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  • #3596

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    REVISING H2 Biology FOR ‘A’ LEVELS IN 3 STEPS

    Topic 6 – Cellular Physiology and Biochemistry

    1. Explain the light dependent reactions of photosynthesis

    2. Describe the three phases of the Calvin cycle

    3. Discuss how limiting factors may affect the rate of photosynthesis

    4. Describe each of the main stages of aerobic respiration.

    5. Explain the production of a small yield of ATP from anaerobic respiration

    6. Describe the formation of ethanol in yeast and lactate in mammals.

    7. Compare the storage and structural forms of starch, glycogen and cellulose

    8. Describe the roles of starch, glycogen and cellulose in plants/animals.

    9. Describe and explain the fluid mosaic model of membrane structure.

    10. Outline the roles and functions of membranes within cells and at
    the surface of cells.

    11. Describe how substances may move across the cell membrane.

    12. Explain the principles of homeostasis.

    13. Explain the need for communication systems within organisms.

    14. Describe and explain the transmission of an action potential along a myelinated neurone.

    15. Describe the structure of a cholinergic synapse and explain how it functions.

    16. Explain what is meant by an endocrine gland, using the islets of Langerhans in the pancreas as an example.

    17. Explain how the blood glucose concentration is regulated by insulin and glucagon.

    18. Describe the main stages of cell signalling

    19. Explain the advantages of a cell signalling system.

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    #3644

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    REVISING H2 Biology FOR ‘A’ LEVELS IN 3 STEPS

    Topic 7 – Diversity and Evolution

    1. Discuss the binomial nomenclature of a species and hierarchical classification.

    2. Describe the classification of species into taxonomic groups.

    3. Explain the various concepts of the species.

    4. Explain how species are formed with reference to geographical isolation, physiological isolation and behavioural isolation.

    5. Explain the relationship between classification and phylogeny.

    6. Explain why variation is important in selection.

    7. Explain, with examples, how environmental factors act as forces of natural selection.

    8. Explain how natural selection may bring about evolution.

    9. Explain why the population is the smallest unit that can evolve.

    10. Explain how homology supports Darwin’s theory of natural selection.

    11. Explain how biogeography and the fossil record support the evolutionary deductions based on homologies.

    12. Explain the importance of the use of genome sequences in reconstructing phylogenetic relationships.

    13. State the advantages of molecular methods in classifying organisms.

    14. Explain how genetic variation may be preserved in a natural population.

    15. Briefly describe the neutral theory of molecular evolution.

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    #3675

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    REVISING H2 Biology FOR ‘A’ LEVELS IN 3 STEPS

    Topic 8 – Isolating, Cloning and Sequencing DNA

    1. Describe the natural function of restriction enzymes.

    2. Outline the procedures for cloning a eukaryotic gene in a bacterial plasmid leading to the production of a named protein.

    3. Explain how the problem of using prokaryotic bacteria to express a eukaryotic protein is solved.

    4. Distinguish between a genomic DNA and cDNA library.

    5. Discuss the polymerase chain reaction.

    6. Discuss the advantages and limitations of the polymerase chain reaction.

    7. Discuss gel electrophoresis.

    8. Outline the process of southern blotting.

    9. Discuss the use of RFLPs in the following:
    a. Genomic mapping
    b. Disease detection
    c. DNA fingerprinting

    10. Discuss the goals, benefits and ethical implications of the Human Genome Project.

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    #3715

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    REVISING H2 Biology FOR ‘A’ LEVELS IN 3 STEPS

    Topic 9 – Applications of Molecular and Cell Biology

    1. Describe the unique features stem cells.

    2. Compare and contrast the various types of stem cells.

    3. Explain the normal functions of stem cells in a living organism

    4. Discuss SCID and explain how it can be treated using gene therapy.

    5. Discuss cystic fibrosis and explain how it can be treated using gene therapy.

    6. Discuss the advantages and disadvantages of the various viral and non-viral delivery methods for gene therapy.

    7. Explain what are the factors that keep gene therapy from becoming an effective treatment for genetic diseases.

    8. Discuss the social and ethical considerations for the use of gene therapy.

    9. Discuss cloning in plants in terms of plant tissue culture techniques.

    10. Using named examples discuss how genetic engineering may improve the quality and yield of crop plants and animals.

    11. Discuss the ethical and social implications of genetically modified crop plants and animals

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    #3733

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    Cell structure and function

    Compare contrast question
    – Take note when compare contrast question is asked, you have to consider BOTH similarities and differences
    – List out the common features that you intend to compare. Do not merely list down all the content that you know.
    – Comparison should point-to-point.

