Marking Scheme (Kedah)

8/4/2019 Marking Scheme (Kedah)

1/15

RESTRICTED*/TERHAD* 964/1/2/MS

STPM 964/1/2/MS [Turn over (Lihat sebelah)

JABATANPELAJARANNEGERIKEDAHJABATANPELAJARANNEGERIKEDAHJABATANPELAJARANNEGERI





JABATANPELAJARANNEGERIKEDAHJABATANPELAJARANNEGERIKEDAHJABATANPELAJARANNEGERIJABATANPELAJARANNEGERIKEDAHJABATANPELAJARANNEGERIKEDAHJABATANPELAJARANNEGERI





964/1/2/MS PERCUBAAN STPM 2011

BIOLOGY (BIOLOGI)MARKING SCHEME (SKEMA PEMARKAHAN)

PAPER 1 AND PAPER 2 (KERTAS 1 DAN KERTAS 2)

PEPERIKSAAN PERCUBAAN BERSAMA

SIJIL TINGGI PERSEKOLAHAN MALAYSIA (STPM) 2011

ANJURAN

JABATAN PELAJARAN KEDAH DARUL AMAN

DAN

PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUASEKOLAH MENENGAH MALAYSIA (PKPSM) KEDAH


2/15

RESTRICTED*/TERHAD* 2 964/1/2/MS

PAPER 1 (KERTAS 1)

1 B 26 D

2 A 27 A

3 A 28 C

4 D 29 D

5 A 30 D

6 C 31 B

7 C 32 D

8 C 33 D

9 B 34 B

10 B 35 C

11 A 36 D

12 D 37 C

13 A 38 D

14 C 39 A

15 B 40 C

16 D 41 A

17 D 42 B

18 D 43 B

19 A 44 C

20 B 45 D

21 C 46 C

22 B 47 B

23 A 48 C

24 A 49 A

25 C 50 A


3/15


PAPER 2 / KERTAS 2

SECTION A

Question 1

(a) 1 oxygen binds to the haem groups//Initially oxygen molecule binds slowly /

harder to bind (therefore the curve of graph is less steep initially). - 1

2 distorts shape of molecule / allosteric//easier for next oxygen to bind / next

oxygen molecule bind faster (therefore the curve is steep in the middle). - 1

3 when haemoglobin is saturated with oxygen, no more oxygen is able to

bind with the haemoglobin (therefore the curve is level off/ less steep at

the end). - 1

3 m

(b) 1 as the partial pressure of carbon dioxide increases, the saturation of

haemoglobin with oxygen decreases / desaturation increases - 1

2 curve shifts to the right (Bohr effect) - 1

2m

(c) 1 increased CO2 levels release hydrogen/hydrogen carbonate

form//decreased pH in the plasma - 1

2 causes the release of the oxygen by the haemoglobin/lowers affinity for

oxygen - 1

2m

(d) 1 rate of respiration increases/ demand for oxygen increases - 1

2 CO2 increases/pH drops - 1

3 body temperature increases - 1

4 affinity of haemoglobin towards oxygen is lowered /More oxygen release

(from haemoglobin) to body cells. - 1

3m

Total 10m

any 3


4/15


Question 2

(a) I : Corpus allatum - 1

II : Prothoracic gland - 1

III : Juvenile hormone // neotonin - 1

IV : Ecdysone - 1

4m

(b) 1 stimulate the corpus allatum to synthesise juvenile hormones - 1

2 stimulates prothoracic gland to synthesise ecdysone - 1

1m

(c) 1 juvenile hormone suppresses the genes that control the production of adult

characteristics - 1

2 its activates the genes that control the retention of larval or nymphal

characteristics - 1

3 ecdysone stimulates the ecdysis process by activating specific genes that

control the synthesis of enzymes involved in the synthesis of exoskeleton - 1

4 it is also stimulates the development of adult characteristics by activatingthe genes involved in the synthesis of adult characteristics - 1

