SECTION A
ANSWER ALL THE QUESTIONS
For Question 1- 30 ,each question is followed by four alternative answers A,B,C or D .Choose one correct answer for each question.
1. Plants cells show turgor pressure when
A the cells are losing water from their vacuoles
B the vacuoles are filled with water
C water is being utilized during photosynthesis
D water is being evaporated from the leaves
2. Which of the following statements is the definition of simple diffusion?
A Movement of molecules from an area of lower concentration to an area of higher
concentration.
B Movement of water from an area of lower solute concentration to an area of higher solute concentration across a semi permeable membrane.
C Movement of molecules from an area of higher concentration to an area of lower
concentration
D Movement of water from an area of lower water concentration to an area of higher
Water concentration across a semi- permeable membrane
3. Which of the following involves osmosis?
A Respiration in unicellular organism.
B Movement of a hypertonic solution into the red blood cell
C Dispersal of hydrogen sulphide gas in the atmosphere
D The release of carbon dioxide from the lungsto the atmosphere.
4. Figure 1 shows a plant cell immersed in salt solution.
Figure 1
What can be inferred from this experiment?
A Loss of water by the plant cell
B Water has moved inside the cell making it turgid
C The salt solution has caused the cell wall to break
D The cytoplasm is hypertonic to the salt solution
5. Four blocks of jelly which are clear and not coloured are put into beakers filled with methyl
blue solution as shown in Figure 2.
Figure 2
Which of the 4 blocks of jelly becomes blue the fastest?
6. How does fresh water fish regulate its body fluid?
A Consuming a lot of food
B The cytoplasm has a high concentration of salt
C Excreting dilute urine frequently
D Absorbing salt through its skin
7 The movement of molecules across the plasma membrane depends on
I the size of the molecules
II the shape of the molecules
III the polarity of the molecules
IV the solubility of the molecules in lipid
A I and II B II and III C II and IV D I, III and IV
8 Which of the following statements about facilitated diffusion are true?
I Facilitated diffusion is a type of passive transport
II Facilitated diffusion moves molecules down the concentration gradient
III Facilitated diffusion does not require the expenditure of energy
IV Facilitated diffusion is made possible by carrier proteins located within the
membranes
A I and II B II and III C I, III and IV D I , II , III and IV
9. Which of the following involves diffusion?
I Oxygen intake in the lungs
II Mineral intake by plant roots
III The absorption of digested food in the intestines
IV The absorption of water into the xylem
A I ,II and III B I and III C II and IV D I , II, III and IV
10. Figure 3 shows the results of an experiment . The solution in the glass tube has risen to level
X after 3 hours .
Figure 3
Which of the following procedure can raise the level of sugar solution in the glass tube
higher than X
A Use a bigger beaker
B Use a smaller visking tube
C Fill the visking tube with distilled water
D Fill the visking tube with a sucrose solution of 0.6 mol/dm3
11 Which of the following equations represents aerobic respiration?
A Carbon dioxide + water + energy Oxygen + glucose
B Oxygen + glucose carbon dioxide + water + glucose
C Glucose + carbon dioxide oxygen + water + energy
D Energy + oxygen carbon dioxide + water + glucose
12 What are the end products of anaerobic respiration in human muscle tissues during vigorous
exercise?
A Lactic acid and energy
B Lactic acid and carbon dioxide
C Ethanol and carbon dioxide
D Carbon dioxide and water
13 Figure 4 shows a longitudinal section through an alveolus and the associated blood capillary
in a lung’
Figure 4
Which of the following describe the concentration of oxygen at X and Y ?
X Y
A Low Low
B Low High
C High Low
D High High
14. Exchange of respiratory gases through the skin is characteristic of
A fish B amphibians C reptiles D mammals
15. During a race ,an athlete experiances
I a reduced avaibility of oxygen to the muscles
II an increase in the breathing rate
III an increased carbon dioxide concentration in the blood
IV an increase in the heartbeat rate
A I and III B II and IV C II, III and IV D I , II , III and IV
16. What are the common characteristic shown by the respiratory surfaces of animals?
I They are thin
II They are moist
III They have a large surface area
IV They are covered by a network of capillaries
A I and II B III and IV C I , II and III D I , II , III and IV
17. Air is inhaled when
I the external intercostals muscles relax
II the rib cage moves upwards and downwards
III the diaphgram muscles relax
IV the pressure in the thoracic cavity decreases
A I and II B II and IV C I, II and III D I, III and IV
18 Which of the following shows the similarity between anaerobic respiration of the yeast and anaerobic respiration in the skeletal muscle?
A Lactic acid is produced
B Energy is produced
C Oxygen is used
D Ethanol is produced
19.
Figure 5
Figure 5 shows the set up of an experimentto study respiration.Which of the following is the colour of the bicarbonate indicator in the boiling tubes U and V after 60 minutes?
U V
A Yellow Yellow
B Yellow Red
C Purple Red
D Red Yellow
20 Which of the following is not an adaptation of the body when in a situationof low atmospheric pressure?
A Increasing the rate of respiration
B Increasing the blood pressure
C Decreasing the pH of blood
D Producing more red blood cells
21 Transport systems have been developed in multicellular organisms because
I oxygen could not get to the body cells quickly enough
II waste products would accumulate in cells , killing them
III the body fluid would clot when the blood vessels are damaged
IV the supply of nutrients to all cells by diffusion would be too slow
A I and II B I, II and IV C II, III and IV D I, II , III and IV
22 During the contraction of the human heart , the highest pressure is developed in the
A right strium B left atrium C right ventrical D left ventrical
23 Which of the following combines with a molecule of oxygen in haemoglobin?
A Calcium B Iron C Magnesium D Potassium
24 Antibodies are manufactured by cells called
A antigens B pathogens C lymphocytes D phagocytes
25 Phagocytosis of a bacterium by a white blood cell includes the following steps.
I The phagocyte strongly adheres to the bacterium
II The breakdown products are absorbed by the bacterium
III The phagocyte releases enzymes to break down the bacterium
IV The phagocyte is attracted to chemicals produced by the bacterium
The correct sequence for this process is
A I , IV, II , III B I , IV , III, II C IV, I , II, III D IV , I , III , II
26 Which of the following are risk factors for heart disease?
I Lack of exercise
II Obesity
III Smoking
IV Chronic stress
A I and III B II and IV C II, III and IV D I, II, III and IV
27 Figure 6 shows the cross section of a type of plant tissue
Figure 6
What is the function of this type of tissue?
A Support B Food storage
C Transport of organic food D Transport of water and mineral salts
28 Transpiration is responsible for
I keeping the stomata open
II increasing the rate of respiration
III keeping the plants cool on hot , sunny days
IV passing mineral salts from the roots to the shoot
A I and IV B III and IV C I, III , IV D II, III and IV
29 Nitrates and sulphates reach the mesophyll cells of the leaf mainly by the way of the
A cuticle B phloem C stomata D xylem
30 The transport of water up the xylem vessels depends on
I capillary
II hydrostatic pressure
III root pressure
IV transpiration pull
A I and IV B III and IV C I, III and IV D II, III and IV
Tuesday, August 10, 2010
Wednesday, November 18, 2009
TIPS TO ANSWER BIOLOGY PAPER
Tips To Answer Biology Paper For Your SPM Examination ??
BIOLOGY PAPER 1
1. Read the question and all the answer choices before answering a question.
2. If you do not sure about the answer, go on to the next question.
3. Give your attention to key words such as except , true/not true, which, etc
4. Make sure yor answer all 50 questions
Biology Paper 2
STRUCTURED QUESTIONS
1. Show your working ( the formulae)
2. Your answers should be precise
3. Diagrams or graphs drawn(using pencil) should be clearly labelled
4. Neatly cross out the wrong answers ,and write your new answer clearly.
ESSAY QUESTIONS
1. Read all the questions before choosing those that you want to answer
2. Read again the choosen questions carefully to make sure you are interpreting
BIOLOGY PAPER 1
1. Read the question and all the answer choices before answering a question.
2. If you do not sure about the answer, go on to the next question.
3. Give your attention to key words such as except , true/not true, which, etc
4. Make sure yor answer all 50 questions
Biology Paper 2
STRUCTURED QUESTIONS
1. Show your working ( the formulae)
2. Your answers should be precise
3. Diagrams or graphs drawn(using pencil) should be clearly labelled
4. Neatly cross out the wrong answers ,and write your new answer clearly.
ESSAY QUESTIONS
1. Read all the questions before choosing those that you want to answer
2. Read again the choosen questions carefully to make sure you are interpreting
Wednesday, August 19, 2009
Answer Scheme Trial SPM (SBP 20090 Paper 2 Section B
No 6
Marking Criteria
Marks
(a) (i)
Able to describe the mechanisms in organelle M that
involved in the formation of starch in the plant.
Sample answer :
P1: The formation of starch in plants is by the process of
photosynthesis which occurs in chloroplasts.
P2: The two stages in photosynthesis are the light and dark reactions.
Light reaction:
P3:Takes place in grana.
P4: Chlorophyll captures light energy which excites the electrons of chlorophyll molecules to higher energy levels.
P5: In the excited state, the electrons can leave the chlorophyll molecules.
P6: Light energy is also used to split water molecules into hydrogen ion (H+) and hydroxyl ions (OH-) (Photolysis of water).
P7: The hydrogen ions then combine with the electrons released by chlorophyll to form hydrogen atoms.
P8: The energy from the excited electrons is used to form energy-rich molecules of adenosine triphosphate /ATP.
P9: Hydroxyl ion loses an electron to form a hydroxyl group. This electron is then received by chlorophyll.
P10: The hydroxyl groups then combine to form water and gaseous oxygen.
Dark Reaction:
P11: Take place in stroma.
P12: Do not require light energy.
P13: The hydrogen atoms are used to fix carbon dioxide in a series of reactions catalysed by photosynthetic enzymes
P14: and caused the reduction of carbon dioxide into glucose.
P15: The glucose monomers then undergo condensation to form starch which is temporarily stored as starch grains in the chloroplasts.
any 10P
(b)(i)
Able to suggest two health problems which that teenager could have if she continuously taking the above menu for a long time.
Sample answer :
F1: Chronic heart disease
P1: Her diet contains large amounts of oil/cream/fat and it causes arteriosclerosis / atherosclerosis / heart problem / cardiovascular disease.
F2: Constipation.
P2: Lack of fruit/vegetables / fibers leads to constipation.
4
(b)(ii)
Able to explain ways to improve daily food intake and stating the reasons for the choice of food.
Sample answer :
P1: Drink scheme milk/low fat milk and eat a boiled egg to reduce the cream and fat from her diet
P2: Replace sausage with an orange to increase her fruit fibre intake
Lunch:
P3: Drink fresh fruit juice because it has lower sugar content/more vitamins
P4: Eat steamed rice/steamed chicken to lower cholesterol intake
P5: Eat vegetable/salad to increase the amount of roughage
Dinner:
P6: Drink clear soup to reduce the cholesterol
P7: Use tomato sauce with her pasta because it contains less fat and more fibers.
P8: Eat a slice of watermelon/any fruits to increase her intake of fiber to prevent constipation
Any 6
TOTAL 20 marks
No 7
(a)
Able to explain what cloning is :
Sample answer :
P1 : Cloning is an asexual reproductive process of producing clones//does not involve gamete
P2 : A clone is a group of cells//organism//a population of organisms produced from a single ancestral cell.
P3 : A clones genetically identical
P4 : The technique can be used to produce high quality of organism / orchids/ oil palm / cocoa plants.
Any 2 P
(b)
Able to describe tissue culture technique.
Sample answer:
P1 : Tissue culture technique
P2 : Tissue culture technique is used to produce (high quality of seedling)oil palm seedlings in vitro/any suitable example.
P3 : The leaves/shoot/stem/root tissues are cut out.(These cut out plant tissues are called explants).
P4 :The pieces of meristematic tissue (explants) are cultured in sterile nutrient medium, in suitable pH and with addition of plant growth substances.( at least 2 factors)
P5 :The flasks containing the tissue are stored in an incubator at 37°C for 2/3 weeks.
P6 : The cell divide by mitosis to produce callus.
P7 : The callus is then cut into small pieces.
P8 : The small pieces of callus tissues are then cultured in sterile nutrient medium.
P9 : When it has grown to a suitable size, the clone is transferred to the nursery.
Any 6 P
(b)
Able to discuss Advantages / strength:
P1 : Genetic engineering involves genes manipulation / transfer / modification in organisms to produce certain products.
P2 : Example; products in pharmacy such as insulin / antibiotics; food products based on plants / animals; agricultural / agrochemical products.
P3 : The products produced are very similar to the original / natural materials because the same genes are used / particularly chosen genes are transfered.
P4 : The production of products is faster especially with the use of microorganisms/bacteria.
P5 : Microorganisms such as bacteria are suitable to be used as gene vectors / they have free DNA in the form of ring / plasmid.
P6 : High reproduction rate of bacteria/microorgansm in optimal culture mediums able to produce a large amount of chosen genes / products / insulin / antibiotics.
P7 : Can be used by thousands of people who need them / widespread usage.
P8 : Able to produce a variety of proteins / recombinant proteins / enzymes used in food industries / medicine / agriculture.
P9 : Genetic engineering technique is used to solve criminal cases through DNA finger printing / DNA fragments analysis.
P10 : Other uses /examples; metal extraction from oxide/any suitable examples
Any 6 P
(c)
Able to describe the effect of cycle malfunctions to the body.
Sample answer
P1: The exposure damage the DNA of the cell
P2: A cell divides through mitosis repeatedly.
P3: Produces cancerous cell
P4: Due to (severe ) distruption to the mechanism
that controls the cell cycle
P5: Cancerous cells divide freely / uncontrollably
heeding the cell cycle control
P6: (these cells ) compete with surrounding normal
cells to obtain nutrient / energy (for growth)
P7: Invade / destroy neighbouring cells
P8: (they can spread to other organ and) initiate
cancers there .
Any 6 P
TOTAL 20 marks
No 8
Marking Criteria
Marks
(a)
Able to describe the movement of water from root to the leaf which aided by :
Root pressure
Capillary action
Transpirational pull
Sample answer:
Root pressure
P1: The cell sap of root hair (usually) hypertonic to the surrounding soil solution
P2: Water diffuses into the root by osmosis.
P3: (As they absorb more water by osmosis), a cell sap becomes more dilute compared to neighbouring cells.
P4: Water (therefore) moves to these adjacent cells which become more diluted themselves, so osmosis continues across the cortex
P5: (At the same time), ions from the soil are actively secreted into the xylem vessels and this causes osmotic pressure to increase
P6: Water flows continuously into the xylem and this create a pressure known as root pressure
P7: Root pressure gives an initial upward force to water and mineral ions in the xylem vessels
Capillary action
P8 : Water moves up through the xylem in the stems by capillarity (with is the upward movement of a fluid in a narrow bore tube)
P9: Capillary action is due to combined force of cohesion (water molecule have attraction for each other) and adhesion (water molecules are attracted to the side of the vessels)
P10: Water molecule form a continuous water column in the xylem vessel (due to cohesion and adhesion force enable water to move up along the xylem vessels)
P11: (As water is pulled upwards) the cohesion of water (which is due to hydrogen bonding holds the water molecule together) prevent the water column in the xylem breaking apart
P12: (At the same time) the adhesion of the water (to the wall of the xylem vessel and tracheids) prevents gravity from pulling the water down the column
Transpirational pull
P13: The lost of water from the mesophyll cells during transpiration is replaces by water which flows in from the xylem vessels in the leaves
P14: This creates a tension / suction force in the water column because water has cohesive properties called transpiration pull
P15: The transpiration pull draws water from the xylem in the leaves/stems/roots.
P16: The continuous flow of water through the plant is known as the transpiration stream
Max 4
b (i)
Able to explain the diffrences of composition fluid P and fluid Q
Sample answer:
F1: Fluid Q/lymph has a larger numbers of lymphocyte compare to fluid P/blood
P1: lymphocyte is produced by the lymph nodes in lymph system
F2: Fluid Q/lymph has lower contents of oxygen compare to fluid P/blood
P2: oxygen has been used up by the cell
b(ii)
Able to describe how lymph is formed from blood until it is brought back into the blood circulatory system.
Sample Answer :
P1: (When the blood flows from arteries into capillaries)there is higher hydrostatic pressure at the arterial end of the capillaries
P2: (This high pressure) forces some plasma to pass through the capillary walls into the intercellular spaces (between the cells)
P3: Once the fluid leaves the capillary walls, it is called interstitial/tissue fluid // The interstitial fluid fills the spaces between the cells and constantly bathes the cells
P4: 90% of the interstitial fluid diffuses back into blood capillary
P5: 10% of the interstitial fluid that has not been reabsorbed into the bloodstream goes into the lymph capillaries.(Once inside the lymph capillaries) the fluid is known as lymph.
