BAHAGIAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN KEMENTERIAN PELAJARAN MALAYSIA

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1 BAHAGIAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN KEMENTERIAN PELAJARAN MALAYSIA PERFECT 2011 SCORE BIOLOGY Teacher s Module PAPER QUESTION 1 1

2 Question 1 : No. Questions Marks Student notes 1 A group of students carried out an experiment to study the effect of the concentration of glucose on the activity of yeast. Diagram 1.1 shows the method used by the students. The initial height of the coloured liquid in the manometer is shown in Diagram 1.2. The experiment was repeated using different concentrations of glucose. Table 1.1 shows the results of the experiment after 10 minutes. clip rubber tubing Rubber stopper Glass tube Manometer with coloured liquid Initial height of coloured liquid Boiling tube containing yeast suspension Diagram 1.1 Initial height of coloured liquid : 1 cm DIAGRAM 1.2 2

3 Percentage concentration of glucose / % Final height of coloured liquid in the manometer after 10 minutes /cm

4 No. Questions Marks Student notes (a) Complete Table 1.2 by recording the height of coloured liquid in the manometer after 10 minutes (b) (i) Based on Table 1.1, state two observations. 1. At 10% concentration of glucose,the final height of coloured liquid after 10 min is cm 2. At 20% concentration of glucose, the final height of coloured liquid after 10 min is 8 cm (ii) State the inference which corresponds to the observation in 1(b(i). 1. Low activity of yeast in lower concentration of glucose, less carbon dioxide is released 2. High activity of yeast in high concentration of glucose, more carbon dioxide is released (c) Complete Table 1.2 for the three variables based on the experiment. Variable Method to handle the variable Manipulated variable: The concentration of glucose Responding variable: Height of coloured liquid// The rate of yeast activity Controlled variable : Volume of yeast suspension /mass of yeast/volume of glucose/ph/light intensity/temperature/time taken Use different concentration of nutrients/glucose Record the height of coloured liquid by using a metre rule // Calculate rate of yeast respiration using formula: = height of coloured liquid time Fix the volume of 100cm of yeast suspension /the mass of 4 g of yeast /ph5 /light intensity at distance of 50cm /temperature at room temperature/time taken for 10 minutes 4

5 (d) State the hypothesis for the experiment. The higher/ lower the concentration of glucose, the higher / lower the rate of yeast activity (e) (i) Based on Table 1.1, construct a table and record the results of the experiment which includes the following aspects: Percentage concentration of glucose Height of coloured liquid The rate of the activity of yeast Percentage concentration of glucose (%) Height of coloured liquid (cm) The rate of the activity of yeast (cm/min) (e) Table 1.1 (ii) Draw a graph of the rate of the activity of yeast against the concentration of glucose (iii ) Based on the graph in 1(e)(ii), state the relationship between the rate of the activity of yeast and the concentration of glucose. Explain your answer. When the concentration of glucose increases/decreases, the rate of yeast activity increases/decreases, more substrate for yeast to use for energy production, more yeast reproduced. (f) Based on the experiment, define anaerobic respiration in yeast operationally. An anaerobic respiration is when yeast using glucose to produce gas that causes the rising of liquid in manometer tube and the process is affected by concentration of glucose 5

6 (g) The experiment is repeated by using 1 ml of 0.1 mol dm - of sodium hydroxide solution is added into the boiling tube. Predict the manometer reading after 10 minutes. Explain your prediction. 1 cm, not increase, sodium hydroxide is alkali, the medium is not suitable for yeast. (h) The following list is part of the apparatus and material used in this experiment. Yeast, metre rule, coloured liquid, electronic balance, glucose solution, measuring cylinder Complete Table 1. by matching each variable with the apparatus and material used in the experiment. Variables Apparatus Material Manipulated Measuring cylinder Glucose Responding Coloured liquid Metre ruler Controlled electronic balance Yeast 6