4m

(d) 1 Level of juvenile hormone is very low, allowing the development of adult

characteristics, pupa moulds into adult insects - 1

1m

Total 10m

any 1


5/15


Question 3

(a) Process Y : meiosis - 1

Process Z : fertilisation - 1

2m

(b) W : gametophyte - 1

X : sporophyte - 1

2m

(c)

W X

multicellular haploid multicellular diploid - 1/0

produces spore (through meiosis) produces gametes (through mitosis) - 1/0

2m

(d) 1 shows an alternation of generation in which the diploid sporophyte

generation is the dominant generation - 1

2 the haploid gametophyte generation is the simple , free living prothallus - 1

3 the sporophyte generation has true stems, leaves and roots - 1

4 the sporophyte generation has vascular tissue but the xylem tissues only

have tracheids and the phloem tissue only have sieve tubes - 1

2m

(e)

Plant Phylum

Dryopteris Filicinophyta - 1

Marchantia Bryophyta - 1

2m

Total 10m

any 2


6/15


Question 4

(a) Given that 250 of 1000 sheep have the genotype rr (recessive allele)

q2

is the frequency of rr - 1

Thus, q

2

= 1000

250

q =1000

250

q = 0.5 (reject : 1105 and2

1) - 1

2m

(b) [ Frequency of heterozygous Rr genotype = 2pq ]** - 1

= 2 0.5 0.5

= 0.5 - 1

2m

(c) (i) Number of sheep with RR genotype = p2

1000 - 1

= 0.25 1000

= 250 - 1

2m

(ii) Number of sheep with heterozygous Rr genotype = 2pq 1000 - 1

= 0.5 1000

= 500 - 1

2m

(d) Dominant allele in the new population.

Genotype RR Rr rr

Number of genotype 250 500 0

Number of allele R 2 250 500 0 - 1

Number of allele r 500

Total number of allele 500 1500 1500

Frequency of dominant allele R =1500

500500 +

= 0.67 - 1

2m

(Note : **compulsory sentence) Total 10m


7/15


Question 5

(a) 1 Krantz anatomy has two rings of cells - 1

2 the outer ring of columnar mesophyll cells surround a ring of - 1

3 inner ring of bundle sheath cells and the vascular bundle - 1

3m

(b) 1 CAM plants contain mesophyll cells and without Krantz anatomy - 1

2 fixation of CO2 take place at night when the stomata are open - 1

3 phosphoenolpyruvate /PEP combines with CO2 to produce oxaloacetate - 1

4 oxaloacetate (OAA) is reduced to malate that required NADPH - 1

5 malate is stored in the vacuole of mesophyll cell - 1

6 during daytime when there is light, malate is oxidised to produce CO2and pyruvate - 1

7 concentration of CO2 increases in mesophyll cells and photorespiration is

prevented - 1

8 CO2 is fixed by RuBP to form PGA / phosphoglyceric acid in Calvin

cycle to produce glucose and starch - 1

Max 7m


8/15


(c)

C3 plants C4 plants

Fixation of CO2 during

photosynthesis occurs once in

mesophyll cells,

Fixation of CO2 during photosynthesis

occurs twice, in mesophyll cells and

bundle sheath cells.

- 1/0

Photorespiration occurs at low CO2

concentration. Oxygen acts as a

competitive inhibitor for CO2.

Photorespiration is inhibited by a high

concentration of CO2 in the bundle

sheath cell.

- 1/0

The first product formed in the

dark phase of photosynthesis is

glycerate-3-phosphate (3C

molecule).

The first product formed in the dark

phase of photosynthesis is

oxaloacetate (4C molecule).- 1/0

The enzymes involved is RuBP

carboxylase.

The enzymes involved are RuBP

carboxylase and PEP carboxylase.- 1/0

The CO2 acceptor is ribulose

biphosphate (RuBP).

The CO2 acceptors are ribulose

biphosphate (RuBP) and

phosphoenolpyruvate (PEP).

- 1/0

Photosynthesis is less efficient

compared to C4 plants. Yield is

usually lower.

Photorespiration is inhibited and

photosynthesis rate is higher. Yield is

higher.

