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
Wednesday, August 19, 2009
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]
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