gerak gempur perak stpm 2012 biology

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SULIT 1964/1

BIOLOGY

20121 hr 45 mins

GERAK GEMPUR BIOLOGI (BIOLOGY)

KERTAS 1 (PAPER 1)

Satu jam empat puluh lima minit (One hour and forty-five minutes)

DAERAH KINTA UTARA, PERAK

Instructions to candidates:

DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO.

There are fifty questions in this paper. For each question, four suggested answers are given.

Choose one correct answer and indicate it on the multiple-choice answer sheet provided.

Read the instructions on the multiple-choice answer sheet very carefully.

Answer all questions. Marks will not be deducted for wrong answers.

Gerak Gempur Biologi STPM 2012

Daerah Kin ta Utara, Perak

This papers contains 15 printed pages

http://edu.joshuatly.com/ http://fb.me/edu.joshuatly



CONFIDENTIAL* Gerak Gempur Bio(P1) STPM 2012

Daerah Kinta Utara, Perak

964/1 [Turn over

This question paper is CONFIDENTIAL until the examination is over. CONFIDENTIAL*

2

Choose the most appropriate answer.

1. Which is the properties of water?

A. Low viscosity with low surface tension.

B. High viscosity with low surface tension.

C. Low viscosity with high surface tension.

D. High viscosity with high surface tension.

2. Which of the following process is not involved in condensation?

A. Formation of glycosidic bond between glucose and fructose.

B. Esterification between glycerol and fatty acids to form a triglyceride.

C. Formation of two units of amino acids from a dipeptide.

D. Synthesis of cellulose.

3. Which of the following statements about lipids are correct?

I. Lipids consist of glycerol combined with alcohols.

II. Both oils and fats are triglycerides.

III. Oils are more saturated with hydrogen than are fats.

IV. Lipids are more highly reduced substances than are carbohydrates.

A. I, II and III only

B. I and III only

C. II and IV only

D. I and IV only

4. Which of the following organelles can only be seen using an electron microscope?

A. Chloroplast.

B. Mitochondrion.C. Nucleus.

D. Ribosomes.

5. Which of the following cell types has the thinnest cell wall?

A. Collenchymal cells.

B. Sclerenchymal cells.

C. Epidermal cells.

D. Meristematic cells.






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6. Which of the following statement is incorrectly to describe compact bone?

A. It consists of living cells embedded in a matrix mainly of chondrin.

B. It is made up of numerous Harversian systems.

C. Osteoblasts are found in lacunae between the lamellae.

D. Canaliculi connect the lacunae to each other.

7. Plasma membrane is selectively permeable to certain substances. From the followingsubstances, which is/are not permeable to plasma membrane?

I. Hydrophobic molecules

II. Polar and tiny molecules

III. Polar and large molecules

IV. Charged ions

A. IV only

B. I and II only

C. III and IV only

D. I, II and III only

8. Which of the following does not increase the rate of diffusion?

A. increasing the concentration of molecules.

B. increasing the temperature of liquid.

C. decreasing the size of molecules.

D. increasing the size of molecules.

9. The following graph shows the Lineweaver-Burke plot for certain reaction catalyzed

by an enzyme. Which value on the graph is used to calculate the Michaelis-Menten

constant?

A. I B. II C. III D. IV






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10. Which of the following statement is incorrect?

A. There are 64 different codons presented in mRNAs.

B. Each codon codes for a specific amino acid.

C. The number of codon is much more than the number of amino acids, so some of the

codons are redundant.D. There are some codons mark the end of a polypeptide sequence called termination

codons.

11. Which of the following statements are correct to describe the translation process(protein

synthesis)?

