AQA Bio A Level Unit 1 Flashcards

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1
Q

Describe the primary structure of all proteins. (2)

A

(a) 1. Sequence/order of amino acids;

  1. (Joined by) peptide bonds;
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2
Q

What is hydrolysis?

A
  1. Breaks a chemical bonds between two molecules;
  2. Using water;
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3
Q

What is a condensation reaction?

A
  1. Creates a chemical bond
  2. Removing a molecule of water
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4
Q

Glycogen - Describe the structure of Glycogen (3)

A
  1. Polysaccharide of α-glucose;
  2. (Joined by) glycosidic bonds;
  3. Branched structure
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5
Q

Glycogen - Glycogen structure
related to function (5)

A
  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
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6
Q

Glycogen - Glycogen compared
with cellulose (4)

A
  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
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7
Q

Starch – Relate 3 properties to its
function (3)

A
  1. Insoluble;
  2. Doesn’t affect water potential;
  3. Helical;
  4. Compact;
  5. Large molecule;
  6. Cannot leave cell
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8
Q

Test for reducing sugar (2)

A
  1. Heat with Benedict’s reagent (1);
  2. colour change from blue to brick-red (1)
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9
Q

Test for a non reducing sugar (3)

A
  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)
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10
Q

Test for starch (2)

A
  1. Add iodine in potassium iodide solution (1);
  2. colour change from brown to blue-black (1)
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11
Q

How are triglycerides formed (3)

A
  1. One glycerol and three fatty acids;
  2. Condensation(reactions) and removal of three molecules of water;
  3. Ester bond(s) (formed);
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12
Q

Sucrose is a disaccharide formed by a condensation reaction between which monosaccharides? (2)

A
  1. Glucose;
  2. Fructose;
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13
Q

How is cellulose formed? (3)

A
  1. Condensation (reactions) and removal of water;
  2. Many β-glucose monomers;
  3. Glycosidic bonds
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14
Q

Test for a lipid (2)

A
  1. (Mix / shake sample) with ethanol, then water;
    Sequence is important
  2. White / milky (emulsion);
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15
Q

Protein - Protein Structure (7)

A
  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;
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16
Q

Test for a protein (2)

A
  1. Add Biuret reagent to the sample (1);
  2. colour change to lilac (1)
17
Q

Enzymes – “Induced Fit” Model (3)

A
  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);
18
Q

Enzymes – Increased temperature
and reaction rate (4)

A
  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
19
Q

Enzymes – Denaturation (5)

A
  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
20
Q

Enzymes – Effect of Changes in pH
(4)

A
  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
21
Q

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

A
  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;
  4. At higher substrate concentrations likelihood of enzyme-substrate collisions increases with competitive inhibitor
    but this is not possible with non-competitive inhibitor;
22
Q

Describe DNA replication (6)

A
  1. DNA helicase unwinds DNA/double helix
    OR
    DNA helicase breaks hydrogen bonds;
  2. Both strands act as templates;
  3. (Free DNA) nucleotides line up in complementary pairs/Adenine-Thymine and Guanine-Cytosine;
  4. DNA polymerase joins nucleotides (of new strand);
    Reject forms hydrogen bonds/joins bases
  5. Forming phosphodiester bonds;
  6. Each new DNA molecule consists of one old/original/template strand and one new strand;
23
Q

Describe the structure of DNA (5)

A
  1. Polymer of nucleotides;
    Accept ‘Polynucleotide’
    Accept for ‘phosphate’. phosphoric acid
  2. Each nucleotide formed from deoxyribose, a phosphate (group) and an organic/nitrogenous base;
  3. Phosphodiester bonds (between nucleotides);
  4. Double helix/2 strands held by hydrogen bonds;
  5. (Hydrogen bonds/pairing) between adenine, thymine and cytosine, guanine;
24
Q

Describe and explain how the structure of DNA allows accurate replication (4)

A

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;

25
Q

Describe how a phosphodiester bond is formed between two nucleotides in a DNA molecule. (3)

A
  1. condensation reaction/loss of water
  2. (between) phosphate and deoxyribose
  3. (catalysed by) DNA Polymerase
26
Q

ATP – Uses and properties as an energy source (5)

A
  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;
27
Q

ATP – Structure compared with DNA nucleotide (3)

A
  1. 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);
28
Q

Water – Properties that make water important for organisms (6)

A
  1. A metabolite in condensation/hydrolysis/photosynthesis/respiration; 2. A solvent so (metabolic) reactions can occur;
  2. High heat capacity so buffers changes in temperature;
  3. Large latent heat of vaporisation so provides a cooling effect (through evaporation);
  4. Cohesion (between water molecules) so supports columns of water (in plants);
  5. Cohesion (between water molecules) so produces surface tension supporting organisms;
29
Q
A
30
Q
A