Biology Paper 1 Questions and Answers - Momaliche Post Mock 2020 Exam

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INSTRUCTIONS TO CANDIDATES

  • Answer All questions 
  1. What is meant by the term sex linkage. (1mk)
  2. Part of one strand of DNA molecule was found to have the following sequence
    G-C-C- G – A – T- T – T – A – C – G – G
    What is the sequence
    1. of the complimentary DNA strand? (1mk)
    2. On a m-RNA strand copied from this DNA portion? (1mk)
  3. State two regions in a plant where the end products of photosynthesis are translocated to? (2mks)
  4. With reference to circulatory system only give two reasons why birds and mammals are more active compared to other organisms? (2mks)
  5.  
    1. What three characteristics are used to divide the phylum Arthropoda into classes? (3mks)
    2. The diagram below shows an organisms from a division in Kingdom plantae. Study it and answer the questions that follow.
                moma Bio PP1 Q5b
      1.  Identify the division from which the plant was obtained. (1mk)
      2. Name the parts labelled X and Y (2mks )
        • X…………………………………………………………..
        • Y……………………………………………………………
  6. What is the relationship between a genus and a species? (1mk)
  7. A drawing of 3 cm was made of a giant spider whose actual length was 7cm. calculate the magnification of the drawing? (3mks)
  8. Explain why osmosis is described as a special type of diffusion? (1mk)
  9. The following table shows the estimated number of organisms recorded in a dam.
     Organisms   Number
     Small fish   3500 
     Microscopic algae   12000 
     Crocodiles   100
     Large fish   950
     Mosquito larvae  8900

    1. Construct a possible food chain for the dam? (1mk)
    2. Construct a pyramid of numbers for the given data? (1mk)
    3. Explain the shape of pyramid obtained? (2mks)
  10.  
    1. Explain why leaves of most plants are thin and broad. (2mks)
    2. State the function of the following enzymes during digestion in the stomach?
      1. Pepsin (1mk)
      2. Renin (1mk)
  11. Explain the following:
    1. Respiratory surface must be moist? (1mk)
    2. Respiratory surface must be thin (1mk)
    3. Palisade cells are cylindrical shaped and arranged with long axis perpendicular to the leaf surface. (1mk)
  12. The diagram below represents the vertical section of a fruit.
              moma bio PP1 Q12
    1. Suggest the possible agent of dispersal of this fruit. (1mk)
    2. Explain two observable features that adapt the fruit to its mode of dispersal. (2 mks)
  13. Explain why the body temperature of a healthy person rises slightly during humid days? (2mks)
  14.  
    1.  
      1. Name the respiratory surface in insects. (1mk)
      2. State any one feature that adapts the structure named in a(i) above to its function. (1mk)
    2. Why are the fish gills highly vascularized? (1mk)
  15. State the function of the following organelles:
    1. Granulated Endoplasmic reticulum (1mk)
    2. Nucleolus (1mk)
  16. State two gaseous exchange sites in plants? (2mks)
  17. The diagram below shows an apparatus used during collection of specimen in biological study.
             moma bio PP1 Q17 apparatus
    1. Identify the apparatus? (1mk)
    2. What is the use of the apparatus named above? (1mk)
  18. List three limitations of fossil records as an evidence of organic evolution? (3mks)
  19. Distinguish between enzyme co-factors and co-enzymes? (2mks)
  20. Give two reasons for the rapid growth during the exponential phase of growth curve? (2mks)
  21. Give two reasons why Carolus Linneaus preferred the use of latin language in the scientific naming of living organisms. (2mks)
  22. State three roles played by active transport in living organisms. (3mks)
  23. List three factors affecting the rate of respiration? (3mks)
  24. Study the diagram below and answer the questions that follow.
    moma bio PP1 Q24 diagram
    1. Identify the cell (1mk)
    2. Label the parts X,Y and W (3mks)
      • X……………………………………………………………………………………………
      • Y……………………………………………………………………………………………
      • W……………………………………………………………………………………………
  25. Explain why it is becoming more difficult to treat malaria using chloroquine? (2mks)
  26. State two ways by which the ileum is adapted for absorption of food materials? (2mks)
  27. Name two processes that contribute to variation during gamete formation? (2mks)
  28. Damage to the mammalian liver may lead to indigestion of fats. Explain this observation. (2 mks)
  29. Name the disease of blood characterized by
    1. Abnormally large number of white blood cells. (1 mk)
    2. Cresent-shaped haemoglobin instead of the normal biconcave shape. (1 mk)
  30. During a strenuous exercise the chemical process represented by the equation below takes place in the human muscle cells.
    C6H12O6 → 2CH3CH(OH)COOH + 150KJ
                             (Substance X)
    1. Name the process represented above .      (1 mk) 
    2. Name substance X       (1mk)
  31. The diagram below represents a stage of growth in two different seeds.
    moa bio PP1 Q31 seedling A
    moma bio PP1 Q31 seedling B
    1. Identify the type of germination exhibited by seedlings A and B. [2 marks]
      Seedling A…………………………………………………………………………………………………………………….
      Seedling B……………………………………………………………………………………………………………………
    2. State the role of oxygen during germination. [1 mark]
    3. Account for the loss of weight in cotyledons in germinating seeds. [1 mark]
    4.  
      1. State the role of juvenile hormone during metamorphosis in insects. [1 mark]
      2. Name the glands that secrete juvenile hormone [1mark]


