Name:. ( ) U6S.. SMJK 964 PEREMPUAN CHINA PULAU PINANG SMJK PEREMPUAN School CHINA Level PULAU Test PINANG

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1 Name:. ( ) U6S.. SMJK 964 PEREMPUAN CHINA PULAU PINANG SMJK PEREMPUAN School CHINA Level PULAU Test PINANG SMJK PEREMPUAN CHINA PULAU BIOLOGY PINANG SMJK PEREMPUAN CHINA PULAU PINANG SMJK PEREMPUAN CHINA PULAU MARK PINANG SCHEME SMJK PEREMPUAN CHINA PULAU PINANG SMJK PEREMPUAN CHINA PULAU 2 PINANG hours & 30 SMJK minutes PEREMPUAN CHINA PULAU PINANG Answer all the structure and multiple choice questions in the answer spaces provided. Part A Structure Questions 40 MARKS Write the answers or correct solutions in the answering space after each question. Question 1: [10 MARKS] (a) The diagram shows the structure of two amino acid molecules. Draw a box around the parts of the two amino acid molecules that would be removed when these amino acids are joined by a condensation reaction. [1 mark] Page 1

2 (b) (i) Many athletes eat food high in carbohydrate for three to four days before a race. This is known as glycogen loading and improves performance. Suggest and explain one advantage of glycogen loading. [2 marks] More glycogen stored (in the liver/muscles) Releases glucose (for respiration) (ii) During vigorous exercise muscles respire anaerobically. This leads to muscle fatigue. Explain how. [2 marks] Production / build up of lactic acid / lactate Decrease in ph / increase in H + ions / acidity Influence, enzyme activity / muscle protein / contraction Ignore reference to denaturation (c) Diagram below shows the Krebs cycle and the reactions preceding it. X H H H X H X H (i) Label on the same diagram all the stages where decarboxylation reactions occur with a letter X. [1 mark] dehydrogenation reactions occur with a letter H. [1 mark] (ii) Explain how NAD is regenerated. [1 mark] oxidized / give up hydrogen Page 2

3 (iii) State how the formation of ATP in the Krebs cycle differs from the formation of ATP in oxidative phosphorylation. [2 marks] ref. to substrate level phosphorylation no proton gradient involved no ATP synthase no ETC Question 2: (a) The diagram shows the taxonomic relationships between some mammals. [10 MARKS] (i) Which two species are most closely related? [1 mark] Brown bear and sun bear (ii) Suggest evidences that could have been used to construct this diagram. [3 marks] Fossil (evidence) Biochemical evidence/example DNA (base) sequences Morphological Anatomical Embryological Behavioural (evidence) (iii) The red panda is a member of the kingdom X. Give three features that distinguish all members of this kingdom from all members of the plant kingdom. [3 marks] Heterotrophic / no chloroplasts Have nervous system Have muscles Growth throughout tissues / no meristems No cell walls / permanent vacuole Store glycogen Page 3

4 (b) Complete the table to show the classification of the red panda, Ailurus fulgens. [3 marks] Question 3: (a) The diagram shows the light-dependent reaction of photosynthesis. [10 MARKS] (i) Using the diagram, explain the role played by water in the production of ATP during photosynthesis. [4 marks] Photolysis / water is split using light energy Provides electrons / hydrogen ions / oxygen Replaces electrons lost by chlorophyll Provides hydrogen ions for ATP production (b) Bee Gnoh and Cindy investigated the effect of a herbicide on the light-dependent reaction. They measured the effects of different concentrations of the herbicide on the production of oxygen and on both electron transport chains. The graph shows the results. Page 4

Using all of the information, suggest how the herbicide causes each of the following: (i) the reduction in the production of oxygen [3 marks] Binds to chlorophyll molecules Stops release / absorption

