unit 1

Question 1

What is hydrolysis?

Answer 1

1. Breaks a chemical bonds between two molecules;
2. Using water;

Question 2

What is a condensation reaction?

Answer 2

1. Creates a chemical bond 2. Removing a molecule of water

Question 3

Glycogen - Glycogen Structure (3)

Answer 3

1. Polysaccharide of α-glucose;
2. (Joined by) glycosidic bonds;
3. Branched structure

Question 4

Glycogen - Glycogen compared with cellulose (4)

Answer 4

1. Cellulose is made up of β-glucose (monomers) and glycogen is made up of α-glucose (monomers);
2. Cellulose molecule has straight chain and glycogen is branched;
3. Cellulose molecule has straight chain and glycogen is coiled;
4. Glycogen has 1,4- and 1,6- glycosidic bonds and cellulose has only 1,4- glycosidic bonds

Question 5

Glycogen - Glycogen structure related to function (5)

Answer 5

1. Insoluble (in water), so doesn’t affect water potential;
2. Branched / coiled / (α-)helix, so makes molecule compact;
3. Polymer of (α-)glucose so provides glucose for respiration;
4. Branched / more ends for fast breakdown / enzyme action;
5. Large (molecule), so can’t cross the cell membrane

Question 6

Test for reducing sugar

Answer 6

1. Heat with Benedict’s reagent (1);
2. colour change from blue to brick-red (1)

Question 7

Starch – Relate 3 properties to its function (6)

Answer 7

1. Insoluble; 2. Don’t affect water potential;
2. Helical; 4. Compact;
3. Large molecule; 6. Cannot leave cell

Question 8

Test for a non reducing sugar

Answer 8

1. Heat with Benedict’s reagent and no colour change (1);
2. boil with acid (HCl) and then neutralise with (NaHCO3) (1);
3. re- heat with Benedict’s reagent and colour change from blue to brick-red (1)

Question 9

Test for starch

Answer 9

1. Add iodine in potassium iodide solution (1);
2. colour change from brown to blue-black (1)

Question 10

How are triglycerides formed

Answer 10

1. One glycerol and three fatty acids;
2. Condensation(reactions) and removal of three molecules of water; 3. Ester bond(s) (formed)

Question 11

Phospholipids compared with Triglycerides (8)

Answer 11

1. Both contain ester bonds (between glycerol and fatty acid);
2. Both contain glycerol;
3. Fatty acids on both may be saturated or unsaturated;
4. Both are insoluble in water;
5. Both contain C, H and O but phospholipids also contain P;
6. Triglyceride has three fatty acids and phospholipid has two fatty acids plus phosphate group;
7. Triglycerides are hydrophobic/non-polar and phospholipids have hydrophilic/polar and hydrophobic/polar region;
8. Phospholipids form monolayer (on surface)/micelle/bilayer (in water) but triglycerides don’t;

Question 12

Describe how an ester bond is formed in a phospholipid molecule.

Answer 12

1. Condensation reaction
2. Between glycerol and fatty acid

Question 13

Test for a lipid

Answer 13

1. (Mix / shake sample) with ethanol, then water; Sequence is important
2. White / milky (emulsion);

Question 14

Protein - Protein Structure (7)

Answer 14

1. Polymer of amino acids;
2. Joined by peptide bonds;
3. Formed by condensation;
4. Primary structure is order of amino acids;
5. Secondary structure is folding of polypeptide chain due to hydrogen bonding; (into alpha helix or beta pleated sheet)
6. Tertiary structure is 3-D folding due to hydrogen bonding and ionic/disulphide bonds between R groups;
7. Quaternary structure is more than one polypeptide chains;

Question 15

Test for a protein

Answer 15

1. Add Biuret reagent to the sample (1);
2. colour change to lilac (1) (or lilac band appears

Question 16

Enzymes – “Induced Fit” Model (3)

Answer 16

1. (before reaction) active site not complementary to/does not fit substrate;
2. Shape of active site changes as substrate binds/as enzyme-substrate complex forms;
   3.Stressing/distorting/bending bonds (in substrate leading to reaction);

Question 17

Enzymes – Increased temperature and reaction rate (4)

Answer 17

1. particles have more kinetic energy
2. therefore they move more
3. so there are more collisions between substrates and active sites
4. so more ES complexes form

Question 18

Enzymes – Effect of Changes in pH (4)

Answer 18

1. Ionic bonds holding tertiary structure break
2. active site distorts and substrate no longer binds to active site
3. charges on amino acids in active site affected
4. fewer ES complexes form

Question 18

Enzymes – Denaturation (5)

Answer 18

1. Heat above the optimum breaks hydrogen bonds
2. this causes the tertiary structure to unfold
3. so the active site changes shape
4. substrate can no longer bind to the active site, as it’s no longer complementary
5. so fewer ES complexes form

Question 19

Enzymes – Concentration of substrate (2)

Answer 19

1. (Rate of) increase in concentration of product slows as substrate is used up OR High initial rate as plenty of substrate/more E-S complexes;
2. No increase after 25 minutes/at end/levels off because no substrate left; Reject ref. to enzyme being used up

Question 20

Enzymes – Describe and explain the temperature graph of enzyme rate (5)

Answer 20

1. Initial rate of reaction faster at 37 °C (than 25 °C); 2. Because more kinetic energy;
2. So more E–S collisions/more E–S complexes formed;
3. Graph reaches plateau /levels off at 37 °C;
4. Because all substrate used up;

Question 21

Enzymes – Comparison of Competitive and Non Competitive inhibition (4)

Answer 21

1. Competitive inhibitor binds to active sites of enzyme but non-competitive inhibitor binds at allosteric site/away from active site;
2. (Binding of) competitive inhibitor does not cause change in shape of active site but (binding of) non-competitive does (cause change in size of active site); 3. So with competitive inhibitor, at high substrate concentrations (active) enzyme still available but with noncompetitive inhibitor (active) enzymes no longer available;
3. At higher substrate concentrations likelihood of enzyme-substrate collisions increases with competitive inhibitor but this is not possible with non-competitive inhibitor;