4.2 Organisation — coverage pack

Answer

• A tissue is a group of cells with a similar structure and function.
• An organ is an aggregation of tissues performing a specific function.

Method: Award one mark for linking a tissue to similar cells carrying out a function and one mark for linking an organ to several tissues carrying out a function.

Answer

• Similar specialised cells form a tissue.
• Different tissues form an organ, such as the stomach.
• Several digestive organs work together as the digestive system to digest and absorb food.

Method: Build the hierarchy in order: cells with similar structure and function form tissues; several tissues form an organ; several organs that cooperate form an organ system. Credit the stomach as an organ and the digestive system as the linked organ system.

Answer

• Similar heart muscle cells form cardiac muscle tissue.
• Cardiac muscle tissue works with other tissues to form the heart.
• The heart is an organ in the circulatory system.
• The circulatory system works with other organ systems in the organism.

Method: Follow the biological hierarchy without skipping levels: specialised cells form tissue, several tissues form the heart organ, the heart works with blood vessels in the circulatory organ system, and organ systems together form the human organism.

Answer

• Protease produces amino acids.
• Lipase produces glycerol and fatty acids.

Method: Recall the two digestion word equations: protein is broken down to amino acids, and lipid is broken down to glycerol and fatty acids.

Answer

• Bile is alkaline and neutralises hydrochloric acid arriving from the stomach.
• This provides suitable alkaline conditions for enzymes in the small intestine.
• Bile emulsifies lipid into small droplets, increasing the surface area available to lipase.

Method: Give both independent actions of bile and link each to faster enzyme action: neutralisation creates suitable pH conditions, and emulsification gives lipase a larger lipid surface to act on.

Answer

• Mix amylase, starch and a buffer of known pH, then sample the mixture at regular time intervals onto iodine reagent.
• The end point is the first sample that remains orange-brown, showing that no starch is detected.
• Repeat at a range of pH values and compare rate using the reciprocal of the time taken.
• Control temperature with a water bath and keep enzyme and starch concentrations or volumes constant.

Method: Use buffer solutions to change only pH. Place drops of iodine in a spotting tile, combine measured amylase and starch solutions, and test a sample at fixed intervals. Record the time when iodine no longer turns blue-black. Calculate relative rate as $1/t$, repeat, and compare means. Keep temperature and the amounts or concentrations of enzyme and substrate constant.

Answer

• The right ventricle pumps blood to the lungs.
• The left ventricle pumps blood around the rest of the body.

Method: Recall the two circuits: the right side supplies the pulmonary circulation, and the left side supplies the systemic circulation.

Answer

• Arteries have thick muscular and elastic walls to withstand and maintain high pressure.
• Veins have a large lumen and valves that prevent backflow at lower pressure.
• Capillaries have walls one cell thick, giving a short diffusion distance for exchange with tissues.

Method: For each vessel, pair a structural feature with its consequence. Thick elastic artery walls handle pressure, vein valves maintain one-way flow, and very thin capillary walls speed diffusion.

Answer

• Mean blood-flow rate $=0.080\,\mathrm{dm^3\,s^{-1}}$.
• The left ventricle must produce greater pressure to pump blood around the whole body, whereas the right ventricle only pumps to the nearby lungs.

Method: Use $\mathrm{rate}=\mathrm{volume}/\mathrm{time}$, so $4.8/60=0.080\,\mathrm{dm^3\,s^{-1}}$. The systemic circulation has a greater distance and resistance than the pulmonary circulation, so the left ventricle needs more muscle to generate higher pressure.

Answer

• Any two from glucose, amino acids, carbon dioxide, urea, hormones or products of digestion.

Method: Choose two named substances that dissolve in plasma; do not give oxygen as the clearest answer because most oxygen is carried by haemoglobin in red blood cells.

Answer

• It contains haemoglobin, which binds oxygen.
• Its biconcave shape gives a large surface-area-to-volume ratio for diffusion.
• It has no nucleus, leaving more space for haemoglobin.

Method: For each mark, link an adaptation to oxygen transport: haemoglobin carries oxygen, the biconcave shape speeds exchange, and loss of the nucleus increases haemoglobin capacity.

Answer

• The nucleated cells are white blood cells involved in defence against pathogens.
• They may engulf pathogens or produce antibodies or antitoxins.
• A low platelet count reduces clotting, so bleeding lasts longer and microorganisms may enter through the wound more easily.

Method: Identify the nucleated blood cells as white blood cells and state a valid defence mechanism. Then connect platelets to clot formation: fewer platelets cause slower clotting, prolonged blood loss and a less effective barrier against pathogen entry.

Answer

• Narrowing reduces blood flow through the coronary artery.
• Less oxygen reaches heart muscle cells for aerobic respiration.

Method: Link the narrowed lumen first to reduced coronary blood flow and then to reduced oxygen delivery to the heart muscle.

Answer

• A stent mechanically holds a narrowed coronary artery open and can restore blood flow quickly.
• Inserting a stent requires a procedure and carries risks such as infection or a blood clot.
• Statins reduce blood cholesterol and slow further fatty deposits but must be taken long term and can have side effects.

Method: Compare mechanism, timescale and risk. A stent directly widens the artery but requires an intervention; statins act more gradually on cholesterol and deposits and require continued medication.

Answer

• A successful donor heart can provide long-term pumping function, but suitable donors are scarce.
• Transplant surgery carries risks and the immune system may reject the organ, so immunosuppressant drugs are needed.
• An artificial heart avoids immediate dependence on a donor and can keep the patient alive or allow the heart to rest.
• Artificial devices can increase risks of infection, bleeding or blood clots and are usually a temporary bridge rather than a biological replacement.
• The best choice depends on donor availability, urgency and the patient's surgical risks.

