K.C.S.E Biology Q & A - MODEL 2013PP1QN05
Describe how turgor pressure builds up.
K.C.S.E Biology Q & A - MODEL 2013PP1QN01
Explain how an increase in temperature affects the rate of active transport.
K.C.S.E Biology Q & A - MODEL 2012PP2QN03
(a) In an investigation, equal amounts of water was placed in. three test-tubes labelled G, H and J. Pondweeds of equal length were dropped in each test tube. The test-tubes were then placed in identical conditions of light and carbon (IV) oxide at different temperatures for five minutes. After the five minutes, the bubbles produced in each test-tube were counted for one minute. The results were as shown in the table below.
(i) Name one requirement for this process that is not mentioned in the investigation.
(ii) Name the gas produced in this investigation.
(iii) Account for the results in test-tubes H and J.
(b) State two ways in which the human intestinal villus is adapted to its function.
(a) (i) chlorophyll;
(ii) oxygen; (1 mark)
(iii) Test tube H is at optimum temperature for enzyme activity; hence high rate of
photosynthesis/more bubbles. In test tube J most enzymes have been denaturedby the high temperature; hence low rate of photosynthesis/fewer bubbles.
(b) — The villus epithelium is thin; for faster diffusion of dissolved food substances;
— The epithelium has goblet cells; which produce mucus to lubricate food passage;
— They have microvilli; which further increase their surface area for absorption;
Have lacteal; for absorption of fatty acid & glycerol/transportation of lipids;
Highly vascularised; for absorption of digested food.
K.C.S.E Biology Q & A - MODEL 2012PP1QN16
The diagram below represents an experimental set-up used by students to investigate a certain process.
Flower Q produced seeds while P did not. Account for the results.
K.C.S.E Biology Q & A - MODEL 2012PP1QN13
State three factors that affect the rate of diffusion.
Temperature; surface area; distance that particles have to travel; diffusion/concentration gradient; size/density of particles; surface area to volume ratio; thickness of membrane; medium of diffusion
K.C.S.E Biology Q & A - MODEL 2012PP1QN07
The diagram below represents a set-up that students used in an investigation.
(a) Name the physiological process that was being investigated.
(b) State the role of potassium hydroxide in flask K.
(c) Account for the observation in boiling tube L and flask N.
(a) Respiration/aerobic respiration;
(b) Flask K Potassium hydroxide removes Carbon IV Oxide from atmospheric air;
(c) L - Lime water remains clear because Carbon (IV) Oxide has been removed;
Flask N lime water forms a white precipitate because the respiring cockroaches produce Carbon (IV) Ovide;
K.C.S.E Biology Q & A - MODEL 2012PP1QN04
In an investigation, a student extracted three pieces of paw paw cylinders using a cork borer.
The cylinders were cut back to 50 mm length and placed in a beaker containing a solution.
The results after 40 minutes were as shown in the table below.
(a) Account for the results in the table above.
(b) What would be a suitable control set-up for the investigation?
(a) The solution was hypotonic/less concentrated compared to the cell sap of pawpaw cylinder cells;
The tissue/cells gained water by osmosis; becoming turgid/longer/stiff;
(b) Pawpaw cylinders of the same size/length; placed in an isotonic solution;
Boiled potato cylinders of the same size; placed in a similar solution;
K.C.S.E Biology Q & A - MODEL 2011PP2QN06
Explain how the osmotic pressure in the human blood is maintained at normal level.
K.C.S.E Biology Q & A - MODEL 2011PP1QN07
The diagrams below show an experimental set-up to investigate a certain process in a plant tissue.
Explain the results obtained after 30 minutes.
K.C.S.E Biology Q & A - MODEL 2010PP2QN06
In an experiment to investigate a certain physiological process, a boiling tube labelled A and a test tube labelled B were covered with cotton wool. The two tubes were simultaneously filled with hot water and fitted with thermometers. The experimental set-up was as in the diagrams below.
Temperature readings were taken at the start and after every two minutes for twenty minutes. The results were as shown in the table below.
(a) Using the same axes, draw graphs of temperature against time.
(b) (i) Work out the rate of heat loss in the boiling tube labelled A and test-tube labelled B between the 5th and 15th minutes.
(ii) Account for the answers in (b) (i) above.
(iii) How does the explanation in (b) (ii) above apply to an elephant and a rat?