MARKING SCHEME

  1. These are genes located on the sex chromosomes and are transmitted together with those determine sex. 1 mk
  2.  
    1. C – G – G – C – T – A – A – A – T – G – C – C 1 mk
    2. C – G – G – C – U – A – A – A – U – G – C – C 1 mk
  3.  
    • The growing and developing regions such as shoots, leaves, flowers, fruits and roots
    • Storage organs or tissues such as tubers, corns, bulbs, rhizomes and seeds.
    • The secretors organs such as nectar glands in some insect pollinated plants such as bananas
      any 3x1=3 mks
  4.  
    1. Deoxygenated and oxygenated blood do not mix
    2. Blood is at a higher pressure once the heart pumps it twice.
      2x1=2 mks
  5.  
    1.  
      • Number of limbs
      • Presence and number of antennae
      • Number of body parts
        3x1=3mks
    2.  
      1. Bryophyta
      2. X – Seta
        Z – Rhizoid rej Rhizoids.
  6. A species is a subset of genus i.e. one genus contains several species. 1x1=1 mk
  7. Magnification =    (Length of drawing)      √1
                            (Length of real object) 
                        = 3cm = 0.429 √1
                           7cm
                         =X 0.43√1
      1x3=3mks
  8. Because it involves movement of solvent (water) molecules from their region of high concentration to region of low concentration across a semi permeable membrane.
    1x1=1mk
  9.  
    1. Microscopic algae → mosquito larvae → small fish → large fish → crocodile
      NB: mark as a whole
      1x1=1 mk
    2.  
      moma bio PP1 ans 9b
      1x1=1 mk
    3.  
      1. Body size of the organism increase at each trophic level from the base as their numbers decrease.
      2. At each trophic level much of the energy obtained is lost in respiration thus fewer organisms can be supported at the succeeding level. 2 mks
  10.  
    1.  
      • Thin
        To reduce distance for diffusion of gases.
        To reduce distance for sunlight to reach the photosynthetic cells. 1x1=1 mks
      • Broad
        To provide large surface area for maximum light absorption.   1x1=1 mks
    2.  
      • Pepsin – Breaks down proteins into peptides.   1x1 = 1 mks
      • Renin – Digests protein caseinogens in milk to casein (curd)    1x1=1mks
  11.  
    1. Moist o dissolve the diffusing gases across the respiratory surface. 2 mks
    2. Thin to reduce distance covered by diffusing gases i.e. for the gases to diffuse through short distance. 1mks
    3. Many palisade cells in a small area to enable them receive maximum sunlight.
      1x1=1 mk
  12.  
    1. Water dispersal. 1x1= mk
    2.  
      1. Has a water proof endocarp for buoyancy. 2mks.
      2. Mesocarp is fibrous and spongy to trap air.
  13. This is because during humid day, there is low rate of sweating; since less water is lost from the body surface, leading to less heat loss through sweat hence body temperature tend to rise slightly.     1x2=2 mks
  14.  
    1.  
      1. Tracheoles.
      2. Thin epithelium for respiratory gas to diffuse through a short distance;
        • Moist to dissolve respiratory gases.
        • numerous to increase surface area for gaseous exchange.
          1x1=1 mk
    2. To create a steep diffusion gradient after transport of oxygen gas. 1 mk.
  15.  
    1. Helps in the transport of proteins. 1 mk
    2. Manufacture of ribosomes 1 mk
  16.  
    • Stomata;
    • Lenticels of woody plants
    • Cuticles        3x1=3 mks
  17.  
    1. Pitfall trap
