| Answers after the questions |
1. A student set up materials in an experiment as shown below.
| Fresh potato | Boiled potato | ||
| Water |
- State the physiological process being investigated. (1mk)
(b) If the experiment set up was left over-night, state observation in the set up A and B. (2mks)
(c) Account for the observations in each set up. (3mks)
(d) If another experiment C was set such that nothing is placed in the potato cup, state and explain the results that would have been obtained. (2mks)
2. An experiment was carried out to investigate, haemolysis of human cells. The red blood cells were placed in different concentration of sodium chloride solution. The percentage of haemolysed cells was determined. The results were shown in the table below.
| Salt conc. (g/100cm3) | 0.33 | 0.36 | 0.38 | 0.39 | 0.42 | 0.44 | 0.48 |
| Red blood cells haemolysed % | 100 | 91 | 82 | 69 | 30 | 15 | 0 |
(a) (i) On the grid provided plot a graph of haemolysed red blood cells against salt concentration. (6mks
(ii) At what concentration of salt solution was the proportion of haemolysed cells equal to non-haemolysed cells? (1mk)
(iii) State the percentage of red blood cells haemolysed at salt concentration of 0.45. (1mk)
(b) Account for the results obtained at:
(i)0.33% salt concentration (3mks)
(ii) 0.48% salt concentration (3mks)
(c) What would happen to the red blood cells if they were placed in 0.50% salt solution. (3mks)
(d) Explain what would happen to onion cells if they were placed in distilled water. (3mks)
3. Explain how various environmental factors affect the rate of transpiration in plants. (20mks)
4. (a) State the meaning of the following terms.
(i) Digestion (2mks)
(ii) Ingestion (2mks)
(b) Describe the process through which a piece of ugali undergoes in man from the time of ingestion up to the time of absorption. (16mks)
- The diagram below represents a unit of gaseous exchange in man. Study it carefully and answer the questions that follow.
a) Name the blood vessel that brings blood to the lungs and the vessel which takes blood away from the lungs. (2mks)
b) Name the structure above. (1mk)
c) Label A and E. (2mks)
d) In what form is carbon (IV) oxide transported in structure labeled E. (1mk)
e) Name the gas G. (1mk)
6. Gastrin is a hormone produced by mammals.
(a) (i) Where is the hormone produced? (1mk)
(ii) What is the function of gastrin? (1mk)
(b) What stimulateds the production of gastrin. (1mk)
(c) The diagram below shows part of the human intestine.
i) Identify the parts labeled A and B (1mk
(ii) To which circulatory system does the part labeled B belong. (1mk)
d) State any two adaptations of the human large intestine to its function. (2mks)
7. The diagram below represents part of a xylem tissue.
a) (i) Name the parts labeled P and Q (2mks)
(ii) Give the function of the part labeled P. (1mks)
b) State the function of the phloem tissue. (1mk)
c) (i) State how the functioning of the phloem tissue is affected if the companion cell is destroyed. (1mk)
(ii) Give a reason for your answer. (1mk
d) State any two structural differences between phloem and xylem tissues. (2mks)
8. In an experiment to determine the effect of exercise on the concentration of lactic acid in blood, the following data was obtained. Study the data and use it to answer the questions that follow.
The lactic acid concentration was measured before, during and after the exercise.
| Time minutes | 0 | 10 | 20 | 25 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 100 |
| Lactic acid conc. (arbituary units) | 0.5 | 0.5 | 5 | 13 | 12 | 8 | 6 | 4 | 3 | 2 | 1 | 0.9 |
a) Using a suitable scale, plot a graph of the concentration of lactic acid against time. (6mks)
b) From the graph you have drawn determine
(i) The period of exercise . Explain. (2mks)
(ii) The time when oxygen debt occurred Explain. (2mks)
(iii) The duration it took to pay back the oxygen debt.Explain (2msk)
c) On the same set of axes plot a hypothetical curve for oxygen intake during the experiment period of 90 minutes. (2mks)
d) Why does lactic acid level usually continue to rise in the blood after exercise ceases. (2mks)
e) Suggest the two importance of anaerobic respiration to animals. (2mks)
d) What is oxygen debt? (2mks)
9. What is the role of the liver in the maintance of a constant level of materials in the body. (20mks)
10.The diagram below represents a simple respiratory pathway in cells
a) Name the process marked X and Y. (2mks)
b) State two differences between process X and Y. (2mks)
c) State the name of substance B and condition under which it is formed. (2mks)
d) Explain how body size affects the rate of respiration in animals. (2mks)
