Mastering KCSE Biology with Comprehensive Topical Questions and 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
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K.C.S.E Biology Q & A - MODEL 2002PP1QN17
State four advantages of vegetation propagation.
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K.C.S.E Biology Q & A - MODEL 2002PP1QN17
Give a reason for grafting in plants
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K.C.S.E Biology Q & A - MODEL 2002PP1QN17
What structures are produced by sisal for vegetative propagation?
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K.C.S.E Biology Q & A - MODEL 2002PP1QN16
The diagram below represents part of phloem tissue.
​a) Name the structures labeled R and S and the cell labeled T.
R S Cell labeled T b) State the function of the structure labeled S c) Explain why xylem is a mechanical tissue
answers
​(a) R. Sieve pore
S- cytoplasmic strand, cytoplasmic filaments rej. Proto plasmic strand) Cell labeled T (b) Translocation (L is tied with structures) (c) Thickened and lignified. K.C.S.E Biology Q & A - MODEL 2002PP1QN15
Ascaris lumbricoides in an example for an endo – parasite
a) The name Ascaris refers to b) State the habitat of the organism c) State three ways in which the organism is adapted to living in its habitat.
answers
(a) Genus
(b) Ileum/ colon/ duodenum/ intestines/ of humans or intestines of pig (c) Lack of elaborate elementary canal ( simple guts) can tolerate raw corn Thick cuticle pellicle, reject the outer covering lays many eggs Mouthparts for sucking partly digested food
K.C.S.E Biology Q & A - MODEL 2002PP1QN14
Distinguish between divergent and convergent evolution giving example in each case.
answers
​Divergent basic structural form is modified to serve different functions; e.g. vertebrate forelimbs, break structure in birds/ feet in birds’ convergent different structures are modified to pass or similar functions e.g. wings and birds and insects/ eye of human and octopus, vertebrates for humans e.g. squeal, legs of vertebrae and insects .
K.C.S.E Biology Q & A - MODEL 2002PP1QN14
State two ways in which Home sapiens differs from Homo habilis
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K.C.S.E Biology Q & A - MODEL 2002PP1QN14
What is organic evolution
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K.C.S.E Biology Q & A - MODEL 2002PP1QN13
The chart below shows the number of chromosomes before and after cell division and fertilization in a mammal.
a) What type of cell division takes place at Z
b) Where in the body of a female does process Z occur c) On the chart, indicate the position of parents and gametes d) Name the process that leads to addition or loss of one or more chromosomes. e) State three benefits of polyploidy in plants to a farmer
answers
​a) Meiosis
b) Ovary c) parent must be the 2n top; any ‘n’ is a gamete d) Non – dysfunctions e) increased yields / highbred Vigor, Resistance decreases Resistance to drought. K.C.S.E Biology Q & A - MODEL 2001PP1QN15
The diagram below represents a set up to investigate the conditions necessary for seed germination.
The set up was left for 7 days
(a) What conditions were being investigated in the experiment? (b) State three reasons for soaking seeds in set ups A and B (c) What were the expected results after seven days?
answers
(a) water, temperature moisture (Acc. Warmth)
(b) Mobilize/ hydrolyze stored food/ active enzymes/ breaking of dormancy softening the testa / seed coat ( acc. As a solvent/ transport media.) (c) Setup A – those in set up A will germinate Setup B- those in set up B will not germinate Setup C- those in set C will not germinate K.C.S.E Biology Q & A - MODEL 2001PP1QN13
The diagram below represents the nitrogen cycle
(a) State the process labeled
A D (b) Name the compound represented by B (c) Name the group of organisms labeled C (d) (i) name the group of plants which promote process A (ii) State the part of the plant where process A takes place (e) How would excess pesticides in the soil interfere with process A
answers
​(a) A – Nitrogen fixation
D – absorption (b) Nitrate/ nitrates/ NO2 (c) Denitrifying bacteria/ Denitrifiers (d) (i) Leguminous plants, (acc. Legume/ acc examples e.g beans peas) (ii) Roots nodules; rej root or nodules alone; acc; root (e) – Killing / reducing of composers - Killing/reduction of nitrogen fixing bacteria/ nitrogen fixing microorganisms - Destruction of leguminous plants K.C.S.E Biology Q & A - MODEL 2001PP1QN12
The graph below shows the effect of substance concentration of the rate of enzyme reaction.
(a) (i) Account for the shape of the graph between A and B
(ii) B and C (b) How can the rate of reaction be increased after point B? (c) State two other factors that effect the rate of reaction of enzyme reaction
answers
(a) (i) More active sites of enzymes available, for a large number of molecules of substrate; hence increase in the rate of reaction ( rapid of fast increase in the rate of reaction)
(ii) B and C Enzymes/ substrate are in equilibrium / All active sites are occupied; hence rate of reaction is constant. (b) Raising concentration of enzymes (c) PH, temperature, inhibitors/ cofactors K.C.S.E Biology Q & A - MODEL 2001PP1QN11
The diagram below represents a mammalian nephron
(a) Name the
(i) Structure labeled P (ii) Portion of the nephron between point X and Y (b) Name the process that takes place at point Q (c) Name one substance present at point R but absent at point S in a healthy mammal (d) The appearance of the substance you have mentioned in (c) above is a symptom of a certain disease caused by a hormone deficiency. Name the (i) Disease (ii) Hormone (e) State the structural modifications of nephrons found in the desert mammals
answers
(a) (i) Efferent arteriole/ vessels
(ii) Loop of henle (b) Ultra – filtration ( acc. Pressure filtration) rej. Filtration (c) Glucose ( acc. Blood sugar) (d) (i) Disease – diabetes mellitus ( acc. Sugar diabetes) (ii) Hormone – insulin (e) Small Bowman’s Capsule/ Groleruli`; Rej few Bowman’s capsule Loop of Henle
K.C.S.E Biology Q & A - MODEL 2000PP1QN19
How are leaves of mesophytes suited to their functions
answers
​Broad/ wide/ flat lamina provides large surface area for absorption of (O) and sunlight, thin to ensure short distance of CO2 reach photosynthesis/ palisade cells; presence of stomata guard cells for efficient diffusion of O2 gaseous exchange / H2O vapour transpiration/ CO2 into the leaf transparent cuticle epidermal cells; for light penetration into palisade cell which contains chloroplast next to upper epidermis; these receives maximum light for photosynthesis. Chloroplasts have chlorophyll, which traps light energy.
Leaves have vein, xylem and phloem to transport products of photosynthesis to other part of the plant. Air spaces on spongy mesopyll, easily circulates gases/ CO2 diffuse into palisade cells. Mosaic arrangements of leaves; enable leaves to trap sunlight.
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.
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
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K.C.S.E Biology Q & A - MODEL 2000PP1QN16a
What is the significance of sexual reproduction?
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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 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. |
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