[3.3] Organisms Exchange Substances with their Environment

Question 1

Describe the relationship between the size and structure of an organism and its surface area to volume ratio (SA:V).

Answer 1

• As size increases, SA:V tends to decrease.
• More thin/flat/folded/elongated structures increase SA:V.

Question 2

How is SA:V calculated? Give an example that illustrates how to calculate SA:V in cubes.

Answer 2

Divide surface area by volume

For example, in cubes:

SA = side length x side width x number of sides
V = length x width x depth

SA:V = SA / V

Question 3

Suggest an advantage of calculating SA:mass for organisms instead of SA:V?

Answer 3

Easier/quicker to find and more accurate because irregular shapes.

Question 4

What is metabolic rate? Suggest how it can be measured.

Answer 4

• Metabolic rate = amount of energy used up by an organism within a given period of time.
• Often measured by oxygen uptake as used in aerobic respiration to make ATP for energy release.

Question 5

Explain the relationship between SA:V and metabolic rate.

Answer 5

As SA:V increases (smaller organisms), metabolic rate increases because:

• Rate of heat loss per unit body mass increases.
• So organisms need a higher rate of respiration.
• To release enough heat to maintain a constant body temperature.

Question 6

Explain the adaptations that facilitate exchange as SA:V reduces in larger organisms.

Answer 6

1. Changes to body shape e.g. long/thin.
   • Increases SA:V and overcomes long diffusion distance/pathway.
2. Development of systems such as specialised surface/organ for gaseous exchange e.g. lungs.
   • Increases internal SA:V and overcomes long diffusion distance/pathway.
   • Maintains a concentration gradient for diffusion by ventilation/good blood supply.

Question 7

Explain how the body surface of a single-celled organism is adapted for gas exchange.

Answer 7

• Thin, flat shape and large surface area to volume ratio.
• Short diffusion distance to all parts of the cell so rapid diffusion of O₂ and CO₂.

Question 8

Describe the tracheal system of an insect.

Answer 8

1. Spiracles = pores on surface that can open/close to allow diffusion.
2. Tracheae = large tubes full of air that allow diffusion.
3. Tracheoles = smaller branches from tracheae, permeable to allow gas exchange with cells.

Question 9

Explain how an insect’s tracheal system is adapted for gas exchange.

Answer 9

1. Tracheoles have thin walls.
   • Short diffusion distance to cells.
2. High numbers of highly branched tracheoles.
   • Short diffusion distance to cells and large surface area.
3. Tracheae provide tubes full of air.
   • Fast diffusion.
4. Contraction of abdominal muscles changes pressure in body causing air to move in and out.
   • Maintains concentration gradient for diffusion.
5. Fluid in end of tracheoles drawn into tissues by osmosis during exercise due to the lactate produced in aerobic respiration which lowers water potential.
   • Diffusion is faster through air than fluid to gas exchange surface.

Question 10

Explain structural and functional compromises in terrestrial insects that allow efficient gas exchange while limiting water loss

Answer 10

1. Thick waxy cuticle/exoskeleton.
   • Increases diffusion distance so less water loss.
2. Spiracles can open to allow gas exchange AND close to reduce water loss.
3. Hairs around spiracles trap moist air.
   • Reduces water potential gradient so less water loss.

Question 11

Explain how the gills of fish are adapted for gas exchange.

Answer 11

1. Gills made of many filaments covered with many lamellae.
   • Increase surface area for diffusion.
2. Thin lamellae wall/epithelium.
   • So short diffusion distance between water and blood.
3. Lamellae have a large number of capillaries.
   • Remove O₂ and bring CO₂ quickly so maintains a concentration gradient.

COUNTER CURRENT FLOW

1. Blood and water flow in opposite directions through/over lamellae.
2. So oxygen concentration always higher in water than blood.
3. So maintains a concentration gradient of O₂ between water and blood.
4. For diffusion along whole length of lamellae.

(If parallel flow, equilibrium would be reached so oxygen wouldn’t diffuse into blood along the whole length of the lamellae)

Question 12

Explain how the leaves of dicotyledonous plants are adapted for gas exchange.

Answer 12

1. Many stomata.
   • Large surface area for gas exchange when opened by guard cells.
2. Spongy mesophyll contains air spaces.
   • Large surface area for gases to diffuse through.
3. Thin.
   • Short diffusion distance.

