Question 1i. P- Mitochondria
R- Choloroplast
ii. P- Site of energy production
R- Has chorophyll to trap sunlight and undergo
photosynthesis.
a) Q is where the ribosome attached to. Protein that is
synthesised by the ribosome cannot be transported when Q is absent.
Enzyme production cannot be occured as the protein is not
transported to golgi apparatus.Question 2
Essay Section B
a) Nitrates and phosphates that contain in fertilizers flow to
the rivers and lakes. This will cause algal bloom that prevent the
penetration of sunlight to the lake. The submerged plants receive
less amount of sunlight and cannot undergo photosynthesis. As a
result, the plants will die and oxygen will be less leading to the
death of other aquatic organism such as fishes. The fertiliser will
be used up that causes the alga to die by itself. Dead substance
will lead to high amount of bacteria to undergo decomposition on
death substance. Much bacteria will cause less amout of oxygen that
lead to high BOD level.this will cause dead zone as no life can
survive. The example is at Gulf of Mexico Dead Zone.b) i) Treating
sewage can remove pathogens and bacteria that carry various
diseases such as malaria. It can also improve the environment
through proper drainage and disposal of waste water. Treating
sewage can maintain water ecosystem ingtergrity so that aquatic
organism can stay alive. It can prevent flood as no sediments and
materials deposited in lakes so that water can move easily.
Treating sewage can preserving water quality for human uses.ii)
Organic fertilizer is beneficial to water environment rather than
inorganic fertilizer because it wont build up harmful residues or
cause pollution due to run-off from irrigation or rain. Organic
fertilizer have adequate amount of phosphorus but inorganic
fertilizers have excess amount of phosphorus that lead to
eutrophication. Organic substance have neutral pH value rather than
inorganic substance that is acidic so, it can mantain the pH value
of water. Organic substance does not contain any chemical so
aquatic plants can live and absorb dissolved carbon dioxide in
water.Question 2
a) Yes. Milk contains high amount of calcium and phophorus that
is needed for the formation of strong teeth and bones. It also
contains protein that function for the formation of new tissues.
Wholemeal bread contains high amount of carbohydrate that provide
energy for the daily activities. It also contains minerals such as
iodine, zinc and iron that is important in body regulation. Butter
contains protein to help the adult to grow and lipid to provide
stored energy.b) When the food is chewed in the mouth, salivary
galnd screted salivary amylase. Amylase will digest carbohydarate
into maltose. The food enter the oesophagus and it is called as
bolus. In the stomah, hydrocloric acid kills the bacteria in the
food. The food enter the small intestine, amylase secreted by
pancreas digest starch to maltose. At the end of the ileum, the
intestinal juice that contains maltase, sucrase and lactase enzymes
will digest the carbohydrates into its constituent monosaccharides.
The monosaccharides will be absorbed at the villi to the
bloodstream and transported to all parts of the body.
QUESTION 3
In X, antibodies are produced by the body while in Y, antibodies
are obtained directly. The reaction of X is slow while the reaction
of Y is immediate. X is injected with vaccine while Y is injected
with antiserum that contains specific antibody. X is long lasting
immunity while y is short lasting immunity. In X, it is injected
before a person is infected with disease while in Y, it is injected
when person is infected or have high risk of getting the disease.
In x, it needs second injection or booster dose because first
injection ussually is slow and low level of antibody while in Y, it
does not need second injestion.PAPER 3
1. Diagram 1 shows an experiment that had been carried out to
investigate the effect of air movement on the rate of transpiration
hibiscus plant by using a potometer. Time is taken for an air
bubble to move from X to Y (10 cm distance) by using stopwatch.
Rajah 1 menunjukkan satu eksperimen yang dijalankan untuk
mengkaji kesan pergerakan udara terhadap kadar transpirasi pokok
bunga raya menggunakan satu potometer. Masa di ambil untuk satu
gelembung udara untuk bergerak dari X ke Y [jarak 10 cm] dengan
menggunakan jam randik
The potometer is placed near a fan with air speed adjusted at
different velocity in the same room with temperature 30oC. The
result of the experiment is shown in the table 1.
Potometer ini diletakkan dalam bilik yang sama, berhampiran
dengan kipas angin dimana halaju angin berbeza. Keputusan
eksperimen ditunjukkan dalam Jadual 1Fan speedStopwatch readingTime
taken by air bubble to move from X to Y (minute)
136
228
3
18
415
Base on the Diagram 1 and Tables 1 answer all questions:1(a)(i)
State two observations on the time of air bubble moves.
