Suggested answer P3
Question 1
No Aspect or criteria marks remark
1(a) [KB0603- Measuring Using Number]
Able to Record all the mass of water absorps by Hibiscus leafy shoot in Table 1 correctly.
Sample answers:
Condition Mass of water absorps by Hibiscus leafy shoot
Present of anhydrous calcium chloride
18.0 g /grams
Absent of anhydrous calcium chloride
3.6 g/ grams
2 correct
mass with units
3
Able to record all/ 2 mass of water absorps by Hibiscus leafy shoot without unit //
record 1mass of water absorps by Hibiscus leafy shoot correctly and one inaccurate mass
2
Able to record 2 mass at idea level 1
No response or wrong response 0
1(b)(i) [KB0601- Observation]
Able to state any two observations correctly based on the aspects : P1: Manipulated variable:
Present or absent of anhydrous calcium chloride / condition in transparent polythene bag
P2: Responding variable:
Mass of water absorps by Hibiscus leafy shoot H: correct relationship.
Sample answer:
1. In the present of anhydrous calcium chloride, the mass of leafy shoot of potted plant after experiment is 2229.3g/grams / the mass of water absorps by Hibiscus leafy shoot is 18.0 g/grams.
2. In the absent of anhydrous calcium chloride, the mass of leafy shoot of potted plant after experiment is 244.3g/grams / the mass of water absorps by Hibuscus leafy shoot is 3.6 g/grams.
3
Able to state any one observation correctly or any two incomplete observations (any 2 aspects)
2
No response of wrong response 0 1(b)(ii) [KB0604- Making Inferences]
Able to make correct inferences based any 2 aspects as follows: P1: More or less humidity
P2: More/less water is absorps by roots // evaporation of water vapour is high/low // rate of transpiration is high / low
Sample answer:
1. In the present of anhydrous calcium chloride,
The humidity is low, so more water is absorps by roots //
evaporation of water vapour is high , hence the rate transpiration is high
2. In the absent of anhydrous calcium chloride,
The humidity is high, so less water is absorps by roots //
evaporation of water vapour is low , hence the rate transpiration is low.
3
Able to state one correct inference and one inaccurate inference or two inaccurate inferences
Sample answers:
1. In the present of anhydrous calcium chloride, the rate of transpiration is high/ more // humidity is low
2. In the absent of anhydrous calcium chloride, the rate of transpiration is low/ less // humidity is high
2
Able to state two inferences at idea level Sample answer:
1. More/ less water taken by plant 2. Plant lost water
3. Transpiration occurs
1
No response of wrong response
0
1(c) KB0610 – Controlling variables Able to state all 6 items from the table correctly
Variable
Pembolehubah
Method to handle the variable
Cara mengendalikan pembolehubah
Manipulated variable / Pembolehubah
dimanipulasikan:
Hadir atau tanpa kalsium klorida
Menukarkan beg politena lutsinar dengan kehadiran kalsium klorida kontang atau dengan tiada kalsium klorida kontang.
// Menggunakan keadaan
kontang // keadaan eksperimen
eksperimen yang berlainan dalam eksperimen.
Responding variable / Pembolehubah
bergerak balas:
Jisim air yang diserap oleh pasu tumbuhan berpucuk // Kadar transpirasi tumbuhan
Menghitung jisim air yang diserap oleh pasu tumbuhan berpucuk menggunakan rumus:
Jisim sebelum eksperimen – jisim selepas 9 jam
//Hitung kadar transpirasi tumbuhan menggunakan rumus:
Jisim air diserap oleh tumbuhan (g)
9 ( jam)
Constant variable /
Pembolehubah malar:
Jenis tumbuhan berpucuk // suhu // keamatan cahaya
Gunakan pucuk bunga Raya //gunakan suhu/keamatan cahaya yang sama dalam eksperimen ini.
Able to state any 4-5 items correctly
2
Able to state any 1-3 items correctly
1
No response or wrong response 0
1(d) KB0611 – Stating a hypothesis
Able to write a complete hypothesis statement correctly based on the following aspects :
P1 = manipulated variable
Present /absent of anhydrous calcium chloride P2 = responding variable
Low/high humidity // high /low rate of transpiration // high/low evaporation of water by leafy shoot // high /low water absorps by roots.
