An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V2) Cambridge International A Level Human Biology paper. Not affiliated with or reproduced from Cambridge.
Paper 1R
Answer all questions. Show all the steps in any calculations and state the units. Calculators and rulers are permitted.
8 題目 · 80 分
題目 1 · multiple_choice
10 分
A student investigated the effect of exercise on cardiac output. The table below shows the results obtained: - Resting: Heart rate = 72 beats/min, Stroke volume = 70 cm³ - After 10 minutes of running: Heart rate = 150 beats/min, Stroke volume = 120 cm³
(i) Calculate the change in cardiac output (in cm³/min) from rest to after running. (ii) Explain the mechanism by which the heart rate is increased during exercise. (iii) State one structural difference between an artery and a vein, and explain how this difference relates to its function.
Which of the following options correctly provides the calculation for (i), the correct physiological mechanism for (ii), and the correct structural-functional relationship for (iii)?
A.A: (i) Increase of \(12960 \text{ cm}^3/\text{min}\); (ii) Chemoreceptors detect increased blood \(CO_2\) / decreased pH and send impulses to the medulla, which sends impulses along the sympathetic nerve to the SA node; (iii) Arteries have thicker elastic walls to withstand high blood pressure.
B.B: (i) Increase of \(18000 \text{ cm}^3/\text{min}\); (ii) Baroreceptors detect increased oxygen and send impulses to the cerebellum, which sends impulses along the vagus nerve; (iii) Veins have thicker muscular walls to pump blood back to the heart.
C.C: (i) Increase of \(12960 \text{ cm}^3/\text{min}\); (ii) Motor cortex sends voluntary signals directly to the AV node via motor neurons; (iii) Arteries have valves to prevent the backflow of blood under low pressure.
D.D: (i) Increase of \(10200 \text{ cm}^3/\text{min}\); (ii) Chemoreceptors detect decreased carbon dioxide levels and stimulate the parasympathetic nervous system; (iii) Veins have wider lumens to decrease resistance to blood flow.
(ii) During exercise, respiration increases, releasing more carbon dioxide (\(CO_2\)) into the blood. This decreases blood pH. Chemoreceptors in the aorta and carotid arteries detect this and send impulses to the medulla oblongata, which sends signals via the sympathetic nervous system to the sinoatrial (SA) node to increase heart rate.
(iii) Arteries have thicker muscular and elastic walls than veins to withstand and maintain high blood pressure as blood is pumped from the heart.
評分準則
Total: 10 marks - (i) 3 marks: 1 mark for resting cardiac output calculation, 1 mark for exercise cardiac output calculation, 1 mark for subtracting to find the correct difference of \(12960 \text{ cm}^3/\text{min}\). - (ii) 4 marks: 1 mark for identifying increased \(CO_2\)/decreased pH, 1 mark for detection by chemoreceptors, 1 mark for coordination by medulla oblongata, 1 mark for sympathetic nerve pathway and stimulation of the SA node. - (iii) 3 marks: 1 mark for identifying structural difference (thicker muscular/elastic wall in arteries or presence of valves in veins), 2 marks for explaining how this relates to function (withstanding high pressure vs preventing backflow under low pressure).
題目 2 · multiple_choice
10 分
This question concerns blood circulation and tissue fluid formation.
(i) Describe the path of a red blood cell from the renal vein to the pulmonary artery, listing the structures in sequence. (ii) Explain the process of tissue fluid formation and how excess tissue fluid is returned to the circulatory system. (iii) Calculate the percentage increase in blood flow to skeletal muscle during exercise compared to rest, using the following data: - Blood flow to skeletal muscle at rest: \(1200 \text{ cm}^3/\text{min}\) - Blood flow to skeletal muscle during heavy exercise: \(12500 \text{ cm}^3/\text{min}\)
Which of the following options correctly matches the requirements?
A.A: (i) Renal vein \(\rightarrow\) vena cava \(\rightarrow\) right atrium \(\rightarrow\) right ventricle \(\rightarrow\) pulmonary artery; (ii) High hydrostatic pressure at the arterial end forces fluid out, while osmotic pressure draws some back at the venous end, and excess is drained by lymphatic capillaries; (iii) \(941.7\%\) increase.
B.B: (i) Renal vein \(\rightarrow\) renal artery \(\rightarrow\) left atrium \(\rightarrow\) left ventricle \(\rightarrow\) pulmonary artery; (ii) Low hydrostatic pressure at the arteriole end sucks fluid in, while lymphatic vessels pump lymph actively using cardiac rhythm; (iii) \(1041.7\%\) increase.
C.C: (i) Renal vein \(\rightarrow\) vena cava \(\rightarrow\) left atrium \(\rightarrow\) left ventricle \(\rightarrow\) pulmonary artery; (ii) Active transport of water out of capillaries, with excess fluid returning via the hepatic portal vein; (iii) \(90.4\%\) increase.
D.D: (i) Renal vein \(\rightarrow\) vena cava \(\rightarrow\) right atrium \(\rightarrow\) right ventricle \(\rightarrow\) pulmonary vein; (ii) Diffusion of plasma proteins out of capillaries creates a concentration gradient, draining into lacteals; (iii) \(10.4\%\) increase.
