ANSWER SCHEME PRA PSPM 2 SB025 SESI 2022/2023 Question Answer scheme Marks 1. a) i. ii. iii. b. Disruptive selection It favours both extreme phenotypes/ lightest coloured oyster and darkest coloured oyster// Act against intermediate phenotype/ intermediate-coloured oysters / intermediate phenotypes were eliminated Doubling the chromosome number/ chromosome doubling of the sterile hybrid// Fusion of the unfertilized gamete of orange and pomelo 1 1 1 1 1 1 Total = 6 2. *Can be anabolic or catabolic reaction Marking criteria for the diagram 1 m given for correct labelling in a correct sequence: Enzyme//Active site, substrate, enzyme-substrate complex and products. 1 m given for shape of active site does not exactly complementary to the substrate. 1 m given for changing of shape of the active site as the substrate binds and as the products are released from the enzyme active site. - The binding of substrate into/ to the active site of the enzyme induces (a slight) conformational change of the enzyme active site // There is also a slight alteration to the shape of the substrate. - The active site becomes fully complementary with the substrate, - forming an enzyme-substrate complex. - Reaction occurs producing products. - Product is released and enzyme active site returns/ changes back into the original shape/ conformation 3 1 1 1 1 1 (max=3) substrate Enzyme Active site Enzyme-substrate complex Products *This part is not necessary.
Total= 6 3. a) b) 4. - The transfer of electrons from X/ NADH to Coenzyme Q is blocked // Electrons from X/ NADH cannot enter the electron transport chain - ATP production is reduced / decreased (Note: ATP production does not stop) - because ATP cannot be produced from the oxidation of X/ NADH // because ATP can only be produced from the oxidation of FADH2 - 8 ATP (per glucose molecule is produced in the cell) (= 4 ATP from substrate-level phosphorylation + 4 ATP from 2 FADH2 via oxidative phosphorylation) - Cytochrome c oxidase - The transfer of electrons to oxygen is blocked // The electron transport chain is blocked // The transfer of electrons along the chain is blocked - ATP production stops - In the absence of oxygen, pyruvate (from glycolysis) will undergo lactate / lactic acid fermentation // lactate / lactic acid fermentation occurs. - Pyruvate is reduced directly by NADH - to form lactate / lactic acid, - without/does not release of carbon dioxide - NADH is oxidized to NAD+ - To regenerate the supply of NAD+ for continuation of/next glycolysis. - Only 2 ATP are generated via substrate-level phosphorylation in glycolysis - Lactate / lactic acid is converted back to pyruvate in the liver. 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 (any 6) Total= 13 5. a) b) c) By closing the stomata (during day) − Carbon dioxide acceptor in C3 plant is RuBP/ ribulose bisphosphate while carbon dioxide acceptor in C4 plant is PEP/ phosphoenolpyruvate and RuBP/ ribulose bisphosphate − Enzyme involved in carbon fixation of C3 plant is RuBP carboxylaseoxygenase/Rubisco while in C4 plant is RuBP carboxylase-oxygenase/Rubisco and PEP carboxylase Malate is transported out of the vacuole into the cytoplasm in the morning (which then will undergo decarboxylation) 1 1 1 1 6. − Involve only photosystem I / PSI/ P700 − produces only ATP − Photon/ Light energy is absorbed by (antenna pigments of) PS I / P700. 1 1 1
− Chlorophyll a P700 undergoes photoactivation. − It releases photoexcited electrons to primary electron acceptor/ The electrons released from PSI are accepted by primary electron acceptor // Electrons in P700 are excited to a higher energy level − (Thus) creates electron deficiency in PS I/ P700 − From primary electron acceptor, electrons are passed through/ along electron transport chain // ferredoxin/Fd, cytochrome complex and plastocyanin/Pc and/before return to PS I/ P700 − During the transfer of electrons/ through electron transport chain, energy is released and been used to form ATP from ADP + Pi // energy is released and is used for synthesis of ATP by chemiosmosis 1 1 1 1 1 (max 5) 7. a) b) c) d) Cooperativity in O2 binding and release // Cooperative binding of O2 (with haemoglobin) ▪ (Actively respiring tissue produce more carbon dioxide,) carbon dioxide/ CO2 combines/ reacts with water to form carbonic acids/ H2CO3 / Increase the formation of carbonic acid/ H2CO3 ▪ Dissociation of carbonic acid forms hydrogen ions/ H+ , (in the red blood cell)/ hydrogen ions/ H+ bind to haemoglobin to form haemoglobinic acid // Carbonic acid dissociates into bicarbonate ions (HCO3 - ) and H+ ▪ causes blood plasma to become more acidic/ lower (blood) pH ▪ Haemoglobin affinity for oxygen decreases/ Reduce the affinity of haemoglobin for oxygen// (more) oxyhaemoglobin dissociate to release (more) oxygen (for respiring tissues) ▪ More oxygen to meet demand for (aerobic / cellular) respiration/more oxygen for respiring tissues ▪ Chemoreceptors in the carotid and aortic bodies are sensitive to small changes in the concentration of PCO2 and PO2 in the blood // Peripheral chemoreceptors in carotid and aortic bodies are stimulated (by the drop in partial pressure of O2 / PO2) ▪ (Decrease in PO2 / increase of PCO2 level stimulates) the chemoreceptors to send (nerve) impulses to the inspiratory centre in the medulla oblongata ▪ The inspiratory centre sends out (nerve) impulse to diaphragm via the phrenic nerves and external intercostals muscle via thoracic/ intercostal nerve ▪ causing diaphragm and external intercostal muscles to contract ▪ Thoracic cavity expand ▪ This increases the rate of inspiration // Inspiration takes place/ occurs 1 1 1 1 1 1 1 (Any 3) 1 1 1 1 1 1 (a)