P6: The lymph capillaries unite to form larger lymphatic vessels.
P7: From the lymphatic vessels, lymph eventually passes into the thoracic duct/the right lymphatic duct.
P8:The thoracic duct empties its lymph into the right subclavian vein. (Hence, lymph drains back into the blood).
Any 6 P
Max 6
TOTAL 20
No 9
Marking Criteria
Marks
(a)
Able to state the Sequence of events that occur when the hand touches a hot object.
Sample answer :
P1: the heat on the object stimulates the nerve endings (receptors) in the skin.
P2: impulses are triggered.
P3: This impulses travel along the sensory/afferent neurone to the spinal cord.
P4: in the spinal cord, the impulses are transmitted first across a synapse to the interneurone and then across another synapse to the motor/efferent neurone. ( at least 2 type of neurone)
At synapse
P5: When an impulse reach a presynaptic membrane, it triggers the synaptic vesicles to release neutrotransmitter into the synaptic cleft.
P6 The neurotransmitter diffuse across the synaptic cleft
P7: and bind to receptors which are attached to the postsynaptic membrane.
P8: The binding of the neurotransmitter to the receptors leads to the generation of a new impulse
P9: Impulses leave the spinal cord along the motor/efferent neurone to the effector
P10: the effector is the biceps muscle which then contracts. This brings about a sudden withdrawal of the hand.
Any 8
(b) (i)
Able to explain how geotropism is brought about in a plant root and shoot.
Sample answer :
Diagram :
Shoot
Root
Correct Diagram ( shoot grows upward, root grow downward): 1 m
Shoot
P1 : The auxin that is produced at the tip of shoot.
P2 : auxin moves downward/ accumulate on the underside of the shoot tip due to the pull of gravity.
P3 : the high concentration of auxin accelerates the growth
P4 : stimulating greater cell elongation on the underside relative to the cells on the upper side.
P5 : this differential elongation causes the shoot to bend away from gravity / grow downwards.
Root
P6 : The auxin that is produced at the tip of root.
P7 : auxin moves downward/ accumulate on the underside of the root tip due to the pull of gravity.
P8 : the high concentration of auxin inhibits the growth
P9 : slowing down cell elongation on the underside relative to the cells on the upper side.
P10 : this differential elongation causes the shoot to bend towards gravity / grow downwards.
Any 7 P
(b)(ii)
Able to explain the advantages
Sample answer
Advantages
Explanation
F1
ensures the root grow/penetrate deep into soil
P1
To anchor the plant firmly in the ground
F2
The roots always contact with the soil water/mineral/nutrients
P2
To maintain turgor pressure/ growth/ photosynthesis
F3
Ensure that the leaves of shoot growth towards sunlight
P3
To absorb maximum amount of light energy for photosynthesis.
Any 2 F with correspond P
4
TOTAL
20
No 6
Marking Criteria
Marks
(a) (i)
Able to describe the mechanisms in organelle M that
involved in the formation of starch in the plant.
Sample answer :
P1: The formation of starch in plants is by the process of
photosynthesis which occurs in chloroplasts.
P2: The two stages in photosynthesis are the light and dark reactions.
Light reaction:
P3:Takes place in grana.
P4: Chlorophyll captures light energy which excites the electrons of chlorophyll molecules to higher energy levels.
P5: In the excited state, the electrons can leave the chlorophyll molecules.
P6: Light energy is also used to split water molecules into hydrogen ion (H+) and hydroxyl ions (OH-) (Photolysis of water).
P7: The hydrogen ions then combine with the electrons released by chlorophyll to form hydrogen atoms.
P8: The energy from the excited electrons is used to form energy-rich molecules of adenosine triphosphate /ATP.
P9: Hydroxyl ion loses an electron to form a hydroxyl group. This electron is then received by chlorophyll.
P10: The hydroxyl groups then combine to form water and gaseous oxygen.
Dark Reaction:
P11: Take place in stroma.
P12: Do not require light energy.
P13: The hydrogen atoms are used to fix carbon dioxide in a series of reactions catalysed by photosynthetic enzymes
P14: and caused the reduction of carbon dioxide into glucose.
P15: The glucose monomers then undergo condensation to form starch which is temporarily stored as starch grains in the chloroplasts.
any 10P
(b)(i)
Able to suggest two health problems which that teenager could have if she continuously taking the above menu for a long time.
Sample answer :
F1: Chronic heart disease
P1: Her diet contains large amounts of oil/cream/fat and it causes arteriosclerosis / atherosclerosis / heart problem / cardiovascular disease.
F2: Constipation.
P2: Lack of fruit/vegetables / fibers leads to constipation.
4
(b)(ii)
Able to explain ways to improve daily food intake and stating the reasons for the choice of food.
Sample answer :
P1: Drink scheme milk/low fat milk and eat a boiled egg to reduce the cream and fat from her diet
P2: Replace sausage with an orange to increase her fruit fibre intake
Lunch:
P3: Drink fresh fruit juice because it has lower sugar content/more vitamins
P4: Eat steamed rice/steamed chicken to lower cholesterol intake
P5: Eat vegetable/salad to increase the amount of roughage
Dinner:
P6: Drink clear soup to reduce the cholesterol
P7: Use tomato sauce with her pasta because it contains less fat and more fibers.
P8: Eat a slice of watermelon/any fruits to increase her intake of fiber to prevent constipation
Any 6
TOTAL 20 marks
No 7
(a)
Able to explain what cloning is :
Sample answer :
P1 : Cloning is an asexual reproductive process of producing clones//does not involve gamete
P2 : A clone is a group of cells//organism//a population of organisms produced from a single ancestral cell.
P3 : A clones genetically identical
P4 : The technique can be used to produce high quality of organism / orchids/ oil palm / cocoa plants.
Any 2 P
(b)
Able to describe tissue culture technique.
Sample answer:
P1 : Tissue culture technique
P2 : Tissue culture technique is used to produce (high quality of seedling)oil palm seedlings in vitro/any suitable example.
P3 : The leaves/shoot/stem/root tissues are cut out.(These cut out plant tissues are called explants).
P4 :The pieces of meristematic tissue (explants) are cultured in sterile nutrient medium, in suitable pH and with addition of plant growth substances.( at least 2 factors)
P5 :The flasks containing the tissue are stored in an incubator at 37°C for 2/3 weeks.
P6 : The cell divide by mitosis to produce callus.
P7 : The callus is then cut into small pieces.
P8 : The small pieces of callus tissues are then cultured in sterile nutrient medium.
P9 : When it has grown to a suitable size, the clone is transferred to the nursery.
Any 6 P
(b)
Able to discuss Advantages / strength:
P1 : Genetic engineering involves genes manipulation / transfer / modification in organisms to produce certain products.
P2 : Example; products in pharmacy such as insulin / antibiotics; food products based on plants / animals; agricultural / agrochemical products.
P3 : The products produced are very similar to the original / natural materials because the same genes are used / particularly chosen genes are transfered.
P4 : The production of products is faster especially with the use of microorganisms/bacteria.
P5 : Microorganisms such as bacteria are suitable to be used as gene vectors / they have free DNA in the form of ring / plasmid.
P6 : High reproduction rate of bacteria/microorgansm in optimal culture mediums able to produce a large amount of chosen genes / products / insulin / antibiotics.
P7 : Can be used by thousands of people who need them / widespread usage.
P8 : Able to produce a variety of proteins / recombinant proteins / enzymes used in food industries / medicine / agriculture.
P9 : Genetic engineering technique is used to solve criminal cases through DNA finger printing / DNA fragments analysis.
P10 : Other uses /examples; metal extraction from oxide/any suitable examples
Any 6 P
(c)
Able to describe the effect of cycle malfunctions to the body.
Sample answer
P1: The exposure damage the DNA of the cell
P2: A cell divides through mitosis repeatedly.
P3: Produces cancerous cell
P4: Due to (severe ) distruption to the mechanism
that controls the cell cycle
P5: Cancerous cells divide freely / uncontrollably
heeding the cell cycle control
P6: (these cells ) compete with surrounding normal
cells to obtain nutrient / energy (for growth)
P7: Invade / destroy neighbouring cells
P8: (they can spread to other organ and) initiate
cancers there .
Any 6 P
TOTAL 20 marks
No 8
Marking Criteria
Marks
(a)
Able to describe the movement of water from root to the leaf which aided by :
Root pressure
Capillary action
Transpirational pull
Sample answer:
Root pressure
P1: The cell sap of root hair (usually) hypertonic to the surrounding soil solution
P2: Water diffuses into the root by osmosis.
P3: (As they absorb more water by osmosis), a cell sap becomes more dilute compared to neighbouring cells.
P4: Water (therefore) moves to these adjacent cells which become more diluted themselves, so osmosis continues across the cortex
P5: (At the same time), ions from the soil are actively secreted into the xylem vessels and this causes osmotic pressure to increase
P6: Water flows continuously into the xylem and this create a pressure known as root pressure
P7: Root pressure gives an initial upward force to water and mineral ions in the xylem vessels
Capillary action
P8 : Water moves up through the xylem in the stems by capillarity (with is the upward movement of a fluid in a narrow bore tube)
P9: Capillary action is due to combined force of cohesion (water molecule have attraction for each other) and adhesion (water molecules are attracted to the side of the vessels)
P10: Water molecule form a continuous water column in the xylem vessel (due to cohesion and adhesion force enable water to move up along the xylem vessels)
P11: (As water is pulled upwards) the cohesion of water (which is due to hydrogen bonding holds the water molecule together) prevent the water column in the xylem breaking apart
P12: (At the same time) the adhesion of the water (to the wall of the xylem vessel and tracheids) prevents gravity from pulling the water down the column
Transpirational pull
P13: The lost of water from the mesophyll cells during transpiration is replaces by water which flows in from the xylem vessels in the leaves
P14: This creates a tension / suction force in the water column because water has cohesive properties called transpiration pull
P15: The transpiration pull draws water from the xylem in the leaves/stems/roots.
P16: The continuous flow of water through the plant is known as the transpiration stream
Max 4
b (i)
Able to explain the diffrences of composition fluid P and fluid Q
Sample answer:
F1: Fluid Q/lymph has a larger numbers of lymphocyte compare to fluid P/blood
P1: lymphocyte is produced by the lymph nodes in lymph system
F2: Fluid Q/lymph has lower contents of oxygen compare to fluid P/blood
P2: oxygen has been used up by the cell
b(ii)
Able to describe how lymph is formed from blood until it is brought back into the blood circulatory system.
Sample Answer :
P1: (When the blood flows from arteries into capillaries)there is higher hydrostatic pressure at the arterial end of the capillaries
P2: (This high pressure) forces some plasma to pass through the capillary walls into the intercellular spaces (between the cells)
P3: Once the fluid leaves the capillary walls, it is called interstitial/tissue fluid // The interstitial fluid fills the spaces between the cells and constantly bathes the cells
P4: 90% of the interstitial fluid diffuses back into blood capillary
P5: 10% of the interstitial fluid that has not been reabsorbed into the bloodstream goes into the lymph capillaries.(Once inside the lymph capillaries) the fluid is known as lymph.
P6: The lymph capillaries unite to form larger lymphatic vessels.
P7: From the lymphatic vessels, lymph eventually passes into the thoracic duct/the right lymphatic duct.
P8:The thoracic duct empties its lymph into the right subclavian vein. (Hence, lymph drains back into the blood).
Any 6 P
Max 6
TOTAL 20
No 9
Marking Criteria
Marks
(a)
Able to state the Sequence of events that occur when the hand touches a hot object.
Sample answer :
P1: the heat on the object stimulates the nerve endings (receptors) in the skin.
P2: impulses are triggered.
P3: This impulses travel along the sensory/afferent neurone to the spinal cord.
P4: in the spinal cord, the impulses are transmitted first across a synapse to the interneurone and then across another synapse to the motor/efferent neurone. ( at least 2 type of neurone)
At synapse
P5: When an impulse reach a presynaptic membrane, it triggers the synaptic vesicles to release neutrotransmitter into the synaptic cleft.
P6 The neurotransmitter diffuse across the synaptic cleft
P7: and bind to receptors which are attached to the postsynaptic membrane.
P8: The binding of the neurotransmitter to the receptors leads to the generation of a new impulse
P9: Impulses leave the spinal cord along the motor/efferent neurone to the effector
P10: the effector is the biceps muscle which then contracts. This brings about a sudden withdrawal of the hand.
Any 8
(b) (i)
Able to explain how geotropism is brought about in a plant root and shoot.
Sample answer :
Diagram :
Shoot
Root
Correct Diagram ( shoot grows upward, root grow downward): 1 m
Shoot
P1 : The auxin that is produced at the tip of shoot.
P2 : auxin moves downward/ accumulate on the underside of the shoot tip due to the pull of gravity.
P3 : the high concentration of auxin accelerates the growth
P4 : stimulating greater cell elongation on the underside relative to the cells on the upper side.
P5 : this differential elongation causes the shoot to bend away from gravity / grow downwards.
Root
P6 : The auxin that is produced at the tip of root.
P7 : auxin moves downward/ accumulate on the underside of the root tip due to the pull of gravity.
P8 : the high concentration of auxin inhibits the growth
P9 : slowing down cell elongation on the underside relative to the cells on the upper side.
P10 : this differential elongation causes the shoot to bend towards gravity / grow downwards.
Any 7 P
(b)(ii)
Able to explain the advantages
Sample answer
Advantages
Explanation
F1
ensures the root grow/penetrate deep into soil
P1
To anchor the plant firmly in the ground
F2
The roots always contact with the soil water/mineral/nutrients
P2
To maintain turgor pressure/ growth/ photosynthesis
F3
Ensure that the leaves of shoot growth towards sunlight
P3
To absorb maximum amount of light energy for photosynthesis.
Any 2 F with correspond P
4
TOTAL
20
Perfect Score Modul 1
PERFECT SCORE BIOLOGY
2009
MODUL I
STRUCTURAL QUESTION
Question 1:
(a) Explain the feeding mechanism in Amoeba sp. [4 marks]
· Amoeba sp. engulfs food by phagocytosis process
· Use a pseudopodia (false feet) to enclose the food particles
· The food particles are packaged in a food vacuole which fuses with lysosomes that contain lysozymes to be digested
· The nutrient is absorbed and the undigested material is left behind
(b) Explain the role of contractile vacuole in osmoregulation. [2 marks]
· Excess water diffuse into the contractile vacuole by osmosis
· When the contractile vacuole is filled to its maximum size , its contract to expel its content from time to time.
Question 2:
The Diagram 1 shows an experiment set-up to investigate the effect of pH on enzyme activity. Enzyme Q is found inside the mammalian digestive system.
Diagram 1
After 24 hours, the following results were obtained:
pH
Reduction in length of egg white strip (mm)
1
7
2
9
3
6
4
3
5
1
6
0
(a) Using the results in the table, plot a graph to show the reduction in length of the egg white strip against pH of the medium. [3 marks]
P – axis with scales ----1
T – points plotted accurately ------ 1
B – shape of line – smooth, single curve ------1
(b) Explain the observed reduction in the length of the egg white strip. [3 marks]
· Increasing the pH value of the medium will decrease the action of enzyme Q on egg white.
· The optimum pH of enzyme Q is 2.
· Enzyme Q works at its maximum rate at this pH/ At this pH, the reduction of length of the egg white strip is the greatest
(c) A type of herbicide was sprayed on the weed plants to control their population. The herbicide causes the proteins and ATPase enzyme in the cell to denature. Explain the effect of this herbicide in the transportation process of potassium ions in weed plants. [3 marks]
· The active sites of the carrier proteins and ATPase enzyme are denatured / destroyed.
· Potassium ion cannot bind with carrier protein
· Potassium ion is not transported into the cells of the weed plants/ No active transport of potassium ions.
· No energy / ATP produced /available
Question 3:
Mitosis and meiosis are a phenomenon in the cell cycle which happens gradually and continuously. The figure shows two cells taken from different parts of the flowering plant
Cell A is taken from the anther while cell B is taken from the root tip.
(a) What are the differences between the importance of cell division in Cell A and Cell B
Cell B to increase number of cell / growth /repair damage tissue whereas cell A to produce haploid gamete
Cell B involves in asexual reproduction whereas cell A involves in sexual reproduction
Cell B maintain the number of chorosome whereas in cell A the chromosome number in daughter cell is haploid
In cell B, the genetic content of chromosome is identical to parents cell, but in cell A there is a variation in genetic contents.