Nyatakan dua pemerhatian tentang masa dalam eksperimen ini
Observation 1
When the fan speed is 1, the time taken by air bubble to move
from X to Yis 36 minutes.
Observation 2
When the fan speed is 4, the time taken by air bubble to move
from X to Y is 15 minutes.
[3 marks]
(a)(ii)State one inference for each observation made in (a)
(i).
Nyatakan satu inferen bagi setiap pemerhatian yang dibuat pada
1(a)(i)Inference for observation 1 The time taken by air bubble to
move from X to Y is the longest because water molecule evaporate at
surface of the leaves slowly under slow air movement causing the
air bubble to move in long time.Inference for observation 2
The time taken by air bubble to move from X to Y is the shortest
because water molecule evaporate at surface of the leaves rapidly
under fast air movement causes it to move in short time.
[3 marks]
(b)(i)Draw a table based on the following criteria:
Bina satu jadual berdasarkan criteria berikut
Fan speed Time taken
Transpiration rate[Transpiration rate = Distance [cm/min] Time
Fan speed
Time taken for air bubbles to move (min)
Transpiration rate(cm /min)
1
36
0.28
2
28
0.36
3
18
0.56
4
15
0.67
[3 marks](c)
State the relationship between the increase of air movement and
the rate of transpiration. Explain your answer
Nyatakan hubungan antara pertambahan keamatan cahaya dengan
kadar transpirasi. Terangkan jawapan anda
When the air movement increases, the rate of transpiration
increases. This is because the water
Molecules evaporate at surface of the leaves rapidly under high
air movement causing the rate of transpiration increases.
[3 marks]
(d)
State the variables in the experiment and explain how the
variables are operated
Nyatakan pembolehubah dan cara mengendalikan
pembolehubahVariables in the experiment
How the variables are operated
Manipulated variable
Air movementUse different speed of fan that is 1, 2, 3 and4
Responding variable:
Rate of transpiration
Calculate and record the rate of transpiration by using formula
distance/time (cm/min)
Fixed variables :
Temperature
Fix the temperature at 30 degree celcius.
[3 mar(e)
State the hypothesis for this experiment.
Nyatakan hipotesis eksperimen ini
The faster the air movement, the higher the rate of
transpiration.
[3 marks]
(f)
Based on the experiment, define transpiration operationally
Berdasarkan eksperimen, beri definisi transpirasi secara
operasi
Transpiration is a process of loss of water vapour shown by the
time taken by air bubble to travel from X to Y affected by the air
movement.
[3 marks](g)
The experiment is repeated by placing the set-up in the dark
without fan. Predict transpiration rate of the plant shoot under
this condition. Explain your prediction.
Jika eksperimen diulang dengan meletakkan radas dalam gelap,
ramalkan kadar transpirasi tumbuhan ini. Terangkan ramalan anda
0.1 cm/min.The rate of transpiration is lower due to low light
intensity and lower air movement.
[3 marks](f)
Classify the apparatus and materials used in this experiment.
Kelaskan alat radas dan bahan yang digunakan dalam eksperimen
ini.ApparatusMaterialsThermometer
Stopwatch
RulerWater
Hibiscus plant
Capillary tube
[3 marks]
[33 marks]
Paper 3(Sec B)Problem statement:
Which sources of water sample has higher BOD
level?Variables:
MV: Type of water sample
RV:BOD value
CV: Amount of water sample
Hypothesis:
C has higher BOD level compared to A and B.
Apparatus and material:
Reagent bottle with stoppers, cupboard, syringe, stopwatch,
label paper, measuring cylinder, beaker, water sources from A, B
and C, methylene blue solution.
Procedure:
1. Collect 200ml of water sample from A, B and C.2. Label 3
reagents of bottle as A, B and C respectively.
3. Measure 100ml of water sample A using measuring cylinder and
put in the reagent bottle.
4. Add 1ml of 0.1% methylene blue solution at the base of water
sample using a syringe.
5. Close the bottles quickly without shaken it.
6. Repeat step 3-5 by using water sample from B and C.
7. Place all bottles in a cupboard and start the stopwatch.
8. Examine the bottles from time to time.
9. Record the time taken for the water samples to decolourise
the methylene blue solution.
10. Record the results in a table.
Data presentation
Water sampleTime taken to decolourise(hour)BOD value
A
B
C
10 cm
Air bubble
X
Y
Fresh plant shoot
Diagram 1