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rc
MAJLIS AI} IANAH RAKYAT
ANSWER
SCIMME
BIOLOGY
PAPER 1
4551/1
TRIAL SPM
2013
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QUESTION
NO
ANSWER
QUESTION
NO
ANSWER
1.
B 26.
D
2.
B
27.
B
3.
B
28.
D
4
B
29.
A
5.
D
30.
C
6.
A
31.
A
7.
B
32.
D
8.
A 33. c
9.
c
34.
c
10.
C
35. A
11.
C
36.
D
L2.
A
37.
A
13. D
38.
c
t4.
c
39.
D
15.
A 40.
B
15.
c
4L.
A
L7.
c 42.
B
18.
A
43.
A
19. A
44.
B
24. D
45.
D
21.
B 45.
D
22.
A
47.
C
23.
A
48.
D
24.
B
49.
D
25.
B
50.
C
MARA TRIAL 2013
EXAMINATION
ANSWER
SCHEME
PAPER
1
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eBr.rl-G
sc trhitrs
PAPER
2
QUESTTON
1
1
(a)(i)
Able
to name
the
structure
correctly
Answer:
P
:
Rough Endoplasmic
Reticulum/
Rough
ER
tNI
a
: Mitochondria
tI',U
(Reject
:
RER)
1
1
2
(ii)
Able
to
state the
function
of
the
structures.
P
:
To transport proteins
synthesized/made
by
ribosome
tF'tr
a
:
To
generate/produce/provide/supply
energy
tFI
(Reject:
give
energy/
site
for
cellular
respiration)
I
1
,)
(b)
Able to explain
how
the
abundance
of organelle
R enable
the
cells
to
carry
out
photosynthesis
efficiently.
Answer
:
Fl
: Contains
more
chlorophyll
tC}
El
: To
absorb
/capturel
trap
more sunlight
/
light
energy
/light
intensity
/
chemical
energy
IF']
E2
: To
produce
more
glucose
/
photosynthesis
product
/
organic
substances
/
organic
material
tFI
(Reject:food/starch)
(Any
two)
1
1
1
Max2 2
(c)(i) Able to
explain the
role of
R
in
osmoregulation
Answer:
Fl
: R
is
contractile vacuole
El
:
regulate
water contenUosmotic
pressure
E2 :
water
diffuses into R/contractile vacuole
by
osmosis
E3
: R/contractile
vacuole
expand to
marimum
size
E4
:
R/contractile vacuole moves to
the
plasma
membrane
E5 :
R/contractile
vacuole contracts to expel the
(excess)water
(Any
two)
Max2
1
I
t
1
1
1
7
01
:
(water)
ClPl: diffuses
into
the contractile
vacuole by
osmosis
02: contractile
vacuole
Cl'Pz: expand
(when
water
diffuses
in)
I
C:condition
I
lP:process/phenomena
I
lR=resrrlf
I
I
1
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C/P3
:
move
to
plasma
rnembrane
CfP4:
the vacuole
contracts
R : regulate water content/osmotic
pressure
(Any
two)
1
1
1
1
a
(cXii)
Able
to
explain
the function
of
S
to
obtain
food.
Answer:
Fl
: sweeps
food
particles
into
cytostome/
mouth /
groove
/ oral
canal
El:
for
digestion
prooess
/
break
down
into
smaller
particles /
hydrolysis
1:
,1
O
:
cilia
ClPl
:
sweeps
food
particles
into cytostopel
mouth
/
groove
lt
.
oral
canal
Ctyz:
for
digestion
process
/
break down
into
smaller
particles
/
,l'
I
(cXiii)
Able to explain the
effects
of
the
respiratory
inhibitor
to the
,
Parameciumsp
Answer :
Fl
:
(the
respiratory
inhibitor)
inhibits
cellular
respiration
ll
mitochondria
cannot'
function
El
: cannot
generate
energy
EZ
:
cellular activities/active
transport cannot
occur
EZ
:
Paramecium
sp
eventually
dies
(tuty
two)
I
t:
:
I
1
Max2
2
ALTERNATIF
(using
science
text)
O :
inhibitor
ClPl :
inhibits
cellular
respiration
process
//
mitochondria
cannot function
C|PZ
:
cannot
generate
energy
/
cellular
activities
cannot
occur
R :
Paramecium
sp
eventually
dies
(Anv
two
I
1
1
2
TOTAL
L2
'
l'S'
s-
.*P'
:V-
$-
.:\,
.s
dv
sT.
\'
\..\
rti'
jp
(-
at
:k
vl
.a
\
.Cr"
^qi
r.r\L
ql
(-,
-g)
,\.)
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4
QUESTIOI\
;
Items
Marking
Criteria
Marks
aXi)
Able
to
name
the
condition
the
plant
cell.
Answers:
Plasmolysed
/
Flaccid Reject:
Plasmolysis)
I
1
ii)
Able
to explain
the
process
Answers:
F : solution
X
is
hypertonic
to the
cell
sap of
the
plant
celU/cell
sap is
hypotonic
to
solution
X/concentration
of
water
molecules)
outside
the
cell is lower than
the
cell
sap
El: water
diffuses
out
by
osmosis
Reject:
move)
E2 :
vacuole
/
cytoplasm
shrinks
/
plasma
membrane
is
pulled
away
from the
cell
wall
1
1
1
Max
2 2
ALTERNATIF
using
science
text)
O1:
solution
X
CtPl:
is
hypertonic to
the cell sap
of
the plant
cellllcell sap
is
hypotonic
to
solution Xllconcentration
of water
molecules)
outside
the
cell
is
lower
than
the
cell
sap
02
:
water
CIPZ
:
diffuses
out
by
osmosis
ClP3
:
vacuole
/ cytoplasm
shrinks
/
plasma
membrane is pulled
away
from
the
cell
wall
R
:
the
cell
is
plasmolysed
/
flaccid
1
1
I
I
Max2 2
bxi)
Able to
state
the
condition
Answer :
Deplasmolysed
/
Turgid
Rej
ect : Deplasmolysis)
I
1
b) ii)
Able
to
explain
the
effects
of solution
Y to
the
plant
cell.
