Cell Physiology Questions and Answers - Form 1 Biology Topical Revision

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  1. The table below shows the concentration of some ions in pond water and in the cells sap of an aquatic plant growing in the pond.

    Ions  Concentration in pond water (parts per million)  Concentration in cell sap (parts per million) 
    Sodium  50  30 
    Potassium  150 
    Calcium  1.5 
    Chloride  180  200 
    1. Name the processes by which the following ions could have been taken up by this plant. (2mks)
      1. Sodium ions
      2. Potassium ions
    2. For each processes named in (a) (i) and (ii) above, state one condition necessary for the process to take place. (2mks)
  2. Explain how water in the soil enters the root hairs of a plant. (4mks)
  3. Explain how drooping of leaves on a hot sunny day is advantageous to a plant. (2mks)
    1. What is diffusion? (2mks)
    2. How do the following factors affect the rate of diffusion?
      1. Diffusion gradient (1mk)
      2. Surface area to volume ratio (1mk)
      3. Temperature (1mk)
    3. Outline 3 roles of active transport in the human body (2mks)
  5. State the importance of osmosis in plants (3mks)
  6. An experiment was set up as shown in the diagram below.
    sucrose solution osmosis
    The set up was left for 30 minutes.
    1. State the expected results. (1mk)
    2. Explain your answer in (a) above. (3mks)
  7. Explain why plant cells do not burst when immersed in distilled water. (2mks)
  8. Distinguish between diffusion and osmosis. (2mks)
  9. Define the following terms in relation to a cell
    1. Isotonic solution
    2. Hypotonic solution
    3. Hypertonic solution (3mks)
  10. Addition of large amounts of salt to soil in which plants are growing kills the plants. Explain. (6mks)
  11. Explain why
    1. Red blood cells burst when placed in distilled water while plant cells remain intact.
    2. Fresh water protozoa like amoeba do not burst when placed in distilled water. (2mks)
  12. Two equal strips A and B were from a potato whose cell was 30% of sugar. The strip A was placed in a solution of 10% sugar concentration while B was placed in 50% sugar concentration
    1. What change was expected in strip A and B
    2. Account for the change in strip A
  13. An experiment was set-up as shown below and left for one hour
    peeled potato
    1. State the expected result at the end of one hour
    2. Explain the observations made in this experiment
  14. State what would happen in each of the following:-
    1. A plant cell placed in: -
      1. Strong salt solution 
      2. Distilled water
  15. State three physiological processes that are involved in movement of substances across the cell membrane
  16. Potato cylinders were weighed and kept in distilled water overnight. They were then reweighed.
    potato cylinders
    1. Calculate the average mass of a potato cylinders after reweighing. Show your working.
    2. Explain why mass of the cylinders had increased.
  17. The diagrams below show a red blood cell that was subjected to a certain treatment.
    red blood cell after treatment
    1. Account for the shape of the cell at the end of the experiment.
    2. Draw a diagram to illustrate how a plant cell would appear if subjected to the same treatment.
  18. The diagram below shows the results obtained when red blood cells are placed in different solutions:
    red blood cell in different solutions
    1. What name is given to the process that occurs when the cell is placed in solution Y?
    2. Describe the process that would occur in a plant cell when placed in a similar solution as that of solution X
  19. The figure below shows the results obtained when red blood cells are put in different solutions:-
    red blood cell in different solutions2
    1. What is the name given to the process that occurs when the cell is put into solution B?
    2. Compare the results obtained when the cell is put in solution B to the results that would be obtained if a plant cell was put in the same solution.
  20. An experiment was carried out to investigate the effect of different concentrations of sodium chloride on human red blood cells. Equal amounts of blood were added to equal volumes of the salt solution but of different concentrations. The results are shown in the table below:

    Set -up


    Number of red blood cells


    Sodium chloride concentration

    At start of experiment

    At the end of the experiment




    No change in number




    Fewer in number

    1. Account for the results in the set-up
    2. If the experiment was repeated using 1.4% sodium chloride solution, state the expected results with reference to:
      1. the number of red blood cells
      2. the appearance of red blood cells if viewed under the microscope
    3. Name support tissues in plants characterized by the following
      1. Cells being turgid
      2. Cells being thickened by cellulose
      3. Cells being thickened by lignin
  21. The diagram below illustrates the behaviour of red blood cells when placed into two different solutions X and Y.
    red blood cell in different solutions3
    1. Suggest the nature of solutions X and Y.
    2. Name the process A and B.
    3. What would happen to normal blood cell if it were placed in a isotonic solution.
  22. Name two plant processes in which diffusion plays an important role.
  23. Two fresh potato cylinders of equal length were placed one in distilled water and the other in concentrated sucrose solution:
    1. Account for the change in length of the cylinder in:
      1. Distilled water
      2. Sucrose solution
      1. What would be the result in terms of length if a boiled potato was used?
      2. Explain your answer in(b)(i) above.
    2. State two uses of the physiological process being demonstrated in the experiment
  24. The two cells shown below are obtained from two different potato cylinders which were immersed in two different solutions P and Q.
    potato cylinders in different solutions
      1. Name the structure labelled A.
      2. State the function of structure B.
    2. If eight of cell I were observed across the diameter of the filed of view of 0.5 mm, Work out the actual diameters of each cell in micrometers.
    3. Suggest the identity of the solution Q.
    4. Account for the change in cell I above.
    5. State any one importance of the physiological process being demonstrated above in animals.                                                                           


      1. Diffusion
      2. Active transport
      1. Diffusion-A concentration gradient between sodium ions in sap and those in the pond.
      2. Active transport-energy in form of ATP must be available/Oxygen and food in the living tissue for respiration provide energy.
  2. A film of water surrounds the soil particle. Root hairs of the plants penetrate between the soils particles/are close to the soil particles; cell sap of the root hair cells is more concentrated in solutes/has less water than the soil solution. Thus water moves into root hair cell by osmosis i.e across the cell a wall and the semi permeable membrane.
  3. The leaves expose a smaller surface area to the sun. Thus reducing transpiration/excessive water loss.
    1. Diffusion is defined as the net movement of a substance from a region where its concentration is high to a region where its concentration is low.
      1. Diffusion gradient-the greater the diffusion gradient, the greater the rate of diffusion
      2. Surface area to volume ratio-the greater the S.A.V.R the higher the temperature the greater the rate of diffusion.
      3. Temperature –The higher the temperature the greater the rate of diffusion
      1. Absorption of mineral salts from the soil by root hairs
      2. Re-absorption of glucose molecules in the kidney tubule.
      3. Absorption of digested food in the ileum e.g glucose, amino acids.
    1. Uptake of water from the soil into root hairs of plant roots
    2. Movement of water from the veins of leaves through the leaf cells to the atmosphere during transpiration.
    1. The visking tubing was fully filled with solution. Level of water in beaker decreased.
    2. Sucrose solution in visking tubing created high concentration gradient. Water molecules moved from distilled water to the visking tubing by osmosis.
  7. Plant cells have cells membrane and cell wall. When the cell is placed or immersed in distilled water, the water is absorbed by osmosis. As cell becomes turgid, the cell created an inward force, wall pressure that prevents the cell from bursting.
    Diffusion  Osmosis 
    Involves movement of particles of molecules of liquid or gas. Involves movement of solvent
    It may be through a membrane or in air. It takes place through a semi-permeable.
    Not affected by PH changes. Rate affected by pH changes.
    1. Isotonic solution- a solution which has the same concentration as the cell sap.
    2. Hypotonic solution- a solution which is less concentrated than the cell sap.
    3. Hypertonic solution- A solution which is more concentrated than the cell sap.
  10. Plants normally grow in soils whose solute concentration is lower than that of the cell sap. This enables the plants to take up water by osmosis. Addition of large amounts of salt to the soil increases the solute concentration of soil water beyond that of the cell sap. The result is that the plants lose water to the soil by osmosis. Since water is very important for maintaining the structural and metabolic activities of plants, its deficiency leads to death of the plants.
    1. The red blood cells take in water by osmosis. They swell and exert pressure on the fragile plasma membrane which then breaks. Plant cells take in water and swell but do not burst. This is because their tough cell wall can only stretch to a limited extent. Once fully stretched, the cell wall resists further expansion of the cell and no more water is taken up.
    2. Fresh water protozoa take in water by osmosis. The excess water is then actively pumped into the contractile vacuole which discharges the water to the outside.
    1. A- The strip increased in length/ size; B -     Decreased in length/ size;
    2. The sugar solution was hypotonic to the cell sap strip A; it gained water by osmosis hence increasing in length;
    1. The potato cup will be filled with solution;
    2. The solution in the potato cells is hypertonic to the water; hence water moves into the cell by osmosis; this makes the solution in the neighbouring cells to be hypertonic to the outer cells; hence water moves from cell to cell until it eventually enters the potato cup;
      1. Will lose water by osmosis and become plasmolysed;
      2. If a plant cell is placed in distilled water it will gain water by osmosis and become turgid and the presence of the cell wall prevents it from bursting
  15. Diffusion; Osmosis; Active transport ; 
    1. 3.0 + 3.1 + 3.2 = 9.3 g;
      Average = 9.3/3 = 3.1g;
    2. The cell sap had a higher concentration of solutes than distilled water, water therefore moves from the environment to the cell by osmosis ;
    1. red blood cells placed in a hypertonic solution and as a result lost water to the surrounding through osmosis hence shrunk/crenated ;
    2. Appearance of that cell if subjected to the same condition