- 1/0

Max 5m

Total 15 m


9/15


Question 6

(a) Ultrafiltration

1 metabolite wastes are ammonia salts, creatinine, excess water sodium and

chloride ions - 1

2 blood containing nitrogenous wastes such as urea enter the kidney via therenal artery - 1

3 high hydrostatic pressure in glomerulus forces the blood content to filter

into Bowmans capsule - 1

4 through ultrafiltration - 1

5 the glomerular filtrate contains glucose, amino acids, vitamins,

hormones, urea, uric acid, creatinine, ions and water - 1

Max 3m

Reabsorption6 from the Bowmans capsule, the glomerular filtrate flows into the

proximal convoluted tubule where active reabsorption of useful

substances occurs - 1

7 all glucose, amino acids and a large proportion of mineral salts are

actively reabsorbed into the peritubular capillaries - 1

8 a large proportion of water (over 80%) is reabsorbed by osmosis - 1

9 some urea is reabsorbed from the filtrate by diffusion - 1

10 waste substances such as creatinine, drugs which are not filtered in the

glomerulus, are carried by blood to the proximal convoluted tubule andactively secreted into its lumen - 1

11 in the loop of Henle, Na+

move out of ascending limb into medulla tissue - 1

12 water is reabsorbed from descending limb - 1

13 counter current multiplier - 1

14 the remaining salt is reabsorbed in the distal convoluted tubule and

collecting duct into the surrounding blood capillaries by osmosis - 1

Max 6m

Active Secretion

15 the urea, toxic substances and drugs are actively secreted into the distal

convoluted tubule from the surrounding capillaries - 1

16 the filtrate flows into the collecting duct as urine and it contains water

and excess mineral salts not required by the body, waste substances like

urea, ammonium ions, uric acid and creatinine - 1

Total 11m


10/15


(b) 1 ADH is secreted by the pituitary gland into the blood - 1

2 ADH increases the permeability of the distal convoluted tubule and

collecting ducts towards water molecules - 1

3 this is created by increasing aquaporins arranged on the plasma

membrane facing the lumen of tubule - 1

4 more water is reabsorbed from the distal convoluted tubule and collecting

duct into the blood capillaries - 1

5 a small amount of concentrated urine is produced - 1

6 the water potential of blood increases back to the normal range - 1

Max 4m

Total 15m


11/15


Question 7

(a) Self

1 the ability to distinguish self from non self depends on the lymphocytes

in the body which have the ability to detect any foreign substance/protein

that is incompatible with it - 12 if the foreign substance is compatible to the body cells / has class I MHC

protein or antigen found on the surface of most nucleated cells, no

immunity response is triggered - 1

3 the foreign substance is self substance and is not rejected - 1

Non-self

4 class I MHC protein is detected as an antigen. The foreign substance is

considered as a non-self substance - 1

5 this trigger an immune response by producing antibodies that attack the

foreign tissue, eventually destroy it - 1

5m

(b) 1 HIV enters the body through body fluid or blood transfusion - 1

2 once inside the body, the glycoprotein of HIV binds to the CD4 receptors

found on the surface membrane of the helper T lymphocytes - 1

3 both membranes then fuse while the capsid is enzymatically removed to

release the viral RNA and reverse transcriptase into the cytoplasm of the

helper T lymphocytes - 1

4 viral RNA undergoes reverse transcription followed by replication to

produce a double-stranded viral DNA with the help of the enzyme

reverse transcriptase - 1

5 the viral DNA will enter the nucleus and incorporate itself into the hosts

DNA as a provirus - 1

6 HIV may persist in a latent state for years - 1

7 once activated, the cells DNA will undergo transcription and translation

to yield huge amounts of viral mRNA and viral proteins - 1

8 new viral proteins and viral RNA copies are assembled in the cytoplasmto form new viruses, which then bud out of the host cell further infecting

other helper T lymphocytes - 1

9 the number of helper T lymphocytes will gradually decrease - 1

10 this exposes the individual to greater risks of contacting other diseases

and finally leading to death - 1

10m

Total 15m


12/15


Question 8

(a) Gene expression needs to be controlled because:

