A2 Biology Past Paper Notes

A Cell Structure

Features of prokaryotes:
1. DNA is not surrounded by a nuclear envelope/membrane; there is no true nucleus.
2. DNA is circular (a loop).
3. DNA is not complexed with histone proteins; it's naked DNA.
4. (Only) 70S/smaller/18nm ribosomes; ribosomes are not attached to membranes.
5. No double membrane-bound organelles; no mitochondria/chloroplasts.
6. Absence of named organelles; e.g., Golgi apparatus, ER/RER/SER if the previous marking point was not given; no membrane-bound organelles.
7. Capsule/slime layer.
8. Very small diameter; $0.5$ to $5.0µm$ .
9. Cell wall of murein/peptidoglycan.
10. Pili/pilus.
11. No 9+2 microtubule arrangement.
12. Flagellum not covered by the cell surface membrane.
13. Presence of plasmids.

TEM

Features that show a given electron micrograph is from the TEM:
1. Higher magnification and higher resolution than a light microscope.
2. Images are in 2D/no surface contours.
3. Organelle ultrastructure can be seen (give named examples such as the internal structure of chloroplast).
4. Very thin section.
Range of most eukaryotes is $10um$ to $100um$ .
Eyepiece graticule is calibrated because it can be used to make measurements.
Lysozyme is used to break down peptidoglycan.

Light Microscope

Advantages of using a light microscope instead of an electron microscope:
1. Can observe living tissue and living processes.
2. Different types of stains can be used to observe specific tissue.
3. Portable and easy to move.
4. Colour can be seen.
5. Lower costs and maintenance.

B Biological Molecules

If glucose is hydrolyzed and boiled with Benedict's solution, it tests negative.
Enzymes do NOT have quaternary structure.
The peptide bond is the last to break as the temperature of the enzyme is increased. Next is disulfide.
Tertiary structure maintains globular shapes of the enzyme.
When 2 cysteine amino acids join, they are held together by peptide bonds only.
What occurs during protein denaturation by pH changes:
- Breaking of peptide bonds leads to the loss of the active site - incorrect
- Disruption of the ionic bond b/w amino R groups - correct
- Loss of alpha helix b/w amino acid - correct
- Loss of tertiary structure which causes loss of function - correct.
What level of protein structures are always involved when competitive and noncompetitive inhibitors bind to enzymes?
- Competitive and noncompetitive: tertiary structure
Globular proteins always have hydrophilic amino acids towards the outside and hydrophobic amino acids towards the inside.
Explain how amino acids can be close together in an active site by explaining the structure of the protein:
1. Further coiling and folding of the polypeptide chain.
2. Giving tertiary structure.
3. Held in position by R group interaction.
4. Brings distant amino acids close.

Hydrogen Bonds

Explain the role of H bonds in maintaining secondary structure:
1. H bonds maintain the alpha helix and beta-pleated sheet.
Describe the hydrogen bonding that occurs between water molecules:
1. A hydrogen bond is a weak bond between the oxygen atom of one $H2O$ molecule and the hydrogen atom of another $H2O$ molecule.
2. Oxygen is highly electronegative, more than H.
3. Oxygen has two lone pairs so it can form 2 hydrogen bonds.
4. Asymmetrical electron distribution.
5. Oxygen has delta + and hydrogen has delta - charge.
Explain the role of H bonds in maintaining tertiary structure:
1. H bonds stabilize further folding of the polypeptide.
2. Between R groups with amine and carboxyl groups.
3. Helps maintain globular shape, 3D shape.
What is the effect of replacing glutamine (polar) with valine (non-polar) in alpha/beta globin?
1. Glutamine is polar and valine is non-polar.
2. Change in tertiary shape.
3. Change in the quaternary structure of hemoglobin.
4. Hemoglobin is less soluble.
5. Hemoglobin is less efficient at transporting oxygen.

