AQA A Level Biology Topic 3 - Organisms Exchange Substances with Their Environment

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Comprehensive practice flashcards covering SA:V ratios, gas exchange, digestion and absorption, mass transport in animals and plants, and relevant practical techniques based on AQA A Level Biology Topic 3 notes.

Last updated 2:31 PM on 8/30/26
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37 Terms

1
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How does an organism's surface area to volume ratio (SA:V) change as its size increases?

As size increases, surface area to volume ratio (SA:V) tends to decrease because volume increases faster than surface area.

2
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How is the surface area to volume ratio (SA:V) of an object calculated?

Divide the total surface area ( side length×side width×number of sides\text{side length} \times \text{side width} \times \text{number of sides} ) by the volume ( length×width×depth\text{length} \times \text{width} \times \text{depth} ).

3
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Why can calculating SA:mass be advantageous compared to calculating SA:V for some organisms?

It is easier, quicker to find, or more accurate when dealing with organisms that have irregular body shapes.

4
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What is metabolic rate, and how is it commonly measured?

Metabolic rate is the amount of energy used up by an organism within a given period of time. It is often measured by oxygen uptake, as oxygen is used in aerobic respiration to release energy and produce ATP.

5
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What is the relationship between SA:V ratio and metabolic rate in smaller organisms?

As SA:V increases in smaller organisms, metabolic rate increases because the rate of heat loss per unit body mass increases. Higher respiration rates are required to release enough heat to maintain a constant body temperature.

6
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What structural adaptations facilitate exchange in larger organisms as SA:V reduces?

  1. Changes to body shape (e.g., long or thin) to increase SA:V and reduce diffusion pathways.
  2. Development of specialized exchange organs (e.g., lungs) that increase internal SA:V, shorten diffusion distance, and maintain concentration gradients via ventilation and blood supply.
7
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How is the body surface of a single-celled organism adapted for gas exchange?

It features a thin, flat shape and a large surface area to volume ratio, providing a short diffusion distance to all parts of the cell for rapid diffusion of O2O_2 and CO2CO_2.

8
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What are the three main structural components of the insect tracheal system?

  1. Spiracles: Pores on the body surface that open and close to allow gas diffusion.
  2. Tracheae: Large tubes full of air that facilitate diffusion.
  3. Tracheoles: Smaller permeable branches extending from tracheae to facilitate gas exchange directly with cells.
9
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How does abdominal pumping in insects assist gas exchange?

Contraction of abdominal muscles changes pressure inside the insect's body, causing air to move in and out of the tracheal system, which maintains a concentration gradient for diffusion.

10
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How does anaerobic respiration during exercise alter gas exchange in insect tracheoles?

Lactate produced during anaerobic respiration lowers the water potential (1cell τ\frac{1}{\text{cell}}\text{ }\tau / water potential\text{water potential}) of tissue cells. Fluid at the end of tracheoles is drawn into tissues by osmosis, allowing faster gas diffusion through air directly to the gas exchange surface.

11
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What adaptations allow terrestrial insects to balance efficient gas exchange with limiting water loss?

  1. A thick waxy cuticle/exoskeleton to increase diffusion distance for evaporation.
  2. Spiracles that can close to reduce water loss.
  3. Hairs around spiracles to trap moist air and reduce the water potential gradient.
12
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How are fish gills structurally adapted for gas exchange?

Gills consist of many filaments covered with thin lamellae (providing a large surface area and short diffusion path) and contain a dense network of capillaries to maintain concentration gradients.

13
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How does counter current flow in fish gills maintain efficient gas exchange?

Blood and water flow in opposite directions across the lamellae. This ensures the oxygen concentration in water is always higher than in adjacent blood, maintaining an O2O_2 concentration gradient along the whole length of the lamellae.

14
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What structural compromises allow xerophytic plants to limit water loss?

Thicker waxy cuticles (increasing diffusion distance), sunken stomata in pits/rolled leaves/hairs (trapping water vapour to reduce the water potential gradient), and spines or needles (reducing the SA:V ratio).

15
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What essential features adapt the alveolar epithelium for gas exchange?

Flattened cell structure (1 cell thick) for short diffusion distance, folded walls for large surface area, high permeability to O2O_2 and CO2CO_2, moist surfaces for gas dissolution, and a large capillary network maintaining concentration gradients.

16
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What muscle actions and pressure changes occur during human inspiration?

The diaphragm contracts and flattens; external intercostal muscles contract while internal intercostal muscles relax, pulling the ribcage up and out. This increases thoracic cavity volume, reduces pressure below atmospheric levels, and draws air into the lungs.

17
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Why is quiet expiration at rest considered a passive process?

Internal intercostal muscles do not normally need to contract during resting expiration; the process is driven primarily by the elastic recoil of the alveolar tissue.

18
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How does lung tissue fibrosis affect gas exchange and ventilation?

Fibrosis causes thickened alveolar tissue, which increases diffusion distance. It also reduces lung elasticity, leading to decreased tidal volume and forced vital capacity (FVC).

19
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Describe the digestion of starch in mammals.

Amylase (from salivary glands and pancreas) hydrolyses glycosidic bonds in starch to form maltose. Membrane-bound maltase on the ileum epithelium then hydrolyses maltose into glucose.

20
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What is the specific role of bile salts in lipid digestion?

Bile salts (produced by the liver) emulsify lipids into smaller lipid droplets. This increases the surface area of lipids available for lipase action, speeding up lipid hydrolysis.

