SULIT No. Matrik........................................... No. Kad Pengenalan................................. SB025 2 SULIT 1 (a) FIGURE 1 represents the phenotypes distribution of original population of Galapagos finches before natural selection takes place. FIGURE 1 (i) An increase in the relative abundance of large seeds over small seeds led to an increase in beak depth in a population of Galapagos finches. Draw the new curves on the above graphs in FIGURE 1, representing the distribution of phenotypes after natural selection has taken place. [1 mark] (ii) Differentiate between stabilizing selection and natural selection in 1 (a) (i) [2 mark] Stabilizing selection Directional selection Favour individuals at intermediate phenotypes Favour individuals at one extreme phenotypes Act against individuals with both extreme phenotypes Act against individuals with another extreme phenotype Operate in stable environment Operate in response to environmental changes *Reject answer in table (b) Mimulus lewisii (purple monkey- flower) attract bumblebees while Mimulus cardinalis (scarlet monkey flower) attract hummingbirds for pollination. Identify the pre zygotic barrier involved in the situation above. [1 mark] Mechanical isolation
SULIT No. Matrik........................................... No. Kad Pengenalan................................. SB025 3 SULIT (c) Sympatric speciation can occur when certain behavioural barriers result in the reproductive isolation of populations. Suggest ONE behavioural barrier that may result in reproductive isolation. [1 mark] Different mating rituals/courtship behaviour (Accept any suitable answer related to courtship pattern) 2 (a) A student was required to investigate the effect of increasing the concentration of sucrose on the rate of activity of enzyme. The rate of enzyme activity was recorded and plotted in the graph below. FIGURE 2 (i) Based on the graph, the rate of enzyme activity increasing until it reaches the maximum rate as the substrate concentration increases. Explain why adding more substrate will not increase the rate of enzyme activity. [2 marks] -At high substrate concentration, all active sites of the enzyme are occupied/ the enzyme becomes saturated with substrate. -No more substrates can fit at any one time even though there is plenty of substrate available// The active site of an enzyme can only bind with a certain number of substrate molecules at a given time. -Any extra substrate will have to wait for the ESC to release its product and become available again. (ii) (a) On FIGURE 2, draw a curve to show the rate of enzyme activity if noncompetitive inhibitor is added. [1 mark]
SULIT No. Matrik........................................... No. Kad Pengenalan................................. SB025 4 SULIT (b) Explain your answer in (ii)(a). [3 marks] -Rate of enzyme activity with non-comptitive inhibitor is lower than rate of enzyme activity without non-competitive inhibitor. -Non- competitive inhibitor binds to allosteric site causes the active site of enzyme changed its conformation. -Adding more substrate cannot increase the rate of enzyme activity because substrate can no longer binds to the active site. (b) Catalase is an enzyme which is similar to hemoglobin. Catalase able to breakdown hydrogen peroxide into water and oxygen gas. (i) State the type and function of the cofactor found in catalase. [2 marks] Type of cofactor is prosthetic group. The function is to transfer atom or chemical groups from the active site of the enzyme to some other substances. 3 (a) FIGURE 3.1 shows steps involved in glycolysis. FIGURE 3.1 (i) Where in the cell does glycolysis occur? [1 mark] Cytosol (ii) Why is it necessary to phosphorylate glucose using ATP and name the enzyme involved? [2 marks] To make glucose more chemically reactive//increases energy level of molecule//supplies energy to the molecule. Hexokinase (iii) Describe process L. [2 marks] -Each G3P is oxidized by the transfer of electron to NAD+ -NAD+ reduced to NADH -Energy released (from this exergonic reaction) is used to attach inorganic phosphate (Pi) to the oxidized substrate Glucose 6- phosphate Fructose 6- phosphate 2 molecules of glyceraldehyde 3-phosphate 2 molecules of 1,3- bisphosphoglycerate 2 molecules of pyruvate L 1
SULIT No. Matrik........................................... No. Kad Pengenalan................................. SB025 5 SULIT (iv) Human skeletal muscle can respire both aerobically and fermentation. Explain why fermentation is advantageous to human skeletal muscle? [2 marks] -NAD + can be regenerated to oxidise more glucose/allow glycolysis to continue -Form ATP when no oxygen (b) FIGURE 3.2 is a diagram of part of a mitochondrion, showing components and processes involved in aerobic respiration. FIGURE 3.2 (i) Name the component labelled U. [1 mark] ATP synthase (ii) Explain what happen to the electron when it reaches component labelled Q [2 marks] -Electron is transferred to final electron acceptor, oxygen. -Oxygen combines with 2 electrons and 2 hydrogen ions forming water//reduced oxygen combines with proton to form water. (iii) How many reduced coenzymes will be produced if two molecules of acetyl-CoA enter stage R? [1 mark] 6 (iv) Oxidative phosphorylation and stage R occur in mitochondria. Explain what would happen to a cell that has less mitochondria. [2 marks] Less oxidative phosphorylation and Krebs cycle/R occur Less ATP produced Cell become less active
SULIT No. Matrik........................................... No. Kad Pengenalan................................. SB025 6 SULIT 4 (a) FIGURE 4.1 below shows a carbon fixation pathway of a plant. FIGURE 4.1 (i) Name the type of plants that fix carbon dioxide as in Figure 2 above. [1 mark] CAM plants. (ii) Name the cell where this process occurs. [1 mark] Mesophyll cell. (iii) Explain what happen to malate at step 3. [2 marks] Malate is oxidised, (NADP+ is reduced to NADPH) and decarboxylated to form PEP. (iv) Why cycle Z cannot occur when there is no light? [2 marks] Light dependent reaction does not occur. No ATP and NADPH (+ H+ ) produced.
