Name …………………………………………….……… ADM/NUMBER…………..
231/ 1 Candidate’s Signature…………..
BIOLOGY
Paper 1 Date …………………………..
(Theory)
Time :2 hours
MOCKS 1
Kenya Certificate of Secondary Education
Instructions to candidates
Write your name and class in the spaces provided above.
Append your signature and write the date of examination in the spaces provided above.
Spelling errors especially of biological terms shall be penalized
Candidates should answer the questions in English.
Answer ALL questions in the spaces provided.
For Examiner’s Use Only
| Question | Maximum Score | Candidate’s Score |
| 1 – 29 | 80
|
|
This paper consists of 12 printed pages.
Candidates should check the question paper to ascertain that
all the pages are printed as indicated and no questions are missing.
BIOLOGY FORM ONE SUMMARIZED NOTES
Biology Form one to four exams, notes and revision materials
BIOLOGY FORM FOUR SUMMARIZED NOTES
BIOLOGY FORM TWO SUMMARIZED NOTES
Biology notes form 1-4; Free KCSE downloads
BIOLOGY FORM THREE SUMMARIZED NOTES
Biology notes (Updated form one to four free notes)
Biology Simplified Notes Form 1 to 4 Free
Biology Free notes and Exams for Form one to Four
BIOLOGY KCSE MARKING AND SETTING TIPS IN LINE WITH EMERGING TRENDS
Biology topical questions and answers
Biology KCSE Past Papers and all marking schemes free downloads
Biology free lesson plans for all topics (Form one to four)
CBC Senior School Subjects (Grade 10, 11, 12)
CRE free lesson plans for all topics (Form one to four)
History and Government free lesson plans for all topics (Form one to four)
Free Secondary School Exams and Marking schemes (Form 1 to 4)
Syllabus For All Secondary Schools Per Subjects (Latest Syllabus)
English free lesson plans for all topics (Form one to four)
FORM 4 EXAMINATIONS AND MARKING SCHEMES: ALL SUBJECTS FOR KCSE CANDIDATES- OVER 1,000 PAPERS
Form 4 Term 1-3 Free Exams and marking schemes; All subjects downloads
FORM 3 ALL SUBJECTS EXAMS, ASSIGNMENTS: FREE TERM 1-3 EXAMS & ANSWERS
Updated Secondary School Notes form one to four Free Downloads (All subjects Comprehensive Notes)
Physics free lesson plans for all topics (Form one to four)
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Enzyme X
CO2(g)+ H2O(l)H2CO3(aq)
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1 2 small animal 3
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Jar 2………………………………………………………………………………………………………
Jar 3………………………………………………………………………………………………………
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START OF EXPERIMENT
END OF EXPERIMENT
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| Intended message | Actual message | |
| I | Metereis a top school | Metre is a top school |
| II | The microscope is my tool | The microscope is my loot |
Identify the type of gene mutation represented in each case
I………………………………………………………………………………………………… [1 mark]
II……………………………………………………………………………………………….. [1 mark]
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the alleles for black and white skin colour in guinea pigs (Caviaporcellus). Give the term used
todescribe this phenomenon [1 mark]
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Surface view Cross section
State two morphological features of cell represented in the diagram above [2 marks]
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| Stomatal opening (μm) | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| Windy | 40 | 63 | 74 | 86 | 94 | 110 | 124 |
| Still air | 0 | 6 | 12 | 19 | 23 | 27 | 30 |
(a) (i) Compare the rates of transpiration in windy and still air conditions [1mark]
(ii) Explain your observation in a(i) above [2marks]
(b) How does stomatal opening affect transpiration rate? [1mark]
| Percentage oxygen in aeration stream | ||||||
| 0 | 5 | 10 | 15 | 20 | 100 | |
| Sugar loss | 15 | 20 | 42 | 45 | 45 | 48 |
| Potassium gain | 5 | 55 | 70 | 73 | 75 | 70 |
I II
At 6am At 2pm
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[2 marks]
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NAME:………………………………………………………………………………………….…………ADM/NO. ………………………….
