which is where the nephrons are located. The other parts (B, C, D) represent other structures like the ureter, bladder, and urethra, which do not contain nephrons.
(b) Water / Urea / Ions (e.g., sodium, chloride) / Salts
Explanation: Urine is composed of water and various waste products that the body needs to excrete. The primary component is water, which acts as a solvent. Urea is a major nitrogenous waste product from the breakdown of proteins. Ions such as sodium, potassium, chloride, and other salts are also found in urine, their concentrations varying based on the body’s needs to maintain homeostasis.
(c)(i)
Explanation: The scientist’s results show that drinking water leads to a much higher volume of urine produced (560 cm³) compared to drinking a salt solution (254 cm³). This happens due to the body’s osmoregulatory mechanisms. When pure water is consumed, the blood becomes more dilute (its water potential increases). This change is detected by osmoreceptors in the hypothalamus. In response, the pituitary gland releases less Antidiuretic Hormone (ADH). With less ADH, the walls of the kidney’s collecting ducts become less permeable to water. This means less water is reabsorbed back into the blood, resulting in a larger volume of dilute urine being produced. Conversely, when a salt solution is consumed, the blood becomes more concentrated (its water potential decreases). This stimulates the release of more ADH. More ADH makes the collecting ducts more permeable, allowing more water to be reabsorbed into the blood to correct the concentration. Consequently, a smaller volume of more concentrated urine is produced.
(c)(ii)
1. Exercise / Physical activity
2. Temperature / Environmental temperature
Explanation: Several other factors can influence urine volume. Exercise or physical activity causes sweating, which loses water from the body. To conserve fluid, the body produces less urine. The temperature of the environment also plays a role; in hot conditions, the body sweats more to cool down, leading to reduced urine production. In cooler temperatures, less sweating occurs, potentially resulting in higher urine output. Other factors not listed here could include the volume of other fluids consumed before the test, certain foods in the diet (like those with high water or salt content), and the intake of diuretics such as caffeine.
Most-appropriate topic codes (Edexcel IGCSE Biology):
• 2(i)B: Kidney function and osmoregulation — parts (a), (c)
• Appendix 4: Mathematical skills — part (a)
• 2(g): Gas exchange (human lungs) — part (b)
▶️ Answer/Explanation
(a)
Answer: 430 cm³
Explanation:
To find the minimum volume of urine, we use the formula for concentration:
concentration = \(\frac{\text{amount of waste}}{\text{volume}}\)
We rearrange this formula to solve for volume:
volume = \(\frac{\text{amount of waste}}{\text{concentration}}\)
Substituting the given values:
volume = \(\frac{600 \text{ milliseconds}}{1400 \text{ milliseconds per dm}^3} = 0.42857 \text{ dm}^3\)
Now we convert from dm³ to cm³. Since 1 dm³ = 1000 cm³:
0.42857 dm³ × 1000 = 428.57 cm³
Rounding to a sensible figure gives us 430 cm³. This is the minimum volume of urine that must be produced each day to excrete all the waste without exceeding the kidney’s maximum concentrating ability.
(b)
Answer: Lungs
Explanation:
The lungs are another major excretory organ in the human body. While their primary function is gas exchange, they also excrete waste products. Specifically, they remove carbon dioxide (CO₂) from the bloodstream. Carbon dioxide is a waste product of cellular respiration. When we breathe out, we are excreting this CO₂ from our bodies. Water vapor is also excreted through the lungs during exhalation.
(c)
Explanation:
Osmoregulation is the process of maintaining the correct water and salt balance in the blood and body fluids. The nephron, the functional unit of the kidney, plays a central role in this process through a series of coordinated steps:
- Detection: Specialized cells called osmoreceptors, located in the hypothalamus of the brain, continuously monitor the water potential (concentration) of the blood. If the blood becomes too concentrated (for example, due to dehydration or eating salty food), the osmoreceptors detect this change.
- Hormonal Response: The hypothalamus stimulates the pituitary gland to release more Antidiuretic Hormone (ADH) into the bloodstream.
- Action on the Nephron: ADH travels in the blood to the kidneys. Its main target is the collecting duct of the nephron. ADH makes the walls of the collecting duct more permeable to water.