Question 13

Explain structural and functional compromises in xerophytic plants that allow efficient gas exchange while limiting water loss.

Answer 13

Xerophyte = plants adapted to live in very dry conditions.

1. Thicker waxy cuticle.
   • Increases diffusion distance so less evaporation.
2. Sunken stomata in pits + rolled leaves + hairs
   • Trap water vapour/protect stomata from wind so reduce water potential gradient between leaf and air so less evaporation.
3. Spines/needles.
   • Reduce surface area to volume ratio.

Question 14

Describe the gross structure of the human gas exchange system.

Answer 14

Question 15

Explain the essential features of the alveolar epithelium that make it adapted as a surface for gas exchange.

Answer 15

1. Flattened cells/1 cell thick.
   • Short diffusion distance.
2. Folded.
   • Large surface area.
3. Permeable.
   • Allows diffusion of O₂ and CO₂.
4. Moist.
   • Gases can dissolve for diffusion.
5. Good blood supply from large network of capillaries.
   • Maintains concentration gradient.

Question 16

Describe how gas exchange occurs in the lungs.

Answer 16

1. Oxygen diffuses from alveolar air space into blood down its concentration gradient.
2. Across alveolar epithelium and then across capillary endothelium.

(Opposite process occurs for carbon dioxide)

Question 17

Explain the importance of ventilation.

Answer 17

Brings in air containing higher concentration of oxygen and removes air with a lower concentration of oxygen so maintains concentration gradients.

Question 18

Explain how humans breathe in and out (ventilation).

Answer 18

INSPIRATION (BREATHING IN)

1. Diaphragm muscles contract and flatten.
2. External intercostal muscles contract, internal intercostal muscles relax (antagonistic).
3. Ribcage pulled up and out.
4. Increasing volume and decreasing pressure below atmospheric in thoracic cavity.
5. Air moves into lungs down pressure gradient.

EXPIRATION (BREATHING OUT)

1. Diaphragm relaxes and moves upwards.
2. External intercostal muscles relax and internal intercostal may contract.
3. Ribcage moves down and in.
4. Decreasing volume and **increasing pressure* above atmospheric in thoracic cavity.
5. Air moves out of lungs down pressure gradient.

Question 19

Suggest why expiration is normally passive at rest.

Answer 19

• Internal intercostal muscles do not normally need to contract.
• Expiration aided by elastic recoil in alveoli.

Question 20

Suggest how different lung diseases reduce the rate of gas exchange.

Answer 20

• Thickened alveolar tissue increases diffusion distance.
• Alveolar wall breakdown reduces surface area.
• Reduced lung elasticity which means lungs expand/recoil less which reduces concentration gradients of O₂ and CO₂.

Question 21

Suggest how different lung diseases affect ventilation.

Answer 21

1. Reduce lung elasticity which means lungs expand/recoil less.
   • Reducing volume of air in each breath (tidal volume).
   • Reducing maximum volume of air breathed out in one breath (forced vital capacity).
2. Narrow airways which reduce airflow in and out of lungs.
   • Reducing maximum volume of air breathed out in 1 second (forced expiratory volume).
3. Reduce rate of gas exchange.
   • Increased ventilation rate to compensate for reduced oxygen in the blood.

Question 22

Suggest why people with lung disease experience fatigue.

Answer 22

Cells receive less oxygen so rate of aerobic respiration reduced so less ATP made.

Question 23

Suggest how you can analyse and interpret data to the effects of pollution, smoking and other risk factors on the incidence of lung disease.

Answer 23

1. Describe overall trend - positive/negative correlation between risk factor and incidence of disease.
2. Manipulate data - e.g. calculate percentage change.
3. Interpret standard deviations - **overlap* suggests differences in means are likely to be due to chance.
4. Use statistical tests to identify whether difference/correlation is significant or due to chance.
   • Correlation coefficient - examining an association between 2 sets of data.
   • Student’s t-test - comparing means of 2 sets of data.
   • Chi-squared test - for categorical data.

Question 24

Suggest how you can evaluate the way in which experimental data led to statutory restrictions on the sources of risk factors.

Answer 24

1. Analyse and interpret data and identify what does and doesn’t support the data.
2. Evaluate method of collecting data:
   • Sample size - large enough to be representative of population?
   • Participant diversity - representative of population?
   • Control groups - used to enable comparison?
   • Control variables - valid?
   • Duration of study - long enough to show long-term effects?
3. Evaluate context - has a broad generalisation been made from a specific set of data?
4. Other risk factors that could have affected results.