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解題
(i) Red blood cell travels from the renal vein \(\rightarrow\) vena cava \(\rightarrow\) right atrium \(\rightarrow\) tricuspid valve \(\rightarrow\) right ventricle \(\rightarrow\) semilunar valve \(\rightarrow\) pulmonary artery.
(ii) High hydrostatic pressure at the arterial end of capillaries forces water and small solutes out into intercellular spaces to form tissue fluid. Large plasma proteins remain in the blood. At the venous end, lower hydrostatic pressure and higher osmotic pressure draw some fluid back in. Excess fluid is collected by lymphatic capillaries as lymph, which is returned to the subclavian veins.
Total: 10 marks - (i) 3 marks: 1 mark for vena cava, 1 mark for right atrium and ventricle in order, 1 mark for exiting via pulmonary artery. - (ii) 4 marks: 1 mark for high hydrostatic pressure forcing fluid out at arteriole end, 1 mark for plasma proteins remaining and causing osmotic drawing of water at venous end, 1 mark for lymphatic system absorbing excess fluid, 1 mark for returning lymph to the vena cava/blood circulatory system. - (iii) 3 marks: 1 mark for calculating correct difference of \(11300 \text{ cm}^3/\text{min}\), 1 mark for correct formula application, 1 mark for final percentage rounded to 1 decimal place (\(941.7\%\)).
題目 3 · multiple_choice
10 分
Cystic fibrosis is an autosomal recessive disorder caused by a mutation in the CFTR gene.
(i) Two parents who are both carriers of cystic fibrosis (\(Ff\)) have a child. Determine the probability that the child will be a carrier of cystic fibrosis and female. (ii) Describe the procedure of amniocentesis as a method of prenatal screening, including when it is carried out. (iii) Compare amniocentesis with chorionic villus sampling (CVS) in terms of timing and relative risk of miscarriage.
Which of the following options is correct?
A.A: (i) Probability is \(0.25\) (\(25\%\)); (ii) Amniocentesis uses ultrasound to guide a needle into the uterus to extract amniotic fluid containing fetal cells at 15-20 weeks; (iii) CVS is performed earlier (10-15 weeks) but has a slightly higher risk of miscarriage.
B.B: (i) Probability is \(0.50\) (\(50\%\)); (ii) Amniocentesis involves taking a blood sample from the mother's arm to isolate fetal DNA at any stage of pregnancy; (iii) CVS is performed later in pregnancy and is completely non-invasive.
C.C: (i) Probability is \(0.125\) (\(12.5\%\)); (ii) Amniocentesis extracts fluid from the umbilical cord to check for proteins; (iii) CVS is performed at 30 weeks and has no risk of miscarriage.
D.D: (i) Probability is \(0.375\) (\(37.5\%\)); (ii) Amniocentesis extracts cells directly from the placenta via a vaginal catheter; (iii) CVS is performed at 5 weeks and carries a 5% risk of miscarriage.
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解題
(i) Parents are \(Ff \times Ff\). Genetic cross yields: 1/4 \(FF\) (healthy), 2/4 = 1/2 \(Ff\) (carrier), 1/4 \(ff\) (affected). Probability of being a carrier = 1/2. Probability of being female = 1/2. Combined probability = \(1/2 \times 1/2 = 1/4 = 0.25\) (or \(25\%\)).
(ii) Amniocentesis is done between 15 and 20 weeks of pregnancy. Ultrasound is used to locate the fetus and placenta. A fine needle is inserted through the abdominal wall into the uterus to withdraw a sample of amniotic fluid containing fetal cells for genetic analysis.
(iii) CVS is performed earlier in pregnancy (typically 10-15 weeks) than amniocentesis (15-20 weeks). CVS has a slightly higher risk of miscarriage (approx 1-2%) compared to amniocentesis (approx 0.5-1%).
評分準則
Total: 10 marks - (i) 3 marks: 1 mark for genetic cross showing 50% / 1/2 carrier probability, 1 mark for gender probability of 1/2, 1 mark for calculating correct combined probability of 25% or 0.25. - (ii) 4 marks: 1 mark for timing (15-20 weeks), 1 mark for ultrasound guidance, 1 mark for abdominal needle insertion, 1 mark for extracting amniotic fluid containing fetal cells. - (iii) 3 marks: 1 mark for comparing timing (CVS earlier than amniocentesis), 1 mark for comparing miscarriage risk (CVS higher risk), 1 mark for comparing sample source (chorionic villi vs amniotic fluid).
題目 4 · multiple_choice
10 分
This question is about hormonal regulation during the menstrual cycle.
(i) Identify the trigger for the sharp luteinising hormone (LH) peak around day 13-14 of the cycle, and state its consequence. (ii) Describe the distinct roles of estrogen and progesterone in preparing and maintaining the endometrium for potential implantation. (iii) If a woman's progesterone level on day 21 is \(32 \text{ nmol/L}\), and the threshold indicating successful ovulation is \(25 \text{ nmol/L}\), calculate the percentage by which her progesterone level exceeds this threshold.
Which of the following options is correct?