(Any 4) Total=9 8. a) b) c) T Atria - In P wave, atrial depolarization occurs. - Sinoatrial/SA nodes generate impulse and spread to both atria// Impulses spread across both atria - Both atria contract - Impulse spreads to AV node - In QRS complex, ventricular depolarization occurs. - Impulse spreads from the atrioventricular/AV node to both ventricles. // Impulses spread across both ventricles - Ventricles contract after the QRS complex appears. // Both ventricles contract // Repolarization of atria/ both atria relax - In T wave, ventricular repolarization occurs. - Electrical changes trigger relaxation of atria and ventricles. // No transmission of impulse in both atria and ventricles, so both atria and ventricles relax // Recovery of ventricles at the end of contraction. 1 1 1 1 1 1 Any 2 1 1 1 Any 2 1 1 Total=7 9. a) b) c) i. ii. iii. Secretion and reabsorption − Blood cells and large proteins cannot exit/ pass through the glomerulus // Glomerulus prevents the exit of large molecules // Blood cells and large proteins are too large to pass through the glomerulus// Glomerulus is impermeable to blood cells and large proteins − Due to the fenestrations / fenestrae / fenestrated endothelium // basement membrane / basal lamina // filtration slits of podocytes // slit diaphragms of glomerulus (Type I // II) Diabetes / Diabetes mellitus − Urine of patient A contains no glucose but urine of patient B contains glucose// Patient B has excess of glucose in their blood − Patient B fails to produce hormone insulin// Excess glucose in patient B is secreted into the urine / secretion removes excess glucose // Glucose is not fully reabsorbed (at proximal tubules) − (Water / Osmotic content of blood of patient A increases // Blood osmotic pressure decreases) Posterior pituitary gland secretes less/ low level of antidiuretic hormone/ ADH / vasopressin // Less ADH is produced − Collecting duct become less permeable to water // Less water is reabsorbed into blood by osmosis − Large volume of diluted urine is produced 1 1 1 1 1 1 1 1 1 (Any 2) Total=7
10. i) ii) iii) iv) v) − Action potential reaches/ arrives at presynaptic knob and depolarisation of presynaptic membranes occurs − Voltage-gated Ca2+ channels (on presynaptic membrane) opens allowing Ca2+ to diffuse into the knob presynaptic membrane − Elevated levels of Ca2+ // The influx of Ca2+ triggers the fusion of neurotransmitter vesicles to pre-synaptic membrane and neurotransmitters release into synaptic cleft by exocytosis // causes synaptic vesicles to fuse with the presynaptic membrane and release neurotransmitters into the synaptic cleft by exocytosis − Binding of neurotransmitter to the receptor on ligand-gated Na+ /ion channel/ II causes the channel to open − Na+ influx/diffuse/flow into the postsynaptic membrane causing depolarization of postsynaptic membrane − If the depolarization reaches threshold level, an action potential is produced in the postsynaptic neuron − Cocaine attaches/ binds/ blocks to the dopamine transporter on the presynaptic membrane − blocking the normal recycling uptake/ reabsorption of dopamine back into the presynaptic terminal/ synaptic knob of presynaptic neuron − Dopamine accumulates at synaptic cleft and continuously bind to receptors at the postsynaptic membrane − Depolarization of postsynaptic membrane occurs continuously (and produces more nerve impulses/ action potential) − Produces the feeling of euphoria // Increases the activation of the pleasure/ reward pathway and causes euphoria − Tropomyosin activated after Ca2+ binds to troponin and exposing myosin binding site on actin molecule // Ca2+ (released from sarcoplasmic reticulum) bind to troponin and then the tropomyosin is shifted away to expose the myosinbinding site (on actin filament) − ATP on myosin head hydrolysed into ADP and Pi and the energy released converts myosin head to high-energy configuration − The (high-energy) myosin head binds to the myosin binding site on actin filament, forming cross-bridge Note: Comparison//differentiation answers must be written in paragraph form. Avoid using table format like below. The table below only highlighting the point of comparison // differentiation. Steroid Protein-type // Non-steroid Hormone can pass through/diffuse the cell membrane of target cell Hormone cannot pass through the cell membrane of target cell The hormone binds with a specific receptor in the cytoplasm of target cell The hormone binds with a specific receptor on the cell surface membrane of target cell 1 1 1 (Any 2) 1 1 1 1 1 1 1 1 (Any 4) 1 1 1 (Any 2) 1 1