SCHOOL:…………………………………………………………..………………SIGNATURE………………………DATE…………….…
231/2
BIOLOGY
PAPER 2
(Theory)
JULY 2023
TIME: 2HOURS
MOCKS 1 2023
Kenya Certificate of Secondary Education
INSTRUCTIONS TO CANDIDATES
For examiners use only:
| Section | Question | Maximum score | Candidates score |
| A | 1 | 8 | |
| 2 | 8 | ||
| 3 | 8 | ||
| 4 | 8 | ||
| 5 | 8 | ||
| B | 6 | 20 | |
| 7 | 20 | ||
| 8 | 20 | ||
| TOTAL SCORE | 80 | ||
SECTION A ( 40 MARKS)
(ii) State a reason for each precautions stated in b(i) above [2marks]
(c) List two structural factors that affect the process under investigation [2marks]
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(a) Why was it necessary to keep the plant in the dark for 24 hours? [1mark]
(b) Give the function of each of the following in the experiment
(i) Sodium hydroxide [1mark]
(ii) Sodium hydrogen corbonate [1mark]
(c) Explain the expected observations in leaf.
(i) M when tested for starch [2mark]
(ii) N when tested for starch? [2mark]
(d) Apart from light intensity, name one other aspect of light that affects photosynthesis [1mark]
SECTION B (40 MARKS)
Answer question 6 and either question 7 or 8
The results are as shown below
| Flask | Temperature (OC)-recorded daily | ||||||||
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 (DAYS) | |
| X | 22 | 25 | 30 | 35 | 38 | 38 | 37 | 33 | 26 |
| Y | 20 | 20 | 20 | 22 | 25 | 30 | 33 | 39 | 45 |
| Z | 20 | 20 | 19 | 20 | 20 | 19 | 20 | 20 | 19 |
NAME………………………………………………………………………….. ADM/NUMBER……………………….
SCHOOL………………………………………………………………… Date……………………………………
BIOLOGY (231/3)
Paper 3 (PRACTICAL)
JULY 2023
TIME: 13/4 hour
MOCKS 1 2023
Kenya Certificate of Secondary Examinations
Instructions to candidates
(a) Write your name and Admission number in the spaces provided.
(b) Answer all the questions in the spaces provided.
(c) You are required to spend the first 15 minutes of the 13/4 hours allowed for this paper reading the whole paper carefully before commencing your work.
(d) This paper consists of 6 printed pages.
(e) Candidates should check the question paper to ascertain that all the pages are printed as indicated and that no questions are missing.
For Examiner’s Use Only
| QUESTION | MAXIMUM SCORE | CANDIDATE SCORE |
| 14 | ||
| 13 | ||
| 13 | ||
| 40 | ||
(a).Cut each specimen into two equal halves. From each specimen, crush one half and leave the other half as a solid piece. Place the solid half of specimen P into a test tube labeled K. Place the solid half of specimen Q into a test tube labeled L.
Put about 2cm3 hydrogen peroxide into each of the test tubes.
(i) State the observations made in the two test tubes. [2marks]
Test tube K ………………………………………………………………………………………
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Test tube L ………………………………………………………………………………………..
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(ii)Place the crushed specimen P into test tube labeled M and also place the crushed specimen Q into test tube labeled N. Add 2cm3 hydrogen peroxide into test tube M and N. Record the observation for each test tubes M and N in comparison to K and L [2marks]
Test tube M ……………………………………………………………………………………
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Test tube N …………………………………………………………………………………….
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(iii) Write down an equation for the reaction that was responsible for your observations in the experiments above. [1mark]
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(iv) Name the process represented by the equation in (iii) above. [1mark]
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(b) Explain how crushing affected the results of the experiments. [2marks]
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(c) Apart from the process named in (a) (iv) above, name three other functions of specimen Q[3marks]
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(d) Explain the importance of the process named in (a) (iv) above in living organisms [3marks]
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Make a solution of the flour provided by adding water and stirring properly. Sieve or decant to obtain a solution from the mixture.
