2.1 - 2.5, 8.1 Molecular Biology Extended Response Q. Flashcards

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1
Q

Distinguish between fibrous and globular proteins with reference to one example of each protein type. ( 6 Marks )

A
  • fibrous proteins are strands whereas globular proteins are rounded
  • fibrous insoluble whereas globular soluble
  • globular more sensitive to change in pH/temperature/salt
  • fibrous proteins have structural roles
  • globular for catalysis and transport
  • name fibrous- keratin, fibrin, collagen, actin, myosin
  • name globular- insulin, hemoglobin, immunoglobin
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2
Q

Role of condensation and hydrolysis in relationship of amino acids and dipeptides. ( 4 Marks )

A

Condensation - dehydration synthesis, water produced (amino acid + amino acid) - Amino Acid -> dipeptide
Hydrolysis - water needed to break bond - Dipeptide -> amino acid

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3
Q

Example of proteins with different numbers of polypeptide chains.

A

1 - Lysozyme - enzyme in secretion i.e nasal mucus - kills bacteria by digesting peptidoglycan in their cell walls
2 - Integrin - membrane protein - makes connections between structures
3- collagen - structural protein - high tensile strength found in skin and blood vessels
4 - Hemoglobin - transport protein in red blood cells; it binds oxygen in the lungs and releases it in the tissue

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4
Q

Describe the structure of protein. ( 9 Marks )

A

PRIMARY STRUCTURE - 1. chains of amino acids, 2. each position occupied by one of 20 different amino acids. 3. linked by peptide bonds

SECONDARY STRUCTURE - 1. interaction of amine and carboxyl group, 2. hydrogen bonds form, 3. a helix forms and polypeptide coils up or ß pleated sheet forms

TERTIARY STRUCTURE - 1. folding up of the polypeptide, 2. stabilised by disulfide bridges/hydrogen/ionic/hydrophobic bond

QUATERNARY STRUCTURE - 1. several polypeptides subunits join 2. conjugated proteins combine with non - proteins molecules i.e metals, nucleic acid,carbohydrates, lipids

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5
Q

List four functions of protein - named example ( 4 Marks )

A
transport - hemoglobin 
hormones - insulin, TSH 
receptors - neurotransmitter receptor 
defense - immunoglobin 
structure - collagen 
active transport - sodium potassium pump
facilitated diffusion - sodium channels
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6
Q

List three functions of Lipids ( 3 marks )

A
  • protection of internal organs
  • structural component of cell membranes
  • glycolipids act as receptors
  • heat insulation
  • energy storage
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7
Q

Describe the significance of water to living organisms ( 6 Marks )

A

surface tension - allows organisms i.e insects to move on water
polarity/ adhesion - helps plants transport water
transparency - allows plants to photosynthesize in water
excellent solvent - capable of dissolving substances for transport in organism
high heat of vaporization - coolant
structure - hydrostatic pressure in plant cells

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8
Q

Describe the use of carbohydrates and lipids for energy storage in animals ( 5 marks )

A

carbohydrates
- stored as glycogen in liver
- short term energy storage
- easily digested so energy quickly available
- more soluble in water so easier transport
Lipids
- stored as fat in animals
- long term energy storage
- more energy per gram
- lipids insoluble in water so less osmotic effect

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9
Q

Describe the significance of polar and non-polar amino acids ( 5 Marks )

A

Polar amino acids

  • hydrophilic
  • can make hydrogen bonds
  • found in hydrophilic channels
  • found in surface of water-soluble protein

Non-polar amino acids

  • hydrophobic
  • found in protein in interior of membranes
  • found in interior of water-soluble proteins
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10
Q

Outline the thermal, cohesive and solvent properties of water. ( 5 Marks )

A
  • high specific heat capacity - large amount of heat causes small increase in temp
  • high latent heat of vaporization - large amount of heat energy needed to evaporate water
  • hydrogen bonds between water make them cohesive ; stick together - gives water high surface tension, explains how water rises up the xylem
  • water molecules are are polar - making water a good solvent
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11
Q

Explain the reason for converting lactose to glucose and galactose during food processing. ( 3 Marks )

A
  • allows people with lactose intolerance consume milk
  • galactose and glucose tase sweeter than lactose
  • galactose and glucose more soluble than lactose ( smoother texture )
  • bacteria ferment glucose and galactose more rapidly shortening production time
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12
Q