1 to save energy, because not all genes are needed at the same time

(In humans, some genes are needed early in life (for development) are

needed in adulthood) - 12 in a multicellular organism different cell types need different gene

products - 1

3 genes are also regulated for cell differentiation - 1

4 cells have different functions and respond to different types of stimuli - 1

Max 3m

(b) 1 inducible system means that in the natural state, the gene is switched off - 1

2 and only switched on when needed - 1

2m

(c) Gene expression in lac operon

1 repressor protein is encoded by regulatory gene - 1

2 without the presence of lactose - 1

3 it binds to operator - 1

4 the binding will block attachment site of RNA polymerase on promotergene - 1

5 thus the lactose operon is deactivated / no transcription of structural

genes (no enzymes is produced to utilize the lactose) - 1

6 when lactose is provided - 1

7 the allolactose (lactose isomer) binds to repressor protein - 1

8 protein conformation is changed - not suite to its binding site on operator - 1

9 RNA polymerase able to bind to its site on promoter - 1

10 thus lactose operon is activated - 111 transcription of structural gene occurred// mRNA is produced//enzyme is

produced - 1

Max 10m

Total 15m


13/15


Question 9

(a) 1 an ecosystem comprises biotic components and abiotic components that

mutually interact and function together (to form a stable and balanced

natural system) - 1

2 the biotic component of an ecosystem involves all the living organismsthat interact with each other in the ecosystem - 1

3 the biotic components of an ecosystem comprises at least three trophic

levels: producers (autotrophs), consumers (heterotrophs) and the

composers - 1

4 The energy flow in an ecosystem is continuous and produces an output of

heat energy - 1

5 Nutrients such as water, sulphur and phosphorus are recycled by

biogeochemical cycles - 1

6 example: Pond ecosystem that consists of biotic components such ascarp, Daphnia and abiotic components such as pond water, soil and

sunlight - 1

Max 5m

(b) 1 food chain reflect the transfer of energy in an ecosystem - 1

2 original source of energy is from the sun / solar energy - 1

3 primary producers / photosynthetic organisms trap (1-5% of light energy

received for biomass production - 1

4 primary consumers / herbivores feed on plants / primary producers - 1

5 energy is transferred from first trophic level to the second trophic level - 1

6 90% of the energy is lost to the environment // only 10 % of the biomass

produced consumed by the herbivores // 10% of the energy is transferred

to the next trophic level - 1

7 there is further energy loss (from the herbivores) through respiration and

excretion - 1

8 secondary consumers feed on primary consumers - 1

9 energy is transferred from second to third trophic level - 1

10 tertiary consumers feed on secondary consumers - 1

11 there are seldom more than 5 trophic levels - 1

12 large amount of energy is lost during each transfer (within and between

the organism) //Each trophic level received less energy than the level

below it - 1

13 at the fourth or fifth level. only a small amount energy is left/insufficient

to support further trophic level - 1

14 limiting the number of members in the higher trophic level - 1

Max 10

Total 15m


14/15


15/15


(c) Differentiate between plant from the group conifers and angiosperms

Pinus sp. Caesalpinia sp.

Sexual reproductive structure is the

cone/ strobilus

Sexual reproductive structure is the

flower

1/0

Cone parts are arranged in spirals Flowers parts advanced in

angiosperms are arranged in whorls

1/0

Ovules are not enclosed by the ovary Ovules are enclosed by the ovary 1/0Seeds are exposed or naked Seeds are enclosed by the fruit wall//

pericarp

1/0Vessel elements absent in xylem

tissues

Vessel elements present in xylem

tissues

1/0

Companion cells absent in the phloemtissues Companion cells present in thephloem tissues . 1/0

No double fertilization. Double fertilization occurs 1/0Pollination by wind. Pollination by several agents. 1/0

Max 7m

Total 15m

Marking Scheme (Kedah)

Documents

Transcript of Marking Scheme (Kedah)