Mastering KCSE Biology with Comprehensive Topical Questions and Answers
K.C.S.E Biology Q & A - MODEL 2000PP1QN18
Describe the role of hormones in the human menstrual cycle
answers
​Inferior lobe of pituitary gland secretes F.S.H which causes grafian follicle develops in the ovary. It also stimulates ovary tissue/ ovary/ follicle walls secret estrogen which repairs, heals uterine wall, oestrogen stimulates inferior lobe of pituitary gland produce L.H. for ovulation. It also causes grafian follicle change into corpus interim L.H stimulates corpus luteum secret progesterone which causes proliferation of the uterine walls; in preparation of implantation; oestrogen/ progesterone inhibits the production of F.S.H ( by anterior lobe of pituitary) thus no more follicle develop; and oestrogen production reduces; 14 days later progesterone level rises inhibits production of L.H from anterior lobe of pituitary gland produce L.H for ovulation. It also causes grafian follicle change into corpus interim L.M stimulates corpus luteum secret progesterone which causes proliferation of the uterine walls in preparation of implantation; oestrogen/ progesterone inhibits the production of F.S.H ( by anterior lobe of pituitary) thus no more follicle develop; and oestrogen production reduces; 14 days later progesterone level rises inhibits production of L.H from anterior lobe of pituitary gland/ The corpus luteum stops secreting progesterone, and menstruation occur when the level of progesterone drops; ( anterior lobe of pituitary starts secreting F.S.H again.
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K.C.S.E Biology Q & A - MODEL 2000PP1QN17
The numbers of different types of animals supported by a square kilometer in two terrestrial ecosystems are shown in the table below
(a) (i) Which domestic animal is better adapted to both ecosystems?
(ii) Give a reason why the animal named in (a) (i) above is better adapted to the two ecosystems. (b) Why are cattle and sheep fewer in the bush land than in the savannah? (c) (i) Name suitable methods that were used to estimate the population of: Domestic animals Wild animals (ii) Give a reason why the method named for wild animals in (c) (i) above is suitable (d) state three methods which could be used to determine the diet of wild animals in an ecosystem (e) Name four biotic factors that could have regulated the animal population in both ecosystems (f) State four human activities that affect population of animals in game parks (g) What is the importance of national park to a nation?
answers
​(a) (i) Goat
(ii) It is a grazer and a browser (b) Insufficient grass in bush/ aren’t adapted to eating twigs/ not browsers/ are grazers (c) (i) Domestic animals - total counts Wild animals – total counts; aerial counts/ quadrat/ Belt transect/ capture/ recapture (ii) Analyzing gut counts, studying dentition/ breaks/ claws/ parts (d) Observation Examine droppings Dissecting a sample of animals/ study structure/ nature of digestive System/ size of caecum/ length of intestine/ chamber (e) Irrigation Competition; diseases Predation; human activity/ man accept any correct Parasitism (f) Poaching, cropping/ culling/ licensed spot hunting (g) Pollution; translocation Burning trees, charcoal- deforestation
K.C.S.E Biology Q & A - MODEL 2000PP1QN16b
State three advantages of asexual reproduction
answers
K.C.S.E Biology Q & A - MODEL 2000PP1QN16a
What is the significance of sexual reproduction?
answers
K.C.S.E Biology Q & A - MODEL 2000PP1QN15
The concentration of the lactic acid in blood during and after an exercise was determined. The results are shown in the graph below
(a) (i) By how much did the lactic acid increase at the end of 13 minutes?
(ii) After how many minutes was the lactic acid concentration 71mg/100cm3? (iii) What would be the concentration of lactic acid at the 60th minute? (b) Give a reason for the high rate of production of lactic acid during the Exercise (c) Give a reason for the decrease in the concentration of lactic acid after the exercise
answers
​(a) (i) 78/78 mg/ 100cm3
(ii) 8.5th and 29.5th / 8min 30 sec and 29 min 30 sec (iii) 47 mg/100cc; Acc. 47 (b) The demand for oxygen is more than the supply leading to anaerobic respiration. Acc. Lactic acid converted to glucose/Glycogen (c) Lactic acid is oxidized ( to form CO2 and H2O) Acc. Lactic acid is converted to glucose/ glycogen K.C.S.E Biology Q & A - MODEL 2000PP1QN14
(a) Name the crop infested by phytophthora infestant and the disease it causes Crop / Disease
(b) State four control measures against the disease
answers
​(a) Crop
Potatoes / tomato Disease Tomato/potato bright/ Acc. Tomato rot (b) Use of fungicides Eradication of infected crop/ uprooting/ burning of infected plants Use biological control Use of disease resistant varieties Crop rotating K.C.S.E Biology Q & A - MODEL 2000PP1QN13
The temperature of a person was taken before, during and after taking a cold bath. The results are shown in the graph below
(a) Explain why the temperature fell during bath
(b) What changes occurred in the skin that enabled the body temperature to return to normal?
