“Never give up, fix mistakes and keep stepping.” SCHEME REVISION FOR PSPM 2 SET 1 – SET 11 “Struggle that you do today is the single way to build a better future.”
SB025 SB025 Biology Biologi Semester 2 Semester 2 Session 2022/2023 Sesi 2022/2023 2 Hour 2 Jam KOLEJ MATRIKULASI KEDAH KEMENTERIAN PENDIDIKAN MALAYSIA KEDAH MATRICULATION COLLEGE MINISTRY OF EDUCATION MALAYSIA GEMPUR PSPM 2 SKEMA JAWAPAN Questions Marks Allocate 1 6 2 6 3 13 4 9 5 9 6 7 7 7 8 14 9 9 Total 80 Skema jawapan ini mengandungi 17 halaman bercetak This answer scheme consists of 17 printed pages
1) FIGURE 1 show two types of species, Rana aurora and Rana boylii that breed in January-March and March-May respectively. a) Identify the type of reproductive isolation related to FIGURE 1. [1 mark] Temporal isolation b) Define the type of reproductive isolation in your answer in (a). [2 marks] -Two species that breed during different times of the day or different seasons or year -cannot mix their gametes. c) State one factor that involved in the formation of a new species. [1 mark] Reproductive isolation/ Genetic drift/ Hybridization/ Adaptive radiation d) How founder effect can lead to the formation of a new species? [2 marks] -Founder effect occurs when a small population is isolated/ migrated from a larger parent population. -The new, smaller population becomes pioneer individuals -Frequencies of certain alleles change over time -Produce /has different gene pool/ allele frequencies from the original population. Rana aurora Rana boylii FIGURE 1
2) (a) FIGURE 2 represents a model for the mechanism of enzyme action. FIGURE 2 (i) Name the model of mechanism of enzyme action shown above. [1 mark] Induced fit model (ii) Explain mechanism of enzyme action based on model above. [2 marks] - Active site of enzyme does not exactly complementary with the substrate - The binding of substrate to the active site induces a slight change in the shape of the active site. (b) FIGURE 3 shows two mechanism of enzyme inhibition FIGURE 2 (i) State the type of inhibitor X and Y. [2 marks] Inhibitor X: Competitive inhibitor Inhibitor Y: Non-competitive inhibitor (ii) How the rate of enzyme reaction with inhibitor X can be increases? [1 mark] By increasing substrate concentration Inhibitor X Inhibitor Y
3. (a) FIGURE 4 shows a summary of an aerobic respiration. i. State THREE processes A,C and D. [3 marks] A: Glycolysis C: Krebs cycle D: Oxidative phosphorylation / ETC and Chemiosmosis ii. Based on processes stated in a) which one produces largest amount of energy in the form of ATP? [1 mark] Process C// Oxidative phosphorylation / ETC and Chemiosmosis FIGURE 4 34 ATP
b) FIGURE 5 represents a cellular process that harnesses the reduction of oxygen to generate high-energy phosphate bonds in the form of adenosine triphosphate (ATP).. i) Based on FIGURE 5 what is the molecule that act as final electron acceptor? [1 mark] Oxygen ii) How many ATP produced per molecules of A and B? [1 marks] A produces 3 ATP while B produces 2 ATP iii) State the importance of process in FIGURE 5 in mammalian respiring tissue. Transfer electron from electron carrier to generate (electrochemical) proton(H+ ) gradient// transfer electron down a series of redox reactions that release energy to form ATP. [1 mark] A B C FIGURE 5 B A A A
c) Briefly explain what happen to pyruvate in human cell with and without oxygen? [6 marks] With the presence of oxygen, pyruvate will undergoes pyruvate oxidation 1 to form acetyl CoA. 1 Then, it will enter the Krebs cycle and continue to oxidative phosphorylation. 1 With the absence of oxygen pyruvate is reduced 1 by oxidising NADH 1 to form lactate/lactic acids// undergoes lactate fermentation 1 4 FIGURE 6 shows three different pathways that occur in plant A, B and C during light independent reaction. FIGURE 6 (a) Identify cycle X and product Y. [2 marks] Cycle X : Calvin cycle Product Y : glucose / sugar / carbohydrate
(b) Name the compound which initially accepts carbon dioxide and the enzyme involved in plant B. [2 marks] CO2 acceptor : Phosphoenolpyruvate Enzyme : PEP carboxylase (c) How many molecules of 3-phosphoglycerate will be produced if 6 molecules of carbon dioxide being fixed during cycle X. [1 mark] 12 (molecules of 3-phosphoglycerate) d) State one (1) advantage of using pathway B and C in photosynthesis. [1 mark] To avoid/ minimize photorespiration. (e) State the difference of CO2 fixing enzyme between pathway A and C. [1 mark] CO2 fixing enzyme in pathway A is RuBP carboxylase-oxygenase / Rubisco whereas in plant C is PEP carboxylase. (f) Explain the roles of ATP and NADPH in cycle X. [2 marks] - ATP is used as a source of energy for reduction of CO2 // to phosphorylate 3- phosphoglycerate to 1,3-bisphosphoglycerate and regeneration of RuBP. - NADPH is reducing agent that is used for reduction of 1,3-bisphosphoglycerate to glyceraldehyde 3-phosphate/ G3P.
5(a) Figure 7 below shows an oxygen dissociation curve. Curve A and B represents the oxygen dissociation curve of different pigments while curve B and C represents the oxygen dissociation curve of the same pigment but in different conditions. FIGURE 7 i. Why curve A is shifted to the left compared to curve B? [2 marks] - The respiratory pigment of curve A consists of one subunit only / has no cooperative binding. - Affinity of pigment A to oxygen is higher than pigment B ii. Explain the condition occurs in curve C and how this condition shifts the curve to the right. [3 marks] - Condition of curve C in high in partial pressure of carbon dioxide / CO2 and low/ decrease in pH. - This condition decreases the affinity of the pigment / haemoglobin towards oxygen. - The pigment / haemoglobin releases more oxygen (compared to B).