7 Question 2 : No. Questions Marks Student notes 2 Lemna minor is a species of free-floating aquatic plants from the duckweed family Lemnaceae. The plants grow mainly by vegetative reproduction: two daughter plants bud off from the adult plant. An experiment is carried out to investigate the effect of abiotic factor such as ph on Lemna sp. growth. Experiment is done under controlled conditions: 12 hours a day light exposure and using the same Knop s solution. Petri dish is filled with 20 ml Knop s solution with different ph value and 5 Lemna sp. each. The Knop s solution is treated by adding acid or alkali to achieve the ph value needed. ** Knop s solution is a solution which contains essential nutrient for plants growth. Knop s solution Petri dish Lemna minor Figure 1 After 7 days, the observation is made and the result shown in Table 1.1. ph value Petri dish Number of Lemna sp. 4 7

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9 1 1 Table 1.1 No. Questions Marks Student notes (a) State the number of Lemna sp. in the spaces provided in Table 1.1 (b) (i) Based on Table 1, state two different observations. Able to state any two observations correctly according to 2 criteria: ph ( Manipulated Variable) Number of Lemna sp (Responding Variable) Sample answers: 1. At ph 2 (Knop solution), the number of Lemna sp is 4 2. At ph 8 (Knop solution), the number of Lemna sp is 11. At ph 12 ( Knop solution), the number of Lemna sp is 1 4. At ph 12 (Knop solution), the number of Lemna sp grow is less than at ph 2/4/6/8/10 5. At ph 8 (Knop solution), the number of Lemna sp is more than at ph2/4/6/10/12 (ii) *1,2 & is a horizontal observation *4 & 5 is a vertical observation State the inferences which corresponds to the observations in 1(b)(i). Able to make one logical inference for each observation based on the criteria suitable abiotic factor Favourable for Lemna sp growth Sample answers: 9

10 1. Strong acidic condition is not favorable for Lemna growth. 2. Weak/slight alkaline // neutral condition is most favorable for Lemna growth.. Strong alkaline is not favorable for Lemna growth. 4. Strong alkaline condition is the least favorable for Lemna growth compare with other conditions. 5. Neutral/Slight alkaline condition is the best/moss favorable condition for Lemna growth. *1,2 & is a horizontal inference *4 & 5 is a vertical inference (c) Complete Table 1.4 to show the variables involved in the experiment and how the variables are operated. Variables Manipulated: ph Responding: Number of Lemna sp Fixed: Light exposure / Volume of Knop solution How the variables are operated Add/Use acid or alkali to the Knop solution to get different ph condition// Use ph solution: ph2, ph4, ph6, ph8, ph10,ph12 // change/alter the medium condition Count and record the number of Lemna sp. plants after 7 days. Fix 12 hours light exposure every day / Maintain the volume at 20ml (d) State the hypothesis for this experiment. Able to state a hypothesis to show a relationship between the manipulated variable and responding variable and the hypothesis can be validated, based on criteria: manipulated variable responding variable relationship Sample answer : 1. In low ph, number of Lemna sp is less than in a higher ph. 2. The higher ph the higher number of Lemna sp.. In a neutral condition the number of Lemna sp. 10

11 plants is the highest /the most. 4. The more alkali the medium is the less number of Lemna sp. (e) (i) Construct a table and record the results of the experiment. Your table should contain the following title. ph of water Number of Lemna sp. Able to draw and fill a table labeled with complete unit Sample answers with all columns and rows ph of water Number of Lemna sp (e) (ii) Plot a graph showing the number of Lemna sp against the ph in the graph below Able to plot a graph with criteria: A(axis): correct title with unit and uniform scale P (point) : transferred correctly S (Shape): able to joint all points, smooth graph, bell shape. (iii) Referring to the graph in (e) (ii), describe the relationship between the Lemna sp growth and the condition of the medium. (f) Able to state clearly and accurately the relationship between the condition of medium and Lemna growth based on the criteria: P1- Alkali, acidic or neutral (abiotic factor) P2- Lemna sp. growth Sample answer: (Associates each of the condition with the Lemna growth) 1. In the acidic medium the Lemna sp. growth is less, and increase when the medium become neutral but decrease when in alkali condition. 2 Lemna sp. grow very well in neutral medium and less growth rate in alkali or acidic medium Based on the experiment, define operationally the abiotic factor in an ecosystem. 11