Question 25

Explain the difference between correlations and causal relationships.

Answer 25

• Correlation = change in one variable reflected by change in another.
• Causation = change in one variable causes a change in another variable.
• Correlation does not mean causation - may be other factors involved.

Question 26

Explain what happens during digestion.

Answer 26

Large biological molecules are hydrolysed to smaller molecules that can be absorbed across cell membranes.

Question 27

Describe the digestion of starch in mammals.

Answer 27

1. Amylase hydrolyses starch to maltose.
2. Membrane-bound maltase hydrolysis maltose to glucose.
3. Hydrolysis of glycosidic bond.

Question 28

Describe the digestion of disaccharides in mammals.

Answer 28

1. Membrane-bound disaccharides hydrolyse disaccharides to 2 monosaccharides.
   • Maltase - maltose –> glucose + glucose
   • Sucrase - sucrose –> fructose + glucose
   • Lactase - lactose –> galactose + glucose.
2. Hydrolysis of glycosidic bond.

Question 29

Describe the digestion of lipids in mammals, including the action of bile salts.

Answer 29

1. Bile salts emulsify lipids causing them to form smaller lipid droplets.
   • This increases surface area of lipids for increased/faster lipase activity.
2. Lipase hydrolyses lipids into monoglycerides and fatty acids.
3. Hydrolysis of ester bond.

Question 30

Describe the digestion of proteins by a mammal.

Answer 30

1. Endopeptidases hydrolyse internal peptide bonds within a polypeptide forming smaller peptides.
   • So more ends/surface area for exopeptidases.
2. Exopeptidases hydrolyse terminal peptide bonds at ends of polypeptides forming single amino acids.
3. Membrane-bound dipeptidases hydrolyse peptide bond between a dipeptide forming 2 amino acids.
4. Hydrolysis of peptide bond.

Question 31

Suggest why membrane-bound enzymes are important in digestion.

Answer 31

• Membrane-bound enzymes are located on cell membranes of epithelial cells lining ileum.
• By hydrolysing molecules at the site of absorption they maintain concentration gradients for absorption.

Question 32

Describe the pathway for absorption of products of digestion in mammals.

Answer 32

Lumen of ileum -> cells lining ileum -> blood

Question 33

Describe the absorption of amino acids and monosaccharides in mammals.

Answer 33

CO-TRANSPORT

1. Na⁺ actively transported from epithelial cells to blood by Na⁺/K⁺ pump establishing a concentration gradient of Na⁺ (higher in lumen than epithelial cell).
2. Na⁺ enters epithelial cell down its concentration gradient with
   monosaccharides/amino acids against its concentration gradient via a co-transporter protein.
3. Monosaccharides/amino acids moves down a concentration gradient into blood via facilitated diffusion.

Question 34

Describe the absorption of lipids by a mammal, including the role of micelles.

Answer 34

1. Micelles contain bile salts, monoglycerides and fatty acids.
   • Make monoglycerides and fatty acids more soluble in water.
   • Carry/release fatty acids and monoglycerides to cell/lining of ileum.
   • Maintain high concentration of fatty acids to cell/lining.
2. Monoglycerides/fatty acids absorbed into epithelial cell by diffusion as they’re lipid soluble.
3. Triglycerides reformed in epithelial cells and aggregate into globules.
4. Globules coated with proteins form chylomicrons which are then packaged into vesicles.
5. Vesicles move to cell membrane and leave via exocytosis.
   • Enter lymphatic vessels and eventually return to blood circulation.

Question 35

Describe the role of red blood cells and haemoglobin in oxygen transport.

Answer 35

1. Red blood cells contain lots of haemoglobin - no nucleus, bioconcave, high SA:V, short diffusion path.
2. Haemoglobin associates with/blinds/loads O₂ at gas exchange surfaces where partial pressure of O₂ is high.
3. This forms oxyhaemoglobin which transports O₂ (each RBC can carry 4O₂ - one at each haem group).
4. Haemoglobin dissociates from/unloads O₂ near cells/tissues where partial pressure of O₂ is low.

Question 36

Describe the structure of haemoglobin.

Answer 36

• Protein with a quaternary structure.
• Made of 4 polypeptide chains.
• Each chain contains a haem group containing an iron ion (Fe²⁺).