The-hormone receptor complex binds to (regulatory site of) DNA The hormone-receptor complex binds to and activates a G protein Transcription and translation occur. No transcription and translation occur There is no involvement of G protein and adenylyl cyclase (thus no cAMP is produced) // No production of cAMP // Involve hormone only The binding of hormone to receptor activates G protein and adenylyl cyclase (thus cAMP is produced) // cAMP is produced from ATP // Involve hormone and cAMP There is no cAMP to activate enzyme cascade reactions // There is no cascade event cAMP acts as a second messenger to activate the enzyme cascade reactions // Cascade event occurs 1 1 1 1 (Any 3) Total=14 11. a)i) ii) b) c) - Activated helper T/ TH cells secrete IL-2 to activate B lymphocytes/ B cells (to proliferate) - Activated B lymphocytes/ B cells proliferate and differentiate into plasma cells and memory B cells - Antibodies bind to antigens on the surface of a foreign cell/ pathogen - This activates the complement system - The activated complement system then forms a membrane attack complex that forms pores in the foreign cell’s/pathogen membrane - water and ions move in/ enter the cell - the foreign cell lyses/destroyed// causes the cell to swell and lyse - The presence of IgM in the blood - IgM is the main antibody produced in primary immune response - HIV destroys/ infects helper T cell/TH cell (or CD4 helper T cell, or CD4 cell). - No activation of cytotoxic T cell (for cell mediated immune response) and B cell (for humoral immune response) 1 1 1 1 1 1 1 (Any 3) 1 1 1 1 Total=9 Total 80
NAME:____________________________________ TUTORIAL GROUP:_________ ANSWERS SCHEME INSTRUCTION: ANSWER ALL QUESTIONS 1. FIGURE 1(a) shows a population of rabbits living in a rocky environment. In an environment surrounded with black and white rocks, the black fur and white fur rabbits could easily camouflage themselves from the predators. The grey rabbits therefore would be seen by predators and that phenotype would become lower in frequency. FIGURE 1(a) i. Based on the information above, draw suitable curve in FIGURE 1(b) to illustrates the size of the rabbit population as a result of natural selection that occur in their environment. [2 marks] FIGURE 1(b) Correct curve – 1 Correct label and selection pressure – 1 ii. What is the type of natural selection that you draw in FIGURE 1(b)? Disruptive selection [Dependent to question 1(i)] BIOLOGY UNIT KOLEJ MATRIKULASI MELAKA BIOBOOSTER ANSWER SCHEME SEMESTER 2 SESSION 2022/2023 CHAPTER 3 -7 TOTAL MARKS: 43 Black White Grey Population Size Phenotypes Population after selection Population before selection Selection pressure
iii. Gives one similarity and two differences between the founder effect and bottleneck effect. Similarity i. Both change alleles/genotypes frequencies by chance// both phenomena cause some alleles completely eliminated/disappear from the populations ii. Both reduce genetic variation and lead to speciation. iii. Both result in new population carries different alle frequency that does not represent original population. Differences (note to student : don’t answer in table form) Founder Effect Bottleneck Effect i Phenomenon which small group of individual isolated from the original population. Phenomenon which occurs when rapid/sudden decrease in population size. ii. Occur due to migration of individual from original population. Occur due to natural disaster. 2. (a) FIGURE 2(a) shows a model for enzyme action mechanism. FIGURE 2(a) Explain the enzyme action based on the mechanism shown in FIGURE 2(a). [3 marks] - Active site of enzyme is not exactly complementary to the substrate. - The binding of substrate induce the conformational changes to active sites of enzymes. - Active site become fully complementary to the substrate. - (Products is formed and) enzyme’s active site change back to its original conformation Product Substrate Enzyme
(b) FIGURE 2(b) shows three different enzyme reaction, P and Q represent two types of inhibitor. FIGURE 2(b) i. Name inhibitors P and Q. [1 mark] P : Non-competitive inhibitor Q : Competitive inhibitor ii. Malonate is an inhibitor of the enzyme succinate dehydrogenase. How would you determine whether malonate is a competitive or noncompetitive inhibitor? [2 marks] o (In the presence of malonate) increase the concentration of the substrate/ succinate o If the rate of reaction increase, malonate is a competitive inhibitor. 3. (a) FIGURE 3(a) shows a schematic diagram of cellular respiration. Glucose FIGURE 3(a) (i) Name Phase 1. [1 mark] Energy investment phase (ii) Describe how ATP is produced at Phase 2. [2 marks] ● By substrate-level phosphorylation. ● an inorganic phosphate is transferred from substrate / 1,3 bisphosphoglycerate/ phosphoenolpyruvate to ADP (to form ATP). ADP ATP Phase 1 ADP ATP Phase 2 Pyruvate
(b) FIGURE 3(b) shows pathway P and cycle Q in aerobic respiration. FIGURE 3(b) (i) Name pathway P and cycle Q. [2 marks] Pathway P : Pyruvate oxidation Cycle Q : Krebs cycle/ Tricarboxylic acid cycle / Citric acid cycle (ii) Describe the significance of NADH and FADH2, production in cycle Q? [2 marks] ● To transfer high energy electrons/carries electron to the electron transport chain ● to generate more ATP (via oxidative phosphorylation) (c) FIGURE 3(c) shows the components of the last stage of aerobic respiration in the inner membrane of mitochondria. FIGURE 3(c) Cycle Q Y X