    Question: Name 3 organelles which are surrounded by double membranes. Compare & contrast between the double membranes of these 3 organelles. [7]
    KEWL

    1. What are the keywords in Qns?  Compare contrast; DOUBLE MEMBRANE!!
    (Hence do not waste time writing about the functions of each organelle)
    2. What are the keywords in Ans?  list out features (see below in brown)
    3. When writing in prose, ensure that the comparison is point to point. Do not list out all the features of one organelle in a paragraph and another organelle in another paragraph. Do not expect the marker to do comparison for you. You have to demonstrate ability to point out the comparison for the marker.

    Double membrane
    Nucleus
    Mitochondria
    Chloroplast
     Membrane made of basic phospholipids bilayers with proteins interspersed within them
     Intermembranal space in between the two membranes
     Compartmentalisation: Double membrane separates the interior of the organelle from its surroundings.

    Pores?
    Nucleus – Perforated with pores
    Mitochondria – No pores on membrane
    Chloroplast – No pores on membrane

    Inner membrane?
    Nucleus
     Inner membrane is continuous with the outer membrane
     inner membrane not folded & is smooth
    Mitochondria
     Inner membrane is not continuous with outer membrane
     Inner membrane folded into cristae  increase surface area for enzymatic action to take place
    Chloroplast
     Inner membrane is not continuous with outer membrane
     Inner membrane is generally smooth but gives rise to form fluid-filled sacs called thylakoids which stack up to form grana

    Outer membrane?
    Nucleus
     outer membrane is smooth but is continuous with RER
    Mitochondria
     Outer membrane is smooth and is not continuous with any other organelle
    Chloroplast
     Outer membrane is smooth and is not continuous with any other organelle

    Presence of membrane proteins?
    Nucleus
     relatively less proteins embedded in the inner membra
    Mitochondria
     presence of stalked particles and transport proteins on the inner membrane
    Chloroplast
     Relatively less proteins embedded in the inner membrane
     presence of stalked particles and transport proteins on the thylakoid membranes

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    #3756

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    TOPIC 1: CELLULAR FUNCTIONS – Part 1

    LEARNING OUTCOME

    (a) Describe and interpret drawings and photographs of typical animal and plant cells as seen under the electron microscope, recognising the following membrane systems and organelles: rough and smooth endoplasmic reticulum,Golgi body, mitochondria, ribosomes, lysosomes, chloroplasts, cell surface membrane, nuclear envelope, centrioles, nucleus and nucleolus. (Knowledge of the principles of TEM and SEM are not required.) (For practical assessment, students may be required to operate a light microscope, mount slides and use a graticule.)

    ESSAY ANSWER

    Cell surface membrane
    The cell surface membrane is a single membrane found at the periphery of
    the cell. Vesicles may be seen forming and fusing at it. Some cell membranes may have foldings.

    Rough endoplasmic reticulum
    The rough endoplasmic reticulum (RER) is bound by a single membrane. It
    consists of a system of flattened cisternae that are interconnected with each other and is continuous with the outer membrane of the nuclear envelope. The outer surface of the RER is studded with 80S ribosomes.

    Smooth endoplasmic reticulum
    The smooth endoplasmic reticulum (SER) is bound by a single membrane. It
    consists of a system of tubular cisternae that are interconnected with each other and is continuous with RER. It lacks ribosomes on its surface.

    Golgi body
    The golgi body is bound by a single membrane and consists of a series of
    flattened cisternae. The cis face is where new cisternae are constantly formed by fusion of vesicles from rough endoplasmic reticulum and smooth
    endoplasmic reticulum. The trans face is where the golgi body buds to form
    vesicles such as lysosomes and transport vesicles. The golgi body is not
    continuous with any other organelle.

    Lysosome
    Lysosomes are spherical vesicles bound by a single membrane. Primary
    lysosomes are formed from the golgi body while secondary lysosomes are
    formed when primary lysosomes fuse with endocytic vesicles or with other
    organelles.

    Mitochondria
    Mitochondria are bound by a double membrane. The outer membrane is
    smooth but the inner membrane is folded into cristae. The two membranes
    enclose an intermembrane space. Within the inner membrane is the matrix,
    which contains circular mitochondrial DNA, ribosomes and RNA.

    Chloroplast
    Chloroplasts are bound by a double membrane. Within the chloroplast is a
    system of flattened sacs known as the thylakoid membrane. The membranous
    system forms stacks called granum with intergranal lamellae between grana.
    The thylakoid membrane encloses the thylakoid space. The stroma of the
    chloroplast contains the circular DNA of the chloroplast, 70S ribosomes and enzymes.