(a) Chemoreceptors is the main components that detects the pH changes in blood. Explain what will happen to breathing mechanisms when chemoreceptor detects the drop of pH in blood. [4 marks] - Chemoreceptor send impulse to inspiratory center in medulla oblongata. - Inspiratory center send impulse to diaphragm through phrenic nerves and external intercostal muscle through thoracic nerve. - Diaphragm and external intercostal muscle contract. - Inspiration occur. 6. The sinoatrial node, atrioventricular node and Purkinje fibres have important roles in the cardiac cycle. a. State the stage of the cardiac cycle that is directly initiated by the wave of impulse sent out by the sinoatrial node. [1 mark] Atria systole ventricular diastole (Not accept atria contraction, atria systole only or ventricular diastole only) b. Part of the control of cardiac cycle involves the contraction of the ventricle walls after the walls of the atria have finished contracting. Relate how this control is achieved with the roles of the nodes and fibres. [2 marks] Atrioventricular node delay impulse for 0.1 second given time for atria to complete the contraction//empty the blood from atrium. Impulse must pass through atrioventricular node before send to Purkinje fibre. impulse from sinoatrial node cannot directly passing to ventricle (block by the ring of non-conducting fibrous tissue) c. Name the valves of the heart that open soon after the Purkinje fibre have received an impulse from the atrioventricular node. [1 mark] Semilunar valve
d. When the heart chamber is fill up with blood, the ‘dup’ sound can be listened through the stethoscope. State the cause for the sound production. [1 mark] Vibration of the closing semilunar valve// closing of the semilunar valve e. The transport tissues of plants are phloem and xylem. The role of xylem is the transport of water and mineral ions from the soil solution to the different parts of the plant body. The role of phloem is the translocation of assimilates and other substances from sources to sinks. The source of mineral ions for the plant is the soil solution. These mineral ions are transported from the roots in the xylem. Mineral ions are also found in the phloem sap within phloem sieve tube. Suggest why mineral ions are found within phloem sieve tubes and state how they are transported within phloem sieve tubes. [2 marks] Either 1 of the following: Mineral ions found in phloem is entry from xylem at the phloem source. Mineral ions dissolved in water as solutes when it transfers in xylem to phloem. (Explanation: When the water from xylem is flow to the phloem, it can bring the solutes together entering the phloem through plasmodesmata) And The mineral ions found in phloem sap transported down the (hydrostatic) pressure gradient within the phloem sap. // transported from high hydrostatic pressure to low hydrostatic pressure region in phloem. // transported from sources to sinks with the difference in pressure gradient.
7. Figure 8 shows the structure of a functional unit of kidney FIGURE 8 i. Name and describe the process that occur in C. [3 marks] − Ultrafiltration − A/afferent arteriole has larger diameter than B/efferent arteriole − A high hydrostatic pressure is generated in C/glomerulus − The pressure generated force the fluid including all solute, except blood cell and larger plasma protein through the pore of capillary wall of C/glomerulus into D/Bowman’s capsule. ii. Which part has the highest reabsorption capacity? Explain your answer. [2 marks] − E/ proximal convoluted tubule − Compose of epithelial cell which are one cell thick to enhance reabsorption/diffusion // each epithelial cell is rich in mitochondria for active transport // highly coiled to increase surface area for reabsorption // inner side of epithelial cell has microvilli to enhance reabsorption.
iii. A man underwent organ transplant surgery. Unfortunately during the operation, he suffered severe bleeding and lost a lot of blood. Based on your understanding, relate this issue to the amount of urine that will be produced. [2 marks] − Less amount of urine produce − Blood loss will lower the blood pressure hence lessen blood flow through the filtration membrane/C to be transform into urine. 8. FIGURE 9 shows changes in permeability of the membrane to both sodium ion (Na+ ) and potassium ion (K+ )during single action potential. FIGURE 9 a i) Explain briefly steep increase in sodium ion permeability between 0.5ms and 0.7ms. [3 marks] A stimulus triggers some voltage-gated sodium ion channels to open and sodium ions diffuse into the axon inside axon become more positive/ cause depolarisation/ membrane potential reach threshold causing more voltage-gated sodium ion channels to open ii) Name the phase in which pottasium ion permeability is increase. What happens to the membrane potential at this phase. [2 marks] Repolarisation// falling phase of action potential Membrane potential becomes more negative/ less positive
iii) Multiple sclerosis is a disease that in which parts of myelin sheath surrounding the neurone are destroyed. Give a reason why this results in slower responses to stimuli. [1 mark] No saltatory conduction/ impulse/action potential unable to ‘jump’ from one nodes of Ranvier to another nodes. b) Explain how cocaine can cause continuos stimulation of the brain limbic system. [2 marks] Cocaine binds to dopamine transporter protein which block the reuptake/ reabsorption of dopamine into synaptic knob. Causing dopamine to stay/ remain in synaptic cleft and continously/ repeatedly binds to the receptors at postsynaptic membrane. c) Progesterone plays an important roles in the menstrual cycle and pregnancy. Explain the mechanism of action for progesterone towards the target cells. [6 marks] i. Progesterone is a steroid hormone/ lipid-soluble hormone ii. Progesterone can pass through the plasma membrane of target cell and bind to a receptor protein in the cytoplasm/ nucleus of target cell . iii. Forming hormone-receptor complex iv. Hormone-receptor complex interacts/ bind with DNA in the nucleus and activates specific genes. v. Specific genes undergoes transcription to produce mRNA vi. mRNA undergoes translation in cytoplasm to produce specific protein 9. Explain the immune suppression by HIV infection (9 Marks) i. Human immunodeficiency virus (HIV) is the pathogen that causes acquired immune deficiency syndrome (AIDS). 1m ii. HIV attacks the adaptive immune response. 1m iii. HIV infects helper T cell by binding specifically to the CD4 accessory protein. 1m iv. HIV RNA genome is reverse-transcribed. 1m v. the product DNA is integrated into the host cell’s genome . 1m vi. The viral genome can direct the production of new viruses. 1m vii. The body responds to HIV with an immune response enough to destroy viral infection. 1m viii. But due to a very high mutation rate ,some HIV are able to escape. 1m
