exchange2

Question 1

Describe process of Expiration.

Answer 1

• Largely passive process, requiring little energy.
  1. Internal intercostal muscles contract, while the external intercostal muscles relax.
  2. Ribs move downwards and inwards; decreasing volume of thorax.
  3. Diaphragm muscles relax, pushed up again (by abdomen contents that were compressed during inspiration) —> into dome shape —> decreasing volume of thorax.
  4. Decreased volume of thorax —> increased pressure in lungs.
  5. Pulmonary pressure > atmospheric pressure —> air is forced out of the lungs.

NB=> during normal, quiet breathing, the recoil of elastic tissue in the lungs is the main cause of air being forced out. Only under more strenuous conditions (exercise) do the various muscles play a major role.

Question 2

Pulmonary ventilation rate = ?

Answer 2

Pulmonary ventilation rate = tidal volume x breathing rate.

Question 3

Define Digestion.

Answer 3

= Process in which large biological molecules are hydrolysed to smaller molecules that can be absorbed across cell-surface membranes and assimilated.

Question 4

Draw and label structure of alimentary canal.

Answer 4

Include:

• Salivary glands.
• Oesophagus.
• Stomach.
• Ileum.
• Large intestine.
• Rectum.
• Anus.
• Pancreas.

Question 5

Describe salivary glands.

Answer 5

• Situated near mouth - pass the amylase enzyme in their secretions.

Question 6

Describe stomach.

Answer 6

• Muscular sac with an inner, enzyme-producing layer —> stores and digests food (especially proteins) and has glands that produce enzymes which digest proteins.

Question 7

Describe ileum.

Answer 7

• Long muscular tube —> food further digested in ileum by enzymes produced by its walls and by glands that pour secretions into it.
• Has an adapted structure for absorption.

Question 8

Describe large intestine’s function.

Answer 8

• Absorbs water, most of the water is absorbed is from secretions of the many digestive glands.

Question 9

Describe pancreas.

Answer 9

• Large gland situated below stomach —> produces secretions of pancreatic juice, containing proteases to hydrolyse proteins, lipase to hydrolyse lipids and amylase to hydrolyse starch.

Question 10

Outline basic process of digestion.

Answer 10

1. Physical Digestion:
   - Food (if large) broken down into smaller pieces by chewing - then ingested - and then by churning muscles in stomach wall.
   - Provides a large SA for chemical digestion.
2. Chemical Digestion:
   - All digestive enzymes function by hydrolysis —> several different enzymes all work together to hydrolyse a molecule.

Question 11

How are carbohydrates digested in mammals?

Answer 11

1. Saliva from salivary glands thoroughly mixed with food in mouth during chewing.
2. Salivary amylase starts hydrolysing (alternate) glycosidic bonds in starch —> maltose. Contains mineral salts to keep pH neutral (optimum pH).
3. Food ingested and then enters stomach. Amylase denatured, preventing further starch hydrolysis.
4. Food passed into small intestine after some time (churning) —> mixes with pancreatic juices.
5. Pancreatic amylase continues starch hydrolysis to maltose. Alkaline salts produced by pancreas/intestinal wall keep pH neutral —> amylase can still function.
6. Intestine wall muscles push food along ileum. Epithelial lining produces the membrane-bound disaccharidases maltase (part of epithelial cell membrane), sucrase and lactase.

Question 12

How are lipids digested?

Answer 12

1. Lipids hydrolysed by lipase enzymes, produced in the pancreas, that hydrolyse the ester bond found in triglycerides —> fatty acids and monoglycerides.
2. Lipids (fats + oils) firstly split up into tiny droplets called micelles by bile salts (produced by liver) —> emulsification —> increases SA of the lipids so speeds up the action of lipases.

Question 13

How are proteins digested?