    Nucleus
    The nucleus is a large, spherical or ovoid structure present in most eukaryotic cells. It is bound by a double membrane known as the nuclear envelope. Within the nucleus is genetic material in the form of chromatin. Darkly-staining heterochromatin is generally found at the inner periphery of the nucleus. Euchromatin is lightly-staining. The nucleus also contains a dense region known as the nucleolus.

    Nuclear envelope
    The nuclear envelope is a double membrane that has nuclear pores to allow
    for transport of materials into and out of the nucleus. The outer membrane is continuous with the RER. The inner membrane is associated with
    heterochromatin.

    Nucleolus
    The nucleolus is a large and dense region inside the nucleus. It consists of a fibrous part (pars fibrosa) and a granular part (pars granulosa).

    Ribosomes
    Ribosomes are non-membrane bound and consist of one large subunit and
    one small subunit. Ribosomes may be found free in the cytoplasm, attached
    to the rough endoplasmic reticulum or attached to the nuclear envelope.
    80S ribosomes are found in the cytoplasm while 70S ribosomes are present
    in the matrix of mitochondria and stroma of chloroplasts.

    Centrioles
    Centrioles are non-membrane bound and are a pair of rod-like structures
    each made up of 9 triplets of microtubules. Centrioles are positioned
    perpendicular (90°) to each other.

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    #3779

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    TOPIC 1: CELLULAR FUNCTIONS – Part 2

    LEARNING OUTCOME

    (b)Outline the functions of the membrane systems and organelles listed in Cellular Functions in Part 1.

    ESSAY ANSWER

    Cell surface membrane
    The cell surface membrane acts as a boundary to separate the cell contents
    from the external environment. It serves to control exchange of substance
    between the cell and its external environment. The cell surface membrane is also a site for chemical reactions, recognition of external stimuli and
    attachment of other cells, the extracellular matrix and the cytoskeleton.

    Rough endoplasmic reticulum
    The rough endoplasmic reticulum (RER) is associated with ribosomes that
    synthesis proteins via translation. It is also part of the endomembrane
    system, functioning as an intracellular transport network. Proteins that are synthesised by ribosomes bound to it are packaged for transport to other parts of the cell. Some chemical modifications are also made to proteins. The RER provides a large surface area for biochemical reactions to occur.

    Smooth endoplasmic reticulum
    The smooth endoplasmic reticulum (SER) is the site of lipid and steroid
    synthesis. It is also part of the endomembrane system, being involved in the packing of certain proteins from the RER. Some chemical modifications are also made of proteins. The SER is also involved in detoxification reactions and is a site of Cl- and Ca2+ storage.

    Golgi body
    The golgi body is part of the endomembrane system. It chemically modifies
    proteins and lipids such as by attachment of carbohydrate chains to proteins and lipids to form glycoproteins and glycolipids. It also packs these substances into vesicles for storage, for transport to other parts of the cell or for secretion. Lysosomes are formed from the golgi body.

    Lysosome
    Lysosomes contain hydrolytic enzymes within an acidic environment. They
    are involved in digestion of materials taken up by endocytosis and
    phagocytosis, autolysis, autophagy and remodelling of the extracellular
    matrix.

    Mitochondria
    Mitochondria are the sites of ATP synthesis via aerobic respiration. ATP is produced via substrate-level phosphorylation in the matrix via the link
    reaction and Krebs cycle and at the inner mitochondrial membrane via
    oxidative phosphorylation. Lipid metabolism and synthesis also occur in
    mitochondria.

    Chloroplast
    Chloroplasts are the sites of photosynthesis. The light dependent reactions occur at the thylakoid membranes. Light energy is harvested and used to synthesis ATP via photophosphorylation. The light independent reactions occur in the stroma. Carbon dioxide is fixed and reduced to form
    carbohydrates. Starch is also temporarily stored in the stroma.

    Nucleus
    The nucleus contains genetic information in the form of DNA. It controls
    cellular activities including cell division and protein synthesis.

    Nuclear envelope
    The nuclear envelope is perforated by channels known as nuclear pores which regulate the exchange of substance between the nucleus and the rest of the cell.

    Nucleolus
    The nucleolus is the site of transcription of rRNA genes to form rRNA. rRNA is then combined with ribosomal proteins to form ribosomal subunits.

    Ribosomes
    Ribosomes are the site of protein synthesis via translation. Ribosomes read the codons on mRNA and allow base-pairing with complementary tRNA anticodons. The large ribosomal subunit carries peptidyl transferase activity allowing the formation of peptide bonds between amino acids joined to tRNAs.