Answer
:
F
: solution
Y is
hypotonic to
the
cell
sap
of
the
plant
cell/lcell
sap
of
the plant
cell
is hypertonic
to solution
Y//concentration
of
water
molecules)
outside
the
cell
is
higher than
the
cell sap
E1: water
diffuses
into
the
plant
cell
by osmosis
Reject:
move)
E3
:
vacuole
expands
/
plasma
membrane
is
pushed
towards
the
cell
wall
I
the
plant
cell
become
turgid
Max 2
I
1
1
2
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ALTEITNATIF
(using
science text)
O1 :
solution Y
C/Pl : is hypotonic
to
the
cell
sap
of
the
plant
celV/concentration
of
water
(molecules)
outside
the
cell is
higher than
the
cell
sap
ot?:
water
CtY/.:
diffirses into
the
plant
cell
by
osmosis
R
:
vacuole
expands
/
plasma
membrane
is
pushed
towards
the
cell
wall
/
the
plant
cell
become
turgid
1
1
I
Max
2
2
(c)
Able to
explain
why the mango can
be
kept for longer
period.
Sample
answers3
:
F
:
sugar
solution
is hlpertonic
to
the
cell
sap of mango/lcell
sap of
mango is
hlpotonic
to
the sugar
solution
El
:
water diffrrses out
by
osmosis
EZ
: bacteria cannot
grow
(without
water) /
die /
dehydrated
1
I
1
Mor 3
3
ALTTRNATIF
(using
science text)
01
:
sugar
solution
C/Pl
:
hypertonic to the cell
sap of
mango
OZ : water
Cn2:
diffiises
out by osmosis
R
:
bacteria cannot
grow
(without
water) I
die
/
dehydrated
1
?i
1
1
Ma,x
3
3
(d)
Able to explain
the importance of drinking isotonic
water
Answer
Fl
:
isotonic
drink
is a
solution which
has
equal
concentration
/
osmotic
pressure
with the fluid
in
the
body
cells
El
:
to balance
the
water
content in body
cells
/
fluid
El
:
to
prevent
dehydration
llto
replace the
water
loss
I
t
I
3
TOTAL
t2
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QUESTIONS
3
Item
Marking
Criteria
Marks
a)
Able to
name
structure
X and
Y.
Answer:
Structure
X: Alveolus
Structure
Y:
Gills
tNl
tNI
1
I
2
b)
Able
to explain
onesimilar
characteristic
in
structure X
and
Y
to
increase
efficiency
of
gaseous
exchange.
Answer:
F 1 :
have
many
/
numerous
structures
El
:
To
increase
total surface
area
F2
: surrounded
by
network
of blood
capillaries
E2
:
for
efficient
transport
of
gases
F3 :
the wall
is
one cell
thick//
thin wall
E3
: easier
for
gases
diffusion
Rej
ect:
Moist surface)
Any
lF
and
respective E
I
1
I
1
I
1
Max
2 2
ALTERNATIE
using
science
text)
N
=
name
C
=
characterestic
F
:
function
N
:
Structure
X
and
structure
Y
C1 : have
many
/
numerous
tiny
structures
Fl
: To increase
total
surface
area
C2 :
surrounded
by
dense
network of
blood
F2
: for
efficient
transport
of
gases
C3 : the wall
is one cell
thick
F3
:
easier for
gases
to diffi.rse
capillary
1
1
1
1
I
I
Max 2
2
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(c)
Able to
explain
effects
of
smoking
in
the
effrciency
of
gaseous
I
exchange in stru-cture
X.
I
Answer:
I
I
Fl
:TarNicotine/Chemical
substance I
I
El:
Deposited on
the
wall
of alveolus
I
1
E2 :Block
the
diffirsion
of
gasses
actoss
the
wall
of,alveolus
I
1
F2
:HeatlHigh
temperature ,'
E3 :causes
the
wall
of
alveolus
become
dry
E4
:Less
gasses
dissolve
E5
:
competes
with
oxygen
to
gombine
with
ha,-emoglobin
/
has
higher
affinity
towards oxygen
E6
: less
oxygen
supplied
to
body
cells
Any
lF
and
resPective
E
, I
Mor2
2
(d)
Aspect
Part P
(aerobic) Part
Q
(anaefobic)
1+1
,
t,.
1+1
4
Products
of
the
respiration
Energy
/
ATP,
carbon
dioxide
and
water
Energy
I ATP,'ethanol
and
carbon
dioxide
Explanation
of
the
process
Complete
brcakdown
/
oxidised
of
glucose
ll
Require
oxygen
lncomplete
I
partiallY
breakdown
of
glucose
ll
'Doesn?t
require
oxygen
(e)
Able
to
explain
the importance
of
warming
up
Sample
answer
3
Fl :
increase
the
rate
of
heartbeat
Et :
more
oxygen
can be
transport(to
the
muscle
cells)
//
transportation
of
oxygen
is
faster
E2
:
increase
the
rate
of
cellular
respiration
(in
muscle
cells)
FZ
:
increase
the temperature
in
the
muscles
E3
:
loosen
up
the muscles
and
joints
I
prevent muscle
cramp
Any
F
with
corresponding
E
1
1
I
1
1
Max
2
2
12
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tuESTI,St{
4
Item
Markins
Criteria
Marks
(aXi)
Able
to
fill
the
table
with
the correct
name
of
molecules
and
enzymes
involved
Answer:
4
4
Malto
s e/S ucrose/Lactose
Maltase/Sucrase/Lactase
F
at
I lipid
/
triglycerides
lipase
(ii)
Able
to explain
the
effects
of diet
rich
in
fat.