12 Able to explain the abiotic factor operationally base on criteria: Lemna sp (organism) affected (growth) ph of medium (abiotic factor in ecosystem) Sample answer: 1. Abiotic factor is ph of the medium that affect the Lemna sp growth in an ecosystem. (g) (h) The effluent from laundry shop flows into a pond nearby, predict the population of Lemna sp in the pond. Explain your answer. Able to predict the result accurately base on 2 criteria. Expected population of Lemna sp The reason of the answer Not suitable for growth Sample answer: P1- No Lemna sp found/ very small population of Lemna sp, P2- Because water is contaminated with soap/detergent contain alkali, P- Which is not suitable/favourable for Lemna to grow Classify the biotic and abiotic factors from the list provided below. Humidity, light intensity, decomposer, parasites, symbiotic organism, soil texture, invertebrates, topography Able to classify all 4 pairs of the abiotic and biotic factors in ecosystem Sample answer Abiotic factors Humidity Light intensity Soil texture Topography Biotic factors Decomposer Parasite Symbiotic organism invertebrates 12

13 Question : No. Questions Marks Student notes 1. A group of students conducted an experiment to study the effect of light intensity on the population distribution of Lichen on the tree trunk. He placed a 10 cm x 10 cm transparent quadrat on the East-facing surface of the tree trunk. He counted the number of squares that contained half or more than half of the areas covered by the Lichen. Square with less than half of the covered areas were not included. The procedures were repeated for the surfaces that face the direction of North (N), south (S) and west (W). Figure 1 shows how a quadrat is placed on the tree trunk. Each small square represent 1 cm 2. Figure 1 Table 1 shows the areas covered by the Lichen on the different surface of the tree trunk. Direction/position of Total surface area covered by surface Lichen East 60 cm 2 1

14 South 10 cm 5 cm 2 North 10 cm 45 cm 2 West 10 cm 52 cm 2 Table 1 14

15 a) Count the total surface area of Lichen for each quadrat and record the answer in the spaces provided in Table 1. b) (i) State two different observation based on the diagram in Table 1. Observation 1: At the surface facing east (MV), the total surface area of Lichen is 60 cm 2 (RV). Observation 2: At the surface facing south (MV), the total surface area of Lichen is 5 cm 2 (RV). (ii) State the inferences from the observation in 1 (b) (i). Inference from observation 1: At the east aspect is most suitable for the growth of Lichen because it receives more light intensity, so higher rate of photosynthesis. Inference from observation 2: At the south aspect is least suitable for the growth of Lichen because it receives less light intensity, so lower rate of photosynthesis. (c) Complete Table 2 based on this experiment. Variable Manipulated variable Direction facing on the tree trunk // Responding variable Total surface area coverage by Lichen Method to handle the variable Use different direction on the tree trunk such a east, north, south and west. Count and record the total surface area coverage by lichen by using the quadrat. Constant variable 15

16 Quadrat size Type of organism Sampling time Fix the size of quadrat at 10 cm X 10 cm. Fix the organism use in the experiment that is Lichen Sampling experiment is carried out at same time Table 2 d) State the hypothesis for this experiment: 1. The total surface area of Lichen on the tree trunk (RV) is higher (R) when the light intensity is high (MV). 2. When the Lichen is facing east (MV), the total surface area covered by Lichen/population of Lichen (RV) is increase (R).. The higher the light intensity (MV), the higher (R) the total surface area covered by Lichen / the higher the population of Lichen (RV). e) (i) Construct a table and record all data collected in this experiment. Your table should have the following aspect: Title with correct unit Position of direction Total surface area covered by Lichen Position of direction Total surface area covered by Lichen (cm 2 ) East 60 South 5 West 52 North 45 (ii) Use the graph paper provided to answer this question. Using the data in 1 (e) (i), draw a bar chart graph to show the relationship between the population of Lichen against the directions facing on the tree. The population of Lichen is represented by the total surface area covered in the quadrat. 16