    Centrioles
    Centrioles are involved in the organisation of spindle fibres during cell
    division, and serve as an anchorage for cilia and flagella.

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    #3791

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    TOPIC 1: CELLULAR FUNCTIONS – Part 3

    LEARNING OUTCOME

    (c)Describe the formation and breakage of a glycosidic bond.

    ESSAY ANSWER

    Formation of an -1,4 glycosidic bond

    A condensation reaction between the hydroxyl group at carbon 1 of 1 -
    glucose and the hydroxyl group at carbon 4 of another -glucose results in the formation of an -1,4 glycosidic bond. 1 water molecule is lost in this reaction.

    Formation of an -1,6 glycosidic bond

    A condensation reaction between the hydroxyl group at carbon 1 of 1 -
    glucose and the hydroxyl group at carbon 6 of another -glucose results in the formation of an -1,6 glycosidic bond. 1 water molecule is lost in this reaction.

    Formation of an -1,4 glycosidic bond

    A condensation reaction between the hydroxyl group at carbon 1 of 1 -
    glucose and the hydroxyl group at carbon 4 of another -glucose results in the formation of -1,4 glycosidic bond. In order the hydroxyl group on carbon 1 to line up alongside with the hydroxyl group on carbon atom 4, every alternate - glucose must be rotated 180o. 1 water molecule is lost in this reaction.

    Breakage of -1,4-glycosidic bond

    A hydrolysis reaction breaks the glycosidic bond between the -glucose
    residues. One water molecule is used in this reaction. This restores the
    hydroxyl group in carbon positions 1 and 4.

    *Note: This answering technique may be applied to the breakage of -1,6
    glycosidic bond and -1,4 glycosidic bond.

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    #3807

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    TOPIC 1: CELLULAR FUNCTIONS – Part 4

    LEARNING OUTCOME

    (d)Analyse the molecular structure of a triglyceride and a phospholipid, and relate these structures to their functions in living organisms.

    ESSAY ANSWER

    Triglycerides
    Triglycerides consist of one glycerol molecule and three fatty acid
    molecules. The glycerol backbone is held to the fatty acids chains by ester linkages formed through condensation reactions.

    The presence of numerous C-C and C-H bonds as well as being compact and
    insoluble allows triglycerides to act as energy stores. Breaking the C-C and CH bonds releases energy. Due to the lower ratio of oxygen, fats yield more energy than an equal mass of carbohydrates. Oxidation of fats also yields metabolic water (twice that of an equal amount of carbohydrates).

    The low density of triglycerides provides buoyancy which is important for
    aquatic animals.

    The hydrophobic, non-polar nature of triglycerides allows them to be stored in large amounts without affecting the water potential of cells. Triglycerides can also be used to reduce water loss from the surfaces of plants and animals, as insulation to prevent loss of heat and for electrical insulation in neurons.

    Phospholipids
    Phospholipids consist of one glycerol molecule, two fatty acid molecules,
    one phosphoric acid molecule and, often, a variable additional group. The
    glycerol backbone is joined to the fatty acid chains by ester linkages to form a fatty acid tail. The glycerol backbone is joined to phosphoric acid to form a phosphate head.

    Phospholipids are major components of cell membranes. They are
    amphipathic due to their hydrophilic phosphate heads and hydrophilic fatty
    acid tails. In an aqueous environment, phospholipids arrange themselves into a bilayer. The heads face outwards, towards the aqueous environment on both sides of the bilayer. The tails face inwards, away from the aqueous environments. This allows them to form effective barriers between cellular contents and the external environment. The variable length and degree of saturation of the fatty acid chains allows the fluidity of the cell membrane to be adjusted. Longer chains and chains with fewer C=C bonds result in more rigid membranes while shorter chains and chains with more C=C bonds result in more fluid membranes.

    Phospholipids also regulate the type of molecules entering or leaving the
    cell, as the hydrophobic fatty acid tails allows only lipid-soluble molecules to pass through the cell membranes.

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    #3865

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    TOPIC 1: CELLULAR FUNCTIONS – Part 5

    LEARNING OUTCOME

    (e)Describe the structure of an amino acid and the formation and breakage of a peptide bond.

    ESSAY ANSWER

    Structure of amino acid
    Amino acids comprise of a central carbon atom (alpha carbon) to which are
    attached an amine group, a carboxylic acid group, a hydrogen group and a
    variable R-group. The R-group may be polar, non-polar or charged. Charged
    R-groups may be acidic (negatively charged) or basic (positively charged).