SULIT No. Matrik........................................... No. Kad Pengenalan................................. SB025 7 SULIT (b) FIGURE 4.2 below shows a graph of rate of CO2 uptake against light intensity for two different plants. FIGURE 4.2 (i) Match curve A and B with their example. [2 marks] Soybean : B Sugarcane : A (ii) Based on FIGURE 4.2, explain why Plant A have higher rate of CO2 uptake at high light intensity than Plant B? [2 marks] -Because Plant A have enzyme PEP carboxylase. -PEP carboxylase have higher affinity towards CO2/ can fix CO2 at even low CO2 concentration.
SULIT No. Matrik........................................... No. Kad Pengenalan................................. SB025 8 SULIT 5 (a) FIGURE 5.1 shows the oxygen dissociation curve of haemoglobin. FIGURE 5.1 (i) Identify the percentage of oxygen released from haemoglobin when partial pressure of oxygen is 50 mmHg at curve B. [1 mark] 13-14% (ii) The curve B will shift to curve C under vigorous exercise. Explain the phenomenon. [3 marks] During vigorous exercise, body will produce a lot of CO2 High PCO2 makes the blood more acidic// pH low/ [H+ ] increases Affinity of haemoglobin to O2 is reduced Shift the oxygen dissociation curve to the right (b) FIGURE 5.2 shows the factors and main events of breathing cycle in human FIGURE 5.2 Explain the mechanism of process Z. [3 marks] Stretch receptor detect the stretching of bronchioles & bronchus Sent impulse through vagus nerve to inhibit inspiratory center. Diaphragm and external intercostal muscle relax. (Volume of lung decreased and the pressure will increased) (causing Z to takes place)
SULIT No. Matrik........................................... No. Kad Pengenalan................................. SB025 9 SULIT 6 (a) FIGURE 6.1 shows the changes of pressure and volume during cardiac cycle. FIGURE 6.1 (i) Identify phase L and phase M. [2 marks] Phase L : Ventricular systole Phase M : Atrial and ventricular diastole (ii) What happens to semilunar and atrioventricular valves at X? Explain your answer. [3 marks] -Semilunar valves open and AV valves close -Semilunar valves open to allow blood to pump out to aorta and pulmonary artery -Atrioventricular valves close to prevent backflow of blood into the atria (iii) What happens at QRS? [1 mark] Depolarization of ventricles occurs (causing ventricles to contract)
SULIT No. Matrik........................................... No. Kad Pengenalan................................. SB025 10 SULIT (b) FIGURE 6.2 shows the movement of molecule X and Y based on pressure-flow hypothesis. FIGURE 6.2 (i) What will happen to molecule X and Y when reach the sieve tube near the sink cell? [2 marks] -Molecule X/ sucrose will be unloaded into the sink cell -Molecule Y/ water will diffuse back into xylem by osmosis (ii) A farmer wants to increase the sugar content of the mango fruits in his orchard. Explain how he can solve this problem based on pressure flow hypothesis. [2 marks] -Pruning/ reduce number of flowers/ fruit in each branch// Reduce number of sink cells -Sucrose will be unloaded to the remaining flower/ fruits
SULIT No. Matrik........................................... No. Kad Pengenalan................................. SB025 11 SULIT 7 FIGURE 7 shows the cells lining the collecting duct in a human kidney. ADH molecules from posterior pituitary gland binds to the receptor protein of the collecting duct cells. FIGURE 7 (a) Based on FIGURE 7, explain how the effect of hormone ADH helps to restore normal water balance in the body. [2 marks] -Increase permeability of collecting duct to water -More water reabsorbs into the blood (b) Selective reabsorption occurs in the proximal convoluted tubule. Describe TWO ways in which the cells of the proximal convoluted tubule are adapted for reabsorption. [2 marks] -Presence of (numerous) microvilli to increase surface are for reabsorption -Presence of many mitochondria to provide ATP for active transport (c) Albuminuria is a sign of kidney disease that indicates presence of albumin protein in the urine. A damaged kidney lets some albumin pass into the urine. Suggest which part of the nephron is damaged if protein is found in the urine. Give reason for your answer. [2 marks] -Glomerulus. -Filtration membrane on glomerulus prevent filtration of protein/ Site for ultrafiltration.