H = Relationship / Link (comparing)
Sample answer:
1. In the present of anhydrous calcium chloride / low humidity, the rate of transpiration / water loss / evaporation of water by leafy shoot / water absorps by roots is higher compare to condition without anhydrous calcium chloride / high humidity.
3
Able to write hypothesis with any two aspects: P1 and P2 // P1 and H // P2 and H
Able to write hypothesis with any one aspect: P1 or P2
1
No response or wrong response 0
1(e)(i) KB0606 – Communicating
Able to draw a table and show all the 3 aspects as Follows correctly :
T – Heading in the table are labeled with correct units D – All data are correct
C – The correct calculation of transpiration rate
Sample answer:
Keadaan
Jisim pasu pucuk berdaun
selepas eksperimen
(g)
Jisim air diserap
oleh pucuk berdaun
Bunga Raya.
(g)
Kadar transpirasi
g/jam
Dengan kehadiran
kalsium klorida kontang
229.9 18.0 2.0
Tanpa kehadiran
kalsium klorida kontang
244.3 3.6 0.4
3
Able to state any two aspects correctly 2
Able to state any one aspect correctly 1
No response or wrong response 0
T
D
1(e)(ii) KB0607 – Using spatial and time relationship
Able to plot a bar chart graph correctly with the following aspects: P – all axis with uniform scale and correct units
T – all point is transferred correctly
B – Bar chart graph was plotted / separate /combine bar Sample answer:
3
Able to state any two aspects correctly
2
Able to state any one aspect correctly
1
No response or wrong response
0
1(e)(iii) KB0608 – Interpreting Data
Able to explain the relationship between the condition of the experiment and the rate of transpiration of leafy shoot correctly based on the following aspects:
P1: Hypothesis statement P2: Humidity high / low
P3: More / less water evaporates through stomatal pores of leaves / more water absorps by roots
1. In the transparent polythene bag with the present of anhydrous calcium chloride gives higher rate of transpiration (in Hibiscus leafy shoot) compare to the polythene bag without anhydrous calcium chloride.
This is because the humidity is low / surrounding air at the leaf is dry. Hence more water vapour evaporates from the stomatal pore of leaves / more water is absorps by roots.
3
Able to state any two aspects correctly
2
Able to state any one aspect correctly.
1
No response or wrong response
0
1(f) KB0605 – Predicting
Able to predict the mass of water absorps by roots of plant correctly based on the following aspects:
P1: Mass of water absorps by roots increase / more /higher than 3.6 g /grams
P2: High Rate of transpiration
P3: More water evaporates through stomatal pores of leaves
Sample answer:
1. The mass of water absorps by roots increases / more / higher than 3.6 grams because the rate of transpiration is higher / increases (surrounding air is dried) causing more water evaporates from the
leaf stomatal pores .
3
Able to state any two of the aspects correctly
2
Able to state any one of the aspect correctly.
1
No response or wrong response
0
1(g) KB0609 – Defining by operation
Able to describe the concept of transpiration based on the following aspects:
P1: Process water absorps by roots of Hibiscus leafy shoot / reduce in mass of water absorps by plant
P2: Shown by (final) the mass of potted leafy shoot after experiment.
P3: Affected by humidity / condition of plant.
Sample answer:
1. Transpiration is a process whereby water is absorps by roots of Hibixcus leafy shoot / reduction in the mass of water absorps by plant . It is shown by the (final) mass of water absorps by plant after experiment. The rate of transpiration is affected by humidity / condition of plant.