Collagen

Function of collagen in the wall of arteries:
1. Withstands pressure.
2. Prevents overstretching.
3. Prevents bursting/rupture.
Structure of collagen vs Hemoglobin:
1. 1 polypeptides are not identical (v. 2 identical, α / β, polypeptides).
2. Triple helix or three polypeptides/helices (v. 4 polypeptides).
3. Only composed of amino acids or no prosthetic group/haem/iron.
4. (fibrous so) not globular.
5. No complex folding (v. complex folding); A no tertiary structure
6. Glycine is repeated every 3rd position/more glycine.
7. Repeating triplets of amino acids/a large number of repeating amino acid sequences (v. greater variety).
8. AVP; e.g. different primary structure.
9. Variation in amino acid sequences (v specific sequences)
10. All polypeptides, helical (v. α different to β, polypeptides).
11. hydrogen bonds between polypeptides (v. Van der Waals).
12. covalent bonds between molecules (to form fibrils) (v. none).
13. $300nm$ long polypeptides (v $5–10nm$ )
14. Each polypeptide over 1000 amino acids (each 141 / 146 amino acids).

C Enzymes

What is meant by specificity of an enzyme:
1. An enzyme acts on only one substrate.
2. The shape of the active site is complementary to the substrate.
How does an enzyme lower $E_a$ :
1. Provides an alternative energy pathway.
2. Brings reactants close together to form ESC.
3. Puts strain on the reactant.
4. So bonds break easily.
5. Transfer of charges between groups.
How to describe and explain enzyme results (D = description, E = explain):
1. D - at low substrate concentration, the ROR is proportional to the substrate concentration.
2. E - not all active sites are occupied.
3. E - so the substrate concentration is limiting.
4. E - so fewer ES complexes are formed.
5. D - at high concentration, there is less steep increase in the rate of the reaction.
6. D - it levels out/plateaus at (XYZ) concentration.
7. E - the enzyme concentration is limiting.
8. E - since all active sites are occupied/saturated.
9. E - $V_{max}$ reached.

Optimum Enzyme Rate

Advantages of enzymes with high optimum:
1. The production/process requires a high temperature.
2. High temperature so more collisions and more ESC.
3. A higher ROR increases.
4. More stable and can be used again and again.
5. Less prone to denaturation.
6. Works well over a range of temperatures and pH, so works at a higher rate.
Outline experiment that should be carried out to find out if an inhibitor is competitive or non-competitive:
1. Carry out the experiment with and without an inhibitor.
2. At many different concentrations of substrates.
3. Keeping other variables constant (e.g., pH and temp).
4. Draw a graph of the rate of reaction against substrate concentration.
5. If the inhibitor is competitive, then $V_{max}$ is the same as without the inhibitor.
Describe the mode of action of an enzyme with induced fit:
1. The active site changes shape OR molds around the substrate OR better fit.
2. The active site becomes fully complementary to the substrate.
3. Formation of ESC.
4. Lowers activation energy.
5. By providing an alternate energy pathway/puts strain on bonds.
6. Breaking of bonds.
7. The active site returns to the initial stage AND can be reused.
After the XYZ substance/device is used why does enzyme B have a lower $Km$ and why does enzyme C have a higher $V{max}$ :
1. B: increases the affinity of the enzyme for the substrate
2. B: makes the shape of the active site more complementary
3. B: Makes the position of the active site more accessible
4. C: increases the rate of ESC formed
5. C: increases the rate of catalysis after binding
6. C: may lower the activation energy required than normal

