Seconday 5 (MYE) - Essay (9)

9. (a) Diagram 17 shows the various structures involved in the regulation of temperature in the human body. A worker enters a cold room that stores frozen meat. Explain how the regulation of the worker’s body temperature occurs. [10 marks]

Diagram 17













· When a worker enters a cold room, the slight drop in body temperature is detected by the hypothalamus when blood is transported to the brain [1]
· This causes impulses to be sent to the various body parts to carry out a negative feedback mechanism to bring the body temperature back to normal level [1]
· In the skin, the erector muscles contract to raise the hair shafts, trapping an insulating layer of warm and still air to prevent heat loss [1]
· Sweating does not occur and body heat is conserved [1]
· Vasocontriction occurs when the smooth muscles around the afferent arterioles contract to decrease the amount of blood flowing through the skin, hence reducing heat loss through radiation [1]
· Shivering occurs to generate heat energy [1]
· The adrenal gland will be stimulated to secrete more adrenaline [1]
· This causes an increase in the rate of breathing and heartbeat [1]
· The rate of glycogen conversion in the liver will also increase [1]
· The thyroid gland will also be stimulated to secrete thyroxine [1]
· This increases the rate of metabolism [1]
· More glucose will be metabolized in the muscle tissues to produce heat energy to warm the body [1]


(b) The urinary system consists of the kidneys which play an essential role in controlling the composition and volume of the body fluids. The kidneys receive blood from the abdominal aorta via the renal artery and the renal veins return blood to the inferior vena cava.

*Concentration of substances in solution (g/dm3)


Glucose
Blood plasma entering glomerulus = 1.0
Glomerular filtrate = 1.0
Urine produced by kidneys = 0.0

Amino acid
Blood plasma entering glomerulus = 1.5
Glomerular filtrate = 1.5
Urine produced by kidneys = 0.0

Protein
Blood plasma entering glomerulus = 8.0
Glomerular filtrate = 0.0
Urine produced by kidneys = 0.0

Urea
Blood plasma entering glomerulus = 0.3
Glomerular filtrate = 0.3
Urine produced by kidneys = 20.0

Sodium ions
Blood plasma entering glomerulus = 3.2
Glomerular filtrate = 3.2
Urine produced by kidneys = 3.5

Table 1 shows the concentration of some substances in the blood plasma, glomerular filtrate and urine of an adult. Based on the table, describe how the process of urine formation occurs in an individual. [10 marks]

· When blood flows from the afferent arteriole to the narrower efferent arteriole, the high hydrostatic pressure forces [1] the blood plasma across the filtration membrane into the Bowman’s capsule via ultrafiltration [1]
· The content of the blood plasma such as glucose, amino acids, urea, sodium ions and other ions will be present in the glomerular filtrate [1] except for red blood cells and the large protein molecules [1]
· The filtrate will be passed through the proximal convoluted tubule [1] where reabsorption of useful molecules takes place [1]
· Amino acids and glucose are both actively reabsorbed [1] from the proximal convoluted tubule into the blood in the peritubular capillaries [1]
· As the filtrate is passing along the renal tubule, some sodium ions will be secreted from the blood into the distal convoluted tubule [1]
· A large quantity of urea will be secreted from the renal tubules to be removed in the urine [1]

Seconday 5 (MYE) - Essay (8)

8. (a) A trip to research institute MARDI is organised to study tissue culture techniques. Based on the short notes collected randomly from different officers at the institute, write a report of how tissue culture is carried out on a named plant with illustration.

Short notes:
· Growth hormones
· Callus
· Steriled culture media
· Subculture
· Explants
· Nursery [10 marks]









All apparatus and materials used in this technique must be sterilized and autoclaved. [1]
The surface of a tobacco leaf is sterilized with ethanol, then is cut into small pieces. [1]
The small pieces of plant tissues are called explants. [1]
The explants are then placed inside a test tube/Petri dish containing steriled nutrient agar medium with glucose, amino acids, vitamins, mineral salts, growth hormones, etc. [1]
The cultures are incubated at 37C and agitated, if necessary. [1]
After three to four weeks, the explants develop into callus (a group of undifferentiated cells). [1]
The callus can now be propagated indefinitely by sub-culturing onto new nutrient media. [1]
The callus is finally transferred onto a fresh agar medium containing different concentrations of auxin and cytokinin for root and shoot generation. [1]
The rooted plantlets (clones) are then transferred to soil and kept in a controlled environment. [1]
From one original plant, thousands of genetically identical clones can be produced. [1]