SULIT No. Matrik........................................... No. Kad Pengenalan................................. SB025 12 SULIT 8 (a) FIGURE 8.1 shows two stimuli that trigger the changes in membrane potential of an axon. FIGURE 8.1 (i) Briefly explain why Stimulus 1 failed to initiate formation of action potential. [2 marks] -Stimulus 1 is not strong enough/ weak stimulus -Cause less voltage-gated Na+ channel to open -Depolarization occur but does not reach threshold (ii) State the characteristic of impulse related to situation in (a)(i). [1 mark] -All-or-none/ all-or-nothing event (iv) Differentiate transmission of impulse along the axon and across the synapse. [2 marks] -Impulse is electrically transmitted along the axon while impulse is chemically transmitted across the synapse -Transmission of impulse along the axon does not require neurotransmitter while transmission of impulse across the synapse require neurotransmitter -Transmission of impulse along the axon is faster while transmission of impulse across the synapse is slower
SULIT No. Matrik........................................... No. Kad Pengenalan................................. SB025 13 SULIT (b) FIGURE 8.2 shows a neuromuscular junction. FIGURE 8.2 (i) Botox (Botulinum toxin) is a neurotoxin used in cosmetic and medical treatments to prevents X from fusing to the presynaptic membrane. Briefly explain how the transmission of impulse at the neuromuscular junction is affected and its effect on muscle contraction. [3 marks] -No acetylcholine released into synaptic cleft by exocytosis and cannot bind to receptor on ligand gated channel -No action potential forms at motor end plate/ postsynaptic membrane -Muscle contraction does not occur/ muscle paralysis occurs (ii) State the function of ATP in muscle contraction. [1 mark] -Its hydrolysis by ATPase activates myosin head/ change myosin head to high energy configuration so it can bind to actin// -Its binding to myosin head cause detachment from actin/ dissociation of cross bridge// -Provide energy to pump Ca2+ from sarcoplasm back into sarcoplasmic reticulum
SULIT No. Matrik........................................... No. Kad Pengenalan................................. SB025 14 SULIT (c) FIGURE 8.3 shows part of mechanism of action for glucagon on its target cell. FIGURE 8.3 Explain how the binding of glucagon to receptor in target cell plasma membrane leads to generation of cAMP. [3 marks] -Hormone-receptor complex is formed -Hormone-receptor complex activates G protein -G protein activates adenylyl cyclase -Adenylyl cyclase catalyses the conversion of ATP to cAMP
SULIT No. Matrik........................................... No. Kad Pengenalan................................. SB025 15 SULIT 9 (a) FIGURE 9 shows one type of immune response. FIGURE 9 (i) State the type of immune response in FIGURE 9. [1 mark] Humoral immune response (ii) Identify the substance that is released by cell P to activate B lymphocyte. [1 mark] Interleukin- 2 (b) Explain the neutralisation in antigen and antibody interaction. [2 marks] -Antibodies bind to antigens / proteins on the surface of a pathogen -blocking its ability to bind with a host cell. (c) Why are people often ill for several days after infected by flu viral for the first time? [2 marks] -The primary immune response to an antigen is slow// takes time for lymphocyte to recognize and undergo clonal selection/ proliferation and differentiation -It takes several days to produce enough antibody (d) AIDS is an impairment in immune response that leave the body susceptible to infection and cancers that a healthy immune system would easily defeat. Describe how HIV cause AIDS? [2 marks] Loss of helper T cells leads to the inactivation of humoral and cell mediated immune response Antigen Antigen Presenting Cell Q P
UNIT BIOLOGI KOLEJ MATRIKULASI PULAU PINANG PROGRAM BIOFORTUNE SEMESTER 2 SESI 2022/2023 ANSWER SCHEME
Biofortune 2.0 Session 2022/2023 1 Event A 1 FIGURE 1 shows an event that can result in genetic drift. FIGURE 1 (a) Define genetic drift. [1 mark] Answers Marks A random change in allele frequency in a small population 1 (b) Identify the type of genetic drift in event A. [1 mark] Answers Marks Bottleneck effect 1 (c) How does the event A lead to the formation of a new species? [2 marks] Answers Marks i. Sudden decrease in the original population size due to natural disaster /human action 1 ii. The (small) surviving population will have different gene pool/allele frequencies from the 1