5 Using all of the information, suggest how the herbicide causes each of the following: (i) the reduction in the production of oxygen [3 marks] Binds to chlorophyll molecules Stops release / absorption of electrons by chlorophyll Named effect on compounds in (first) transport chain Stops movement of electrons down first transport chain (So) break down of water reduces (ii) the death of plants [3 marks] Less ATP for light-independent reaction/named reaction (So) less sugars produced for respiration Rate of respiration is greater than rate of photosynthesis Question 4: (a) Figure 4 shows a sarcomere in a relaxed muscle. [10 MARKS] (i) Draw and label the appearance of this sarcomere in a contracted muscle. Sarcomere drawn with actin filaments closer together [2 marks] (ii) Explain the role of each of the following in muscle contraction. Tropomyosin: [2 marks] (In relaxed muscle) tropomyosin covers binding site Tropomyosin is attached to another protein / troponin Calcium (ions) bind to troponin Causes tropomyosin to move (from binding site) / change shape Rachet mechanism exposes myosin head binding site Myosin: [2 marks] Myosin filaments have a head Bridges form between actin and myosin (actomyosin cross-bridge) Bridges break and reform further along Pulling actin filaments past myosin ATP provides energy (in correct context) Page 5

6 (b) During exercise, muscle contraction increases and so does heart rate. Complete the chart to show what causes this increase in heart rate. [4 marks] Carbon dioxide Decrease / become (more) acidic Chemoreceptors Sino-atrial node / SAN Ignore pacemaker Page 6

7 Part B Essay Questions 60 MARKS Answer four of the six questions below. Arrange your answers in numerical order and tie the answer sheets for this section at the end of the examination. 5. (a) The vampire bat is a small flying mammal that feeds on blood. The rapid intake of blood during a feed leads to a large increase in body mass of the bat. It also leads to the production and release of large amounts of dilute urine. (i) Explain the advantage of the increased rate of urine production during feeding. [2 marks] 1. (Rapid loss of fluid) decreases weight / maintains normal weight 2. Enables bat to fly / enables bat to take in more blood (ii) After a blood meal has been digested, the vampire bat produces concentrated urine. Explain how urine is concentrated in mammals. [7 marks] 1. Ascending limb (of loop of Henle) pumps sodium / chloride ions out (of filtrate) into surrounding tissue / tissue fluid 2. Ascending limb is impermeable to water 3. Decreases water potential in medulla / creates / maintains water potential gradient causing 4. water is reabsorbed from descending limb / distal convoluted tubule / collecting duct 5. Decrease in water potential of blood detected by hypothalamus 6. ADH released from pituitary gland 7. ADH makes cells of collecting duct / distal convoluted tubule more permeable to water / open aquaporins 8. More water leaves filtrate by osmosis Accept correct reference to water absorption in the pct (b) Vampire bats are able to maintain a constant body temperature when the environmental temperature falls. Explain how. [6 marks] 1. Receptors in skin 2. Impulses to / from heat gain centre (of hypothalamus) 3. Vasodilation of shunt vessels / Vasoconstriction in arteriole walls (Reject Capillaries) reduces blood flow to skin 4. Piloerection traps a layer of insulating air next to skin 5. Shivering (caused by muscle spasms) produces heat (heat produced) from respiration 6. Release of adrenaline increases metabolic rate 7. Sweating slows / is reduced which reduces heat loss by evaporation 8. Heat loss occurs by conduction, convection and radiation.. 6. (a) Describe and explain four ways in which the structure of a capillary adapts it for the exchange of substances between blood and the surrounding tissue. [4 marks] 1. Wall / endothelium one cell thick, reduces diffusion distance 2. Flattened / squamous cells, reduced diffusion distance 3. Narrow (lumen) / small diameter, reduces flow rate / more time for diffusion / exchange 4. Narrow / small diameter, large SA/V ratio / short diffusion distance 5. Narrow / small diameter, RBC in contact with wall / pass singly 6. Gaps / pores / in the wall between cells / fenestrations, faster filtration / movement out / large molecules through Page 7