(b) Knowledge on the concept of mitosis is applied to cloning of plants and animals in agriculture. What are the advantages and disadvantages of cloning to mankind? Support your answer using named examples. [6 marks]

· Advantages
· 1. Selection of quality & desired traits
· e.g. a disease resistance gene is extracted from tobacco and is then inserted into the cotton plant to be resistant against a particular pest
· 2. Large & fast production of desired traits
· e.g. a copy of the human insulin gene encoding for insulin synthesis is inserted into E. coli bacteria. The bacteria can now produce insulin for commercial use
· 3. Pollinating agents are not required (vegetative reproduction)
· e.g. reproduction with seeds takes a long time. Clones mature fast and produce seeds within a short time
· 4. Genetic manipulation can be carried out
· e.g. genetic engineering allows genetically modified bacteria to clean up oil spills in ocean / heavy metals in soil

· Disadvantages
· 1. No variation / Preventing natural selection
· e.g. clones are unable to adapt to drastic changes of the environment
· e.g. a new pathogen can wipe out the entire clones as they have the same level of resistance to the pathogen
· 2. Transfer of undesired genes
· e.g. clones might undergo mutation, and the mutated gene could be transferred to other plants by viruses
· e.g. herbicide-resistant gene is transferred from clones to weed, which makes them resistant to the herbicide applied (herbicide is no longer effective) – ‘super weed’
· 3. Short term or long term side effects
· unknown, could lead to development of bio-weapons
· 4. Playing God (ethical & religious issues)
· e.g. can vegetarians consume crops with insertion of bovine growth hormone gene?

(c) Contrast the division process that takes place in the apical meristem to that of the anther. [4 marks]

MITOSIS (Apical meristem)
Synapsis / Cross over: No
Number of divisions: 1
Number of daughter cells produced: 2
Number of chromosomes in daughter cells: Diploid (2n), same as the mother cell
Genetic contents in daughter cells: Identical to the mother cell
Genetic variation : No

MEIOSIS (Anther)
Synapsis / Cross over: Yes
Number of divisions: 2
Number of daughter cells produced: 4
Number of chromosomes in daughter cells: Haploid (n), half of the mother cell
Genetic contents in daughter cells: Different from the mother cell
Genetic variation : Yes

ANY FOUR

Seconday 5 (MYE) - Essay (7)

7. (a) Diagram 15 shows a series of experiments conducted by a student. All the experiment set-ups are placed near sunlight. Observations are made after two days.

Diagram 15

(a) (i) What are the objectives of this experiment? [2 marks]

· 1(a) and 1(b) aimed at studying the parts of the coleoptile that is sensitive to light stimulus
· 1(c) and 1(d) aimed at studying the diffusion of auxin from the tip to the elongating region to cause a growth response

(ii) Two days have lapsed after the experiment. Draw the expected observations. Explain your observations and finally draw a conclusion for each of the above experiments. [9 marks]

· Each drawing with explanation is worth 1 mark
· 1(a) - the coleoptile bends towards unilateral light stimulus to cause a growth response [1] because auxin that is produced at the shoot tip diffuses to the shaded side to cause a more rapid growth in that region [1]
· 1(b) – when the tip is removed, auxin is absent to cause any growth response [1]. Therefore, there is no response observed [1]
· Conclusion: Auxin is produced at the tip which makes the shoot tip sensitive to light stimulus [1]
· 1(c) - There is no growth response observed [1] because auxin that is produced at the tip cannot diffuse down to the elongating region to cause a growth response [1]
· 1(d) - the coleoptile bends towards unilateral light stimulus to cause a growth response [1] because auxin that is produced at the shoot tip diffuses through the agar block [1] to the shaded side of the coleoptile to cause a more rapid growth in that region [1]
· Conclusion: Auxin has diffused down the coleoptile through the agar block, but not the mica plate, to stimulate growth response towards unilateral light [1]
Any 9

(b) Diagram 16 shows an experiment to test the action of a detergent with enzymes on oil residue on clothes. Three similar shirts P, Q and R with the same oil residue of equal size were washed at three different temperatures which are 15 °C, 65 °C and 35 °C, respectively.

