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PAPER 3PAPER 3SPM SPM
PAPER 3PAPER 3SPM SPM
Example 1 : Question 1_2011Example 1 : Question 1_2011
QUESTION 1QUESTION 1QUESTION 1QUESTION 1
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OBSERVATIONSOBSERVATIONSOBSERVATIONSOBSERVATIONS
1(a)(i) Observation• P1 : environment condition• P2 : reading of stopwatch
• ECF : Error Carry Forward (for air column reading)
1 (b) (i) [3m] 1. The first/second/third reading at
moving air condition is ……………min
2. The first/second/third reading at still air condition is ……………min
Horizontal observation
Scoring Obs. 1 Obs. 2 Marks
Acc. Acc. 3
Acc. Inacc. 2
Inacc. Inacc. 2
Acc. Wrong 1
Acc. Idea 1
Inacc. Idea 1
Idea Idea 1
Inacc. Wrong 0
Idea Wrong 0
INFERENCESINFERENCESINFERENCESINFERENCES
1(a) (ii) [3m]1. At moving air, the rate of
transpiration is high.
2. At still air, the rate of transpiration is low.
Horizontal inference
CONSTRUCT A TABLECONSTRUCT A TABLECONSTRUCT A TABLECONSTRUCT A TABLE
1 (b) Table 1
environment
First Second Third
Moving air
Still air
1(c) (i) [3m]
1. At moving air =………………
2. At still air = ………………
1 (c) (ii)
environment
First Second Third ROT
Moving air
Still air
VARIABLESVARIABLESVARIABLESVARIABLES
1(d) (i)
Variable Method to handle the variable
MV
Environment condition
Changed the …………………….. at ……………and…………………
Variable Method to handle the variable
RV
Time taken for the air bubble to move a distance of 5 cm
The rate of transpiration
Measure and record …………….. by using a stopwatch
Calculate the……………………….. by using the formula :
Distance from X to YTime taken
Variable Method to handle the variable
CV
Length of distance
Type of plant
Temperature
Fix …………………………………..at ……… cm
Use the same …………………to carry out the xtvt
Fix the ………………………………… at room temperature
1 (c)• Scoring :
6 √ = 3 marks
4 – 5 √ = 2 marks
2 – 3 √ = 1 marks
1 √ = 0 marks
HYPOTHESISHYPOTHESISHYPOTHESISHYPOTHESIS
1 (d)Must have:
• MV : environment condition• RV : rate of transpiration• CV : Relationship
1 (d)(ii) [3m]1. At moving air condition, the rate
of transpiration is high.
2. At …………… air condition, the rate of transpiration is …………
RELATIONSHIPRELATIONSHIPRELATIONSHIPRELATIONSHIP
1(f)• Criteria:
– Relationship (MV & RV from the graph)
– Explanation 1– Explanation 2
1(f) [3m]
1. As the time of xtvt increases, the rate of transpiration increases because more water loses through evaporation and the movement of air bubble increases.
PREDICTIONPREDICTIONPREDICTIONPREDICTION
1(g)
• Criteria:– Correct prediction.
– 2 Reason.
1(g) [3m]
1. The rate of transpiration will increase because the evaporation of water is high and the time taken for air bubble to move a distance of 5 cm is higher.
OPERATIONAL OPERATIONAL DEFINITIONDEFINITION
OPERATIONAL OPERATIONAL DEFINITIONDEFINITION
1(e)
• Transpiration is the time taken of the air bubble to move a distance of 5 cm and affected by the condition of air.
CLASSIFYINGCLASSIFYINGCLASSIFYINGCLASSIFYING
1(i)
Scoring:7 ticks = 3 marks4-6 ticks = 2 marks2-3 ticks = 1 mark0-1 ticks = 0 mark
Variables Apparatus Material
Manipulated
Fan Speed of fan
Responding
Stopwatch Air bubble in the capillary tube
Controlled Beaker Volume of water
PAPER 3PAPER 3SPM SPM
PAPER 3PAPER 3SPM SPM
Question 1, 2007Question 1, 2007
QUESTION 1QUESTION 1QUESTION 1QUESTION 1
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CLASSIFYINGCLASSIFYINGCLASSIFYINGCLASSIFYING
1(a)Materials (M) Apparatus (A)
1. 2% sodium hydrogen carbonate solution 2. Hydrilla sp
1. 60wbulb2. Ruler3. Boiling tube4. Paper clip5. Retort stand
1(b) Distance between light
source and Hydrilla sp /cm Total number of
bubbles released in 5 minutes
60 10
50 12
40 15
30 20
OBSERVATIONSOBSERVATIONSOBSERVATIONSOBSERVATIONS
1(c)(i)1. At a distance of 60 cm, the total number of
bubbles released is 10.
