Unit 3 - Exchange of Substances Flashcards
(49 cards)
1
Q
What is the relationship between SA:V ratio and metabolic rate for a small organism.
(3 marks)
A
- Smaller organism so larger surface area to volume ratio
- More heat loss per gram of weight
- Faster rate of respiration to release more heat
2
Q
Explain the advantage for larger animals of having a specialised system that facilitates oxygen uptake.
(2 marks)
A
- Larger organisms have a smaller surface area to volume ratio
- System is needed to overcome the long diffusion pathway
3
Q
Explain why stomata open due to an increase in light intensity.
(1 mark)
A
Allows carbon dioxide to enter for photosynthesis
4
Q
Describe how carbon dioxide in the air outside a leaf reaches the mesophyll cells inside the leaf.
(4 marks)
A
- Carbon dioxide enters via stomata
- Stomata are opened by guard cells
- The carbon dioxide diffuses through air spaces
- Down a diffusion gradient
5
Q
Describe and explain an advantage and disadvantage to having a higher stomatal density.
(2 marks)
A
- One strength is more carbon dioxide intake
- One weakness is more water loss
6
Q
What are the adaptations to desert plants that help avoid water loss.
(6 marks)
A
- Hairs around stomata that trap water vapor so the water potential gradient is decreased
- Stomata are in pits that trap water vapour also decreasing the water potential gradient
- Thick waxy layer that increases diffusion distance to reduce transpiratipm
- Curled leaves that trap water vapour so the water potential gradient is decreased
- Spines to increase surface area
7
Q
Describe the counter-current mechanism in fish.
(3 marks)
A
- Water and blood flow in opposite directions
- Blood always meets water with a higher oxygen concentration
- Concentration gradient is maintained along the whole length of the lamella
8
Q
Explain two ways in which the structure of fish gills is adapted for sufficient gas exchange.
(2 marks)
A
- Many lamellae and gill filaments so large surface area
- Thin surface so short diffusion pathway
9
Q
Describe and explain how the structure of the insect gas exchange system provides cells with sufficient oxygen and limits water loss.
(7 marks)
A
- Spiracles lead to tracheae that lead to tracheoles
- Open spiracles allow diffusion of oxygen from the air
- Tracheoles are highly branched so have a large surface area for exchange
- Tracheole walls are permeable to oxygen
- Chitin in tracheae impermeable to reduce water loss
- Spiracles can close to prevent water loss
10
Q
How does abdominal pumping aid gas exchange in insects?
(2 marks)
A
- Raises the pressure in the body
- Carbon dioxide is pushed out of the body as it moves down the pressure gradient to the atmosphere
11
Q
Describe and explain one feature of the alveolar epithelium that makes the epithelium wall adapted as a surface for gas exchange.
(6 marks)
A
- Cells are one cell thick
- Reducing the diffusion pathway
- Cells are permeable
- Allows diffusion of oxygen/carbon dioxide
- Cells are moist
- Increases the rate of diffusion
12
Q
Describe how the lungs breathe in and out.
(8 marks)
A
(In)
- The diaphragm contracts so it flattens
-The external intercostal muscles contract so the ribs move upwards and outwards
- The volume of the thoracic cavity increases so pressure decreases
- Air moves into the lungs down a pressure gradient
(Out)
- The diaphragm relaxes so it becomes dome shaped
- The internal intercostal muscles contract so the ribs move downwards and inwards
- The volume of the thoracic cavity decreases so pressure increases
- Air moves out of the lungs down a pressure gradient
13
Q
Describe the pathway taken by an oxygen molecule from an alveolus to the blood.
(2 marks)
A
- Across the alveolar epithelium
- Into the capillary endothelium
14
Q
Describe the structure of the human gas exchange system.
(1 mark)
A
- Trachea, bronchi, bronchioles, alvioli
15
Q
Describe the digestion of proteins.
(4 marks)
A
- Hydrolysis of peptide bonds
- Endopeptidases hydrolyse internal bonds to break polypeptides into smaller chains
- Exopeptidases hydrolyses the terminal bond of the chain to form a dipeptide and an amino acid
- Dipeptidases hydrolyse dipeptides to form two amino acids
16
Q
Compare endopeptidase and exopeptidase.