Diagram 16 - Clothes

Explain your observations. ` [9 marks]

· At 15C – the low temperature did not promote rapid enzymatic reaction to break down the oil residue on clothes P [1]
· This is because the enzymes were inactive which caused low frequency of effective collision between the enzyme and its substrate to form an enzyme-substrate complex [1]
· Therefore, the rate of reaction was low [1]
· At 35C – the optimal temperature allowed rapid enzymatic reaction to break down the oil residue on clothes R [1]
· This is because the enzymes were actively moving to collide with its substrate to form an enzyme-substrate complex [1]
· Therefore, the rate of reaction was at optimal level [1]
· At 65C – the high temperature has denatured the enzymes [1]
· This caused the enzymatic reaction to stop and the oil residue on clothes Q was not greatly removed [1]
· Therefore, the rate of reaction was very low [1]

Seconday 5 (MYE) - Essay (6)

6. (a) Based on your knowledge in Biology and Diagram 13, explain the uptake of water from the soil to the leaves. Your answer should include all the mechanisms/processes that facilitate the uptake of water in plants. [8 marks]

Diagram 13











The cytoplasm of root hair cells is usually hypertonic to the surrounding soil water [1]
Hence, water enters the root hair cells by osmosis [1]
The root hair cells are now hypotonic to the adjacent root cortex cells [1]
Water then diffuses into these cells until it reaches the cortex region [1]
Water flows through the parenchyma cells (cortex) until it reaches the endodermis through either the apoplastic pathway or the symplastic pathway [1]
The Casparian strips (impermeable to water) prevent water from moving through the cell wall of the endodermal cells [1]
The water concentration gradient which exists across the cortex creates a transpiration pull to direct water into the xylem vessels [1]
Water finally reaches the mesophyll cells in the leaves where it will be lost through the stomata into the surroundings [1]
The upward movement of water through the xylem vessels in the stems to the stomata is also assisted by the root pressure, adhesion and cohesion [1]
Any 8


(b) For each type of food in Diagram 14 below explain how the concept of osmosis can be used in food preservation. [8 marks]

Diagram 14 - FISH, CARROTS

· Fish can be preserved by dehydration [1]
· Fish is smoked or dried under the sun to remove water [1]
· Fish can also be dried by adding salts which causes water to diffuse out from fish into hypertonic surrounding [1]
· Dehydrated fish has low water content that prevents growth of microbes [1]

· Carrot strips are soaked in a vinegar syrup solution [1]
· The cell sap of the carrot cells is hypotonic to the external solution [1]
· Water diffuses out of the carrot cells into the hypertonic surroundings [1]
· The dried and shrunk carrot strips prevent the growth of microbes [1]


Most green plants do not have woody stems to hold them up, but depend on turgidity of cells.

(c) Based on the statement above, explain what will happen to a herbaceous plant or a freshly picked flower if it is kept out of water. [4 marks]

· Support in most green plants such as herbaceous plants is supported by turgidity of the parenchyma and collenchyma cells [1]
· Water diffuses into the vacuole of the plant cells throught osmosis [1]
· The developed turgor pressure of fluids in the central vacuole pushes the protoplasm and plasma membrane against the cell wall [1]
· This creates support for the stems, roots and leaves [1]

Seconday 4 (MYE) - Essay (9)

9. Diagram 14 shows phase X and phase Y in the cell cycle of an organism. Phase X consists of three sub-phases, P, Q and R. Phase Y involves two processes, U and V.

