Download Paper 2 - Soalan-Percubaan-STPM

JABATAN PELAJARAN NEGERI SEMBILAN
TRIAL STPM 2011
MARKING SCHEME
BIOLOGY 964
Paper 1 (964/1)
Paper 2 (964/2)
PEPERIKSAAN PERCUBAAN BERSAMA STPM
PAPER 1(96411)
Marking Scheme- BIOLOGY
+-
ANSWER
B
NO
1
2
-"~
-4
5
6
7
--
~-
A
B
D
8
c
9
A
1
D
A
B
D
~____1__8
D
,---
B
D
22
r---23
j24
c-
____2s
-- --
_f____
I
-T
_
-
c
-
B
B
A
I
32
i\ - .
--12--1
19
20
21
-
c
cA ----
c
B
A
13
14
____
15 ___._I'
16
17
· - - - - - - · - --
--t
-t
c ___[
-
49
so
c
A
29
30
31
""
~~
34
35
36
37
38
39
40
41
42
43
44
45
46
47
~---- 48
10
I1
ANSWER
--
28
c
A
B
NO
26
27
D
c
D
B
D
--
c
~-
D
B
c
c
D
A
D
B
-·
c
-·
. ·-- 1--
-- 1--------
--
A
B
A
1-------t---
I
1
.MARKING SCHEME PAPER 2
SECTION A
ANSWER
NO
-
Amino acid
1(a)
'J(b) .
~~) (i)
Hydrogen bond
r--
(ii)
•
Secondary structure/ 0-pleated sheet
•
Many hydrogen bonds are formed between C=O
and NH- group from the peptide bond regions.
SUBTOTAL
TOTAL
1
1
1
1
1
I
1
These maintain the stable structure of 0-pleated
structure
•
1(d)
Polypeptide chain is folded to form
~-pleated
sheet
Hydrogen bonds break;
the secondary structure II
~-pleated
sheet will unfold II
l
Fibrous protein :Keratin I collagen I myosin I silk I
1
Globular protein : any named enzyme e.g amylase I any
1
•
•
Is linked by peptide
bonds
Contains sequence
( any 1)
2
Polysaccharide
Polypeptide
•
1
(Any 1)
named protein hormone I antibodies
1(f)
2
l
become denatured
I ! (e)
1
I
Has glycosidic
j
2
bonds
•
of difference
Has identical/same
monomers
1
Some have
1
monomers/amino
I
L__
acids
•
Linear/straight
chain
•
branches
(any 2)
total
10
·--
2
NO
ANSWER
SUBTOTAL
TOTAL
2 (a)
•
Both X and Y axes with correct labels and unit
Both curves with labels correct
•
•
%of Hb saturation for PC0 2 (40mmHg)- 60%, 75%
% ofHb saturation for PC0 2 (80mmHg)- 95%, 98% ·
(c)
•
(d)
•
At higher partial pressure of C0 2 , the affinity of Hb
for 0 2 is reduced
Hb is comprised of 4 protein subunits I 4 polypeptide
chains
Has 2 a chains and 2 p chains
Each globular subunit contains a haem molecule
At the centre of each haem molecule is a ferrous ion
•
(b)rul
(i i)
1 - - · -+--
•
•
•
'-------··
1/0
110
2
2
Max2
3
•
(e)
•
•
•
RBC has no nucleus -more space for more
haemoglobin thus more 02 can be transported
Round, disc-like shape, flexible/elastic- allow cells
to be squeezed through narrow capillaries
Thin plasma membrane- give a short diffusion
distance into and out of cells
Biconcave- increases the surface area to volume
ratio/ providing larger surface area for diffusion of
gases I gaseous exchange
l
I
I
l
Max 3
Total
0
0
(a)
(b)
•
•
Allometric growth
Different parts of the body grows at different rates
compared to the overall growth rate of the organisms
•
P-Thymus/ Lymphoid tissues
Q-Brain/head
R-Absolute growth rate of human
S-Reproductive organs
•
•
•
l
2
l
I
l
l
l
•
(c)
•
•
•
•
P-Lymphoid tissues grow rapidly in early childhood
compared to adolescent.