ix. Mutated viruses has altered proteins on the cell surface. 1m x. This reduce interaction with antibodies and cytotoxic T cells. 1m xi. Viruses replicate and mutated further / HIV evolves within host body. 1m xii. Viral replication and (helper T) cell death triggered by the virus . 1m xiii. lead to loss of helper T cells. 1m xiv. impairing(damaging) both humoral and cell- mediated immune responses 1m xv. Resulting in acquired immune deficiency syndrome(AIDS) that leaves the body susceptible(at risk) to infections and cancers 1m Total 15m
SULIT NO MATRIK: SB025 1a) Give TWO differences between natural and artificial selections. [2M] Natural Selection Artificial Selection Selection agent is environment Selection agent is human Random event/ by chance// on any wild organism Planned event//selected organism/intervention Selected varieties are robust/strong/tough//fittest variety Selected varieties are defective because of inbreeding (depression) b) Explain how reproductive isolation leads to speciation. [4M] - Caused by pre-zygotic barriers/mechanisms or post-zygotic barrier/mechanisms - Pre-zygotic barriers prevent formation of hybrid zygote - Occur before fertilization of gametes/formation of zygote - Post-zygotic barriers prevent the zygotes from developing into normally functioning/ fertile individuals. - Occur after hybrid mating. 2a) Figure 1 shows an enzymatic reaction. Figure 1 i) State the main function of M. [1M] A site on enzyme for the binding of substrate. ii) How do X assist for the proper functioning of enzyme? [2M] - By binding loosely and temporary to the enzyme (forming weak hydrogen bond with enzyme) and change its active site, - so, substrate can fit easily, and reaction take place to produce product. b) Malonate competes with substrate succinate for the active site of enzyme succinate dehydrogenase and interrupt Krebs cycle process during cellular respiration. Explain mechanism of action of malonate and succinate towards the enzyme succinate dehydrogenase. [3M] - Malonate is a competitive inhibitor that has similar structure with the substrate succinate. - When malonate binds to the active site of enzyme (succinate dehydrogenase), it prevents the substrate molecule from binding to the active site, - and reduces the enzymatic reaction rate. 3a) FIGURE 2 shows an outline of two stages of aerobic respiration. X X M
SULIT NO MATRIK: SB025 Figure 2 i) Name two phases involved during process X. [2M] - Energy investment (phase) - Energy payoff (phase) ii) State the number of ATP produce from one molecule of glyceraldehyde-3-phosphate in process X. [1M] 2 ATP iii) What is the high energy molecule produce during process Y and explain how does it produced? [3M] - NADH + H+ - Pyruvate’s remove carboxyl group and given off as a molecule of CO2 by decarboxylation. - The remaining two-carbon fragment is oxidized, and the electrons transferred to NAD+ reduced into NADH + H+ b) FIGURE 3 shows process of oxidative phosphorylation which occurs in the mitochondria. Figure 3 i) Identify Complex C. [1M] Cytochrome c reductase ii) Explain the relationship of H+ at complex A, C and D with the production of ATP. [4M] - Complex A, C and D pump proton/ H+ from matrix of mitochondria into intermembrane space to create proton gradient known as proton-motive force. - The force drives protons/ H+ back to matrix of mitochondria through ATP synthase. - Proton/ H+ diffuse through ATP synthase from intermembrane space to the matrix of mitochondria down concentration gradient, - release energy to phosphorylate ADP forming ATP. iii) What happen to the electron transport chain and ATP production if oxygen level insufficient in cell? [2M] - Electron transfer will be stop - No ATP produced by chemiosmosis 4a) FIGURE 4 shows the schematic diagram of light reaction of photosynthesis. Complex A Complex B Complex C Complex D NADH + H+ NAD+ Figure 3
SULIT NO MATRIK: SB025 Figure 4 i) State the part of chloroplast where does this reaction take place. [1M] Thylakoid membrane ii) Name the substances labelled X and Y. [2M] X: 2H+ Y: Water / H20 iii) What is the function of NADPH produced in FIGURE 4? [1M] As a reducing agent/as an electron donor/To reduce 3-phosphoglycerate/ to glyceraldehyde-3-phosphate/G3P iv) The light reactions involved are cyclic and non-cyclic photophosphorylation. Give two (2) differences between cyclic and non-cyclic photophosphorylation. [2M] Cyclic Non-cyclic Photolysis of water does not occur, so no O2 formed as by-product Photolysis of water occur and O2 formed as by- product Only ATP produced ATP and NADPH produced Only photosystem I involved Photosystem I and photosystem II involved First electron donor is photosystem I First electron donor is water Final electron acceptor is photosystem I Final electron acceptor is NADP+ Electron flow is cyclic and return to reaction center Electron flow is non-cyclic and do not return to reaction center. b) FIGURE 5 shows carbon fixation pathways in two plant cells. Figure 5 Photosystem II Photosystem I Y X ½O2
SULIT NO MATRIK: SB025 i) Name the type of plant that use both cycles A and B. [1M] C4 plants ii) Name the enzymes which catalyse the fixation of CO2 in both cycles. [2M] Enzyme for A: PEP carboxylase // Phosphoenol pyruvate carboxylase Enzyme for B: Rubisco// RuBP carboxylase-oxygenase // Ribulose-1,5-bisphosphate carboxylase-oxygenase 5a) Figure 6 shows oxygen dissociation curve of haemoglobin. Figure 6 i) What is shown by oxygen dissociation curve? [1M] The oxygen dissociation curve is a curve that shows the relative amount of oxygen molecules bound to haemoglobin with different partial pressure of oxygen (PO2). ii) Why is the shape of oxygen dissociation curve being sigmoid? [1M] Due to cooperative binding of haemoglobin with oxygen // due to cooperativity characteristic of haemoglobin toward oxygen. iii) Briefly explain the event occur as haemoglobin reaches tissue at 40mmHg. [2M] - When the PO2 falls to 40 mm Hg, haemoglobin is 75% saturated with oxygen. - Only 25% of the oxygen molecules is given to the tissues by haemoglobin when it reaches the resting tissues. - Resting tissues has lower PO2 because they undergo cellular respiration. b) Explain the role of chemoreceptors when concentration of carbon dioxide is high in the body. [5M] - A rise in partial pressure of carbon dioxide (PCO2) causes carbon dioxide reacts with water and forms carbonic acid / H2CO3, then dissociates into bicarbonate ions (HCO3 - ) and hydrogen ions (H+ ). - High concentration of H+ , lowers the pH blood detected by peripheral chemoreceptor, the aortic bodies, and carotid bodies. - and low pH of cerebrospinal fluid (CSF) detected by central chemoreceptor in the medulla oblongata. - These central and peripheral chemoreceptors send nerve impulses to the inspiratory centre (of the respiratory control centre) in the medulla oblongata. - The inspiratory centre sends nerve impulses via the phrenic nerve to diaphragm and via the intercostal nerve to the external intercostal muscles. - This increases the rate of inhalation as the rate of contraction of external intercostal muscles and diaphragm increase. - The excess carbon dioxide is eliminated in exhaled air and pH returns to a normal. 6a) Figure 7 show three types of lateral movement of water from the soil to the root xylem.