I. codon of mRNA is coded from direction 3’ 5’

II. initiate codon of mRNA for translation is AUG

III. attachment of mRNA to ribosome takes place in cytoplasm

IV. tRNA, which has a region called codon, carries amino acid to ribosome

V. translation process takes place in cytoplasm


B. I, III and V only

C. II, IV and V only

D. II, III and V only

12. A culture of bacteria had all its DNA labelled with the hcavy isotope of nitrogen, 15 N.

The culture was then allowed to reproduce using nucleotides containing normal 14 N. The

DNA was examined using a centrifuge after one generation and again after the second generation. The diagram shows the position of the DNA band at Z in the centrifuge tube

when the DNA was first labelled.

In which pattern would the DNA be found after the first and after the second cellgenerations?

After first generation After secong generation

A. All at Y Half at X and half at Y

B. All at Z Half at Y and half at Z

C. Half at X and half at Y Quarter at X and Z and half at Y

D. Half at X and half at Z Quarter at X and Z and half at Y






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13. Energy from the photons cause the electron excited from the pigment molecules in thylakoid

membrane and start a series of photosynthesis reactions. Three reactions occur during light

reaction are

A. reduction of O2, oxidation of NADPH, production of ATP.

B. oxidation of H2O, reduction of NADP+, production of ATP.

C. fixation of CO2, release of O2, synthesis of glucose.

D. release of O2, fixation of CO2, hydrolysis of ATP.

14. Which of the following represents the photorespiration phenomenon?

A. RuBP + CO2 3-PGA

B. RuBP + O2 3CO2 + 5ATP + 2NADPH

C. RuBP + CO2 glucose + 12ADP

D. RuBP + O2 glucose + 12ADP

15. Which of the following event is not occur in a plant cell that undergoes anaerobic respiration

using a glucose molecule as substrate?

A. Two ATPs are yielded.

B. Lactate as final product.

C. One molecule of glucose is converted into two molecules of pyruvate throughglycolysis process.

D. No net gain of NADH.

16. How many of the hydrogen carrier can be generated by each acetyl Co-A molecule thatenters citric acid cycle/Krebs cycle?

I. 3 NADH

II. 4 NADH

III. 1 GTP

IV. 1 FADH2 V. 2 FADH2


B. I, III and IV only

C. II, III and V only

D. III, IV and V only






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17. Which statement below is incorrectly to describe the organism that involved in

chemosynthesis nutrition?

A. They are called photoheterotrophs.

B. Synthesis organic substances from CO2 and water.

C. Energy source is from chemical reaction such as oxidation of iron.

D. Nitrifying bacteria use ammonia and ammonium as energy source.

18. Which of the following organism does not involve in holozoic nutrition?

A. Fish

B. Housefly

C. Fungi

D. Amoeba

19. Carbon dioxide can be transported by blood as form of

I. carbonic acid

II. bicarbonate ions

III. carbamino ions

IV. haemoglobinic acid


B. I, II and IV only

C. II, III and IV only

D. I, II, III, IV

20. What is meant by vital capacity when measuring human lung volumes by using spirometry?

A. Volume of air breathed in and out during a normal single breath.

B. Maximum volume of air that can be breathed in or breathed out of the lungs.

C. Volume of air which cannot be expelled even during forced expiration.D. Total volume of air taken into the lungs in one minute.






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21. Graph 1 shown as below is an oxygen dissociation curve for myoglobin and haemoglobin.

What conclusion about myoglobin can be made from Graph 1?

A. It binds one oxygen molecule but haemoglobin binds with four oxygen molecules.

B. It will release oxygen to haemoglobin if partial pressure of oxygen is low.

C. It will only release oxygen when the partial pressure of oxygen is low.

D. It has less affinity towards oxygen molecule compare of those in haemoglobin.

22. Which of the following events will cause the opening of stomata?

I. The thickened inner cell wall of the guard cells.

II. Water move into guard cells by osmosis.

III. Conversion of glucose into starch in guard cells.

IV. Osmotic pressure in guard cells is much higher than that of the epidermal cells.

A. I, II and IV only


C. II and IV onlyD. II, III and IV only

23. Which statement is not true of auxin?

A. It stimulates the division of cell in a stem.B. It stimulates the elongation of coleoptile.

C. It promotes the formation of lateral shoot.

D. It inhibits the elongation of root at the high concentration.

Graph 1






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24. If plant X is a short day plant with a critical day length of 8 1/2 hours, in which of these

conditions will plant X not flower?