11. The diagram below represent the structure of a nephron. Study it and answer the questions that follow.
a) (i) State the physiological process by which solutes are selectively re-absorbed back into blood at the part labelled B. (1mk)
(ii) How is the part labeled B adapted to carry out the physiological process named in 3 (a) (i) above. (1mk)
b) In which part of the kidney is the part labelled A abundantly found. (1mk)
c) On the diagram above , indicate the direction of flow of blood using arrows at the part labelled C. (1mk)
e) State the functions carried out by the following hormones in the functioning of the nephron.
(i)Aldosterone. (1mk)
(ii)Anti diuretic hormone. (1mk)
12. The data below shows the rate of photosynthesis at different temperature in attached leaves of three East African plants. (Crotolaria, Gynandropsis and Amaranthus species) respectively which were grown outside with the same illustration while water and carbon (IV) oxide are not limiting factors in this experiment.
Rate of photosynthesis was expressed interms of carbon (IV) oxide uptake in mg/mm2/hr at various temperatures as tabulated below.
| Temperature oC | Rate of photosynthesis (mg/mm2/hr) | ||
| Gynandropsis sp | Crotolaris sp | Amaranthus sp | |
| 5 10 15 20 25 30 35 40 45 50 | – 22 50 60 80 85 80 73 66 2 | 20 40 49 64 48 45 42 31 15 – | – 10 27 42 55 54 50 45 40 11 |
a) Represent the results graphically (rate of photosynthesis against temperature)
b) Using the graph in (a) above indicate optimum temperature for the Gynandropsis and Amaranthus species. (2mks)
Gynandropsis ………………………………………………
Amaranthus ………………………………………………
c) Give a reason why Gynandaropsis and Amaranthus could not function photosynthetically at 5oC. (1mk)
d) What are the possible ecological habitats for the following plants. (2mks)
(i) Amaranthus
(ii) Crotolaria
e) At what temperature was the amount of carbon (IV) oxide around the leaf of Gynandropsis highest? (1mk)
f) What raw material is required in the light stage of photosynthesis. (1mk)
g) Name the parts of chloroplasts in which the following stages of photosynthesis take place. (2mks)
(i) Light stage
(ii) Dark stage
h) State one structural similarity and difference between chloroplast and mitochondria. (2mks)
Similarity
Difference
i)What is the compensation point of photosynthesis? (1mk)
13 (a) Explain why plants lack elaborate excretory organs like those found in animals. (3mks)
(b) Name five methods of excretion in plants. (5mks)
(c) State any six excretory products in plants and give economic uses. (12mks)
14. During a laboratory investigation, a scientist extracted gastric juice from the mammalian stomach. He used it to carry out tests on a food sample B which was suspected to contain proteins. He divided the food sample B into three portions and treated them as below.
I. On the 1st portion of B, he added Gastric juice and mixed them thoroughly before adding sodium hydroxide followed with copper (II) sulphate drop by drop.
II. On the 2nd portion of B, he added boiled gastric juice and mixed them thoroughly before adding sodium hydroxide followed with copper (II) sulphate drop by drop.
III. On the 3rd portion of B, he added Gastric juice, sodium bi-carbonate and mixed them thoroughl before adding sodium hydroxide followed with copper (II) sulphate drop by drop.
a) State the observations he made in each set up. (3mks)
– 1st portion
– 2nd portion
– 3rd portion
b) Why was the experiment on the 1st portion included in the tests? (1mk)
c) Name the property of the chemical being investigated in these tests. (1mk)
d) Account for the observations made in 2 (a) above. (3mks)
15. The diagram below illustrates circulation in certain organs of the mammalian body.
Identify the blood vessels represented by A, B and C. (3mks)
b) Explain why blood from the small intestines goes to the liver before it goes to any other organ of the body. (2mks)
c) Compare the blood in vessels B and C. (1mk)
d) Outline how a glucose molecule in vessel A finally reaches the heart. (2mks)
16. The table below shows how the internal temperature two animals X and Y varied with the external temperature. The temperature was measured regularly and recorded for 12 hours in a day. Study the table and answer the questions that follow.
