133/1-fv Joint Examination 2023
PRESIDENT’S OFFICE
         REGIONAL ADMINISTRATION AND LOCAL GOVERNMENT
ADVANCED CERTIFICATE OF SECONDARY EDUCATION EXAMINATION
              FORM FIVE JOINT EXAMINATION SOUTHERN ZONE
                                  (MTWARA AND LINDI)
133/1                                 BIOLOGY 1
                            MARKING SCHEME
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  1. (a) (i) Binomial nomenclature                                        (02 marks)
         (ii) * Scientific name must compose of two names that is generic name and
                a specific name
              * The generic name must begin with a capital letter, then specific name
                must follow with a small letters
              * The scientific name should be latinized
              * The scientific name should be underlined separately
                                                                (Any four04 marks)
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  2. (a) (i) He used 1% copper II sulphate solution                          (01 mark)
         (ii) Procedure
              He used the following procedures
               * 2ml of solution was put into a test tube.            2ml of sodium
                 hydroxide(NaOH) solution added.                            (02 marks)
               * Then to test tube, three drops of 1% copper II sulphate solution added
                 while shaking after each addition.                         (02 marks)
              Observation
               * The solution turned into purple colour or violet.          (02 marks)
       (b) The basis for the positive results
              – In alkaline solution, the nitrogen atoms in the peptide bonds form a purple
                complex with copper II ions.                                    (03 marks)
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  3. (a) (i) Name of the gland is Pancrease                                     (01 mark)
         (ii) Five (5) digestive enzymes produced are;
               * Trypsinogen
                  · activated to an enzyme trypsin by non-digestive enzyme called
                    enterokinase
                  · Trypsin breaks protein into peptide.
               * Chymotrypsinogen
                  · which is activated to chymotrypsin by trypsin that also split protein
                    into peptide
               * Procarboxypeptidease
                  · This activated to enzyme carboxypeptidedase by trypsin
                  · It convert large peptide (polypeptide) into smaller peptides and
                    amino acids.
               * Pancreatic amylase
                  · This complete the digestion of starch into maltose which started in
                    the mouth
               * Pancretic Lipase
                  · Hydrolyses lipids(fat and oil) into fatly acid and glycerol
               * Nuclease
                  · catalyse the break down of nucleic acid into their constituents
                    nucleotides
                                                                     (Any five 05 marks)
       (b) (i) In region A light intensity is the limiting factor
           (ii) In region B both light intensity and the other factor(s) are limiting
                In region C light intensity is no longer a limiting factor.
          (iii) Point D represent saturation point for light intensity under these condition
                that is the point beyond which an increase in light intensity will cause no
                further increase in the rate of photosynthesis
          (iv) In point E the maximum rate of photosynthesis attainable under the
                condition of the experiment.                             (@01 = 04 marks)
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  4. (a) The difference between nervous coordination and hormonal coordination
                 S/No Nervous coordination                Hormonal coordination
                  (i) Electrical impulse are the          Hormones are the chemical
                      messenger in the nervous            messengers in the endocrine
                      system                              system that target cells
                                                          through the blood stream.
                  (ii)     Nerve        impulses    are   Hormones are transmitted
                           transmitted through neurons    through blood vessels
                  (iii)    Nerve impulses make us         The        hormones     enter
                           the neurotransmitters ate      into the target cells by
                           synaptic clefts and sodium     diffusing through the plasma
                           and potassium channels and     membrane or by binding to
                           enter the target cells         the cell receptor.
                  (iv)     Response are localized         Response are widespread
                  (v)      The signal transmission        The signal transmission
                           process is fast                process is slow.
                                                       (Any five points @01 = 05 marks)
       (b) Adaptations of the nerve tissue
             (i) They have nerve cells(neurones) that receive information from sensory
                 parts and send or transmit it to the CNS for interpretations and then to the
                 effector for a response.
            (ii) The cells of the nervous tissue are tightly pecked for effective working of
                 the tissue
           (iii) They have neuroglia (glia cells) that provide protection and support to the
                 tissue
           (iv) Nerve cells or Neurones have nodes of ranvier and fatty myelin sheath
                 which facilitate rapid transmission of impulses.
            (v) The cells in the nervous tissue produce neurotransmitter chemicals which
                 act as conveyors that carry impulses from one neurone to another across
                 the synaptic gap.