I. Exposed to red light for 9 hours.

II. Kept in darkness for 8 hours.III. Exposed to far red light for 12 hours.

IV. Kept in darkness for 15 hours before exposed to red light for 1 hour

A. I and II

B. III and IV

C. I, II and IVD. II , III and IV

25. Which of the following correctly explains the distribution of ions on either side of themembrane of an axon in its resting state?

A. A high concentration of organic anions outside and a low concentration of K+ ions inside.B. A high concentration of organic anions inside and a low concentration of Na+ ions outside.

C. A high concentration of K+ ions and organic anions outside and a high concentration of Na+

ions inside.

D. A high concentration of Na+ outside and a high concentration of K+ ions and organic anionsinside.

26. Which of the following are true of B cell?I. It forms immunity through the humoral response.II. It forms immunity through the cell mediated mechanism.

III. It is produced and it achieves maturity in the bone marrow

IV. It is produced in the bone marrow and it achieves maturity in the thymus gland

A. I and II

B. I and III

C. II and IV

D. III and IV

27. Which one of the following is true of T cells?

I. T cells produce complement proteinsII. T cells mature in the bone marrow

III. T cytotoxic cells kill body cells that are infected with viruses

IV. T suppressor cells inhibit the acivities of T cytotoxic cells

A. I and II

B. I and IVC. II and III

D. III and IV






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28. Which is not correct about the concept of self and non-self ?

A. MHC protein markers on the surface of cells act as specific antigens.

B. Each organism carries a set of antigens which is unique to each individual.C. In the embryo, T lymphocytes which can bind to the receptors (proteins and

polysaccharides) of their own body cells are induced to proliferate to form a clone of T

cells.D. The immune system can distinguish “self” (the body’s own tissue) from “non-self”

(foreign tissue)

29. Which one of the following represents the general pattern of the alternation of generations?

A. gametophyte meiosis sporophyte sporesB. gametophyte meiosis spores sporophyte

C. sporophyte meiosis gametophyte spores

D. sporophyte meiosis spores gametophyte

30. During development, each microspore divides by mitosis to produce

I. microsporangium

II. a tube nuclues

III. a generative nuclues

IV. a synergids nuclues


B. I and IV only

C. II, III and IV only

D. II and III only

31. If the diploid state is represented by (D) and the haploid state by (H), which one of the

following sequences correctly describes the chromosome number in the named plant structures?

Polar Nucleus Pollen grain Pollen tube nucleus Cell in the testa Cell in the pericarp

A. (D) (D) (D) (D) (D)B. (D) (H) (H) (H) (H)

C. (H) (D) (H) (H) (D)

D. (H) (H) (H) (D) (D)






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32. Which of the following is true of an oviparous animal?

A. An individual hatches from the egg outside the female parent’s body

B. An individual hatches from the egg in the uterus of the female parentC. An individual is born before maturity and continues to develop in the sac of the female

parent.

D. An individual develop in the uterus of the female parent and the embryo obtains thenutrient from the placenta.

33.

The above statement refers to which stage of embryonic development?

A. BlastomereB. Cleavage

C. Gastrul ation

D. Organogenesis

34. Which of the following is the absolute growth curve of maize plant?The following are events that occur during seed germination.

I. Synthesis and secretion of enzymes

II. Activation of the aleurone layer III. Flow of sugars to the embryo

IV. Release of gibberellin

V. Hydrolysis of starch

Which of the following is the correct sequence of events during seed germination?

A.

II

lV

V

III

IB. IV I V II IIIC. IV II I V III

D. V III IV II I

The formation blastula from a zygote involves a succession of rapid cell divisions. This special

type of cell division creates a multicellular embryo.