A.A: (i) Estrogen level peaks, triggering LH surge which causes ovulation; (ii) Estrogen repairs/thickens uterine lining, progesterone maintains lining and increases vascularisation; (iii) \(28.0\%\) above the threshold.
B.B: (i) Progesterone level falls, triggering LH surge which causes menstruation; (ii) Estrogen breaks down the uterine lining, progesterone causes follicle development; (iii) \(7.0\%\) above the threshold.
C.C: (i) FSH level peaks, triggering LH surge which stimulates FSH production; (ii) Estrogen maintains the corpus luteum, progesterone repairs the endometrium; (iii) \(22.0\%\) above the threshold.
D.D: (i) HCG peaks, triggering LH surge which prevents ovulation; (ii) Estrogen stimulates milk production, progesterone stimulates uterine contractions; (iii) \(128.0\%\) above the threshold.
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解題
(i) High levels of estrogen secreted by the developing follicle stimulate a surge of LH from the anterior pituitary (positive feedback). This LH surge causes ovulation (the release of the secondary oocyte from the Graafian follicle).
(ii) Estrogen stimulates the proliferation and repair of the endometrium after menstruation, making it thicker. Progesterone, produced by the corpus luteum, maintains this lining, stimulates vascularisation (development of blood vessels), and makes it secretory to support embryo implantation.
Total: 10 marks - (i) 3 marks: 1 mark for high estrogen level trigger, 1 mark for LH surge release, 1 mark for causing ovulation. - (ii) 4 marks: 1 mark for estrogen repairing/thickening endometrium, 1 mark for progesterone maintaining thickness, 1 mark for increasing blood vessel density/glandular secretion, 1 mark for inhibiting further FSH/LH release to prevent multi-follicular growth. - (iii) 3 marks: 1 mark for difference (\(7 \text{ nmol/L}\)), 1 mark for percentage calculation, 1 mark for final correct answer of \(28\%\).
題目 5 · multiple_choice
10 分
A student conducted a calorimetry experiment to estimate the energy content of a peanut sample. - Mass of peanut sample: \(2.5 \text{ g}\) - Volume of water in boiling tube: \(50 \text{ cm}^3\) (density \(1.0 \text{ g/cm}^3\)) - Initial temperature of water: \(20 \text{ }^\circ\text{C}\) - Final temperature of water: \(68 \text{ }^\circ\text{C}\) - Specific heat capacity of water: \(4.2 \text{ J/g}^\circ\text{C}\)
(i) Calculate the energy released per gram of the peanut (in J/g). (ii) Describe two systematic errors in this experimental setup and how to improve accuracy. (iii) Explain how the lipids contained in the peanut are chemically digested in the human gut.
Which of the following options correctly responds to these tasks?
A.A: (i) \(4032 \text{ J/g}\); (ii) Heat loss to air and incomplete combustion are errors; improve by using an insulated bomb calorimeter with oxygen supply; (iii) Lipids are emulsified by bile and digested by lipase into fatty acids and glycerol.
B.B: (i) \(10080 \text{ J/g}\); (ii) Thermometer inaccuracy and water boiling are errors; improve by using less water; (iii) Lipids are digested by pepsin in the stomach and absorbed directly into the blood capillaries.
C.C: (i) \(1612.8 \text{ J/g}\); (ii) Distance of peanut from the tube is the main error; improve by holding the burning peanut closer; (iii) Lipids are emulsified by amylase and broken down into glucose in the mouth.
D.D: (i) \(4032000 \text{ J/g}\); (ii) Water evaporation is the only error; improve by using a cork stopper; (iii) Lipids are absorbed unchanged by active transport across the large intestine.
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解題
(i) Energy absorbed by water: \(Q = m \times c \times \Delta T = 50 \times 4.2 \times (68 - 20) = 50 \times 4.2 \times 48 = 10080 \text{ J}\). Energy per gram of peanut = \(10080 \text{ J} / 2.5 \text{ g} = 4032 \text{ J/g}\).
(ii) Main systematic errors include: heat loss to surrounding air, heat absorbed by the boiling tube, and incomplete combustion of the peanut. Improvements: use insulation around the boiling tube, use a metal calorimeter (better conductor), or use a bomb calorimeter supplied with pure oxygen to ensure complete combustion and minimal heat loss.
(iii) Lipids enter the duodenum where they are emulsified by bile (produced in liver, stored in gallbladder) into small droplets to increase surface area. Pancreatic and intestinal lipase then hydrolyse lipids into fatty acids and glycerol.
評分準則
Total: 10 marks - (i) 3 marks: 1 mark for correct temperature change (\(48\text{ }^\circ\text{C}\)), 1 mark for total energy calculation (\(10080\text{ J}\)), 1 mark for division by peanut mass yielding \(4032\text{ J/g}\). - (ii) 3 marks: 1 mark for identifying heat loss or incomplete combustion, 1 mark for identifying heat absorbed by tube, 1 mark for suggesting a bomb calorimeter or draft shields/insulation as improvement. - (iii) 4 marks: 1 mark for bile emulsifying lipids to increase surface area, 1 mark for bile secretion details (liver/gallbladder), 1 mark for lipase hydrolysing ester bonds, 1 mark for products being fatty acids and glycerol.