                                                       (Any five points @01 = 05 marks)
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  5. (a) The structure of Plasma membrane
                                                                              (06 marks)
       (b) Difference
                     Criteria   Cell wall                  Cell membrane
                     Stability  very stable due to         Less stable due to lack of
                                presence of thickening     thickening material
                                materials such as legnin
                                and cellulose
                   Permeability less          permeable    More permeable as it has
                                as    it   depend     on   smaller pores together
                                plasmodesmata              with protein receptors
                                                                              (04 marks)
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  6. (a) (i) Competitive inhibitor take place when substrate and inhibitors compete
              for the active site of an enzyme.                           (02 marks)
         (ii) Enzyme inhibitors are important in biochemical pathways because they
              help in regulation of biochemical pathway example end product
              inhibition can switch off certain metabolic reaction if products are in
              higher concentration.                                       (02 marks)
       (b) (i) Prosthetic group is the type of enzyme cofactors that slightly and
                permanently bind to an enzyme to help catalysis eg FAD while
                coenzymes are type of cofactor which are not remain attached to the
                enzyme between reaction eg NAD, NADP, COA               (03 marks)
           (ii) Chemical composition of ATP
                 * Adenine (organic base)                                (01 mark)
                 * Ribose (pentose sugar)                                (01 mark)
                 * 3-phosphate molecules                                 (01 mark)
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  7. (a) (i) Features of Respiratory surface
               * It has thin epithelium to facilitate rapid diffusion of the gases.
               * It has permeable membrane to allow the passage of gases
               * It has larger surface area to volume ratio for maximum diffusion of the
                 gases
               * It has moist to allow the passage of gases in a solution form
               * It has blood capillaries for transportation of diffused gases.
                                                   (Any three points @01 = 03 marks)
         (ii) Factors influencing Breathing
               * Concentration of carbon dioxide
                 If there is high concentration of carbon dioxide blood the rate of
                 gaseous exchange increase in order to provide the tissue with enough
                 oxygen
               * Physical activities
                 Rate of gaseous exchange increase when there is increased physical
                 activities as here is large increase of carbon dioxide and oxygen is
                 also needed for tissue respiration
               * Age
                 Breathing rate of young people is higher than old people because
                 young people are generally more active
                                                                       +
               * And other related factors such as smorking, H concentration in
                 the blood, stretching of the thoracic and pulmonary walls and oxygen
                 saturation in the blood.
                                                   (Any three points @01 = 03 marks)
       (b) Ways through which oxygen is transported
              – As a mixture of gases from the atmosphere to the nostril to the lungs.
              – As dissolved in solution from the lungs to other part of the body
              – By combination with Haemoglobin from the lungs to other parts of the
                body.
                                                   (Any two points @02 = 04 marks)
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  8. (a) Summary of essential features of each stage.
             (i) Summary of essential features of glycolysis
                  * Phosphorylation of sugar. This activate the sugar making it more
                    reactive, the process uses ATP
                  * Splitting of the 6C sugar into 3C sugar phosphate
                  * Oxidation by dehydrogenation                            ( 03 marks)
            (ii) Essential features of Kreb’s Cycle
                  * The citrate is oxidized in many stages to produce carbon dioxide and
                    energy in form of ATP and FAD
                  * Acetyl group(2C) enter the cycle by combining with 4C compound
                    (oxaloacetate) to form a 6C compound(Citrate)           ( 03 marks)
           (iii) Essential features of Electron transport chain
                  * Energy in the form of hydrogen atoms from the Kreb’s cycle is
                    converted to ATP
                  * Hydrogen atom are carried by hydrogen carries NAD and FAD to
                    other carriers
                  * As the hydrogen is passed from one carrier and energy is released in
                    form of ATP                                             ( 03 marks)
       (b) (i) Role played by oxygen in Respiration
                  * Oxygen is an electron acceptor within the electron transport chain
                     which synthesizes ATP from nutrients.                     ( 02 marks)
           (ii) Kreb’s Cycle products
                 Note that prior to Kreb’s Cycle 2 molecules of Acetyl COA are produced
                 from oxidative decarboxylation of 2 pyruvate so two cycles are required
                 per glucose molecules.
                 Therefore, in each citric acid cycle forms the following products
                 (i) 4 molecules of CO2 are released. Removal of CO2 or decarboxylation
                     of citric acid takes place at two places
                     · In the conversion of isocitrate (6C) to α−ketoglutarate (5C)
                     · In the conversion of α−Ketoglutarate to Succinyl COA(4C) ( 01
                       mark)
                (ii) 2 ATP are produced directly in the conversion of Succcinyl COA to
                     Succinate.                                                 ( 01 mark)
               (iii) 6 reduced NAD are produced by reduction of NAD+ to NADH in the
                     following reactions.