Answer
Fl:
fatllipid/molecule
N
contain high
level
of
cholesterol
El
:
deposited
on
(the
inner
wall
of)
artery
E2
:
increase
blood
pressure
/ cause arteriosclerosis
/
hypertension
Any two Mor 2
1
1
1
2
ALTERNATIF
(using
science
text)
O
: fatllipid/rnolecule
N
CtPl
: contains
high
level
of cholesterol
CtPz
: deposited
on
(the
inner
wall
of) artery artery
R
:
increase
blood
pressure
/
cause
arteriosclerosis
/
hypertension
1
1
1
Max2 2
(b)
(i)
Able to
state
the
food
which
supplies the
most
energy.
Answer:
Rice
INI
1
1
(ii)
Able
to
state the
food
which helps to
prevent
scurvy.
Answer:
Orange
tNI
I
I
(iii)
Able
to
calculate
the
total amount
of
energy.
Answer:
Rice:
1530
kI x
2
:3060
kJ
Fish
:
320
kJ
x 1.5
:
480 kJ
rotal
energy:;lf8lj
.
480
kr
I
1
1
a
J
(c)
Able
to
explain
why
egg has double amount
of
energy than
fish
Sample answeri
Egg
contain
more fat than
fish
1
I
TOTAL
12
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Question
5
Item
Answer
Mark
(aXi)
Able
to explain the
role of
polythene
bag according
to the
]
different
in reducing the
temperature.
Answer:
PlantA
Fl
:
polythene
bag traps water
vapour
(from
kanspiration)
El :
more water vapour
around
the leaves
E2
: increase
in
humidity
(inside
the
polythene
bag)
y
two
Plant
B
F2:
rate
of transpiration increases /higher
An
I
i
EZ
:
more water move
up through
>rylem
2
2
(b)
Able
to explain
why
Plant
B
shows
higher
loss in mass
,
,
,
Fl
:
more loss
in mass
of
the
plant
B
El
:
the leaves in Plant
B
are
not
covered
(by polythene
bag)
E2
:,more
water
valrcur
are released
to the
surounding
through
F
with
any E
1i
... :
1
,1,
Malr
2
2
ALTERNAJIF
(using
science
text)
Ol
: Plant
B
ClPl
:
the leaves
in
are
not
covered
(by
polythene
bag)
CtPz:
more water
vapour
are
released to the
surrounding
I
rate
of transpiration
is
higher
R
:
more loss
in
mass
of
the
plants
I
I
I
Max2
2
(c)
Able
to
suggest
one
way
and explain
Answer:.
Suggestion
: remove /
cut
offthe
leaves
llplace
the
pot
under
a
shaded area
/
under
a big
tree
Explanation :
no
/
less
stomata,
less
water
loss
low temperature /
high
humidity
/
less
light
intensity/slow
air movement
1
I
2
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(c)
10
Abf
to
.xr,'
I
ai
ntlre
.ff..
i-,*trn
*
;it. *b
cn.:li
oi,
i
i*
.., ,i*t
and
iight
intensity
on
the
yield
of
crops.
Answer
:
Temperature
F
1
:
increase
the
enzymatic
activity
/
maximum
rate
/
optimum
El
:yield
of
crops
increases
Carbon
dioxide
F2 :the
rate
of
photosynthesis
increased
/
maximum
/
photosynthesis
product
increase
El
:
increase
the
yield
of
crops
ALTERNATIF
(using
science
text)
Ol
:Temperature
CtPl
: increase
the
enrymatic
activity /
maximum
rate
/
optimum
R
:
yield
of
crops
increases
02
: Carbon
dioxide
CtP2
:
the
rate of
photosynthesis
increased
/
maximum
/
photosynthesis
product
increase
R:
increase
the
yield
of
crops
=
\r
6
6
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18
Item
Number
MARKING
CRITERIA
MARKS
F1 :
Typhoid
/
Cholera
Pl
:
caused
by
bacteria
I vibrio
cholerae
I
Salmonetta
typhii)
spread
through
contaminated
food
and
drink
ll
the
feaces
of
infected
people
reach
food tluough
unwashed
hand
Food
poisoning
caused
by
bacteria
I
Salmonella
sp.)
spread
through
contaminated
food,
drinks
and utensil
Malaria
disease
caused
by
protozoa
Plasmodium
sp.) transmitted
by vector
Anophele.r
sp.
mosquito
F4:
Dengue
fever
P4:
caused
by
virus)
transmitted
by
vector
Aedes
mosquito
Any
2 F
and 2P
accordingly
1
1
1
Max
4
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19
Item
Number
MARKTNG
CzuTERIA
MARKS
7
(c)
Able
to explain
similarities
and differences
between
interaction
in
Diagram
7.2(i) and Diagram
7.2(ii)
www.
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i
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Anv
3
*
Criteria
PQRST
Diagrarn 7.2
(i)
Diagram 7.2
(ii)
T
Al :
The
interaction
is
intraspecifi
c
competition
81: The
interaction
is
interspecifi c
competiiion
Qr
M
:
Competition
is between
members
ofthe
same
species
llbetureen
species X
andspecies
X or between
species
Y
and species
Y.
B2
:Competition
is between
members
from different species//
between species
X
and
species
Y.
Q2
A 3 : The
strong /
dominant
member
win
/ remain but the
weak one die /become
extinst.
83 :
The strong
/
dominant
specibs win
the competition
but
the weaker species
become
extinct /
migrate
to other
area
I
die
Q3
44
: The
population is
always
in
dynamic
equilibrium /
stable
/
constant
only
84
:
Population
of
weaker
sPecies
decrease
and
the
dominant
species increases.