8 (b) Describe and explain how tissue fluid is formed and how it is returned to the blood. [6 marks] 1. Arterial end blood high (hydrostatic) pressure 2. Fluid / water / soluble substances forced out 3. Proteins / large molecules remain behind 4. Water potential become more negative 5. Friction / resistance to flow 6. Reduces hydrostatic pressure 7. Water moves in by osmosis 8. Excess water taken up by lymph capillaries 9. Returned to blood stream (via blood vessels) in the neck / into vein (c) Sucrose is moved through phloem by mass flow. Describe how. [5 marks] 1. (Sucrose loaded) into phloem by active transport; 2. Involving companion cells / transfer cells 3. Reduced water potential 4. Water moves in by osmosis 5. High / hydrostatic pressure 6. Lower at sink / pushed towards sink.. 7. (a) The rate of an enzyme-catalysed reaction is affected by temperature. Explain how. [6 marks] Below optimum temperature: 1. Increase in temperature increases kinetic energy (of enzyme and substrate) 2. More collisions (between substrate and active site / enzyme) 3. More enzyme-substrate complexes formed 4. More activation energy Above optimum temperature: 1. Denaturation occurs 2. H bonds broken 3. Change (permanent) in tertiary structure / shape of active site (of enzyme) 4. Enzyme and substrate unable to bind / no longer complementary / less enzymesubstrate complexes formed Credit 1 mark to labelled illustration (b) Immobilised enzymes are frequently used in the food industry. Explain how these enzymes are used in the production of high fructose corn syrup. [9 marks] 1. The enzyme amylase is used to hydrolyse starch into maltodextrins / modified starch. 2. Amyloglucosidase / glucoamylase converts maltodextrins into dilute glucose syrup. 3. The dilute glucose syrup is refined, filtered and concentrated to form high glucose syrup. 4. Glucose isomerase is immobilised onto beads of calcium aginate in a glass column, 5. As the glucose flows down the column, some of the glucose is converted into low fructose syrup. 6. The syrup collected will have some glucose and fructose. 7. The syrup is purified to obtain high (concentrated) fructose syrup. 8. The enzymes are removed from the syrup and can be reused again. 9. Fructose is sweeter than glucose and is used in drinks and food sweeteners. Page 8

9 8. (a) Explain what is meant by each of the following ecological terms. [2 marks] (i) Population (All the) organisms of one species in a habitat/area (ii) Ecosystem All the organisms / populations (in an area) and their abiotic/physical environment The community and the abiotic / physical environment (b) Explain what a transect is and describe when it is appropriate to study a habitat by using a transect. [2 marks] Line (through habitat) along which organisms are sampled / measurements are taken 1. When there is a change in the habitat / environment 2. Example of where used. zonation on a rocky shore development of a succession (c) (i) A farmer changed the use of his land from raising animals to growing cereal crops. This is more efficient in terms of total food production. Explain why. [2 marks] Shorter food chain Less energy lost (transferring) between (trophic) levels (ii) The farmer was paid to stop using one of his fields for farming. Over the next 20 years the species in the field changed. Describe a method that could be used to measure these changes. [7 marks] 1. Use coordinates / random numbers / permanent quadrats / a large number of quadrats (reject transect) 2. This is a common quantitative sampling method when the area to be studied is too large. 3. It is a convenient way to measure ecological variables (like abundance, species diversity or biomass). 4. The user defines a fixed area, called a plot or quadrat, within which plant characters are measured. 5. It is suitable for immobile organisms such as plant communities. 6. Usually, a rectangular quadrat frame is used to define the sampling area. 7. Plant characters of only those plants inside the quadrat are measured. 8. The size of quadrat depends on the vegetation type in the area. 9. Usually quadrats are 1 m by 1 m in size. 10. A plot size should be large enough to include significant numbers of individuals, but small enough so that plants can be separated, counted and measured without duplication or omission of individuals. 11. Quadrat sampling attempts to define the characteristics of the plant community for an area much larger than the actual area sampled and therefore care must be taken to obtain samples that represent the entire habitat and that can eliminate the human factor. 12. Quadrats are randomly placed in the area under investigation. Count number of (each) species of plant / percentage cover. 13. Repeat at regular time intervals / stated time interval. Page 9