This is because the risk of infection is high in early
life when immunity has not yet been acquired
S-The reproductive organ grow slowly
In early life but rapidly at puberty
To ensure that the reproductive system starts to
develop after other parts of the body have completed
their development
10
4
l
l
l
l
I
Max 4
1----
L
Total
I
-
10
4
·~~-----~~--------------~--~~~~~--,-----,
4 (a)
GameteS : 3
Gamete T: I
2
6
(b) (i)
Meio~is
I
Meiosis II
=
GameteS
Gamete S
Gamete T
Gamete T
Non-disjunction
(ii)
7
Aneuploidy
( c)
Total
5
ANSWER SCHEME PAPER 2 BIOLOGY TRIAL STPM 2011
Section B
NO
-
ANSWER
SUBTOTAL
I TOTAL
5(a)
~
}G
I
Drawing and structures correct
5 labels(all):
1. Intrinsic(A)(D)/lntegral(A)(D)//Transmembrane(A)(D)/
Pore protein( A)/Extrinsic protein (G)
2. Phospholipid bilayer (B)
3. Cholesterol (E)
4. Glycoprotein (C)
5. Glycolipid (F)
Structure and function:
• the biological cell membrane acts as barrier and are
selectively permeable
• the membrane consists of a fluid bilayer o f phospholipids
and various protein molecules act as ion c hannels, CatTier
protein or pumps embedded in it
• the phospholipids bilayer has a hydrophobi c fatty acids tail
and a hydrophilic head
• the phospholipids bilayer is permeable to very small
uncharged molecules like oxygen and carbon dioxide,
steroid based hormone, fatty acids and alcohol (simple
diffusion)
-~----+-=------'cc_:_.:_:ccL.__.--;:---;------;------;----c:-----Transportation of substances into the cell:
• simple diffusion of water molecules across the semi
permeable cell membrane is called osmosls.
• some integral mt;mbrane protein fom1 hydrophilic ion
cham1els enable diffusion of various charged ions e.g.K+,
Na+, Ca+, and HC0 3- down their concentration gradient
• some of this ion protein channels can open or close and are
called gated channels e.g. voltage-gated channels and
ligand-gated channels
• large sized hydrophilic molecules such as glucose are
twnsportcd across the cell membrane th;ough facilitated
dr>IusJOn usmg a canier pwtem
,_____ "_ir1__i_:lC!lttatscl_ c]~ffusron, the_QJ_11dmg_ __o__L_s_t.llStdt1C_es_t_o__tl1e
Diagram
:1
I
Label:
210
3 Max
5(b)
1
1
1
l
T
l
I
4
-
I
1
1
i
I
I
-~_}__-~'----~
6
ANSWER
NO
•
•
•
•
f--
6(a)
I
specific protein carrier causes the change in its shape and
the substance is released into the cell
Ill active transport, the shape of protein carrier changes
using energy (ATP) to transport substances across the cell
membrane
in endocytosis the substances are transported into the cell
through the invagination of the cell membrane
pinocytosis occurs when the cell membrane invaginates to
actively transport a small amount of fluid into the cell
all these structures and its related process enable the cell
membrane to function as semi-permeable membrane as well
as enable the cell membrane to regulate the movement
of substances in and out of the cell
• Phloem tissue consists mainly of sieve tubes and companion
cells
• Sieve elements are arranged end to end to form a long
cylindrical sieve tubes
• The end walls of sieve tubes are perforated fanning a sieve
plates with sieve pores
• Mature sreve tubes are Living cells with no nucleus,
ribosomes or Golgi apparatus
• Companion cells connected to the sieve elements through
plasmodesmata contains a large nucleus, dense cytoplasm
and numerous of mitochondria
(b) • Translocation IS the movement of orgamc solutes I
sucrose/an1ino acids/soluble Qroducts of Qhotosynthesis
• From the leaves/source through the sieve tubes to be carried
to other garts of the Qlant/sink/root.