Diagram 14 - Cell Cycle












(a) Explain the processes that take place during phase X. [10 marks]

· Phase X is interphase which is subdivided into G1 phase (P), S phase (Q) and G2 phase (R). [1]
· At G1 phase, proteins and new organelles are synthesized. [1]
· Chromosomes are extremely fine and are not visible. They are called chromatin. [1]
· The cell has a high metabolic rate and is preparing to divide if the external condition is conducive for growth. [1]
· At S phase, replication of DNA occurs. [1]
· Chromosomes are duplicated and each chromosome will now has two sister chromatids. [1]
· Both sister chromatids contain identical copies of the DNA. [1]
· At G2 phase, the cell continues to grow. [1]
· The cell remains metabolically active by accumulating energy. [1]
· The cell completes its final preparations for division by synthesizing more organelles and proteins. [1]

(b) State the stages within process U and explain the events that take place.
[10 marks]

· Process U refers to mitosis which consists of prophase, metaphase, anaphase and telophase. [1]
· At prophase, the chromosomes condense and become tightly coiled/ [1]
· The two sister chromatids joined together at the centromere. [1]
· Spindle fibres begin to form between the centrioles. [1]
· Nucleolus and nuclear membrane start to break down/ [1]
· Diagram of prophase [1]
· At metaphase, the centrosomes reach at the opposite poles [1]
· The chromosomes align themselves at the metaphase plate (equator) with the centromere attaching to the spindle fibres. [1]
· Diagram of metaphase [1]
· At anaphase, the spindle fibres contract and the centromere divides. [1]
· The sister chromatids are pulled apart to the opposite poles. [1]
· At telophase, the chromosomes reached at the opposite poles. [1]
· Chromosomes start to uncoil to become the non-visible chromatins. [1]
· Spindle fibres break down where the nucleolus and nuclear membrane reform/ [1]
ANY 10

Seconday 4 (MYE) - Essay (8)

8. (a) Diagram 13 shows the process of cell division in the apical meristem of the root of a plant. Explain the process of cell division that takes place and contrast with the one that occurs in the anther of the plant. [6 marks]

Diagram 13 - Root meristem

· Mitotic cell division takes place actively in the meristematic tissues of the root tip and shoot tip of plants [1]
· Active cell division in meristematic tissues allows growth and elongation of a plant to take place at a faster rate [1]








MITOSIS (Meristematic tissue)

Synapsis / Cross over: No
Number of divisions: 1
Number of daughter cells produced: 2
Number of chromosomes in daughter cells: Diploid (2n), same as the mother cell
Genetic contents in daughter cells: Identical to the mother cell
Genetic variation : No

MEIOSIS (Anther)

Synapsis / Cross over: Yes
Number of divisions: 2
Number of daughter cells produced: 4
Number of chromosomes in daughter cells: Haploid (n), half of the mother cell
Genetic contents in daughter cells: Different from the mother cell
Genetic variation : Yes

ANY FOUR

(b) A trip to research institute MARDI is organised to study tissue culture techniques. Write a report of how tissue culture is carried out on a named plant with illustration. [10 marks]












· All apparatus and materials used in this technique must be sterilized and autoclaved. [1]
· The surface of a tobacco leaf is sterilized with ethanol, then is cut into small pieces. [1]
· The small pieces of plant tissues are called explants. [1]
· The explants are then placed inside a test tube/Petri dish containing steriled nutrient agar medium with glucose, amino acids, vitamins, mineral salts, growth hormones, etc. [1]
· The cultures are incubated at 37C and agitated, if necessary. [1]
· After three to four weeks, the explants develop into callus (a group of undifferentiated cells). [1]
· The callus can now be propagated indefinitely by sub-culturing onto new nutrient media. [1]
· The callus is finally transferred onto a fresh agar medium containing different concentrations of auxin and cytokinin for root and shoot generation. [1]
· The rooted plantlets (clones) are then transferred to soil and kept in a controlled environment. [1]
· From one original plant, thousands of genetically identical clones can be produced. [1]

(c) Breast cancer makes up more than 10% of cancer cases reported in Sabah. It is the second most common cause of death among Malaysian women after heart disease. Approximately one out of 20 women in Malaysia will get breast cancer. Explain how cancer is due to uncontrolled mitosis. [4 marks]

· When a cell divides by mitosis repeatedly, without control and regulation, it can produce cancer cells. [1]
· This is due to severe disruption to the mechanism that controls cell cycle, which eventually causing the cells to divide freely and uncontrollably without heeding the cell cycle control system. [1]
· Cancer cells compete with the surrounding normal cells to obtain sufficient nutrients and energy for their own growth. [1]
· They can also intrude on and spread to other tissues which then lead to the malfunction of the tissues and ultimately death. [1]

Seconday 4 (MYE) - Essay (7)

7. Diagram 11 shows an experiment to test the action of a detergent with enzymes on oil residue on clothes. Three similar shirts P, Q and R with the same oil residue of equal size were washed at three different temperatures which are 15 °C, 65 °C and 35 °C, respectively.