題目 6 · multiple_choice
10 分
This question concerns thermoregulation when a person enters a cold environment (5 °C).
(i) Explain the physiological responses of the skin to conserve heat, focusing on vasoconstriction and erector pili muscles. (ii) Describe how the hypothalamus acts in a negative feedback loop to coordinate these responses. (iii) A researcher measures sweat production: at 20 °C, it is \(12 \text{ g/hour}\); at 35 °C, it is \(180 \text{ g/hour}\). Calculate the factor increase in sweat production.
Which of the following options is correct?
A.A: (i) Vasoconstriction narrows arterioles supplying skin surface capillaries to reduce blood flow and heat loss by radiation; erector pili muscles contract to raise hairs, trapping a layer of insulating air; (ii) Hypothalamus detects low temperature and sends nerve impulses to effectors to reverse change; (iii) 15-fold increase.
B.B: (i) Vasodilation widens surface capillaries to allow warm blood to shield skin; erector pili muscles relax to lay hairs flat; (ii) Hypothalamus stimulates pituitary gland to release insulin to generate metabolic heat; (iii) 1.5-fold increase.
C.C: (i) Shivering of skin surface cells releases heat directly; erector pili muscles sweat to cool the skin; (ii) Hypothalamus acts as an effector that actively heats the body using thermal energy; (iii) 150-fold increase.
D.D: (i) Sweat glands produce sebum to insulate the skin; arteriole valves close to trap blood in the dermis; (ii) Hormones from thyroid gland directly heat the sweat glands; (iii) 168-fold increase.
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解題
(i) In a cold environment: Vasoconstriction occurs where arterioles supplying skin surface capillaries narrow. This reduces blood flow near the skin surface, minimizing heat loss by radiation. Erector pili muscles contract, causing skin hairs to stand upright. This traps a layer of still, warm air next to the skin, which acts as an insulator.
(ii) Thermoreceptors in the skin and hypothalamus detect a drop in temperature. The hypothalamus coordinates responses, sending impulses via nerves to effectors (vasoconstriction of arterioles, shivering of muscles, contraction of hair erectors) to raise temperature back to the set point (37 °C), shutting off the response once normal temperature is restored.
(iii) Factor increase = \(180 / 12 = 15\) times.
評分準則
Total: 10 marks - (i) 4 marks: 1 mark for vasoconstriction narrowing arterioles, 1 mark for reducing blood flow to surface capillaries to decrease radiation heat loss, 1 mark for erector pili muscles contracting to raise hairs, 1 mark for trapping an insulating layer of still air. - (ii) 3 marks: 1 mark for thermoreceptor detection, 1 mark for hypothalamus coordination, 1 mark for feedback mechanism returning blood temperature to set point. - (iii) 3 marks: 1 mark for dividing exercise/high sweat rate by resting/low rate, 1 mark for correct calculation of 15, 1 mark for stating factor increase unit/concept correctly.
題目 7 · multiple_choice
10 分
This question is about biological molecules and enzyme activity.
(i) State the chemical tests and positive results used to detect reducing sugars, proteins, and lipids. (ii) In an investigation of amylase activity, the time taken for starch to disappear at pH 7 is 30 seconds. Calculate the rate of reaction at pH 7 in \(\text{s}^{-1}\) using the formula: \(\text{Rate} = 1 / \text{time}\). (iii) Explain why the rate of starch digestion is significantly lower at extreme pH levels (such as pH 5 or pH 9) compared to pH 7.
Which of the following options is correct?
A.A: (i) Reducing sugars: Benedict's test (heat, turns blue to brick-red); Proteins: Biuret test (turns blue to violet); Lipids: Emulsion test (ethanol, turns cloudy white); (ii) \(0.033 \text{ s}^{-1}\); (iii) Extreme pH levels alter the ionic charges on amino acids, changing the shape of the active site (denaturation), so the substrate (starch) can no longer fit.
B.B: (i) Reducing sugars: Iodine solution (turns yellow to blue-black); Proteins: Benedict's test (turns red); Lipids: Biuret test (turns purple); (ii) \(30.0 \text{ s}^{-1}\); (iii) High and low pH values make the starch molecules repel the amylase active site without changing the enzyme.
C.C: (i) Reducing sugars: Emulsion test (turns green); Proteins: Iodine test (turns orange); Lipids: Biuret test (turns yellow); (ii) \(0.004 \text{ s}^{-1}\); (iii) Low pH activates inhibitors that block the enzyme, while high pH speeds up the reaction too much.
D.D: (i) Reducing sugars: Biuret test (turns blue to pink); Proteins: Emulsion test (turns cloudy); Lipids: Iodine test (turns black); (ii) \(0.120 \text{ s}^{-1}\); (iii) pH changes the activation energy of the starch molecules so that they decompose spontaneously.
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解題
(i) Reducing sugars: Benedict's reagent + heat \(\rightarrow\) changes from blue to green/yellow/brick-red precipitate. Proteins: Biuret reagent \(\rightarrow\) changes from blue to lilac/purple/violet. Lipids: Ethanol emulsion test \(\rightarrow\) white cloudy emulsion formed.