Able to state any two of the aspects correctly 2
Able to state any one of the aspect correctly // at idea level / text book definition of transpiration
1
No response or wrong response 0
1(h) KB0602 – Classifying
Able to classify all / 6 correct ticks for high and low transpiration as follows:
Sample anwer:
High transpiration Low transpiration Dry wind
Sunny day More leaves
High humidity Cool temperature Low speed of fan
3
Able to classify 4-5 ticks correctly 2
Able to classify 1-3 ticks correct 1
No response of wrong response 0
Tentatif answer
Question 2
Question Score Explanation Remarks
2 (i) 3 Able to state the problem statement correctly based on:
P1 : light intensity/ distance of light bulb to the Lemna sp. P2 : the population size of Lemna sp
H : Question form
Sample answer:
1. What is the effect of light intensity on the population size of Lemna sp.?
2. Does light intensity affect the population size of Lemna sp.?
3. How does the distance of light bulb to the Lemna sp. affect the population size of Lemna sp.?
2 Able to state the problem statement based on any 2 aspects Sample answer :
1. What is the effect of light intensity on the population size of Lemna sp.
2. Does (different) light intensity affects Lemna sp.?
KB061202 ( KB061203 – Making Hypothesis )
Question Score Explanation Remarks
2 (ii) 3 Able to state the hypothesis based on the following aspects: P1 = Manipulated variable
P2 = Responding variable R = Relationship / Link Sample answer :
1. The higher/increase the light intensity, the higher/increase the population size of Lemna sp.
2. The less/ smaller the distance of light bulb to the Lemna sp, the bigger/ larger population size of Lemna sp.
2
Able to state the hypothesis based on any 2 aspects // P1 & P2 // P1 and R // P2 and R
Sample answer :
1. Different light intensity produce /gives different population size of Lemna sp.
1 Able to state the hypothesis based on any 1 aspect // P1 // P2 Sample answer :
1. Light intensity affects Lemna sp.
0 Wrong answer or no response
KB061203 ( Variables)
Question Score Explanation Remarks
2 (iii) 3 Able to state all the three variables correctly Sample answers :
1. Manipulated variable
Light intensity // the distance of light bulb to the Lemna sp.
2 Responding variable
Population size of Lemna sp.
3 Constant variable 1. pH
2. Temperature
3. number of Lemna sp.
1. Light intensity affect lemna sp.
2 Able to state any 2 correct variables
1 Able to state any 1 correct variable
0 Wrong answer or no response
KB061204 - ( Apparatus and materials)
Question Score Explanation Remarks
2(iv) 3 Able to list out all the important apparatus and materials correctly or any 2 Materials (M) and 4 Apparatus (A) Sample answers:
Apparatus:
Termometer, light bulb/table lamp, ruler/ metre rule, beaker (500ml), Marker pen
Materials:
(100) Lemna sp, pond water
2 Able to list out any 1- 2M and 3 A
1 Able to list out any 1M and 1-2A
0 Wrong answer or no response
KB061205 - ( Experimental Procedures)
Question Score Explanation Remarks
2(iv) 3 Able to explain the correct procedures based on: K1. Setting apparatus
K2. Operating Constant variable. K3. Operating Responding variable K4. Operating Manipulated variable K5. Precaution
Sample answers:
1. Filled up five beakers with 300ml of pond water and labelled as P, Q, R, S and T.
2. Placed ten lemna sp. in each of the beakers
3. Placed the beaker P at distance 10cm from the light bulb/source of light/table lamp
4. Swicth on the light and leave it for two weeks. 5. Count the number of Lemna sp. in the beaker after
two weeks.
6. Calculate the population size of Lemna sp and record into a table.
7. Repeat steps 1to 6 by using different distances of light bulb/light source/table lamp such as 20cm,
K1/K2
K1/K2 K1
K1 K1/K3
K1/K3
30cm, 40cm and 50cm
8. All results are recorded in a Table
9. Repeat the experiment twice to get the accurate average readings
* Notes:
1. The distance of light bulb to the lemna sp. ...must have at least 3 distance.. (range 5 – 100cm)
2. To get K1 there must be more than 4 steps. K2/K3/K4 and K5 any one step
K1 K5
3 Any 5K’s
2 Any 3 – 4K’s
1 Any 1 - 2 K’s
0 Wrong answer or no response
KB061206 ( Presentation of Data)
Question Score Explanation Remarks
2(v) 2 Able to present all the data with units correctly OR based on P1: Title with correct units
P2: Correct sample
Sample answer: Beaker Distance of
light bulb to Lemna sp. (cm)
Number of Lemna
(unit)
Population size of Lemna sp.
(unit) Initial Final
P Q R S T
1 Able to present data based on any one P’s