Diagram 11 - Clothes with stains

(a) Explain your observations. ` [9 marks]

· At 15C – the low temperature did not promote rapid enzymatic reaction to break down the oil residue on clothes P [1]
· This is because the enzymes were inactive which caused low frequency of effective collision between the enzyme and its substrate to form an enzyme-substrate complex [1]
· Therefore, the rate of reaction was low [1]
· At 35C – the optimal temperature allowed rapid enzymatic reaction to break down the oil residue on clothes R [1]
· This is because the enzymes were actively moving to collide with its substrate to form an enzyme-substrate complex [1]
· Therefore, the rate of reaction was at optimal level [1]
· At 65C – the high temperature has denatured the enzymes [1]
· This caused the enzymatic reaction to stop and the oil residue on clothes Q was not greatly removed [1]
· Therefore, the rate of reaction was very low [1]


Diagram 12 - Graph

(b) Diagram 12 shows the effect of an increase in temperature on the reaction rate of an enzyme. Based on the above graph, explain why a rise in temperature increases the rate of reaction until it reaches the optimal temperature. [4 marks]

· At low temperature, the enzyme is inactive because due to the low kinetic energy of the enzyme and substrate molecules [1]
· As the temperature increases, the enzyme and substrate molecules are moving faster and the effective collision between the molecules becomes more frequent [1]
· This encourages the formation of enzyme-substrate complex which leads to the formation of products [1]
· At optimal temperature, the enzyme catalyses a reaction at the maximum rate [1]

(c) Explain what causes the decrease in the rate of reaction in Diagram 12 when the temperature exceeds the optimal temperature. [4 marks]

· Beyond the optimal temperature, any increase in temperature causes the rate of reaction to decrease sharply until it finally stops completely at about 60C [1]
· At high temperature, the chemical bonds holding the enzyme molecules in their precise configuration begin to break [1]
· This alters the 3D shape of the enzyme molecules and eventually destroys the active sites [1]
· This means that the substrate can no longer fit into the active sites of the enzyme as the enzyme has denatured [1]

(d) Enzymes are widely used in many industrial processes. State two enzymes that can be used in food production and their functions. [3 marks]

· Protease - Meat/Fish products [1]
· To tenderize meat/remove skin of fish [1]

· Cellulase – Seaweed products [1]
· To extract agar from seaweed [1]

Seconday 4 (MYE) - Essay (6)

6. (a) Based on your knowledge in Biology, explain the following processes:

(i) excessive use of fertiliser on crop plants

· The excessive use of fertiliser causes the soil to become hypertonic to the cell sap of the plant root cells [1]
· This causes water to diffuse out of the root cells by osmosis [1]
· As a results, the root cells undergo plasmolysis and the plant will wilt [1]

(ii) absorption of glucose in the small intestines

· After digestion, the concentration of glucose in the ileum increases [1]
· Glucose is a large molecule that is transported across the epithelial cells of ileum by facilitated diffusion [1]
· Glucose molecules move across the transport protein down the concentration gradient into the blood capillaries [1]
[6 marks]

(b) Blood is a medium to transport respiratory gases in human beings. Explain how the transport of oxygen and carbon dioxide takes place between the alveolus and the body cells. [6 marks]

· In the alveolus, the concentration of oxygen is higher than that of the blood capillary [1]
· Therefore, oxygen diffuses from the alveolus into the blood capillary down its concentration gradient by simple diffusion [2]
· In the blood capillary, the concentration of carbon dioxide is higher than that of the alveolus [1]
· Therefore, carbon dioxide diffuses from the blood capillary into the alveolus down its concentration gradient by simple diffusion [2]

(c) For each type of food in Diagram 10 below explain one method in which it can be preserved for a long period of time. [8 marks]