(i) Name the stage shown in FIGURE 3(c). [1 mark] Oxidative Phosphorylation (ii) Explain how ATP is synthesized at Y? [2 marks] ● H+ ions/hydrogen ions diffuse from intermembrane space into the mitochondrial matrix (down concentration gradient) through ATP synthase ● and release energy to activate ATP synthase (enzyme). ● (Activated) ATP synthase catalyzes the formation of ATP (from ADP and inorganic phosphate.) (iii) Electron transport chain is completed when molecule R is reduced to molecule S. What is the importance of molecule R. [1 mark] Accept electrons to form water// as final electron acceptor (iv) Name TWO products of the process in FIGURE 3(c)? [2 marks] ATP Water molecule/H2O 4. (a) FIGURE 4 below shows a schematic diagram of electron flow in light dependent reaction. FIGURE 4 (i) Name the reaction that replaces the loss of electron in A? [1 mark] Photolysis of water (ii) What is the form of energy represented by C? Briefly describe how C is produced? [3 marks] C is ATP Photosystem X Photosystem Y
Photoexcited electrons are passed along electron transport chain / plastoquinone, cytochrome complex and plastocyanin Energy released is used to synthesis ATP from ADP + Pi by chemiosmosis (iii) What is the role of the substance that is used to generate D? [1 mark] Last electron acceptors // NADP+ accepts electron and proton/H+, (and reduced to NADPH (+ H+ )). (b) Explain the mechanism how CAM plant overcome photorespiration. [4 marks] i. At night, carbon dioxide binds with phosphoenolpyruvate to produce oxaloacetate ii. Catalysed by PEP carboxylase iii. Oxaloacetate reduced into malate by NADPH + H+ iv. Malate is stored in vacuole of mesophyll cell throughout night v. During day, when stomata is closed, malate is transported to chloroplast of mesophyll cell vi. Malate undergoes decarboxylation to form pyruvate vii. Carbon dioxide released is used in Calvin cycle Any 4 points 5. (a) FIGURE 5(a) shows the oxygen dissociation curve of respiratory pigment R and pigment S. FIGURE 5(a) (i) Name pigment R and pigment S. [2 marks] Pigment R: Myoglobin Pigment S: Hemoglobin (ii) Name the tissue where pigment R can be found. [1 mark] Skeletal muscle (iii) What is the difference between the oxygen dissociation curve for pigment R and S. [1 mark]
- Oxygen dissociation curve of pigment R is hyperbolic while oxygen dissociation curve of pigment S is a sigmoid shape. - Pigment R reaches saturation point at a low partial pressure of oxygen while pigment S reaches saturation point at a higher partial pressure of oxygen // Pigment R has a higher affinity to oxygen than in pigment S. Any 1 (b) FIGURE 5(b) shows the oxygen dissociation curve for hemoglobin, P, and Q. FIGURE 5(b) (i) Based on FIGURE 5(b), calculate the amount of oxygen being used by tissues for oxygen dissociation curves of P and Q at 30 mm Hg partial pressure of oxygen. [2 marks] P : (99 – 60)% = 39% Q : (99 – 40)% = 59% (ii) Explain how strenuous exercises can affect the oxygen dissociation curves and the oxygen supply to body tissues. [3 marks] - due to the increase of partial pressure of CO2 // blood pH lowered in body - oxygen dissociation curve shift to the right. - more oxyhemoglobin is dissociated to release oxygen to body tissues//affinity of hemoglobin towards oxygen is low
NAME:____________________________________ TUTORIAL GROUP:_________ ANSWERS SCHEME INSTRUCTION: ANSWER ALL QUESTIONS 1. (a) FIGURE 1(a) shows a structure of a human heart. The arrow shows the contraction of the heart chamber. FIGURE 1(a) (i) Give TWO differences between blood flowing out of B and C. [2 marks] - B/aorta contains oxygenated blood (from left ventricle) while C/pulmonary artery contains deoxygenated blood (from right ventricle). - Blood from B/aorta will be sent to all parts of the body while blood from C/pulmonary artery will be sent to the lungs. (ii) Explain what happen to valve in the heart after the stage shown in Figure 1(a) [1 mark] - Bicuspid and tricuspid valves closed and semilunar valve open BIOLOGY UNIT KOLEJ MATRIKULASI MELAKA BIOBOOSTER ANSWER SCHEME SEMESTER 2 SESSION 2022/2023 CHAPTER 8 - 11 37 TOTAL MARKS:
(b) FIGURE 1(b) shows the ECG in human heart. FIGURE 1(b) (i) Which step shown in FIGURE 1(b) that leads to“dub” sound in the heart. T [ 1 mark ] (ii) How ECG wave is affected when the blood pH decrease? The distance within wave 1 and wave 2 is shorter. [1 mark ] (c) In translocation process, explain how sink cell is always has low concentration of sucrose compare to source cell. [2 marks] - Sucrose is always produce at source cell but not in sink cell. - Sucrose is use for growth and metabolism or converted into starch at the sink cell. 2. FIGURE 2 shows a part of human nephron. FIGURE 2 X Y Z
(a) Name cell X, cell Y and structure Z. [3 marks] X: Simple squamous epithelium Y: Podocyte Z: Glomerulus (b) Cell Y is not tightly arranged. Explain how this arrangement facilitate the transportation of substances from structure Z into Bowman’s capsule. [2 marks] • Filtration slits (between foot process/ pedicel on cell Y) prevent large molecules / red blood cell / plasma protein to pass through. • Small molecules / water / glucose / NaCl are able to pass through the filtration slits. (c) Alcohol consumption will make the collecting duct less permeable to water which will affect kidney output. Explain briefly this condition. [2 marks] • Alcohol inhibits the release of antidiuretic hormone/ ADH • cause more diluted and large volume of urine produced. 3. a) FIGURE 3(a) shows a sarcomere of humans skeletal muscle. FIGURE 3(a) i) Identify structure R [1 mark] Thin/ actin filament ii) Name the structure on P that will interact with R [1 mark] Actin binding site iii) Are the muscle above contracting? Explain your answer. [3 marks] • No. • No cross bridge between myosin head and actin • H zone is visible • Actin filament does not overlap in H zone