    Peptide bond formation
    A peptide bond is formed when a condensation reaction occurs between a
    hydroxyl group from the carboxylic acid group of one amino acid, and a
    hydrogen atom from the amine group from another amino acid. One water
    molecule will be lost in this reaction.

    Peptide bond breakage
    The peptide bond is broken through a hydrolysis reaction. One water
    molecule is used up for this reaction. This restores the hydroxyl group at the carboxylic acid group of one amino acid, and the hydrogen atom at the
    amine group of the other amino acid.

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    #4001

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    TOPIC 1: CELLULAR FUNCTIONS – Part 6

    LEARNING OUTCOME

    (f)Explain the meaning of the terms primary structure, secondary structure, tertiary structure and quaternary structure of proteins, and describe the types of bonding (hydrogen, ionic, disulphide and hydrophobic interactions) which hold the molecule in shape.

    ESSAY ANSWER

    Primary structure
    Primary structure refers to the linear sequence of amino acids held together by petide bonds in a polypeptide chain. The number, type and sequence of amino acids incorporated into the primary structure are very specific and vary in different proteins. There are 20 different fundamental amino acids. The amino acid sequence (and the R groups) dictates the biological function of the protein. It determines the furthest levels of organisation and hence different properties of
    proteins. A change in one amino acid in the chain may completely alter the properties of the polypeptide.

    Secondary structure
    Secondary structure refers to the shape of a polypeptide chain that is formed as a result of intramolecular hydrogen bonding between the amino acids of the polypeptide chain. Disulfide linkages / bonds can be found in some secondary structure proteins like keratin. Two common secondary structures present in proteins are  helix and  pleated sheet.

    In a helix, polypeptide chain is normally coiled up into an a helix held together by many intra-H bonds formed. H of the NH group of 1 amino acid is hydrogen bonded to O atom of C=O group 4 amino acids ahead of it. The hydrogen bond is the result of electron-sharing in the NH group which leaves the H slightly positive and the CO group which leaves O slightly negative, hence they attract. a helix has 1 complete turn for every 3.6 amino acids forming a stable structure. Few proteins consist entirely of a helix. Majority have other forms of interaction between their side chains.

    Beta pleated sheet consists of adjacent chains which are parallel that run in same or antiparallel that run in opposite directions. Chains are joined by H bonds between C=O and NH groups of 1 chain & NH & C=O groups of the adjacent chain. This structure gives the protein its stability, high tensile strength and flexibility, but not elasticity.

    Tertiary structure
    Tertiary structure refers to the compact globular structure (globular proteins) as a result of further coiling and extensive folding of polypeptide chain (a helix and/or  pleated sheet). As the polypeptide folds into its functional conformation, amino acids with hydrophobic (nonpolar) R groups (side chains) congregate in clusters via hydrophobic interactions at the core of the protein, out of contact with water. Meanwhile, hydrogen bonds between polar R groups (side chains) and ionic bonds between positively and negatively charged R groups (side chains) also help to stabilise tertiary structure, by facing towards the aqueous environment. Hence, the tertiary structure is a result of the intramolecular bonds and interactions which are disulfide linkages /
    bonds, ionic bonds, hydrophobic and hydrophilic interactions and
    hydrogen bonds.

    Quaternary structure
    Quaternary structure refers to the highly complex protein consists of an
    aggregation of 2 or more extensively coiled polypeptide chains held
    together by intermolecular bonds or interactions such as disulfide
    linkages, hydrophobic interactions, ionic and hydrogen bonds. For
    example, haemoglobin has 4 polypeptide chains linked by various bonds.
    Tertiary and quaternary structures are important because the conformation in which the protein is folded is responsible for the biological activity of the protein.

    Types of bonding and interactions that hold protein molecules in shape:

    Peptide bonds
    Peptide bonds are involved in the primary structure of proteins. A peptide bond is a covalent bond formed between the carboxyl group of one amino acid and the amino group of another by means of a condensation reaction. Once two amino acids are joined together via a peptide bond, a dipeptide is formed. The dipeptide possesses a free amino group at one end and a free carboxyl group at the other. Continued condensation leads to the addition of further amino acids (monomers) resulting in the formation of a long chain polymer called a polypeptide.

    Disulfide linkages / bonds
    Disulfide linkages / bonds are strong, covalent bonds formed from
    oxidation of sulfhydryl (-SH) groups of 2 neighbouring cysteine’s R
    groups. These chemical bonds can be broken by reducing agents even
    though they are the strongest of all the chemical bonds.