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35. Which of the following statements are true about neurosecretion secreted by the

neurosecretory cell in insects?

I. It is stored in corpus allatumII. It inhibits the effect of juvenile hormone

III. It is also known as prothoracictrophic hormone

IV. It stimulates the secretion of ecdysone from prothoracic gland

A. I, II and III

B. I, III and IVC. II, III and IV

D. I, II, III and lV

36. The delay in development of insect pupa due to a decrease in light intensity is known as

A. hibernation

B. aestivation

C. diapause

D. metamorphosis

37. Two parents, each of blood group A, have a daughter of blood group O. What is the

probability that their next child will be a boy who has blood group O?

A. 0.125

B. 0.25

C. 0.50

D. 0.75

38. Turner syndrome is an example of

A. allopolyploidy

B. autopolyploidy

C. aneuploidy

D. euploidy






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39. The Hardy-Weinberg equation does not apply if

I. there is migration of population

II. there is natural selection

III. mutations occur

IV. there is non-random mating

V. there is a large population

A. I, II, III and IV

B. I, II, III and V

C. I, III, IV and V

D. II, III, IV and V

40. From data below:

The frequency of allele “A” is

A. 0.75

B. 0.87

C. 0.34

D. 0.25

41. The lactose operon consists of

I. a regulator gene

II. an operator geneIII. three structural genes

IV. a promoter gene

A. II and III only

B. II and IV only

C. I, II and III only

D. II, III and IV only

Number of individuals with “AA” is 3500.

Number of individuals with “Aa” is 1000.

Number of individuals with “aa” is 80.






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42. Which of the following is/are the suitable characteristic/s for plasmid used as cloning

vector?

I. More than one restriction sites.

II. Replicate very fast

III. Insignificant genetic markers for resistance to antibiotics.

IV. Can be easily isolated and purified from bacterial culture.



C. I, II and IV only

D. I, II, III and IV

43. Which of the following regarding taxa and their examples not correct?

A. Phylum: Chordata

B. Class: MammaliaC. Order: Hominidae

D. Species: sapiens

44. Based on the table below, match phyla of organisms to their characteristics.

Phylum Characteristic

I Cnidaria P Body divided into head, muscular foot and visceral massII Artropoda Q Diploblastic body, polymorphism

III Mollusca R Segmented legs, chitinous exoskeleton

IV Nematoda S Body covered with thin and elastic cuticle, pseudocoelom

I II III IV

A P Q S R B Q R P S

C R S Q P

D S R P Q






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45. Mechanisms of intrinsic isolation leading to speciation include

I. geographical barriers

II. ecological isolation

III. temporal isolation

IV. gamete isolation


B. I, II and IV only

C. I, III and IV only

D. II, III and IV only

Questions 46-47 S. townsendii is a new hybrid plant species between S. maritime and S. alterniflora.

46. Why the first hybrid generation is sterile?

A. because of the mutation in chromosome number.

B. because of non-disjunction during meiosis I.

C. because of the reciprocal translocation between two non-homologous chromosomes.D. because of the absence of homologous pairs, meiosis not probable to take place.

47. What are the number of chromosomes represented by P and Q?

P Q

A. 63 126

B. 126 252

C. 252 504

D. 126 63

x S. alterniflora

(2n = 70)

S. maritime

(2n = 56)

Sterile hybrid

(2n = P)

S. townsendii

(4n = Q)






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48. Diagram 1 & 2 represent a biomass and a number pyramid for a particular food chain.

Which of the following correctly identifies the diagram?