b) FIGURE 3(b) shows a type of hormonal action. FIGURE 3(b) i) Name the mechanism of hormone A and give ONE reasons of your answer. [2 marks] cAMP activation Because hormone A cannot pass through the cell membrane/ bind to receptor on cell surface// needs to activate G protein and enzymes/ adenylyl cyclase// required second messenger/ cAMP ii) Compare the mechanism of hormone action between hormone A and estrogen? [ 2 marks] Hormone A Estrogen Similarities Both act on specific target cell// Both bind to specific receptor// Both form hormone-receptor complex Differences Non-steroid hormone Steroid hormone Unable to diffuse the plasma membrane Able to diffuse the plasma membrane Bind in the receptor on plasma membrane Bind in the receptor in cytoplasm Hormone-receptor complex activates a G protein which in turn activates adenylyl cyclase Hormone-receptor complex binds to DNA for gene activation Involve two messengers/ hormone and cAMP Involve one messenger/hormone only
iii) What happen to molecule E if molecule D is absent? [1 mark] If molecule D absent, molecule E unable to activate phosphorylase kinase/enzymes c) FIGURE 3(c) shows the response of a flowering plant to different photoperiods. Shaded bars indicate periods of darkness and unshaded bars indicate periods of light. FIGURE 3(c) i) State the type of plant above. Explain your answer. [2 marks] Long day plant. Plant that flower when the period of darkness is shorter/ daylight is longer. ii) What flowering response in c (i), when the dark period is interrupted with a flash of red light? Explain your answer [2 marks] Plant will flower The light interruption shortens the dark period and causes the conversion of Pr into Pfr 4. FIGURE 4 shows part of mechanism of an immune response. FIGURE 4 0 4 8 12 16 20 24 Non flowering Flowering
(a) Name the cell Q and structure R. [ 2 marks] Q: Plasma cell R: Antigen binding site (b) (i) Name the class of antibody secreted by cell Q when a person is exposed to a pathogen for the first time. [1 mark] IgM (ii) Give ONE advantage of the antibodies produced after second infection in immune response S. [1 mark] - Greater affinity towards antigen - Faster response - Long lasting - Higher in concentration (c) Substance X is secreted by macrophage to trigger the immune response. Name the substance X and explain the effect of substance X on immune response T. [ 3 marks] - Substance X is Interleukin -1/ IL-1 - Substance X activate the Helper T cell to secrete interleukin -2/ IL-2 to activate cytotoxic T/ Tc cell - The activated Tc cell bind to infected cell and destroy /kills it. (d) (i) The Human Immunodeficiency Virus (HIV) is retrovirus that infect cells of the human immune system. Which cell does HIV attack? [1 mark] Helper T cell (ii) Suggest why HIV still can persists in human body even though after immune response is triggered towards HIV. [1 mark] Because HIV has very high mutation rate
BLUE 3 SB025 2022/2023 ANSWER SCHEME 1 (a) The graph in FIGURE 1 below shows the result of a type of selection for a population. FIGURE 1 (i) State the type of natural selection shown in FIGURE 1. [1 mark] Disruptive selection (ii) Name ONE example for natural selection stated in 1 (a) (i). [1 mark] Varieties of beak size and shapes of Galapagos finches (iii) Artificial selection has been practiced by humans for many centuries. Give ONE difference between natural selection and artificial selection. [1 mark] Natural selection Artificial selection Selection agent is environment// does not involve human intervention Selection agent is human// involve human intervention Random event or occur by chance Planned event, on selected organism Selected varieties are strong, robust or tough varieties produced Selected varieties are defective because of inbreeding 1/0 1/0 1/0 Any 1 Selection against the intermediate Number of individuals
BLUE 3 SB025 2022/2023 ANSWER SCHEME (b) “A species of organism is separated into two populations by a mountain range. Mating between individuals of the two populations produces sterile progeny.” (i) Based on the statement above, identify the mode of speciation. [1 mark] Allopatric (speciation) (ii) State ONE similarity between the mode stated in 1 (b) (i) and the other mode of speciation. [1 mark] - The resultant of new species in both processes are incapable of interbreeding with the pre-existing species. - Both processes caused two different populations become reproductively isolated. Any 1 (iii) Give ONE factor that involved in the formation of a new species. [1 mark] Reproductive isolation / genetic drift / hybridization / adaptive radiation 2 (a) FIGURE 2 shows an enzyme and substrate action mechanism. FIGURE 2 i) Name the mechanism of enzyme action in FIGURE 2. [1 mark] Induced fit model 1 Enzyme + substrate Transition state Enzyme + products
