Bondo Biology ppr 2 Marking scheme

BONDO SUB-COUNTY SECONDARY SCHOOLS JOINT EVALUATION TEST – 2015

231/2

BIOLOGY

Paper 2

July/August- 2015

MARKING SCHEME

1.         (a)        (i)         Caterpillars;ü

                                    Aphids;ü

                                    Mice;ü

                                    Slug;ü

                        (ii)        Primary consumers;ü              3mks

            (b)        (i)         Plant                Caterpillars                  Insectivorous birds                  Hawk

                        (ii)        Plant                Aphids                        Beetle              Insectivorous birds                  Hawk

                                    Plant                Slug                 Frogs               Snakes                         Hawk

                                                                                    (2mks)

            (c)        –           Lightening provide high activation energy; which causes Nitrogen to combine with

oxygen forming various oxides of Nitrogen; The oxides combine with rain water to form nitric acid; The acid sinks in soil and reacts with various salt ions to form nitrate salts;                             (3mks)

                                    Total   08 marks

2.         (a)        W is Soda lime; Absorbs any carbon (IV) oxide present / produced by the animal; (2mks)

            (b)        –           To maintain the temperature of the flask where the animal is at constant / room

level;    (1mk)

            (c)        (i)         –           There will be a rise in the level of coloured water;

                        (ii)        –           The animal respires using up oxygen in the flask and producing carbon (iv)

oxide which is absorbed by the soda lime; The air pressure in the flask falls; causing the atmospheric pressure to push fluid up the capillary tube;

                                    (3mks)

            (d)       –           Body size;

                        –           Sex of individual;

                        –           Health;

                        –           Basal metabolic rate;

                                                Mark first two           Total 08mks

3.         (a)        –           Round seed – Mm;

M  
m  
m  
m  
M  

                        –           Wrinkled seed – mm;              (2mks)

M  

            (b)        –           Round seed parent                  and                              // M and m ;

                        –           Wrinkled seed parent                          and                      // all m; (2mks)

            (c)                                            Round Seed                              Wrinkled seed

                        Genotype                        Mm                          x                  mm

m  
m  
m  
M  

                        Gametes         

                        F1                    Mm                  Mm                              mm                  mm ;

                                    Genotypes       Mm                  and      mm

                                    Phenotype       Round                                     Wrinkled

                                                            Seed                            Seed; (3mks)

            (d)       A situation where there are more than two genes occupying the same gene locus, but only

two form a pair in a diploid cell;         (1mk)

                                    Total = 08 marks

4.         (a)        Excess amino acids are deaminated / amino group is removed; and converted to ammonia;

Ammonia combined with carbon (IV) oxide in ornithine cycle to form Urea; carbohydrate group is converted to glucose for respiration / glycogen for storage;   (3mks)

            (b)        –           Glomerulus;

                        –           Bowman’s capsule;

                        –           Proximal convoluted tubule;

                        –           Distal convoluled tubule; (2mks)

                                    Mark first two

            (c)        (i)         –           Carboxyhaemoglobin;

                        (ii)        –           Carboxyhaemoglobin does not dissociate easily; thereby reducing the

                                                capacity of haemoglobin to transport oxygen; (3mks)

                                                            Total – 08 marks

5.         (a)        –           Broad and flat lamina / leaf to provide large surface area for absorption of gases;

                        –           Thinness allows gases to pass through short distance;

                        –           Presence of stomata ensures efficient diffusion of gases

                        –           Presence of air spaces for easy diffusion of gases; (3mks)

                                                            Mark first three

            (b)        –           Has ring of cartilage which is hollow for passage of air; and keep it open allt he

time;   

                        –           Has Cilia that move mucus / particles to the top of trachea / pharynx;

                       –            Has mucus to trap dust / solids particles and micro – organism (Acc. Microbes /

                                    Pathogens / bacteria / Virus / microscopic fungi) (3mks)