                     · Isocitrate to α−Ketoglutarate
                     · α−Ketoglutarate to Siccinyl COA
                     · Malate to oxaloacetate                                   ( 01 mark)
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               (iv) 2 FADH are produced by the reduction of FAD+ to FADH2 in the
                    conversion of succinate to fumarate.              ( 01 mark)
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  9. (a) (i) Objects are seen more clearly at night by not looking at them because in
              dim light only are activated
               * Rods are more distributed over the retina but non of them found at the
                 fovea where as cones are at greater concentration.
                 While looking directly to the object at night photoreceptor cells from
                 fovea are yo be poor vision in dim light that why you can see less
                 clearly
               * While Not looking directly to the object at night the photoreceptor
                 cells away from the region of fovea centralis are to be used of which
                 are rods, these rods have good vision in dim light that why you can
                 see more clearly.                                         ( 05 marks)
         (ii) Advantages of Myelinated Axon
               * It increase the transmission speed of the impulse since impulse jump
                 from one node of Ranvier to another.
               * It protects the inner part of the axon from the mechanical injury ( 03
                 marks)
       (b) Synthetic plant hormones can be effectively used in different ways for large
           scale production in agriculture as follows.
             (i) Gibberellins
                  * Promote fruit setting
                  * Enhance seed production
                  * Bolting and flowering in long-day plants
                  * Breaking seed dormancy
            (ii) Auxins
                  * Root initiation in cutting
                  * Initiation of flowering
                  * Promote fruit setting without fertilization
                  * Selective of weed killer
                  * Prevention of leaf and premature fruit fall
           (iii) Ethylene
                  * It promote fruit ripening
                  * It induce flowering in pineapple
                  * It stimulates the latex flow in rubber
                  * It induces the sprouting of storage organs
           (iv) Cytokinin
                  * Delay of leaf senescence
                  * Break seed dormancy
                  * Prolong shelf life
                                                             (Any seven @01 = 07 marks)
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 10. (a) The process of photosynthesis is important to the life of living on earth in the
         following ways
              – Photosynthesis is essential for the existence of all life on earth surface
                because plays a role in the food chain.
              – Photosynthesis responsible for oxygen production which is needed by
                most organism
              – Photosynthesis balances the amount of carbon dioxide.         ( 03 marks)
       (b) Plant obtain light energy by absorbing light through pigment called
           chlorophyll. The chlorophyll consist of reaction centres
              – The reaction centres are called photosystems
              – There are two types of photosystem
             (i) Photosystem I
                 Photosystem I contain chlorophyl called p700. In the reaction center of
                 PSI, the light absorbing pigment is a specialized chlorophyll a molecule
                 that absorb red light of 700nm wavelength most efficiently, hence the
                 name p700.
            (ii) Photosysytem II
                 Photosystem II contains chlorophll a called p680. The PSII reaction
                 center chlorophyll molecules absorb maximally at 680nm and are
                 therefore named P680
            Mechanism of converting light to energy to NADPH+H+ and ATP
              – Absorption of light excites an electrons to higher energy level there by
                converting the sunlight energy to potential chemical energy
              – The photosynthesis pigment are organized into photocenters in the
                thylokoid membrane each of which contain hundreds of pigments
              – The many pigment molecules in each photocenter act as antennae to
                absorb light.
              – The reaction center chlorophyll then transfer its high energy electron to
                an acceptor molecules in an electron transport chain.
              – High energy electrons are then transferred through a series of membrane
                carries coupled to the synthesis of ATP and NADPH
              – During the process electrons lost by p680(PSII) are taken up by P700(PSI)
                and do not get back to P680 that is undirectional and hence it is called
                non-cyclic phosphorylation.
              – Electrons given out by P700 are taken by the primary acceptor and are
                ultimately passed on to NADP. The electron combine with H+ and reduce
                NADP to NADPH+H+ . The hydrogen ions also called protons are made
                available by spiliting up water.
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              – The excited electrons from the PSI and taken by PSI electron acceptor
                and instead of sending them to NADPH deposits them the electron chain
                between PSII and PSI and they flow down along of the 2 scheme
                                                                         ( 12 marks)
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