Intracellular vs Extracellular Enzymes

Advantages of end-product inhibition when an enzyme is intracellular:
1. Maintains balance.
2. Efficient metabolism.
3. Avoids osmotic problems that could be caused by the build-up of a product.
Disadvantages of end-product inhibition when the enzyme is used extracellular and commercially
1. Loss of product
2. Slow rate of reaction
3. The product is required continuously
Why are immobilized enzymes more active over a range of pH:
1. The alginate covering is protective, so the $H^+/OH^−$ ions do not penetrate alginate beads, so the shape of the active site is not disrupted.
Explain what is meant by a higher $K_m$ :
1. The enzyme has a lower affinity for the substrate.
2. Needs a higher concentration of substrate to reach $V_{max}$
3. Less likely to be saturated by substrate.
Suggest one other advantage of using enzymes obtained from microorganisms, rather than enzymes extracted from barley seeds, in the production of sugar syrups.
1. easier to extract
2. idea that microorganisms can be cultured in large quantities and produce large amounts of enzyme
3. higher rate of reaction
4. active over a greater temperature range

Immobilized Enzymes

Advantages of immobilized enzymes compared to enzymes in free solutions:
1. The enzyme can be reused.
2. (may be able to) obtain more product (per unit time).
3. Can use higher temperatures (to obtain more product) / still active at higher temperatures; A thermostable.
4. Does not denature (as easily as free) if temperatures increase.
5. The enzyme can be easily recovered.
6. Downstream processing is easier.
7. The product is not / less, contaminated; A less purification is needed.
8. Longer shelf-life of enzyme.
9. Reduces product inhibition.
10. The enzyme is more stable / less likely to denature or described; A thermostable / can work at high temperatures A in context of change in pH I ‘can withstand changes in temperature.’
Explain how the addition of the competitive inhibitor results in the same value for $V{max}$ but a higher value for $Km$ :
1. The competitive inhibitor occupies the enzyme’s active site / competes with the substrate.
2. Reduces the frequency of successful collisions / fewer ES complexes are formed.
3. Reduce the reaction rate at low substrate concentration.
4. The idea that the curve with the inhibitor is to the right of the curve without the inhibitor.
5. At high substrate concentration, the effect of the inhibitor is reduced.
6. Therefore the $V_{max}$ is the same as it is determined by the concentration of the enzyme
7. The idea of the intercept to the curve gives a higher value for $K_m$ .

CHAPTER 4: Cell Membranes and Transport

The greater the SA: V ratio, the lower the time for diffusion.
Why is it called the fluid mosaic model?
1. Phospholipid (and protein) molecules move about/ diffuse.
2. Protein (molecules) scattered.

Cell Membrane Roles

Role in cell membrane of:
- Glycoproteins: receptors/receptor molecules; for hormones/neurotransmitters/named hormone/neurotransmitter (e.g., insulin, acetyl choline, noradrenaline); idea of ( ) antigens / (cell surface) markers / cell recognition / cell adhesion; help to stabilize membrane structure / forms H bonds with water molecules; carrier proteins
- Cholesterol: maintains / regulates the fluidity of the membrane / prevents the membrane being too rigid or fluid / mechanical stability (qualified) / prevents ions / polar / water-soluble / named molecule, passing / leaking through the membrane.

Cell Signaling

General points to write when asked “how is XYZ a cell signalling mechanism”
1. The “molecule” [given in qs] acts as a cell signalling molecule
2. It moves through the bloodstream/extracellular space /intracellular space
3. To reach the target cell which is the [given in qs]
4. It will bind to complementary, specific, receptors [on the cell membrane - depends on qs]
5. This will lead to a response which is [given in qs]
6. a detail of change, such as activating G proteins / secondary messenger / enzyme cascade /chain of reactions

Phospholipids

Phospholipids:
1. can form a bilayer
2. link between a hydrophobic core, and a barrier to water-soluble substances; A polar/ ionic
3. the hydrophilic/phosphate head forms H bonds with water; A facing, water/watery environment/aqueous environment/cytoplasm/cytosol
4. ref. contribution to the fluid nature of the membrane
5. further detail; e.g., mainly saturated fatty acids, less fluid e.g., mainly unsaturated fatty acids, more fluid
6. ref. to control over membrane protein orientation; e.g., hydrophobic – hydrophobic interaction for ‘floating’ proteins