BLUE 3 SB025 2022/2023 ANSWER SCHEME ii) What happen when substrate bind to active site of enzyme. [2 marks] The binding induces a slight change in the shape/conformation of active site of enzyme 1 Allowing the substrate fit precisely to active site of enzyme // Active site of enzyme become fully complementary with (shape of) substrate 1 iii) What will happened to the rate of reaction if substrate is added excessively? [1 mark] Rate of reaction become constant // remained unchanged 1 (b) Farmers can use insecticides to remove parasites from their sheep. Some of these insecticides act as non-competitive inhibitors of the enzyme acetylcholinesterase. Suggest and explain how an inhibitor of acetylcholinesterase could work. [2 marks] -Inhibitor binds to allosteric site of enzyme / acetylcholinesterase 1 -Substrate/Acetylcholine, cannot bind to active site of enzyme/ acetylcholinesterase// Active site of enzyme change conformation/shape 1 -No / few enzyme-substrate complexes form // rate of (enzymatic) reaction decrease 1
BLUE 3 SB025 2022/2023 ANSWER SCHEME 3 (a) The Krebs cycle with its sequential reactions are shown in the FIGURE 3 below. FIGURE 3 (i) State the name of reaction IV. [1 mark] Oxidative decarboxylation (ii) Identify T and explain the formation of compound T [2 marks] - T is citrate - T/citrate is formed when (2C) acetyl group/compound S combine with (4C) oxaloacetate by condensation process (iii) State the importance of Krebs cycle. [2 marks] - To produce electron carriers/reducing agents which are NADH and FADH2 to enter electron transport chain -To generate ATP via substrate-level phosphorylation
BLUE 3 SB025 2022/2023 ANSWER SCHEME (b) The complete oxidation of one glucose molecule is summarised in the FIGURE 4 below. FIGURE 4 (i) Name P and R. [2 marks] P: Glycolysis R:Oxidative phosphorylation// electron transport chain and chemiosmosis (ii) How many ATP are produced at P and R? [2 marks] P: 2 net ATP molecules / 4 ATP molecules R: 34 ATP molecules (iii) How many NADH and FADH2 are produced at P and Q? [2 marks] P Q NADH 2 6 FADH2 0 2 1/0 1/0 P
BLUE 3 SB025 2022/2023 ANSWER SCHEME (iv) Where does R occur? [1 mark] Inner membrane of mitochondria/ Inner mitochondrial membrane (v) What happens to the electron at R? [1 mark] Electron will be transferred/ passed through a series of electron carrier/ electron transport chain and involve redox reaction// Oxygen/O2 molecule is the final electron acceptor that accepts the electron which then combines with hydrogen ions/H+ to form water 4 (a) A study was conducted to observe the effect of photorespiration in potato cells in different time interval of day. TABLE 1 below shows the data from the observation. Time interval 8.00 am 12.00 pm 4.00 pm Concentration level of glucose (mM/ml chlorophyll) High Low High Photorespiration level Low High Low TABLE 1 Based on TABLE 1, explain how high level of photorespiration in potato cells during time interval 12.00 pm can affect concentration level of glucose. [4 marks] -During time interval 12.00 pm / hot and dry day, stomata close (to conserve water) 1 -causes concentration of O2 is higher than concentration of CO2 (in leaf potato cells). 1 -RuBP carboxylase-oxygenase / Rubisco catalyse the fixing of O2 to Ribulose bisphosphate (instead of CO2) // RuBP carboxylase-oxygenase / Rubisco act as oxygenase (instead of carboxylase) 1 -forming phosphoglycolate (2C) and 3-phosphoglycerate / PGA (3C) (for each fixation of O2). 1
BLUE 3 SB025 2022/2023 ANSWER SCHEME -As amount 3-phosphoglycerate / PGA reduce (by half), concentration level of glucose also reduce/low (by half) 1 (b) Some plants such as cactus has alternative mechanism of carbon fixation to reduce photorespiration. Explain carbon fixation that occurs in cactus during the day. [5 marks] -During the day, stomata are closed (to conserve water) 1 -Malate that stored in the vacuole during night time is transported back into chloroplasts 1 -Malate undergo decarboxylation to form pyruvate and release CO2 1 -CO2 enters Calvin cycle to form 3-phosphoglycerate (catalyze by RuBP carboxylase-oxygenase / Rubisco) 1 -3-phosphoglycerate then convert to glyceraldehyde-3-phosphate / G3P that later exit Calvin cycle to form glucose 1 -Pyruvate then is phosphorylated by ATP and convert again to phosphoenolpyruvate / PEP. 1 5 (a) FIGURE 5 shows the oxygen dissociation curves for hemoglobin at different partial pressure of carbon dioxide (Pco2). FIGURE 5
BLUE 3 SB025 2022/2023 ANSWER SCHEME (i) State the percentage of oxygen dissociated from hemoglobin at partial pressure of oxygen (Po2) 40 mmHg and 70 mmHg when Pco2 is 15 mmHg. [2 marks] Po2 40 mmHg : 31% - 36% *for reference: percentage oxygen saturation range from (62%-67%) Calculation: 98% (fully saturated)- (range 62%-67%) = 31% - 36% 1 Po2 70 mmHg : 3% - 8% *for reference: percentage oxygen saturation range from (90%-95%) Calculation: 98% (fully saturated)- (range 90%-95%) = 3% - 8% 1 (ii) Briefly explain the Bohr effect based on FIGURE 5. [3 marks] -Due to increasing in Pco2/ concentration of CO2 / hydrogen ion concentration 1 -Affinity of hemoglobin to oxygen reduced/ decrease/ low 1 - More oxyhemoglobin dissociate/ unloading to release oxygen. 1 -Shift the oxygen dissociation curve/ graph (downwards and) to the right. 1 Any 3 (b) Describe 3 characteristics of hemoglobin as respiratory pigment. [3 marks] i. Hemoglobin bind (loosely @ reversibly) with (4) oxygen/O2 molecules forming oxyhemoglobin Hb(O2)4/ HbO8 1 ii. When partial pressure of oxygen is high (as in lung capillaries), hemoglobin has a high affinity for oxygen to form oxyhemoglobin 1 iii. When partial pressure of oxygen is low (as in respiring tissues), the oxyhemoglobin dissociates and oxygen is liberated / released // When partial pressure of oxygen is low, hemoglobin has low affinity for oxygen 1