2. At a distance of 30 cm, the total number of bubbles
released is 20 .
3. The number of bubbles released at 60 cm is less than the
total number of bubbles released at 30 cm // inversely
Scoring Obs. 1 Obs. 2 Marks
Acc. Acc. 3
Acc. Inacc. 2
Inacc. Inacc. 2
Acc. Wrong 1
Acc. Idea 1
Inacc. Idea 1
Idea Idea 1
Inacc. Wrong 0
Idea Wrong 0
INFERENCESINFERENCESINFERENCESINFERENCES
1(c)(ii)1. At a distance of 60 cm , the light intensity / (rate
of) photosynthesis is low / less.
2. At a distance of 30 cm , the light intensity / (rate of) photosynthesis is high/more.
3. Light intensity (Rate of) photosynthesis at 60 cm is less than light intensity / (rate of) photosynthesis at 30 cm.
VARIABLESVARIABLESVARIABLESVARIABLES
1(c)
Variable Method to handle the variable
MV
Distance between light source and Hydrilla sp
Light intensity
Changed the …………………….. at 60 cm, 50 cm, 40 cm and 30 cm Use different …………………………….
Variable Method to handle the variable
RV
Total numbers of bubbles released in 5 minutes
Rate of photosynthesis
Measure and record …………….. by using a stopwatch
Calculate the……………………….. by using the formula :
Number of bubblesTime taken
Variable Method to handle the variable
CV
Time taken to count the total number of bubbles
Type of plant
Power of bulb
Fix …………………………………..at 5 minutes
Use the same ………………… / Hydrilla sp
Fix the ………………………………… at 60W
1 (c)• Scoring :
6 √ = 3 marks
4 – 5 √ = 2 marks
2 – 3 √ = 1 marks
1 √ = 0 marks
HYPOTHESISHYPOTHESISHYPOTHESISHYPOTHESIS
1 (d)Must have:
• MV : light intensity/distance between
light source and Hydrilla sp.• RV : R.O.P• CV : Relationship
1(e)
1. As the light intensity increases, the rate of photosynthesis increases / the total number of bubbles released increases // inversely .
2. As the distance between the light source and Hydrilla sp. decreases, the rate of photosynthesis increases / the total number of bubbles released
increases // inversely .
CONSTRUCT A TABLECONSTRUCT A TABLECONSTRUCT A TABLECONSTRUCT A TABLE
1(f)(i) Distance between light source
and Hydrilla sp. / cm 60 50 40 30
Number of bubbles released in 5 minutes 10 12 15 20
Light intensity / cm-1 // 1/cm 0.017 0.020 0.025 0.033
1(f) (i)Distance between light
source and Hydrilla sp. / cm
Number of bubbles released in 5
minutes
Light intensity / cm-1 // 1/cm
DRAW A GRAPHDRAW A GRAPHDRAW A GRAPHDRAW A GRAPH
1(f)(ii)
P (paksi) : Correct title with unit on both horizontal, vertical axis and uniform scale on the axis.
T (titik) : All points plotted/transferred correctly .
B (bentuk) : Able to joint any 2 points to form a smooth graph , and positive gradient
RELATIONSHIPRELATIONSHIPRELATIONSHIPRELATIONSHIP
• Criteria:– Relationship (MV & RV from the
graph) – Explanation 1– Explanation 2
1(g)Relationship:P1 = Able to state the relationship between the manipulated variable and responding variable.
Explanation:P2 = Able to state rate of photosynthesis increases.P3 = Able to state more gas/oxygen is produced
Sample answer: When the light intensity increases, the total number of bubbles increases because the rate of photosynthesis increases. More gas/oxygen is produced.
PREDICTIONPREDICTIONPREDICTIONPREDICTION
1(h)
Sample answer :
1. The total number of bubbles released is higher / increase because two sprigs have more leaves and the rate of photosynthesis increased.
2. The total number of bubbles released is the same because two sprigs share (the same) concentration of carbon dioxide.
OPERATIONAL OPERATIONAL DEFINITIONDEFINITION
OPERATIONAL OPERATIONAL DEFINITIONDEFINITION
1(i)P1 : Hydrilla sp. in sodium hydrogen carbonate solution P2 : Produces bubbles / gasP3 : Factor
Sample answer :
1. Photosynthesis is a process occurring in Hydrilla sp in sodium hydrogen carbonate solution and produces bubbles / gas. The number of bubbles produced / rate of photosynthesis is influenced by the light intensity / distance from light source.