I
I
I
I
Mur
3
3
S
I :
Both
interaction
involve
organism
living
together
in
the
same
habitat
52
:
Both interactions
compete
for
the sa:ne
resourc eVwater/nutri
ent/space
53
:
Both
interactions
compete for
the
limitedre sources/water/nutrient/space
P
:
Process
q
=
Quality/Quantity
R
=
Reaction
S:
Structure
T
=
Terminolory
At
least
one
similarities
1
I
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20
ITEM
NUI\{BER
MARKII{G
CRITERIA MARKS
7
d)
Able explain
how scientific
names derived on Linneaus
binomial system
P
1
:
Each
organism
has
two
names
in
Latin
P2
:
The
first
name
must begin
with capital letter
P3
:
The
first
name
refer to
the
genus
name)
P4
:
The
second
name
begin
with small letter
P5
:
The
second
name refer
to
the
species
name)
P6:
Both
genus
and
species narne must
be written in itallic
P7
:
alternatively,
the words
can be underlined
P8:
For example
Panthera
is the
genus
of both organisms,
but
they
are
of
different
species)
P9
:
Lion belongs
to
species
lgq
while
tiger belongs
to
species
tigris
1
1
1
1
1
1
1
1
1
Max 6 6
TOTAL
20
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2t
Question
8
ITEM
I\[T]MBER
MARKTNG
CRITERIA
MARKS
8
(a)
Able to explain the body s
response
towards
the
entry
of
bacteria
into
the
body
Fl
: the
second line of
body
defense
rhechanism
Pl :
Pathogen
/ bacteria
succeed
in
penetrating the
skin
/
enter
the body through
skin
P2
: Chemicals /
proteins
/
antigens
(produced
by
pathogens
)
P3
:
Attracts
the
phagocytes /
neutrophil
/
macrophage
/
phagocytes
(to
injected
area
)
P4
:
The
phagocytes
surround
the
pathogen
using
pseudopodiurn
:
P5
The
phagocytes engulf
the pathogen
forming
vacuoles
/
phagosomes
P6 :
Lysosomes
secretes
lysozymes
and
release
the
lyso4mres
into
the vacuole
/
phagosomes
,.
P7
:
The
lyso4mes
kitl /
destroy
/
digest
/
hydrolise
the
phagosytes
:
P8
:
The
phagocytes
release /
egest
the
digested
product from
the
cell
P9:
The
process
is known
as
phagocytosis.
1
I
I
1
I
1
1
1
I
1
Mar
4
I
I
4
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22.
ITEM
NUMBER
IVIARKTNG
CRTTERIA
ALTERNATIF
using
science
text)
Name
of the
process/mechanism
:
the
second
line
of body
defense
01 :Pathogen
lbactena
t
ClPl
:
penetrating
the
skin
/ enter
the
body
through
skin
02 :
Chemicals
/
proteins
/ antigens
produced
by
pathogens
)
ClPz:
Attracts
the
phagocytes
/
neutrophyls
/
macrophage
I
phagocytes
to
injected
area)
03
:
phagocytes
C/P3
: surround
the
pathogen
using
pseudopodium.
C/P4:
engulf
the pathogen
forming
vacuoles
/
phagosomes
04
:
The
phagocytes)
ClPs
: release
/
egest
the digested
product
from
the cell
R: The
process
is
known as
phagocytosis.
05
: Lysosomes
ClP6 :
secretes
lysozymes
and release
the lysozymes into the
vacuole
/
phagosomes
C|PT
:
kill
/
destroy
/
digest /trydrolise the
phagocytes
lVIARKS
I
1
Max 4
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23
Item
Number
MARKXNG CRITERIA
MARKS
8(b)
Able
to
explain
how
the
lymph
is drained
back
into
the
blood
P
I
:
(From
the
lymphatic
vessel),
the
llmph
is
drained
into
thoracic duct
P2
: and
the
right lymphatic
duct.
P3
:
The
thoracic
duct receives lymph
from
the
left side
of
the
head / neck /
chest
/
left
upper limb
llentte body
below
the
rib
P3
:
The right
lymphatic
duct
receive lymph
from
the
right
armt
I shoulder area il
ndht
side
of
the
head
/
neck
P4 : The
right lynphatic
duct
empties
it s
lymph
into
the
right
subclavian vein
(
entering
back into
the
blood
circulating
system
)
P5
:
The
thoracic duct empties
its lymph into
the
left
zubclavian.
P6
:
The
valves
prevent
back
flow
of
lymph
P7
:
Muscular contracti
on
ll
peristalsis movement / intestinal
//
pressure
changes
dr:ring
inhalation
and
exhalation
breathing action
push
the
lymph to
rflow
forward
1
I
I
1
I
I
I
1
1
Mur
6
6
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24
Item
Number
MARKING
CRTTERIA MARKS
8(b)
ALTERNATIF
(using
science
text)
01
:
lymph
C/P 1
:
(From
the lyrnphatic
vessel),
drained
into thoracic
duct
C/P2:
and the
right lymphatic
duct.
02
:
(The
thoracic
duct)
ClPz: receives
lymph
from
the
left
side
of the
head
I
neck
I
chest
I
left upper
limb
llentfue
body below the
rib
03 :
(The
right
lyrnphatic
duct)
ClP3
:
receive
lymph
from
the
right
arm /
shoulder
arca
/l
right
side
of
the
head
/
neck
ClP4: The
right lymphatic
duct empties
it s
lyrnph
into
the
right
subclavian
vein
(
entering back into the
blood
circulating
system
)
04 :
(The
thoracic
duct)
C|PS
:
empties
its
lymph into the left
subclavian.
05
:
The
valves
ClP6
:
prevent
back flow
of lymph
06
:
Skeletal muscle/smooth muscle
C|PT
:
contraction
llperistalsis
movement
i
intestinal //
pressure
changes
during
inhalation
and exhalation
breathins action
1
1
I
1
I
1
1
1
1
CiPS:
push
the lymph
to flow forward
1
Max
6
6
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25
Item
Number
MARKTNG
CRIT'ERIA
MARKS
8c)
i)
Students
are
able
to
explain
the formation
of
lymph
Fl
:
Diameter
of
arteriole end is larger
compared
to
the
blood capillary
F2
:
cause
high
hydrostatic
pressure
at the
arteriole
end of
blood
capillary
i
F3
:
Some
of the'bloodplasmais;filtered/forced
into
the
spaces
between
the cellVintercellular
spaces
(Reject:
diffirse)
F4 :
forming
interstitial
fluid
,
',i
1
F5
: contains
of
the
interstitial fluid
are
nutrients/
oxygen/
amino' acid/
glucosel
water/,
ions.