SULIT NO MATRIK: SB025 Figure 7 i) Based on the above figure, identify which pathway that occur more frequent and describe that pathway. [2M] - A - In apoplastic pathway, water & solutes moves through the cell wall & extracellular spaces (non-living parts of the roots). b) Figure 8 shows the event that occur to the heart during one phase in cardiac cycle. Figure 8 i) Name the node that responsible to transmit the impulse causing the phase in Figure 8 to occur. Explain that phase. [3M] - AV Node. – 1M - Atria diastole and ventricles systole/ contracts because impulse from AV node was transmit to the whole ventricle, via bundle of His, bundle branches & Purkinje fibres - The semilunar valves are forced open due to pressure generated during contraction of the ventricles - blood is pumped out of the ventricles into the pulmonary artery and aorta. ii) What sound produced during the phase in Figure 8 above and explain how that sound produce. [2M] - Lub sound - due to recoiling of blood against closing of AV valve 7) Figure 9 shows a functional unit found in the kidney.
SULIT NO MATRIK: SB025 Figure 9 a) Name the type of functional unit that helps camels to conserve more water and how it differs from Figure 9. [2M] - Juxtamedularry nephron - - Long loop of Henle b) Name the process that contributes to the reabsorption of water in Y. [1M] - Counter current multiplier mechanism d) Describe how hormone effect structure Z when the water content in the body is higher than normal? [4M] - Low blood osmolarity are detected by osmoreceptor in hypothalamus. - Posterior pituitary secrete less antidiuretic hormone (ADH) - Low level of ADH decrease permeability of Z/collecting duct - Water reabsorption decrease - Producing large volume of urine 8a) Explain how two neurons communicate at synapse. [7M] - Arrival of an action potential at axon terminal, depolarizes the presynaptic membrane. - Depolarization opens voltage-gated calcium (Ca2+) channels, triggering an influx / diffuse of Ca2+ into synaptic knob. - Rise in Ca2+ causes synaptic vesicles to fuse with the presynaptic membrane. - Synaptic vesicle releasing the neurotransmitter into the synaptic cleft by exocytosis. - The neurotransmitter diffuses across the synaptic cleft and attaches to a specific receptor site on the ligand gated ion channel in the postsynaptic membrane. - Triggers opening ligand gated ion channel, allowing Na+ diffuse / influx into the postsynaptic neurons and leads to depolarization of the postsynaptic membrane. - This leads to a depolarization of the postsynaptic membrane and depolarization response is known as an excitatory postsynaptic potential (EPSP). - If the EPSP reach the threshold level, an action potential is generated and is transmitted along the postsynaptic neuron. b) Describe refractory period as characteristics of nerve impulse along the membrane of an axon. [3M] - It is a short time immediately after an action potential in which the neuron cannot respond to another stimulus. - It has 2 parts: - Absolute Refractory Period: The interval during which a second action potential absolutely cannot be initiated, no matter how large a stimulus is applied. - Relative Refractory Period: The relative refractory period is the interval immediately following the absolute refractory period during which initiation of a second action potential is inhibited but not impossible (the stimulus is greater than the threshold stimulus). c) Plants require different day lengths to induce flowering. Figure 10 shows an experiment, samples of plants, plant A, plant B, plant C, plant D, plant E and plant F from a species were exposed to a range of light and dark treatments. Shaded bars indicate period of darkness and unshaded bars indicate periods of light. The result of each treatment on flowering is shown in Figure 10.
SULIT NO MATRIK: SB025 Figure 10 i) Based on the results of plant A and plant B, state the photoperiodic group this species belongs to. Explain your answer. [2M] - short day plant - it only flower if the period of uninterrupted darkness is more than a certain critical night / length/when the day length is shorter than the critical day length. ii) How would short light interruption affect flowering in long-day plants? [2M] -the long day plant does not flower when grown under long night condition -unless the long night is interrupted with a brief flash of red light 9) FIGURE 11 shows part of the humoral immune response in human body. Figure 11 a) Identify the structure labelled X and Y. [2M] X: Macrophage Y: Plasma cells b) State the name and function of substance that released by structure X. [2M] Name: Interleukin-1 Function: Activate helper T cell to proliferate by mitosis to produce a clone of activated helper T cell. c) Explain the response of B cell when a virus invades our body, until secretion of antibodies. [3M] - B cells bind with activated helper T cells which has taken up and display the same antigen. - The activated B cell proliferates and differentiates into memory B cells and plasma cells. - Plasma cells secrete antibodies that specific for the antigen to initiate the response.
SULIT NO MATRIK: SB025 d) Briefly explain the consequences if the helper T cell is attacked by HIV. [2M] - Helper T cell cannot activate B cell and T cell/ cannot trigger immune response - Immune system of infected person become weaken. END OF QUESTIONS PAPER “Everyday is a chance to get better. Don’t waste it”. “You don’t have to be great to start, but you have to start to be great.”