Endocytosis

Some cells take in bacteria by endocytosis. Explain how endocytosis occurs at the cell surface membrane.
1. Attachment (of bacteria) to receptor(s).
2. Ref. the ability to attach to an antibody (bound to antigen on the bacterium).
3. Infolding of the membrane; A membrane engulfs.
4. Form (round bacterium)
5. Fusion of the membrane.
6. Formation of a vacuole/vesicle.
Functions of a lysosome in the endocytosis of bacteria:
1. Break down / digest / destroy, bacteria / pathogens; A autophagy
2. Break down / digest / destroy, (worn out / defective organelles / named organelle (in animal cell).
3. Catalyses / speed up, hydrolysis.
4. Any two named substrates; e.g., (any named) polysaccharides / proteins / (phospho)lipids / (named) nucleic acids
5. Idea that recycle / reuse, biological molecules within the cell
6. (macrophage / phagocyte) cut up to present antigen

Multicellular Organisms

The surface area to volume ratio decreases as animals increase in size. Use this fact to suggest why multicellular animals require transport systems.
1. the idea that diffusion (via, the body surface / to cells), cannot satisfy needs / it's too slow;
2. long distances (to reach some cells/tissue)
3. takes materials close to cells
How does the phospholipid molecule make it suitable for its function.
1. Hydrophilic phosphate head and hydrophobic fatty acid tail
2. Forms a bilayer with head outside and tail inside
3. Head faces aqueous environment and tail faces each other to form a hydrophobic core
4. Forms H-bonds with water
5. Stabilizes membrane
6. Fatty acid may be saturated/unsaturated
7. Unsaturated makes membrane fluid
8. Barrier to polar substance

Ions

Role of ions in a living organism: calcium
1. bone/teeth, formation/strengthening; R calcium in bone R calcium for healthy bones and teeth
2. enamel/shell, formation/strengthening;
3. reference to muscle/nerve/synapse, function e.g., muscle contraction, generation of nerve impulse;
4. blood clotting;
5. calcium pectate, in cell wall/middle lamella;
6. spindle formation;
7. for fertilization/fusion of egg and sperm;
iron
1. forms part of, haem/haemoglobin/myoglobin; A transport of oxygen in haemoglobin
2. reference cytochrome(s)/electron carrier(s);
3. important in chlorophyll synthesis;
4. prosthetic group of some/named, enzymes/catalase
potassium
1. activates enzymes;
2. cofactor in, photosynthesis/glycolysis;
3. reference to nerve/muscle, function e.g., conduction of nerve impulse, muscle contraction;
4. maintains osmotic balance/water potential of cells;
5. stomatal, opening/closure/turgidity of guard cells;
6. reference to $Na^+/K^+$ pump mechanism - qualified;

CHAPTER 5: Cell and Nuclear Division

Prophase and metaphase
1. chromatids/chromosomes/chromatin, condense/become shorter/become thicker/coil/supercoil/AW;.
2. centrioles, move to/reach, opposite poles.
3. nucleolus disappears
4. spindle is formed; A ‘more developed’ A description in terms of spindle fibres
5. ref to the assembly of microtubules; A ‘makes’ microtubules R 9+2
6. the nuclear envelope disintegrates/breaks down.
7. chromosomes, move to / at, the equatorial plate/equator/metaphase plate/AW
8. centromeres attach to, spindle/fibres;
  9.ref to the random arrangement of chromosomes; A ‘not in pairs.
Features characteristic of metaphase:
- Chromosomes line up at the equator / equatorial plate/metaphase plate.
- Centromeres attached to spindles.
- Centrioles reach the poles.
- Spindle is fully formed.
Describe the cell during prophase:
1. Chromosomes have condensed/coiled up so they are more visible
2. Two sister chromatids are held together by a centromere
3. Each chromatid has 1 DNA molecule
4. DNA is associated with histone proteins
Cytokinesis:
1. A cell plate forms / cell wall/cellulose is laid on or A formed (only for plants)
2. Cytoplasm divides into two

Importance of Mitosis

Importance of Mitosis:
1. replacement of cells;
2. repair of tissue; R repair of cells
3. growth / increase in cell numbers;
4. asexual reproduction / vegetative propagation; R cloning
5. maintains / same, number of chromosomes; A two sets of chromosomes / diploid / 2n
6. genetically identical to parents;
7. A produces daughter cells that are genetically identical A ref. clone(s)
8. ref to rejection.