Biofortune 2.0 Session 2022/2023 2 original population // the (small) surviving population has less genetic variation from the original population // (small) surviving population, some allele may be over represented / less represented or eliminated (d) State another type of genetic drift and briefly describe the event that causes it. [2 marks] Answers Marks Founder effect 1 Occurs when small population is isolated/ migrated from the large/original/main population 1 2 FIGURE 2 shows three different enzyme reactions. M and N represent enzyme reaction with the presence of inhibitors. FIGURE 2 (a) State the type of inhibitors that present in M and N. [2 marks] Answers Marks M : competitive (inhibitor) 1
Biofortune 2.0 Session 2022/2023 3 N : noncompetitive / non competitive / non- competitive (inhibitor) 1 (b) What happen when inhibitor at N bind to the enzyme? [4 marks] Answers Marks Inhibitor (at N) binds to allosteric site and causes a change in the conformation/shape of enzyme’s active site 1 Substrate cannot bind/fit to the active site 1 Enzyme-substrate complex do not form 1 Reduce rate of reaction 1 3 (a) FIGURE 3 .1 shows a schematic representation of aerobic respiration in eukaryotic cell. FIGURE 3.1
Biofortune 2.0 Session 2022/2023 4 (i) Name the process how ATP is produced in stage A. [1 mark] Answers Marks Substrate level phosphorylation 1 (ii) Briefly explain how FAD is reduced to FADH2 in stage B. [2 marks] Answers Marks i. Two hydrogens/ electrons are transferred to FAD (forming/ reduce to FADH2) // FAD received/accepted two hydrogens/electrons 1 ii. Succinate is oxidized/dehydrogenation to fumarate// succinic acid is oxidized/dehydrogenation to fumaric acid 1 (iii) Explain how H2O and ATP are produced in stage C. [8 marks] Answers Marks i. Electron from NADH / FADH2 flow/transferred/pass through electron transport chain/ ETC *if not mention ETC, must include all components in ETC 1 ii. by redox reaction 1 iii. Oxygen / ½ O2 act as final electron acceptor / will accept electrons / e- and combine/react with hydrogen ions/ proton/ H+ to form water// ½ O2 + 2H+ + 2e- → H2O 1 iv. The energy released (from redox reaction) in electron transport chain is used to pump hydrogen ions/ proton/ H+ 1 v. from matrix of mitochondria into intermembrane space 1 vi. and create proton gradient / proton motive force 1
Biofortune 2.0 Session 2022/2023 5 vii. H+ diffuse back from intermembrane space into matrix of mitochondria down the concentration gradient (through ATP synthase) 1 viii. ATP synthase catalyse formation of ATP from ADP and inorganic phosphate/Pi//ATP synthase catalyse phosphorylation of ADP to ATP 1 ix. (ATP is produce/synthesised) by chemiosmosis 1 x. One NADH will produce 3 ATP // One FADH2 will produce 2 ATP Max 8 (b) FIGURE 3.2 shows fermentation process that occur in a muscle cell. Molecule Q is the product of glycolysis, while molecule R is the product of fermentation. Briefly explain how molecule R is produced. FIGURE 3.2 [2 marks] Answers Marks i. Pyruvate/Q reduced by NADH (+ H+ ) to lactate/ R // Pyruvic acid reduced by NADH (+ H+) to lactic acid 1 ii. NADH (+ H+) is oxidized/dehydrogenation to NAD+ 1
Biofortune 2.0 Session 2022/2023 6 4 (a) FIGURE 4 shows a graph of a concentration of two different molecules involve in Calvin cycle. FIGURE 4 (i) Where does Calvin cycle occur in C3 plants? [1 mark] Answers Marks Stroma of chloroplast (of mesophyll cells) 1 (ii) What is the product from light dependent reaction required to produce one glucose molecule from Calvin cycle? [2 marks] Answers Marks 18 ATP + 12 NADPH (H+) *mention both products will get 1 mark. Another 1 mark is one correct number of both products 1+1
Biofortune 2.0 Session 2022/2023 7 (iii) Why concentration of 3-phosphoglycerate become high during light independent reaction? [1 mark] Answers Marks Carbon fixation occur. // Ribulose bisphosphate/RuBP bind/attach/fix with CO2 occur. 1 (iv) Why the Calvin cycle cannot occur when there is no light? [2 marks] Answers Marks i. Light dependent reaction does not occur. 1 ii. No ATP and NADPH produced. // Regeneration of ATP and NADPH does not occur. 1 (v) Predict what will happen to glucose concentration when there is no light. [1 mark] Answers Marks (Glucose concentration) decrease/ reduce. 1 (b) In C4 plants, carbon dioxide fixation occurs twice catalysed by two different enzymes, RuBP carboxylase-oxygenase (Rubisco) and PEP carboxylase. Differentiate between these two enzymes in this plant. [2 marks] Answers Marks RuBP carboxylase-oxygenase (Rubisco) PEP carboxylase i. Catalysed the fixation of CO2 with RuBP to form 3- phosphoglycerate. Catalysed the fixation of CO2 with phosphoenolpyruvate / PEP to form oxaloacetate. 1/0 ii. Has low affinity to CO2 when CO2 concentration is low.// Has high affinity to CO2 when CO2 concentration is high. Has high affinity to CO2 even when CO2 concentration is low. 1/0