a) Using the same grid, draw graphs of external temperature, and internal temperature of animals X and Y
(Y-axes) against time (X-axes). (7mks)
b) Account for the variation of internal and external temperatures for the animals X and Y. (2mks)
c) Identify the classification of organisms whose internal temperature varies as X and Y (2mks)
d) Explain two ways used by organism Y to make its internal temperature vary as shown despite of changes in external temperature. (4mks)
18 a) Give the functions of the skin in organisms. (6mks)
b) How is the mammalian skin modified to enable it perform its functions? (l4mks)
19. The diagram below shows how gaseous exchange occurs across the gills in fish.
(a) According to the diagram water and blood flow in opposite direction across the gills.
(i) Give the term used to describe this flow. (1 Mark)
(ii) Explain the advantage of the above flow named in a(i) above. (2 Marks)
(b) What difference would be observed if water and blood flows across the gills in the same direction? (2 Marks)
(c) In which structures in the gills does gaseous exchange take place? (1 Mark)
(d) Name two organs in man which display the flow system named in a(i). (2 Marks)
20. An experiment was carried out to investigate the effect of different concentrations of Sodium Chloride on human red blood cells. Equal volumes of blood were added to equal volumes of salt solutions of different concentrations. The results were as shown below:-
| Set up | Sodium Chloride concentration | Shape of red blood cells at the end of experiment | Number of red blood cells at the end of experiment |
| A B | 0.9% 0.3% | Normal Swollen | No change in number Fewer in number |
- If the experiment was repeated with 1.4% Sodium Chloride solution, state the results you would expect with reference to:-
(i) Number of red blood cells. (1 Mark)
(ii) Appearance of red blood cells when viewed under the microscope. (1 Mark
Account for the fewer number of red blood cells in 0.3% Sodium Chloride salt solution. (3 Marks)
c) Give the biological term which can be used to describe 0.9% Sodium chloride solution.(1 Mark)
d) Define plasmolysis. (1 Mark)
Marking Scheme
1. i) Osmosis.
ii) A – solution in potato cup increases. Level of water in the beaker decrease;
B- Remain the same;
iii) A – Surrounding the cube is a region with high concentration of water molecules while in the sugar crystals, there are very few water molecules;
The sugar crystals exert on Osmosis pressure by Osmosis water molecules move across the potato tissue, which acts as a semi-permeable membrane. The level rises;
B- No change since boiling denatures the membrane structure of potato cells;
iv) C- No water moves into the potato cup/remains the same; since there is no concentration gradient;
2(a) i) Graph.
- 0.402 ± 0.01
- 11% ± 1%.
b) i) All cells have been haemolysed; cells contains one hypertonic to salt solution; water enters cells by osmosis; cells swell and eventually burst.
ii) No cells were haemolysed; cell contents were isotonic to salt solutions (aments of water entering the cell was equal to that leaving the cell); no net movement of water into cells;
c) The cells would become crenated; the cell contents would be hypotonic to salt solutions; water would leave cells by osmosis; membranes would shrink.
d) Contents of Onion epidermal cells would be hypertonic to water; water would enter cells by Osmosis; cells would become turgid;
3. – Temperature;- High temperature faster rate of transpiration; high temperature increases the capacity of atmosphere to hold water and moisture; also heat increase internal temperature of the leaf hence water evaporation; 4 accept converse
– Atmospheric pressure; Low atmospheric pressure, high rate of transpiration 2
– Humidity; Low humidity higher rate of transpiration; low humidity increases the saturation defiant; hence water moves form leaves to drier atmosphere; 4
– Wind; When it is windy the rate of transpiration is higher; wind sweeps away vapour that has accumulated at the surface of leaf; increasing saturation deficit; hence faster rate of transpiration 5
– Light intensity;High light intensity faster rate of transpiration high light intensity increase photosynthesis rate hence stomata opens; 4
– Amount of water in soil; More water in the soil increases the rate of transpiration; it wets the xylem (ensure xylem is wet throughout); 3 Max 20
4. a) Define digestion and ingestion.
i) Digestion- It is break down of complex insoluble; √ food substance into simple soluble food substance;√
ii) Ingestion- is introduction of food through the mouth into the digestive system;
b) Describe the digestion of Ugali.