F6
:
except
large'
moleculeslRBC/plasma
protein
/fibrinogen/globulin/platelets
:
F7:
Some
/
85-90 ,ofthe interstitial
fluid
arereabsorbed
intothe
blood
capillary'(at
venule
end)-
F8:
Some/
lff/o
is
absorbed
into
the'lynphatic
capillary/
'
vessels(forming
lympU
lymphatic
fluid).
;'
Any
6F
;'
.,
'
,,.
i
ii) Students
are able
to
explain body
defense
mechanism
F9:
Lymph nodes
produce
lymphocytes'
F10
:Lymphocytes
produce
antibodi
e s/lysine/op
s
onin/ag
gl
utinin/antitoxin
Fl I :
Deshoy
pathogen/
neutalize
the
toxin
produced
by,the
:
pathogen.
F13 : PhagocyteVneutrophiVmonocytes engulf
and
.
'
i
destroy/digestlhydrolyze
pathogen
:Any
4F
I
I
1
I
1
I
1
Ma>r 6
ll
t,
l,
I'
r
I
**o
6
4
TOTAL
20
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26
MARKING
CRITERIA
MARI(S
ALTBRNATI{
(using
science
text)
O1
:
arteriole
C/P1
:
Diameter
of
arteriole
end
is
larger
compared
to the
blood
capillary
ClPz:
cause
high
hydrostatic
pressure
at
the arteriole
end
of
blood
capillary
02
: blood
plasma
ClP3
:
Sorne
of
the
blood
plasma
is
filtered/forced into
the
spaces between
the cells/intercellular
spaces
c/p4:
forming
interstitial
fluid
(Reject:
diffuse)
ClPs
:
contains
of
the
interstitial
fluid
are
nutrients/
oxygen/
amino
acid/
glucose/
waterl ions.
C/P6: except
large
molecules/RBC/plasma
protein
/
fi brino
gen/gl
o bul in/pl
atel et
s
03:
interstitial
fluid
C|PT
:
Some
I 85-90
(of
the
interstitial
fluid) are
reabsorbed
into
the
blood
capillary
(at
venule
end).
C/P8 : Some/
10
is
absorbed
into
the
lymphatic capillary/
Vessels (forming
lymph/
lymphatic
fluid).
Any
6F
ii) Students
are
able
to explain body defense mechanism
04
: Lymph
nodes
ClPg
:
produce
lymphocytes
05 : Lymphocytes
C/P 10
:
produce
antibodies/lysine/opsonin/
agglutinin
/antitoxin
CIP
1 1
:
Destroy
patho gen/
neutralize
the
toxin
produced
by
the pathogen.
06 : Phagocytes/neutrophil/monocytes
C/PIZ
:
engulf
and
destroy/digestlhydrolyze
pathogen
C/Pl3
: Phagocytosis
occur
07
: Lymph nodes
ClPl4
:
filtered
pathogen
Any
4F
1
1
1
1
1
1
Max 6
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27
Question
9
ITEM
I\TUMBER
MARKING
CRITERIA
MARKS
e(a)
P1
x
Parents
PhenotSrpe
:
Genotype :
Meiosis
Gamete:
Random
Fertilization
Genotype
Fl:
Phenotype
Fl:
Rabbit P
White
fur, black
eyes
Rabbit
Q
Black fur,
red
eyes
Bbrr
P4
P5
and
P6
P7
P8
Black
fur,
black
eyes
Notes;
n
must have
word meiosis
and
seggregotion
lines
ii)
gametes
must be
circled
fiil must
hqve
word
random
fertilization
and lines
Pl:
The
genotype
for Rabbit
P is
bbn
and the
genotype
for
Rabbit
Y
is
BBRr
P2 :
genotlpe
bbrr consist
of
b,recessive
allele
for
white
fur
and
r,
recessive
allele
for
black
eyes.
P3
:
genotlrye
BBRr consist of
B,
dominant
allele
for
black
fur
and
R
and r
for
dominant
and
recessive
allele
for
red eyes and black
eyes
respectively
P4
:During
meiosis
P5
: Gamete
produced
for
Rabbit
P
arebr.
P6:
Gamete
produced
for
Rabbit
Q
are
BR
and Br.
P7
: Random fertilisation
occur
P8
:
bet'ween
gametes
br from
Rabbit
P
and
gamete
Br from
Rabbit
Q
produced
genotype
of
Rabbit
Bbrr.
P9: Dominant
allele,B
is inherited
from
Rabbit
Q,
while
recessive
allele,b
is
inherited
from
Rabbit
P
Pl0:
Recessive alleles,
r
is
inherited from
both parent,
P
and Q
Pl 1: The Dresence
of
(dominant)
allele,Band(recessive)
allele
b,
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28
i i
sho ;v:the
phe,rotype
black
fur.
i
P12:lihe
presence
of
(recessive)
allele,r and
(recessive)
allele
r,
showsthe
phenotype
black
eyes
Any 6
1
Max
6
ITEM
NUMBER
MARKING
CRITERIA
MARKS
e(bxi)
AbIe to
discuss
the
advantages
of
using
genetic
engineering
Technology
in agriculture
Sample
answer
The advantages
:
F1
: Resistance toward disease/pest resistant
increase
Pl :
because
it
alters the
genetic
information
in
plants.