(ii) Rate of reaction = \(1 / 30 \approx 0.033 \text{ s}^{-1}\).
(iii) Amylase has an optimum pH around pH 7. Extreme pH values (too acidic or too alkaline) alter the charge on the amino acids making up the enzyme. This disrupts the hydrogen and ionic bonds maintaining the tertiary structure of the protein, causing the active site to change shape (denaturation). The substrate (starch) can no longer fit into the active site, stopping the reaction.
評分準則
Total: 10 marks - (i) 3 marks: 1 mark for Benedict's test (heat + brick-red color), 1 mark for Biuret test (lilac/purple color), 1 mark for emulsion test (ethanol + cloudy white emulsion). - (ii) 3 marks: 1 mark for identifying the time (30 s), 1 mark for applying the formula, 1 mark for correct calculation of \(0.033 \text{ s}^{-1}\) (allow \(0.03\) recurring). - (iii) 4 marks: 1 mark for optimum pH concept, 1 mark for pH changing amino acid charges, 1 mark for breaking ionic/hydrogen bonds altering the active site shape, 1 mark for denaturation preventing substrate binding.
題目 8 · multiple_choice
10 分
This question is about cell structure and microscopy.
(i) A student views a human cheek epithelial cell under a light microscope. The actual diameter of the cell is \(60 \text{ }\mu\text{m}\). In the microscope drawing, the cell measures \(24 \text{ mm}\). Calculate the magnification of the image. (ii) Compare the structure and function of squamous, ciliated, and cuboidal epithelial tissues. (iii) Explain how a mature red blood cell is adapted for its function.
Which of the following options is correct?
A.A: (i) \(\times 400\); (ii) Squamous is thin/flat for rapid diffusion (e.g. alveoli), ciliated has hair-like projections to sweep mucus (e.g. trachea), cuboidal is cube-shaped for secretion/absorption (e.g. kidney tubules); (iii) Biconcave shape increases surface area to volume ratio, absence of nucleus leaves more room for haemoglobin, flexible membrane allows squeezing through narrow capillaries.
B.B: (i) \(\times 2.5\); (ii) Squamous is thick for protection, ciliated produces bile, cuboidal contracts to move food; (iii) Large nucleus for rapid protein synthesis, multiple mitochondria for active transport of oxygen, spherical shape to roll smoothly.
C.C: (i) \(\times 40000\); (ii) Squamous is found in bones to provide strength, ciliated lines the stomach to digest food, cuboidal lines the bladder to stretch; (iii) Flagellum for swimming, high water content to dilute toxins, thick cell wall to resist pressure.
D.D: (i) \(\times 0.0025\); (ii) Squamous secretes hormones, ciliated senses sound waves in the ear, cuboidal covers the outer skin surface; (iii) Lacks cytoplasm to reduce weight, microvilli to absorb nutrients, granules containing digestive enzymes.
(ii) Squamous epithelium: flat, thin, single-layered cells; specialized for rapid diffusion (e.g., alveoli). Ciliated epithelium: columnar cells with hair-like cilia; sweep mucus and trapped pathogens (e.g., trachea). Cuboidal epithelium: cube-shaped cells; involved in absorption and secretion (e.g., kidney tubules).
(iii) Red blood cell adaptations: Biconcave shape increases surface-area-to-volume ratio for rapid diffusion of oxygen; lack of nucleus and other organelles leaves more space for haemoglobin to transport oxygen; flexible cell membrane allows them to squeeze through thin capillaries.
評分準則
Total: 10 marks - (i) 3 marks: 1 mark for unit conversion (e.g. \(24000 \text{ }\mu\text{m}\)), 1 mark for correct formula substitution, 1 mark for correct magnification (\(\times 400\)). - (ii) 3 marks: 1 mark for squamous description/function, 1 mark for ciliated description/function, 1 mark for cuboidal description/function. - (iii) 4 marks: 1 mark for biconcave shape (SA:V ratio), 1 mark for lack of nucleus (room for haemoglobin), 1 mark for presence of haemoglobin (oxygen binding), 1 mark for flexibility (capillary transit).
Paper 2R
Answer all questions. Write answers in spaces provided. Cross-references, calculations, standard form and ratio formatting are required where indicated.
8 題目 · 90 分
題目 1 · 選擇題
11.25 分
An investigation was carried out to measure the pressure changes in the left atrium, left ventricle, and aorta of a healthy human subject during one complete cardiac cycle. The peak pressure in the left ventricle was recorded as 120 mmHg, while the peak pressure in the left atrium was 10 mmHg. The pressure in the aorta fluctuated between 80 mmHg and 120 mmHg. At what point during the cardiac cycle do the semi-lunar (aortic) valves open, and what is the correct state of the bicuspid (mitral) valve at this exact moment?
A.When left ventricular pressure exceeds 80 mmHg; the bicuspid valve is closed.
B.When left ventricular pressure exceeds 10 mmHg; the bicuspid valve is open.
C.When left atrial pressure exceeds left ventricular pressure; the bicuspid valve is closed.
D.When aortic pressure exceeds left ventricular pressure; the bicuspid valve is open.
