Mastering KCSE Biology with Comprehensive Topical Questions and Answers
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?
answers
​(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;
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K.C.S.E Biology Q & A - MODEL 2011PP2QN06
Explain how the osmotic pressure in the human blood is maintained at normal level.
answers
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.
answers
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. A B (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.
ANSWERS
(b) (i) A: 56—48.5 7.5C
7.5C/10 Minutes ; = 0.75C Per Minute; ±0.05 B: 48-34=14C 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.
answers
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
answers
​(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 Hydrochloric (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) Plants (ii) Animals
ANSWERS
(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
ANSWER
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
answers
​(a) L1
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. L2 Inner cells lost water by osmosis; leading to (flaccidity) decrease in length; epidermal cells did not lose water due to waterproof leading to currature (b) 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
K.C.S.E Biology Q & A - MODEL 2007PP1QN03
Plant cells do not burst when immersed in distilled water. Explain
ANSWERS
K.C.S.E Biology Q & A - MODEL 2006PP1QN12
An experiment was set up in the experiment as show below.
The set up was left for 30 minutes.
a) State the expected results. b) Explain your answer in (a) above
answers
​(a) Visking tubing will become turgid; accept will increase in volume / bulges/ swells/ becomes bigger/ expands.
(b) Sucrose solution is hypertonic/ water is hypotonic; water moves from beaker into visking tube by osmosis though semi permeable visking tubing, making visking tubing turgid. Or water moves from beaker into visking tubing by osmosis, through semi permeable visking tubing; with hypertonic solution.
​K.C.S.E Biology Q & A - MODEL 2005PP1QN07
State the importance of osmosis in plants.
answers
K.C.S.E Biology Q & A - MODEL 2004PP1QN16
Outliner three roles of active transport in the human body
answers
K.C.S.E Biology Q & A - MODEL 2004PP1QN16
how do the following factors affect the rate of diffusion?
i) Diffusion gradient ii) Surface area volume ratio iii) Temperature
answers
(i) The higher diffusion gradient between ( two points) the rate of diffusion; acc converse.
(ii) The higher the surface area:: Volume ratio, the faster is the rate of diffusion ; acc converse (iii) Increasing temperature increases the rate of diffusion; acc converse.
K.C.S.E Biology Q & A - MODEL 2004PP1QN16
What is diffusion
answers
K.C.S.E Biology Q & A - MODEL 2002PP1QN18
Two person X and Y drunk volumes of concentrated solution of glucose. The amount of glucose in their food was determined at intervals. The results are shown in the table below:
​a) On the grid provided, plot graphs of glucose level in blood against time on the same axes.
b) What was the concentration of glucose in the blood of X and Y at the 20th minute? X = 120 + -3) Y = 140 +-3) c) Suggest why the glucose level in person X stopped rising after 30 minutes while it continued rising in person Y. d) Account for the decrease in glucose level in person X after 30 minutes and person Y after 60 minutes (3 minutes) e) Name the compound that stores energy released during oxidation of glucose. f) Explain what happens to excess amino acids and development of plants.
answers
(a) – For exchanged axis award maximum 3 marks for points x identity
The scale must however be correct. For graphs on separate axis mark both and award the highest mark. (a) Axis = 2 (b) Scale = 1 (c) ( plotting) = 1 (d) curves) = 1 (b) X = 120 + -3) Y = 140 + -3 (c) Person X is capable of regulating glucose while person y is likely to be diabetic. X – Insulin (d) X insulin released, excess glucose is converted into glycogen ( in liver) must be mentioned if insulin is not mentioned Y Insulin not released, thus the decline is due to glucose being released in urine. (e) A.T.P / Adenosine triphosphate (f) Deaminated; resulting in formation of ammonia Ammonia combines with CO2 to form urea ( and H20); Urea is passed out in Urine carbohydrate group is oxidized/ stored as glycogen
K.C.S.E Biology Q & A - MODEL 2001PP1QN06
Adult elephants flap their ears twice as much as their calves in order to cool their bodies when it is hot. Explain.