Role of Microtubules

Outline the role of microtubules in mitosis:
1. Forms part of a spindle fibre
2. Attaches to the centromere/chromosome
3. Moves sister chromatids to opposite poles
Outline the changes that occur as a stem cell transforms into a mature RBC:
1. Stem cell divides rapidly
2. synthesizes/produces/makes hemoglobin
3. synthesizes/produces/makes carbonic anhydrase
4. Loss of nucleus
5. Loss of organelles e.g., ribosome, mitochondria
6. Becomes biconcave
Suggest the importance of mitotic cell cycle timing and control:
1. Coordination of growth
2. Minimizes exposure to mutation
3. Prevents tumor formation
4. Which would spread to other tissues
5. Allows producing cells only when required
Outline how mutations can cause healthy cells to become tumor cells:
1. The idea that a mutation occurs for a gene controlling cell division
2. Proto-oncogene to oncogene
3. The tumor suppressor gene is switched off
4. Ref. to the disruption of the cell cycle / shortened interphase
5. (results in) uncontrolled cell division.

Cancer

How does smoking cause lung cancer?
1. Tar is carcinogenic.
2. The gene controls mitosis.
3. Mutation in genes
4. Gene expression is affected
5. Tumor suppressor gene is turned off
6. A cell grows uncontrollably
7. Cancer cells do not respond to signals
8. Forms a tumor
Role of centrioles in animal cells
1. Important in mitosis
2. Replicated before cell division
3. Organizes microtubules
4. Moves to opposite poles
5. Formation of spindle
Structure of chromatids in metaphase:
1. two chromatids;
2. identical / sister, chromatids;
3. joined by a centromere; A kinetochore
4. one from
5. (reach chromatid) DNA complexed with protein
6. histone proteins / histones;
7. telomeres at end of chromatids
State two differences between the chromosome at metaphase and the chromosome at late anaphase:
1. two chromatids versus, one chromatid / one daughter chromosome;
2. sister chromatids joined at centromere versus chromatids separated
3. the distance between sister chromatids zero versus increasing distance between chromatids
4. share a centromere versus do not share a centromere / centromere divides
5. two DNA molecules versus one DNA molecule;
6. at, equator / metaphase plate versus towards / at, poles; R centre R ends
7. linear / straight versus V shape
Outline the function of telomeres:
1. Permit continued replication
2. Prevent loss of genes
3. Protect ends of chromosomes
Suggest a reason why daughter cells are not identical immediately after cytokinesis unequal sharing out of cytoplasm / organelles / named organelles, (at cytokinesis) ; A uneven sizes

CHAPTER 6: PROTEIN SYNTHESIS AND NUCLEIC ACIDS

State what is meant by a STOP codon:
1. A codon that terminates polypeptide synthesis/translation
2. Does not code for an amino acid
3. Has no complementary anticodon
4. Causes the release of a polypeptide chain
Compare the peptide bond formed during translation with the types of bond made during tertiary structure formation SIMILARITY
1. peptide bond and disulfide bond are covalent bonds
Differences
1. tertiary structure bonds are between, R groups / side chains, (of different amino acids
2. H bond / ionic bond / hydrophobic interaction, versus covalent peptide bond; A tertiary structure bonds apart from disulfide are not covalent
3. peptide bond, stronger / more thermostable, than, tertiary structure bonds / two named bonds;
4. the peptide bond is between, carboxylic acid / COOH, and, amino / $NH_2$ , group (of the adjacent amino acid
5. AVP ; detail of a tertiary structure bond e.g. hydrophobic between non-polar R groups disulfide between sulfur-containing R groups ionic between carboxyl and amino groups of R groups H bond between oxygen on —CO groups and H on either the —OH or —NH groups of R groups