answers
​(a) Heat loss by conduction/ convection from the blood vessels
The body skin to the cold water, the cooler blood leaving skin enters general circulation cooling the whole body. (b) Vasoconstriction; thus less blood flowing to the skin reducing heat loss. Sweating eases heat produced through metabolism Accept shivering producing heat K.C.S.E Biology Q & A - MODEL 2000PP1QN12
The chart below represents the result of successive crosses, staring with redflowered plants and white flowed plants and in which both plants are pure breeding.
(a) What were parental genotype? Use letter R to represent the gene for red colour and r for white colour
(b) (i) What was the colour of the flowers in the first filial generation? (ii) Give a reason for your answer in b (i) above (c) If 480 red flowered plants were obtained in the second filial generation, how many F2 plants and white flowers? Show your working.
answers
(a) RR and rr
(b) (i) red (ii) complete dominant; i.e Rd dominant/ white recessive (c) Ratio of filial generation: 3: 1 (I.e. in every 4 flowers 3 are red 1 is white Therefore 480 red flowers means ¾ of the total number Total number of flowers (480 x 4)/3 = 640 So ¼ of 640 flowers are white in F2 plants ¼ x 640 = 160 flowers K.C.S.E Biology Q & A - MODEL 2000PP1QN11
The diagram below represents a section of a leaf.
(a) Name the parts labeled X, and Y
(b) Using arrows indicate on the diagram the direction of flow of water during the transpiration stream (c) State two ways in which the leaf is suited to gaseous exchange
answers
​(a) X– Spongy mesopyll ( cell) layer
(b) Y – Cuticle (c) Broad/ flat leaf ( lamina) to provide large surface area or absorption of gases Thickness: allow gases to pass though fast Presence of stomata for efficient diffusion of gases Presence of air spaces for easy defuses
K.C.S.E Biology Q & A - MODEL 2000PP1QN10
Explain how birds of prey are adapted to obtaining their food.
answers
K.C.S.E Biology Q & A - MODEL 2000PP1QN09
State two advantages of metamorphosis to the life of insects.
answers
K.C.S.E Biology Q & A - MODEL 2000PP1QN07
Why is oxygen important in the process of active transport in cells?
answers
K.C.S.E Biology Q & A - MODEL 2000PP1QN07
Give reason why each of the following is important in the study of evolution:
a) Fossils records b) Comparative anatomy.
answers
(a) Fossils records
Gives evidence of types of plants/ animals/ organism that exist at a certain geological age. Long ago Gives evidence of morphological/ anatomical. Structure/ changes that have occurred over a long period of time. (b) Comparative anatomy. Gives evidence of relationship among organisms Gives evidence of common ancestry of a group of organisms; e.g. structural/ functional relationship among organization
​K.C.S.E Biology Q & A - MODEL 2000PP1QN06
Give a reason why lumbar vertebrae have long and broad transverse process
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 2000PP1QN03
State two ways in which some fungi are beneficial to humans
answers
K.C.S.E Biology Q & A - MODEL 2000PP1QN02
Give a reason why two species in ecosystems cannot occupy the same niche.
answers
K.C.S.E Biology Q & A - MODEL 2000PP1QN01
What is the function of the following cells in the retina of human eye ? Cones
answers
(a) Cones Discrimination of colours/ details/ accurate/ vision colour perception/ sensitivity to high intensity/ bright
(b) Rods Dim light vision/ low light intensity
K.C.S.E Biology Q & A - MODEL 1999PP1QN17
Explain how the various activities of man have caused pollution of air.
answers
K.C.S.E Biology Q & A - MODEL 1999PP1QN16
Describe the:
a) Process of inhalation in mammals. b) Mechanisms of opening and closing of stomata in plants.
