Unit 1 Flashcards
1
Q
What is hydrolysis?
A
- Breaks a chemical bonds between two molecules;
- Using water;
2
Q
What is a condensation reaction?
A
Creates a chemical bond
2. Removing a molecule of water
3
Q
Glycogen - Glycogen Structure (3)
A
- Polysaccharide of α-glucose; 2. (Joined by) glycosidic bonds; 3. Branched structure
4
Q
Glycogen - Glycogen compared with cellulose (4)
A
- Cellulose is made up of β-glucose (monomers) and glycogen is made up of α- glucose (monomers);
- Cellulose molecule has straight chain and glycogen is branched;
- Cellulose molecule has straight chain and glycogen is coiled;
- Glycogen has 1,4- and 1,6- glycosidic bonds and cellulose has only 1,4- glycosidic bonds
5
Q
Glycogen - Glycogen structure related to function (5)
A
- Insoluble (in water), so doesn’t affect water potential;
- Branched / coiled / (α-)helix, so makes molecule compact;
- Polymer of (α-)glucose so provides glucose for respiration;
- Branched / more ends for fast breakdown / enzyme action;
- Large (molecule), so can’t cross the cell membrane
6
Q
Starch – Relate 3 properties to its function (6)
A
- Insoluble;
- Don’t affect water potential;
- Helical; 4. Compact;
- Large molecule; 6. Cannot leave cell
7
Q
Test for reducing sugar
A
Heat with Benedict’s reagent (1);
2. colour change from blue to brick-red (1)
8
Q
Test for a non reducing sugar
A
- Heat with Benedict’s reagent and no colour change (1);
- boil with acid (HCl) and then neutralise with (NaHCO3) (1);
- re- heat with Benedict’s reagent and colour change from blue to brick-red (1)
9
Q
Test for starch
A
- Add iodine in potassium iodide solution (1); 2. colour change from brown to blue-black (1)
10
Q
How are triglycerides formed
A
- One glycerol and three fatty acids;
- Condensation(reactions) and removal
of three molecules of water; 3. Ester bond(s) (formed);
11
Q
Phospholipids compared with Triglycerides (8)
A
- Both contain ester bonds (between glycerol and fatty acid);
- Both contain glycerol;
- Fatty acids on both may be saturated or unsaturated;
- Both are insoluble in water;
- Both contain C, H and O but phospholipids also contain P;
- Triglyceride has three fatty acids and phospholipid has two fatty acids plus phosphate group;
- Triglycerides are hydrophobic/non-polar and phospholipids have hydrophilic/polar and hydrophobic/polar
region; - Phospholipids form monolayer (on surface)/micelle/bilayer (in water) but triglycerides don’t;
12
Q
Describe how an ester bond is formed in a
phospholipid molecule.
A
- Condensation reaction
- Between glycerol and fatty acid
13
Q
Test for a lipid
A
- (Mix / shake sample) with ethanol, then water;
Sequence is important - White / milky (emulsion);
14
Q
Protein - Protein Structure (7)
A
- Polymer of amino acids;
- Joined by peptide bonds;
- Formed by condensation;
- Primary structure is order of amino acids;
- Secondary structure is folding of polypeptide chain due to hydrogen bonding; (into alpha helix or beta pleated sheet)
- Tertiary structure is 3-D folding due to hydrogen bonding and ionic/disulphide bonds between R groups;
- Quaternary structure is more than one polypeptide chains;
15
Q
Test for a protein
A
- Add Biuret reagent to the sample (1); 2. colour change to lilac (1) (or lilac band appears
16
Q
Enzymes – “Induced Fit” Model (3)
A
- (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);
17
Q
Enzymes – Increased temperature and reaction rate (4)
A
- particles have more kinetic energy
- therefore they move more
- so there are more collisions between substrates and active sites
- so more ES complexes form
18
Q
Enzymes – Denaturation (5)
A
- Heat above the optimum breaks hydrogen bonds
- this causes the tertiary structure to unfold 3. so the active site changes shape
- substrate can no longer bind to the active site, as it’s no longer complementary
- so fewer ES complexes form
19
Q