Semi-Conservative Replication

Explain how the structure of DNA enables it to replicate semi-conservatively.
1. base pairing/A-T and C-G
2. ref to complementary/explained with ref to H bonds
3. (free) nucleotides pair with both, strands/each strand/polynucleotides/sides
4. both strands act as templates
5. to produce two DNA molecules that are identical to one another
Differences between mRNA and DNA:
Role of mRNA after leaving the nucleus:
1. translation; R if transcription given as well, unless in correct context A use of, nucleotide / base, sequence, to make, amino acid chain / polypeptide / protein I protein / polypeptide, synthesis
2. moves towards / combines with, ribosome;
3. ref to small and/or large sub-units
4. codon(s); only accept in correct context
5. transfer / t, RNA, bringing, amino acid(s), to mRNA / ribosome

tRNA

(Protein Synthesis) Role of tRNA in protein synthesis:
1. (tRNA) carries amino acid to ribosome;
2. ref. to the specificity of amino acid carried .
3. ref. anticodon (on tRNA): codon (on mRNA) binding;
4. ref. complementary / base pairing; A A-U, C-G
5. ref to tRNA binding sites within ribosome;
6. two tRNAs bound to mRNA / ribosome, at the same time
7. amino acids held close to each other / AW;
8. (for) peptide bond formation;
9. (tRNA) can be reused / binds another amino acid;
During interphase: cells metabolically active
- Protein synthesis
- Transcription
- Translation
- Gene expression
- DNA / semi-conservative, replication
- Respiration
- Synthesizing macromolecules

Cancer Risk

Factors that increase the risk of cancer:
1. chemical carcinogens;
2. virus, qualified
3. ionizing radiation / X-rays / gamma rays/particles from radioactive decay / ultraviolet light / alpha particles / beta particles; allow two named radiation examples for two marks
4. free radicals
5. hereditary predisposition
6. tobacco smoking
7. obesity
How can a mutation in a gene lead to the formation of an altered polypeptide where one amino acid is replaced with a different amino acid
1. (in DNA / gene) altered sequence, of, nucleotides / bases;
2. base substitution or base / nucleotide, replaces another, base / nucleotide;
3. (mRNA synthesised) during transcription;
4. (mutation leads to) altered / AW, mRNA / messenger RNA;
5. (only) one (mRNA) codon changed / a different codon; (A one DNA, triplet / codon, changed I ref. to codons changed)
6. tRNA, with / has, a different anticodon;
7. tRNA brings a different / a changed / the incorrect, amino acid during translation / to the ribosome;
8. codon-anticodon, binding / complementary.
Role of anticodon in translation:
1. The anticodon is complementary to the codon on mRNA, hence, it binds to the mRNA
2. The anticodon is specific to an amino acid. This ensures that the correct amino acid sequence is formed during translation.