Biofortune 2.0 Session 2022/2023 8 Has high affinity to O2 when CO2 concentration is low.// Binds to O2 when CO2 concentration is low. Has no affinity towards O2. Binds to CO2 even when CO2 concentration is low. iii. Found in bundle sheath cell (of C4 plant.) Found in mesophyll cell (of C4 plant.) 1/0 iv. Catalyze (second) carbon fixation in Calvin cycle. Catalyze (first) carbon fixation in C4 pathway/ Hatch-Slack pathway. 1/0 Any 2 *answer must be in sentences and NOT TABLE 5 (a) FIGURE 5 below shows oxygen dissociation curve for human oxyhaemoglobin in the lung. FIGURE 5
Biofortune 2.0 Session 2022/2023 9 (i) How haemoglobin is saturated with oxygen in the lung. [1 mark] Answers Marks Partial pressure of oxygen in the lung is high// Affinity of haemoglobin /hemoglobin to oxygen is high// Oxygen bind to haemoglobin to form oxyhaemoglobin /oxyhemoglobin 1 (ii) Why is the shape of oxygen dissociation curve for haemoglobin is sigmoid? [1 mark] Answers Marks Due to cooperative binding of haemoglobin with oxygen// due to cooperativity characteristic of haemoglobin toward oxygen// binding of one oxygen to one haem group will induce other haem group to change their conformation to bind to oxygen easily/to increase their affinity to oxygen 1 (iii) Sketch a curve on the graph in FIGURE 5 to show the position of the oxygen dissociation curve at pH 7.2. [1 mark] Answers Marks Sigmoid shape curve (and label) 1
Biofortune 2.0 Session 2022/2023 10 *the curve will level off either overlapping or lesser than the original curve (b) Explain how a change in pH from 7.4 to 7.2 affects the supply of oxygen by haemoglobin to the tissues. [3 marks] Answers Marks i. Due to Bohr effect // Due to increasing in partial pressure /concentration of carbon dioxide/ hydrogen ion concentration 1 ii. Affinity of haemoglobin to oxygen dearease/low 1 iii. More oxyhaemoglobin dissociate unloading/releasing oxygen (to tissue) 1 (c) State THREE methods how carbon dioxide is transported in the blood. [3 marks] Answers Marks i. About 60% is transported in the form of hydrogencarbonate ions / bicarbonate ions. 1 ii. About 30% combine with haemoglobin in the red blood cell to form carbaminohaemoglobin. 1 iii. About 10% dissolved in blood plasma (only a very small amount carbonic acid.) 1
Biofortune 2.0 Session 2022/2023 11 6 FIGURE 6 below shows the structure of a human heart. FIGURE 6 (a) Name the structure W and state its function. [1 mark] Answers Marks W: Sinoatrial node // SA node Function: Generates/Produce impulse // set as pacemaker // initiate heartbeat 1 (b) Which blood vessel containing deoxygenated blood? [1 mark] Answers Marks vena cava // pulmonary artery 1 (c) How ‘lub’ and ‘dup’ sound produced? [2 marks] Answers Marks i. ‘Lub’ because of the (recoiling of blood against) closing of atrioventricular valves / AV valve / tricuspid and bicuspid/mitral valve 1
Biofortune 2.0 Session 2022/2023 12 ii. ‘Dup’ because of the (vibration of) closing of semilunar valves 1 (d) The impulse which coordinates the heartbeat is delayed slightly at X. It then passes along Y to the heart apex. Explain the importance of the following events: (i) The slight delay at X. [1 mark] Answers Marks To allow the atria to empty completely before the ventricles contract/systole // to allow atria complete its contraction/systole before the ventricles contract/systole 1 (ii) The conduction of impulse to the heart apex. [2 marks] Answers Marks Ventricles contracts from apex upwards 1 Blood will be pumped (out) through the (pulmonary) artery / aorta 1
Biofortune 2.0 Session 2022/2023 13 7 FIGURE 7 below shows a structure of a human nephron. FIGURE 7 (a) Differentiate between the solute concentration of glomerular filtrate at B and C. [1 mark] Answer Mark Glomerular filtrate at B is high (solute) concentration/hypertonic while glomerular filtrate at C is low solute concentration/hypotonic// (Solute) concentration in (glomerular filtrate) B is more than (solute) in C. 1