SB 025 SB 025 Biology 2 Biologi 2 Semester 2 Semester 2 Session 2022/2023 Sesi 2022/2023 2 hours 2 jam D’ PUNCAK KOLEJ MATRIKULASI PAHANG ===================================================================
1. (a) In the Poaceae family, two species exist, Poa fawcettiae (smooth blue snowgrass) and Poa pratensis (Kentucky bluegrass). P. fawcettiae survives in the alpine area,1600 m to 2000 m above sea level and it flowers in the summer. P. pratensis meanwhile survives at temperate lowlands and it flowers in spring. (i) Define biological species concept. [2 marks] A species refers as a population whose member has the potential to interbreed in nature producing viable and fertile offspring. When they become reproductively isolated from other population, they do not produce interbreed with members of other species. (ii) State two types of isolation that result in the formation of the two species above. [2 marks] Habitat isolation Temporal isolation (iii) If a thunder storm wiped out a large proportion of the P. pratensis population, identify the factor that leads to formation of new species from the remaining population? [1 mark] Genetic drift b) A scientist studied wild birds that lived by a lake. He observed that one bird species had a beak that was adapted to extract small insects from the water. Which process would have occurred in the development of this specialized beak? [1 mark] Natural selection
2. FIGURE 1 shows the enzyme action in a chemical reaction. FIGURE 1 (a) State molecule X which binds to active site of enzyme in FIGURE 1 above. [1 marks] Cofactor (a non-protein substance) (b) Identify molecule that will bind to structure Y in FIGURE 1 above and how its affect the chemical reaction. [2 marks] Non-competitive inhibitors (have different shape from substrate and bind to allosteric site). The binding causes conformational change of active site. Substrate cannot bind to the active site enzyme-substrate complex does not form. Reduce rate of reaction. (c) A crime incident was reported to have occurred because the victim had been poisoned by liquid cyanide in his drink. This was caused by chemical reactions in cellular respiration was stopped due to exposure of enzyme to cyanide. Explain. [3 marks] Cyanide is a non-reversible inhibitor. It binds tightly and permanently to the enzyme to allosteric site or active site Changes the conformation of the active site of enzyme permanently // cause permanent damage to enzymes (enzymes will denatured) Substrate can no longer bind to the active site of enzyme.
3. FIGURE 2 shows the process of cellular respiration in muscle tissues. FIGURE 2 (a) Identify what is compound X? [1 mark] Compound X : Acetyl CoA (b) Briefly describe how A was produced from one molecule of glucose? [2 marks] 2 molecules of Glyceraldehyde-3-phosphate is oxidized causes 2NAD+ is reduced to 2NADH + 2H+ (c) What happen to Compound Y in the absence of oxygen in plant cell? [2 marks] Pyruvate will remain in cytosol Pyruvate undergo alcohol fermentation to produce ethanol. (d) Explain briefly what happen to Acetyl-CoA when it enters the Krebs cycle [8 marks] 1. Acetyl-CoA removes CoA group, remaining acetyl (2C) combines with oxaloacetate, OAA (4C) to form citrate (6C). 2. Isomerization process. Citrate then rearranged to its isomer isocitrate by removal and addition of 1 water molecule. 3. Oxidative decarboxylation process. Isocitrate oxidized, NAD+ reducing to NADH + H+. 4. The 6C compound loses a CO2 and form α-ketoglutarate (5C). 5. Oxidative decarboxylation process. α-ketoglutarate loses a molecule of CO2 forming a 4C Glucose Compound X Krebs cycle A Compound Y
6. molecule (decarboxylation). 7. The remaining 4C is oxidized before combines with Coenzyme A, to form succinyl CoA (4C). NAD+ is reduced to NADH + H+ 8. ATP produced by substrate level phosphorylation. Succinyl-CoA will lose CoA forming succinate. CoA is displaced by phosphate group which is transferred to GDP forming GTP. 9. GTP then transfer the phosphate to ADP to form ATP. 10. Oxidation. Succinate oxidized to fumarate (4C), Two hydrogen transferred to FAD cause it reduced to FADH2 11. Hydration (addition of water molecule). Fumarate rearranged into malate (4C) by adding H2O molecule. 12. Oxidation. Malate is oxidized, reducing NAD+ to NADH + H+ , regenerating oxaloacetate(4C). 13. OAA can now combine with another molecule Acetyl coenzyme A beginning a new cycle. 4. (a) FIGURE 3 shows a reaction that occurs in the thylakoid membrane. FIGURE 3 i. Explain the importance of the reaction that takes place at A. [2 marks] Synthesise ATP via chemiosmosis / proton motive force across the membrane. To reduce NADP+ into NADPH and H+ . The photoexcited electron are replaced by electron from water.