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解題
The aortic valve (a semi-lunar valve) opens when the pressure inside the left ventricle exceeds the pressure in the aorta. The diastolic pressure of the aorta is 80 mmHg, so the ventricle must generate a pressure greater than 80 mmHg to force this valve open and eject blood. At this moment, the left ventricle is contracting (ventricular systole). To prevent the backflow of blood into the left atrium, the bicuspid (mitral) valve must be firmly closed. This occurs during the early phase of ventricular ejection.
評分準則
Award 3.75 marks for correctly identifying that the left ventricular pressure must exceed the diastolic aortic pressure (80 mmHg) to open the aortic valve. Award 3.75 marks for explaining that the bicuspid (mitral) valve must be closed at this point to prevent backflow of blood into the left atrium. Award 3.75 marks for explaining the overall pressure gradient dynamics during ventricular systole.
題目 2 · 選擇題
11.25 分
In an experiment to study tissue fluid formation, a researcher measures the hydrostatic pressure and oncotic (osmotic) pressure at both the arteriole end and the venule end of a capillary bed in a skeletal muscle tissue. The hydrostatic pressure at the arteriole end is 35 mmHg and at the venule end is 15 mmHg. The oncotic pressure of the blood plasma remains constant at 25 mmHg throughout the capillary, while the hydrostatic pressure of the interstitial fluid is 2 mmHg and its oncotic pressure is 5 mmHg. What is the net filtration pressure (NFP) at the arteriole end and the net absorption pressure (NAP) at the venule end, and what physiological consequence would occur if the plasma protein concentration were severely reduced due to starvation?
A.NFP = +13 mmHg, NAP = -7 mmHg; reduced plasma proteins would lead to decreased tissue fluid formation and tissue dehydration.
B.NFP = +13 mmHg, NAP = -7 mmHg; reduced plasma proteins would lead to increased tissue fluid retention and edema.
C.NFP = +17 mmHg, NAP = -3 mmHg; reduced plasma proteins would lead to decreased tissue fluid formation and tissue dehydration.
D.NFP = +17 mmHg, NAP = -3 mmHg; reduced plasma proteins would lead to increased tissue fluid retention and edema.
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解題
At the arteriole end, the net filtration pressure (NFP) is calculated as Outward forces (Capillary Hydrostatic Pressure + Interstitial Oncotic Pressure) minus Inward forces (Plasma Oncotic Pressure + Interstitial Hydrostatic Pressure). Thus, NFP = (35 + 5) - (25 + 2) = 40 - 27 = +13 mmHg (favoring filtration). At the venule end, the outward forces are (15 + 5) = 20 mmHg and inward forces are (25 + 2) = 27 mmHg. The net force is 20 - 27 = -7 mmHg, indicating an absorption pressure of 7 mmHg. If plasma proteins are severely reduced (e.g., due to starvation), the plasma oncotic pressure decreases. This reduces the inward force, leading to less tissue fluid being reabsorbed at the venule end, causing accumulation of fluid in tissue spaces (edema).
評分準則
Award 3.75 marks for calculating NFP at the arteriole end as +13 mmHg and the net absorption pressure at the venule end as -7 mmHg (or 7 mmHg inward). Award 3.75 marks for identifying that reduced plasma protein concentration decreases plasma oncotic pressure. Award 3.75 marks for explaining that decreased oncotic pressure reduces reabsorption, leading to tissue fluid accumulation and edema.
題目 3 · 選擇題
11.25 分
Sickle-cell anemia is an autosomal co-dominant disorder caused by a mutation in the HBB gene that encodes the beta-globin chain of hemoglobin. Individuals with the homozygous genotype Hb^S Hb^S suffer from severe sickle-cell disease, while heterozygotes (Hb^A Hb^S) have the sickle-cell trait and are generally healthy but exhibit resistance to malaria. A genetic counselor is consulting a couple where both partners have the sickle-cell trait. What is the probability that their first child will be a female who is resistant to malaria, and what is the expected ratio of phenotypes among their offspring?
A.Probability = 0.25; phenotypic ratio of 1 normal hemoglobin : 2 sickle-cell trait : 1 sickle-cell disease.
B.Probability = 0.50; phenotypic ratio of 3 normal hemoglobin : 1 sickle-cell disease.
C.Probability = 0.25; phenotypic ratio of 3 normal hemoglobin : 1 sickle-cell disease.
D.Probability = 0.50; phenotypic ratio of 1 normal hemoglobin : 2 sickle-cell trait : 1 sickle-cell disease.
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解題
The genetic cross is between two heterozygous individuals (Hb^A Hb^S x Hb^A Hb^S). The probability of producing an offspring with the sickle-cell trait (Hb^A Hb^S), which provides resistance to malaria, is 0.50 (2 out of 4). The probability of the child being female is 0.50. Using the product rule for independent events, the probability of having a female child who is resistant to malaria is 0.50 * 0.50 = 0.25. The expected phenotypic ratio is 1 normal hemoglobin (Hb^A Hb^A) : 2 sickle-cell trait (Hb^A Hb^S) : 1 sickle-cell disease (Hb^S Hb^S).
