2.1 - 2.5, 8.1 Molecular Biology Extended Response Q. Flashcards
Distinguish between fibrous and globular proteins with reference to one example of each protein type. ( 6 Marks )
- 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
Role of condensation and hydrolysis in relationship of amino acids and dipeptides. ( 4 Marks )
Condensation - dehydration synthesis, water produced (amino acid + amino acid) - Amino Acid -> dipeptide
Hydrolysis - water needed to break bond - Dipeptide -> amino acid
Example of proteins with different numbers of polypeptide chains.
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
Describe the structure of protein. ( 9 Marks )
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
List four functions of protein - named example ( 4 Marks )
transport - hemoglobin hormones - insulin, TSH receptors - neurotransmitter receptor defense - immunoglobin structure - collagen active transport - sodium potassium pump facilitated diffusion - sodium channels
List three functions of Lipids ( 3 marks )
- protection of internal organs
- structural component of cell membranes
- glycolipids act as receptors
- heat insulation
- energy storage
Describe the significance of water to living organisms ( 6 Marks )
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
Describe the use of carbohydrates and lipids for energy storage in animals ( 5 marks )
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
Describe the significance of polar and non-polar amino acids ( 5 Marks )
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
Outline the thermal, cohesive and solvent properties of water. ( 5 Marks )
- 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
Explain the reason for converting lactose to glucose and galactose during food processing. ( 3 Marks )
- 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
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 )
- less denaturation/ enzymes last longer at low temperatures
- lower energy cost
- reduces bacterial growth
- control rate of reaction
Outline one industrial use of lactase. ( 5 Marks )
- 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
Outline how enzymes catalyse reactions. ( 7 Marks )
- 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;
Effects of pH on enzyme activity. ( 3 Marks )
- 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
Compare induced fit model of enzyme activity with lock and key model ( 4 Marks )
- 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
Effects of a competitive inhibitor on enzyme activity. ( 6 Marks )
- 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
Structure of triglycerides ( 6 Marks )
- 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
Explain how proteins act as enzymes, including control by feedback inhibition in allosteric enzymes. ( 9 Marks )
- 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