(2 marks)
A
- Endopeptidases hydrolyse internal peptide bonds
- Eeopeptidases hydrolyse the terminal bond
17
Q
Describe the action of membrane-bound dipeptidases and explain their importance.
(2 marks)
A
- Hydrolyses the peptide bond to release amino acids
- So that amino acids can can cross the ilium cell membrane by facilitated diffusion
18
Q
Describe the complete digestion of starch by a mammal.
(5 marks)
A
- Hydrolysis
- Of glycosidic bonds
- Starch is converted into maltose by amylase
- Maltose is converted into glucose by membrane-bound maltase
19
Q
Describe the function of bile salts and micelles.
(3 marks)
A
- Bile salts emulsify lipids to form smaller droplets which increase the surface area for lipase
- So faster hydrolysis of lipids
- Micelles carry fatty acids and glycerol to the illium cell membrane
20
Q
Describe lipid digestion.
(3 marks)
A
- Lipase hydrolyses triglycerides
- Ester bonds break
- Forms monoglycerides and fatty acids
21
Q
Explain the advantages of emmulsification and micelle formation.
(4 marks)
A
- Droplets increase the surface area for lipase action
- So triglycerides are digested faster
- Micelles make the triglycerides more water-soluble in the small intestine
- And carry the fatty acids and glycerol through the membrane
22
Q
Describe and explain the features you would expect to find in a cell specialised for absorption.
(4 marks)
A
- Folded membrane so a large surface area
- Large number of carrier proteins so fast rate of absorption
- Large number of mitochondria to produce more ATP for active transport
- Membrane-bound digestive enzymes to maintain the concentration gradient
23
Q
Describe the processes involved in the absorption and transport of digested lipid molecules from the ilium into lymph vessels.
(4 marks)
A
- Micelles contain bile salts and fatty acids
- Make the fatty acids and monoglycerides more soluble in water
- Fatty acids and monoglycerides are absorbed by simple diffusion
- Triglycerides are reformed in cells and packaged into vesicles
- Vesicles move to the cell membrane
24
Q
How is the golgi apparatus involved in the absorption of lipids.
(3 marks)
A
- Modifies tryglicerides
- Combines triglycerides with proteins to form chylomicrons
- Chylomicrons are packaged in vesicles
25
Explain how monosaccharides and amino acids are absorbed into the blood.
(5 marks)
- Some glucose is absorbed by facilitated diffusion when concentration in the lumen is higher
- Sodium ions are constantly actively transported from the ilium cell to the blood
- This maintains the concentration gradient for sodium to constantly enter the the ilium cell via facilitated diffusion
- Sodium ions enter the cell by facilitated diffusion and bring with it a molecule of glucose by co-transport
- Glucose leaves the ilium cell and enters the blood via facilitated diffusion
26
Binding of one molecule of oxygen to haemoglobin makes it easier for a second oxygen molecule to bind, explain why.
(2 marks)
- Binding of the first oxygen changes the tertiary structure of haemoglobin
- This uncovers another binding site
27
Explain how changes in the shape of haemoglobin result in the S-shaped oxyhaemoglobin disassociation curve.
(2 marks)
- The first oxygen binds causing a change in the tertiary structure of haemoglobin
- The change in shape allows more oxygen to bind easily
28
Describe the advantage of the Bohr effect during intense exercise.
(2 marks)
- Increases dissociation of oxygen
- For more aerobic respiration at the respiring tissues
29
Describe and explain the effect of increasing carbon dioxide concentration concentration on the dissociation of oxyhaemoglobin.
(2 marks)
- Increases dissociation of oxygen
- By decreasing the pH of the blood
30
Explain why the oxygen-haemoglobin curve shifts left when diving.
(2 marks)
- Haemoglobin has a higher affinity for oxygen
- Allows aerobic respiration when at a lower partial pressure of oxygen (underwater)
31
Explain why the oxygen-haemoglobin dissociation curve of animals living at high altitudes (low oxygen) shift to the left.
(3 marks)
- High altitudes have a low partial pressure of oxygen
- Haemoglobin has a higher affinity for oxygen
- So enough oxygen can be unloaded at the respiring tissues
32
Explain why the oxygen-haemoglobin dissociation curve of mice shifts to the right.