Biofortune 2.0 Session 2022/2023 14 (b) How drugs and toxins are excreted out through the nephron? [1 mark] Answer Mark Drugs and toxin secreted out by active transport at proximal (convoluted) tubule. 1 (c) Diabetes mellitus type II occur due to insensitivity of pancreas to secrete insulin even when glucose concentration in blood is high. Briefly explain how this condition could affect the reabsorption process occur at A and the urine output produced. [2 marks] Answer Mark i. Glucose is not fully reabsorbed (into peritubular capillary/blood by active transport) 1 ii. Excessive glucose excreted out into the urine//cause glycosuria/glucosuria//Glucose is found in the urine. 1 (d) Diabetes insipidus is a disorder occur due to posterior pituitary unable to secrete ADH. Briefly explain how this condition effects the blood osmolarity and the urine output. [3 marks] Answer Mark i. Less water will be reabsorbed at collecting duct//collecting duct is less permeable to water. 1 ii. Causes blood osmolarity/osmotic pressure will increase 1 iii. Urine will be diluted and large amount / high volume 1
Biofortune 2.0 Session 2022/2023 15 8 (a) Explain the mechanism of impulse transmission across the synapse. [8 marks] Answers Marks i. An action potential /impulse arrives, trigger/ induce/stimulate voltage-gated calcium (ion)/Ca2+ channel opens (in synaptic knob). 1 ii. This causes an influx/flow/diffusion of Ca2+ through/into presynaptic membrane. 1 iii. This trigger synaptic vesicles to fuse/merge with presynaptic membrane. 1 iv. Releasing neurotransmitter into the synaptic cleft by exocytosis. 1 v. Neurotransmitter binds to receptor of ligand (-) gated (ion) channel in postsynaptic membrane causes these channels to open. 1 vi. Na+ ion diffuse / influx into postsynaptic membrane and causes depolarization 1 vii. When EPSP/excitatory post synaptic potential reach / When depolarization reaches threshold, triggers an action potential 1 viii. After the action potential produced, neurotransmitter degraded/breakdown/hydrolysed by specific enzyme 1 ix. The (degraded) neurotransmitter transported back / reuptake/moved back/reabsorbed to synaptic terminal/knob 1 *if mentioned name of neurotransmitter penalized once 9 max 8
Biofortune 2.0 Session 2022/2023 16 (b) FIGURE 8.1 shows below shows the structure that is involved in muscle contraction. FIGURE 8.1 (i) State the ion that is involved in muscle contraction [1 mark] Answers Marks Ca2+ / Calcium ion 1 (ii) Briefly explain the consequential event after the ion in 8(b)(i) binds to F. [2 marks] Answers Marks i. Troponin/F change its conformation/shape 1 ii. Tropomyosin shifts/pushed away/slide away/dislodge/move 1 iii. expose the myosin binding site/active site on actin 1 *answer must follow sequence Any 2 F
Biofortune 2.0 Session 2022/2023 17 (c) FIGURE 8.2 below shows two forms of phytochrome. FIGURE 8.2 (i) Identify phytochrome S. [1 mark] Answers Marks Pr / P660 1 (ii) What happen when level of phytochrome T increases in long-day plant? [2 marks] Answers Marks Releasing/Production of (hormone) florigen/ stimulate conversion of inactive hormone precursor to florigen 1 Promotes/ induce/stimulate flowering 1 9 (a) Influenza virus is a virus that infects the respiratory tract cell and causes flu. Explain how the activated helper T cell stimulate immune response against the infected cells. [6 marks] Answers Marks i. Activated helper T cell/TH cell secretes interleukin2/IL-2 to activate cytotoxic T cell/TC cell 1 S T Red light Far red light
Biofortune 2.0 Session 2022/2023 18 ii. Stimulate proliferation and differentiation / clonal selection of activated TC cell 1 iii. Activated TC cell binds at the (Class I) MHC-antigen complex of the infected cells 1 iv. Activated TC cell secretes perforin and granzyme 1 v. Perforin forms pores on the infected cell 1 vi. cause water and ions / extracellular fluid flow into infected cell 1 vii. and the infected cell lysed / cytolysis 1 *point vii is dependent on point vi viii. Granzyme digest proteins in infected cell and stimulate apoptosis / programmed cell death / cell distruction 1 * term activated if not mentioned in any points penalize once 8 max 6 (b) HIV is a virus that causes acquired immunodeficiency syndrome (AIDS), which attacks the adaptive immune response. Briefly explain how HIV integrates with a host cell’s DNA. [3 marks] Answers Marks i. After HIV enters a host cell, its RNA, viral proteins and reverse transcriptase enzymes are released into the cytoplasm 1 ii. Reverse transcriptase catalyses the synthesis of a DNA strand by using the viral RNA as a template and hydrolysing it 1 iii. Reverse transcriptase (enzyme) then catalyses the synthesis of a second DNA strand complementary to the first strand 1 iv. Inside the nucleus, the double-stranded DNA is incorporated as a provirus into the DNA of the cell 1 4 Max 3