ii. Atrazine and bentazon are herbicide that prevents the flow of electrons from B which will cause severe damage to chloroplast and reduces oxygen to a chemically reactive superoxide radical. Justify the danger of exposing plant with both herbicide even though there are adequate amount of light energy. [2 marks] Photophosphorylation / Light dependent reaction / Photosynthesis will not occur. No ATP and NADPH produced. Calvin cycle cannot occur. No glucose can be produced. Plant dies. (b) Photorespiration is a major problem that occurred in C3 plants during hot and dry days. Describe what happen during photorespiration and its effect on overall photosynthesis output. [5 marks] Answers Marks Explain photorespiration During hot and dry days, C3 plants closes its stomata. 1 To conserve water. 1 Carbon dioxide / CO2 level drops but oxygen / O2 level increases. 1 Oxygen / O2 act as competitive inhibitor to carbon dioxide / CO2. 1 Enzyme RuBP carboxylase-oxygenase / Rubisco will fix oxygen / O2 to ribulose bisphosphate / RuBP instead of carbon dioxide / CO2. 1 (ANY 3) Effect on photosynthetic output Reduced formation of 3-phosphoglycerate (in Calvin cycle). 1 Producing one molecule of phosphoglycolate and one molecule of 3-phosphoglycerate / PGA. 1 Phosphoglycolate will undergo a series of reactions using oxygen / O2 and ATP while releasing carbon dioxide / CO2. 1 (ANY 2)
5. FIGURE 4 shows the oxygen dissociation curve with different oxygen carriers. FIGURE 4 (a) State the location where oxygen carrier A can be found [1 mark] Skeletal muscle cell (b) Which oxygen carrier has higher affinity for oxygen? Give the reason. [2 marks] Myoglobin/A Myglobin/A only release oxygen when partial pressure of oxygen in skeletal muscle is very low (c) State the difference between A and B [1 marks] Myoglobin /A Haemoglobin /B Consist of one polypeptide chain/ Tertiary protein structure Consist of four polypeptide chains/ Quaternary protein structure Contain one heme group Contain four heme groups Can bind with one oxygen molecule Can bind with four oxygen molecules No allosteric cooperativity/ cooperative binding Can show allosteric cooperativity/ cooperative binding Pigments found in muscle cell for oxygen storage Pigments found in erythrocytes of blood for transportation of O2 and CO2 Has higher affinity towards oxygen than haemoglobin/B Has lower affinity towards oxygen than myoglobin/A The oxygen dissociation curve (ODC) shows hyperbolic curve // ODC of myoglobin does not show cooperativity characteristic The oxygen dissociation curve (ODC) shows sigmoid curve // ODC of haemoglobin shows cooperativity characteristic 20 40 60 80 100 120 140 A O 2 P B O 2 P Po2 (mmHg) Haemoglobin saturation (%)
(d) Explain how carbon dioxide is transported in blood as bicarbonate ion. [5 marks] Answers Marks i. 60% of carbon dioxide transported in the form of hydrogen carbonate ions / bicarbonate ion. 1 ii. CO, enters the red blood cell and combine reacts with water to form carbonic acid (H2CO3). 1 iii. The reaction is catalysed by the enzyme carbonic anhydrase. 1 iv. Carbonic acid dissociates into a bicarbonate ion (HCO3 - ) and a hydrogen ion (H+ ). 1 v. HCO3 - diffuses into the blood plasma where it is carried in the bloodstream to the lungs. 1 vi. Chloride ion diffuse into red blood cell to maintain the electrochemical neutrality// chloride shift occurs. 1 vii. H + binds to Hb to form haemoglobinic acid (HHb) which prevents the acidification of the blood/prevents the Bohr effect. 1 TOTAL 7 Max 5 6. (a) FIGURE 5a shows one of the phases in the cardiac cycle. FIGURE 5a A
(i) Explain the events that occur at A. [3 marks] Blood pressure in the right ventricle is higher than right atria. Tricuspid valves are closed to prevent backflow of blood to atria. Semilunar valve are open to allow blood to flow into pulmonary artery and aorta. Lub sound is produced. (b) FIGURE 5b below shows how a hypothesis of translocation of organic substance which occurs in a plant. FIGURE 5b (i) Explain in term of water potential how an increase in the concentration of sucrose in C causes translocation to occur in structure C. [4 marks] The concentration of substances P (sucrose) in C (sieve tube elements) increases, thus lowering the water potential in C. Hence water in B (xylem) which has a higher water potential diffuses into C by osmosis. This causes an increase in hydrostatic pressure of C. 7. FIGURE 6 shows a structure of nephron and comparison of the contents of various solutes in blood plasma entering kidney tubules and discharge in urine. FIGURE 6 Solute Concentration of solute (g/litre) Blood plasma Glomerular filtrate Urine Glucose 2.0 2.0 0.0 Protein 70.0 0.0 0.0 Urea 0.3 0.3 20.0 P B C
(a) Name the process that occur at structure Q and R. [1 mark] Reabsorption (b) Describe TWO factor that contribute to efficient process at structure P. [2 marks] High hydrostatic pressure at glomerulus because diameter of efferent arteriole is smaller than dimeter of afferent arteriole. Highly coiled and long glomerulus to increase surface area for ultrafiltration High permeability of glomerulus due to present of porous capillary/ tiny pore at endothelium of glomerulus. Present of filtration slit because inner wall of Bowman’s capsule has podocytes. [ANY 2] (c) Explain why: i. Protein not found in glomerular filtrate? [1 mark] Protein is large molecule that cannot be filtered out of the blood plasma. ii. Glucose is not found in the urine excreted? [1 mark] All glucose is fully reabsorbed at proximal convoluted tubule into blood capillary (d) Hydrochlorothiazide is diuretic drug that is prescribed to treat high blood pressure. How would consuming the drug affect the volume and concentration of urine? [1 mark] High volume and diluted urine. (e) Name the hormone which regulate amount of water reabsorbed by kidney? [1 mark] Antidiuretic hormone 8. FIGURE 7 below shows the formation of action potential. FIGURE 7
(a) Name phases A and B [2 marks] A: Resting potential B: Depolarization / Rising phase of action potential (b) Describe the membrane potential at point D . [2 marks] Voltage-gated sodium channels are closed Some voltage-gated potassium channels are slowly closed Inside axon becomes more negative, fall slightly below -70 mV. Later, voltage-gated potassium channel closed. The membrane potential returns to the resting state. Any 2 (c) Briefly explain the generation of action potentials from B to C. [4 marks] B is Depolarization/ Rising phase of action potential. Depolarization open most voltage-gated sodium channels. while voltage-gated potassium channels remain closed. More Na+ influx/inflow make inside axon membrane becomes more positive than outside. C is Repolarization / Falling phase of action potential Most voltage-gated sodium channels are closed While most voltage-gated potassium channels are open. Causing K+ ions efflux/outflow As a result, inside axon becomes negative again Any 4 (d) Explain the mechanism of cocaine action at synapse [6 marks] - When the cocaine present in synaptic cleft, the cocaine bind to neurotransmitter dopamine reuptake transporter - Block the reabsorption of dopamine (neurotransmitter) back into the presynaptic membranes. - More dopamine stays in synaptic clefts - and continually binds to the receptors on the postsynaptic membrane. - Depolarization of postsynaptic membranes repeatedly causes continuous impulse transmission.