- Water Reabsorption: As the filtrate flows down the collecting duct, which passes through the hypertonic renal medulla, water moves out of the duct by osmosis, down its concentration gradient. This water is reabsorbed back into the surrounding blood capillaries.
- Result: Because more water is reabsorbed, a smaller volume of more concentrated urine is produced. This helps to conserve water in the body and increases the water content of the blood, diluting it back to its normal concentration.
- Negative Feedback: Once the blood concentration returns to normal, the osmoreceptors are no longer stimulated. This leads to a reduction in ADH secretion from the pituitary gland. The collecting ducts become less permeable again, and urine output returns to normal. This is an example of a negative feedback loop, which is crucial for maintaining homeostasis.
Conversely, if the blood is too dilute, less ADH is released, the collecting ducts remain impermeable, less water is reabsorbed, and a large volume of dilute urine is produced to remove excess water from the body.
▶️ Answer/Explanation
(a) An explanation that makes reference to two of the following points:
- less oxygen (1)
- (less) respiration (1)
- (less) energy / ATP (1)
Detailed Explanation: When red blood cells burst (haemolysis), they can no longer carry oxygen effectively. Oxygen is essential for cellular respiration, which releases energy in the form of ATP. Reduced oxygen leads to less respiration and less ATP. Since foetal development requires energy for growth and cell division, a lack of ATP directly hinders development.
(b) An explanation that makes reference to three of the following points:
- DNA unzips / separates / one strand copied (1)
- complementary / base pairing (1)
- template (1)
- mRNA produced (1)
- transcription (1)
Detailed Explanation: Transcription occurs: the DNA double helix unwinds, one strand acts as a template, RNA nucleotides pair complementarily (A-U, C-G), and mRNA is formed by RNA polymerase, carrying the code from the gene to the ribosome.
(c) no nucleus / DNA / no ribosomes / mitochondria (1)
Detailed Explanation: Mature red blood cells lack a nucleus (and therefore DNA) and ribosomes. Protein synthesis requires DNA for instructions and ribosomes for assembly, so they cannot make proteins.
(d) can pass across placenta (1)
Detailed Explanation: The passage states antibodies pass across the placenta, a selective barrier. Red blood cells cannot cross normally, indicating antibodies are smaller.
(e)(i) An answer that makes reference to the following points:
- parent genotype dd x Dd (1)
- gamete d (and d) and D or d (1)
- offspring genotype Dd and dd (1)
Detailed Explanation: Mother: dd, Father: Dd. Gametes: mother → d only; father → D or d. Possible offspring: Dd (Rhesus positive) or dd (Rhesus negative).
(e)(ii) 50% / 0.5 / half / 50:50 (1)
Detailed Explanation: From a Punnett square, 2 out of 4 possibilities are Dd (Rhesus positive), so probability = ½ or 50%.
(f) An explanation that makes reference to three of the following points:
- memory cells (1)
- remain in mother’s blood (1)
- recognise / identify antigen / binds with antigen (1)
- more antibodies produced / produced faster / sooner (1)
- secondary immune response (1)
Detailed Explanation: During the first pregnancy, memory cells are made. In a second pregnancy, these memory cells quickly recognise the Rhesus antigen and trigger a rapid, strong secondary immune response, producing large amounts of antibody quickly.
(g) in the uterus / womb (1)
Detailed Explanation: “In utero” means inside the uterus (womb), referring to procedures performed on the foetus before birth.
(h) (adult) (rhesus) negative / negative (1)
Detailed Explanation: The foetus is being attacked by anti-Rhesus antibodies. Transfusing Rhesus negative blood (lacking the antigen) ensures the new cells are not attacked, allowing them to survive and carry oxygen.
▶️ Answer/Explanation
(a)
(i) C
Explanation: Ultrafiltration occurs in the Bowman’s capsule, which is labelled as structure C in the diagram. This is where high pressure forces small molecules like water, glucose, and urea out of the blood and into the nephron, forming the filtrate.
(ii) D
Explanation: Glucose is reabsorbed back into the blood in the proximal convoluted tubule, labelled as structure D. This process, called selective reabsorption, uses active transport to move useful substances like glucose from the filtrate back into the bloodstream.