(c) (i) State the role of the cotton wool in this experiment. ,
(ii) Name two structures in mammals that play the role stated in (c) (i) above.
(d) State three advantages of having constant body temperature in mammals.
(b) (i) A: 56—48.5 7.5C
7.5C/10 Minutes ; = 0.75C Per Minute; ±0.05
14C/10 Minutes ; =1.4C Per Minute; ±0.05
(ii) B has a larger surface area to volume ratio; making it to lose heat to the surrounding faster; (the converse is true)
(iii) A rat has larger surface area to volume ratio compared to an elephant; making the rat to lose heat at a faster rate than an elephant;
(c)(i) Insulation/insulate against heat loss; (to surrounding);
(ii) Subcutaneous fat layer / adipose tissue;
Fur / hair; .
(d) Are active always; (even under very cold conditions)
Are able to escape from predators/search for mates/food; (because they are active always)
Can survive in a wide variety of habitats: (both cold and hot)
K.C.S.E Biology Q & A - MODEL 2010PP1QN07
Distinguish between haemolysis and plasmolysis.
K.C.S.E Biology Q & A - MODEL 2009PP2QN06
An experiment was carried out to investigate the effect of temperature on the rate of reaction catalyzed by an enzyme. The results are shown in the table below
(a)On t he grid provided draw a graph of rate of reaction against temperature
(b) When was the rate of reaction 2.6 mg of product per unit time?
(c) Account for the shape of the graph between
(i) 5 C and 40 C
(ii) 45 C and 60C
(d) Other than temperature name two ways in which the rate of reaction between 5C and 40C could be increased
(e) (i) Name one digestive enzymes in the human body which works best in acidic condition
(ii) How is the acidic condition for the enzyme named in (e) (i) above attained?
(f) The acidic conditions in (e) (ii) above is later neutralized
(i) Where does the neutralization take place?
(ii) Name the substance responsible for neutralization
(b) 33C and 51.5 ( ± 0.5C)
32.5 - 33.5 and 51.0 – 52.0
(c)(i) As temperature is increased rate of reaction is increased/ more products are formed (per unit time) because enzymes become more active
(ii) As temperatures increases rate of reaction decreases less products are formed (unit per time) because enzymes become denatured by high temperatures.
(b) Increase in enzyme concentration and substance concentration
Increasing number of enzyme
(e) (i) Pepsin, remain/ chymosin
(ii) Wall of stomach/ gastric gland/ oxyntic/ pariental/ cell produced
(f) (i) Duodenum
(ii) Bile juice/ SANS any correct salt e.g. NaHCO3
K.C.S.E Biology Q & A - MODEL 2009PP1QN13
State one role that is played by osmosis in
(i) absorption of water from the soil by root hair cells/ movement of water between plant cells/ from cell to cell/ opening one closing of stomata/ support in herbaceous plants due to turgidity / feeding in insectivorous plant.
(ii) Water reabsorption by blood capillaries from renal tubules/ absorption of water in colour dicututary/ canal/ gut movement of water from cell to cell in animals.
K.C.S.E Biology Q & A - MODEL 2009PP1QN13
Distinguish between diffusion and active transport
In diffusion molecules move from a highly conc. Region to a lowly conc. Region while in active transport molecules move from a lowly concentration region to a highly concentration region; on diffusion molecules move along conc. gradient while in active transport molecules move against conc. gradient. No energy is required in diffusion while energy is required in active transport/ active requires carrier molecules while carrier molecule not required in diffusion;
K.C.S.E Biology Q & A - MODEL 2008PP2QN05
A freshly obtained dandelion stem measuring 5 cm long was split lengthwise to obtain two similar pieces The pieces were placed in solutions of different concentrations in Petri dishes for 20 minutes.
The appearance after 20 minutes is as shown
(a) Account for the appearance of the pieces in solutions L1 and L2
(b) State the significance of the biological process involved in the experiment
Inner cells gained water by Osmosis; hence increased in length; epidermal cells did not gain water because they are covered by a water proof cuticle leading to currature.
Inner cells lost water by osmosis; leading to (flaccidity) decrease in length; epidermal cells did not lose water due to waterproof leading to currature
Support in (herbaceous) plants
Absorption of water
Opening and closing of stomata
Movement of water from cell to cell
Leading in infectious plants
Folding of leaves in the Mimosa