CHAPTER 8 Transport in mammals

Explain, with reference to function, the difference in the thickness of muscle of the left and right ventricle.
1. The left ventricle pumps blood to the rest of the body and the right ventricle pumps blood to the lungs
2. Right ventricle has smaller muscles because travel is short distance
3. Less resistance
4. Less force/pressure is required
Explain why the mammalian circulatory system is described as close double circulation.
- Double – blood passes through the heart twice during one circulation;
- Closed – blood travels inside blood vessels.
Disadvantage of having no nuclei in RBCs:
1. Cannot carry out protein synthesis/replication/repair
2. Short life span
3. Cannot divide/replace themselves.
Explain how the structure of red blood cells is suited to their function of transporting oxygen to body tissues.
1. Small size, to squeeze through capillaries
2. Small size so haemoglobin molecules near the surface
3. No nucleus/lack of organelles, so more room for haemoglobin
4. Biconcave shape increases surface area for oxygen
5. Flexible membrane, to squeeze through capillaries.
Explain how heart action is initiated and controlled (reference should be made to the sinoatrial node, the atrioventricular node and the Purkyne tissue).
1. myogenic;
2. SAN, is pacemaker / sends out impulses / waves of excitation / initiates, heart beat / action potential / contraction;
3. AVN delays, impulse / contraction (of ventricles);
4. specific time ref (0.1 - 0.2 secs) or to allow ventricles to fill / atria to empty;
5. relays impulse to Purkyne tissue / bundle of His;
6. Purkyne tissue conducts (impulse) to base / apex of heart / septum/ ventricles;
7. ref to the papillary muscles contracting;
8. ventricle (muscle) contracts / ventricular, contraction / systole, from base upwards;
9. (blood) into arteries / named artery;
Explain how the structure of haemoglobin aids the uptake of oxygen in the lungs.
1. 4 polypeptides/4 globins/4 amino acid chains;
2. outwardly pointing hydrophilic groups, maintain solubility;
3. each with a haem group;
4. ref to iron/ $Fe^(2+)$ ion;
5. temporary attachment to oxygen; A readily attaches/binds combines with R oxygen binds to haem
6. 4 molecules of oxygen;
  7.oxyhaemoglobin;
7. ref to cooperative binding;
Explain how $CO_2$ stimulates the release of oxygen from the blood.
1. carbon dioxide reacts with water to form carbonic acid;
2. catalysed by carbonic anhydrase;
3. dissociates to hydrogen carbonate and hydrogen ions;
4. hydrogen ions combine with haemoglobin;
5. forms haemoglobinic acid;
6. so releasing oxygen; ignore ref to Bohr shift (question says ‘explain’) A from equations
7. Transport of $CO2$ CO2 + H2O à H2CO3 à H+ + HCO3-
8. (catalyzes very) fast / AW, reaction ;
9. (carbon dioxide as) hydrogen carbonate ions / bicarbonate ions;
10. diffuse / move / leaves, out of the (red blood) cell;
11. in(to) the plasma; R ‘into blood'
12. so that blood can transport more than could be transported as carbon dioxide (in 6. solution) / 80 – 90% $CO_2$ transported this way;
13. reaction maintains concentration gradient for $CO_2$ from, tissues / tissue fluid, to blood;
14. if carbon dioxide is transported, then the pH would decrease;
15. therefore maintains pH / prevents pH decreasing / acts as a buffer;

Oxygen from Hemoglobin

Describe and explain how carbon dioxide ( $CO_2$ ) and hydrogen ions (H^+$) play a role in the unloading of oxygen from haemoglobin.
1. diffusion of, carbon dioxide / CO_2$$;
2. into red blood cell from correct source;
3. description of carbonic acid formation followed by H+ production;
4. ref. carbonic anhydrase ) fast reaction;
5. haemoglobin has a higher affinity for hydrogen ions than oxygen; A haemoglobin releases oxygen more easily in acidic conditions
6. ref. to, allosteric effect / change in tertiary structure in (oxy)haemoglobin, causes, release / AW, of oxygen;
7. formation of haemoglobinic acid;
Composition of blood at the venule end, compared to that at the arteriole end:
- Blood at venule end has:
  - less pressure
  - less oxygen
  - less glucose
  - fewer / more, amino acids / fatty acids
  - less water / lower water potential / lower solute potential / higher osmotic pressure / higher concentration of solutes and / or rbcs:
  - A ‘blood is more concentrated’ fewer ions
  - more of named cell product
  - more urea
The percentage saturation of haemoglobin with oxygen decreases as the partial pressure of carbon dioxide increases. Explain how this happens.
1. hydrogen ions / protons; A H+
2. either react or combine with haemoglobin / form haemoglobinic acid / form HHb; A ‘picks up’ / absorb or carbon dioxide combines with hemoglobin / forms carboxyhem