(Haemoglobins are a group of chemically similar molecules found in many different organisms).

Question 37

Describe the loading, transport and unloading of oxygen in relation to the oxyhaemoglobin dissociation curve.

Answer 37

AREAS WITH LOW pO₂ (RESPIRING TISSUES):

1. Haemoglobin has a low affinity for O₂.
2. So O₂ readily unloads/dissociates with haemoglobin.
3. So % saturation of haemoglobin with oxygen is low.

AREAS WITH HIGH pO₂ (GAS EXCHANGE SURFACES):

1. Haemoglobin has a high affinity for O₂.
2. So O₂ readily loads/associates with haemoglobin.
3. So % saturation of haemoglobin with oxygen is high.

Question 38

Explain how the cooperative nature of oxygen binding results in an s-shaped (sigmoid) oxyhaemoglobin dissociation curve.

Answer 38

1. Binding of the first oxygen changes tertiary/quaternary structure of haemoglobin.
2. This uncovers haem group binding sites, making further binding of oxygen easier.

Question 39

Describe evidence for the cooperative nature of oxygen binding.

Answer 39

1. A low pO₂ as oxygen increases there is little/slow increase in % saturation of haemoglobin with oxygen.
   • When first oxygen is binding.
2. At higher pO₂ as oxygen increases, there is a big/rapid increase in % saturation of haemoglobin with oxygen.
   • Showing it has got easier for oxygens to bind due to changes in haemoglobin’s structure.

Question 40

What is the Bohr effect?

Answer 40

Effect of CO₂ concentration on dissociation of oxyhaemoglobin - curve shifts to right.

Question 41

Explain the effect of CO₂ concentration on the dissociation of oxyhaemoglobin.

Answer 41

1. Increasing blood CO₂ due to increased rate of respiration.
2. Lowers blood pH - more acidic.
3. Reducing haemoglobins affinity for oxygen as shape/tertiary/quaternary structure changes slightly.
4. So more/faster unloading of oxygen to respiring cells at a given pO₂.

(Curve shows this because at a given pO₂ % saturation of haemoglobin with oxygen is lower)

Question 42

Explain the advantages of the Bohr effect (e.g. during exercise).

Answer 42

More dissociation of oxygen -> faster aerobic respiration -> more ATP produced.

Question 43

Explain why different types of haemoglobin can have different oxygen transport properties.

Answer 43

• Different types of haemoglobin are made of polypeptide chains with slightly different amino acid sequences - different primary structures.
• Resulting in different tertiary/quaternary structures/shape so different affinities for oxygen.

Question 44

Explain how organisms can be adapted to their environment by having different types of haemoglobin.

Answer 44

CURVE SHIFT LEFT -> HAEMOGLOBIN HAS HIGHER AFFINITY FOR O₂

1. More O₂ associates with haemoglobin more readily.
2. At gas exchange surfaces where pO₂ is lower.
   • e.g. organisms in low O₂ environments - high altitudes, underground, foetuses.

CURVE SHIFTS RIGHT -> HAEMOGLOBIN HAS LOWER AFFINITY FOR O₂

1. More O₂ dissociates from haemoglobin more readily.
2. At respiring tissues where more O₂ is needed.
   • e.g. organisms with high rates of respiration/metabolic rate (may be small or active)

Question 45

Describe the general pattern of blood circulation in a mammal.

Answer 45

1. Deoxygenated blood in right side of heart pumped to lungs; oxygenated returns to left side.
2. Oxygenated blood in left side of heart pumped to rest of body; deoxygenated returns to right.

Question 46

Suggest the importance of a double circulatory system.

Answer 46

Double circulatory system = blood passes through heart twice for every circuit around body.

• Prevents mixing of oxygenated/deoxygenated blood.
  
  • So blood pumped to body is fully saturated with oxygen for aerobic respiration.
• Blood can be pumped to body at a higher pressure
  
  • Substances taken to/removed from body cells quicker/more efficiently.

Question 47

Draw a diagram to show the general pattern of blood circulation in a mammal, including the names of key blood vessels.

Answer 47

Question 48

Name the blood vessels entering and leaving the heart and lungs.

Answer 48

ENTERING

• Vena cava - transports deoxygenated blood from respiring body tissues to heart.
• Pulmonary artery - transports deoxygenated blood from heart to lungs.