(a) (i) Using the reagents provided test for the presence of starch, proteins and lipids in the solution from specimen Z. Record the procedures, observation, and conclusions in the table below. [9marks]
| FOODSUBSTANCE | PROCEDURE | OBSERVATION | CONCLUSION |
| Starch
| |||
| Proteins |
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| Lipids |
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(ii) From the conclusions made in (a) (i) above, suggest the regions of the alimentary canal where the digestion of specimen Z would take place. [3marks] …………………………………………………………………………………………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………….
(b) State one use of any two food substances found in specimen Z. [2 marks]
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(a) Use the following features in the order in which they are listed, to prepare a dichotomous key: [8 marks]
Type of leaf
Shape of the lamina
Succulent or non-succulent
Leaf margin
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(b) (i) Name the likely habitat of specimen C. [1mark]
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(ii) Give a reason for your answer in (b) (i) above. [1mark]
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(c) State the significance of the shiny upper surface of specimen A. [2marks]
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MOCKS 1 2023
Piece of flesh labeled P
Piece of Liver labeled Q
Groundnuts flour labelled Z( each student requires about 20gms of the flour)
Leaves – (a) Broad leave with smooth margins e.g Mango labeled A
(b) Grass leaf Labeled B.
(c) Tradescantia leaf / any succulent leaf labeled C
(d) Broad leaf with serrated leaf margine.g Tobacco leaf labeled D
(e) Compound leafe.gJacaranda labeled E
Iodine solution
Absolute ethanol
10% sodium hydroxide solution
1% copper (ii) sulphate solution
Hot water bath
Means of labeling
Four test tubes
MOCKS 1 2023
| Transpiration | Guttation |
| Loss of water from a plant surface in from of water vapour | Loss of water from a plant surface in form of liquid droplets |
| Take place/ water is lost through stomata, lenticels, thin cuticle | Takes place/ water is lost though hydathodes; |
Trypsin secreted as trypsinogen
9 a) carotid artery
Population size = first marked/ captured x second capture
Marked re-captured
= 100 x 80
40 = 200 grasshoppers
Population density = 200 grasshoppers
5km2
= 40 grasshoppers/km2
b)Draw an arrow alongside diagram above to indicate direction of flow of nerve impulse in the neurone.
Man (rej men) human/ human being/, bear, hedge dog, skunk, squirrel, Raccoon, mouse, rat. Chimpazee, orangutan, Armadilo, Monkey (Accept the plural’s/ humans beings, pigs, badgers etc)
reducing/eliminating oxygen gas and sugar;
MOCKS 1 2023
b)(i)
Cut shoot under water;
Apply petroleum jelly to cork, glass/ bung glass
Open reservoir:
(ii)
To ensure no air enters leafy shoot xylem;
To ensure the apparatus is airtight;
To remove air bubbles from tube
(c)
Stomata size and number; Hairy leaf; Leaf size and shape
Leaf fall; Cuticle;
Population size = First capture X Second capture
Marked recaptured
= 70 X 40
27
= 104 garsshoppers
Large samples are more representative of the population
Sweep nets
¾ x 600 = 450 or 600 – 150 = 450 fruit flies ;
Short life cycle
Produce many offsprings
Has clear contrasting characteristics / exist in many mutant forms
Large chromosomes
Few chromosomes
Wide spread throughout the world
Not known to be a vector of human diseases
c)
– Have numerous chloroplasts to absorb light for photosynthesis
– Chloroplast are sensitive to low light intensities, thus photosynthesiseunder low light
Intensities
– The leaves are deeply dissected into thread like straws to increase the surface area for absorption of light
of water
5.(a) To destarch;
(b)
(i) To absorb carbon (iv) oxide gas;
(ii) To increase the concentration of carbon (iv) xide gas;
(c)
(i) M : brown colour of iodine solution is retained ;photosynthesis did not take place due to absence of carbon (iv) oxide gas hence no starch thus negative results;
(ii) N : blue-black colour was observed; all conditions necessary for photosynthesis were available hence accumulation of starch thus positive results;
(d)light duration; light wavelength;
6a)