[3 marks]
(b) Suggest a technique that can produce plants with the same good characteristics in a short period of time. Briefly explain the technique which you have mentioned
[4 marks]
· Tissue culture technique / invitro technique
· Suitable pH, use of sterile instruments and a correct balance of nutrients. [Conditions for the success of this technique ]
· The piece of cell or explants is cultured in a conical flask containing a suitable culture medium for growth, for example, a culture containing growth hormones.
· The suspension start to divide by mitosis to form a callus that is a group of tissues that have not differentiated into stems leaves and roots.
· The callus is then transferred into an agar medium and when the roots sprout, they are transferred into the soil.
· [After approximately two weeks] a new orchid clone similar to the parent is produced.
Question 4:
The figure below shows the process which occurs in structure P.
(a) Name process K. Explain the part played by sunlight in process K.
· Photolysis / Hydrolysis of water.
· Light energy will react with the chlorophyll molecule to free an electron.
· The water molecule will break up into hydrogen ion and hydroxide ion
[3 marks]
(b) Explain how process L can produce the substance Z.
· Hydrogen atoms are produced during photolysis of water
· Which will be used to reduce carbon dioxide during the dark reaction to produce Z.
[2 mark
(c) Diagram below shows type of stomach
(i) What is the ruminant's true stomach? Give a reason for your answer.
[ 2 marks]
· The abomasums is the ruminant's true or glandular stomach.
· Here also are the gastric juices that contain enzymes which assist in food digestion.
(ii) Describe the functions of stomach in human.
[3 marks]
· Temporary food storage until it passes to the intestines.
· Hydrochloric acid is produced by stomach secretion, suitable medium for pepsin.
· Movement of stomach muscle enable breaking up food particle and mixing it with gastric juices
· Digestive enzymes such as pepsin break down protein to peptides.
· Rennin solidifies milk protein/ coagulates for pepsin to act upon them.
Question 5:
The figure below shows the respiratory organ of an animal and human being.
(a) Based on the figure above, explain how the structure M and N can increase the efficiency of gaseous exchange in each of the organisms above.
[4 marks]
Adaptive features of structure N:
· Thin alveolar wall (one cell thick) which helps in the diffusion of gases,
· The wall of the alveolus is moist, respiratory gases can dissolve in it, easily diffuse the alveolus.
· Covered by network of blood capillaries provide large totals surface for gaseous exchange.
Adaptive features of structure M:
· Contains a fluid which can dissolve oxygen or carbon dioxide. Surrounded by body tissue.
· M is also connected straight to the trachea/ spiracle.
· Tracheol is one cell thick for easily diffusion of gaseous
(b) The amount of carbon dioxide and oxygen in the atmosphere is always stable due to the mutual dependence between photosynthesis and respiration. Explain why.
[4 marks]
The amount of oxygen in the atmosphere is absorbed by plants and animals to carry out respiration. Thus, the oxygen content in the atmosphere will be lowered.
During respiration, plants and animals also give out carbon dioxide. Thus the carbon dioxide content in the atmosphere will be increased especially at night.
Photosynthesis which is carried out by green plants takes in carbon dioxide from the atmosphere. Therefore the carbon dioxide content will be reduced especially during the day.
This process also gives out oxygen; therefore oxygen concentration in the atmosphere will be increased.
The two processes, photosynthesis by plants and respiration by plants and animals are mutually dependent so as to maintain the oxygen & carbon dioxide content in the atmosphere
Question 6
Diagram shows one of the environment phenomena which has become a current topic of discussion
(a) Based on the diagram, name this phenomena
Green house effect
(1 mark)
(b) Explain how the phenomena mention in (a) happens.
(4 marks)
· Ultra violet(uv) from solar radiation is absorbed by the earth and some of them is reflected back to the atmosphere in the form of heat/infra red.
· Heat or infrared radiation cannot be reflected back to the atmosphere.
· Because it is trapped by green house gases such as CO2, nitrogen dioxide and methane.
· Heat/infrared warmed the surface of earth
· Earth temperature increases.
(b) Suggest measures can be employed to overcome the phenomenon.
[4 marks]
· Replanting forest can preserve the concentration of CO2 in the atmosphere as green plants absorb carbon dioxide for photosynthesis.
· Avoid open burning can reduce the concentration of carbon dioxide
· Use incinerator to reduce the emission of carbon dioxide
· Replace fossil fuel with solar energy/ wind energy/ hydro energy to reduce emission of CO2 from the combustion of fossil fuel.
Question 7:
Diagram 7 shows the human circulatory system.
(a) Human circulatory system consists of circulations R and S. Describe each circulation.
[4 marks]
· Circulation R is pulmonary circulation
· The deoxygenated blood is pumped into the lungs via pulmonary artery for gases exchanged
· and the oxygenated blood back to the heart via pulmonary vein.
· Circulation S is systemic circulation
· the oxygenated blood is pumped to all parts of body cells via aorta.
· After oxygen is supplied, the deoxygenated blood is transported back into the heart via vena cava.
(c) The pacemaker of a patient’s heart fails to function. An electronic pacemaker is used to replace the original pacemaker. Explain how the electronic pacemaker works.
[3 marks]
· It detects the heart’s own rhythms / initiates impulses
· Impulses spread to both of the atria
· Causing both of the atria contract simultaneously.
Question 8:
Diagram 8 shows a longitudinal section of the reproductive parts of a flower during
fertilization.
(a) In the space below, draw a section through the ovule showing all the cells in S.
Label the cells involved in fertilization.
[3 marks]
Polar cell
Polar cell
Egg cell
Drawing: clear diagram with 8 nucleus – 1 mark Label : 2 label = 2 mark
(ii) Describe the fertilization process that occurs.
· One of the Q/ male nucleus fertilizes an egg to form the diploid zygote
· One of the Q/ male nucleus fertilizes 2 polar nuclei to form the triploid endosperm
[2 marks]
(c) In Diagram 8, the structure Y has to be kept dormant for future research purposes.
(i) Explain how Y can be prevented from germinating.
· Keep Y in dry place/ low temperature
· Because moisture initiate germination// enzyme is in inactive state
[2 marks]
(ii) If Y is to be germinated, suggest one method to stimulate the germination of Y
· Dropping/ spraying sucrose / sugary solution on Y
[1mark]
Question 9
Diagram 9.1 shows the prenatal development of a human embryo.
R
(a) Explain how the adaptation of R in order to function effectively
[3 marks]
· Consist of embryonic tissue that developed into plenty of small villi attached to endometrium, increasing the surface area for efficient absorption.
· Network of blood capillaries in the placenta separated from the mother network capillaries in endometrium, prevent the foetal blood and the mother’s blood from mixing.
· Blood capillaries of placenta and blood capillaries of foetus is much closed, enable the diffusion of digested food and waste product.
(b) If the mother is addicted to the drugs, does it possible for the baby to become addicted too? Explain your answer.
[2 marks]
· Yes.
· Drugs in the mother blood able to diffuse across the placenta into the baby blood, it is possible the baby to become addicted.
Question 10:
(a) Diagram 10.1 below show two types of variation in students,
Blood group
Number of student
Body mass (kg)
Diagram 10.1
Describe the differences between these two types of variation.
[3 marks]
· Blood group is discontinuous variation whereas the body mass is continuous variation.
· The differences in type of blood is appears for a few discrete categories but body mass has no distinct categories.
· Blood group is controlled by single gene and not influenced by environment but controlled by combined effect of many of genes and affected by environmental factors.
(b) Diagram 10.2 and diagram 10.3 show the changes occur in the structure of chromosomes which causes mutation.
P
Q
R
S
T
U
P
Q
S
T
U
P
Q
R
S
T
U
P
Q
S
R
T
U
(i) State one factor which causes the mutation.
[1mark]
Mutagen (such as benzene, drugs, beta-ray)
(ii) Compare the changes of chromosome structure that causes mutations in Diagram 10.2 and Diagram 10.3
[4 marks]
· A segment in chromosome in Diagram 10.2 breaks off and lost but in Diagram 10.3 the segment breaks off but reattaches avertedly.
· The mutation in Diagram 10.2 can causes lost of genes and can be lethal (dead) but mutation in Diagram 10.3 can causes changes in phenotype.
Question 11:
(a) (i) What P, Q, R ?
P : Starch
Q : Glycogen
R : Cellulose
[ 1 mark]
(ii) What is the process that breaks down the polysaccharides into smaller units? Explain the process briefly.
Hydrolysation
polysacharrides can be broken down by adding water
the H+ group and the OH- group will combine to each smaller unit / monosaccharide
smaller unit is formed
[ 3 marks
( b) What is the chemical reaction P that takes place to form the bonds between X and Y?
What is Q? Explain the process briefly.
P-condensation.
Q-water
o When two molucle / X and Y joined together by condensation a water molecule is formed
o Each molucle will give out H+ group and the OH- group
[ 4 marks]
( c)(i) What is denaturation?
Denaturation is a process to break down a peptide bond causing the protein structure to change. This is caused by heat, pH, ultraviolet rays.
[ 2 marks]
(ii) What is the result of denaturation of proteins?
The protein becomes inactive and cannot function.
[ 1 mark]
Question 12:
An experiment is carried out to study the translocation of auxin in plants.
In experiment I, the coleoptile tip was cut and placed on an agar block for 5 hours. This agar block X produced a growth curvature in another decapitated coleoptile.
In experiment II, the coleoptile tip was cut and placed in an inverted position on an agar block Y, for 5 hours.
However no growth curvature was obtained when the agar block was placed on a decapacitated coleoptile.
[ 2 marks ]
Answer :
(a) The auxin from the agar block diffuses down to increase the rate of growth on the left side causing it to bend to the right.
(b)
(i) Indole acetic acid/auxin.
(ii) Tip of the stem.
(b) -X.
-The agar block X produces a growth curvature indicating that auxin has diffused into one side of the coleoptile to increase the rate of growth.
-The agar block Y produces no curvature indicating the absence of auxin in it.
(c) -From the tip of the coleoptile to its base.
- The presence of auxin in X and the absence of auxin in Y suggest that translocation of auxin is from the tip to the base but not the reverse.
Question 13:
Figure below shows part of the circulatory system and the lymphatic system in the human body.
Fluid X
Body tissue
Tissue fluid
heart
Lymp node
Vessel Q
ANSWERS
a) (i) Lymphatic fluid ( 1 mark )
(ii) Fluid X was lack of lymphocytes whereas after lymph nodes more lymphocytes.
( 1 mark )
b) (i) When the fluid seeps through the blood capillaries into interstitial fluid, the pressure is higher in the interstitial fluid causes the fluid to be forced into X.
( 2 marks )
(ii) There is no red blood cells and plasma protein.
( 1 mark )
c) (i) If Q is blocked so the pressure in X will increase and causes the fluid to move back into the interstitial and accumulate in the interstitial fluid.
( 2 marks )
(ii) swollen. ( 1 mark )
d) The blood circulatory system will maintain its contents such as water and minerals salt by osmoregulation.
The water and mineral salt is controlled by hormone from pituitary glands and hence control the concentration of urine to be excreted by kidney.
The lymphatic system uses the movement of muscle valve and the pressure of the fluid to concentrate the water and mineral salt content.
( 3 marks )
Question 14:
A
B
Figure 14.0
Figure 3.0 shows the reversible process digestion of substance P by lipase produced the products Q and R.
(a) Name the processes A and B.
A: Hydrolysis/ breaks down
B: Condensation
(b) Explain how the organic substance is absorbed into the villus.
[4 marks]
Lipids or triglycerides is digested/hydrolysed/ broken down
by lipase to produce fatty acid and glycerol.
Fatty acid and glycerol diffuse into the lacteal by diffusion.
(c) A gall bladder of a patient is removed due to a gall stone. Explain the effects on the health. [4 marks]
· Bile cannot be secreted by gall bladder to emulsify excess of lipids into the tiny droplets
· Lipase cannot reacts on lipids effectively / less lipids is digested to fatty acid and glycerol.
· Bile cannot neutralise the acidic foods from the stomach
· Alcaline medium that is an optimum meduim for the digestion of lipids, carbohydrate and protein cannot be created.
· The digestion of lipids, carbohydrate and protein are affected.
· More acid in the duodenum, it leads to duodenum gastric.
(d) The products P and Q are transported by the lymphatic system to the cells for assimilation. Explain the assimilation of glucose and amino acid in body cells.
Glucose is oxidised to produce energy, carbon dioxide and water by cellular respiration.
Amino acid is used to synthesis protoplasm (the component of cell). By this way new cells will be synthesised causing growth.
Amino acid also can be used to synthesis enzyme, hormone or antibody.
[4 marks]
Question 15:
Figure 11 shows a part of the ileum structure.
(a) Explain how P is adapted enable the absorption of nutrients effectively.
P is one cell thick so nutrients can diffuse efficiently
P has microvilli so it can increase large total surface area for absorption.
[2 marks]
(b) 10 cm of small intestine (ileum) of a patient is cut due to a cancer.
Explain the effects of removal 10 cm of small intestine on the digestion of food.
[4 marks]
Total surface area of ileum is decreased.
So secretion of enzymes lipase/ maltase/ lactase/ sucrase/ peptidase are decreased too.
Digestion of lipids/ maltose/ lactose/ sucrose/ polypeptides are decreased or not efficient.
Less fatty acid and glycerol/ glucose/ fructose/ galactose/ amino acid produced.
Absorption of nutrients affected
Due to the total surface area of ileum is less than normal person.
Question 16
Diagram 16 shows a ball and socket joint with tissues P, Q and which are responsible for locomotion in a human.
DIAGRAM 16
(a) State the organelle found abundantly in the muscle cell.
[1 mark]
Mitochondria
(b) Explain your answer in (a)
Muscle cells are an active cell
It needs lot of energy for the contraction of skeletal muscle
[2 marks]
(c) Name the muscle fibres that involve in the contraction of antagonistic muscle.
Actins and myosin
[2 marks]
(d) Explain how bending of arm is brought by the structures P, Q, R and joints.
[10 marks]
During bending an arm, R / biceps muscle contracts, while the triceps relaxes.
Contraction of biceps muscle produce energy
That generated by mitochondria
Energy is transferred to the ulna through the Q / tendon.
And connects the biceps muscle to the ulna
So ulna is pulled upwards
At the same time P binds humerus and ulna
Prevents both of the bone from dislocate/ hold two bones together
Humerus and ulna provide surface for the attachment of skeletal muscles
Joints allows the bone to move in one plane
Synovial membrane secretes synovial fluid
That acts as lubricant to reduce friction between two bones
Question 17
(a) Diagram below shows two phenotypes of pea seed produced by two varieties of pea plant.
Smooth and green pea seed
Wrinkled and yellow pea seed
A farmer has crossed a smooth and green pea plant with a wrinkled and yellow pea plant. 1600 pea seeds produced were collected and sorted according to different phenotypes as shown in Table below.
Phenotype
Smooth and green pea seed
Smooth and yellow pea
Wrinkled and green pea seed
Wrinkled and yellow pea
Number of seed
398
401
399
402
(a) State the ratio of the offspring produced.
1:1:1:1
[1 mark]
(b) (i) S represents the dominant allele for smooth while s for wrinkled. G is the dominant allele for green and g ressesive allele for yellow. State the genotype of the parents pea plant.
SsGg and ssgg
[2 marks]
State the probabilities of the alles in the gametes produced by the pea plants.
SG, Sg, sG and sg
[2 marks]
(c) Based on your answer in b(ii), explain the formation of alleles in the gametes by applying the Mandel’s Second Law
[3 marks]
During gametes formation, each member of a pair of alles Ss may combine randomly with either member of another pair of alles Gg.
During meiosis, only one of each pair of alleles Ss, and Gg can be present in a single gamete.
So alleles in gametes produced may be SG, sG, Sg and sg
(i) Draw a schematic diagram to show the product of this cross.
Parents: SsGg x ssgg
Gametes: SG : Sg : sG : sg all sg
Offspring: SsGg : SsGg : ssGg : ssgg
[3 marks]
(d)(i) A heterozygous plant genotype TtRr undergoes meiosis to produce gametes cell. Alleles T and t for tall and short are located at the same locus on the homologous
chromosomes. During anaphase I, the alleles Tt are not separated but Rr are separated completely.
Draw the diagram for the two gamete cells which will be formed in the space provided below.
[2marks]
(d)(ii) State the probabilities of the alleles in the gametes produced.
TtR, r or Ttr, R
[2 marks]
(e) The individual has three chromosome 21 due to non-disjunction during meiosis where the two homologous chromosome 21 fail to separate normally during anaphase I or II of meiosis. State type of mutation faced by the individual.
Chromosomal mutation [1 mark]
(f) Explain how mutation stated in (e) is formed.