17 (f) Based on the graph in 1 (e)(ii), explain the relationship between the population distribution of Lichen and the light intensity. P1 Population of Lichen / Total surface area covered by Lichen P2 Position direction of quadrat P Degree of light intensity Sample answer: 1. Population of Lichen / The total surface area covered by Lichen is higher at east direction which receives high light intensity. 2. Population of Lichen / The total surface area covered by Lichen is low at south direction which receives low light intensity.. Population of Lichen / The total surface area covered by Lichen is higher at east direction than at the south direction because Lichen at east direction receives high light intensity so rate of photosynthesis is higher. (g) State the operational definition for population distribution of Lichen. P1 Total surface area covered by Lichen P2 Size of quadrat P Abiotic factor that influence the population distribution Sample answer: 17

18 1. Population distribution is defined as total surface area covered by Lichen (P1) within the quadrat size of 10 cm x 10 cm at different direction of compass (P2) which influence by the light intensity (P). (h) Lightning strike the tree and cause the tree to fall. The Lichen under study is then exposed to direct sunlight from 7.00 a.m. to 6.00 p.m. daily. Predict what will happen to the total surface area covered by Lichen after a month. Explain your prediction. P1: Prediction of total surface area of Lichen P2: Effect of light intensity P: Effect on the Lichen Sample answer: Size of total surface area covered by lichen is increase / more than 60 cm 2 because Lichen receive more sunlight / light intensity, so more photosynthesis by Lichen and more growth to Lichen. (i) The following is a list of biotic and abiotic factors. ph of water, pigeon orchid, humidity, bird, temperature, Elodea sp. Classify these factors in the Table. Abiotic factors ph of water Humidity Temperature Biotic factors Pigeon orchid Bird Elodea sp 18

19 Question 4 : No. Questions Marks Student notes 4 An experiment was carried out to investigate the water pollution level or BOD in three different locations from a suspected polluted Rivers. Three water samples are collected from these three locations and labelled as P, Q and R as in Diagram ml of each sample is put in a reagent bottle and added with 1 ml of 0.1% methylene blue solution. All the bottles are kept in dark cupboard. Observations are made every minute to see the changes in the methylene blue colour. Sample P Sample Q Sample R Table 1 shows the results of this experiment. Diagram 1 Each sample is added with methylene blue solution Water sample P Q R Time taken for methylene blue solution become colourless 10 minutes 2 minutes 42 minutes Table 1 19

20 No. Questions Marks Student notes (a) Record the time taken for methylene blue solution become colourless in the boxes provided in Table 1. (b) (i) Based on Table 1, state two different observations. Able to state any two observations correctly according to the criteria: o Sample o Time taken o Become colourless Sample answers: 1. Time taken for methylene blue to become colourless for sample P is 10 minutes. 2. Time taken for methylene blue to become colourless for sample R is 42 minutes. Time taken for methylene blue to become colourless for sample Q is 2 minutes 4. Time taken for sample P is 10 minutes that is shorter than time taken for sample R that is 42 minutes to become colourless (ii) State the inferences which corresponds to the observations in 1(b)(i). Able to make one logical inference for each observation based on the criteria o o o Sample Oxygen concentration Duration of time for methylene blue to become colourless Sample answers: 1. In sample P, oxygen concentration is low, the methylene blue become colourless very fast/ less time taken 2. Oxygen concentration in sample R is high, the methylene blue become colourless slow/ longer time taken. Oxygen concentration in sample P is lower than oxygen concentration in sample R, the time taken for methylene blue to become colourless is shorter. 20