    2. For catching crawling animals.
  18.  
    • Distortion of parts of fossils during sedimentation hence can give wrong impression of the structure;
    • There was several missing links of fossils records as some parts or whole organism decomposed, some scavenged upon and conditions may not be conducive for fossilization (O.W.T.T.E).
    • Destruction of fossils by geological activities like earthquakes, faulting and mass movement.
      3x1=3 mks
  19. Enzymes cofactors are non-proteinous substances which activate enzymes; while co-enzymes are organic non-protein molecules that work is association with particular enzymes. (Mark as a whole)   2 mks
  20.  
    • Cells have adjusted to the new environment
    • Food and other factors are not limiting hence no competition for resources.
    • Rate of cell increase is higher than cell death.
    • There is an increase in the number of cells dividing
      First two 1x2=2 mks
  21.  
    • Latin language was widely spoken and used by scientists during his time;
    • Local names used previously could not be understood by everyone thus Latin language enhanced scientific communication worldwide. 1x2=2 mks
  22.  
    • Excretion of waste products from the body cells.
    • Absorption of digested food from alimentary canal of animals in the blood stream.
    • Absorption of some minerals salt from soil by plant roots.
    • Accumulation of substances into the body to offset osmotic imbalance in arid and saline environment.
    • Reabsorption of sugars and some salts by the kidney. 3x1=3 mks
  23.  
    • Oxygen concentration
    • Presence or absence of hormones
    • Substrate concentration
    • Surface area to volume ratio/body size of an organism. 3x1=3 mks
  24.  
    1. Mature human ovum 1mk
    2. X – follicle cell
      Y – viteline membrane 1mk
      W – Plasma membrane 1mk
  25. Some malaria plasmodium developed resistance; to chloroquine drug; through mutation; those resistant individuals transmit the characteristic to their offspring through reproduction thus establishing a new population of resistant forms. 2 mks
  26.  
    • It is long to provide large surface area for absorption.
    • It is numerous to bring digested food into close contact with walls of the ileum for easier absorption.
    • Highly coiled to slow down movement of food, allowing more time for absorption.
    • Higher surface has large number of villi and micro-villi which increase the surface area for absorption of end products of digestion.
    • Presence of thin layer of cells through which digested food diffuses.
    • Presence of tense network of blood capillaries in villi into which nutrients are absorbed.
    • Presence of lacteals in the villi for absorption of fatty acids and glycerol. 2x1=2 mks
  27.  
    • Independent assortment
    • Crossing over 2x1=2 mks
  28. Damage lead to failure of liver to produce bile; bile carries out emulsification of fats; 2 mks
  29.  
    1. Leukemia. 1 mk.
    2. Sickle cell Anaemia. 1 mk.
  30.  
    1. Anaerobic respiration. (1 mk)
    2. Lactic acid. 1 mk.
  31.  
    1. A- Hypogeal germination.
      B- Epigeal germination. 2 mks.
    2. Its used in oxidation of stored food substances in the seedto release energy for growth. (1 mk)
    3. Due to breakdown of stored food to yield energy for growth. (1 mk)
    4.  
      1. It leads to formation of larval cuticle. (1 mk)
      2. Corpora allata. (1 mk)

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