LEAVING

• Pulmonary vein - transports oxygenated blood from lungs to heart.
• Aorta - transports oxygenated blood from heart to respiring body tissues.

Question 49

Name the blood vessels entering and leaving the kidneys.

Answer 49

• Renal arteries - oxygenated blood to kidneys.
• Renal veins - deoxygenated blood to vena cava from kidneys.

Question 50

Name the blood vessels that carry oxygenated blood to the heart muscle.

Answer 50

Coronary arteries - located on surface of the heart, branching from aorta.

Question 51

Describe the gross structure of the human heart.

Answer 51

Question 52

Suggest why the wall of the left ventricle is thicker than that of the right.

Answer 52

• Thicker muscle to contract with greater force
• To generate higher pressure to pump blood around entire body

Question 53

Explain the pressure & volume changes and associated valve movements during the cardiac cycle that maintain a unidirectional flow of blood.

Answer 53

ATRIAL SYSTOLE

1. Atria contract.
   
   • Volume decreases, pressure increases.
2. Atrioventricular valves open when pressure in atria exceeds pressure in ventricles.
3. Semi-lunar valves remain shut as pressure in arteries exceeds pressure in ventricles.
4. So blood pushed into ventricles.

VENTRICULAR SYSTOLE

1. Ventricles contract.
   
   • Volume decreases, pressure increases.
2. Atrioventricular valves shut when pressure in ventricles exceeds pressure in atria.
3. Semi-lunar valves open when pressure in ventricles exceeds pressure in arteries.
4. So blood pushed out of heart through arteries.

DIASTOLE

1. Atria and venricles relax.
   
   • Volume increases, pressure decreases.
2. Semi-lunar valves shut when pressure in arteries exceeds pressure in ventricles.
3. Atrioventricular valves open when pressure in atria exceeds pressure in ventricles.
4. So blood fills atria via veins and flows passively to ventricles.

Question 54

Explain how this graph showing pressure changes during the cardiac cycle can be interpreted to identify when valves are open/closed.

Answer 54

SEMI-LUNAR VALVES CLOSED

• Pressure in aorta higher than in left ventricle
• To prevent backflow of blood from artery to ventricles

SEMI-LUNAR VALVES OPEN

• When pressure in left ventricle is higher than in aorta
• So blood flows from left ventricle to aorta

ATRIOVENTRICULAR VALVES CLOSED

• Pressure in left ventricle higher than left atrium
• To prevent backflow of blood from left ventricle to left atrium

ATRIOVENTRICULAR VALVES OPEN

• When pressure in left atrium is higher than in left ventricle
• So blood flows from left atrium to left ventricle

Question 55

Describe the equation for cardiac output.

Answer 55

Cardiac output = stroke volume x heart rate

• Cardiac output = volume of blood pumped out of heart per min
• Stroke volume = volume of blood pumped in each heartbeat
• Heart rate = number of beats per min

Question 56

How can heart rate be calculated from cardiac cycle data?

Answer 56

Heart rate = 60 (seconds) / length of one cardiac cycle (seconds)

Question 57

Explain how the structure of arteries relates to their function.

Answer 57

FUNCTION

• Carry blood away from heart at high pressure.

STRUCTURE

1. Thick smooth muscle tissue.
   
   • Can contract and control/maintain blood flow/pressure.
2. Thick elastic tissue.
   
   • Can stretch as ventricles contract and recoil as ventricles relax, to reduce pressure surges/even out blood pressure/maintain high pressure.
3. Thick wall.
   
   • Withstand high pressure/stop bursting.
4. Smooth/folded endothelium.
   
   • Reduces friction/can stretch
5. Narrow lumen.
   
   • Increases/maintains high pressure.

Question 58

Explain how the structure of arterioles relates to their function.

Answer 58

FUNCTION

• Divison of arteries to smaller vessels which can direct blood to different capillaries/tissues.

STRUCTURE

1. Thicker smooth muscle layer than arteries.
   
   • Contracts to narrow lumen (vasoconstriction) to reduce blood flow to capillaries.
   • Relaxes to widen lumen (vasodilation) to increase blood flow to capillaries.
2. Thinner elastic layer.
   
   • Pressure surges are lower as further from heart/ventricles.

Question 59

Explain how the structure of veins relates to their function.

Answer 59

FUNCTION

• Carry blood back to heart at lower pressure.

STRUCTURE

1. Wider lumen than arteries.
   
   • Less resistance to blood flow.
2. Very little elastic and muscle tissue.
   