The number of chromosomes change during meiosis produced the gametes cell has 24 chromosomes and the another one 22 chromosomes. An abnormal gamete 24 chromosomes fuses with the normal gamete 23 chromosomes produced zygote that has 47 chromosomes.
[3 marks]
2009
MODUL I
STRUCTURAL QUESTION
Question 1:
(a) Explain the feeding mechanism in Amoeba sp. [4 marks]
· Amoeba sp. engulfs food by phagocytosis process
· Use a pseudopodia (false feet) to enclose the food particles
· The food particles are packaged in a food vacuole which fuses with lysosomes that contain lysozymes to be digested
· The nutrient is absorbed and the undigested material is left behind
(b) Explain the role of contractile vacuole in osmoregulation. [2 marks]
· Excess water diffuse into the contractile vacuole by osmosis
· When the contractile vacuole is filled to its maximum size , its contract to expel its content from time to time.
Question 2:
The Diagram 1 shows an experiment set-up to investigate the effect of pH on enzyme activity. Enzyme Q is found inside the mammalian digestive system.
Diagram 1
After 24 hours, the following results were obtained:
pH
Reduction in length of egg white strip (mm)
1
7
2
9
3
6
4
3
5
1
6
0
(a) Using the results in the table, plot a graph to show the reduction in length of the egg white strip against pH of the medium. [3 marks]
P – axis with scales ----1
T – points plotted accurately ------ 1
B – shape of line – smooth, single curve ------1
(b) Explain the observed reduction in the length of the egg white strip. [3 marks]
· Increasing the pH value of the medium will decrease the action of enzyme Q on egg white.
· The optimum pH of enzyme Q is 2.
· Enzyme Q works at its maximum rate at this pH/ At this pH, the reduction of length of the egg white strip is the greatest
(c) A type of herbicide was sprayed on the weed plants to control their population. The herbicide causes the proteins and ATPase enzyme in the cell to denature. Explain the effect of this herbicide in the transportation process of potassium ions in weed plants. [3 marks]
· The active sites of the carrier proteins and ATPase enzyme are denatured / destroyed.
· Potassium ion cannot bind with carrier protein
· Potassium ion is not transported into the cells of the weed plants/ No active transport of potassium ions.
· No energy / ATP produced /available
Question 3:
Mitosis and meiosis are a phenomenon in the cell cycle which happens gradually and continuously. The figure shows two cells taken from different parts of the flowering plant
Cell A is taken from the anther while cell B is taken from the root tip.
(a) What are the differences between the importance of cell division in Cell A and Cell B
Cell B to increase number of cell / growth /repair damage tissue whereas cell A to produce haploid gamete
Cell B involves in asexual reproduction whereas cell A involves in sexual reproduction
Cell B maintain the number of chorosome whereas in cell A the chromosome number in daughter cell is haploid
In cell B, the genetic content of chromosome is identical to parents cell, but in cell A there is a variation in genetic contents.
[3 marks]
(b) Suggest a technique that can produce plants with the same good characteristics in a short period of time. Briefly explain the technique which you have mentioned
[4 marks]
· Tissue culture technique / invitro technique
· Suitable pH, use of sterile instruments and a correct balance of nutrients. [Conditions for the success of this technique ]
· The piece of cell or explants is cultured in a conical flask containing a suitable culture medium for growth, for example, a culture containing growth hormones.
· The suspension start to divide by mitosis to form a callus that is a group of tissues that have not differentiated into stems leaves and roots.
· The callus is then transferred into an agar medium and when the roots sprout, they are transferred into the soil.
· [After approximately two weeks] a new orchid clone similar to the parent is produced.
Question 4:
The figure below shows the process which occurs in structure P.
(a) Name process K. Explain the part played by sunlight in process K.
· Photolysis / Hydrolysis of water.
· Light energy will react with the chlorophyll molecule to free an electron.
· The water molecule will break up into hydrogen ion and hydroxide ion
[3 marks]
(b) Explain how process L can produce the substance Z.
· Hydrogen atoms are produced during photolysis of water
· Which will be used to reduce carbon dioxide during the dark reaction to produce Z.
[2 mark
(c) Diagram below shows type of stomach
(i) What is the ruminant's true stomach? Give a reason for your answer.
[ 2 marks]
· The abomasums is the ruminant's true or glandular stomach.
· Here also are the gastric juices that contain enzymes which assist in food digestion.
(ii) Describe the functions of stomach in human.
[3 marks]
· Temporary food storage until it passes to the intestines.
· Hydrochloric acid is produced by stomach secretion, suitable medium for pepsin.
· Movement of stomach muscle enable breaking up food particle and mixing it with gastric juices
· Digestive enzymes such as pepsin break down protein to peptides.
· Rennin solidifies milk protein/ coagulates for pepsin to act upon them.
Question 5:
The figure below shows the respiratory organ of an animal and human being.
(a) Based on the figure above, explain how the structure M and N can increase the efficiency of gaseous exchange in each of the organisms above.
[4 marks]
Adaptive features of structure N:
· Thin alveolar wall (one cell thick) which helps in the diffusion of gases,
· The wall of the alveolus is moist, respiratory gases can dissolve in it, easily diffuse the alveolus.
· Covered by network of blood capillaries provide large totals surface for gaseous exchange.
Adaptive features of structure M:
· Contains a fluid which can dissolve oxygen or carbon dioxide. Surrounded by body tissue.
· M is also connected straight to the trachea/ spiracle.
· Tracheol is one cell thick for easily diffusion of gaseous
(b) The amount of carbon dioxide and oxygen in the atmosphere is always stable due to the mutual dependence between photosynthesis and respiration. Explain why.
[4 marks]
The amount of oxygen in the atmosphere is absorbed by plants and animals to carry out respiration. Thus, the oxygen content in the atmosphere will be lowered.
During respiration, plants and animals also give out carbon dioxide. Thus the carbon dioxide content in the atmosphere will be increased especially at night.
Photosynthesis which is carried out by green plants takes in carbon dioxide from the atmosphere. Therefore the carbon dioxide content will be reduced especially during the day.
This process also gives out oxygen; therefore oxygen concentration in the atmosphere will be increased.
The two processes, photosynthesis by plants and respiration by plants and animals are mutually dependent so as to maintain the oxygen & carbon dioxide content in the atmosphere
Question 6
Diagram shows one of the environment phenomena which has become a current topic of discussion
(a) Based on the diagram, name this phenomena
Green house effect
(1 mark)
(b) Explain how the phenomena mention in (a) happens.
(4 marks)
· Ultra violet(uv) from solar radiation is absorbed by the earth and some of them is reflected back to the atmosphere in the form of heat/infra red.
· Heat or infrared radiation cannot be reflected back to the atmosphere.
· Because it is trapped by green house gases such as CO2, nitrogen dioxide and methane.
· Heat/infrared warmed the surface of earth
· Earth temperature increases.
(b) Suggest measures can be employed to overcome the phenomenon.
[4 marks]
· Replanting forest can preserve the concentration of CO2 in the atmosphere as green plants absorb carbon dioxide for photosynthesis.
· Avoid open burning can reduce the concentration of carbon dioxide
· Use incinerator to reduce the emission of carbon dioxide
· Replace fossil fuel with solar energy/ wind energy/ hydro energy to reduce emission of CO2 from the combustion of fossil fuel.
Question 7:
Diagram 7 shows the human circulatory system.
(a) Human circulatory system consists of circulations R and S. Describe each circulation.
[4 marks]
· Circulation R is pulmonary circulation
· The deoxygenated blood is pumped into the lungs via pulmonary artery for gases exchanged
· and the oxygenated blood back to the heart via pulmonary vein.
· Circulation S is systemic circulation
· the oxygenated blood is pumped to all parts of body cells via aorta.
· After oxygen is supplied, the deoxygenated blood is transported back into the heart via vena cava.
(c) The pacemaker of a patient’s heart fails to function. An electronic pacemaker is used to replace the original pacemaker. Explain how the electronic pacemaker works.
[3 marks]
· It detects the heart’s own rhythms / initiates impulses
· Impulses spread to both of the atria
· Causing both of the atria contract simultaneously.
Question 8:
Diagram 8 shows a longitudinal section of the reproductive parts of a flower during
fertilization.
(a) In the space below, draw a section through the ovule showing all the cells in S.
Label the cells involved in fertilization.
[3 marks]
Polar cell
Polar cell
Egg cell
Drawing: clear diagram with 8 nucleus – 1 mark Label : 2 label = 2 mark
(ii) Describe the fertilization process that occurs.
· One of the Q/ male nucleus fertilizes an egg to form the diploid zygote
· One of the Q/ male nucleus fertilizes 2 polar nuclei to form the triploid endosperm
[2 marks]
(c) In Diagram 8, the structure Y has to be kept dormant for future research purposes.
(i) Explain how Y can be prevented from germinating.
· Keep Y in dry place/ low temperature
· Because moisture initiate germination// enzyme is in inactive state
[2 marks]
(ii) If Y is to be germinated, suggest one method to stimulate the germination of Y
· Dropping/ spraying sucrose / sugary solution on Y
[1mark]
Question 9
Diagram 9.1 shows the prenatal development of a human embryo.
R
(a) Explain how the adaptation of R in order to function effectively
[3 marks]
· Consist of embryonic tissue that developed into plenty of small villi attached to endometrium, increasing the surface area for efficient absorption.
· Network of blood capillaries in the placenta separated from the mother network capillaries in endometrium, prevent the foetal blood and the mother’s blood from mixing.
· Blood capillaries of placenta and blood capillaries of foetus is much closed, enable the diffusion of digested food and waste product.
(b) If the mother is addicted to the drugs, does it possible for the baby to become addicted too? Explain your answer.
[2 marks]
· Yes.
· Drugs in the mother blood able to diffuse across the placenta into the baby blood, it is possible the baby to become addicted.
Question 10:
(a) Diagram 10.1 below show two types of variation in students,
Blood group
Number of student
Body mass (kg)
Diagram 10.1
Describe the differences between these two types of variation.
[3 marks]
· Blood group is discontinuous variation whereas the body mass is continuous variation.
· The differences in type of blood is appears for a few discrete categories but body mass has no distinct categories.
· Blood group is controlled by single gene and not influenced by environment but controlled by combined effect of many of genes and affected by environmental factors.
(b) Diagram 10.2 and diagram 10.3 show the changes occur in the structure of chromosomes which causes mutation.
P
Q
R
S
T
U
P
Q
S
T
U
P
Q
R
S
T
U
P
Q
S
R
T
U
(i) State one factor which causes the mutation.
[1mark]
Mutagen (such as benzene, drugs, beta-ray)
(ii) Compare the changes of chromosome structure that causes mutations in Diagram 10.2 and Diagram 10.3
[4 marks]
· A segment in chromosome in Diagram 10.2 breaks off and lost but in Diagram 10.3 the segment breaks off but reattaches avertedly.
· The mutation in Diagram 10.2 can causes lost of genes and can be lethal (dead) but mutation in Diagram 10.3 can causes changes in phenotype.
Question 11:
(a) (i) What P, Q, R ?
P : Starch
Q : Glycogen
R : Cellulose
[ 1 mark]
(ii) What is the process that breaks down the polysaccharides into smaller units? Explain the process briefly.
Hydrolysation
polysacharrides can be broken down by adding water
the H+ group and the OH- group will combine to each smaller unit / monosaccharide
smaller unit is formed
[ 3 marks
( b) What is the chemical reaction P that takes place to form the bonds between X and Y?
What is Q? Explain the process briefly.
P-condensation.
Q-water
o When two molucle / X and Y joined together by condensation a water molecule is formed
o Each molucle will give out H+ group and the OH- group
[ 4 marks]
( c)(i) What is denaturation?
Denaturation is a process to break down a peptide bond causing the protein structure to change. This is caused by heat, pH, ultraviolet rays.
[ 2 marks]
(ii) What is the result of denaturation of proteins?
The protein becomes inactive and cannot function.
[ 1 mark]
Question 12:
An experiment is carried out to study the translocation of auxin in plants.
In experiment I, the coleoptile tip was cut and placed on an agar block for 5 hours. This agar block X produced a growth curvature in another decapitated coleoptile.
In experiment II, the coleoptile tip was cut and placed in an inverted position on an agar block Y, for 5 hours.
However no growth curvature was obtained when the agar block was placed on a decapacitated coleoptile.
[ 2 marks ]
Answer :
(a) The auxin from the agar block diffuses down to increase the rate of growth on the left side causing it to bend to the right.
(b)
(i) Indole acetic acid/auxin.
(ii) Tip of the stem.
(b) -X.
-The agar block X produces a growth curvature indicating that auxin has diffused into one side of the coleoptile to increase the rate of growth.
-The agar block Y produces no curvature indicating the absence of auxin in it.
(c) -From the tip of the coleoptile to its base.
- The presence of auxin in X and the absence of auxin in Y suggest that translocation of auxin is from the tip to the base but not the reverse.
Question 13:
Figure below shows part of the circulatory system and the lymphatic system in the human body.
Fluid X
Body tissue
Tissue fluid
heart
Lymp node
Vessel Q
ANSWERS
a) (i) Lymphatic fluid ( 1 mark )
(ii) Fluid X was lack of lymphocytes whereas after lymph nodes more lymphocytes.
( 1 mark )
b) (i) When the fluid seeps through the blood capillaries into interstitial fluid, the pressure is higher in the interstitial fluid causes the fluid to be forced into X.
( 2 marks )
(ii) There is no red blood cells and plasma protein.
( 1 mark )
c) (i) If Q is blocked so the pressure in X will increase and causes the fluid to move back into the interstitial and accumulate in the interstitial fluid.
( 2 marks )
(ii) swollen. ( 1 mark )
d) The blood circulatory system will maintain its contents such as water and minerals salt by osmoregulation.
The water and mineral salt is controlled by hormone from pituitary glands and hence control the concentration of urine to be excreted by kidney.
The lymphatic system uses the movement of muscle valve and the pressure of the fluid to concentrate the water and mineral salt content.
( 3 marks )
Question 14:
A
B
Figure 14.0
Figure 3.0 shows the reversible process digestion of substance P by lipase produced the products Q and R.
(a) Name the processes A and B.
A: Hydrolysis/ breaks down
B: Condensation
(b) Explain how the organic substance is absorbed into the villus.
[4 marks]
Lipids or triglycerides is digested/hydrolysed/ broken down
by lipase to produce fatty acid and glycerol.
Fatty acid and glycerol diffuse into the lacteal by diffusion.
(c) A gall bladder of a patient is removed due to a gall stone. Explain the effects on the health. [4 marks]
· Bile cannot be secreted by gall bladder to emulsify excess of lipids into the tiny droplets
· Lipase cannot reacts on lipids effectively / less lipids is digested to fatty acid and glycerol.
· Bile cannot neutralise the acidic foods from the stomach
· Alcaline medium that is an optimum meduim for the digestion of lipids, carbohydrate and protein cannot be created.
· The digestion of lipids, carbohydrate and protein are affected.
· More acid in the duodenum, it leads to duodenum gastric.
(d) The products P and Q are transported by the lymphatic system to the cells for assimilation. Explain the assimilation of glucose and amino acid in body cells.
Glucose is oxidised to produce energy, carbon dioxide and water by cellular respiration.
Amino acid is used to synthesis protoplasm (the component of cell). By this way new cells will be synthesised causing growth.
Amino acid also can be used to synthesis enzyme, hormone or antibody.
[4 marks]
Question 15:
Figure 11 shows a part of the ileum structure.
(a) Explain how P is adapted enable the absorption of nutrients effectively.
P is one cell thick so nutrients can diffuse efficiently
P has microvilli so it can increase large total surface area for absorption.
[2 marks]
(b) 10 cm of small intestine (ileum) of a patient is cut due to a cancer.
Explain the effects of removal 10 cm of small intestine on the digestion of food.
[4 marks]
Total surface area of ileum is decreased.
So secretion of enzymes lipase/ maltase/ lactase/ sucrase/ peptidase are decreased too.
Digestion of lipids/ maltose/ lactose/ sucrose/ polypeptides are decreased or not efficient.
Less fatty acid and glycerol/ glucose/ fructose/ galactose/ amino acid produced.
Absorption of nutrients affected
Due to the total surface area of ileum is less than normal person.
Question 16
Diagram 16 shows a ball and socket joint with tissues P, Q and which are responsible for locomotion in a human.
DIAGRAM 16
(a) State the organelle found abundantly in the muscle cell.
[1 mark]
Mitochondria
(b) Explain your answer in (a)
Muscle cells are an active cell
It needs lot of energy for the contraction of skeletal muscle
[2 marks]
(c) Name the muscle fibres that involve in the contraction of antagonistic muscle.