Steady increase in temperature reading followed by a steady decline;the starch in soaked seeds was hydrolysed by water to reducing sugars; which was oxidized to produce energy needed for germination of seeds ;after all the stored food had been used up ,the decline in energy production led to decline in temperature reading;
Steady /continuous rise in temperature reading; boiling denatured the respiratory enzymes, bacteria(decomposers) respired as they fed on boiled seeds to bring about decomposition; due to their increasing number following steady reproduction ,the heat energy produced increased exponentially;
Constant temperature reading ;Methylated spirit disinfected the seeds hence no bacteria that would have respired to release heat energy;
To conserve the heat energy produced during respiration
To allow free circulation of air in and out of the flask;. (Bacteria need oxygen gas for respiration)
To show that germinating seeds produce energy during respiration;
To show that decomposers produce heat energy during respiration;
Locomotion
Wings for flight
Legs for walking
Segmentation to allow movement
Moulting /ecdysis to permit locomotion
Protection
Hard cuticle for protection against mechanical injury/ infestation
Polymorphic forms / colour for camouflage against predators
Cuticle is waterproof /chitinous to prevent water loss
Short necked stock over produced resulting to competition for food ; variation occurred in population ; long necked giraffes resulted among short necked ; both stock struggled for existence ; long necked giraffes were better adapted than short necked /fittest for survival; long necked giraffes reached reproductive age and passed on desirable trait to offspring while short necked giraffes failed to do so / were eliminated, long neck was naturally selected ; gradual changes accumulated in new stock forming a population of long necked giraffes ;
Fertilization is the fusion of the male and female nuclei in the embryo sac; after pollination the pollen grains absorbs nutrients from the stigma; and develops an outgrowth called the pollen tube; the pollen tube grows down the style ,to the embryo sac taking along the male gametes with it; the pollen grains usually adhere to the sigma as a result of the stigma cells secreting a sticky substance; which also stimulates the pollen grain to germinate sending down its pollen tube;the growth of the pollen tube into the stigma ,through the style to the ovary is by pushing its way between the cells where it gets nourishment from the surrounding tissues;this process is quite rapid and takes place in a matter of minutes; as the pollen grain germinates ,the tube nucleus occupies a position at the tip of the growing pollen tube; the generative nucleus divides by mitosis into two male gamete nuclei, which follow behind the tube nucleus as the pollen grows down the style; the pollen tube enters an ovule through the micropyle and when it reaches the centre of the ovule it penetrates the wall of the embryo sac and burst open; meanwhile the tube nucleus disintegrates leaving a clear way for the entry of the male nuclei; one of the male nuclei fuses with the egg cell nucleus to form a diploid zygote; which develops in to the embryo;while the other male nucleus fuses with the polar nucleus to form a triploid primary endosperm;
NAME………………………………………………………………………….. ADM/NUMBER……………………….
SCHOOL………………………………………………………………… Date……………………………………
BIOLOGY (231/3)
Paper 3 (PRACTICAL)
JULY 2023
TIME: 13/4 hour
MOCKS 1 2023
Kenya Certificate of Secondary Examinations
Instructions to candidates
(a) Write your name and Admission number in the spaces provided.
(b) Answer all the questions in the spaces provided.
(c) You are required to spend the first 15 minutes of the 13/4 hours allowed for this paper reading the whole paper carefully before commencing your work.
(d) This paper consists of 6 printed pages.
(e) Candidates should check the question paper to ascertain that all the pages are printed as indicated and that no questions are missing.
For Examiner’s Use Only
| QUESTION | MAXIMUM SCORE | CANDIDATE SCORE |
| 14 | ||
| 13 | ||
| 13 | ||
| 40 | ||
(a).Cut each specimen into two equal halves. From each specimen, crush one half and leave the other half as a solid piece. Place the solid half of specimen P into a test tube labeled K. Place the solid half of specimen Q into a test tube labeled L.