BLUE 3 SB025 2022/2023 ANSWER SCHEME iv. When partial pressure of CO2 increases (and pH decreases) there is a decrease of the affinity of hemoglobin for oxygen 1 v. Hemoglobin binds with CO2 to form carbaminohemoglobin 1 vi. Hemoglobin shows cooperative in binding with O2 / shows cooperativity// When O2 binds to one subunit, increasing affinity of other three subunits for O2 // binding of oxygen to one subunit induces the remaining subunits to change conformation slightly so their affinity for oxygen increases // when one oxygen dissociate from one subunit induces the remaining subunit to dissociate 1 vii. hemoglobin binds with carbon monoxide/CO to form carboxyhemoglobin 1 Any 3 (c) Myoglobin is the red pigment in mammalian muscle. State ONE structural difference between myoglobin and hemoglobin. [1 mark] Myoglobin Hemoglobin Myoglobin consists of a one/ single polypeptide chain Hemoglobin consists of 4 polypeptide chain Contain one heme group Contain 4 heme group Combine with oxygen forming oxymyoglobin Combine with oxygen forming oxyhemoglobin 1/0 1/0 1/0 Any 1
BLUE 3 SB025 2022/2023 ANSWER SCHEME 6 FIGURE 6 below shows the nervous control for regulation of heartbeat. FIGURE 6 a) During vigorous activity, human blood pH decrease, and the heartbeat rate will increase. By using information from FIGURE 6 above, explain the mechanism. [4 marks] • A decrease in blood pH is detected by chemoreceptors 1 • (Chemoreceptors) send impulse to cardioacceleratory center in (medulla oblongata) 1 • Cardioacceleratory center will be stimulated and sent impulse to SA node via sympathetic nerve. 1 • At nerve ending norepinephrine (neurotransmitter) is released to stimulate SA node. 1 • The rate of cardiac muscle contraction increases, (and heartbeat will increase) 1 Max: 4 marks
BLUE 3 SB025 2022/2023 ANSWER SCHEME b) FIGURE 7 below shows a region of tissues with a part lymphatic system. FIGURE 7 i) What is Z? State ONE characteristic of Z. [2 marks] • Z is lymphatic capillary. 1 • Tiny, ‘dead-end’ tubes (with no opening)// 1 • Has thin wall with only one cell thick// 1 • Has one-way valves (‘flap valves’) 1 Any 1 ii) Why lipid needs to be transported through lymphatic vessel before enter blood circulation? [1 mark] • because lipid need to be breakdown into smaller size// save quantity to the blood circulatory system. • because to prevent the blockage of blood vessels.
BLUE 3 SB025 2022/2023 ANSWER SCHEME 7 Explain how our body regulate the blood water content during fasting period. [7 marks] - When water intake is low, the blood volume decreases 1 - The osmotic blood pressure increases 1 - Detected by osmoreceptors in the hypothalamus 1 - The posterior pituitary gland is stimulated to release Antidiuretic hormone (ADH) 1 - ADH is transported in blood streams to kidney tubules 1 - ADH increases the permeability of collecting duct (and distal convoluted tubule) towards water 1 - More water is reabsorbed into the peritubular capillaries/ vasa recta by osmosis 1 - Blood osmotic pressure decreases and returns to normal 1 - A small volume of concentrated urine is produced 1 Max:7 8 (a) FIGURE 8 shows the propagation of an action potential in an axon at different regions. FIGURE 8 (i) State the condition of the axon membrane in P, Q and R regions [1 mark] P region: polarized / resting state Q region: repolarize P Q R Direction of action potential
BLUE 3 SB025 2022/2023 ANSWER SCHEME R region: depolarize (ii) Describe the event that occur at the axon membrane in Q region. [2 marks] • voltage-gated Na+ channel close // axon membrane is impermeable to Na+ • voltage-gated K+ channel open • K+ diffuse out from axoplasm • inside axon become more negative/less positive // repolarization occur Any 2 (iii) What causes the change of membrane potential in region R? [1 mark] Influx/ inflow/ diffuse in of Na+ into axoplasm (down concentration gradients) (b) FIGURE 9 shows a synapse of a neuromuscular junction. FIGURE 9 (i) Name structures X, Y and Z. [1 mark] X: Synaptic vesicle Y: Presynaptic membrane Z: Ligand-gated ion channel (ii) Describe the event that occur in the neuron when Ca2+ diffuse into presynaptic membrane. [2 marks] • synaptic vesicle is stimulated to fuse with presynaptic membrane • synaptic vesicle release acetylcholine into synaptic cleft by exocytosis Ca2+ Ca2+
BLUE 3 SB025 2022/2023 ANSWER SCHEME (iii) Anticholinergic are drugs that block and inhibit the activity of the acetylcholine in the nervous system. Predict the effect of the presence of these drugs in the neuromuscular junction. [1 mark] - acetylcholine cannot bind to the receptor on the ligand-gated ion channel // - depolarization of post synaptic membrane do not occur // - contraction of muscle cannot occur // muscle will be relaxed Any 1 (c) The hormone testosterone is synthesized in the testes, acts on cells in the developing male fetus to promote the expression of genes involved in male sexual differentiation. Discuss the mechanism of action by this hormone. [6 marks] • Testosterone hormone diffuse into cytoplasm through plasma membrane of target cell. • In cytoplasm, binding of testosterone to its receptor protein forming hormone-receptor complex. • Hormone-receptor complex passes through nuclear pore and enter nucleus. • Hormone-receptor complex then binds with the specific gene on DNA thus activate the gene. • Promotes the transcription of specific gene on DNA to mRNA • mRNA moves from nucleus to cytoplasm through nuclear pore. • mRNA binds with ribosome to be translated into new protein / enzyme. • Specific protein / enzyme will produce the specific response / effect to the hormonal signal. Total: 8 marks Max: 6 marks