F2
:
The
quality
of
agricultural
product
increases
P2
: because
food/fruit/crop
yield
product
have
higher
vitamin/mineral/nutritional
content
F3 :
Foo
dl tn;Jrtl
Crop
yield
can
be
harvested
shortly
after
planting/in
shorter
time
P3
:
because
it
shorten
maturity
period
F4
:
Food/ fruit/ crop
yield supply is sufficienVmore
food
can
be supply to the
people
P4
:
because the
quatity
of
crops
yeild
increases
ANY 3
F
AND 3P ACCORDINGLY
1
1
1
1
1
1
1
1
Max
6 6
TOTAI-,
20
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29
MARKING CRITERIA
State
fotr
differences
between
Down s
syndrome and
haemophilia
based on
these
criteria.
1. Cause
2.
hlunober
of
chromosomes
3. Characteristics
4. Inheritance
Cause
by
chromosomal
mutation
of
chromosomes
numbers
2l
Cause
by
a recessive
allele
found
on the
X
sex chromosome.
Abnormalll4T
Number
of
chromosomes
NormaU/46
Blood
unable
to clot
ave flatlbroadllbroad
faces//slanted
eyell
protoding
tongue//tend
to
be
mentally
retarded
Characterestics
Not
inherited
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SULIT
45s113
A}.iSWER
SCFIEivfE
PAPER
3
TRIAL
BIOI-CGl ]013
No.
Vlark
Scheme
Score
1
(u)
Able
to
record
all
3 readings
for
the
time taken
to obtain 30
heartbeats
correctly
Samole answer
3
Numbers
of
laps
Bilangan pusingan
The
time taken
to
obtained
30
heartbeat
(s)
Masa untuk
memperoleh
30 denyutan
iantuns
(s)
1
20
2
12
J
10
www.
myschoolch
ild
ren.com
Able
to
record
any 2
readinss
correctly
2
Able
to record
any
1 reading
correctly I
No
response
or incorrect
response
0
1(bxi)
Able
to
state
2 different
observations correctly following these
criteria:
C
l.-manipulated
variables
C2-Responding
variables
C3-correct value with unit
Sample answer
Vertical
1. The time taken
to obtain
30
heartbeats
after running
one round
is 20 seconds.
2.The time
taken
to obtain 30 heartbeats
after
running
two
rounds
is 12 seconds.
3. The time taken
to obtain 30 heartbeats
after
running
three
rounds
is 10 seconds.
Horizontal
4. The time taken
to
obtain 30 heartbeats after running one
lap
is the
longest
compared
to after
running
two
and
tluee
laps.
3
AbIe
to
state one
observations
correctly
and
one
-
two
inaccurate
observation
Sample
answer
Uertical)
1.The
time
taken to
obtain
30 heartbeats
after
running one
lap
is the longest.
2.
The time taken to obtain
30 heartbeats
after
running
three
laps
is the
shortest
(Horizontall
3. The
time
taken to
obtain
30 heartbeats
after running
one
lap
is
longer
than
when
running
three
laps
that
is
12 seconds.
2
455L CI
2013 Hak Cipta
Bahagian
Pendidikan Menengali
MARA
SULIT
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SULIT
4551t3
ANSETR
SCHEME
PAPER
-?
TRTAL
BIOLOGY
2013
Able
to
state the
obsenation
at idea
level
l.The
longest
time
is 20
seconds.
2. The
shortest
time is
10
seconds.
3. The
time taken
to
obtain
30 heartbeat
decreases.
1
No response
or
incorrect
response
0
1(bxii)
Able
to
make
two
inferences
correctly
Note
:
inference
must
matsh to
the observation
Cl
- number
of laps
ran
by
an
athlete and
the
condition
of
time
taken
to
obtain
30 heartbeat
C2 -
rate
of heartbeat
increases
/ decreases
C3
-
o*Jgg,l ngeded
by
muscle,3efl
large/small
Sarnple
answer
(Verticall
1.
When
the
boy
mns
one lap, the
time taken
to
obtain
30
heartbeat is
the longes ,
because
the rate
of heartbeats
is
decrepsesAower due to
muscle
cells'needs
a small amount
bf'oxygen
.
.
:
2. When
the
boy nms threq laps,
the
time
taken
to
obJain
30
heartbeat is
shortest becausb
the
iati: of
hiartUeats is increases/tiigher due to muscle
cells
needs
a
large
ambunt
of oxygen
fiforizontail.
3.The
time
taken
to obtain 30
heartbeats
after
nrnning
one
lap
is
longer
because
the
cells need
small
amowrt
of oxygen
comfared
to
when
running three
laps
cl
+
c2+c3
3
AbIe
to
make
one
correct inference and one - two inaccurate
inference
Eampls
answer
(Verticall
l.The
time
taken
to obtain 30 heartbeats after
running
one
lap
is the longest
because
rate
of
heartbeat
is the
lowest.
2.
The
time
taken
to obtain 30 heartbeats
after
running three
laps is the
shortest
because
rate
of
heartbeat
is the highest
(Horizontal)
l.The
time
taken
to
obtain 30
heartbeats after
running
one lap /three
laps
is
longer/shorter
than
when running
three/one lap because
rate
of
heatbeat for one/three
laps lower/higher than
three/one lap.
Any 2C
2
AbIe
to
state the
inference
at idea
level
SAmple
anlwers
1.Rate
of
heartbeats
increases
1
No
response
or
incorect
response
0
4551
@2013
Hak Cipta Bahagian
Pendidikan
Menengah
MARA
SULIT
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SULIT
4s51/3
ANS\frIER
SC}MME
PAPFR
3
TRIAL
BIOLOGY
2OI3
No.
Mark Scheme
, Score
I
(d)
I
Able
to
state
a
hypothesis
correctty
following
aII
criteria
:
C
I
: number
of laps
C
2
:
time,
taken
to,
obtain
30 heartb
eatsllrute
of
heartbeat
C3
: Clear/concreterelationship
,,;
Sampleanswer:r
,
1. As
the
number
of
laps increases,
the
time
tiken
to obtain
30
heartbedts
decreases
2.
