Kenya Certificate Of Secondary Education(KCSE 2013) Biology Paper 3 with Marking Scheme

1.      
   1. The photograph shows the inner surface of the upper left side of the rib cage.
      
      
      1. Name the bone covered by the fatty tissue labeled K. (1 mark)
      2. Explain the role of part labelled M in the inhalation. (5 marks)
   2. The photograph below shows a mammalian vertebra.
      
      
      1. State the view of the vertebra presented. (1 mark)
      2. Name and state one function of the part labeled T.
      3. How are the parts labelled S and V adapted to their functions? (4 marks)
   3. The actual width of the vertebra below in cm is shown by a section of the ruler in the photograph.
      
      
      1. Determine the width of the vertebra on the photograph. (1 mark)
      2. Calculate the magnification of this image. (2 marks)
      3. Determine the actual length of the vertebra from point A to B. Show your working. (2 marks)
2. You are provided with a food sample labelled solution C. Using the reagents provided, carry out tests to identify the food substances present in the sample.
   
3. Below are photographs showing some observable features of leaves.
   
   Using the features in the order given below, construct a dichotomous key that can be used to identify the specimen.
   
   • simple and compound leaves;
   • leaf venation;
   • leaf margin;
   • arrangement of leaves on the stem;
   • pinnate or trifoliate nature of leaves. (10 marks)

MARKING SCHEME

1.    
   1.    
      1. Sternum; (1 mark)
      2. The internal intercostal muscles relax; pulling the ribs upwards; and outwards;
         This increases the volume of the rib cage while pressure decreases;
         Forcing air into the lungs; (5 marks)
   2.    
      1. Anterior/dorsal view; (1 mark)
      2. Name -  Neural canal; (1 mark)
         Function -  Passage of the spinal cord. (1 mark)
      3. V:  It is thick and solid; for bearing the weight of the body (back) (2 marks)
         S:  It is long; to provide a large surface area for attachment of muscles; (2 marks)
   3.    
      1. Image width = 9.8 cm
      2. Magnification = Image length / width
                                 Actual lrngth / width
         =9.8 ± 0.1
           4.6 ± 0.1
         Mg = x 2.13
      3. Actual length = 10.4 ± 0.1
                                2.13
         = 4.8826 cm

1.    
   1. Simple leaves  ...............................................................          go to 2;
   2. Compound leaves  ........................................................           go to 4;
2.    
   1. Leaves net-veined/reticulate  ...........................................       go to 3;
   2. Leaves parallel veined ...............................................              Commelinaceae;
3.    
   1. Leaves with serrated margins  ....................................             Malvaceae;
   2. Leaves with smooth (entire) margins ..........................            Nystaginaceae;
4.    
   1. Leaves opposite ..........................................................             go to 5;
   2. Leaves alternate  ........................................................             Bignoniceae;
5.    
   1. Leaves pinnate  ..........................................................             Papilionaceae;
   2. Leaves trifoliate .........................................................             Compositae; (10 marks)