BLUE 3 SB025 2022/2023 ANSWER SCHEME 9 (a) FIGURE 10 shows an immune response inside a lymph node. FIGURE 10 (i) Identify structure X. [1 mark] X: Accessory protein / CD4 1 (ii) Determine the type of immune response shown by the cells in FIGURE 10 above. [1 mark] Humoral immune response/ antibody-mediated immune response 1 (ii) Name cell A and explain what happen after it encounters with a bacterium. [2 marks] B cell/ B lymphocyte 1 B cell internalise the antigen (of the pathogen) and display the antigen fragment on the cell surface// B cell with specific receptor bind with the antigen then antigen enter B cells through endocytosis and degraded into antigen fragment. 1 Cell A X Cell B S receptor receptor Cell C Cell D
BLUE 3 SB025 2022/2023 ANSWER SCHEME Become antigen-presenting cell 1 Any 2 (iii) Describe briefly how production of S contributes to the immune response shown in FIGURE 9. [3 marks] To stimulate proliferation and differentiation of B cell 1 Activated B cell proliferate and differentiate into memory B cell and plasma (B) cell 1 Plasma B cell secretes antibody specific for pathogen 1 Memory B cell remain in the body even after the infection is recovered 1 Any 3 Note: S is Interleukin- 2 secreted by activated helper T cell (iv) Give TWO differences between the immune response stated in (ii) with another type of immune response in human body. [2 marks] Humoral immune response involves B cell while cell-mediated immune response involves cytotoxic T cell 1 Site of cell maturation for humoral immune response is bone marrow while site of cell maturation for cell-mediated immune response is thymus gland 1 In humoral immune response plasma cell secrete antibody into the lymph to attack extracellular pathogen while in cell-mediated immune response cytotoxic T cell attack infected cell, cancer cell, and transplanted cell 1 Any 2
SULIT SB025/2 SB025/2 Biology Biologi Paper1 Kertas 1 Semester II Semester II Session 2022/2023 Sesi 2022/2023 1 hour 1 jam No. Matrik No. Kad Pengenalan No. Tempat Duduk (Isikan maklumat dengan lengkap) KOLEJ MATRIKULASI SELANGOR SELANGOR MATRICULATION COLLEGE PEPERIKSAAN SEMESTER PROGRAM MATRIKULASI MATRICULATION PROGRAMME EXAMINATION BIOLOGI Kertas 1 1 jam JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU. DONOTOPENTHISQUESTIONPAPERUNTILYOUARETOLD TODOSO. Untuk Kegunaan Pemeriksa No. Soalan Markah Pemeriksa KP / KKP 1 2 3 4 5 6 7 8 9 JUMLAH Kertas soalan ini mengandungi 7 halaman bercetak. This question paper consists of 7 printed pages. © Kolej Matrikulasi Selangor SULIT
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
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 enzymesubstrate complex to release its product and become available again. (ii) (a) On FIGURE 2, draw a curve to show the rate of enzyme activity if non-competitive inhibitor is added. [1 mark] Any 2
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. -Substrate can no longer binds to the active site. -Adding more substrate cannot increase the rate of enzyme activity (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 one molecule to another. 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 Glucose 6- phosphate Fructose 6- phosphate 2 molecules of glyceraldehyde 3-phosphate 2 molecules of 1,3- bisphosphoglycerate 2 molecules of pyruvate L Any 3
5 SULIT (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. (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// ½ O2 + 2e- + 2H+ H2O 1
6 SULIT (iii) How many reduced coenzymes will be produced if two molecules of acetylCoA enter stage R? [2 marks] 6 NADH 2 FADH2 (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 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 pyruvate. (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.
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] -(At high light intensity, plants close stomata, concentration of CO2 low) -Because Plant A have enzyme PEP carboxylase. -PEP carboxylase have higher affinity towards CO2/ can fix CO2 at even low CO2 concentration.
SULIT SB025 SB025 Biology2 Biologi 2 Semester II Semester II Session 2022/2023 Sesi 2023/2023 1 hour 1 jam No. Matrik No. Kad Pengenalan Kelas (Isikan maklumat dengan lengkap) KOLEJ MATRIKULASI SELANGOR SELANGOR MATRICULATION COLLEGE PRA PEPERIKSAAN SEMESTER PROGRAM MATRIKULASI MATRICULATIONPROGRAMMEPRE-EXAMINATION JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU. DONOTOPENTHISQUESTIONPAPERUNTILYOU ARETOLDTODOSO. Untuk Kegunaan Pemeriksa No. Soalan Markah Markah Pemeriksa KP / KKP Penuh 5 6 7 8 9 JUMLAH Kertas soalan ini mengandungi 9 halaman bercetak. This question paper consists of 9 printed pages. © Kolej Matrikulasi Selangor SULIT INSTRUCTIONS TO CANDIDATE: This question paper consists of 5 questions. Answer all questions in the space provided inthis question paper. The use of non-programmable scientific calculator is permitted.