The
more
the
nuurber
of laps,
the
shorter the
time
taken
to
obtain
30 heartbeats
(vice
versa)
,
.
,
;
;
.i,
.
,,r,.
,
1
3.
As the number
of
laps
increasgsr
the,rate
dheartbeat
elso,,incpases
i
i
(viceversa)
..:,,,:::
I
i:,
;.'..
.0,'iL,,r,..
_
.;
.
,
..,.,. ,i,,.i;
,
All 3C
3
Able
to make
a hypothesis
relating
the
manipulated
variable and
the responding
variable
inaceurately
Sample
answer
:
'
::
i
l.The
time
taken
to obtain
30
heartbeats
llThe
rate
of
heartbeat
is
affectedbynunrberoflaps
.:
.,
..i.,.,..,:,ri.
:i,].
Any
2C
,
tL
Able
to
make a
hypothesis
at
idea
level
sqmpleanswgr:
'i'
ij''-'
i'
l.TG
time
tut r
to
obtain
30 heartbeats
decreases
,
,',
,
'
2.The
rate
of
heartbeat
increases
Only
1C
I
No
response
or
incorrect
response
0
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SULIT
4s5U3
/\NSWER
SCTIEME
PAPER
3
TIUAL
BIOLOGY 2013
Ni;.
Mark Scheme
Score
1
e) i)
Able
to
construct a
table
and
record all
the data correctly
Sample
answer
Note :
T)
Titles
with
correct
unit
-
I
mark
D)
Record
all
the data
correctly
-
I
mark
C)
Calculate
and
record all
the
rate
of
heartb eat
-
I
mark
3
Any
two correct 2
Any
one correct 1
No
response
or
incorrect
response
0
No.
Mark
Scheme
Score
1
e) ii)
Able
to draw a
graph
correctly
A
: Uniform
scale
on both axis
-
1
mark
PT :
Able
to
plot
all
3
points
correctly
S
:
Able to
connect all
point
smoothly
Refer
to
graph
provided
-
1
mark
-
I mark
J
Aoy
two
criteria correct
2
Uniform scales
on
either
the
horizontal-axis or
the
vertical-axis
1
No
response
or
incorrect
response
0
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SULIT
455U3
A].iSWER
SCI{EME
PAPER
3
TRIAI,
BIOI,OGY
2013
l(0
Able
to explain the
relationship
between
the
number of
laps and
the
rate
of heartbeat
based on
the following
aspect:
R
-
able
to
state the
relationship
El
-
able
to
explain
u y the
rate
of
heartbeat affected
by
the
number
of
round
EZ
-
able to
give
a
reason
Sample answer
,t
-
.
As the ntrmber
of laps increases
,
the
rate
of
heartbeat
increases,
,
,'
to
pump
more
blood
(into
circulation).Thefore
rate of cellular
respiration
increases//generate more energy/ATP
Note:
Rmustcotrect,..:
:. i.,.
.;.i.
.i
.,,..Rt,NryZBn
3
Able to interpret
the
relationship
incoppJetety
R'h
lE
,
2
Able to interpret
the
relationship at idea
lpvel,
,
R
only
I
No response or.incor.rect
response
0
No.
Mark Scheme
Score
1(e)
Able
to
define
the
heartbeat
rate
operationally
based on
the
result
of this
experiment.
Pl- Fact
classification
PZ
-
t:me taken
to
obtain
30 heartbeat
(RV)
// observation
P3
-
(RV)
influence
ll affected
by
MV)
Sample
answer :
The rate
of
heartbeat is
the
number
of
heartbeat
per
minute
(P1)
which
can be determine by time
taken to obtain
30
heartbeats
(P2),
and
it
is
affected
by the
number
of
laps
(P3).
All
PL,P2
and
P3
3
Any
two
correct
Any
2P
2
Any one
correct
Only
lP
I
No response
or incorrect
response
0
7
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1
(ir)
Able to
predi.t
ttr.
.,
the
following
criteria:
Pl
:
The rate
of
heartbeat
will
be less
than
1.5
s'1.
P2
=
less
glucose
and
oxygen
is
transported
to the muscle
cells
/
tissues
P3= rate
of cellular respiration
decreases //less energy
is
producedi/
less
muscle
contractions.
Sample
answers
The
rate
of
his heartbeat
will be
Iess
than
1.5
s-1', because
less
glucose
and oxygen
is
transported
to the muscle.
Therefore the
rate
of cellular
respiration
decreases//,
less energy
is
produced//
less muscle
contractions
1
Able
to
predict
the
outcome
of the
experiment
incompletely
P+
1E
2
Able
to
predict
the outcome
of the
experiment
at idea
Ievel
P
only
1
No response
or incorrect
response
0
1
(i)
Able
to
classify
all
Samole answer
6
activities
correctly
J
Aerobic
resoiration
Anaerobic
Respiration
Archery
Watering
Plant
Walking
Sprinting
100m
Fermentation
in
yeast
Plant
in
waterlogged
condition
Able to list
4-5
condition
correctly
2
Able
to
list 2-3 condition
correctly
I
No
response
or
incorrect
response
0
SULIT
45sl13
ANSWER
SCI-IEME PAPER
3
TRJ,{L
BIOLOGY 2013
t)
E
t3)
t'l t
v
rD
(D
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SULIT
ANIS ,ER
SCI{EI,,,{E
PAPER. 3
TRI,A..L
BIT}LOGY 2013
Graph
of
the rate
of
heartbeat
against the
number of
laps
45sU3
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s' :rLIT
Question
2
4s5113
,,].NS WER
SCI]E]V{E
P.\,PER 3
TR{AL BIOI,OGY 2013
No.
Mark
Scheme
Score
2(i)
Able
to
state the
problem
statement
relating
the manipulated
variable
with
the
responding
variable correctly
that
include criteria:
'
P
1-Manipulated
variables
-
Sources of
water
pollutiotr
X, Y and
Z
.