Simple laboratory experiments show that when enzyme lactase is mixed with lactose the initial rate is highest at 48 degrees. In food processing, lactose is used at 5 degrees. Why lower temp in food processing? ( 2 Marks )

A
  • less denaturation/ enzymes last longer at low temperatures
  • lower energy cost
  • reduces bacterial growth
  • control rate of reaction
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13
Q

Outline one industrial use of lactase. ( 5 Marks )

A
  • lactose intolerance high in some human pop. - asian, African , native American
  • lactase used to produce lactose free milk
  • lactase breaks down lactose to glucose and galactose
  • source of lactase usually yeast
  • milk passed over immobilized lactase
  • increases sweetness no need for extra sugar
  • can add bacteria L.acidophilus which same effect on lactose as in yoghurt
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14
Q

Outline how enzymes catalyse reactions. ( 7 Marks )

A
  • they increase rate of reaction
  • remain unchanged at end of reaction
  • lower activation energy - activation energy is energy needed to overcome energy barrier that prevents reaction
  • substrate joins with enzyme at active site
  • forms enzyme-substrate complex
  • active site - enzyme specific for particular substrate
  • enzyme binding with substrate brings reactants closer together to facilitate chemical reactions (such as electron transfer)
  • induced fit model / change in enzyme conformation (when enzyme-substrate/ES complex forms);
    making the substrate more reactive;
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15
Q

Effects of pH on enzyme activity. ( 3 Marks )

A
  • enzymes have an optimal pH
  • lower activity above and below optimum pH / graph showing this
  • too acidic / base pH can denature enzyme
  • change shape of active site / tertiary structure altered
  • substrate cannot bind to active site / enzyme-substrate complex cannot form
    hydrogen / ionic bonds in the enzyme / active site are broken / altered
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16
Q

Compare induced fit model of enzyme activity with lock and key model ( 4 Marks )

A
  • in both models substrate binds to active site
  • substrate fits active site exactly in lock and key, whereas fit is not exact in induced fit
  • substrate / active site changes shape in induced fit, whereas active site does not change shape in lock and key
  • in both models an enzyme - substrate complex is formed
  • in lock and key binding reduces activation energy, whereas in the induced fit change to substrate reduces activation energy
  • lock and key model explains narrow specificity, whereas induced fit allows broader specificity
  • induced fit explains competitive inhibition, whereas lock and key does not
17
Q

Effects of a competitive inhibitor on enzyme activity. ( 6 Marks )

A
  • competitive inhibitor has similar shape/structure to the substrate therefore it fits to the active site
  • no reaction is catalyzed so the inhibitor remains bound
  • substrate cannot bind as long as the inhibitor remains bound
  • only one active site per enzyme molecule
  • substrate and inhibitor compete for the active site
  • therefore high substrate concentrations can overcome the inhibition
    as substrate is used up ratio of inhibitor to substrate rises
  • named example of inhibitor - trypsin and trypsin inhibitor
18
Q

Structure of triglycerides ( 6 Marks )

A
  • composed of C, H and O (must be stated)
  • relatively more C and H/less O than carbohydrates
  • composed of fatty acids and glycerol
  • fatty acids are carboxyl groups with hydrocarbon chain attached/ diagram showing it separately or as part of a triglyceride ester bonds/diagram showing C-O-C=O
  • three fatty acids/hydrocarbon chains linked to each glycerol (must be stated)
  • 12-20 carbon atoms per hydrocarbon tail
  • saturated if all the C-C bonds are single/unsaturated if one or more double bonds
  • whole molecule is nonpolar/hydrophobic
19
Q

Explain how proteins act as enzymes, including control by feedback inhibition in allosteric enzymes. ( 9 Marks )

A
  • enzymes are globular proteins
  • there is an active site
  • substrate(s) binds to active site
  • shape of substrate (and active site) changed / induced fit
  • bonds in substrate weakened
  • activation energy reduced
  • sketch of energy levels in a reaction to show reduced activation energy
    in feedback inhibition a (end) product binds to the enzyme
  • end-product is a substance produced in last / later stage of a pathway
  • modulator / inhibitor / effector / product binds at the allosteric site / site away from the active site
  • binding causes the enzyme / active site to change shape
  • substrate no longer fits the active site
  • the higher the concentration of end-product the lower the enzyme activity
    enzyme catalyzes the first / early reaction in pathway so whole pathway is inhibited
  • prevents build-up of intermediates
  • allosteric inhibition is non-competitive