2.1.1 Biological Molecules

Transcription

1 2.1.1 Biological Molecules Relevant Past Paper Questions Paper Question Specification point(s) tested 2013 January 4 parts c and d p r 2013 January 6 except part c j k m n o 2012 June 1 part ci d e f g 2012 June 4 l m r 2012 January 1 a b c 2012 January 3 part bi j 2011 June 3 c f j k l m o p q 2011 January 2 parts a and b a r s 2011 January 7 d e f g i 2010 June 1 p q r 2010 June 4 part c d f 2010 January 2 a c q 2010 January 4 b c f h i n 2009 January 3 parts a and d k l m s Condensed Notes By Specification Point a) Describe how hydrogen bonding occurs between water molecules, and relate this, and other properties of water, to the roles of water in living organisms. Hydrogen bonds form between oxygen and hydrogen atoms.

2 When oxygen is covalently bonded to hydrogen (as it is in a water molecule), the shared electrons are slightly closer on average to the oxygen atom than they are to the hydrogen atom. This is because oxygen is more electronegative (better at attracting electrons) than hydrogen. This results in the oxygen having a slight negative charge (δ ) and the hydrogen having a slightly negative charge (δ + ). The δ oxygen atom on one water molecule is attracted to a δ + hydrogen atom on another water molecule, forming a hydrogen bond. Hydrogen bonds are always drawn a dashed lines. They form between the oxygen atom on one water molecule and a hydrogen atom on another water molecule. There is always a straight line between the two atoms involved in the hydrogen bond and one of the hydrogen atoms that is covalently bonded to the oxygen. Water has a high latent heat of vaporisation (a large amount of energy is required to change it from a liquid to a gas). This means that evaporation is an efficient cooling mechanism, for example sweating, panting and transpiration are all effective ways that organisms cool themselves. Water has a high specific heat capacity (a large amount of energy is needed to raise its temperature), meaning that it forms a thermally stable environment for aquatic organisms and the internal temperature of organisms changes only slowly. Ice is less dense than liquid water and therefore floats on top of it. This insulates the water below, preventing it from freezing. This means that aquatic organisms can still swim in the water. Water is an effective solvent, meaning that it can act as a medium for reaction and that it can be used as an internal transport medium (e.g. blood plasma, phloem sap). Water exhibits cohesion and adhesion. These properties have many uses in living organisms, including allowing water to move up the transpiration stream. Water is transparent, which allows photosynthesis to occur underwater. b) Describe, with the aid of diagrams, the structure of an amino acid. H 2 NCHRCOOH An amino acid consists of a central carbon atom to which a hydrogen atom, an amine group (NH 2 ), a carboxyl group (COOH) and an R group are attached. The R group is different in every amino acid. c) Describe, with the aid of diagrams, the formation and breakage of peptide bonds in the synthesis and hydrolysis of dipeptides and polypeptides. A peptide bond is the covalent bond that is formed when two amino acids are joined together. A peptide bond forms between the amine group of one amino acid and the carboxyl group of another. The H atom from the amine group combines with the OH from the carboxyl in a condensation reaction (a reaction in which water is produced). A polypeptide is polymer formed of a chain of amino acids.

3 Hydrolysis is when water is added to break a bond in a molecule. d) Explain, with the aid of diagrams, the term primary structure. The sequence of amino acids, joined by peptide bonds. e) Explain, with the aid of diagrams, the term secondary structure with reference to hydrogen bonding. Secondary structure consists of alpha helices and beta pleated sheets. These features of secondary structure are held together by hydrogen bonds. f) Explain, with the aid of diagrams, the term tertiary structure, with reference to hydrophobic and hydrophilic interactions, disulphide bonds and ionic interactions. Tertiary structure is the further folding of the secondary structure. The tertiary structure is held together by a variety of interactions including hydrophobic and hydrophilic interactions, disulphide bridges and ionic interactions. If the protein is water soluble (e.g. proteins that act in the cytoplasm or within organelles), then hydrophilic R groups are on the outside of the protein and hydrophobic R groups are on the inside of the molecule. Ionic interaction: An R group that is a positive ion and an R group that is a negative ion are electrostatically attracted towards each other. Disulphide bridge: The sulphur atoms of two cysteine R groups bond together covalently. g) Explain, with the aid of diagrams, the term quaternary structure, with reference to the structure of haemoglobin. Haemoglobin is made up of four subunits, each of which is a polypeptide. These four polypeptides are made up of two pairs of identical polypeptides: the alpha chains and the beta chains. Each polypeptide also has one prosthetic group called haem, which contains Fe 2+. h) Describe, with the aid of diagrams, the structure of a collagen molecule. Collagen is a protein. It is made up of different amino acid units joined by peptide bonds in unbranched, helical chains. There are three of these chains in every collagen molecule and there are cross links between the chains. i) Compare the structure and function of haemoglobin (as an example of a globular protein) and collagen (as an example of a fibrous protein). Function of collagen: provide strength (e.g. in ligaments that hold bones together). Properties of collagen that enable function: High tensile strength. Is inelastic. Is insoluble. Is flexible.