- Digestion of ugali begins in the mouth; √ ugali is chewed by the teeth to increase large surface area √ for action of salivary amylase/ptyalin; √ The food mixes with saliva produced by salivary glands;√
- Saliva contains mucus and enzyme ptyalin. Mucus moistens, softens and lubricates the food;√ ptyalin speeds up the conversion of starch to maltose; √ ugali is made into bolus in the mouth;√
- The bolus moves along the oescophagus and prestalsis/by contraction and relaxation of circular and longitudinal muscles into the stomach;√
- The digestion continues until ugali become acidic since the stomach does not contain carbohydrase/carbohydrate digesting enzymes no digestion of ugali takes place here.√
- Ugali now moves into duodenum by peristalsis in form of acidic chime; √ where it mixes with the bile from the liver and pancreatic juice from the pancreases;√ bile being alkaline neutralizes the stomach acid;√ and provides a suitable alkaline medium for the enzymes to act on carbohydrates;√
- Pancreatic juice contains three enzymes out of the which enzyme amylase speeds conversion of starch to maltose;√
- When food reaches the ileum; it mixes with intestinal juice which contains several enzymes. Maltase – speeds up conversion of maltose to glucose;√
- Lactose which speeds up conversion of lactose to glucose√ and galactose; sucrase which speeds up conversion of sucrose into fructose and glucose;
- Absorption – glucose, the end product of all carbohydrates diffuses through the epithelium of villi and capillary walls and enters into blood stream and is carried to the liver via hepatic portal veins;√
- Assimilation- in the liver excess glucose is converted into glycogen and stored;
- – The rest of the glucose is carried by the blood tissues where is oxidized during tissue respiration to release energy;√ (21 max 18 mks Total 20mks)
5. (a) – Pulmonary artery
– Pulmonary vein
(b) Alveolus
(c) A– cavity of alveolus
E – Red blood cell
(d) Hydrogen carbonate ions;
Carbamino haemoglobin;
6. (a) (i) Walls of stomach;
(ii) Stimulates the secretion / production of gastric juice; √
(b) Presence of food in the stomach;
(c) A – Blood capillaries; B – Lacteal;
(d) – Produces plenty of mucus to lubricate coarse/indigestible material during peristalsis;
– Wide human accommodates /store indigestible food
– Elongate to increase surface are for absorption of water.
– has muscles to facilitate peristalsis when they contract;
7. (a) (i) P – Tracheids Q – pits
(ii) P– water conducting elements of xylem
(b) Function of phloem – translocation/ transport of organic substances from the leaves to the of the plant;
(c) (i) Translocation of food will not occur acc. Slow translocation
(ii) Reason – it contains a lot of mitochondria which provide energy for translocation;
(d)
| Phloem | Xylem |
| Made of living cellsHave companion cellsHave cytoplasmic strands Lack lignin deposits | made of dead cellslack companion cellslack cytoplasmic strandshave lignin deposits; (any 2×1=2mks) |
8. (a) Photocopy – scale – 1m
Labeling axes – 1
Plotting – 2m
Curves – 2m (curves must be labeled) rej. Dotted line for curves
A GRAPH OF LACTIC ACID CONCENTRATION AGAINST TIME
(b) (i) 10-15 minutes ; period of rapid increase in lactic acid concentration (2mk
(ii) 10-20 seconds : period when lactic acid level starts to increase; (2mks)
(iii) 75minutes i.e. from 25th minutes to the 100minutes, this is the time lactic cid took to decrease from the highest level to normal; (2mks)
(c) It would have the same basic shape; but would peak slightly ahead of the lactic acid curve in time;
(e) Because it is still diffusing out of the muscles, where it was made a few minutes earlier;
(e) Allows for energy production even cases of oxygen deficiency; thus enables animals to survive active exercise and to inhabit even in areas with limited oxygen supply;
(f) Oxygen debt is the amount of oxygen to get rid of the lactic acid; that has accumulated due to anaerobic respiration; (2mks)
9. Regulation of blood sugar level; under the influence of insulin; and glucagons (hormones). When there is excess sugar; the hormone insulin stimulate/causes liver cell to convert it to glycogen; some converted to fats/lipids for storage;