21
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Distinguish between the actions of endopeptidases, exopeptidases, and dipeptidases in protein digestion.

Endopeptidases hydrolyse internal peptide bonds within polypeptides to produce smaller peptides. Exopeptidases hydrolyse terminal peptide bonds at the ends of chains to release single amino acids. Membrane-bound dipeptidases hydrolyse peptide bonds between dipeptides to produce two amino acids.

22
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How are monosaccharides and amino acids absorbed into ileum epithelial cells via co-transport?

  1. Na+Na^+ is actively transported out of epithelial cells into blood by the Na+/K+Na^+/K^+ pump, maintaining a lower Na+Na^+ concentration inside the cell.
  2. Na+Na^+ diffuses into the cell down its gradient via a co-transporter protein, bringing glucose or amino acids into the cell against their concentration gradient.
  3. Monosaccharides/amino acids pass into the blood by facilitated diffusion down their concentration gradient.
23
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What is the function of micelles in lipid absorption?

Micelles formed from bile salts, monoglycerides, and fatty acids make lipids soluble in water, transport fatty acids and monoglycerides to the ileum cell surface, and maintain a high local concentration gradient for simple diffusion into epithelial cells.

24
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Describe the quaternary structure of haemoglobin and its oxygen carrying capacity.

Haemoglobin is a globular protein made of 4 polypeptide chains. Each chain contains a Haem group with an iron ion (Fe2+Fe^{2+}). Each haemoglobin molecule can bind up to 4 O2O_2 molecules forming oxyhaemoglobin.

25
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What accounts for the S-shaped (sigmoid) oxyhaemoglobin dissociation curve?

Cooperative binding: the binding of the first oxygen molecule alters the tertiary/quaternary structure of haemoglobin, uncovering remaining Haem binding sites and making subsequent oxygen binding easier.

26
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What is the Bohr effect and how does increased CO2CO_2 concentration alter oxygen delivery?

Increased blood CO2CO_2 lowers pH, slightly altering haemoglobin's tertiary/quaternary structure and lowering its affinity for oxygen. The dissociation curve shifts right, causing oxygen to unload more readily to respiring tissues at a given pO2pO_2.

27
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Why is a double circulatory system advantageous in mammals?

It prevents the mixing of oxygenated and deoxygenated blood (maximizing oxygen transport efficiency) and allows blood to be pumped to body tissues at higher pressure after leaving the lungs.

28
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Describe heart volume, pressure, and valve positions during ventricular systole.

Ventricles contract, decreasing volume and increasing pressure. Atrioventricular (AV) valves shut when ventricular pressure exceeds atrial pressure. Semilunar (SL) valves open when ventricular pressure exceeds arterial pressure, allowing blood to exit into arteries.

29
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What is the formula used to calculate cardiac output?

Cardiac output=stroke volume×heart rate\text{Cardiac output} = \text{stroke volume} \times \text{heart rate}

30
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How does the structural composition of arteries relate to high-pressure blood transport?

They contain thick smooth muscle layers to control flow, thick elastic tissue to stretch and recoil (smoothing pressure surges), thick overall walls to prevent bursting, and narrow lumens to maintain elevated pressure.

31
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How is tissue fluid formed at the arteriole end of capillaries?

High hydrostatic pressure inside capillaries (generated by ventricular contraction) exceeds the hydrostatic pressure of tissue fluid, forcing water and dissolved small molecules out through capillary walls while large plasma proteins remain.

32
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How is tissue fluid reabsorbed into capillaries at the venule end?

Hydrostatic pressure inside the capillary drops while the concentration of remaining plasma proteins lowers capillary water potential. Water moves back into capillaries from tissue fluid by osmosis down a water potential gradient.

33
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What precautions should be observed when carrying out a biological specimen dissection?

Cover open cuts with waterproof dressings, cut away from the body on a hard surface using a sharp scalpel, carry scalpels blade down, wear disposable gloves, disinfect work surfaces and hands, and dispose of biological waste in dedicated bins.

34
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Explain the cohesion-tension theory of water transport through xylem.

  1. Water evaporates from leaf mesophyll into air spaces and diffuses out through stomata (transpiration), lowering mesophyll water potential.
  2. Water is drawn from xylem into mesophyll cells.
  3. This creates tension ('pull') in the xylem column.
  4. Hydrogen bonding between water molecules creates cohesion, pulling water upwards as an unbroken column while water adheres to xylem walls.
35
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How is transpiration rate calculated from potometer data?

Calculate volume of water uptake (Volume=cross-sectional area of capillary tube (227r2 or 11×radius2×pi)×distance moved by bubble\text{Volume} = \text{cross-sectional area of capillary tube } (\frac{22}{7} r^2 \text{ or } \frac{1}{1} \times \text{radius}^2 \times \text{pi}) \times \text{distance moved by bubble}) and divide the volume by the time taken.

36
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What is translocation, and how does the mass flow hypothesis describe solute transport in phloem?

Translocation is the movement of solutes (like sucrose) from source to sink. Companion cells actively load sucrose into phloem sieve tubes at the source, lowering water potential so water enters from xylem by osmosis. This creates high hydrostatic pressure, forcing mass flow toward the sink where sucrose is unloaded.

37
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How do ringing experiments provide evidence for translocation in phloem?

Removing a ring of outer bark (including phloem) causes a bulge of fluid containing high sugar concentrations to accumulate on the source side of the ring. Tissues below the ring die due to lack of organic solutes, confirming phloem carries organic nutrients down.