    Ionic bonds
    Ionic bonds are weak electrostatic interactions between oppositely
    charged ions. Ionic bonds are formed at a certain pH when there is interaction between ionised amino (NH3+) and carboxylic groups (COO-) of the R groups of the basic amino acids (e.g. lysine) and the acidic amino acids (e.g. aspartic acid) respectively. The oppositely charged R groups containing the COO- and NH3+ are attracted to one another. Besides that, the NH3+ and COO- can also occur at both ends of a polypeptide chain. Ionic bonds are formed at a suitable pH but can be broken if pH changes.

    Hydrophobic interactions
    Hydrophobic interaction is a type of weak chemical bond formed when
    uncharged and non-polar groups cluster away from water. The polypeptide
    folds so as to shield hydrophobic R groups from the aqueous environment.
    Hydrophilic R groups project out of the protein and interact with water
    (hydrophilic interactions) while hydrophobic R groups are shielded inside protein and do not interact with water. Hence, in the aqueous cytoplasm, proteins assume the globular shape with hydrophobic groups on the inside and the hydrophilic groups outside to associate with the aqueous medium. Thus the hydrophobic interaction is actually caused by the action of water molecules which exclude non-polar substances as water molecules hydrogen-bond to one another and to hydrophilic parts of the protein.

    Note:
    Hydrophobic and hydrophilic interactions are not bonds; they are, as
    termed, interactions.

    Hydrogen bonds
    Hydrogen bonds are bonds formed between H atoms (such as those
    attached to N or O atoms as in –NH or OH groups) and N or O atoms within
    the polypeptide chain. H atoms have a small positive charge while N and O atoms have a small negative charge. This is because N and O atoms have
    higher electronegativities than hydrogen atom and therefore are able to
    draw the bonding electrons nearer to them. Oppositely charged atoms (e.g. between hydrogen and nitrogen or between hydrogen and oxygen) are attracted to each other and form a hydrogen bond. While each hydrogen bonds is very weak, the large number of such bonds means they are strong and they play an important role in the shape and stability of a polypeptide molecule.

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    TOPIC 1: CELLULAR FUNCTIONS – Part 7

    LEARNING OUTCOME

    (g) Analyse the molecular structure of a protein with a quaternary structure e.g. haemoglobin, as an example of a globular protein, and of collagen as an example of a fibrous protein, and relate these structures to their functions.

    ESSAY ANSWER

    Globular protein – Haemoglobin
    Haemoglobin is an example of a globular protein. It is required for the
    transport of oxygen in blood. The adult haemoglobin molecule has a quaternary structure as it is made up of 4 separate polypeptide chain subunits; consisting of two  chain subunits and two  chain subunits. The 2  chains each consists of 141 amino acids while the 2  chains each contains 146 amino acids. Both  and  chain subunits consist primarily of helix secondary structure.

    Haemoglobin molecule is a good transport protein for oxygen for the
    following reasons:

    1. Globular structure
    The haemoglobin molecule is globular in shape. Hence many haemoglobin
    molecules can be packed into a red blood cell for transport of oxygen.

    2. Four subunits
    Each haemoglobin molecule consists of four subunits each capable of
    binding one oxygen molecule. This greatly facilitates transport of oxygen by haemoglobin.

    3. Haemoglobin is an allosteric protein
    This means that the binding of O2 to one of the subunits is affected by its interaction with the other subunits. The binding of one O2 molecule to one of the subunits induces the remaining unfilled subunits to change their shape slightly so that their affinity for O2 increases. Thus the loading of the first O2 molecule results in rapid loading of 3 more O2. The binding of O2 to haemoglobin is said to be cooperative. Conversely, when one subunit unloads oxygen, the other 3 more readily unload as a conformation change lowers their affinity for O2.

    4. Haem binding site
    Each haemoglobin subunit consists of a protein (globin) and a non-protein (haem group) component. The structure of globin is globular except for a deep hydrophobic cleft. This is the haem binding site. It is lined with hydrophobic amino acid residues to provide a hydrophobic environment for the haem group, which is largely hydrophobic.

    5. Porphyrin ring and an iron ion (Fe2+)
    The haem group consists of a porphyrin ring and an iron ion (Fe2+).
    The Fe2+is at the centre of the planar porphyrin ring. The haem is so oriented that its Fe2+ on one face is complexed to an amino acid residue, leaving the other face accessible for oxygen binding. The Fe2+ can combine reversibly with oxygen and hence enhances the release of oxygen in metabolically active tissues such as muscle.