Answer 13

• Large, complex molecules, hydrolyses by peptidases (proteases).
  1. Endopeptidases - hydrolyse peptide bonds between amino acids in the central region of a protein molecule —> forms a series of shorter peptide molecules.
  2. Exopeptidases - hydrolyse peptide bonds on the terminal amino acids of the peptide molecules formed by endopeptidases —> forms single amino acids and dipeptides.
  3. Dipeptidases - hydrolyse peptide bond between the two amino acids of dipeptide —> membrane-bound - part of epithelial cell-surface membranes.

Question 14

How is the ileum adapted for absorption of digestion products?

Answer 14

Villi:

1. Thin-walled —> short diffusion distance.
2. Contain muscle —> can move to mix with the ileum contents to help maintain favourable concentration gradients.
3. Many capillaries/rich network on other side of epithelial cells —> blood can carry away absorbed molecules —> maintains a favourable concentration gradient.
4. Many channel/carrier proteins.
5. Microvilli —> finger-like projections of the cell-surface membrane that further increase the SA for absorption.

Epithelial cells also have many mitochondria to provide ATP for active transport.

Question 15

How are amino acids and monosaccharides absorbed?

Answer 15

• Amino acids and monosaccharides are all absorbed by diffusion and co-transport.

Question 16

How are triglycerides absorbed?

Answer 16

1. Micelle structures formed —> monoglycerides remain in association with the bile salts that initially emulsified the lipid droplets.
2. Micelles come into contact with villi epithelial cells as they move through ileum. Micelles break down, releasing the monoglycerides and fatty acids —> non-polar molecules - easily diffuse across the cell-surface membrane into the epithelial cells.
3. Once inside epithelial cells, monoglycerides and fatty acids transported to the ER where they are recombined to form triglycerides.
4. Starting in ER —> golgi apparatus, triglycerides associate with cholesterol and lipoproteins to form chylomicrons (particles adapted for lipid transport).
5. Chylomicrons move out of the epithelial cells by exocytosis —> enter lymphatic capillaries called lacteals, found at centre of each villus.
6. Chylomicrons then pass into blood via lymphatic vessels —> triglycerides in chylomicrons hydrolysed by enzyme in blood capillaries’ endothelial cells, from where they diffuse into cells.

Question 17

What are haemoglobins?

Answer 17

= Group of chemically similar molecules adapted for O2 transport, found in a wide variety of organisms.

Question 18

Describe the structure of haemoglobin.

Answer 18

1 structure —>

2 structure - alpha helix —>

3 structure - specific and each chain folded into a precise shape —> important factor in its ability to carry O2.

4 structure - 4 polypeptide chains linked together, with each chain associated with a prosthetic Fe2+ group.

Each Fe2+ can combine with a single O2 molecule —> 4xO2 molecules can be carried by each human haemoglobin molecule in humans.

Question 19

Describe loading/unloading of Hb with oxygen.

Answer 19

• Hb binding with O2 = loading/associating, takes place in lungs (in humans).
• Hb releasing O2 = unloading/disassociating, takes place in the respiring tissues (in humans).

NB => Hbs with a high affinity for O2 take up O2 more easily but release it less easily and vice versa.

Question 20

What is the role of Hb? What must it do to be efficient?

Answer 20

• To transport O2 around the body.
• To be efficient at transporting O2, Hb must:
  1. Readily associate with O2 at gas exchange surface.
  2. Readily dissociate from O2 at respiring tissues.

=> seems contradictory - Hb can change its affinity (chemical attraction) for O2 under different conditions —> shape changes due to presence of CO2.

—> in the presence of CO2, the new shape of the Hb molecule binds more loosely to O2 —> Hb releases its O2.

Question 21

Is oxygen loaded/unloaded at the gas exchange surface? Explain why.

Answer 21

O2 loaded, because:

High [O2], low [CO2] —> high affinity for O2.

Question 22

Is oxygen loaded/unloaded at the respiring tissues? Explain why.

Answer 22

O2 unloaded, because:

Low [O2], high [CO2] —> low affinity for O2.

Question 23

Why do different Hbs exist?

Answer 23

• Different organisms have differing O2 requirements —> their DNA codes for a (slightly) different amino acid sequence in their Hb —> slightly different 3/4 structures.
• Happens as a result of evolution within species as adaptations to different environments and conditions.