(c) The Mass flow hypothesis state that :
• The hydrostatic pressure gradient formed between the leaf
cells (source) and root cells (sink) drives
• the passrve mass flow of water and dissolve solutes
downwards
• In leaves, orgamc substances /sucrose synthesized Ill the
mesophyll cells are actively transQorted into sieve tube
• Low water QOtential is created in sieve tubes,
• Thus water enters the sieve tubes tLu·ough osmosis from the
;:cylem
• The entry of water generates a high hydrostatic Qressure in
the sieve tubes
• In the roots/sink, su~ars/sucrose are activelv transQorted into
the tissue for cellular respiration or converted to starch for
storage
• As sugars are ren1oved, the water [JOtential of sieve tubes in
roo!,;_b;_ increased, causing the water molecules to diffuse out
through osmosis into xyler~1 roots_______.__
SUBTOTAL
TOTAL
1
1
1
1
(Any 8)
Max 8
TOTAL
15
l
1
1
1
1
Max4
1
1
2
1
1
I
1
I
1
I
1
L - __ _
_j
7
NO
ANSWER
• Low hvdrostatic pressure is created in the sieve tubes of the
roots
• Antibody is globular protein that reacts with specific antigen
" it is made up of 2 heavy chains and 2 light chains
• the structure are held together by disulphide bonds
" each chain has constant and variable parts
• the variable part is the binding site of the antibody onto a
sgecific antigen
(b) "Humoral I antibody-mediated immune response involves B
lymphocyte cell
• pathogens bearing foreign antigens invade the body
• macrophages 111 the body engulf I phagoc ytocizes these
pathogens I digest the antigen into fragments by hydrolytic
enzymes
•MHC class II molecules bind to the macrophages to the
antigen fragment to fonn MHC complex
• this foreign antigen , MHC complex is then displayed on the
cell surface
• this macrophage is called antigen-presenting cell (APC)
• helper T cell binds to APC and secretes IL-2
• The activated B cell then divides by mitosis and a clone of B
cell is produced
• B cells differentiate, becoming plasma cells and memory B
cells
• The plasma cells secrete antibodies
SUBTOTAL
I
TOTAL
TOTAL
15
I
I
I
I
I
5
7(a)
I
1
1
I
l
I
I
I
I
I
10
I
1----
r-- 8(a)(i)
1ardy-Weinberg's Law
• The frequency of (dominant and recessive) alleles ll1 a
population will rem am constant from one generation to
generation/ allele and genotype frequencies do not change
from generation to generation 111 a population at genetic
equilibrium
• provided these four conditions are met:
1.
the population is large
]I.
mating is random
Ill. no mutation occurs
IV. no migration occurs
(all four conditions must be correct)
(ii) Sickle cell anaemia
• is a result of point/gene mutation/base substitution
• Glutamine is replaced by Valine.
• This causes the abnormal haemoglobin to be pulled into a
I
sickk shape
~___j__._c::ausingJE~S~_()XjigCn to be carried to tbe tisstJe
9
TOTAL
IS
1
1
2/0
2
I
I
I
I
'
1--
Max 3
I
I
'
_ _ _ _ _ __j
8
ANSWER
NO
-
SUBTOTAL
TOTAL
(b)
r' +
p+q=l
2pq + q' = 1
1
requency of allele h = q, frequency of allele H = p,
q' = 0.085%
q =
q =
p =
p =
p =
v
=
0.00085
0.00085
0.029*
I - q
I - 0.029
0.971 *
%genotype HH
0.971 2 X 100
= 94.28%
q·
=
4% = 0.04
v
q =
q = 0.2
p
p
p
=
=
=
..
(c)(u)
~
1
Max4
I - q
I - 0.2
0.8
% genotype Hh = 2 X 0.8 X 0.2 X 100
= 32%
Normal blood cell
sickle cell anemia blood cell
(hh)
(HH, Hh)
4%
64% + 32%
4%
96%
24
I
I
I
I
I
I
Max 3
I
I
2
--
TOTAL
9(a)
I
I
1
0.04
% genotype HH = 0.8 2 X 100
= 64%
-,
1
=
% genotype Hh = 2 X 0.971 X 0.029 X 100
= 5.63%
(c)(i)
1
~-
Amniocentesis
• indirect screening for genetic defects in a foetus.