• When the blood sugar level is below normal: the hormone glucagon causes liver cells to convert glycogen to glucose;
• Regulation of amino acids; excess amino acids; are deaminated; by the liver (cells) leading to formation of urea; which is transported by the blood to the kidney; for elimination;
• Production of heat: the liver is involved in the thermoregulation due to many metabolic, activities; taking place in the liver cells a lot of heat is generated which is distributed to the entire/whole body;
• Detoxication of toxic substances; (such as drugs and hydroxide peroxide)
• Elimination of haemoglobin; and formation of bile; breakdown worn out red blood cells; the bile salts (sodium tyrochocolate and sodium glycocholate; in the bile eEMULSIFYfats (in the duodenum)
• Storage of blood in its veins; thus regulating the volume of blood circulating in the body
• Elimination of sex hormones after they have performed their function/work; storage of vitamin AD and B12 some mineral salts; thus regulating their levels in the blood TOTAL 23 MAX 20
10. (a) X – glycolysis Y – Kreb’s cycle
(b)
| Process X | Process Y |
| occurs in cytoplasmindependent of oxygenproduces less energyraw material is glucoseEnd products are energy, CO2, lactic acid or ethanol | occurs in mitochondriaIs oxygen dependentproduces more energyRaw material is pyruvateEnd products are energy, CO2 and water |
(c) lactic acid; under anaerobic conditions
(d) small body size leads to alarge surface area to volume ratio; hence more loss of heat to the environment; leading to increased rate of respiration to replace the lost heat;
11. (a) (i)Active transport/diffusion
Tied (ii) Numerous Mitochondria in its wall to generate energy/microvilli/coiling increase surface area/thin epithelium for quick diffusion.
(b) Cortex
(c) on the diagram
(d) plasma proteins; Blood cells; accept specific examples e.g. albumins, red blood cells;
(e) (i) Regulate re-absorption of Sodium salts;
(ii) Regulate re-absorption of water
12. (a) Allocation of marks on graph
(b) Gynandropsis – opt To = 30oC
Amaranthus – opt. to = 25oC
(c)At 5oC, the enzymes that catalyse the process of photosynthesis are inactivated.
(d) Amaranthus – Terrestrial; Crotolaria – terrestrial;
(e) 50oC;
(f) water;
(g)(i) Granum; (ii) Stroma
(h) Similarity: Both have double membrane; 1mk – Both have fluid filled matrix;
Difference : inner membrane of mitochondrion is folded to form cristae while inner membrane of chloroplast is smooth;
– chloroplast is biconcave shaped while mitochondria is oval/sausage shaped (any 1×1=1mk)
(i) Point at which the rate of photosynthesis equals to the rate of respiration.
13. (a) – plants wastes accumulate slowly;
– plants produce less toxic wastes;
– some excretory products are recycled by plants e.g. CO2, SO2)
– plant tissues are tolerant to toxic wastes;
– plant wastes are stored in temporary structures which fall off e.g. leaves (any 3×1=3mks)
(b) – Diffusion;
– Transpiration;
– Exudation;
– Deposition of wastes/ leaf fall/ flower fall/ storage in bark;
– Recycling;
– Guttation ; (any 5×1=5mks)
(c)
| Excretory products | Uses |
| Caffeine;Popain;Tannin;Nicotine;Latex;Quinine;Atropine;Morphine;Colchine;Cocaine;Cannabis;Khat/ miraa; | Body stimulant; Meat tenderizer; Leather tanning; Stimulant; insecticide; Manufacture of tyre/rubber products; Anti-malarial drugs; Increase heart beat; dilate eye pupil; Cancer treatment; Used in genetics to induce polypoidy; Anesthesia/painkiller/stimulant; Pain killer; Stimulant; (Any 6×2=12mks) |
14. (a) 1st portion. – Blue; colour was observed
2nd portion – Purple; colour was observed
3rd portion – Purple; colour was observed
(b) A control experiment;
(c) Proteins are highly sensitive to temperature and pH changes; (award if either temp of pH is stated singly)
(d) 1st portion – Enzyme pepsin broke down proteins into peptones;
2nd portion – Enzyme pepsin works in acidic medium; (not in basic medium)
15. (a) A – Hepartic portal vein; B – Hepartic vein; C – Hepartic artery;
(b) – So that any toxic substances absorbed together with food nutrients from the ileum be detoxified;
– So that food substances e.g. glucose, amino acids can be regulated. Only the required quantity of glucose is left in circulation as excess is either stored as glycogen, fat and excess may be respired.
