NAME ………………………………………………ADM. NO ………………CLASS:……….
DATE……………..
231/3
BIOLOGY
PAPER 3
PRACTICAL
MARCH/APRIL
TIME: 13/4HRS
KENYA CERTIFICATE OF SECONDARY EDUCATION
Instruction to Candidates
| QUESTION | MAXIMUM SCORE | CANDIDATE’S SCORE |
| 1 2 3
| 15
14
11
| |
| TOTAL SCORE | 40 |
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
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Test tube A:…………………………………………………………………….
Test tube B:…………………………………………………………………….
| Food Substance | Procedure | Observation | Conclusion |
|
|
| ||
|
|
| ||
|
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A –
B –
C –
D –
F –
1 (a) Animals with jointed appendages ……………………go to 3
(b) Animals without jointed appendages …………………go to 2
2 (a) Animals with a slender long body ……………………Nematoda
(b) Animals with a thick short body ……………………..Mollusca
3 (a) ………………………………………………………..go to 5
(b) Animals without wings ………………………………go to 4
4 (a) Animals with numerous legs ……………………….Myrioponda
(b) ………………………………………………………Hymenoptera
5 (a) Animals with short antenna …………………………Diptera
(b) Animals with a pair of long antenna ………………..go to 6
6 (a) Animals with cuticulized forewings ………………..Dictyoptera
(b) Animals with a pair of membranous wings………….Hymenoptera
Identity Steps followed
A ……………………………… …………………………………………..
B ……………………………… …………………………………………..
C ……………………………… …………………………………………..
D ……………………………… …………………………………………..
E ……………………………… …………………………………………..
F ……………………………… …………………………………………..
G ……………………………… …………………………………………..
EXAMS TERM 2
FORM 3 BIOLOGY P3 MARKING SCHEME
Liquid becomes cloudy/turbid suspension formedÖ/oil broken up into small droplets which are dispersed throughout the liquid. Ö (The oil becomes emulsified)
Test tube Y
Oil floats on the water/two separate/immiscible layers are formedÖ
(ii) Emulsification Ö
(iii) Increased surface area for action of enzyme lipase Ö (answer tied to a (ii))
(iv) BileÖ
(v) DuodenumÖ (tied to (a) (ii) and (iv) )
(b) (i) Blue black Ö
(ii) Starch Ö
(iii) Contents of F remain unchanged. Blue black colour in E disappears/fades/changes to pale/light yellow/light brown/orange. (Answer tied to b (ii) )
(iv) Enzyme/Amylase in potatoÖ breaks down starch/converts/hydrolyses/changes/digests Östarch into maltose/reducing sugarsÖ/simple sugars that do not give a blue black colour with iodine.Ö(1mk)
Procedure
(c) (i) Add equal amount of Benedict’s solution to paste and boil in a hot water bath
| Food being tested | Procedure | Observation | Conclusion |
| Reducing sugarÖ | To the food substance add equal amounts of Benedict’s solution and heat/boil (in a hot water bath) Ö | Colour changes from blue to green to yellow to orange and finally brown Öor colour changes to brown | Reducing sugars present Ö |
4/2 max 2mks
(ii) Starch Öin potato is converted to maltose/glucose/reducing sugar Öby enzyme amylase/maltose/diastaseÖ. Rejptylin
Q2. The photographs labelled J, K and L are all related to mammalian kidney.
(a) Name the hormone produced by the structure labelled P.: Adrenaline ;aldosterone (1 mk)
(d) (i) Give two components of blood that that are not filtered at structure S. – Blood cells / Plasma protein (2 mks)
(ii) Give reason why the components you have named in d (i) above are not filtered. (2mks)
They have very large molecules; structures that can filter through the pores in the glomerulus.
(e) Give two nutrients reabsorbed at the part labelled S – Glucose / Amino acids (2 mks)
(f) What three adaptations would be expected in the structure L in a desert animal like a camel. (3 mks)
| Superior / hypogynous Ovary ; | Inferior / Ovary;Epigynous ovary |
| Monocarpous ; | Polycarpous / apocarpous / free capels |
;
C – Anther ;
D- Sepal / Calyx ;
The ovary develop into a fruit ;
The ovules develop into seeds
The ovary wall develop into a pericarp;
Reason – The floral structure (anthers ) are in five in Q and 12 ( multiple of 4) in R ;
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