Actins and myosin
[2 marks]
(d) Explain how bending of arm is brought by the structures P, Q, R and joints.
[10 marks]
During bending an arm, R / biceps muscle contracts, while the triceps relaxes.
Contraction of biceps muscle produce energy
That generated by mitochondria
Energy is transferred to the ulna through the Q / tendon.
And connects the biceps muscle to the ulna
So ulna is pulled upwards
At the same time P binds humerus and ulna
Prevents both of the bone from dislocate/ hold two bones together
Humerus and ulna provide surface for the attachment of skeletal muscles
Joints allows the bone to move in one plane
Synovial membrane secretes synovial fluid
That acts as lubricant to reduce friction between two bones
Question 17
(a) Diagram below shows two phenotypes of pea seed produced by two varieties of pea plant.
Smooth and green pea seed
Wrinkled and yellow pea seed
A farmer has crossed a smooth and green pea plant with a wrinkled and yellow pea plant. 1600 pea seeds produced were collected and sorted according to different phenotypes as shown in Table below.
Phenotype
Smooth and green pea seed
Smooth and yellow pea
Wrinkled and green pea seed
Wrinkled and yellow pea
Number of seed
398
401
399
402
(a) State the ratio of the offspring produced.
1:1:1:1
[1 mark]
(b) (i) S represents the dominant allele for smooth while s for wrinkled. G is the dominant allele for green and g ressesive allele for yellow. State the genotype of the parents pea plant.
SsGg and ssgg
[2 marks]
State the probabilities of the alles in the gametes produced by the pea plants.
SG, Sg, sG and sg
[2 marks]
(c) Based on your answer in b(ii), explain the formation of alleles in the gametes by applying the Mandel’s Second Law
[3 marks]
During gametes formation, each member of a pair of alles Ss may combine randomly with either member of another pair of alles Gg.
During meiosis, only one of each pair of alleles Ss, and Gg can be present in a single gamete.
So alleles in gametes produced may be SG, sG, Sg and sg
(i) Draw a schematic diagram to show the product of this cross.
Parents: SsGg x ssgg
Gametes: SG : Sg : sG : sg all sg
Offspring: SsGg : SsGg : ssGg : ssgg
[3 marks]
(d)(i) A heterozygous plant genotype TtRr undergoes meiosis to produce gametes cell. Alleles T and t for tall and short are located at the same locus on the homologous
chromosomes. During anaphase I, the alleles Tt are not separated but Rr are separated completely.
Draw the diagram for the two gamete cells which will be formed in the space provided below.
[2marks]
(d)(ii) State the probabilities of the alleles in the gametes produced.
TtR, r or Ttr, R
[2 marks]
(e) The individual has three chromosome 21 due to non-disjunction during meiosis where the two homologous chromosome 21 fail to separate normally during anaphase I or II of meiosis. State type of mutation faced by the individual.
Chromosomal mutation [1 mark]
(f) Explain how mutation stated in (e) is formed.
The number of chromosomes change during meiosis produced the gametes cell has 24 chromosomes and the another one 22 chromosomes. An abnormal gamete 24 chromosomes fuses with the normal gamete 23 chromosomes produced zygote that has 47 chromosomes.
[3 marks]
Wednesday, July 22, 2009
Program Peningkatan Prestasi Biologi T.5 2009
1. Bengkel Menjawab Kertas 2 oleh Penceramah Jemputan -11 Julai 2009
2. Kelas Tambahan Selasa dan Khamis- Ulangkaji Tajuk Tingkatan 4
3.Membuat latihan Modul Perfect Score 2009
4.Bengkel Menjawab Kertas 3 oleh Penceramah Jemputan 26 Julai 2009
5 Latih Tubi soalan - soalan SPM sebenar / mirip SPM / Soalan Percubaan SPM lain-lain negeri
6.Klinik Biologi untuk pelajar pilihan - waktu PJK
7.PRAS - 1 Mac 2009 , 19 April 2009, 13 Sept 2009 dan 8 Nov 2009
8.Program Bersama Menjana Kecemerlangan Oktober & November 2009
2. Kelas Tambahan Selasa dan Khamis- Ulangkaji Tajuk Tingkatan 4
3.Membuat latihan Modul Perfect Score 2009
4.Bengkel Menjawab Kertas 3 oleh Penceramah Jemputan 26 Julai 2009
5 Latih Tubi soalan - soalan SPM sebenar / mirip SPM / Soalan Percubaan SPM lain-lain negeri
6.Klinik Biologi untuk pelajar pilihan - waktu PJK
7.PRAS - 1 Mac 2009 , 19 April 2009, 13 Sept 2009 dan 8 Nov 2009
8.Program Bersama Menjana Kecemerlangan Oktober & November 2009
Friday, July 17, 2009
PERFECT SCORE BIOLOGY 2009
PERFECT SCORE Biology 2009
MODUL II
ESSAY QUESTIONS
Question 1:
Diagram 1 shows a reaction catalyzed by sucrase.
(a) Based on Diagram 1, explain the characteristics of an enzyme. (6 marks)
· Enzyme is highly specific
· M only can break down N because it has an active site which is fits to N molecule.
· The enzyme reaction is reversible.
· Substrate / complex molecule can be broken down by hydrolysis and
· products can be converted to complex compound by condensation.
· Enzyme does not destroy after the reaction / can be reused after the reaction
· After the substrate is broken down into products, it leaves the active site of enzyme.
(b)
Enzymes are widely used in our daily life and industries.
Based on the above statement, explain how enzymes are used in the food industry. [4 marks]
· Lipase is used to react with fats in milk to produce cheese.
· Protease is used to tenderise meat so that it can be consumed easily.
· Cellulase used to separate agar from seaweed.
· Zymase in yeast is used widely in the preparation of alcoholic drinks such as wine and beer.
· Zymase acts with the glucose in grape juice during fermentation, thus producing ethanol, which is used to make wine.
· Amylase is used to remove starch from fruits in the making of fruit juice.
· In breweries, amylase is used to remove starch before the addition of malt.
Question 2:
A study is carried out on two individuals X and Y. They were asked to drink glucose solution of the same volume and concentration. The glucose and insulin level in the blood is measured.
The graph in Diagram 2 shows the changes in concentration of glucose and insulin in the blood of individuals X and Y.
The concentration of glucose in a normal individual is 0.09 gdm-3
(a) Explain the differences between the graph of individuals X and Y in relation to:
(i) blood glucose concentration
(ii) insulin concentration in the blood [10 marks]
· At stage 1, before glucose intake, blood glucose concentration in individual Y is normal which is 0.09 g/dm3 while blood glucose concentration in individual X is high (0.13 g/dm3).
· Y is normal while X is suffering from diabetes mellitus.
· At stage 2 after glucose intake, blood glucose level in individual Y increases until it reaches at maximum level (0.13 g/dm3) while blood glucose level in individual X increases rapidly and reaches at maximum level 0.20 g/dm3.
· At stage 3 blood glucose level decreases rapidly in individual Y until it reaches at normal level (0.09 g/dm3) while blood glucose level in individual X decreases to 0.165 g/dm3.
· Y is normal while X is suffering from diabetes mellitus.
· At stage 1 insulin concentration in the individual Y is 0.02 arbitrary units before glucose intake while insulin concentration in the individual X is 0.01 arbitrary units.
· After glucose intake / at stage 2, insulin concentration in the individual Y increases rapidly and reach the maximum level 0.06 arbitrary units while insulin concentration in the individual X is
increases to 0.02 arbitrary units.
· More insulin is required to convert excess of glucose into glycogen back to
normal level while in individual X pancreas fails to secrete high concentration
of insulin so glucose level is high.
· At stage 3 that is after a few hours the concentration of insulin in the individual Y decreases and remain high in the blood (0.02 g/dm3) while in the individual X concentration of insulin is remain low (0.01 arbitrary units).
· Glucose level can be converted to the normal level in the individual Y while
glucose level of X is high and X is suffer from diabetes.
Question 3:
Diagram 3 shows the longitudinal section through the thorax of a man.
Diagram 3
Describe briefly the mechanism of breathing in man . In your answer describe the part played by the
(i) diaphragm
(ii) intercostals muscle
(iii) alveoli (10 marks)
Answer:
(i)
· Diaphragm is a muscular sheet in the body cavity separating the thorax from the abdomen
· At the start of inhalation, the muscles of the diaphragm contract , making it less arched
· This helps to increase the volume of the thoracic cavity and reduce the pressure of the thoracic cavity. Air rushes into the lungs
· When the muscles of the diaphragm relax , it returns to its arched condition , reducing the volume of the thoracic cavity and increasing the pressure of the thoracic cavity. Air is forced out of the lungs
(ii)
· The muscles between the ribs are known as intercostals muscles
· During inhalation the external intercostals muscle contracts and raise the lower ribs
· This helps to increase the volume of the thoracic cavity and reduce the pressure of the thoracic cavity. Air rushes into the lungs.
· During exhalation the external intercostals muscles contract , the ribs return to their original position , reduce the pressure of the thoracic cavity. Air is forced out of the lungs
(iii)
· The alveoli are thin-walled air sacs with the lungs
· These sacs are surrounded by a network of capillaries
· During inhalation the alveoli are filled with air and gaseous exchange occurs between the alveoli and the capillaries
· Oxygen from the alveoli diffuses into the capillaries while carbon dioxide diffuses from the capillaries into the alveoli.
Question 4:
Why do we need to practice good posture? Explain how we can practice good postures while standing, sitting, walking and lifting heavy objects. [10 marks]
Answer:
· It helps to ensure that our body is always well supported
· Then no undue pressure is placed on our muscles and spine
· The internal organ such as the lungs, heart and stomach will be able to function properly
· Muscles working properly prevent us from getting tired easily
· It will prevent backache and shoulder ache
· While standing our body should be erect and upright so that the weight of our body is supported by our feet
· While sitting our back should rest against the back of the chair
· We should maintain an upright position
· While walking , our body should be upright and straight
· Our head should be held high and our eyes looking straight ahead.
· While lifting heavy object , we should bend both our knees and lift objects from the floor
Question 5:
Reproduction technology has contributed a lot to human health and population growth. Describe how fertility is overcome in humans using the following methods:
Sperm banking
Artificial insemination
In vitro fertilization
Surrogate mother [10 marks]
Answer:
Sperm bank
· used when the man/spouse/male suffers from low sperm count / production of weak / unhealthy sperm
· sperms are donated and donor’s identity is recorded/ secret
· genetic biodata of donor and the woman recipient should match for successful fertilization
Artificial insemination
· Practiced by a couple when the man is sterile
· Donor’s sperm are injected into the woman uterus during ovulation
· No sexual intercourse is involved
· The genetic background and health of donor is screened before acceptance
In vitro fertilization
· Practiced when the woman’s ovum cannot be fertilized due to blockage of fallopian tube
· Ovum is extracted (from the woman) and fertilized outside the body / in petri dish / test tube
· Fertilized ovum / zygote grows in culture medium
· Than it is implanted into the woman’s uterus
Surrogate mother
· Method practiced when a woman cannot be pregnant in normal way
· The zygote is obtained by in vitro fertilization
· Another woman is implanted with the couple zygote
· (surrogate mother) carries the baby only until it is born [any 10 points]
Question 6
The figure below shows a fluid mosaic model of the plasma membrane
(a) Explain the functions of the structures X, Y and Z in the movement of substances across the plasma membrane
[4 marks]
· Structure X allows hydrophobic molecules which can dissolve in lipids such as fatty acids, glycerol, steroid hormones, vitamins A, D, E and K to move in and out of the cell.
· Structure Y allows bigger molecules which do not dissolve in lipids like glucose and amino acids to move in and out of the cell.
· Structure Z allows small charged molecules to move in and out of the cell.
(b) All movements of substances have to pass through a plasma membrane. Explain why the plasma membrane is known as a semi permeable membrane
(6 marks)
· The plasma membrane is known as a semi permeable membrane because it only allows the passage of certain molecules and limits the passage of other molecules.
· It consists of two layers of phospholipids with protein molecules scattered in them
· Molecules that can pass through the plasma membrane easily are those that can dissolve in lipid and small uncharged molecules.
· Molecules which cannot move through the plasma membrane but require proteins include large molecules that do not dissolve in lipid and small charged molecules.
· Channel protein does not require the binding of a molecule and conformational change to open.
· Channel protein allows molecules to steadily diffuse across the membrane through diffusion.
· However a carrier protein allows specific molecules to cross the cell membrane by undergoing a conformational change upon the binding of the molecule.
· The conformational change opens a hole through which the molecule can enter or leave a cell.
Question 7:
(a) Explain how the first line of defence can prevent the entry of pathogens into the body
[6 marks]
The first line of defence in the prevention of pathogens entering the body can be the function of
(i) The skin
· The skin serves as a protective layer against the entry of pathogens.
The ph of the skin which is in the range of 3 to 5 provides an acidic environment which is not conducive for the growth of microorganisms.
The enzyme lysozyme which is present in the sweat and sebum on the surface of the skin can kill microorganisms.
(ii) The nose and respiratory tract.
The mucous membrane that lines the respiratory tract can trap microbes, dust and foreign matter.
The ciliated epithelium which lines the respiratory tract and the nasal cavity can remove the mucus when a person coughs or sneezes.
(iii) The stomach.
The hydrochloric acid in the stomach can kill the microbes that is present in the food that has entered the body.
Besides all these, the body has natural reactions such as diarrhoea and diarrhoea that
Can help to remove toxins and microbes very quickly before it is sent to other parts of the body.
(b) Explain the effect of taking fatty foods that have high cholesterol content to human healthy life.
[4 marks]
Our body has better ability to store fats rather than use it to release energy.
As a result fats and excess cholesterol will settle on the arterial wall causing a condition known as arteriosclerosis.
The lumen of the artery becomes narrow and hinders/slowing down the flow of blood, thus causing high blood pressure/hypertension.
If the coronary artery is blocked, the supply of oxygen and nutrients to the heart is affected.
This will cause a heart attack.
If the flow of blood to the brain is blocked, there is a possibility that stroke can occur.
Question 8:
(a) Explain the principle involved in genetic engineering
[2 marks]
· Genetic engineering is a technique of manipulating the genes in an organism by removing a gene from one organism and transferring it to another organism
(b) Discuss the importance of genetic engineering in the medical and agricultural field.
[4 marks]
The importance of genetic engineering in the medical field.
· Production of insulin from bacteria.
· Gene from the pancreas that can produce insulin is removed and inserted into the DNA molecule of bacteria.
· The bacteria that can produce insulin grows very fast in a suitable nutrient medium.
· Insulin can be produced in large quantities in a short time.
Genetic engineering in agriculture.
· Produce food or plants which are genetically modified to produce high quality yield.
· Produce plants with high quality yield as they are resistant to viral attacks, pesticides or predators.
(c) Diagram 8 shows DNA finger printing. DNA finger printing is more scientific, accurate and advanced compared to the thumbprint in criminal investigation.
Discuss the use of DNA fingerprinting in the identification of the parents of a child. Based on your answer, identify the father of the child.
[4 marks]
· DNA fingerprinting is a technique used to identify parents of a child based on DNA banding similarity
· Generally, the DNA profile of a child is the combination of the mother and father.
· A child inherit half of the chromosomes number from father and half from mother.
· DNA fingerprinting gives more details and information regarding the identity of a person while thumbprints give general information as it is based on a few different types.
· A blood sample, seminal fluid or any tissue can be analysed and its result is compared with that of an individual who is a suspect.
· In conclusion, man P is the father of the child as the similarity of his DNA is higher
Question 9:
Diagram 9 shows organism, P, Q and R.
a) Explain briefly why humans carry out the activity as shown in diagram above.
[4 marks ]
b) Explain the important ecological roles of the tropical rainforest.
[6 marks]
c) i) Explain the impacts of the activity shown above on the environment.
[7 marks]
ii) Suggest methods to control the mentioned activity.
[3 marks]
Answers :
a)
- The human population grows rapidly. The demands for food and housing areas have increased.
( 1 mark )
- Vast areas of forest are cleared for agricultural and commercial purposes.
( 1 mark )
- Urbanization and industrialization have caused more forests to be cleared for road construction and housing areas.
( 1 mark )
- Deforestation is also caused by the demands for timber and fuel wood.
( 1 mark )
TOTAL 4 MARKS
b) - The rainforest sustains almost half of the flora and fauna of the world.