    6. Secondary Structure (a helix)
    Each polypeptide chain in haemoglobin consists of both hydrophilic and
    hydrophobic amino acid residues. The secondary structure of the
    polypeptide is folded such that the bulk of the hydrophobic amino acid
    residues are buried in the interior of the globular structure while the
    hydrophilic amino acid residues are on the outside. This makes the
    haemoglobin soluble in an aqueous medium and hence a good transport
    protein for oxygen in blood.

    Fibrous protein – Collagen
    An example of a fibrous protein with a quaternary structure is collagen.
    Collagen is a type of fibrous protein and is found in skin, tendons, cartilage, bones, teeth and connective tissue of blood vessels. It is a structural protein that provides great mechanical and tensile strength resisting tearing and stretching.

    Collagen consists of 3 polypeptide chains (each chain known as  chain) that wind around each other. The complete triple-helix compound is called tropocollagen. Polypeptide chain structure:
    i. Each chain consists of about 1000 amino acid residues.
    ii. Every 3rd residue is glycine (monomer: Glycine-X-Y)
    iii. The polypeptide is a left handed helix (NOT  helix) structure, with a turn every three residues. This produces a structure where glycine passes through the centre of the helix. The helix is tightly coiled, because glycine is a small molecule.

    Collagen molecule is a good structural protein for the following reasons:

    1. Insoluble in water
    Collagen is insoluble in water due to the large molecular size of the
    tropocollagen molecule as well as the nature of the amino acid
    residues. Each of the 3 polypeptide chains in tropocollagen consists
    of about 1,000 amino acid residues and all the 3 polypeptide chains
    consist largely of glycine and proline residues which are hydrophobic
    in nature. Besides that, in collagen; amino acids with hydrophobic R
    groups are found at exterior surface, thus, collagen is insoluble in
    water and metabolically inactive and resistant to chemical changes.

    2. Great tensile strength
    The three polypeptide chains (each known as  chain which is a tightly
    coiled left handed helix) are linked together by intermolecular
    hydrogen bonds, forming a loose, 3-stranded rope – a triple helix
    called tropocollagen. Each complete triple helix of collagen interacts
    with other triple helix of collagen molecules running parallel to each
    other. Covalent bonds are formed between the side chains of lysines
    in chains lying next to each other. These cross-links hold many
    collagen molecules side by side forming collagen fibrils. Many collagen
    fibrils in turn unite to form collagen fibres, giving rise to its high tensile strength.

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    JC 2 Biology Tuition Schedule Jan – March 2014

    4-Jan Saturday Energetics: Photosynthesis – Lecture
    11 Jan Saturday Energetics: Photosynthesis – Lecture
    18-Jan Saturday Energetics: Photosynthesis – Tutorials
    25-Jan Saturday Energetics: Photosynthesis – Tutorials

    8-Feb Saturday Genetic Basis of Variation – Lecture
    15-Feb Saturday Genetic Basis of Variation – Lecture
    22-Feb Saturday Genetic Basis of Variation – Tutorials

    1-Mar Saturday Genetic Basis of Variation – Tutorials
    8-Mar Saturday Genetic Basis of Variation – Tutorials
    15-Mar Saturday Cell Signalling – Lecture
    22-Mar Saturday Cell Signalling – Lecture
    29-Mar Saturday Cell Signalling – Tutorials

    5-Apr Saturday Cell Signalling – Tutorials

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    TOPIC 1: CELLULAR FUNCTIONS – Part 8

    LEARNING OUTCOME

    (h)Explain the mode of action of enzymes in terms of an active site,
    enzyme/substrate complex lowering of activation energy and enzyme specificity.

    ESSAY ANSWER

    Enzymes are biological catalysts. Their main role is to speed up chemical
    reactions but remains chemically unchanged at the end of reaction. Enzymes
    speed up chemical reactions by lowering the activation energy of a reaction (i.e. their mode of action). Activation energy is the energy required to make substrates react. It represents the energy barrier that has to be overcome before a reaction takes place to form products. Most metabolic reactions do not occur at useful rates because their activation energy is too great. The greater the activation energy, the slower the reaction at any particular temperature. If the activation energy of a reaction is decreased (which is the case for an enzyme-catalysed reaction), the rate of reaction would increase.

    This lowering of activation energy is due to formation of enzyme-substrate
    complex which is formed through the binding of the substrate to the active
    site of the enzyme. Mechanisms by which enzymes decrease the activation
    energy of a reaction when the substrate bind to its active site (i.e. forming enzyme-substrate complex) are: – serving as a template to position substrate molecules in the correct orientation for reaction to occur. In the absence of enzymes, there is low probability that the various substrate molecules will collide with each other in the correct orientation for a reaction to occur, and thus the rate of reaction would be very slow.