   • Blood pressure lower.
3. Valves.
   
   • Prevent backflow of blood.

Question 60

Explain how the structure of capillaries relates to their function.

Answer 60

FUNCTION

• Allow efficient exchange of susbtances between blood and tissue fluid (exchange surface).

STRUCTURE

1. Wall is a thin, one cell layer of endothelial cells.
   
   • Reduces diffusion distance.
2. Capillary bed is a large network of branched capillaries.
   
   • Increases surface area for diffusion.
3. Small diameter/narrow lumen.
   
   • Reduces blood flow rate so more time for diffusion.
4. Pores in walls between cells.
   
   • Allow larger substances through.

Question 61

Explain the formation of tissue fluid.

Answer 61

AT THE ARTERIOLE END OF CAPILLARIES

1. Higher blood/hydrostatic pressure inside capillaries due to contraction of ventricles than tissue fluid so net outward force.
2. Forcing water and dissolved substances out of capillaries.
3. Large plasma proteins remain in capillaries.

Question 62

Explain the return of tissue fluid to the circulatory system.

Answer 62

AT THE VENULE END OF CAPILLARIES

1. Hydrostatic pressure reduces as fluid leaves capillary.
2. Due to water loss an increasing concentration of plasma proteins lowers water potential in capillary below that of tissue fluid.
3. Water enters capillaries from tissue fluid by omosis down water potential gradient.
4. Excess water taken up by lymph capillaries and returned to circulatory system through veins.

Question 63

Suggest and explain causes of excess tissue fluid accumulation.

Answer 63

1. Low concentration of protein in blood plasma.
   
   • Water potential in capillary not as low so water potential gradient is reduced.
   • So more tissue fluid formed at arteriole end/less water absorbed at venule end by osmosis.
2. High blood pressure -> high hydrostatic pressure.
   
   • Increases outward pressure from arterial end AND reduces inward pressure at venule end.
   • So more tissue fluid formed at arteriole end/less water absorbed at venule end by osmosis.
   • Lymph system may not be able to drain excess fast enough.

Question 64

What is a risk factor? Give examples for cardiovascular disease.

Answer 64

• An aspect of a person’s lifestyle or substance in a person’s body/environment that have been shown to be linked to an increased rate of disease.
• Examples: age, diet high in salt or saturated fat, smoking, lack of exercise, genes.

Question 65

Describe the function of xylem tissue.

Answer 65

Transports water and mineral ions through the stem, up the plant to leaves of plants.

Question 66

Suggest how xylem tissue is adapted for its function.

Answer 66

1. Cells joined with no end walls forming a long continuous tube
   
   • Water flows as a continuous column.
2. Cells contain no cytoplasm/nucleus
   
   • Easier water flow/no obstructions.
3. Thick cell walls with lignin.
   
   • Provides support/withstand tension/prevents water loss.
4. Pits in side walls.
   
   • Allow lateral water movements.

Question 67

Explain the cohesion-tension theory of water transport in the xylem.

Answer 67

LEAF

1. Water lost from leaf by transpiration
   
   • Water evaporates from mesophyll cells into air spaces and water vapour diffuses through stomata.
2. Reducing water potential of mesophyll cells.
3. So water drawn out of xylem down a water potential gradient.

XYLEM

4. Creating tension (‘negative pressure’ or ‘pull’) in xylem
5. Hydrogen bonds result in cohesion between water molecules (stick together) so water is pulled up as a continuous column.
6. Water also adheres (sticks to) to walls of xylem.

ROOT

7. Water enters roots via osmosis.

Question 68

Describe how to set up a potometer.

Answer 68

1. Cut a shoot underwater at a slant.
   
   • Prevents air entering xylem.
2. Assmeble potometer with capillary tube end submerged in a beaker of water.
3. Insert shoot underwater.
4. Dry leaves and allow time for shoot to acclimatise.
5. Shut tap to reservoir.
6. Form an air bubble.
   
   • Quickly remove end of capillary tube from water.

Question 69

Describe how a potometer can be used to measure the rate of transpiration.

Answer 69

Potometer estimates transpiration rate by measuring water uptake:

1. Record position of air bubble.
2. Record distance moved in a certain amount of time.
3. Calculate volume of water uptake in a given time:
   
   • Use radius of capillary tube to calculate cross-sectional area of water (πr²).
   • Multiply this by distance moved by bubble.
4. Calculate rate of water uptake.
   