( 1 mark )
- Certain plants from the forest provide food and pharmaceutical products to humans. ( 1 mark )
- The rainforest regulates the climate by influencing humidity, rainfall and temperatures. ( 1 mark )
- Green plants inside the forest absorb carbon dioxide from the atmosphere and maintain the concentration of the gas in the atmosphere. ( 1 mark )
- Green plants provide oxygen to all living things. ( 1 mark )
- The rainforest also acts as water catchment areas. ( 1 mark )
TOTAL 6 MARKS
c) - Deforestation causes soil erosion , landslides, flash floods and global
warming. ( 1 mark )
- Causes the soil to become loose and less stable. ( 1 mark )
- Without the protection of green plants, the soil is exposed to the forces of wind and rain. ( 1 mark )
- The top layer of soil is washed away gradually by the rainwater. This is known as soil erosion. ( 1 mark )
- Soil erosion causes the depletion of minerals from the soil, therefore the soil becomes infertile and unsuitable for agriculture.
(1 mark )
- Landslides may happen on steep hillsides during heavy rain. It is because rainwater flows quickly and causes the top layer of the soil to crumble. ( 1 mark )
- Rivers and drains are silted and the flow of water is blocked.
( 1 water )
- Therefore, water flows inland and this causes flash floods in the lower areas during rainy seasons. ( 1 water )
TOTAL Max 7 MARKS
d) - Deforestation can be controlled by passing stricter laws to protect the
forest. ( 1 mark )
- Through education, the public should be taught on the importance of
the forest. ( 1 mark )
- New trees should be planted to replace trees which have been out drawn. ( 1 mark )
TOTAL 3 MARKS
Question 12:
Activities that carried out by the human to provide food, power and industrial needs have a extensive effects on the environment. These effects include atmospheric pollution, water pollution, destruction of habitats and communities and a natural phenomenon known as the greenhouse effect.
d) Explain the fenomena of greenhouse effect and problems caused by this fenomena.
( 10 marks )
e) Suggest two measures which can be employed in the management of development activities and the ecosystem to ensure that the balance of nature is maintained.
( 10 marks )
ANSWERS
a) Able to define and describe the causes and effects of greenhouse effects.
· The greenhouse effect is caused by specific gases which form a thin layer around the atmosphere. ( 1 mark )
· These gases include (water vapor )/ carbon dioxide/ methane/ ozone/ nitrogen oxides and CFCs – any 2 gases ( 1 mark )
· The increased concentrations of atmospheric carbon dioxide is primarily due to
- combustion of fossil fuels ( 1 mark )
- increased deforestation ( 1 mark )
- industrial production ( 1 mark )
Max 2 marks
· The greenhouse gases allows short wavelength radiation from the sun to reach the earth’s surface. ( 1 mark )
· Some of the infra-red radiation fails to pass back through the greenhouse layer.
- more heat is trapped in the atmosphere ( 1 mark )
- results in global warming ( 1 mark )
· Polar ice caps may melt
- causing the sea to rise / extensive floodings
( 1 mark )
- subsequent reduction in land mass
( 1 mark )
- some aquatic populations would increase
( 1 mark )
- some terrestrial populations decrease
( 1 mark )
Max 2 marks
· Climatic changes occur
- rainfall changes ( 1 mark )
- heat increases ( 1 mark ) Max 1 mark
TOTAL 10 marks
b) Able to list the following measures and its examples
· The practice of biological control
Use natural predators to control the population of pest species
( 1 mark )
· Examples:
- The use of caterpillars of the Cactoblastis cactorum to control the population of the prickly pear cactus.
( 1 mark )
- The use of (khaki chambel) to control the population of gold snails which feed on paddy stalks. ( 1 mark )
( or any suitable examples )
· Biological control is better than using chemical control because
( 1 mark )
- the use of pesticide is indiscriminate and harmless organisms can be killed. ( 1 mark )
- the effects of pesticides can be persistent and remain in the environment for long periods. ( 1 mark )
( or any suitable explanation)
Max 4 marks
· The use of renewable energy
· examples of renewable energy
- solar energy ( 1 mark )
- geothermal energy ( 1 mark )
(or any suitable examples)
· Renewable energy occurs naturally in the environment is inexhaustible/does not pollute the environment ( 1 mark )
Max 2 marks
· The use of technology ( 1 mark )
· Use of unleaded petrol ( 1 mark )
- reduce the emission of lead into the atmosphere ( 1 mark )
· Use of microorganism to clean up the environment ( 1 mark )
- certain bacterias are used to break down oil spills at sea.
( 1 mark )
· Install catalytic converters in vehicles ( 1 mark )
- convert the harmful gases released during combustion of fossil fuels to less harmful products. ( 1 mark )
(or any suitable examples)
Max 4 marks
TOTAL 10 MARKS
Question 13:
(a) The shaded area of the graph in Diagram1 shows the intake of oxygen by an athlete before, during and after running for five minutes.
Period of running
Key
Oxygen
demand
Oxygen
consumption
0
5
10
15
20
25
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.0
Oxygen consuption
(litre per minute)
Time (minutes)
Based on the graph , explain how an oxygen debt is built up when an athlete is running and how it is settled after he stops running.
[ 10 marks ]
Answer
- During a vigorous exercise (running), the breathing rate is increased. [1 mark]
- This is to supply more oxygen to the muscles for rapid muscular contraction. [1 mark]
- However, the supply of oxygen to muscles is still insufficient [1 mark]
- and the muscles have to carry out anaerobic respiration to release energy. [ 1 mark ]
- The glucose is converted into lactic acid, with only a limited amount of energy being produced. [ 1 mark ]
- An oxygen debt builds up in the body, when no oxygen use in energy production [ 1 mark ]
- High level of lactic acid in the muscles cause them to ache. [ 1 mark ]
- After running, the athlete breathes more rapidly and deeply than normal for
twenty minutes [1 mark ]
- There is recovery period after 10 minutes until it reaches 20 minutes when oxygen is paid back during aerobic respiration. [ 1 mark ]
- About 1/6 lactic acid is oxidized to carbon dioxide, water and energy. [ 1 mark ]
Max 10 marks
Question 14:
1. The diagram shows a terrestrial ecosystem.
eagle
bird
rat
snake
earthworm
Explain how energy flows through the food chain and how it is lost to the environment.
[ 8 marks]
Answer :
- Energy flows through the food chain in one direction [ 1 mark]
- In the food chain, the plant is the producer, the rat is the primary consumer, the snake is the secondary consumer and the eagle is the tertiary consumer. [ 2 marks] //
In the food chain, the plant is the producer, the earthworm is the primary consumer, the bird is the secondary consumer and the snake/ eagle is the tertiary consumer.
- Each level of food chain is called a trophic level. [ 1 mark ]
- Energy is transferred from one trophic level to another trophic level [ 1 mark ]
- When energy is transferred from one trophic level to another level as much as 90% of the chemical energy in the food consumed by primary consumer is used for its metabolic activities and lost as heat ` [ 2 marks ]
- Only 10% of the energy in an organism is passed on to the organism at the next trophic level [ 1 mark ]
[10 marks]
Question 15:
The figure above shows a type of transport which happens on the plasma membrane of a cell.
Based on the diagram above, name the type of transport in the above figure.
Explain your answer.
[6 marks]
Answer :
- Active transport
- is dynamic and requires energy / ATP for the transfer of molecules or ions
- an area of low concentration to an area of high concentration through the living cell membrane.// go against the concentration gradient
- It requires carrier protein
- The carrier protein must be able to change its shape (and also be able to return to its original state.)
- To combine the active site of the carrier protein with the molecules or ions.
- To carry the complex molecular carriers through the plasma membrane.
Question 16:
The figure shows the organelles involved in the production of extracellular enzymes.
(a) (i) Using a name example, explain the term extracellular enzyme.
[2 marks]
(iii) Based on the organelles shown, explain how extracellular enzymes are produced.
[8 marks]
(b) The figure shows a reaction of an enzyme and its substrate
Explain the mechanism of the enzymatic reaction
[10 marks]
Answer:
( a) (i) - Extracellular enzyme is produced in a cell, then packed and secreted from the cell.
It catalyses its reaction outside the cell. An example is amylase.
- (ii) The instruction for making the extracellular enzyme is transcribed from the deoxyribonucleic acid
o (DNA) to ribonucleic acid (RNA) in the nucleus.
o The RNA then leaves the nucleus through the nuclear pore
o and attaches itself to the ribosome located on the endoplasmic reticulum.
o When the synthesis of the enzymes is completed
o it is encapsulated in a transport vesicle which fuses with the golgi body.
o In the golgi body, the enzyme is further modified before being packed in a secretory vesicle.
o The secretory vesicle transports the enzyme to the plasma membrane, where it fuses with it
o and the enzyme is released outside the cell.
(b) The explanation of enzyme action is known as the ‘lock and key hypothesis’.
· The substrate molecule fits into the active site of the enzyme molecule.
· The substrate is the ‘key’ that fits into the enzyme ‘lock’.
· Various types of bonds such as hydrogen and ionic bonds hold the substrate
· in the active site forming the enzyme-substrate complex.
· Once the complex is formed, the enzyme changes the substrate to its product.
· The product leaves the active site.
· The enzyme is not altered by the reaction and it can be reused.
Question 17:
Discuss the uses of enzymes in the household & industries.
Why are enzyme used in these industries.
Answer
Enzymes are used as biological detergents.
· Protease degrades coagulated proteins into soluble short-chain peptides.
· Lipase degrades fat or oil stains into soluble fatty acid and glycerol.
· Amylase degrades starch into soluble shorter-chain polysaccharides and sugars.
Enzymes are used in the baking industry.
· Protease is used in the breakdown of proteins in flour for the production of biscuits.
· Amylase is used in the breakdown of some starch to glucose in flour for making white bread, buns and rolls.
Enzymes are used in the medical field.
· Trypsin is used to remove blood clots and to clean wounds.
· Various other enzymes are used in biosensors.
Enzymes are used in industries because:
·They are effective.
·They are cheap and easy to use.
·They can be re-used, thus only small amounts are needed.
· They don't require high temperature to work, thus this reduces fuel costs.
Question 18
Diagram above shows the structure of HIV. The virus infects helper T cells of the human ‘s immune system that caused AIDS, Acquired Immune Deficiency Syndrome.
i) Descibe the effects of the human immunodeficiency virus (HIV) on the body’s defence mechanism.
( 6 marks )
ii) Suggest ways to prevent the spread of Acquired Immune Deficiency Syndrome (AIDS). ( 4 marks )
ANSWERS
2 (i) - HIV infects helper T cells.
- Helper T cells are essential to activate other lymphocytes in the body’s defence mechanism against diseases.
- There may be a long incubation period of several years for the virus before the symptoms appear.
- The immune system of the infected person gradually becomes weakened and defenceless against many pathogens.
- HIV also attacks the central nervous system.
- There is weight loss and a decrease in the function of the nervous system.
( 6 marks )
(ii) - By practice sexual abstinence or only having sexual relationship with one partner.
- Use condom during copulation/sexual relationship.
- Set up centres to exchange free sterile syringe and needles for used ones to reduce the practice of sharing needles amongst drug users.
- Strict screening of blood before it is used for transfusion.
-Carry out awareness campaigns to educate public and schoolchildren about the dangers of AIDS.
- Proper counseling provided to HIV positive person so that they do not spread the virus to other healthy people.
( Any 4 answers)
Question 19
Figure shows cells P and Q undergoing cell division.
M
(a) Based on the cell division of P, explain the best technique can be used by an oil palmer to produce a large number of oil palm in a short time.
State one problem to be considered in using technique.
The technique can be used is cloning and culture tissue.
Tip of shoot/root namely explants is cut
And cultured into the medium that contain nutrient/ hormone and the temperature
Is kept constant at the optimum temperature and sterile (in vitro).
Explant divides actively by mitosis to produce large number of cells namely callus.
Callus differentiates to form shoot and root.
Plantlets / clones are formed.
The disadvantage of cloning is no variation. Clones are exactly identical to parent cell. If one clone is infected by disease, the others will be infected too because they have same resistance to the certain disease.
(b) A farmer wants to produce large number of goat A in a short period time. He tries to apply the concept of cell division cell P to produce new individuals of goat A. Describe the technique used. Justify the technique.
The somatic cell (diploid cell) of goat A is taken and the nucleus of goat A is removed.
An egg cell of goat B is removed and the haploid nucleus of goat B is removed.
The diploid nucleus of goat A is inserted into the egg cell of goat B.
The egg cell contain the diploid nucleus is cultured into the medium that contain nutrient and the condition is aseptic.
The cell divides by mitosis to produce large number of cells namely embryo.
Embryo is implanted into the uterus of goat B.
A farmer should use the technique because by this he can select good characteristics of goat A and produces large number of offspring in short period time.
So production yield can be increased.
The disadvantages are no variation between individuals produced. Clones A are identical and have same resistance to the diseases.
(c) Explain how the daughter cell produced differs than parent cell.
The daughter cell produced differs than parent cell in aspect of the number of chromosome and the recombination of DNA.
During prophase 1, chromatids of the homologous chromosomes cross over.
After that the DNA is exchanged/ recombined
Recombination of DNA causes variation among the daughter cell produced.
During anaphase 1 each of homologous chromosome are separated and pulled at the opposite poles
After meiosis 1 each daughter cell carried half number of chromosome than parent cell.
Now genetically the daughter cells produced are different as compared to the parent cell.
Question 20:
Diagram 18.1 shows parent cells undergoing cell division.
DIAGRAM 18.2
The exposure drives the cell cycle malfunctions. Based on the Diagram 18.2 describe the
effect of cell cycle malfunctions to the body.
[ 6 marks]
Formation of an abnormal cell (cancerous cell)
Due to the DNA of the normal cell changes spontaneously (mutation)
So the cell divides by uncontrolled mitosis
To produce an abnormal cells
That has two nuclei (without cytokinesis).
These abnormal cells diffuse into the blood or lymph and transported by the blood circulatory system or lymphatic system
To the other parts of the body cells.
So it spreads to other organ and destroys the normal cells.
(d) Cancer is a disease which causes uncontrolled growth of tissues. Suggest the method to treat cancer. Explain how this treatment stops the growth of cancer cells.
Radiotherapy is a method to treat cancer.
Radiotheraphy contain radioactive rays.
Radioactive rays is exposed and destroys the cancerous cells.
Cancerous cells cannot divide and increase the number.
Question 21:
Figure 19.1 shows the structure of leaf. Figure 19.2 shows the organelle M involves in the formation of starch in plants.
Based on the raw materials (inorganic substances) required in photosynthesis, explain the formation of starch(organic substance) in the plant. [10 marks]
Carbon dioxide and water are the raw materials needed in photosynthesis to form starch with the presence of chlorophyll.
During the light reaction, water is splitted to hydrogen ion and hydroxyl ion and release ATP
When chlorophyll traps/ captures light
The process namely photolysis of water
Hydroxyl ion gives electrone to form hydroxyl group
Hydroxyl groups combine to produce oxygen and water
Hydrogen ion receives electrone to form hydrogen atom.
Hydrogen atom enters the dark reaction
To reduce carbon dioxide into glucose and water
Using energy / ATP provides from the light reaction.
Molecules of glucose undergo condensation to form starch.
Question 22
Why this food can be kept longer compared its normal condition.
- Cooked prawn
- Frozen meat
- Dried prunes 8 marks ]
MODUL II
ESSAY QUESTIONS
Question 1:
Diagram 1 shows a reaction catalyzed by sucrase.
(a) Based on Diagram 1, explain the characteristics of an enzyme. (6 marks)
· Enzyme is highly specific
· M only can break down N because it has an active site which is fits to N molecule.
· The enzyme reaction is reversible.
· Substrate / complex molecule can be broken down by hydrolysis and
· products can be converted to complex compound by condensation.
· Enzyme does not destroy after the reaction / can be reused after the reaction
· After the substrate is broken down into products, it leaves the active site of enzyme.
(b)
Enzymes are widely used in our daily life and industries.
Based on the above statement, explain how enzymes are used in the food industry. [4 marks]
· Lipase is used to react with fats in milk to produce cheese.
· Protease is used to tenderise meat so that it can be consumed easily.
· Cellulase used to separate agar from seaweed.
· Zymase in yeast is used widely in the preparation of alcoholic drinks such as wine and beer.
· Zymase acts with the glucose in grape juice during fermentation, thus producing ethanol, which is used to make wine.
· Amylase is used to remove starch from fruits in the making of fruit juice.
· In breweries, amylase is used to remove starch before the addition of malt.
Question 2:
A study is carried out on two individuals X and Y. They were asked to drink glucose solution of the same volume and concentration. The glucose and insulin level in the blood is measured.
The graph in Diagram 2 shows the changes in concentration of glucose and insulin in the blood of individuals X and Y.
The concentration of glucose in a normal individual is 0.09 gdm-3
(a) Explain the differences between the graph of individuals X and Y in relation to:
(i) blood glucose concentration
(ii) insulin concentration in the blood [10 marks]
· At stage 1, before glucose intake, blood glucose concentration in individual Y is normal which is 0.09 g/dm3 while blood glucose concentration in individual X is high (0.13 g/dm3).