Enzymes – Effect of Changes in pH (4)
A
- Ionic bonds holding tertiary structure break 2. active site distorts and substrate no longer binds to active site
- charges on amino acids in active site affected 4. fewer ES complexes form
20
Q
Enzymes – Concentration of Substrate (2)
A
- (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; - No increase after 25 minutes/at end/levels off because no substrate left;
Reject ref. to enzyme being used up
21
Q
Enzymes – Describe and explain the temperature graph of enzyme rate (5)
A
- Initial rate of reaction faster at 37 °C (than 25 °C);
- Because more kinetic energy;
- So more E–S collisions/more E–S complexes formed;
- Graph reaches plateau /levels off at 37 °C; 5. Because all substrate used up;
22
Q
Enzymes – Comparison of Competitive and Non Competitive Inhibition (4)
A
- Competitive inhibitor binds to active sites of enzyme but non-competitive inhibitor binds at allosteric site/away
from active site; - (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); - So with competitive inhibitor, at high substrate concentrations (active) enzyme still available but with noncompetitive inhibitor (active) enzymes no longer available;
- At higher substrate concentrations likelihood of enzyme-substrate collisions increases with competitive inhibitor
but this is not possible with non-competitive inhibitor;
23
Q
Describe DNA replication (6)
A
- DNA helicase unwinds DNA/double helix
OR
DNA helicase breaks hydrogen bonds; - Both strands act as templates;
- (Free DNA) nucleotides line up in
complementary pairs/Adenine-Thymine and Guanine-Cytosine; - DNA polymerase joins nucleotides (of new strand);
Reject forms hydrogen bonds/joins bases - Forming phosphodiester bonds;
- Each new DNA molecule consists of
one old/original/template strand and one new strand;
24
Q
Describe the structure of DNA (5)
A
- Polymer of nucleotides;
Accept ‘Polynucleotide’
Accept for ‘phosphate’. phosphoric acid - Each nucleotide formed from deoxyribose, a phosphate (group) and an organic/nitrogenous base;
- Phosphodiester bonds (between nucleotides);
- Double helix/2 strands held by hydrogen bonds;
- (Hydrogen bonds/pairing) between adenine,
thymine and cytosine, guanine;
25
Describe and explain how the structure of DNA allows accurate replication (4)
1 two strands therefore semi- conservative replication (possible);
2 base pairing / hydrogen bonds holds strands together
3 hydrogen bonds weak / easily broken, allow strands to separate;
4 bases (sequence) (exposed so) act as template / can be copied;
5 A with T, C with G / complementary copy;
6 DNA one parent and one new strand;
26
Describe how a phosphodiester bond is formed between two nucleotides in a DNA molecule. (3)
1. condensation reaction/loss of water
2. (between) phosphate and deoxyribose 3. (catalysed by) DNA Polymerase
27
ATP – Uses and properties as an energy source (5)
1. Releases relatively small amount of energy / little energy lost as heat;
2. Releases energy instantaneously;
3. Phosphorylates other compounds, making them more reactive;
4. Can be rapidly re-synthesised; 5. Does not leave cells;
28
ATP– Structure compared with DNA nucleotide (3)
. ATP has ribose and DNA has deoxyribose;
2. ATP has 3 phosphates and DNA nucleotide has one phosphate;
3. Base is always adenine in ATP and bases vary in DNA nucleotide (A,C,G or T);
29
Water – Properties that make water important for organisms (6)
1. A metabolite in condensation/hydrolysis/photosynthesis/respiration; 2. A solvent so (metabolic) reactions can occur;
3. High heat capacity so buffers changes in temperature;
4. Large latent heat of vaporisation so provides a cooling effect (through evaporation);
5. Cohesion (between water molecules) so supports columns of water (in plants);
6. Cohesion (between water molecules) so produces surface tension supporting organisms;