   • Divide volume by time taken.

Question 70

Describe how a potometer can be used to investigate the effect of a named environmental variable on the rate of transpiration.

Answer 70

• Carry out a method to measure the rate of transpiration using a photometer, but change one variable (wind, humidity, light or temperature).
  
  • E.g set up a fan OR spray water in a plastic bag and wrap around the plant OR change distance of light source OR change temperature of room.
• Keep all other variables constant.

Question 71

Suggest limitations in using a potometer to measure rate of transpiration.

Answer 71

1. Rate of water uptake might not be same as rate of transpiration.
   
   • Water used for support/turgidity.
   • Water used in photosynthesis and produced during respiration.
2. Rate of movement through shoot in potometer may not be same as rate of movement through shoot of whole plant.
   
   • Shoot in photometer has no roots where as a plant does.
   • Xylem cells very narrow.

Question 72

Suggest how different environmental variables affect transpiration rate.

Answer 72

LIGHT INTESNSITY

• Increases rate of transpiration.
  
  • Stomata open in light to let in CO₂ for photosynthesis.
  • Allowing more water to evaporate faster.
  • Stomata close when its dark so there is a low transpiration rate.

TEMPERATURE

• Increases rate of transpiration.
  
  • Water molecules gain kinetic energy as temperature increases.
  • So water evaporates faster.

WIND INTENSITY

• Increases rate of transpiration.
  
  • Wind blows away water molecules from around stomata.
  • Decreasing water potential of air around stomata.
  • Increasing water potential gradient so water evaporates faster.

HUMIDITY

• Decreases rate of transpiration.
  
  • More water in air so it has a higher water potential.
  • Decreasing water potential gradient from leaf to air.
  • Water evaporates slower.

Question 73

Describe the function of phloem tissue.

Answer 73

Transports organic substances e.g. sucrose in plants.

Question 74

Suggest how phloem tissue is adapted for its function.

Answer 74

1. Sieve tube elements.
   
   • No nucleus/fewer organelles.
     
     • Maximise space for/easier flow of organic substances.
   • End walls between cells perforated (sieve plate).
2. Companion cells
   
   • Many mitochondria.
     
     • High rate of respiration to make ATP for active transport of solutes.

Question 75

What is translocation?

Answer 75

1. Movement of assimilates/solutes such as sucrose.
2. From source cells (where made, e.g. leaves) to sink cells (where used/stored, e.g. roots) by mass flow.

Question 76

Explain the mass flow hypothesis for translocation in plants.

Answer 76

1. At source, sucrose is actively transported into phloem sieve tubes/cells.
2. By companion cells.
3. This lowers water potential in sieve tubes so water enters from xylem by osmosis.
4. This increases hydrostatic pressure in sieve tubes at source/creates a hydrostatic pressure gradient.
5. So mass flow occurs.
   
   • Movement from source to sink.
6. At sink, sucrose is removed by active transport to be used by respiring cells or stored in storage organs.

Question 77

Describe the use of tracer experiments to investigate transport in plants.

Answer 77

1. Leaf supplied with a radioactive tracer e.g. CO₂ containing radioactive isotope ¹⁴C.
2. Radioactive carbon incorporated into organic substances during photosynthesis.
3. These move around plant by translocation.
4. Movement tracked using autoradiography or a Geiger counter.

Question 78

Describe the use of ringing experiments to investigate transport in plants.

Answer 78

1. Remove/kill phloem
   
   • Remove a ring of bark.
2. Bulge forms on source side of ring.
3. Fluid from bulge has higher concentration of sugars than below.
   
   • Shows sugar is transported in phloem.
4. Tissues below ring die as cannot get organic substances.

Question 79

Suggest some points to consider when interpreting evidence from tracer & ringing experiments and evaluating evidence for/against the mass flow hypothesis.

Answer 79

• Is there evidence to suggest the phloem (as opposed to the xylem) is involved?
• Is there evidence to suggest respiration/active transport is involved?
• Is there evidence to show movement is from source to sink? What are these in the experiment?
• Is there evidence to suggest movement is from high to low hydrostatic pressure?
• Could movement be due to another factor e.g. gravity?

Question 80

What is the equation for pulmonary ventilation rate (PVR)?

Answer 80

PVR = tidal volume x breathing rate