· Y is normal while X is suffering from diabetes mellitus.
· At stage 2 after glucose intake, blood glucose level in individual Y increases until it reaches at maximum level (0.13 g/dm3) while blood glucose level in individual X increases rapidly and reaches at maximum level 0.20 g/dm3.
· At stage 3 blood glucose level decreases rapidly in individual Y until it reaches at normal level (0.09 g/dm3) while blood glucose level in individual X decreases to 0.165 g/dm3.
· Y is normal while X is suffering from diabetes mellitus.
· At stage 1 insulin concentration in the individual Y is 0.02 arbitrary units before glucose intake while insulin concentration in the individual X is 0.01 arbitrary units.
· After glucose intake / at stage 2, insulin concentration in the individual Y increases rapidly and reach the maximum level 0.06 arbitrary units while insulin concentration in the individual X is
increases to 0.02 arbitrary units.
· More insulin is required to convert excess of glucose into glycogen back to
normal level while in individual X pancreas fails to secrete high concentration
of insulin so glucose level is high.
· At stage 3 that is after a few hours the concentration of insulin in the individual Y decreases and remain high in the blood (0.02 g/dm3) while in the individual X concentration of insulin is remain low (0.01 arbitrary units).
· Glucose level can be converted to the normal level in the individual Y while
glucose level of X is high and X is suffer from diabetes.
Question 3:
Diagram 3 shows the longitudinal section through the thorax of a man.
Diagram 3
Describe briefly the mechanism of breathing in man . In your answer describe the part played by the
(i) diaphragm
(ii) intercostals muscle
(iii) alveoli (10 marks)
Answer:
(i)
· Diaphragm is a muscular sheet in the body cavity separating the thorax from the abdomen
· At the start of inhalation, the muscles of the diaphragm contract , making it less arched
· This helps to increase the volume of the thoracic cavity and reduce the pressure of the thoracic cavity. Air rushes into the lungs
· When the muscles of the diaphragm relax , it returns to its arched condition , reducing the volume of the thoracic cavity and increasing the pressure of the thoracic cavity. Air is forced out of the lungs
(ii)
· The muscles between the ribs are known as intercostals muscles
· During inhalation the external intercostals muscle contracts and raise the lower ribs
· This helps to increase the volume of the thoracic cavity and reduce the pressure of the thoracic cavity. Air rushes into the lungs.
· During exhalation the external intercostals muscles contract , the ribs return to their original position , reduce the pressure of the thoracic cavity. Air is forced out of the lungs
(iii)
· The alveoli are thin-walled air sacs with the lungs
· These sacs are surrounded by a network of capillaries
· During inhalation the alveoli are filled with air and gaseous exchange occurs between the alveoli and the capillaries
· Oxygen from the alveoli diffuses into the capillaries while carbon dioxide diffuses from the capillaries into the alveoli.
Question 4:
Why do we need to practice good posture? Explain how we can practice good postures while standing, sitting, walking and lifting heavy objects. [10 marks]
Answer:
· It helps to ensure that our body is always well supported
· Then no undue pressure is placed on our muscles and spine
· The internal organ such as the lungs, heart and stomach will be able to function properly
· Muscles working properly prevent us from getting tired easily
· It will prevent backache and shoulder ache
· While standing our body should be erect and upright so that the weight of our body is supported by our feet
· While sitting our back should rest against the back of the chair
· We should maintain an upright position
· While walking , our body should be upright and straight
· Our head should be held high and our eyes looking straight ahead.
· While lifting heavy object , we should bend both our knees and lift objects from the floor
Question 5:
Reproduction technology has contributed a lot to human health and population growth. Describe how fertility is overcome in humans using the following methods:
Sperm banking
Artificial insemination
In vitro fertilization
Surrogate mother [10 marks]
Answer:
Sperm bank
· used when the man/spouse/male suffers from low sperm count / production of weak / unhealthy sperm
· sperms are donated and donor’s identity is recorded/ secret
· genetic biodata of donor and the woman recipient should match for successful fertilization
Artificial insemination
· Practiced by a couple when the man is sterile
· Donor’s sperm are injected into the woman uterus during ovulation
· No sexual intercourse is involved
· The genetic background and health of donor is screened before acceptance
In vitro fertilization
· Practiced when the woman’s ovum cannot be fertilized due to blockage of fallopian tube
· Ovum is extracted (from the woman) and fertilized outside the body / in petri dish / test tube
· Fertilized ovum / zygote grows in culture medium
· Than it is implanted into the woman’s uterus
Surrogate mother
· Method practiced when a woman cannot be pregnant in normal way
· The zygote is obtained by in vitro fertilization
· Another woman is implanted with the couple zygote
· (surrogate mother) carries the baby only until it is born [any 10 points]
Question 6
The figure below shows a fluid mosaic model of the plasma membrane
(a) Explain the functions of the structures X, Y and Z in the movement of substances across the plasma membrane
[4 marks]
· Structure X allows hydrophobic molecules which can dissolve in lipids such as fatty acids, glycerol, steroid hormones, vitamins A, D, E and K to move in and out of the cell.
· Structure Y allows bigger molecules which do not dissolve in lipids like glucose and amino acids to move in and out of the cell.
· Structure Z allows small charged molecules to move in and out of the cell.
(b) All movements of substances have to pass through a plasma membrane. Explain why the plasma membrane is known as a semi permeable membrane
(6 marks)
· The plasma membrane is known as a semi permeable membrane because it only allows the passage of certain molecules and limits the passage of other molecules.
· It consists of two layers of phospholipids with protein molecules scattered in them
· Molecules that can pass through the plasma membrane easily are those that can dissolve in lipid and small uncharged molecules.
· Molecules which cannot move through the plasma membrane but require proteins include large molecules that do not dissolve in lipid and small charged molecules.
· Channel protein does not require the binding of a molecule and conformational change to open.
· Channel protein allows molecules to steadily diffuse across the membrane through diffusion.
· However a carrier protein allows specific molecules to cross the cell membrane by undergoing a conformational change upon the binding of the molecule.
· The conformational change opens a hole through which the molecule can enter or leave a cell.
Question 7:
(a) Explain how the first line of defence can prevent the entry of pathogens into the body
[6 marks]
The first line of defence in the prevention of pathogens entering the body can be the function of
(i) The skin
· The skin serves as a protective layer against the entry of pathogens.
The ph of the skin which is in the range of 3 to 5 provides an acidic environment which is not conducive for the growth of microorganisms.
The enzyme lysozyme which is present in the sweat and sebum on the surface of the skin can kill microorganisms.
(ii) The nose and respiratory tract.
The mucous membrane that lines the respiratory tract can trap microbes, dust and foreign matter.
The ciliated epithelium which lines the respiratory tract and the nasal cavity can remove the mucus when a person coughs or sneezes.
(iii) The stomach.
The hydrochloric acid in the stomach can kill the microbes that is present in the food that has entered the body.
Besides all these, the body has natural reactions such as diarrhoea and diarrhoea that
Can help to remove toxins and microbes very quickly before it is sent to other parts of the body.
(b) Explain the effect of taking fatty foods that have high cholesterol content to human healthy life.
[4 marks]
Our body has better ability to store fats rather than use it to release energy.
As a result fats and excess cholesterol will settle on the arterial wall causing a condition known as arteriosclerosis.
The lumen of the artery becomes narrow and hinders/slowing down the flow of blood, thus causing high blood pressure/hypertension.
If the coronary artery is blocked, the supply of oxygen and nutrients to the heart is affected.
This will cause a heart attack.
If the flow of blood to the brain is blocked, there is a possibility that stroke can occur.
Question 8:
(a) Explain the principle involved in genetic engineering
[2 marks]
· Genetic engineering is a technique of manipulating the genes in an organism by removing a gene from one organism and transferring it to another organism
(b) Discuss the importance of genetic engineering in the medical and agricultural field.
[4 marks]
The importance of genetic engineering in the medical field.
· Production of insulin from bacteria.
· Gene from the pancreas that can produce insulin is removed and inserted into the DNA molecule of bacteria.
· The bacteria that can produce insulin grows very fast in a suitable nutrient medium.
· Insulin can be produced in large quantities in a short time.
Genetic engineering in agriculture.
· Produce food or plants which are genetically modified to produce high quality yield.
· Produce plants with high quality yield as they are resistant to viral attacks, pesticides or predators.
(c) Diagram 8 shows DNA finger printing. DNA finger printing is more scientific, accurate and advanced compared to the thumbprint in criminal investigation.
Discuss the use of DNA fingerprinting in the identification of the parents of a child. Based on your answer, identify the father of the child.
[4 marks]
· DNA fingerprinting is a technique used to identify parents of a child based on DNA banding similarity
· Generally, the DNA profile of a child is the combination of the mother and father.
· A child inherit half of the chromosomes number from father and half from mother.
· DNA fingerprinting gives more details and information regarding the identity of a person while thumbprints give general information as it is based on a few different types.
· A blood sample, seminal fluid or any tissue can be analysed and its result is compared with that of an individual who is a suspect.
· In conclusion, man P is the father of the child as the similarity of his DNA is higher
Question 9:
Diagram 9 shows organism, P, Q and R.
Diagram 9
(a) (i) Describe the type of nutrition in P and Q.
[4 marks]
P: Autotrophic that enables to produce their own food by photosynthesis process using simple inorganic substances to form complex organic substances
Q: Heterotrophic unable to produce their own food, it depends to others organism for food
(ii) Explain one similarity and three differences between the alimentary canal of Q and R.
[6 marks]
Both contain anaerobic bacteria and protozoa in their alimentary canal for digestion of cellulose
Q
R
Has a very large caecum – to provide large compartment for digestion of cellulose
Has small caecum – digestion of cellulose didn’t occur in caecum
Has one stomach digestion of cellulose didn’t occur in stomach
Has four stomachs; rumen, reticulum, omasum and abomasums – increase storage and digestion of cellulose
Does not have diastema – because the main source of food is not grass/cellulose.
Has a diastema to store cut cellulose before chewing by the molar
Question 10
Pregnant mother, infants and teenager require different energy daily. Each target group needs to choose a diet appropriate to its needs.
Based on your biological knowledge discuss the above statement. [10 marks]
Pregnant mother
· Need rich in calcium and phosphorus for formation of strong bone in the growing of foetuses.
· More folic acid and ferum for formation of red blood cells,/formation of normal neural tube development in the embryos
· Rich of protein for the formation of new tissues
· Eat lot of vegetables and grains as source of fibre to prevent constipation
· Reduce a food rich in fats and sugar and caffeinated drinks to ovoid obesity and neural suppression
Infants and children
· Infants need of lot protein, carbohydrate, lipid, vitamins and mineral, for muscle and bone development during growing process.
· Babies need a higher intake of calcium and phosphorus for the formation of bones and teeth.
· Growing children need a diet rich in protein, vitamins, calcium and phosphorus for bone and tooth formation and promotes growth.
· They also require more energy-rich food because they have higher basal metabolism rate than adult.
Teenagers
· Need food rich in proteins and vitamins during the period of rapid growth
· Consumes rich of fruit and vegetable to supply vitamins and mineral, and ovoid constipation
· Female teenager needs food rich in ferum to prevent iron-deficiency anaemia especially after menstruation.
· Reduce intake of fatty and oily food to avoid having pimples and acne
Question 11:
The diagram shows a human activity.
(a) (i) Describe the type of nutrition in P and Q.
[4 marks]
P: Autotrophic that enables to produce their own food by photosynthesis process using simple inorganic substances to form complex organic substances
Q: Heterotrophic unable to produce their own food, it depends to others organism for food
(ii) Explain one similarity and three differences between the alimentary canal of Q and R.
[6 marks]
Both contain anaerobic bacteria and protozoa in their alimentary canal for digestion of cellulose
Q
R
Has a very large caecum – to provide large compartment for digestion of cellulose
Has small caecum – digestion of cellulose didn’t occur in caecum
Has one stomach digestion of cellulose didn’t occur in stomach
Has four stomachs; rumen, reticulum, omasum and abomasums – increase storage and digestion of cellulose
Does not have diastema – because the main source of food is not grass/cellulose.
Has a diastema to store cut cellulose before chewing by the molar
Question 10
Pregnant mother, infants and teenager require different energy daily. Each target group needs to choose a diet appropriate to its needs.
Based on your biological knowledge discuss the above statement. [10 marks]
Pregnant mother
· Need rich in calcium and phosphorus for formation of strong bone in the growing of foetuses.
· More folic acid and ferum for formation of red blood cells,/formation of normal neural tube development in the embryos
· Rich of protein for the formation of new tissues
· Eat lot of vegetables and grains as source of fibre to prevent constipation
· Reduce a food rich in fats and sugar and caffeinated drinks to ovoid obesity and neural suppression
Infants and children
· Infants need of lot protein, carbohydrate, lipid, vitamins and mineral, for muscle and bone development during growing process.
· Babies need a higher intake of calcium and phosphorus for the formation of bones and teeth.
· Growing children need a diet rich in protein, vitamins, calcium and phosphorus for bone and tooth formation and promotes growth.
· They also require more energy-rich food because they have higher basal metabolism rate than adult.
Teenagers
· Need food rich in proteins and vitamins during the period of rapid growth
· Consumes rich of fruit and vegetable to supply vitamins and mineral, and ovoid constipation
· Female teenager needs food rich in ferum to prevent iron-deficiency anaemia especially after menstruation.
· Reduce intake of fatty and oily food to avoid having pimples and acne
Question 11:
The diagram shows a human activity.
a) Explain briefly why humans carry out the activity as shown in diagram above.
[4 marks ]
b) Explain the important ecological roles of the tropical rainforest.
[6 marks]
c) i) Explain the impacts of the activity shown above on the environment.
[7 marks]
ii) Suggest methods to control the mentioned activity.
[3 marks]
Answers :
a)
- The human population grows rapidly. The demands for food and housing areas have increased.
( 1 mark )
- Vast areas of forest are cleared for agricultural and commercial purposes.
( 1 mark )
- Urbanization and industrialization have caused more forests to be cleared for road construction and housing areas.
( 1 mark )
- Deforestation is also caused by the demands for timber and fuel wood.
( 1 mark )
TOTAL 4 MARKS
b) - The rainforest sustains almost half of the flora and fauna of the world.
( 1 mark )
- Certain plants from the forest provide food and pharmaceutical products to humans. ( 1 mark )
- The rainforest regulates the climate by influencing humidity, rainfall and temperatures. ( 1 mark )
- Green plants inside the forest absorb carbon dioxide from the atmosphere and maintain the concentration of the gas in the atmosphere. ( 1 mark )
- Green plants provide oxygen to all living things. ( 1 mark )
- The rainforest also acts as water catchment areas. ( 1 mark )
TOTAL 6 MARKS
c) - Deforestation causes soil erosion , landslides, flash floods and global
warming. ( 1 mark )
- Causes the soil to become loose and less stable. ( 1 mark )
- Without the protection of green plants, the soil is exposed to the forces of wind and rain. ( 1 mark )
- The top layer of soil is washed away gradually by the rainwater. This is known as soil erosion. ( 1 mark )
- Soil erosion causes the depletion of minerals from the soil, therefore the soil becomes infertile and unsuitable for agriculture.
(1 mark )
- Landslides may happen on steep hillsides during heavy rain. It is because rainwater flows quickly and causes the top layer of the soil to crumble. ( 1 mark )
- Rivers and drains are silted and the flow of water is blocked.
( 1 water )
- Therefore, water flows inland and this causes flash floods in the lower areas during rainy seasons. ( 1 water )
TOTAL Max 7 MARKS
d) - Deforestation can be controlled by passing stricter laws to protect the
forest. ( 1 mark )
- Through education, the public should be taught on the importance of
the forest. ( 1 mark )
- New trees should be planted to replace trees which have been out drawn. ( 1 mark )
TOTAL 3 MARKS
Question 12:
Activities that carried out by the human to provide food, power and industrial needs have a extensive effects on the environment. These effects include atmospheric pollution, water pollution, destruction of habitats and communities and a natural phenomenon known as the greenhouse effect.
d) Explain the fenomena of greenhouse effect and problems caused by this fenomena.
( 10 marks )
e) Suggest two measures which can be employed in the management of development activities and the ecosystem to ensure that the balance of nature is maintained.
( 10 marks )
ANSWERS
a) Able to define and describe the causes and effects of greenhouse effects.