(3 marks)
- Mouse haemoglobin has a lower affinity for oxygen
- So more oxygen can be dissociated at the respiring tissues
- As mice have a quicker respiration rate
33
Explain why the oxygen-haemoglobin dissociation curve of more active animals is shifted to the right.
(4 marks)
- Curve to the right so haemoglobin has a lower affinity for oxygen
- Haemoglobin dissociates more readily
- More oxygen is provided to respiring tissues
- For a greater rate of respiration
34
What are the control measures the students must use to reduce the risks associated with carrying and using a scalpel.
(5 marks)
- Carry with a blade protected
- Always cut away from the body
- Always cut onto a hard surface
- Use a sharp blade
- Dispose of the used scalpel
35
Give the pathway a red blood cell takes when travelling in the human circulatory system from a kidney to the lungs.
(3 marks)
- Through the renal vein
- Enters heart via the vena cava into the right atrium
- Leaves the heart from the right ventricle into the pulmonary artery
36
Outline the calculation for cardiac output.
(1 mark)
Cardiac output = stroke volume x heart rate
37
What causes the semi-lunar valve to close?
(1 mark)
Because pressure in the aorta is higher than in the ventricle
38
Explain how the atrioventricular valve is closed.
(2 marks)
- Ventricle contracts and volume decreases
- Pressure in the ventricle increases higher than the pressure in the left atrium
39
Describe the structure and function of an artery.
(5 marks)
- Elastic tissue to allow stretching and recoil to smooth out blood flow and maintain pressure
- Elastic tissue stretches when the ventricles contract due to a higher volume of blood
- Muscular walls for vasoconstriction
- Thick walls withstand the high pressure
- Smooth endothelial cells reduce friction
40
Explain six ways in which the structure of the aorta is related to it's function.
(6 marks)
- Elastic tissue to allow stretching/recoil to smooth out the flow of blood
- Elastic tissue stretches when the ventricle contracts
- Muscular layer for vasoconstriction
- Thick wall withstands pressure
- Smooth endothelial cells reduce friction
- Semi-lunar valve prevents backflow
41
Describe the fish circulatory system.
(3 marks)
- Single circulatory system
- Only 2 chambers
- Only one vein and one artery
42
Explain the formation of tissue fluid and how it is returned to the circulatory system.
(6 marks)
- Contraction of the ventricle causes high hydrostatic pressure at the arteriole end
- Water and dissolved molecules are forced out of the capillary
- Plasma proteins remain in the blood
- Which lowers the water potential
- Water moves back into the blood via osmosis
- Excess fluid is returned by the lymphatic system
43
Explain how diet high in salt can lead to a build-up of tissue fluid.
(2 marks)
- Higher salt content lowers the water potential of the tissue fluid
- Less water can return to the capillary by osmosis
44
Explain how high blood pressure leads to an accumulation of tissue fluid.
(3 marks)
- High blood pressure causes high hydrostatic pressure
- Increases the arterial end pressure
- So more tissue fluid is formed and less can be reabsorbed
45
Describe the cohesion-tension theory of water transport in the xylem.
(5 marks)
- Water is evaporated from the stomata
- This lowers the water potential of the leaf cells
- Water is pulled up the xylem, creating tension
- Water molecules cohere with hydrogen bonds
- Adhesion of water molecules to the xylem walls
46
Give two reasons why the potometer does not truly measure the rate of transpiration.
(2 marks)
- Some water is used in photosynthesis
- The apperatus is not fully sealed
47
Give three precautions the students should have taken when setting up the potometer to obtain reliable measurements of water uptake by the plant shoot.
(3 marks)
- Cut the shoot underwater
- Dry off leaves
- Seal joints with vaseline
48
What are the processes involved in the transport of sugars in plant stems?
(6 marks)
- At the source sucrose is actively transported into the phloem
- By companion cells
- This lowers the water potential in the phloem phloem so water enters from the xylem via osmosis
- This produces a high hydrostatic pressure a the top of the phloem
- Mass flow down the pressure gradient towards the sink
- At the sink sugars are removed and stored
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