21 (c) Complete Table 2 based on this experiment. Variables Manipulated: Water sample Responding variable Time taken to decolourise methylene blue How the variables are operated Water sample is collected from three different locations. Time taken for methylene blue to become colourless is recorded by using a stopwatch. Fixed variable Metlhylene blue concentration / volume/ volume of water sample 0.1% of Methylene blue is used for all experiments/ 1 ml volume/ 200 ml of water sample. Table 2 (d) State the hypothesis for this experiment. Able to state a hypothesis to show a relationship between the manipulated variable and responding variable and the hypothesis can be validated, base on criteria: manipulated variable responding variable relationship Sample answer : 1. The most polluted water has shortest time for methylene blue to become colourless. 2. Sample water P is the most polluted has shortest time for methylene blue to become colourless.. Sample water R is less polluted compare to water samples P and Q, has longest time for methylene blue to become colourless, (e) (i) Construct a table and record all the data collected in this experiment based on the following criteria: Water sample Time taken 21

22 Able to tabulate a table and fill in data accurately base on three criteria: o Table draw with labeled column. o Sample o Time taken with unit. Sample answers : Water Sample Time taken ( minutes) P 10 Q 2 R 42 (f) Based on the data in 1(e) draw a bar chart of time taken for methylene blue solution become colourless against water samples. Able to draw a bar chart base on criteria: o Correct chart o Axis with correct scale o Correct value (g) (h) What is the relationship between time taken, oxygen concentration and BOD value of water in this experiment? Able to state clearly and accurately the relationship between: o time taken o oxygen content o BOD value Sample answer: 1. The shorter time taken for methylene blue to become colourless, less oxygen in the water and BOD value is high. Based on the result of this experiment, state the operational definition for BOD Able to explain BOD base on experiment correctly according to the criteria: o Amount of oxygen in the water sample o used by microorganisms o shown by time taken Sample answer: 22

23 2. BOD is amount of oxygen in the water sample that used by microorganisms and can be shown by time taken of methylene blue to become colourless. (i) This experiment is repeated by using water sample from chicken farm areas. Predict the time taken for methelyne blue to become colourless. Able to predict the result accurately. o Expected time o Compare to which o Reason Sample answer: The time taken for methylene blue to become colourless is 5 minutes, less than water sample P, because chicken farm water can be contaminated with chicken faeces/ or any other answer. (j) Arrange the water samples from the most polluted to the least polluted. Able to arrange the level of polluted water Sample answer: Types of water Polluted P Q R Most Moderate Least Most polluted least polluted P Q R 2

24 Question 5 : No. Questions Marks Student notes 5 Transpiration is the evaporation of water from a plant to the surroundings. The rate of transpiration is affected by environmental factors such as temperature. A group of students carried out an experiment to study the effect of temperature on the rate of transpiration. Diagram 1 shows the set up of the apparatus. An air bubble was trapped in the capillary tube. The apparatus was placed in an air-conditioned room at 20 o C. The time taken for the air bubble to move a distance of 10 cm was recorded. The experiment was repeated for a second time to get average readings. The experiment is repeated by placing the apparatus at three more different temperatures: an airconditioned room at 25 o C, an air-conditioned room at 0 o C and in a non air-conditioned room at 5 o C. Diagram 1 Table 1 shows the reading of stopwatch for air bubble to move a distance of 10 cm at different temperature 24

25 Temperature o C Time taken for air bubble to move a distance of 10 cm (min) First reading Second reading Average Reading

26 Temperature Suhu o C Time taken for air bubble to move a distance of 10 cm (min) First reading Second reading Average Reading No. Questions Marks Student notes (a) Record the time taken for the air bubbles to move a distance of 10 cm and average reading in Table 1. (b) (i) Based on Table 1, state two different observations. 1. When temperature is 20 o C, the average time taken for air bubble to move a distance of 10 cm is 40 minutes 2. When temperature is 5 o C, the average time taken for air bubble to move a distance of 10 cm is 10 minutes. 26