=> some have higher/lower affinities.

Question 24

Only 1/4 O2 per Hb released when at rest. What is the advantage when more active?

Answer 24

• Some O2 kept in reserve —> Hb can provide O2 immediately to aerobically respiring tissues (other 3/4).

Question 25

Why does breathing in CO lead to loss of consciousness —-> death?

Answer 25

• CO replaces O2 by permanently binding to Hb in preference to O2.
• Less O2 released by Hb to aerobically respiring tissues, such as those in the brain —> these cells can no longer (aerobically) respire enough so cannot function at a high enough level —> loss of consciousness.

Question 26

Why is the oxygen dissociation curve shaped like an S?

Answer 26

• O2 does not bind evenly to Hb across different partial pressures of O2.
  1. Closely packed polypeptide subunits, so hard for O2 to bind —> low [O2], little binds —> shallow gradient.
  2. Binding of 1st O2 molecule changes the 4 structure of Hb molecule —> changes shape so that it is easier for other O2 molecules to bind to each subunit.

=> takes a smaller increase in PO2 to bind 2nd O2 molecule than it did to bind 1st O2 => positive cooperativity —> binding of 1st facilitates binding of 2nd —> steeper gradient of curve.

3. After 3rd O2 molecule binds —> more difficulty due to decreased probability of a single O2 molecule finding a free binding site —> graph flattens off.

NB=> Curve to left = higher affinity for O2.
=> Curve to right = lower affinity for O2.

Question 27

Define the Bohr Effect.

Answer 27

= Increasing [CO2], the more readily the Hb releases its O2 (+ less likely to load O2) due to a reduced Hb affinity for O2.

Curve shifts to right.

Always ensures enough for O2 for respiring tissues:

• Increased respiration rate —> more CO2 —> lower pH —> greater change in shape of Hb —> O2 more readily unloaded —> more O2 for respiration.

Question 28

Outline continuous process of oxygen loading and unloading in the body.

What is the consequence of this process?

Answer 28

1. Gas exchange surface, CO2 constantly removed.
2. Slightly higher pH due to low [CO2].
3. Higher in pH changes shape of Hb —> loads O2 readily.
4. Increase in affinity for O2 —> O2 not released when being transported in blood to tissues.
5. CO2 produced by respiring cells in tissues.
6. CO2 acidic in solution so pH of blood falls.
7. Lower pH changes Hb shape —> decreased affinity for O2.
8. Hb releases its O2 into respiring tissues.
   - Increased respiration rate —> more CO2 —> lower pH —> greater change in shape of Hb —> O2 more readily unloaded —> more O2 for respiration.

Question 29

Comment on and explain the lugworm dissociation curve, relative to that of a human.

Answer 29

• Lugworms are inactive and covered by water most of the time.
• O2 diffuses into lugworm’s blood from the water —> uses Hb to transport O2 to tissues.
• Tide goes out, no new O2 supply so [O2] in burrow progressively decreases as it’s used up => must extract enough O2 to survive until tide covers it again.

=> Curve shifted to far left of that of humans —> lugworm Hb fully loaded with O2 even in a low [O2] environment.

Question 30

Suggest how the O2 dissociation curves of more metabolically active organisms looks.

Answer 30

=> Curve shifted to right —> Hb becomes less saturated with O2.

=> O2 more readily released to respiring tissues so Hb supplies more O2 to enable muscles to respire rapidly.

Question 31

Define Xylem Vessels.

Answer 31

= Hollow, thick-walled tubes that transport water in the stem and leaves of plants.

Question 32

Define transpiration.

Passive/active?

Answer 32

= Evaporation of water from the leaves —> pulls water through xylem vessels via transpiration stream.

• Passive process as energy for process provided sun and particles’ own energy.

Question 33

Describe how water moves out of leaf via stomata.