評分準則
Award 3.75 marks for determining the offspring genotypes (25 percent Hb^A Hb^A, 50 percent Hb^A Hb^S, 25 percent Hb^S Hb^S). Award 3.75 marks for calculating the probability of a female child with the trait as 0.25 (0.50 for trait multiplied by 0.50 for female). Award 3.75 marks for stating the correct phenotypic ratio of 1 normal : 2 trait : 1 disease.
題目 4 · 選擇題
11.25 分
During the human menstrual cycle, the pituitary gland and the ovaries secrete hormones that interact via positive and negative feedback loops to regulate follicle development and ovulation. A fertility clinic monitors the hormone levels of a patient undergoing IVF. On day 11 of her cycle, they detect a rapid, sharp rise in estrogen levels secreted by the developing follicle. What physiological response does this high level of estrogen trigger in the pituitary gland, which hormone peaks as a direct result to cause ovulation, and what is the subsequent role of the corpus luteum?
A.It triggers a switch from negative to positive feedback, causing a surge in luteinizing hormone (LH) that stimulates ovulation; the corpus luteum then secretes progesterone to maintain the endometrium.
B.It triggers negative feedback to inhibit follicle-stimulating hormone (FSH) secretion, causing a surge in progesterone; the corpus luteum then secretes estrogen to rebuild the endometrium.
C.It triggers positive feedback, causing a surge in follicle-stimulating hormone (FSH) that stimulates ovulation; the corpus luteum then degenerates immediately to allow menstruation.
D.It triggers negative feedback to inhibit luteinizing hormone (LH) secretion, causing a surge in human chorionic gonadotropin (hCG); the corpus luteum then secretes progesterone to inhibit the pituitary gland.
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解題
High concentrations of estrogen toward the end of the follicular phase trigger a switch in the feedback mechanism from negative to positive feedback at the level of the hypothalamus and anterior pituitary. This causes a sudden, massive surge in luteinizing hormone (LH) secretion, which triggers ovulation (release of the secondary oocyte from the Graafian follicle). Following ovulation, LH stimulates the remains of the follicle to develop into the corpus luteum, which secretes progesterone. Progesterone maintains the thickness and vascularization of the endometrium to prepare for potential implantation.
評分準則
Award 3.75 marks for identifying the transition from negative to positive feedback caused by elevated estrogen levels. Award 3.75 marks for identifying the LH surge as the direct trigger for ovulation. Award 3.75 marks for describing the development of the corpus luteum and its role in secreting progesterone to maintain the endometrium.
題目 5 · 選擇題
11.25 分
A student designed a simple laboratory calorimeter to measure the energy content of a 2.0 g sample of dried peanut. The sample was ignited and used to heat 100 g of water in a copper beaker. The initial temperature of the water was 21.5 degrees Celsius, and the final temperature of the water after the sample had completely burned was 68.5 degrees Celsius. Given that the specific heat capacity of water is 4.2 J/(g degrees Celsius), calculate the energy released per gram of the peanut sample, and identify the primary limitation of this experimental setup that causes the calculated value to be an underestimate.
A.9.87 kJ/g; heat loss to the surrounding air and the copper beaker.
B.9.87 kJ/g; incomplete combustion of the peanut sample due to excess oxygen.
C.19.74 kJ/g; heat loss to the surrounding air and the copper beaker.
D.19.74 kJ/g; incomplete combustion of the peanut sample due to excess oxygen.
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解題
The heat energy absorbed by the water is calculated using the formula Q = m * c * delta T. Here, m = 100 g, c = 4.2 J/(g°C), and delta T = 68.5 - 21.5 = 47.0 °C. Thus, Q = 100 * 4.2 * 47.0 = 19,740 J = 19.74 kJ. The mass of the peanut sample is 2.0 g, so the energy released per gram of peanut is 19.74 kJ / 2.0 g = 9.87 kJ/g. The primary limitation of a simple calorimeter is that a significant amount of heat is lost to the surrounding air, the copper beaker, and through incomplete combustion, meaning that not all chemical energy is transferred to the water, resulting in an underestimated value.
評分準則
Award 3.75 marks for calculating the total heat energy absorbed by the water as 19.74 kJ (or 19,740 J). Award 3.75 marks for dividing by the mass of the food sample to obtain 9.87 kJ/g. Award 3.75 marks for identifying heat loss to surroundings and the beaker as the primary limitation leading to an underestimate.
題目 6 · 選擇題
11.25 分
A clinical researcher investigates human thermoregulation during exposure to a cold environment (4 degrees Celsius). When thermoreceptors in the skin and the hypothalamus detect a decrease in core body temperature, they initiate homeostatic effector responses. Which of the following describes the correct combination of nervous pathway, effector action, and physical mechanism used to minimize heat loss and generate metabolic heat?
A.Sympathetic pathway stimulates cutaneous vasodilation to increase blood flow to the skin, while somatic motor pathways trigger shivering in skeletal muscles to release heat from respiration.
B.Parasympathetic pathway stimulates cutaneous vasoconstriction to reduce blood flow to the skin, while somatic motor pathways trigger voluntary movement.
C.Sympathetic pathway stimulates cutaneous vasoconstriction (reducing blood flow to the superficial capillary loops), while somatic motor pathways trigger involuntary shivering of skeletal muscles.