• a long, sterilized needle is pierced through the abdominal and
uterus walls into the amniotic cavity.
• a small sample of amniotic fluid together with cells sloughed
off from the foetus' body are extracted.
• the foetal cells are cultured in laboratory.
• biochemical analysis of DNA and products of defective genes
I (such as) u-foctoprotcin (AFP) for the spina bifida defect
15--
I
1
1
1
1
I
I
I'
I
I
I
9
NO
ANSWER
and karyotypic study are carried out.
• this test is especially important for mothers above the age of
35
• if the foetus is handicapped, measures such as gene therapy,
genetic counseling or abortion can be carried out
(b) DNA fingerprinting.
• DNA is isolated from samples of blood, sperm or skin
• and is amplified using the polymerase chain reaction (PCR)
teclmique.
• restriction enzyme is used to cut the DNA into fragments that
are different in length and base sequences (the fragments are
placed in agarose gel).
• the fragments are then separated according to s1ze and
charged by using the technique of gel electrophoresis.
• DNA fragments are heated to split double-stranded DNA into
single strands (and then denatured tlu·ough chemical
treatment.)
• then transferred to a nylon membrane (through the process of
Southem blotting).
• radioactive probes are added to bind to complementary base
sequences in the DNA and are placed on an X-ray film.
• radioactive probes in DNA fragments produce dark bands on
the film/autoradiography
SUBTOTAL
TOTAL
1
1
Max 6
1
1
1
1
1
I
1
I
_____ --c,--c;------;.:~-------;---:-.---------------+-'"(a=n:y_y_7:1)--+__::M.:.:a::-x:...:7~
Use of fingerprinting:
• DNA fingerprint can be used in forensic science for testing
1
specimens and identification of individuals in criminal cases
I accidents, immigration cases.
• determination of kinship I patemity.
I
• identification of defective gene.
I
(any 2)
Max 2
--lO(a)
Biogeochemical cycle means:
• Natural cycle of essential chemical elements in (various
forms) by the geological and biological processes
• chemical elements flow from abiotic reserve to the biotic
components and back to the abiotic pool II Involves
interactions between living things and non-living things
• The cycle prevent depletion of the resource.
1
1
3
10
ANSWER
NO
(b)
SUBTOTAL
TOTAL
erosion\!'
Inorganic
mining, industrial
phosphate
I>'
production of fertilisers
Dissolved
in rocks
decomposition by
phosphate
detmposecs v
ions in the soil
inorganiC
phosphates in
r
~
inorganic
f-+
fr shwater,
teans
active uptake by aquatic
organ1sms
J
land
t
uplih V
death, decomposition
death
active ion upt ke
bones, teeth,
,).
shells. excretorywas
tes
Sedimentation V
c> •cue"
i
Organic
phosphates
eaten by
animals V
Organic
phosphates
Inorganic
phosphates on
ocean floors
in animals
in plants
-
~-
D iagrarn co nect
2
210
Max4
All 6 ticked (,J) labels must be conect
c-------
L Sedimentation of phosphates on ocean floor from the rocks.
2. After millions of years, geological land uplift raised the
rocks containing phosphorns above sea leveL
Throngh
erosion, phosphates returned to soil, rivers, lakes
~and oceans from the rocks.
4. Mining of rocks for the manufacture of fertilizers supply
inorganic phosphates to the soiL
5. Inorganic phosphate ions are absorbed by plants,
6. then assimilated to synthesise important compounds/amino
acids.
7. Transfen-ed to herbivores when eaten and then to other
trophic levels in the food chain through assimilation.
8. When organisms die, remains of bones , teeth , shells and
excretory wastes area acted on by bacteria to release
inorganic phosphates into soil through decomposition.
0
I
I
I
I
c-_··---
~-
-
I
1
I
I
I
I
I
1
1
TOTAL
8
15
~
~