10 (d) (i) Suggest two biotic factors that might cause a rapid decline in population size. Explain how each factor might bring about this decline. [2 marks] 1. Predation - Large numbers of predators would decrease population 2. Food supply - Lack of food lead to (starvation and) decrease in numbers 3. Disease/pathogens - Spread rapidly in dense populations; 4. Competition for nest sites - When sites scarce, fewer breed.. 9. (a) Describe the process of oxidative phosphorylation in the mitochondrion. [9 marks] 1. reduced, NAD / FAD 2. passed to ETC 3. inner membrane / cristae 4. hydrogen released (from reduced, NAD / FAD) ; Reject H 2 5. split into electrons and protons 6. protons in matrix 7. electrons pass along, carriers / cytochromes 8. refer to redox reactions 9. refer to energy gradient 10. energy released ; Reject the word produced 11. protons (pumped) into intermembrane space 12. proton gradient 13. protons pass through (protein) channels 14. ATP synthase / stalked particles 15. ATP produced 16. chemiosmosis 17. electron transferred to oxygen 18. addition of proton (to oxygen) to form water / (oxygen) reduced to water Note: If candidate mistakenly writes about photosynthesis only allow marking points #7, #8, #9, #10 and #15 to max 5 (b) The ornithine cycle is a series of reactions that converts ammonia formed by deamination of amino acids to urea that is less toxic. With the aid of a diagram, explain the process of the ornithine cycle. [6 marks] 1. Excess amino acids are deaminated in the hepatocytes whereby the amino group is converted first into toxic ammonia. 2. Ammonia combines with CO 2 and H 2 O to form carbamoyl phosphate. 3. Carbamoyl phosphate reacts with ornithine to form citrulline. 4. Citrulline reacts with ammonia from aspartate to form argininosuccinate. 5. Argininosuccinate then cleaves into arginine and fumarate. 6. Arginine is hydrolysed to form urea and ornithine (which is recycled again). 7. Urea produced by the liver is carried by blood to the kidneys to be excreted in the urine. Page 10

11 Ornithine Cycle Note: Labelled diagram 2 marks 10. (a) Describe the structure of a motor neurone. [5 marks] Most of these points can be taken from an annotated diagram: [max 2] 1. nucleus in cell body 2. (short), dendrites / dendrons 3. axon 4. synaptic knob 5. node of Ranvier 6. Schwann cells / myelin sheath Most be stated (not on diagram): [max 5] 7. (axon) much longer than, dendrite / dendrons 8. cell body contains, mitochondria / rough ER / golgi / groups of ribosomes 9. many mitochondria at, synaptic knob / terminal branch 10. synaptic vesicles contain neurotransmitter / named neurotransmitter 11. nucleus in Schwann cell ; Reject nucleus in myelin sheath 12. motor end plate / (dendrites) have receptors (for neurotransmitters) (b) Explain how an action potential is transmitted along a motor neurone. [10 marks] 1. Na + channels open ; Accept sodium (ion voltage gated) channels 2. Na + enter cell ; Reject enter membrane 3. inside becomes less negative / positive / +40mV / depolarised 4. Na + channels close ; Accept sodium (ion voltage gated) channels 5. K + channels open ; Accept potassium (ion voltage gated) channels 6. K + move out (of cell) ; Reject of membrane 7. inside becomes, negative / repolarised Accept negative figure 8. local circuits / description 9. (myelin sheath / Schwann cells) insulate axon / does not allow movement of ions 10. action potential / depolarisation, only at nodes (of Ranvier) / gaps 11. saltatory conduction 12. self propagate in one-way transmission 13. hyperpolarisation / refractory period related to Point # 12 Page 11

12 Note: Labelled diagram 2 marks Prepared by: Yeap Chee Beng BIOLOGY PANITIA P.C.G.H.S. Page 12

Name:. ( ) U6S.. SMJK 964 PEREMPUAN CHINA PULAU PINANG SMJK PEREMPUAN School CHINA Level PULAU Test PINANG

Name:. ( ) U6S.. SMJK 964 PEREMPUAN CHINA PULAU PINANG SMJK PEREMPUAN School CHINA Level PULAU Test PINANG SMJK PEREMPUAN CHINA PULAU BIOLOGY PINANG SMJK PEREMPUAN CHINA PULAU PINANG SMJK PEREMPUAN CHINA