27 . When temperature is 20 o C,the average time taken for air bubble to move a distance of 10 cm is longer than the average time taken when temperature is 5 o C (ii) State the inferences which corresponds to the observations in 1(b)(i). (c) 1. (When temperature is low), the amount of water lost from the leaf is low(p1). So the rate of transpiration is low (P2) 2. (When temperature is high), the amount of water lost from the leaf is high(p1). So the rate of transpiration is high (P2). When the temperature is higher/lower, the amount of water lost from the leaf is higher/lower. So the rate of transpiration is higher/lower when the temperature is higher/lower Complete Table 2 based on this experiment. Variable Method to handle the variable Manipulated Variable Temperature Responding Variable 1.Rate of transpiration 2. Time taken for air bubble to move a distance of 10 cm Place the apparatus/potometer at different temperature / 20 o C, 25 o C, 0 o C and 5 o C 1.Calculate and record the rate of transpiration by using formula : Distance / time 2. Record the time taken for air bubble to move a distance of 10 cm by using stopwatch Constant Variable 1.Type of plant 2.Distance travelled by air bubble 1.Use the same plant 2.Fix the distance travelled by air bubble at 10cm 27

28 (d) State the hypothesis for this experiment. Able to make a hypothesis based on the following aspects P1 : MV- Temperature P2 : RV rate of transpiration H : Relationship (Higher. Higher) Sample Answer: 1. The higher the temperature, the higher the rate of transpiration//vice versa (e) (i) Construct a table and record all the data collected in this experiment. Your table should have the following aspects: - Temperature - Average time taken for air bubbles to move a distance of 10 cm. - Rate of transpiration Rate of transpiration = Distance Time Able to construct a table based on the following aspects 1. Title with correct unit - 1 mark 2. Data - 1 mark. Rate of transpiration - 1 mark Sample Answer Temperature Suhu o C Average time taken for air by air bubble to move a distance of 10 cm (min) Rate of transpiration cm/min (e) (ii) Using the data in 1(e)(i), draw the graph of the rate of transpiration against the temperature Able to draw the graph correctly Axes : Uniform scales on both horizontal and vertical axis with correct unit 1 mark Points : All points plotted correctly - 1 mark Curve : smooth without touching the axes - 1 mark 28

29 (f) Based on the graph in 1(e)(ii), explain the relationship between the rate of transpiration and temperature. Able to explain the relationship between the rate of transpiration and temperature based on the following aspects. P1 State the relationship P2 kinetic energy of water P evaporation Sample Answer When the temperature increases, the rate of transpiration increases. When the temperature increases, kinetic energy of water molecules (in the leaf) increases, causes the rate of evaporation increase. (g) Based on the result of this experiment, state the operational definition for process of transpiration. Able to define operationally the process of transpiration based on the following aspects: P1 water loss from plant at different places P2 Air bubble in capillary tube move at 10 cm P The rate of transpiration is influenced by temperature Sample Answer Transpiration is a process where water is lost from the plant when it is placed at different temperature which causes the air bubble in capillary tube move a distance of 10 cm. The rate of transpiration is influenced by the temperature. (h) If the surface of the leaves of a plant at temperature of 5 o C are covered with vaselin, predict the time taken for air bubble to move a distance of 10 cm. Explain your prediction. Able to predict the outcome of the experiment based on the following aspects P1 : Correct prediction P2 : Effect P : Reason Sample Answer Time taken for air to move a distance of 10 cm is more than 10 minutes. Rate of transpiration decreases because vaselin covered the stomata/stomata closed 29

30 (i) The following list is a factor that affecting transpiration. Relative humidity Kelembapan relatif cuticle kutikel air movement pergerakan angin stomata stomata light intensity keamatan cahaya Classify the factors into two group in Table. Environmental factor Morphology factors 1. Relative humidity 2. Air movement 1. Cuticle 2. Stomata. Light intensity Table 0

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