Diagram 1 Diagram 2 Habitat Producer

A. Number Biomass Pond Tiny algae

B. Number Biomass Tropical rain forest Terrestrial trees

C. Biomass Number Sea-shore Phytoplankton

D. Biomass Number Pond Tiny algae

49. Organic sulphur in plant and animal proteins can be decomposed by bacteria directly to form

A. SO2

B. SO3

C. H2S

D. SO42-

50. The relationship between gross primary production (GPP), net primary production (NPP)and heat loss through respiration (R) can be represented by the equation

A. GPP = NPP

B. GPP = NPP - R

C. NPP = GPP - R

D. NPP = GPP + R

Diagram 1 Diagram 2




SULIT 1964/2

BIOLOGY

20121 hr 45 mins

GERAK GEMPUR BIOLOGI (BIOLOGY)

KERTAS 2 (PAPER 2)

Dua jam setengah (Two and a half hours)

DAERAH KINTA UTARA, PERAK

For examiner’s use

1

2

3

4

5

6

7

8

9

10

Total

Gerak Gempur Biologi STPM 2012


This papers contains 9 printed pages

Instructions to candidates:

Answer all the questions in Section A in the spaces provided.

Answer any four questions from Section B. For this section,

write your answers on the answer sheets provided. Begin each

answer on a fresh sheet of paper. Answers should be illustrated by

large, clearly labelled diagrams wherever suitable.

Answers may be written in either Malay or English.

Arrange your answers in numerical order and tie the answer

sheets to this booklet.




CONFIDENTIAL* Gerak Gempur STPM (Bio P2) 2012


964/2 [Turn Over*This question paper is CONFIDENTIAL until the examination is over. CONFIDENTIAL*

2

SECTION A [40 marks]

Answer all the questions in this section.

1. a) The diagram shows a molecule of an enzyme called ribonuclease. Each amino

acid in the protein is indicated by a 3-letter symbol e.g. Arg = arginine.

i) How many nucleotides are there in the mRNA molecule that codes

for this enzyme? [1]

X

80

20

60

100

120

1

37







3

ii) The table gives the mRNA code for the four amino acids in the part of enzyme

labelled X.

Amino acid Symbol mRNA code

Alanine Ala GCU

Glutamine Glu GAG

Leucine Leu UUA

Serine Ser AGU

Give the DNA code for the part of the enzyme labelled X. [1]

iii) Where does translation occur in a cell? [1]

iv) Describe briefly what happens during translation. [2]

b) Figure 1 shows a reac tion catalyesed by an enzyme in lock and key hypothesis .

succinic dehydrogenase

hydrogen

Figure 1







4

i) If the enzyme is succinic dehydrogenase, name the parts labelled Q and R. [2]

ii) Describe the mechanism of this enzymatic reaction based on the lock and keyhypothesis. [3]

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________







5

2. The diagram below shows changes in the membrane potential of a neurone during the

production of an action potential.

(a) Label phases A,B,C and the period labelled D as shown in the diagram. [4]

A : ___________________________________________________

B : ___________________________________________________

C : ___________________________________________________

D : ___________________________________________________

(b) Describe how A is maintained. [2]

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________







6

c) Distinguish between B and C. [2]

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

d) Why is the potential of D less than A? [1]

______________________________________________________________________________

______________________________________________________________________________

e) State the importance of D. [1]

______________________________________________________________________________

______________________________________________________________________________







7

3. a) Complete the table below to show the differences between the cells of the immune system

involved in humoral immune response and cell-mediated response. [6]

Description Humoral immune

response

Cell-mediated immune

response

Origin of cells

Primary cells

Site of cell maturation

b) What is the effect of an antigen on B cells? [2]

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

c) Name the substance that is secreted by macrophages and leucocytes to trigger immune

response. [1]

______________________________________________________________________________

d) State the effect of the subtance named in (c) in immune response. [1]

______________________________________________________________________________

______________________________________________________________________________







8

4. The two graphs (S and T) in the following figure show the measurement of growth rates of the same plant, Zea mays.

a) Name the type of curve illustrated by graph S and T. [2]

Curve S -

Curve T -

b) Explain what graphs S and T represent. [2]

______________________________________________________________________________

______________________________________________________________________________

c) Based on its lifespan, what type of plant is Zea mays known as? [1]

______________________________________________________________________________

d) i) Describe the growth pattern in Zea mays. [3]

______________________________________________________________________________

______________________________________________________________________________

ii) Explain how is this type of growth pattern different from that of woody plants. [2]

______________________________________________________________________________

______________________________________________________________________________







9

Section B [60 marks]

Answer any four questions from this section.