Answer 33

1. Humidity of air usually lower than air spaces next to stomata so wp gradient from air spaces through stomata to air.
   => water molecules diffuse out of the air spaces into surrounding air (providing stomata are open).
2. Water evaporating from surrounding mesophyll cells replaces water in air spaces that was lost by transpiration.

NB=> rate of transpiration controlled by changing the size of their stomatal pores.

Question 34

Describe how water moves across the cells of a leaf.

Answer 34

1. Water lost from mesophyll cells by evaporation from cell walls to leaf air spaces is replaced by water from xylem, via cell walls/cytoplasm.
2. Cells’ wp lowered so water moves into them from other cells by osmosis.
3. Wp gradient established that pulls water up from xylem, across leaf mesophyll, into atmosphere.

Question 35

Describe how water moves up the stem in the xylem.

Answer 35

• Cohesion-tension is mainly responsible:
  1. Water evaporates from mesophyll cells due to heat from sun leading to transpiration.
  2. Water molecules form H-bonds with one another —> cohesion.
  3. Water forms a continuous, unbroken column across mesophyll cells and down xylem.
  4. Column of water pulled up xylem as a result of transpiration => transpiration pull.

=> puts xylem under tension —> negative pressure within xylem.

Question 36

Evidence for cohesion-tension theory.

Answer 36

1. Water flow decreases at night (as would be expected - less photosynthesis).
   - Diameter of tree trunk increases - therefore tension has decreased as decrease in negative pressure - xylem not pulled inwards as much.
2. If xylem vessel broken and air enters it —> tree can no longer draw up water —> continuous column of water is broken —> H20 molecules can no longer stick together.
3. Water does not leak out when xylem vessel broken —> but air drawn in. Suggests water not under pressure but that xylem is under negative pressure.
4. Xylem vessels have no end walls —> form a series of continuous, unbroken tubes from root to leaves —> essential to cohesion-tension theory of water flow up the stem.

=> consistent with xylem being under tension.

Question 37

Comment on the nature of the cohesion-tension process, mainly in terms of energy.

Answer 37

• Passive process —> no metabolic energy required.
• Xylem vessels are not made of living cells, so cannot actively move water.
• Energy needed to drive process of transpiration —> heat from sun evaporates water from leaves.

Question 38

If a tree is sprayed with herbicide, suggest why the rate of water flow remains unchanged.

Answer 38

• Water flow is a passive process.
• Xylem not made of living cells (root cortex and mesophyll are) but osmosis is passive.

=> Even if cells have been killed by the herbicide, water flow will continue, at least for a short while.

Question 39

Adaptations of root hair cells

Answer 39

1. Thin cell wall.
2. Extension.
3. Large surface area.
4. Many mitochondria.

Question 40

Adaptations of xylem vessels. Advantages of spiral thickening.

Answer 40

• As they mature, cell walls incorporate lignin and cells die.
• Lignin has mechanical strength and waterproofing abilities —> ensures water carried up plant.
• Lignin forms rings/spirals around vessel.

=> Can cope with tension (negative pressure) because they have thick walls (lignin).

• Hollow, elongated vessels.
• Dead cells better for this purpose as water flow would be slowed as it crossed this membrane/cytoplasm.
• Allows vessel to elongate as the plant grows.
• Allows stems to be more flexible.
• Use less material —> less wasteful and lower plant mass.

Question 41

Define translocation.

Answer 41

= Process in which organic molecules (sucrose/amino acids) and some mineral ions are transported from one part of the plant to another.

Carried out by phloem tissue in flowering plants.

Question 42

Define source.

Answer 42

= Part of the plant where sugars produced during photosynthesis.

=> place where something produced.

Question 43

Define sink.

Answer 43

= Parts of the plant where sugars used directly or stored for future use.

=> place where something used, but not produced.

NB=> sink can be above or below the source => translocation can be above/below source.

Question 44

What is the phloem?

Answer 44

the vascular tissue in plants which conducts sugars and other metabolic products downwards from the leaves.

Question 45

Define Mass Flow.

Answer 45

= Bulk movement of a substance through a given channel/area in a specific area.