ANSWER SCHEME ROAD TO CHAMPION SET 3 1. FIGURE 1 below shows graphs representing three types of natural selection. FIGURE 1 (a) i. State the types of selection I and II. [2 marks] I : Stabilising selection II : Directional selection ii. What is the significance of selection III? [1 mark] Promotes genetic variability that leads to the evolution of new species. (b) Give ONE advantage of natural selection. [1 mark] Favourable genes / traits can be transmitted through generations. // Increases the chance of individuals surviving in the new population (c) Give TWO differences between natural and artificial selection. [2 marks] Natural selection Artificial selection Produces a great biological diversity Produces varieties of organisms such as improved crops and livestock Nature-made selection process // not involved human intervention man-made selection process // involved human intervention occurs in natural populations Mainly occurs in domestic populations
Only allows favourable characters to be inherited over successive generation Only allows selected traits to be inherited over successive generation Facilitates evolution through generating biological diversity Does not facilitates evolution 2. FIGURE 2 shows the mechanism of enzyme action. FIGURE 2 a) Based on FIGURE 2, state TWO characteristics of the enzyme. [2 marks] • The action of enzyme is highly specific to the substrate • Enzyme catalyse reversible reaction • Enzyme are not destroyed in the reaction that they catalysed and can be reused again b) There are few factors that affect the activity of enzymes. Explain how temperature affects the enzyme activity. [2 marks] • Increase in temperature will increase the reaction of enzyme / substrate collide with active sites more frequently • Above optimal temperature, enzyme will become denatured / rate of reaction drops sharply • At the low temperature, the enzyme is not active thus rate of reaction decrease
c) Sketch a graph to show the effect of temperature on the rate of catalysed reaction. [2 marks] Correct curve shape 1 mark State optimum temperature 1 mark 3. a) The main components of electron transport chain and chemiosmosis in the inner mitochondrial membranes are shown in the FIGURE 3. FIGURE 3 i) Name compounds S and T, U and V and structure R. [3 marks] S : FADH2 / reduced FAD T : FAD / oxidized FAD U : NADH / reduced NAD+ V : NAD+ / oxidized NAD+ R : ATP synthase ii) Name the process that uses the principles of chemiosmosis shown in the FIGURE 3. [1 mark] Oxidative phosphorylation
iii) What is meant by the process named in (b)? [3 marks] • ATP is produced from ADP + Pi • by using energy derived from redox reaction on the electron transport chain • where oxygen is the final electron acceptor • Process occurs at the cristae of mitochondrion. • Comprises electron transport chain and chemiosmosis b) Vigorous exercise could induce muscle cramp due to the production of excess lactic acid. Describe the process of producing lactic acid in the muscle. [6 marks] • During vigorous exercise, there is an oxygen deficit in the body where the supply of oxygen through breathing process is not sufficient to meet the body’s demand • the muscle then carries out anaerobic respiration • glycolysis take place in the cytoplasm of the cells • pyruvate is produced • pyruvate is reduced by NADH • forming lactate/ lactic acid as an end product • and 2 ATP molecule are produces • accumulation of lactic acid causes muscle cramp and fatigue 4. (a) The Calvin cycle is divided into three phases labelled as R, S and T as shown in the FIGURE 4. FIGURE 4
i. Name R and T. [2 marks] R : Carbon fixation T : Regeneration of RuBP ii. Describe what happens in phase S. [2 marks] • 3-Phosphoglycerate is phosphorylated by ATP to produce 1,3- bisphosphoglycerate • 1,3-bisphosphoglycerate reduced by NADPH to become glyceraldehyde-3-phosphate / G3P • NADPH oxidised into NADP+ iii. What happens to G3P in phase T? [1 mark] G3P is phosphorylated by ATP to regenerate RuBP (b) Differentiate carbon fixation in C3 and C4 plants. [4 marks] C3 C4 Carbon fixation occurs once Carbon fixation occurs twice Carbon fixation occurs in mesophyll cells Carbon fixation occurs in mesophyll cells and then in bundle sheath cells CO2 acceptor is RuBP CO2 acceptor is PEP in mesophyll cell and RuBP in bundle sheath cells CO2 fixing enzyme is Rubisco CO2 fixing enzyme is PEP carboxylase in mesophyll and rubisco in bundle sheath cells First product form is 3- phosphoglycerate (3C) First product form is oxaloacetate (4C)