Excess amino acids are deaminated;
(c) B – Deoxygenated C – Oxygenated
(d) From the small intestines, it is transported to the liver through Hepartic portal vein; (It is then transported to the heart through the hepartic vein;
16. (a) graph
(b) X – Lacks internal mechanisms to regulate its internal temperature
Y – Has internal means to regulate its internal temperature. hence able to maintain it within narrow range
(c) X – Poikilotherm Y – Endotherm
(d) – Blood vessels vasodilate when temperature is higher than norm to allow for heat loss from blood through radiation, evaporation, etc; when temperature is lower, blood vessels constrict to prevent loss of heat from blood through radiation, evaporation etc;
– When temperature is higher, lies flat to allow for heat loss from the body since insulation layer of air is removed; when temperature is lower, hair strands erect to hold air which insulates the body against heat loss through radiation, evaporation etc.
17. (a) Higher temperature; increases the kinetic energy; of water molecules which makes water turn into vapour on the leaf surfaces faster and hence increase rate of transpiration
- Higher light intensity; influences maximum opening of stomata which increases the surface area; over which transpiration occurs maximumly
- Wind; carries away moisture around the plant and create a higher saturation deficit; which then increases the rate of water loss/transpiration in plants.
- Higher relative humidity; reduces saturation deficit; which causes lowering of water loss/transpiration in plants.
- Higher amount of water in the soil; makes the plant to absorb excess water which increases the need for the plant to get rid of it through transpiration faster;
- If the leaf is broader and has numerous larger open stomata; the surface area over which water loss occurs is increased; causing increase in the rate of transpiration. (award max. 12mks)
18. (a) – It protects the underlying tissues against mechanical injury, UV-light rays and entry of pathogens; (Rej. germs)
- As an excretory organs, it enables organisms to eliminate excess water, ions and traces of urea;
- As a sensory organ, it enables the organisms to be aware of deviations in pressure, touch and temperature from the external environment;
- It is a thermoregulator such that it enables the body to lose excess heat to lower its temperature back to norm or may enable the organisms to store it s heat if the temperature is lower and hence raise it back to the norm;
- It takes part in osmoregulation by enabling the body fluids to get rid of excess water or excess ions;
- It takes part in the regulation of the pH of body fluids by enabling the body to get rid of either Hydrogen ions or bi-carbonate ions;
(b) – Presence of the cornified layer; which tough and has keratin to enable it protect the underlying tissues from mechanical injury; It alsos has sebum; which is antiseptic and enables it to prtect the ody against entry of pathogesn. Presence of melanin; enables it to protect the underlying tissues against damage by t UV-light radiations.
- It has sweat glands with secretory cells; which absorb excess water, excess ions and traces fo urea from blood and secrete them into the sweat duct;
- Has the sweat pores; which open son the skin surface to allow for elimination of sweat containing excess water, excess ions and traces of urea;
- It has nerve endings; which enables it to detect any deviations in temperature, pressure and contact/touch;
- It has hair follicles; which stand erect when temperature is lower than normal to reduce heat loss from the body or lie flat to enable to body lose excess heat and lower temperature back to the norm when the internal temperature is higher;
- Has blood vessels; which vasodilate when temperature is higher than norm to enable the organisms lose than the norm to reduce heat loss from the body;
19. a) i) Counter current system;
ii) Maintain a diffusion gradient so that there is maximum uptake of oxygen; continue diffusing into blood and Carbon (iv) oxide into water;
- Parallel flow lower diffusion gradient; so that less oxygen diffuse into blood/low rate of gaseous exchange
- Gill filaments
- Placenta
Kidney
20 a) i) Remain the same
ii) Crenated
- The solution is hypotonic to red blood cells hence the cells grains water; by osmosis; swelling until they burst.
- Isotonic solution
- Plasmolysis – the process by which plant cells loss water by Osmosis shrink and become flaccid.