Question 46

Outline the mechanism of translocation:

1. Sucrose from photosynthesising tissue into sieve elements.

Answer 46

• Mass flow theory:
  1. Sucrose manufactured from photosynthesis products in cells containing chloroplasts.
  2. Sucrose diffuses down concentration gradient by facilitated diffusion from photosynthesising cells into companion cells.
  3. H+ actively transported from companion cells into spaces within cell walls using, ATP.
  4. H+ diffuse down concentration gradient through carrier proteins into the sieve tube elements.
  5. Sucrose molecules are transported along with H+ in co-transport to sieve-tube elements via co-transporter proteins.

Question 47

Outline the mechanism of translocation:

2. Mass Flow of sucrose through sieve-tube elements.

Answer 47

• Sucrose produced at source —> actively transported into sieve tubes as described.
  6. Sieve-tube wp falls as a result.
  7. Xylem has a much higher wp so water moves from xylem into sieve tubes by osmosis => high hydrostatic pressure in tubes.
  8. At respiring cells (sink), sucrose is used up in respiration/converted to starch for storage.
  9. These cells have a low [sucrose] —> so sucrose is actively transported into them from the sieve tubes lowering their wp.
  10. Due to this lowered wp —> water also moves into these respiring cells from sieve-tubes by osmosis.
  11. Hydrostatic pressure of sieve tubes is therefore lowered => higher hydrostatic pressure at source >sink.
  12. Therefore, a mass flow of sucrose solution down this hydrostatic gradient in the sieve tubes.

—> Mass flow itself is a passive process but it occurs as a result of active transport of sugars —> process as a whole is active so affected by temperature and metabolic poisons.

Question 48

Evidence for Mass Flow hypothesis.

Answer 48

1. Pressure within sieve-tubes shown by release of sap when they’re cut.
2. Higher concentration of sucrose in leaves (source) than in roots (sink).
3. Downward flow in phloem occurs in daylight, but ceases when leaves shaded/at night.
4. Increases in sucrose levels in the leaf —> followed by similar increased in sucrose levels in phloem slightly later.
5. Metabolic poisons and a lack of O2 can both inhibit translocation of sucrose in the phloem.
6. Companion cells have many mitochondria and readily produce ATP.

Question 49

Evidence against Mass Flow hypothesis.

Answer 49

1. Function of sieve plates is unclear - they would seem to hinder mass flow. (Although, may have a structural function, helping tubes to stop bursting under pressure).
2. Not all solutes move at the same speed —> should under mass flow.
3. Sucrose delivered at largely same rate to all regions, rather than going more quickly to regions with lowest [sucrose], which is what mass flow theory suggests should happen.

Question 50

How can we use Ringing experiments to investigate transport in plants?

Answer 50

1. Woody stems have an outer protective layer of bark on the inside of which is a layer of phloem that extends all around the stem —> inside phloem layer is xylem.
2. Outer layer section removed (protective layer and phloem) around complete circumference of a woody stem while still attached to plant.
3. Parts of stem immediately above the missing ring of tissue swells —> liquid there is rich in sugars and other dissolved organic substances.
4. The non-photosynthetic tissues in region below ring (towards roots) —> wither and die - while those above the ring continue to grow.

Conclusion => phloem, rather than xylem, is the tissue responsible for translocating sugars in plants - ring of tissue (left in stem) not affected so if tissue were responsible for translocation, tissues below it would not die.

Question 51

How can we use Tracer experiments to investigate transport in plants?

Answer 51

1. Radioactive isotopes are useful for tracing movement of plants - 14C - plants grown in 14CO2.
2. 14C incorporated into sugars produced in photosynthesis.
3. Can be tracked using autoradiography —> thin cross-sections of the plant stem taken and placing them on a piece of x-ray film.
4. Film blackened where exposed to radiation produced by 14CO2.
5. Blackened regions are found to correspond to where phloem tissue is in the stem —> other tissues do not blacken film.

=> follows that these other tissues do not carry sugars + phloem alone is responsible is responsible for their translocation.