· The greenhouse effect is caused by specific gases which form a thin layer around the atmosphere. ( 1 mark )
· These gases include (water vapor )/ carbon dioxide/ methane/ ozone/ nitrogen oxides and CFCs – any 2 gases ( 1 mark )
· The increased concentrations of atmospheric carbon dioxide is primarily due to
- combustion of fossil fuels ( 1 mark )
- increased deforestation ( 1 mark )
- industrial production ( 1 mark )
Max 2 marks
· The greenhouse gases allows short wavelength radiation from the sun to reach the earth’s surface. ( 1 mark )
· Some of the infra-red radiation fails to pass back through the greenhouse layer.
- more heat is trapped in the atmosphere ( 1 mark )
- results in global warming ( 1 mark )
· Polar ice caps may melt
- causing the sea to rise / extensive floodings
( 1 mark )
- subsequent reduction in land mass
( 1 mark )
- some aquatic populations would increase
( 1 mark )
- some terrestrial populations decrease
( 1 mark )
Max 2 marks
· Climatic changes occur
- rainfall changes ( 1 mark )
- heat increases ( 1 mark ) Max 1 mark
TOTAL 10 marks
b) Able to list the following measures and its examples
· The practice of biological control
Use natural predators to control the population of pest species
( 1 mark )
· Examples:
- The use of caterpillars of the Cactoblastis cactorum to control the population of the prickly pear cactus.
( 1 mark )
- The use of (khaki chambel) to control the population of gold snails which feed on paddy stalks. ( 1 mark )
( or any suitable examples )
· Biological control is better than using chemical control because
( 1 mark )
- the use of pesticide is indiscriminate and harmless organisms can be killed. ( 1 mark )
- the effects of pesticides can be persistent and remain in the environment for long periods. ( 1 mark )
( or any suitable explanation)
Max 4 marks
· The use of renewable energy
· examples of renewable energy
- solar energy ( 1 mark )
- geothermal energy ( 1 mark )
(or any suitable examples)
· Renewable energy occurs naturally in the environment is inexhaustible/does not pollute the environment ( 1 mark )
Max 2 marks
· The use of technology ( 1 mark )
· Use of unleaded petrol ( 1 mark )
- reduce the emission of lead into the atmosphere ( 1 mark )
· Use of microorganism to clean up the environment ( 1 mark )
- certain bacterias are used to break down oil spills at sea.
( 1 mark )
· Install catalytic converters in vehicles ( 1 mark )
- convert the harmful gases released during combustion of fossil fuels to less harmful products. ( 1 mark )
(or any suitable examples)
Max 4 marks
TOTAL 10 MARKS
Question 13:
(a) The shaded area of the graph in Diagram1 shows the intake of oxygen by an athlete before, during and after running for five minutes.
Period of running
Key
Oxygen
demand
Oxygen
consumption
0
5
10
15
20
25
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.0
Oxygen consuption
(litre per minute)
Time (minutes)
Based on the graph , explain how an oxygen debt is built up when an athlete is running and how it is settled after he stops running.
[ 10 marks ]
Answer
- During a vigorous exercise (running), the breathing rate is increased. [1 mark]
- This is to supply more oxygen to the muscles for rapid muscular contraction. [1 mark]
- However, the supply of oxygen to muscles is still insufficient [1 mark]
- and the muscles have to carry out anaerobic respiration to release energy. [ 1 mark ]
- The glucose is converted into lactic acid, with only a limited amount of energy being produced. [ 1 mark ]
- An oxygen debt builds up in the body, when no oxygen use in energy production [ 1 mark ]
- High level of lactic acid in the muscles cause them to ache. [ 1 mark ]
- After running, the athlete breathes more rapidly and deeply than normal for
twenty minutes [1 mark ]
- There is recovery period after 10 minutes until it reaches 20 minutes when oxygen is paid back during aerobic respiration. [ 1 mark ]
- About 1/6 lactic acid is oxidized to carbon dioxide, water and energy. [ 1 mark ]
Max 10 marks
Question 14:
1. The diagram shows a terrestrial ecosystem.
eagle
bird
rat
snake
earthworm
Explain how energy flows through the food chain and how it is lost to the environment.
[ 8 marks]
Answer :
- Energy flows through the food chain in one direction [ 1 mark]
- In the food chain, the plant is the producer, the rat is the primary consumer, the snake is the secondary consumer and the eagle is the tertiary consumer. [ 2 marks] //
In the food chain, the plant is the producer, the earthworm is the primary consumer, the bird is the secondary consumer and the snake/ eagle is the tertiary consumer.
- Each level of food chain is called a trophic level. [ 1 mark ]
- Energy is transferred from one trophic level to another trophic level [ 1 mark ]
- When energy is transferred from one trophic level to another level as much as 90% of the chemical energy in the food consumed by primary consumer is used for its metabolic activities and lost as heat ` [ 2 marks ]
- Only 10% of the energy in an organism is passed on to the organism at the next trophic level [ 1 mark ]
[10 marks]
Question 15:
The figure above shows a type of transport which happens on the plasma membrane of a cell.
Based on the diagram above, name the type of transport in the above figure.
Explain your answer.
[6 marks]
Answer :
- Active transport
- is dynamic and requires energy / ATP for the transfer of molecules or ions
- an area of low concentration to an area of high concentration through the living cell membrane.// go against the concentration gradient
- It requires carrier protein
- The carrier protein must be able to change its shape (and also be able to return to its original state.)
- To combine the active site of the carrier protein with the molecules or ions.
- To carry the complex molecular carriers through the plasma membrane.
Question 16:
The figure shows the organelles involved in the production of extracellular enzymes.
(a) (i) Using a name example, explain the term extracellular enzyme.
[2 marks]
(iii) Based on the organelles shown, explain how extracellular enzymes are produced.
[8 marks]
(b) The figure shows a reaction of an enzyme and its substrate
Explain the mechanism of the enzymatic reaction
[10 marks]
Answer:
( a) (i) - Extracellular enzyme is produced in a cell, then packed and secreted from the cell.
It catalyses its reaction outside the cell. An example is amylase.
- (ii) The instruction for making the extracellular enzyme is transcribed from the deoxyribonucleic acid
o (DNA) to ribonucleic acid (RNA) in the nucleus.
o The RNA then leaves the nucleus through the nuclear pore
o and attaches itself to the ribosome located on the endoplasmic reticulum.
o When the synthesis of the enzymes is completed
o it is encapsulated in a transport vesicle which fuses with the golgi body.
o In the golgi body, the enzyme is further modified before being packed in a secretory vesicle.
o The secretory vesicle transports the enzyme to the plasma membrane, where it fuses with it
o and the enzyme is released outside the cell.
(b) The explanation of enzyme action is known as the ‘lock and key hypothesis’.
· The substrate molecule fits into the active site of the enzyme molecule.
· The substrate is the ‘key’ that fits into the enzyme ‘lock’.
· Various types of bonds such as hydrogen and ionic bonds hold the substrate
· in the active site forming the enzyme-substrate complex.
· Once the complex is formed, the enzyme changes the substrate to its product.
· The product leaves the active site.
· The enzyme is not altered by the reaction and it can be reused.
Question 17:
Discuss the uses of enzymes in the household & industries.
Why are enzyme used in these industries.
Answer
Enzymes are used as biological detergents.
· Protease degrades coagulated proteins into soluble short-chain peptides.
· Lipase degrades fat or oil stains into soluble fatty acid and glycerol.
· Amylase degrades starch into soluble shorter-chain polysaccharides and sugars.
Enzymes are used in the baking industry.
· Protease is used in the breakdown of proteins in flour for the production of biscuits.
· Amylase is used in the breakdown of some starch to glucose in flour for making white bread, buns and rolls.
Enzymes are used in the medical field.
· Trypsin is used to remove blood clots and to clean wounds.
· Various other enzymes are used in biosensors.
Enzymes are used in industries because:
·They are effective.
·They are cheap and easy to use.
·They can be re-used, thus only small amounts are needed.
· They don't require high temperature to work, thus this reduces fuel costs.
Question 18
Diagram above shows the structure of HIV. The virus infects helper T cells of the human ‘s immune system that caused AIDS, Acquired Immune Deficiency Syndrome.
i) Descibe the effects of the human immunodeficiency virus (HIV) on the body’s defence mechanism.
( 6 marks )
ii) Suggest ways to prevent the spread of Acquired Immune Deficiency Syndrome (AIDS). ( 4 marks )
ANSWERS
2 (i) - HIV infects helper T cells.
- Helper T cells are essential to activate other lymphocytes in the body’s defence mechanism against diseases.
- There may be a long incubation period of several years for the virus before the symptoms appear.
- The immune system of the infected person gradually becomes weakened and defenceless against many pathogens.
- HIV also attacks the central nervous system.
- There is weight loss and a decrease in the function of the nervous system.
( 6 marks )
(ii) - By practice sexual abstinence or only having sexual relationship with one partner.
- Use condom during copulation/sexual relationship.
- Set up centres to exchange free sterile syringe and needles for used ones to reduce the practice of sharing needles amongst drug users.
- Strict screening of blood before it is used for transfusion.
-Carry out awareness campaigns to educate public and schoolchildren about the dangers of AIDS.
- Proper counseling provided to HIV positive person so that they do not spread the virus to other healthy people.
( Any 4 answers)
Question 19
Figure shows cells P and Q undergoing cell division.
M
(a) Based on the cell division of P, explain the best technique can be used by an oil palmer to produce a large number of oil palm in a short time.
State one problem to be considered in using technique.
The technique can be used is cloning and culture tissue.
Tip of shoot/root namely explants is cut
And cultured into the medium that contain nutrient/ hormone and the temperature
Is kept constant at the optimum temperature and sterile (in vitro).
Explant divides actively by mitosis to produce large number of cells namely callus.
Callus differentiates to form shoot and root.
Plantlets / clones are formed.
The disadvantage of cloning is no variation. Clones are exactly identical to parent cell. If one clone is infected by disease, the others will be infected too because they have same resistance to the certain disease.
(b) A farmer wants to produce large number of goat A in a short period time. He tries to apply the concept of cell division cell P to produce new individuals of goat A. Describe the technique used. Justify the technique.
The somatic cell (diploid cell) of goat A is taken and the nucleus of goat A is removed.
An egg cell of goat B is removed and the haploid nucleus of goat B is removed.
The diploid nucleus of goat A is inserted into the egg cell of goat B.
The egg cell contain the diploid nucleus is cultured into the medium that contain nutrient and the condition is aseptic.
The cell divides by mitosis to produce large number of cells namely embryo.
Embryo is implanted into the uterus of goat B.
A farmer should use the technique because by this he can select good characteristics of goat A and produces large number of offspring in short period time.
So production yield can be increased.
The disadvantages are no variation between individuals produced. Clones A are identical and have same resistance to the diseases.
(c) Explain how the daughter cell produced differs than parent cell.
The daughter cell produced differs than parent cell in aspect of the number of chromosome and the recombination of DNA.
During prophase 1, chromatids of the homologous chromosomes cross over.
After that the DNA is exchanged/ recombined
Recombination of DNA causes variation among the daughter cell produced.
During anaphase 1 each of homologous chromosome are separated and pulled at the opposite poles
After meiosis 1 each daughter cell carried half number of chromosome than parent cell.
Now genetically the daughter cells produced are different as compared to the parent cell.
Question 20:
Diagram 18.1 shows parent cells undergoing cell division.
Cell division
a) Based on the Diagram 18.1
i) What is the process shown ? State the significant of the process.
[ 4 marks ]
· The process shows parent cell undergoes mitotic division.
· The significance of mitosis are to increase the number of daughter cells.
· So new daughter cells produced causes growth in an organism.
· In asexual reproduction, daughter cells produced are identical to parent cell.
· If epithelial cells of skin damaged, the parent cell (Malpighi layer divides to produce new daughter cell. By this way daughter cell produced can replace or repair the damage cell.
· The production of the daughter cells by mitosis can be applied in cloning or tissue culture. So new individuals can be produced in large numbers and at short time.
b) Diagram 6.2 shows a group of cells that is exposed to ultraviolet ray.
a) Based on the Diagram 18.1
i) What is the process shown ? State the significant of the process.
[ 4 marks ]
· The process shows parent cell undergoes mitotic division.
· The significance of mitosis are to increase the number of daughter cells.
· So new daughter cells produced causes growth in an organism.
· In asexual reproduction, daughter cells produced are identical to parent cell.
· If epithelial cells of skin damaged, the parent cell (Malpighi layer divides to produce new daughter cell. By this way daughter cell produced can replace or repair the damage cell.
· The production of the daughter cells by mitosis can be applied in cloning or tissue culture. So new individuals can be produced in large numbers and at short time.
b) Diagram 6.2 shows a group of cells that is exposed to ultraviolet ray.
DIAGRAM 18.2
The exposure drives the cell cycle malfunctions. Based on the Diagram 18.2 describe the
effect of cell cycle malfunctions to the body.
[ 6 marks]
Formation of an abnormal cell (cancerous cell)
Due to the DNA of the normal cell changes spontaneously (mutation)
So the cell divides by uncontrolled mitosis
To produce an abnormal cells
That has two nuclei (without cytokinesis).
These abnormal cells diffuse into the blood or lymph and transported by the blood circulatory system or lymphatic system
To the other parts of the body cells.
So it spreads to other organ and destroys the normal cells.
(d) Cancer is a disease which causes uncontrolled growth of tissues. Suggest the method to treat cancer. Explain how this treatment stops the growth of cancer cells.
Radiotherapy is a method to treat cancer.
Radiotheraphy contain radioactive rays.
Radioactive rays is exposed and destroys the cancerous cells.
Cancerous cells cannot divide and increase the number.
Question 21:
Figure 19.1 shows the structure of leaf. Figure 19.2 shows the organelle M involves in the formation of starch in plants.
Based on the raw materials (inorganic substances) required in photosynthesis, explain the formation of starch(organic substance) in the plant. [10 marks]
Carbon dioxide and water are the raw materials needed in photosynthesis to form starch with the presence of chlorophyll.
During the light reaction, water is splitted to hydrogen ion and hydroxyl ion and release ATP
When chlorophyll traps/ captures light
The process namely photolysis of water
Hydroxyl ion gives electrone to form hydroxyl group
Hydroxyl groups combine to produce oxygen and water
Hydrogen ion receives electrone to form hydrogen atom.
Hydrogen atom enters the dark reaction
To reduce carbon dioxide into glucose and water
Using energy / ATP provides from the light reaction.
Molecules of glucose undergo condensation to form starch.
Question 22
Why this food can be kept longer compared its normal condition.
- Cooked prawn
- Frozen meat
- Dried prunes 8 marks ]
Answers :
- beyond 60ºC in cooking process, the growth of microorganism is inhibited. [ 1 mark ]
- microorganisms and their spores can be destroyed at temperature of about 121ºC.
[ 1 mark]
- Therefore, cooking the food properly will kill microorganisms and cooked prawn can be last longer compared to uncooked prawn. [ 1 mark ]
- The temperature in refrigerator is low and microorganism become inactive.
[1 mark ]
- The optimum temperature for microorganisms to become active is between 35ºC - 40ºC, [1 mark ]
- Moist medium is suitable for the growth of microorganisms because water is basic needs for microorganism. [ 1 mark]
- Low moisture in refrigerator will slow down the spore germination. Therefore frozen meat can be kept within few months. [1 mark]
- Dark condition and the moist at room temperature will give high chance for the growth of microorganisms in fresh food such as fruits. [1mark]
- When prunes is exposed to ultraviolet rays and high intensity of sunlight , the light can kill the microorganisms in the fruit.
[ 1 mark]
- The dried food contains less moisture and less nutrients which may decrease microorganisms keep growing.
[ 1 marks]
TO BE CONTINUED…
- beyond 60ºC in cooking process, the growth of microorganism is inhibited. [ 1 mark ]
- microorganisms and their spores can be destroyed at temperature of about 121ºC.
[ 1 mark]
- Therefore, cooking the food properly will kill microorganisms and cooked prawn can be last longer compared to uncooked prawn. [ 1 mark ]
- The temperature in refrigerator is low and microorganism become inactive.
[1 mark ]
- The optimum temperature for microorganisms to become active is between 35ºC - 40ºC, [1 mark ]
- Moist medium is suitable for the growth of microorganisms because water is basic needs for microorganism. [ 1 mark]
- Low moisture in refrigerator will slow down the spore germination. Therefore frozen meat can be kept within few months. [1 mark]
- Dark condition and the moist at room temperature will give high chance for the growth of microorganisms in fresh food such as fruits. [1mark]
- When prunes is exposed to ultraviolet rays and high intensity of sunlight , the light can kill the microorganisms in the fruit.
[ 1 mark]
- The dried food contains less moisture and less nutrients which may decrease microorganisms keep growing.
[ 1 marks]
TO BE CONTINUED…
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