ANSWERS
K.C.S.E Biology Q & A - MODEL 2000PP1QN05
State the importance of osmo-regulation in organisms
answers
K.C.S.E Biology Q & A - MODEL 1999PP1QN15
An experiment was carried out to investigate haemolysis of human red blood cells. The red blood cells were placed in different concentrations of sodium chloride solution. The percentage of haemolysed cells was determined. The results were as shown in the table below.
a) i) On the grid provided, plot a graph of harmolysed red blood cells against salt concentration.
ii) at what concentration of salt solution was the proportion of haemolysed cells equal to non-haemolysed cells? iii) State the percentage of cells haemolysed at salt concentration of 0.45% b) Account for the results obtained at: i) 0.33 percent salt concentration. ii) 0.48 percent salt concentration. c) What would happen to the red blood cells if they were placed in 0.50 percent salt solution? d) Explain what would happen to onion epidermal cells if they were placed in distilled water.
answers
a) (i)0.403; 0.404; + 0.002
ii) 0.402; iii) 9-10-11% b) Account for the results obtained at: (i) 0.33 percent salt concentration. Less concentration /hypotonic / dilute than blood cells cytoplasm/ red blood cells; water is drawn in by osmosis the cells swells and eventually burst. (ii)0.48 (ii)0.48 percent salt concentration. Concentration of cytoplasm same as concentration of salt solution/isotonic; therefore no net movement of water; hence no heomolysis. c)Percentage of cells haemolysed would still be zero? Becomes turgid; but does not burst; due to the cell wall. d)The cells would absorb water due to osmosis, swell and become turgid. The cell sap move conc. than surrounding water gate into the cell by osmosis; the cell swells/becomes turgid; but does not burst due to the cell wall K.C.S.E Biology Q & A - MODEL 1998PP1QN18
​ A hungry person had a meal, after which the concentration of glucose and amino acids in the blood were determined. This was measured hourly as the blood passed through the hepatic portal vein and the iliac vein in the leg. The results were as shown in the table below.
(a) Using the same axes draw graphs of concentration of glucose in the heptic portal vein and the iliac vein in the leg against time
(b) Account for the concentration of glucose in the hepatic vein from: (i) 0/1 hour (ii) 1-2 hour (iii) 2- 4 hours (iv) 5 – 7 hours (c) Account for the difference in the concentration of glucose in hepatic portal vein and the iliac vein between 2 and 4 hours. (d) Using the data provided in the table explain why the concentration of amino acids in the hepatic portal vein took longer to increase.
answers
​(a) If axes reversed allow marks for identification of curves only max 2
Correct scales Correctly leveled axes Curves reject broken lines for curves (b) 0-1 hour. i) Acc constant/low/below normal levels in blood; No/little digested foods/glucose from the intestines/gut/alimentary canal/absorption. ii) 1-2 hours Sharp increase in concentration of glucose in blood; (more) absorption of glucose; after digestion of the meal. iii) 2-4 hours. Glucose concentration declining/decreasing; less glucose being absorbed; (more) glucose being converted to glycogen in the liver/tissue/used for (tissue) respiration. iv) 5-7 hours. Concentration of glucose stabilizes/constant/ this is the normal glucose level concentration in the blood. (c) The concentration of glucose in the iliac vein is lower than in the hepatic portal vein because it hasn’t been stored in the liver to be used respiration. Portal vein because most of it was stored/used up by the liver/other tissues/respiration. (d) Proteins take longer to digest.
K.C.S.E Biology Q & A - MODEL 1998PP1QN11
State two ways by which the human immuno deficiency (H.I.V) is transmitted other than through sexual intercourse?
ANSWERS
K.C.S.E Biology Q & A - MODEL 1998PP1QN03
Which organelle would be abundant in:
ANSWERS
Skeletal muscle cell
K.C.S.E Biology Q & A - MODEL 1997PP1QN04
State two ways in which xylem vessels are adapted to their function
ANSWERS
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