    – inducing physical stress in bonds of substrate. Once a substrate
    binds to the active site of an enzyme, certain bonds in the substrate
    molecule may be placed under physical stress. This increases the
    likelihood that the bonds will break.
    – increasing substrate reactivity. When the R groups of amino acid
    residues at the active site of an enzyme are very close to part of the
    substrate, they can change the charge on the substrate, alter the
    distribution of electrons within the bonds of the substrate or cause other
    changes that will increase the reactivity of the substrate.
    The active site is a crevice with a precise shape and proper charge to
    contain and bind the specific substrate molecule. The active site involves
    only a small part (about 3-12 amino acids) of the enzyme molecule which
    consists of the ‘contact’ and ‘catalytic’ residues;

    i. Contact/binding residues are responsible for the specificity of the
    enzyme and form a shape that is complementary to the shape of
    substrate.
    ii. Catalytic residues are responsible for the ability of the enzyme to
    catalyse a particular chemical reaction; acts on the chemical
    bonds in the substrate.

    The remaining amino acids i.e. structural residues, which make up the bulk of the enzyme, function to maintain the correct globular shape of the enzyme molecule. This is essential for the optimal function of the active site.

    The formation of enzyme-substrate complex requires the binding of substrate molecules to the active site of enzymes. It is known that this binding is specific where every enzyme acts on specific substrates (also known as enzyme specificity). There are two hypotheses on the mechanisms by which enzymes act on specific substrates; which are the “Lock and Key” hypothesis, and the “Induced fit” hypothesis.

    Based on “Lock and Key” hypothesis, substrate molecules have a
    complementary shape to the active site of enzyme. Substrate molecules fit
    into active site like a lock to a key forming the enzyme-substrate complex. Binding of substrate to enzyme gives rise to enzyme-substrate complexes. Once products are formed, they no longer fit the active site and are released into surrounding medium and thus, free the active site for further binding with substrate molecules.

    Based on “Induced Fit” Hypothesis, there is dynamic interaction between
    substrate and enzyme. The initial shape of active site of enzyme might not
    be complementary to shape of substrate. However, binding of substrate to
    the active site induces a conformational change in shape of the enzyme,
    which enables more effective binding of the substrate to the active site. The catalytic amino acids are brought into their correct orientations in the active site. This enables the enzyme to perform its catalytic function more effectively.

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    TOPIC 1: CELLULAR FUNCTIONS – Part 9

    LEARNING OUTCOME

    (i)Follow the time course of an enzyme-catalysed reaction by measuring rates of formation of products (e.g. using catalase) or rate of disappearance of substrate (e.g. using amylase).

    ESSAY ANSWER

    The time course of an enzymatic reaction can be monitored by either
    measuring formation of products of reaction or by measuring the
    disappearance of substrate of reaction.

    By measuring formation of products of reaction
    (using catalase and hydrogen peroxide as an example)
    Catalase is an enzyme that catalyses the decomposition of hydrogen peroxide (H2O2) into water and oxygen gas.

    2 H2O2 —–> 2 H2O + O2

    Depending on the experimental setup, rate of formation of oxygen can be
    measured by:
    – Counting oxygen (O2) bubbles formed
    – Measuring volume of water displaced by oxygen
    – Measuring the volume of O2 produced using a gas syringe

    By measuring the disappearance of substrates of reaction
    (using amylase and starch as an example)
    Amylase catalyses the breakdown of starch into maltose.
    starch ——–> maltose

    Depending on the experimental setup, rate of disappearance of starch can be measured by:
    – Iodine in potassium iodide test
    – Benedict’s test

    Iodine in potassium iodide test
    As the reaction progresses, less starch will be present and more maltose will be present. The activity of amylase was observed by using iodine in potassium iodide solution. Iodine reacts with starch to form a blue-black colour. As amylase breaks down starch, less and less starch will be present and the colour of the solution will become lighter and lighter. When all the starch has been digested, the solution will remain light-brown in colour. The colour change was observed using spot-plates (e.g. white tiles).

    Benedict’s test
    As the reaction progresses, less starch will be present and more maltose will be present. The activity of amylase was observed by using the Benedict’s solution. Benedict’s solution contains copper sulfate. Reducing sugars reduce soluble blue copper sulfate containing Cu2+ to insoluble reddishbrown copper oxide containing Cu+, which is seen as a precipitate. As amylase breaks down starch, more and more maltose which is a reducing sugar will be present and the initial blue colouration of the mixture turns green (low concentration of reducing sugar), then yellowish (middle concentration of reducing sugar) and may finally form a brick red precipitate (high concentration of reducing sugar).

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