D.Hormonal pathway secretes adrenaline to cause sweat gland activation, while parasympathetic pathway triggers erector pili muscles to relax and flatten body hair.
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解題
In a cold environment, the hypothalamus coordinates autonomic and somatic motor responses. Sympathetic nerves stimulate the smooth muscle in the walls of arterioles supplying the skin capillaries to contract (vasoconstriction), reducing blood flow to the superficial capillary loops and minimizing heat loss via radiation and convection. Concurrently, somatic motor nerves stimulate involuntary, rapid contraction and relaxation of skeletal muscles (shivering), which increases cellular respiration rates to release metabolic waste heat. Erector pili muscles also contract to raise hairs, trapping a layer of insulating air.
評分準則
Award 3.75 marks for identifying the role of the sympathetic nervous system in cutaneous vasoconstriction. Award 3.75 marks for explaining how vasoconstriction reduces blood flow to the skin surface to decrease heat loss. Award 3.75 marks for describing somatic motor control of involuntary shivering and its role in generating heat through increased respiration.
題目 7 · 選擇題
11.25 分
At a neuromuscular junction, transmission of an action potential from the motor neuron to the muscle fiber involves several steps. A newly discovered neurotoxin, Tox-X, is isolated from a marine organism. Investigations show that Tox-X binds irreversibly to voltage-gated calcium channels on the presynaptic membrane, preventing them from opening. What is the direct consequence of Tox-X on synaptic transmission, and what would be the clinical effect on the skeletal muscle of a patient?
A.Acetylcholine vesicles cannot fuse with the presynaptic membrane, preventing acetylcholine release into the synaptic cleft; this causes flaccid paralysis of the skeletal muscle.
B.Acetylcholine remains bound to the post-synaptic receptors permanently, preventing repolarization; this causes spastic paralysis of the skeletal muscle.
C.Acetylcholinesterase is inhibited, leading to continuous stimulation of the post-synaptic membrane; this causes muscle spasms.
D.Sodium ions cannot enter the presynaptic terminal, preventing the propagation of the action potential; this causes muscular hypertrophy.
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解題
When an action potential arrives at the presynaptic terminal, depolarization normally opens voltage-gated calcium channels. Calcium influx triggers synaptic vesicles containing acetylcholine (ACh) to fuse with the presynaptic membrane and release ACh into the synaptic cleft by exocytosis. If Tox-X blocks these calcium channels, no calcium enters, vesicles cannot fuse, and ACh is not released. Consequently, ACh cannot bind to ligand-gated sodium channels on the postsynaptic muscle membrane, preventing muscle depolarization and contraction, leading to flaccid paralysis.
評分準則
Award 3.75 marks for identifying that preventing calcium influx stops acetylcholine-containing vesicles from fusing with the presynaptic membrane. Award 3.75 marks for explaining that the absence of acetylcholine in the cleft prevents depolarization of the post-synaptic muscle membrane. Award 3.75 marks for concluding that the lack of muscle contraction results in flaccid paralysis.
題目 8 · 選擇題
11.25 分
During high-intensity anaerobic exercise, such as a 100m sprint, human muscle cells experience a temporary shortage of oxygen and rely on anaerobic respiration to generate ATP. Which of the following statements correctly describes the biochemistry of anaerobic respiration in human muscle cells, the resulting physiological condition, and how the 'oxygen debt' is subsequently repaid during recovery?
A.Glucose is partially broken down into ethanol and carbon dioxide, yielding 2 ATP molecules; the accumulated ethanol is transported to the kidneys to be excreted.
B.Glucose is partially broken down into lactic acid, yielding 2 ATP molecules; the oxygen debt represents the extra volume of oxygen required after exercise to oxidize lactic acid to carbon dioxide and water, or convert it back to glucose in the liver.
C.Glucose is completely oxidized to lactic acid, yielding 36 ATP molecules; the oxygen debt is repaid by deep, rapid breathing to replenish blood pH immediately in the lungs.
D.Glycogen is directly converted into carbon dioxide and water without the formation of any intermediates; the oxygen debt is repaid by resting in a hyperbaric chamber to restore hemoglobin saturation.
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解題
During anaerobic respiration in human muscle cells, glucose undergoes glycolysis to produce two molecules of pyruvate, yielding a net of 2 ATP molecules. Pyruvate is then reduced to lactic acid (lactate) by accepting electrons from NADH, regenerating NAD+ to allow glycolysis to continue. Lactic acid accumulation lowers muscle pH, contributing to muscle fatigue and soreness. After exercise, the 'oxygen debt' is the additional volume of oxygen needed to process this lactic acid: it is transported in the blood to the liver, where oxygen is used to oxidize it back to pyruvate (and then to CO2 and H2O) or convert it back to glucose/glycogen via gluconeogenesis.
評分準則
Award 3.75 marks for describing the breakdown of glucose to lactic acid with a net yield of 2 ATP. Award 3.75 marks for explaining that lactic acid accumulation causes muscle fatigue. Award 3.75 marks for explaining that the oxygen debt is repaid by consuming extra oxygen to oxidize lactic acid or convert it to glycogen in the liver.
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