Diagram 10 - FISH, CARROTS

· Fish can be preserved by dehydration [1]
· Fish is smoked or dried under the sun to remove water [1]
· Fish can also be dried by adding salts which causes water to diffuse out from fish into hypertonic surrounding [1]
· Dehydrated fish has low water content that prevents growth of microbes [1]

· Carrot strips are soaked in a vinegar syrup solution [1]
· The cell sap of the carrot cells is hypotonic to the external solution [1]
· Water diffuses out of the carrot cells into the hypertonic surroundings [1]
· The dried and shrunk carrot strips prevent the growth of microbes [1]

Paper 1 Results - Secondary Five

1. Yeow Eu Jin (43)
2. Wong Hui Xim (41)
3. Marcus Foo Chun Hoe (41)
4. Juno Teo Yaw King (40)
5. Daniel Tham (40)
6. Saw Mei Yi (40)
7. Geoffrey Yong Hong Kiat (40)

TCW (39), LDL (39), LSM (37), NT (37), SMY (37), FYZ (36), CGL (36), FLWY (36), PK (36), SWN (36), FKC (36), FSD (35), NAB (35), DK (35), SWJ (35), NYC (35), KSF(35), CLJS (34), VTLY (34),GSJ (34), GKH (34), DC (34), GLHH (33), TJS (33), VSLW (32), TYL (32), DH (32), TSZ (32), SNS (32), CTYZ (31), LT (31), OSH (31), LLWL (31), LX (29), AOKY (29), NCL (28), VYXH (27), QAR (27), CST (27), WCL (26), LJS (25), RR (24), RN (24), NCX (24), ML (23)

Secondary 5 Mid-Year Exam 2010 - Paper 1 (Answers)

1B, 2B, 3A, 4C, 5C, 6B, 7C, 8C, 9C, 10D, 11A, 12C, 13A, 14A, 15D, 16C, 17D, 18C, 19B, 20A, 21C, 22D, 23D, 24D, 25C, 26C, 27C, 28B, 29D, 30C, 31C, 32C, 33C, 34B, 35C, 36B, 37D, 38A, 39D, 40C, 41D, 42A, 43D, 44B, 45D, 46C, 47B, 48D, 49A, 50C

Paper 1 Results - Secondary Four

1. Ee Wui Liang (48)
2. Ian Lim (45)
3. Felicia Chong (45)
4. Lim Jen Yi (45)
5. Boo Jin-Shan (44)
6. Yong Tien Cin (44)
7. Teh Xin Er (44)
8. Chloe Lee Yean Li (44)
9. Athirah Che Azizuddin (43)
10. Loo Sheng Yin (43)
11. Wong Shu Yi (43)
12. Maharani Bebita (43)
13. Andrew Chang (42)
14. Deanna Anuar (42)
15. Tunku Alysha (42)
16. Tiew Kai Hao (42)
17. Gwyneth Tai Lin Xiang (42)
18. Lee Vin Sen (41)
19. Nur Alia Abu Bakar (41)
20. Tracy Pak Loo Kee (41)
21. Soo Hon Wye (41)

LYM (38), AC (38), ESL (38), CAC (38), BL (37), TYY (37), LDS (37), SHR (37), ANYS (35), LHH (34), GHQT (33), TMK (33), TJY (32), WRY (32), WAFO (31), YL (30), JRR (29), ETYL (29), RTTSY (29), AK (29), WQF (29), CLPY (28), SPY (28), AMW (26), TWT (26), LJY (25), JJ (24), MKK (24), EWKY (23), MGH (23), PMR (22), CTH (17), LLH (17), YMF (17), YLY (13),

Secondary 4 Mid-Year Exam 2010 - Paper 1 (Answers)

1D, 2C, 3A, 4B, 5D, 6C, 7C, 8B, 9D, 10B, 11C, 12A, 13C, 14B, 15D, 16C, 17B, 18B, 19B, 20D, 21C, 22A, 23C, 24B, 25A, 26B, 27D, 28C, 29B, 30B, 31C, 32B, 33C, 34C, 35B, 36B, 37D, 38C, 39D, 40C, 41D, 42B, 43B, 44A, 45B, 46C, 47D, 48A, 49D, 50C