4 Similarities: Both contain chains of amino acids joined by peptide bonds. Both contain helical structures. Both contain hydrogen bonds, disulphide bridges, ionic interactions, hydrophobic interactions and hydrophilic interactions. They are both made up of more than one polypeptide (they have quaternary structure). j) Describe, with the aid of diagrams, the molecular structure of alpha glucose as an example of a monosaccharide carbohydrate. Glucose is a monosaccharide. k) State the structural difference between alpha and beta glucose. The structural difference between alpha and beta glucose is that the hydroxyl group (OH) on carbon one is inverted. Position of hydroxyl groups on alpha glucose: C1 down, C2 down, C3 up, C4 down. Position of hydroxyl groups on beta glucose: C1 up, C2 down, C3 up, C4 down. l) Describe, with the aid of diagrams, the formation and breakage of glycosidic bonds in the synthesis and hydrolysis of a disaccharide (maltose) and a polysaccharide (amylose). A polysaccharide is many monosaccharide units joined together by glycosidic bonds. Glycosidic bonds are broken by hydrolysis (the addition of water). m) Compare and contrast the structure and functions of starch (amylose) and cellulose. Orientation of glucose units? Similarities: Insoluble. Contain only carbon, hydrogen and oxygen. Subunits are linked to form chains by 1 4 glycosidic bonds formed by condensation reactions. Differences: Starch is made of alpha glucose monomers, whereas cellulose is made of beta glucose monomers. Starch has 1 6 glycosidic bonds, whereas cellulose does not. In starch, the chains are branched, whereas in cellulose they are not. n) Describe, with the aid of diagrams, the structure of glycogen. Glycogen is a polysaccharide carbohydrate composed of alpha glucose units joined by glycosidic bonds to form branched chains. o) Explain how the structures of glucose, starch (amylose), glycogen and cellulose molecules relate to their functions in living organisms.

5 Glucose: Function is as a respiratory substrate (is respired to release energy that can be used in the formation of ATP). Soluble so that it can be easily transported around an organism. Small molecule meaning that it can diffuse across cell membranes. It is easy to break down to release energy. Glucose molecules can join together to form disaccharides and polysaccharides. Glycogen: Function is to act as an energy storage molecule. Glycogen is insoluble, which means that it does not affect the water potential of the cell that it is in. Glycogen can be quickly broken down and built up. It has a lot of branches for enzymes to attach to. Glycogen is compact, and therefore energy dense. p) Compare, with the aid of diagrams, the structure of a triglyceride and a phospholipid. Similarities: Both contain glycerol. Both contain fatty acids. Both contain ester bonds. Both are made of carbon, hydrogen and oxygen. Differences: Triglyceride has three fatty acids, whereas phospholipid has two fatty acids. Triglyceride has three ester bonds (between glycerol and each fatty acid), whereas phospholipid has two ester bonds. Triglyceride does not contain phosphate, whereas phospholipid does contain phosphate. q) Explain how the structures of triglyceride, phospholipid and cholesterol molecules relate to their functions in living organisms. Roles of lipids: Energy storage/respiratory substrate Make up part of membranes Absorption of fat soluble vitamins Electrical insulation Thermal insulation Protection of organs Source of steroid hormones Source of cholesterol Waterproofing Source of vitamin D Buoyancy From January 2013 Q4 copied from notes remove from their? Add it to this page s question list? r) Describe how to carry out chemical tests to identify the presence of the following molecules: protein (biuret test), reducing and non reducing sugars (Benedict s test), starch (iodine solution) and lipids (emulsion test).

6 Protein: Do biuret I and II tests. Turns lilac colour if protein is present. Reducing sugar: Add Benedict s reagent and heat. If reducing sugar is present then the solution changes colour from blue to red as a red precipitate forms. The amount of reducing sugar present can be estimated by comparing the colour of the precipitate to the colour of solutions containing known amounts of reducing sugar that have been tested. Non reducing sugar: Add hydrochloric acid and boil. Add alkali (e.g. sodium hydrogen carbonate). Then carry out reducing sugar test. Starch: Iodine solution. The iodine solution turns dark blue or black in the presence of starch. Lipids: Emulsion test. Add ethanol to the sample and mix thoroughly by stirring. Add water. Mix the sample with water. Add Sudan III stain. Mix thoroughly by stirring. If lipid is present, the mixture turns cloudy as a white emulsion forms. s) Describe how the concentration of glucose in a solution may be determined using colorimetry. Using known concentrations of a reducing sugar. Hear with Benedict's solution. Use the same volumes of solutions each time. Use an excess of Benedict s. The colour changes from blue to green/yellow/orange/brown/(brick) red. Filter out the precipitate and keep the filtrate. Zero the colorimeter using water. Use a red filter. Measure the transmission. The higher the transmission, the more sugar is present. Obtain a calibration curve by plotting the transmission against the reducing sugar concentration. Measure the transmission of the unknown sugar concentration and read off the graph to find its concentration.