- This produces the feeling of ‘euphoria’ (very happy) & ‘energetic’ until cocaine is naturally removed from the system. - Long effect: Neurons respond to continuous impulse transmission by reducing the number of dopamine receptors in postsynaptic membranes. - Thus, more drugs are needed for the addict to experience the pleasurable effects Any 6 9. FIGURE 8 shows the mechanism of immune response. FIGURE 8 (a) Identify structure S. [1 mark] S: Helper T cell (b) How does cytotoxic T cell react with antigen? [2 marks] - Cytotoxic T cell binds to the (class I MHC antigen complex on the) infected cell - and release perforin to form pores in the infected cell membrane - and release granzymes to initiate apoptosis / death of infected cell
(c) Explain how the activated helper T cell involved in the humoral immune response that causes antigen-antibody response? [6 marks] The activated helper T cell binds with Class II MHC antigen fragment complex on B cell 1 via antigen receptor and accessory protein CD4 1 This interaction with the aid of interleukin-2 secreted by activated helper T cells activates B cells 1 The activated B cell proliferates and differentiates into memory B cells and plasma cells 1 Plasma cells secrete antibody molecules specific for pathogen 1 Antibody move to the infected area: antigen-antibody interaction occurs 1 Total 6
NAME:____________________________________ TUTORIAL GROUP:_________ INSTRUCTION: ANSWER ALL QUESTIONS 1. (a) Compare between continuous and discontinuous variation. [5 marks] • Both show differences among individuals of same species. • Both are caused by meiosis and sexual reproduction • Both occur due to the differences in DNA in different genomes caused by mutations and genetic recombination. Continuous variation Discontinuous variation Individual do not show distinct differences phenotype Individual show distinct differences phenotype Quantitative inheritance/ effect characters can be measured Qualitative inheritance/ effect characters cannot be measured Have an intermediate phenotype No intermediate phenotype Phenotype/characteristics is influenced by environmental and genetic factors// phenotypic expression is influenced environmental factor Phenotype/characteristics is influenced by genetic factors only// phenotypic expression is not influenced by environmental factor Characteristic is controlled by two or more genes/ polygenes Characteristic is controlled by one gene/ different allele at a single locus Shows a normal distribution curve Shows a discrete distribution curve Similarity = 1 Difference = 4 BIOLOGY UNIT KOLEJ MATRIKULASI MELAKA BIOBOOSTER ANSWER SCHEME SEMESTER 2 SESSION 2022/2023 CHAPTER 3-7 80 TOTAL MARKS:
(b) FIGURE 1 shows a type of natural selection. FIGURE 1 (i) Identify the type of natural selection in FIGURE 1. [1 mark] Disruptive selection (ii) What are the results of natural selection mention in 2(b)(i). [2 marks] • Natural selection that favours both extreme phenotypes • Acts against or eliminates intermediate phenotypes (iii) Give another type of natural selection. [1 mark] Directional selection /Stabilizing selection 2. FIGURE 2 shows the mechanism of enzyme action. FIGURE 2 (a) Name the action mechanism shown in FIGURE 2. [1 mark] Induced Fit (model/mechanism) Frequency of individuals Phenotypes Enzyme Enzyme Enzyme Y
(b) How does enzyme speed up rate the rate of reaction? [1 mark] Enzymes (speed up the reaction) by lowering the activation energy. -1m (c) Describe the events that occur when molecules Y binds to the enzyme. [5 marks] i. The active site of an enzyme is flexible/not fully/not exactly complementary/compatible to the conformation of substrate/molecules Y. 1 m ii. Enzyme collides with the substrate molecule. 1 m iii. Substrates/Molecules Y bind to active site of enzyme 1 m iv. to form enzyme - substrate complex. 1 m v. The binding induces the changes in the conformation of active site of enzyme. 1 m vi. The active site becomes fully/exactly complementary/compatible to the conformation of substrate/molecules Y // Substrate fit precisely to the active site of enzyme. 1 m vii. Enable the enzyme to carry out catalytic function // Catalytic reaction occurs. 1 m viii. Product is formed and released. 1 m ix. Active site of enzyme changes back to the original conformation. 1 m Total: 9 marks Max: 5 marks (d) Identify the following molecules that (i) is the non-protein component of an enzyme to enable proper functioning of an enzyme. [1 mark] Cofactor / Metal ion activators / Coenzyme / Prosthetic group (ii) decrease the rate of enzyme reaction when it is bound to the active site of enzyme? [1 mark] Competitive inhibitor
3. FIGURE 3 shows the flow of electrons in specific pathway of cellular respiration. FIGURE 3 (a) Name process X and Y. [2 marks] X: Electron transport chain Y: Chemiosmosis (b) Compare between the characteristic of coenzyme Q and cyt c. [2 marks] Similarity: They are mobile electron carrier Difference: Coenzyme Q is non protein/lipid electron carrier while cyt c is protein electron carrier (c) How NADH + H+ and FADH2 transfer the electron to the electron carrier? [2 marks] • NADH + H+ transfer its electron to NADH dehydrogenase and oxidized to NAD+ . NADH dehydrogenase will reduced. • FADH2 transfer its electron to succinate dehydrogenase and oxidized to FAD. Succinate dehydrogenase will reduced. (d) What will happen to efficiency of ATP synthesis if structure S is more permeable to proton? Why? [2 marks] • The efficiency of ATP synthesis decrease • Because H+ / proton will flow back directly through structure S/inner membrane of mitochondria without passing through the ATP synthase into matrix of mitochondria. (e) Explain what happen if there is a leakage of proton in structure P. [4 marks] • Proton from intermembrane space will diffuse out to the cytoplasm through the outer membrane.