5. FIGURE 5 shows the structure of a stoma. FIGURE 5 a) Identify Y and Z. [2 marks] Y : Subsidiary cell/ Neighbouring cell/ Epidermal cell Z : Guard cell b) Explain the mechanisms of stomata closing based on the potassium ion accumulation hypothesis during drought. [7 marks] • During drought, transpiration rate is high • ABA is released to stop potassium ion exchange • ABA secreted and bind to ABA receptor on guard cell • Ca2+ channel open, Ca2+ enter guard cell • Accumulation of calcium ion inhibit proton pump • K+ move out from guard cell/ H+ are pumped into guard cell. • Water potential in guard cell increase • Water diffuses out from guard cell by osmosis • Guard cell becomes flaccid • Stomata closed
6. The movement of water and dissolved minerals from one cell to another in plant roots are represented by R, S and T pathways as shown in FIGURE 6. FIGURE 6 a) Identify R, S and T pathways. [3 marks] R : Apoplast pathway S : Symplast pathway T : Transmembrane pathway b) Describe S pathway. [2 marks] • Water diffuses by osmosis through the cell wall and partially permeable membrane into the cytoplasm • Water diffuses by osmosis from the cytoplasm of one cell to the cytoplasm of the adjacent cell through plasmodesmata c) State TWO significance of Casparian strip in endodermis to prevent the movement of water and ions through R pathway. [2 marks] • The Casparian strip is a waterproof suberin band on the endodermal cell wall that prevents the apoplast pathway of water movement • Water and minerals then pass through the cell surface membrane on the endodermal cells into the cytoplasm, thus regulating the water and mineral movement into the stele.
Organ R Hormone P Hormone Q Q Organ R Normal blood glucose level High Low 7. FIGURE 7 shows the regulation of human blood glucose level. FIGURE 7 a) Name hormone P, Q and organ R. [3 marks] P : Insulin Q : Glucagon R : Pancreas b) What will happen to a person who lacks hormone P? [2 marks] • Blood glucose level cannot be control • Blood glucose will exceed above normal point • The person will suffer from diabetes mellitus c) How does hormone P act on the liver and state its effects? [2 marks] • Liver will absorb the excess glucose • Effect: Glucose is converted into glycogen which is stored in the cells
8. a) Flowering species P and Q could be induced by different day lengths. The result of an experiment on the effect of light on flowering are shown in the FIGURE 8. FIGURE 8 (i) State the photoperiodic group for species P and Q. [2 marks] P : Short-day plant Q : Long-day plant (ii) Give reasons for your answers in a) (i) [2 marks] • Species P flower when the night length exceeds the critical night length period. • Species Q flowers when the night length is shorter than the critical night length period. iii) Explain why species P is able to flower even being exposed to a single flash of red light followed by a single flesh of far-red light. [2 marks] • The flash of far-red cancels the effect of red light. • Pfr is converted to Pr and low Pfr stimulates flowering in shortday plants
b) Describe the roles of Pr and Pfr phytochromes in regulating flowers in short-day and long-day plants. [8 marks] • Phytochromes have two photo interconvertible forms which are Pr and Pfr • Pr absorbs red light and rapidly converted into Pfr • Pfr absorbs far-red light and slowly converted into Pr • Pfr is active form of the phytochrome • In long-day plant, Pr absorbs red light during the day and is converted to Pfr • When the concentration of Pfr increase, florigen is produced • Long-day plant will flower • In short-day plant, Pfr absorbs far-red light during the night and is converted to Pr • When concentration of Pfr decrease, florigen is produce • Short-day plant will flower 9. Fatimah was stricken by a disease and after a month she was re-infected by the same disease. Explain how Fatimah’s immune system responds to the situation. [9 marks] • Reinfection due to the same antigen triggers the secondary immune response. • The secondary immune response has a stronger and faster response. • A humoral/ cell mediated immune response can be stimulated. • In a humoral immune response, memory helper T cells are stimulated. • The memory helper T cell stimulates the memory B cells. • The memory B cells divide to form clones. • The clones differentiate into plasma cells • The plasma cell produces antibodies that react against the same antigen. • In a cell mediated immune response, memory helper T cells are stimulated. • The memory helper T cell will stimulate the memory cytotoxic T cells. • The activated memory cytotoxic T cells form clones • The clones destroy the same antigens.