Method: A balanced evaluation gives benefits and limitations of both treatments before reaching a conditional judgement. Compare likely duration, donor availability, rejection, major surgery and device-related complications.

Answer

• Health is a state of physical and mental well-being.

Method: The definition must include both physical and mental well-being; simply saying that health is the absence of disease is incomplete.

Answer

• For example, an immune-system defect makes infectious disease more likely because pathogens are not destroyed effectively.
• For example, a virus living in cells can trigger changes that lead to cancer.
• Other valid interactions include pathogen-triggered immune reactions causing allergies or severe physical illness contributing to depression.

Method: Give two distinct interactions. For each, identify the first condition and explain the biological or health consequence that produces the second problem.

Answer

• The first-town incidence is $40$ cases per $1000$ people.
• The recorded incidence is higher in the first town than in the second town.
• A valid comparison needs representative samples collected in the same way and over the same time period, or control of relevant differences such as age distribution.

Method: Calculate $(80/2000)\times1000=40$ cases per $1000$. Compare $40$ with $15$, then identify a sampling or standardisation condition needed so the difference is not an artefact of the samples or collection method.

Answer

• Any two from poor diet, smoking or lack of exercise.

Method: Select two lifestyle factors explicitly linked by the specification to cardiovascular disease.

Answer

• Tobacco smoke contains carcinogens that increase the chance of changes leading to lung cancer.
• Smoking also damages the cardiovascular system and increases cardiovascular-disease incidence.
• A risk factor changes probability rather than guaranteeing disease, and other genetic, lifestyle or environmental factors also contribute.

Method: Link smoking to the specified diseases, including carcinogens for cancer, then distinguish increased risk from certainty. Recognise that disease usually reflects interaction among several factors.

Answer

• $3.0\%$ of smokers and $1.0\%$ of non-smokers were affected.
• The observed proportion was three times as high among smokers.
• The association supports smoking as a risk factor but other variables, such as age, occupation or air pollution, may differ between groups.
• A larger representative sample and control of confounding variables would strengthen the conclusion.

Method: For smokers, $(24/800)\times100=3.0\%$; for non-smokers, $(9/900)\times100=1.0\%$. The relative risk in this sample is $3.0/1.0=3$. Association is not enough to assign every case to smoking because uncontrolled factors and natural variation may contribute.

Answer

• Changes occur in cells.
• The changes lead to uncontrolled cell growth and division.

Method: State both parts of the specification description: cells change, and their growth and division become uncontrolled.

Answer

• Both are growths of abnormal cells.
• Benign tumours remain contained in one area and do not invade other tissues.
• Malignant tumour cells invade neighbouring tissues and can spread through the blood.
• Malignant cells can form secondary tumours elsewhere in the body.

Method: Compare location and behaviour: benign growths are contained, whereas malignant cells invade, travel in blood and establish secondary tumours.

Answer

• Some cancers have genetic risk factors, so the inherited variant can increase probability.
• Lifestyle factors can also increase cancer risk.
• Risk factors do not guarantee the cell changes that cause uncontrolled division.
• Cancer can result from interaction among several genetic, lifestyle and environmental factors.
• A malignant tumour would additionally have to invade neighbouring tissue and may spread to form secondary tumours.

Method: Distinguish a risk factor from a certain cause. Explain that multiple factors affect whether relevant cell changes arise, then identify invasion as the feature that makes a tumour malignant.

Answer

• Its cells contain many chloroplasts.
• It is positioned near the upper surface of the leaf where light intensity is high.

Method: Link the abundance of chloroplasts and the tissue's position near the light-facing surface to absorption of light for photosynthesis.

Answer

• Epidermal tissue protects the leaf while stomata permit controlled gas exchange.
• Palisade mesophyll absorbs light and carries out much of the photosynthesis because it contains many chloroplasts.
• Air spaces in spongy mesophyll allow carbon dioxide and oxygen to diffuse between stomata and photosynthesising cells.

Method: Assign a distinct function to each named tissue, then link gas entry through stomata and diffusion through spongy air spaces to photosynthesis in palisade cells.

Answer

• The tightly packed region is palisade mesophyll, with many chloroplasts for photosynthesis.
• The loosely arranged region is spongy mesophyll, whose air spaces allow rapid gas diffusion.
• The vein contains xylem, which supplies water and mineral ions, and phloem, which transports dissolved sugars.

Method: Identify each region from its position or structure, connect it to its function, and name both vascular tissues with their different transported substances.

Answer

• Xylem transports water and mineral ions.
• Phloem transports dissolved sugars or food molecules.

Method: Distinguish the two vascular tissues by their cargo: water and mineral ions in xylem, dissolved sugars in phloem.

Answer

• Higher temperature increases evaporation and the diffusion rate of water vapour.
• Faster air movement removes water vapour from around the leaf, maintaining a steep concentration gradient.
• Lower humidity makes the water-vapour concentration gradient from leaf to air steeper.
• Greater light intensity causes stomata to open for gas exchange, increasing water loss.

Method: Treat each factor separately and link it to evaporation, removal of humid air, the water-vapour concentration gradient or stomatal opening.

Answer

• Mean uptake $=0.40\,\mathrm{cm^3}$.
• Mean uptake rate $=0.020\,\mathrm{cm^3\,min^{-1}}$.
• A potometer measures water uptake, and some absorbed water is used in photosynthesis, growth or maintaining cells rather than being lost by transpiration.

Method: The mean is $(0.42+0.38+0.40)/3=0.40\,\mathrm{cm^3}$. Use $\mathrm{rate}=\mathrm{volume}/\mathrm{time}$: $0.40/20=0.020\,\mathrm{cm^3\,min^{-1}}$. Water uptake is a proxy because not every absorbed water molecule evaporates from the leaf.