• No accumulation of proton in the intermembrane space// No proton gradient//Proton motive force is not created. • Proton cannot diffuse back into mitochondrial matrix through ATP synthase • ATP cannot be generated// No synthesis of ATP. 4. FIGURE 4 shows the non-cyclic photophosphorylation in chloroplasts. FIGURE 4 (a) Name X and Y. [2 marks] X: Photosystem II / PSII Y: Photosystem I / PSI (b) Give importance of Z. [1 mark] To increase high energy level of electron (c) (i) Identify W. [1 mark] NADPH (+ H+ ) (ii) Describe the role of W during light independent reaction. [ 2 marks] - act as reducing agent - transfer electron to 1,3-bisphosphoglycerate and reduce to glyceraldehyde-3-phosphate
(d) Differentiate between non-cyclic photophosphorylation and cyclic photophosphorylation . [5 marks] Non-cyclic photohosphorylation Cyclic photophosphorylation Produce ATP and NADPH + H+ Only produce ATP Involve PS II and PS I // P680 and P700 Involve PS I/P700 only Water act as the first electron donor The electron donor is PS I/P700 Last electron acceptor is NADP+ Last electron acceptor is PS I/P700 Photolysis of water occur and O2 is released Photolysis of water do not occur and O2 is not released Involve non-cyclic electron flow Involve cyclic electron flow (e) When does cyclic photophosphorylation occur? [1 mark] - When an adequate of ATP during light independent reaction/Calvin cycle 5. FIGURE 5 shows the cross section of a pair of guard cells. FIGURE 5 (a) Based on FUGURE 5, (i) Name structure labelled A and state its function. [2 marks] A: Stoma -1m Allow water evaporation/water loss via transpiration // -1m Allow gaseous exchange -1m (ii) Describe the structure labelled B? [1 mark] Thicker and less elastic inner wall A B Any 1
(b) Explain the stomatal opening by using the role of potassium ions. [3 marks] ▪ ATP produced in the guard cells during light dependent reaction is used to pump the proton/hydrogen ions/H+ out of the guard cells. ▪ Potassium ions/K+ diffuse from the adjacent / subsidiary cells into the guard cells. ▪ Guard cells become hypertonic // water potential becomes more negative / decreases (due to accumulation of K+ ). ▪ Water diffuses into the guard cells by osmosis. ▪ Thus, the guard cells become turgid, widening the stomatal opening. 6. FIGURE 6 shows cells from the root of Trapa natans. Water travels between the cells by two different pathways. FIGURE 6 (a) Name the pathways represented by X and Y. [2 marks] X : Symplast route/pathway Y: Apoplast route/pathway Any 3
(b) Name another pathway beside X and Y pathways. [1 mark] Transmembrane route/pathway (c) Describe briefly the water movements for (a) and (b) from surrounding soil to the endodermis of Trapa natans. [3 marks] X/ Apoplast route: Water move along the cell wall. Y/ Symplast route: Water move through the cytoplasm/ plasmodesmata of the cell Transmembrane route: Water move through vacuole/ protoplast of the cell. 7. Define homeostasis and discuss blood glucose regulation in human after meal. [6 marks] - Homeostasis is the ability of an organism to maintain a (relatively) constant/stable internal environment / dynamic constancy of internal environment (1 mark) - After meal, blood glucose level increase / higher than normal (hyperglycemia) - β-cells secrete insulin - Insulin stimulates uptake of glucose from the blood into body cells / muscle cells, liver and adipose tissue. - Glucose is converted to glycogen by glycogenesis in muscle cell / liver - Glucose is converted to fat in adipose tissue. - Insulin inhibits liver cells from releasing glucose. - The blood glucose level decreases back to normal. (max 5 marks) 8. (a) Explain the mechanism of impulse transmission across into the synaptic cleft. [6 marks] - The arrival of an action potential at axon terminal depolarizes the presynaptic membrane / synaptic knob / synaptic terminal - (The depolarization) opens voltage-gated Ca2+ channels - Triggering an influx of Ca2+ into presynaptic membrane / synaptic knob / synaptic terminal - Causes synaptic vesicles to fuse with the presynaptic membrane. - Synaptic vesicle releasing the neurotransmitter into the synaptic cleft by exocytosis. - The neurotransmitter diffuses across the synaptic cleft. - Attaches/bind to a specific receptor site on the ligand gated ion channel in the postsynaptic membrane. - Triggers opening ligand gated ion channel. - Allowing Na+ inflow into the postsynaptic neurons. - This leads to a depolarization of the postsynaptic membrane.
- The depolarization response is known as an excitatory postsynaptic potential (EPSP). - If the EPSP is large enough to reach the threshold level, an action potential/impulse is generated in the postsynaptic neuron. ( b) FIGURE 7 illustrate a mechanism of hormone M in liver cell. FIGURE 7 (i) State the type and example of hormone M. [2 marks] Peptide/Polypeptide/Protein Hormone M: Glucagon// Epinephrine/Adrenaline (ii) Identify S and state the reason why hormone M required S to relay its signal inside of cell. [2 marks] S: cAMP Reason: Because Hormone M cannot diffuse across the plasma membrane//is water soluble molecule 9. (a) FIGURE 8(a) below shows the basic structure of an antibody. Hormone M S Cellular Response
FIGURE 8(a) (i) Label the structures of J, K and L. [3 marks] J: antigen-binding site K: light chain L: heavy chain (ii) Explain why the conformation of J on the antibody molecule is different from one antibody to another. [1 mark] (The configuration of the antigen-binding sites of) each antibody is specific to a particular antigen. (b) FIGURE 8(b) shows the antibody concentration in human after exposure to antigen A and antigen B. FIGURE 8(b) (i) State the level of development of immune response at Y and Z. X Z Y J L K
[2 marks] Y: Secondary immune response to antigen A Z: Primary immune response to antigen B (ii) State TWO advantages of the antibodies produced at Y compared to Z. [2 marks] - Greater affinity towards antigen - Faster response - Long-lived - Concentration is higher (iii) Based on FIGURE 8(b), why is there a high antibody concentration at Y? [2 marks] - Memory B cells recognize the same antigen/ pathogen faster. - Memory B cells proliferate rapidly into plasma cells to produce antibody.