answers
a) Muscles of diaphragm contract; causing the diaphragm to flatten (from dome position. The external intercostals muscles contract internal intercostals muscles relax pulling the ribcage upward/forward and outward in man. These movements increases the volume of the thoracic cavity; reducing the pressure; of the thoracic cavity; compared to atmospheric pressure; this causes the atmospheric air to rush into the lungs. (Through the nostrils, trachea bronchioles and alveoli).
b) Theory- photosynthesis Guard cells have chloroplasts; in the presence of light; photosynthesis occurs in guard cells, producing sugar in guard cells; osmotic pressure increases/osmotic potential lowers; water from neighboring /adjacent cells enter into guard cells; causing turgidity of guard cells; causing turgidity of guard cells. Theory 1. Guard cells have chloplasts; in the presence of light photosynthesis occur in the guard cells of stomata; producing in the guard cells; osmotic pressure increases/lowers osmotic potential water from the neighboring /adjacent cells, enter into guard cells; causing turgidity of guard cells The inner walls of the guard cells are thicker than outer walls; so during turgidity the inner walls stretch more; causing the guard cells to bulge outward; stomata opens. Theory 2. Guard cells have chloroplasts (Day) in light; photosynthesis occurs in the leaf/guard cells lowering the CO2 concentrations; this increases PH/alkalinity which triggers of enzymatic conversion of starch to sugar (glucose); leading to low osmotic potential/ increased osmotic pressure in guard cells; guard cells absorb water from epidermal cells; thus becoming turgid; the inner walls are thicker than the outer walls; outer walls stench more than inner walls; causing guard cells to bulge outwards, stomata opens; In the absence of light (night); no photosynthesis; CO2 concentration increases due to respiration; PH lowered/ acidity increases; sugar converted to starch; osmotic pressure lowered/ osmotic potential increases; guard cells lose water to adjacent epidermal cell becoming flaccid; stomata close. Day low H+ high PH opens stomata. Starch glucose. Theory 3 Guard cells have chloroplasts; in light AT produced; the energy drives K+ irons from adjacent epidermal cells into guard cells; accumulation of K+ raises osmotic pressure (lower osmotic potential) of guard cells; guard cells absorbs water from adjacent epidermal cells; becoming turgid; the inner walls are thicker than the outer walls so outer walls stretch more than inner walls causing guard cells to bulge outward. Stomata opens. In the absence of light (night ) ATP rapidly decreases; no energy of potassium +ions pump ion; migrate by diffusion from guard cells to adjacent epidermal cells; become flaccid; the thinner outer walls of guard cells shrink 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 1999PP1QN14
The photograph below represents a blood smear obtained from a person suffering from a certain disease.
a) Name the structure labeled X.
b) i) Name the structure labeled L ii) State the function of the source labeled M c) What disease was the person suffering from? d) List three ways by which micro-organisms enter the human body.
answers
a) Trypanosome
b) i) Locomotion ii) c) Sleeping sickness/trypanosomiasis d) -Orally ingested including boring through bites Sexually; cuts and wounds (contaminated) needles syringes/surgical instruments; contaminated blood transfusion.
K.C.S.E Biology Q & A - MODEL 1999PP1QN13b
Describe how population of grasshoppers in a given area can be estimated.
answers
​Use the capture and recapture methods; Catch the grasshoppers count and mark using permanent ink; record and release; and allow time 1 to 2 hours; recapture and count the marked and the unmarked; total population is equal to the number of marked and unmarked grasshoppers in the second sample multiplied by number marked grasshoppers in the first sample; divided by the number of grasshoppers marked in the second sample that were recaptured.
K.C.S.E Biology Q & A - MODEL 1999PP1QN13a
Distinguish between a community and population.
answers
K.C.S.E Biology Q & A - MODEL 1999PP1QN12
A student set up an experiment as shown in the diagram below..
a) i) What is being investigated in the experiment?
ii) On the diagram below indicate the expected results after three days.
​iii) Why was it necessary to have wet cotton wool in the containe?
b) What is the role of the following to a germinating seed/ i) Oxygen ii) Cotyledons.
answers
​.a) i) Region of elongation (rapid) growth in a root.
b) i) Oxygen
Oxidation of stored food; to provide energy (for germination) ii) Cotyledons Store food necessary for germination; protecting the plumule. |
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