      cell in hypertonic solution
    1. Haemolysis
    2. Plant cell will lose water the cell sap to the outside solution by osmosis; the cell becomes plamolysed/ flaccid; but it will retain its shape due to rigid cell wall;
    1. Haemolysis ;
    2. The plant cell will draw in water molecules by osmosis; it will swell and become turgid; but it will not burst because of the presence of cellulose cell wall;
    1. A-no change in; number because 0.9% sodium chlorine solution is isotonic to RBC/blood; B-fewer in number because 0.3% sodium chloride solution is hypotonic to RBC/blood therefore some water was drawn in to RDC by osmosis ;leading to haemolysis/boosting of RBCs
      1. number will not change;
      2. RBC will appear small in size/wrinkled/crenated/shriveled/shrink; 1mk    Reject. Flaccid/flabby/plasmolysed
      1. Parenchyma;
      2. Collenchyma;
      3. Xylem: and sclerenchyma
    1. X – hypotonic solution;
      Y – hypertonic solution;
    2. A – haemolysis;
      B – crenation /laking;
    3. The cell will maintain/retain its normal shape.
  22. Absorption of mineral salts by root hairs from the soil; Translocation of food from leaves to other parts of the plant; movement of salts from one cell to the next;
      1. Increased in length, absorbed water through osmosis, (since cylinder cells were hypertonic/ at higher concentration) and become turgid.
      2. Reduced in length, cylinder host water to the hypertonic sucrose solution/become flaccid.
      1. No change in length
      2. Cells are dead and cannot carry out osmosis.
    3. opening and closing of stomata, Support in plants, Movement of water from cell to cell, Feeding in insectivorous plants, Absorption of water by root hairs, Absorption of water in the intestines, Reabsorption of water in kidney nephron.
      1. Nucleus
      2. Maintain the shape of he cell providing support to herbaceous plants; stores sugar and salts; (mark first one)
    2. (0.5 x 100)/8 ; 62.5µm;
    3. Hypotonic solution;
    4. The potato cell sap were lowly concentrated than the surrounding solution; hence lost water molecules by osmosis through the semi permeable membrane to become plasmolysed;
    5. Re-absorption of water from the kidney tubules/ hence important in osmoregulation;
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