Put about 2cm3 hydrogen peroxide into each of the test tubes.
(i) State the observations made in the two test tubes. [2marks]
Test tube K …………less effervescence
…………………………………………………………………………………………………….
Test tube L …………more effervescence compared to K
…………………………………………………………………………………………………….
(ii)Place the crushed specimen P into test tube labeled M and also place the crushed specimen Q into test tube labeled N. Add 2cm3 hydrogen peroxide into test tube M and N. Record the observation for each test tubes M and N in comparison to K and L [2marks]
Test tube M ………more effervescence compared to K
…………………………………………………………………………………………………
Test tube N ………more effervescence compared to L
………………………………………………………………………………………………….
(iii) Write down an equation for the reaction that was responsible for your observations in the experiments above. [1mark]
Hydrogen peroxide → water + oxygen gas
(iv) Name the process represented by the equation in (iii) above. [1mark]
…………detoxification
(b) Explain how crushing affected the results of the experiments. [2marks]
Crushing increased the surface area ;upon which enzyme catalase can speed up the decomposition of toxic hydrogen peroxide ;
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(c) Apart from the process named in (a) (iv) above, name three other functions of specimen Q [3marks]
Deamination
Blood sugar regulation
Thermoregulation
Regulation of plasma proteins
Haemoglobin regulatiog
……………………………………………………………………………………….
(d) Explain the importance of the process named in (a) (iv) above in living organisms [3marks]
Harmful/ toxic substances e.g hydrogen peroxide are broken down; to less harmful substances such as water and oxygen gas; this creates a conducive environment for physiological processes in the cell;the oxygen gas produced can also be used in respiration;
Make a solution of the flour provided by adding water and stirring properly. Sieve or decant to obtain a solution from the mixture.
(a) (i) Using the reagents provided test for the presence of starch, proteins and lipids in the solution from specimen Z. Record the procedures, observation, and conclusions in the table below. [9marks]
| FOODSUBSTANCE | PROCEDURE | OBSERVATION | CONCLUSION |
| Starch
| To 2ml of solution Z,add iodine solution dropwise;
| Blue-black colour is observed ; | Starch present; |
| Proteins | To 2ml of solution Z,add equal amount of 10% sodiumhydroxide solution followed by 1% copper (ii) sulphate solution;
| Purple/violet colour is observed ; | Proteins present; |
| Lipids | To 2ml of solution Z,add 4ml of absolute ethanol and shake .transfer into another test tube half filled with water;
| Formation of white emulsion ; | Lipids present; |
(ii) From the conclusions made in (a) (i) above, suggest the regions of the alimentary canal where the digestion of specimen Z would take place. [3marks]
Starch –mouth;duodenum;
Proteins –stomach;duodenum;
Lipids –duodenum;ileum;
(b) State one use of any two food substances found in specimen Z. [2 marks]
Starch –digested to glucose which is then oxidized during respiration to generate energy;
Proteins –digested to amino acids which are then oxidized to release energy during starvation;
(a) Use the following features in the order in which they are listed, to prepare a dichotomous key: [8 marks]
Type of leaf
Shape of the lamina
Succulent or non-succulent
Leaf margin
1 a) leaf simple………………………………………………………….go to 2 b) leaf compound………………………………………………………E
2 a) leaf with broad lamina ……………………………………………go to 3
3 a) leaf succulent ……………………………………………………… C
4 a) leaf with serrated margin ……………………………………………..D
………………………………………………………………………………………
(b) (i) Name the likely habitat of specimen C. [1mark]
Desert;
(ii) Give a reason for your answer in (b) (i) above. [1mark]
Succulent leaves to store water
(c) State the significance of the shiny upper surface of specimen A. [2marks]
Shiny to reflect light away; hence reduce the rate of transpiration;
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