P2-Responding
variables Tirne to decolourise
methylene
blue
solution//level
of
water
pollution
'
P3-concrete
relationship
between
the
variables
in
question
form
and
question
symbol
[?]
Sample
answers:
1.
Does/Do
the
time
taken
for methylene
blue
solution
to
decolourise
(P2)is
the
shortest(P3)
in
source
XllY/lZ, compare
to
source
Y
and ZllX
and
ZllX
and
Y(Pl)?
2.
DoeslDo the water
sample
from
source
WlYllZ
is
the
most
polluted
compare to water sample
from
source
Yand
Z/lZ
andX/lX
and
Y?
3. DoeslDo
the
water
sample
from
source
XINI/Z is
the Ieast
polluted
compare
to
water
sample
from
source Yand ZllZ and )V/)(
and Y?
4. Do/Does
the time
taken
for
methylene blue solution to decolourisei/level
of
water
pollution
is
affected by
water sample from
source X,Y and
Z
?
5. What is the
level
of water
pollution
from
the
(source)
X,
Y andZ?
6.
DoeslDo
the level
of
water
pollution
from the
(source)
X,
Yand
Z
are
different?
Pl+P2+P3
J
Able
to
state a
problem
statement
inaccurately
Sample
answers:
l.DoeslDo
the level
of
water
pollution//time
taken
for
methylene blue
solution
to
decolourise
from
the
(source)Xll,Y
ll
Z is
the
lower//shorter?
Any
2P
2
Able
to state
a
of
problem
statement
at idea level
Sample answers:
DoeslDo the
level
of
water
pollution is different?
Any
lP
1
No
response or
incorrect
response
0
4551
O
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SI]LIT
4551/3
ANSWER SC}IEME
PAPER
3
TRIAL BIOLOGY
2013
No.
Mark
Scheme
Score
2(ii)
Able
to
state
the hypothesis
relating
the manipulated variable
to
the
responding variable correctly:
C2.
RV
C3
-
Clear/concrete
relationship
between
MV
and RV
(compare)
:
Sample
answers:
I
.The
water
sarrple
from
source
X
(C
1)
is
the most
polluted/
methylene
blue dgcolourise
faster/fastest
(C2)
compare/than
(C3)
to
Y
andZ
(C1)
(vice
funu)
2.T\e
water
sample
from
source
X(Cl)
is
the
most
polluted/
methylene
blue
decolourisation
is
the fastest.
.
r i..
'
r
r.l
cl
+
c2+c3
3
Able
to
state
a
liypothesis
inaccurately
Sample
answers:
The
water sample
from source X
(Cl)
is
polluted/
shows
faster
decolourisation
of
methylene
blue.
...(incomplete
CI)
Any
2C
,,,
^
2
Ableto
statethe
ideabfthe'hlpothesis.
'
I
:
':'
i
i'
r":
.i
';{
"i'
I
Sample
answers:
1. The
water
sample
from
sorlrce X(Cl)
is
polluted.
2.The
time
taken
for
methylene
blue solution to decolourise
decreases//increases
Any'1i
I
No response
or incorrect response
0
11
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SULIT
455y3
ANS\IER
SG{EME PAPER
3
TRIAT
BI{}LOGY
2OI3
N0.
Mark
Scheme
Score
2(v)
AbIe
to
describe
the
steps
of
the
experiment
procedure
or
method
K'S
1.
The reagent
bottles
ane
labelled
A,
B and
C Kl.
2.
The
water
samples
are
taken
from
three
different
source
of
X
(effluent
water),
Y
(tape
water)
and
Z(spring
water)
KI
3.
250
ml
of the
water
sample
from
source
A
is
measured
by using
the
measwgs,c.ylinder,
thgo
put,into p
fegge.nt
bottte labeiied
e
KI
KI
4.
By using
a
long needle
syringri,
aAd
f
ffi
of methylene
blqe
solution
at
the
bottom
of the
water sample
't
K1
K6
5.
Close
the
bofile immediately
to
prevent
atnospheric oxygen
from
dissolving
into the
water
samples
I
K6
6. do
not
shake
the
bottle
to
prevent
afinospheric
oxygen
from
dissolving
into
the
water
samples
K6
7 . Repeat
step
3
-
5
using
water
samples
Y
and Z K2
8.
Make dure'to
use
the
sarne amount
biwater
admpG in6eiy'--'
repeatedexperimeui
:'
'
i
:
'
''1
K4
9.
Keep the bottles
in the diik
place (ctipboard)
to
prevent
photosynthesis
being carried out by
algae
in
the
water
sample
and
start the stopwatch
K6
KI
10.
Observe
the
bottles
every
l0
minutes
K3
I
l. Measure
and record
the
time taken
for the methylene
blue to
decolor:rise
by using
a stopwatch
K3
12.
Tabulate
the data.
K5
Notes
:
Kl
:
Preparation
of materials
&
apparatus
K2
:
Operating
manipulated
variable
K3
:
Operating responding variable
K4
: Operatingfixed
variable
K5:
Data
communication
Kd
: Precaution
Note:At
least
4Kl
Able
to
state
all
6K
AbIetort t.4-5K
3
2
Abletoltfte2-3K
_.
Able
to state lK or
no response
or incorrect
response
I
0
4551
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No.
Mark
Scheme
Score
2(vi)
Able
to
construct a
table
to record data based
on the
following
criteria:
C 1 :
Manipulated variables
with
parameter
and unit
C2:
Operating
responding
variables
and
responding
variables
with
unit
Sample
answers:
Source
of
water
samples
Time
taken
for
methylene
blue
solution
to decolourise
(hours)
or
(days)
Level
of
water
pollution
x
Y
Z
or
Source
of
water
samples
Time taken
for
methylene
blue
solution
to
decolourise
(hours)
or
(days)
Level
of
water
pollution
1
2
a
J
4
Average
x
Y
Z
2
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