(b)
(i) Osmoregulation
Explanation: The control of blood concentration, specifically water and salt balance, is known as osmoregulation. The kidneys play a vital role in this by adjusting the amount of water reabsorbed or excreted.
(ii) The removal of metabolic waste / waste from chemical reactions (from cells).
Explanation: Excretion is the biological process of removing waste products generated by metabolic activities within cells. In the context of the kidney, this primarily involves removing urea, a toxic substance produced from the breakdown of excess proteins in the liver.
(c)
Explanation: ADH (Antidiuretic Hormone) makes the walls of the collecting duct more permeable to water. In diabetes insipidus, no ADH is produced. This means the collecting duct walls remain impermeable to water. As a result, very little water is reabsorbed from the filtrate back into the blood in this final part of the nephron. This leads to the production of a large volume of very dilute urine. Consequently, the body loses excessive amounts of water, which can lead to dehydration and an increase in blood concentration (it becomes more concentrated).
(d)
(i) It increases the permeability of the collecting duct (wall).
Explanation: Desmopressin acts as a substitute for ADH. Its primary effect on the nephron is to bind to the cells of the collecting duct and cause the insertion of aquaporins (water channels) into their walls. This makes the walls permeable to water, allowing water to be reabsorbed.
(ii) The drug would decrease the volume of urine produced and increase the concentration of the urine.
Explanation: By making the collecting duct permeable, Desmopressin allows more water to be reabsorbed from the filtrate back into the blood. This means less water remains in the tubule to be excreted, so the volume of urine decreases. As more water leaves the filtrate, the remaining substances (like urea) become more concentrated, resulting in the production of more concentrated urine.
▶️ Answer/Explanation
(a) C (ultrafiltration)
A is incorrect because digestion is not a process in the kidneys.
B is incorrect because mutation is not a process in the kidneys.
D is incorrect because vaccination is not a process in the kidneys.
(b) An answer that makes reference to three of the following points:
• treat drinking water / boil water (before drinking) / do not drink water / drink bottled water / eq (1)
• sanitation / no faeces in water / no urine in water / eq (1)
• remove snails / eq (1)
• vaccination (1)
Additional guidance: Allow “do not go in infected rivers or lakes / cover skin when in water / avoid contact with affected water / only wash in clean water”.
(c) An answer that makes reference to two of the following points:
• red blood cells / rbc (1)
• white blood cells / wbc (1)
• lymphocytes (1)
• phagocytes / macrophages (1)
(d) D (4800 mg)
Calculation: \(120\ \text{kg} \times 0.040\ \text{g/kg} = 4.8\ \text{g} = 4800\ \text{mg}\)
A is incorrect because it is the wrong value.
B is incorrect because it is the wrong value.
C is incorrect because it is the wrong value.
(e) \(1920\) people
Working:
• \(8 \times 10^{-4}\% = 0.0008\%\)
• \(0.0008\% \text{ of } 240,000,000 = 0.000008 \times 240,000,000 = 1920\)
Allow 1 mark for: \(19200000 / 1920000 / 192000 / 192200 / 192 / 19.2 / 1.92 / 0.192 / 0.0192\)
Award full marks for correct numerical answer without working.
(f) A (a circle of DNA)
B is incorrect because it is not RNA.
C is incorrect because it is not a protein.
D is incorrect because it is not RNA.
(g) An explanation that makes reference to three of the following points:
• antigen (1)
• memory cells / lymphocytes (1)
• (secondary) immune response (1)
• more antibodies / antibodies made sooner / faster / faster immune response / eq (1)
(h)(i) • (a treatment with) no plasmid / no protein / only water / saline / eq (1)
Allow placebo vaccine / a placebo / plasmid with no gene / plasmid with no DNA / different DNA.
(h)(ii) An answer that makes reference to three of the following points:
• reduced numbers / eq (1)
• by 19 or by 47% / about 50% (1)
• schistosomes / worms, still present in body (1)
• no idea of group size / needs to be repeated (1)
• no idea of age / sex / health (1)
Additional Guidance:
Allow “reduces numbers of worms / worms decrease / lower number of worms after vaccine”.
Allow “more worms in control group”.
Allow “does not completely get rid of them”.
Allow “more testing / more people tested”.