            (c)        –           Has hairs / mucus secretion which trap solid / foreign particles;

                        –           Air is warmed as it enters the lungs to conform with body temperature

6.         (a)        Graph of rate of reaction against PH, of the enzymes

SECTION B

6.         (a)        Scale                            =          1 = ½ x2

                        Labelling of axes         =          1 x ½ x2

                        Plotting                       =          2 = 1×2

                        Smooth Curves            =         2 =       1 x2

                        Identity of curves       =          1          =          ½ x2

                                                                        07                                                        7mks

            (b)        (i)         at PH 5.0:         Enzyme R = 6.0mg / unit time

                                                            Enzyme T = 1.3mg / unit time

                        (ii)        Optimum PH:   Enzyme R = 3.0;

                                                            Enzyme T = 8.0;

                        (iii)       Product formation / reaction is maximum at these PH;  (5mks)

            (c)        Enzyme R:      Identity:          Renin / Pepsin;

                                                Region:            Stomach;

                        Enzyme T:       Identity:          Tylin / Salivary amylace / Trypsin/ Mouth / Deodenum

                                                Region:            Ileum;  (4mks)

            (d)       –           Temperature; enzyme concentration; Inhibitors substrate concentration; (2mks)

                        –           Contain bile salt; which neutralizes acidic enzyme and emulsifies fats; (2mks)

                                                                                    Total 20mks

7.         (a)        Support           –          Maintain shape of body of organism;

                                                –           Provide base for muscle attachment;

                                                –           Project other delicate organs from mechanical damage;

                        Movement

  • Enable animals to search for food;
  • Enable animals to search for favourable habitat;
  • Enable them to move away from predators / enemies / hostile environment
  • Enable them search for better breeding places;
  • Enable them to search for mates;  (8mks)

            (b)        –           Carbon (IV) oxide from the respiring cells of the biceps muscles diffuse into the

                        blood; within the venule bed of capillaries; where it forms weak carbonic acid; in the plasma; and carbominohaemoglobin in the RBCs; From here blood moves into a (branchial) vein; which joins the subclavian vein; that connects with (superior) vena cava;

                        –           From the vena cava the blood containing carbon (IV) oxide enters the right auricle

                        / atrium; The right auricle contracts to push blood into the right ventricle; via the tricuspid valve;

                        –           Blood from right ventricle is pushed into pulmonary artery; via semi – lunar valve;

                        from the pulmonary artery the blood enters the capillary system on the lung alveoli at the arterial bed; At this point carbonic acid and carbonmonohaemoglobin dissociates; to release carbon (iv) oxide; which diffuses across the capillary wall and wall of alveoli into the alveolar cavity;     (12mks)

                                                            Total 20mks

8.         –           The process of cell division is mitosis where two daughter cells each having the same

                        number of chromosome as parent is formed;                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                              

            –           It involves behavior of chromosome which occur in various stages

            –           The first stage is interphase; here there is multiplication of genetic material (so that

daughter cells have the same number as parent cell); there is also synthesis of new cell organelle; and build up of energy to drive the cell through the process;

            –           The cell gets into a prophase stage; where the centrioles separate and move to opposite

poles of the cell; spindle fibres begin to form; the nuclear membrane begin to disappear; the chromosomes shorten and thickens; and chromatids become visible;

            –           At metaphase stage; nuclear membrane disappears completely and chromosomes appear

free in the cytoplasm; The spindle fibres lengthens; and chromosomes align themselves at the equator of the spindle / cell; The chromosomes attach themselves to the spindle by their centromere;

            –           During anaphase stage; the chromatids seperate at the centromere and migrate to opposite

poles; this is due to shorteneing of spindle  nfibres ; spindle fibres begin to disappear; cell plate forms and grows to separate the cell into two;

–           The cell merges into the last stage called telophase; where chromatide collect together at

the two opposite poles of the cell; The nuclear membrane forms around each set of

chromatids; chromatids replicates and become chromosome;

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