5. (a) Draw and labeled the structure of a cell membrane based on Singer’s model. [2]

(b) Explain the roles of the structures of the cell membrane in the transportation of substances

into the cell. [13]

6. (a) State briefly the differences between anaerobic respiration that occur in plants and

animals. [6]

(b) NAD and NADP are important molecules in plant cells. Describe in detail, the role of these

molecules within a palisade mesophyll cell. [9]

7. (a) Explain why no flowering response is shown by some plants that grow near highways.

[5]

(b) Explain the changes that occur during the onset of labour in humans. [10]

8. (a) Outline on 3 examples of internal fertilisation, namely oviparity, ovoviviparity and viviparity. [7]

(b) i) State the similarities and differences in characteristics between Marchantia sp. and Dryopteris sp. [8]

9. (a) Give an account of the events which occur after fertilization from cleavage to

organogenesis in humans. [8]

(b) With the aid of labeled diagram describe the function of placenta. [7]

10. (a) Define the following terminologies:

i) Speciation

ii) Gene pool [ 3]

(b) Describe the importance of following mechanisms in genetic divergence:

i) geographical isolationii) pre-zygotic and post-zygotic isolation. [ 12]




Gerak Gempur Bio STPM 2012 (Schematic answers P2) Daerah Kinta Utara, Perak

1

Answer Gerak Gempur Biologi STPM Kinta Utara 2012

BIOLOGY STPM PAPER 1 (964/1)

1 C 11 D 21 C 31 D 41 D

2 C 12 A 22 A 32 A 42 C3 C 13 B 23 C 33 B 43 C

4 D 14 D 24 C 34 C 44 B

5 D 15 B 25 D 35 C 45 D

6 A 16 B 26 B 36 C 46 D

7 C 17 A 27 D 37 A 47 A

8 D 18 C 28 C 38 C 48 A

9 B 19 A 29 C 39 A 49 C

10 C 20 B 30 D 40 B 50 C


Section A (Structured Questions)

Q1 SUGGESTED ANSWERS M ARK

1. ( A) I) 3 NUCLEOTIDES X 123 AMINO ACIDS = 369 NUCLEOTIDES 1

II) 3’ CGA AAT TCA CTC 5’ 1

III) RIBOSOMES IN CYTOPLASM OF CELLS 1

IV) INITIATION : MRNA BINDS TO SMALL RIBOSOME SUBUNIT WITH ITS 5’ END.

AMINO ACID-TRNA

COMPLEX PAIR WITH AN INITIATION CODON ON THE

MRNA.L ARGE SUBUNIT RIBOSOME BINDS TO SMALL SUBUNIT TO FORM COMPLETE

RIBOSOME ELONGATION FIRST AA-TRNA COMPLEX HAS PAIRED WITH INITIATION CODON (AUG) ON

THE P SITE, ANOTHER AA-TRNA COMPLEX THEN PAIR WITH THE MRNA

CODON ADJACENT TO THE INITIATION CODON ON A SITE.BOTH AMINO ACID LINKED TOGETHER BY PEPTIDE BONDS. ANTICODON ON AA-TRNA MOLECULES PAIR WITH COMPLEMENTARY CODON

ON MRNA.RIBOSOME MOVES ONE BY ONE CODON ALONG THE MRNA TOWARDS 3’

END.TERMINATION :POLYPEPTIDE SYNTHESIS END WHEN THE RIBOSOME REACH AT THE

TERMINATION CODON : UAA,UAG OR UGA ON MRNATHE COMPLETED POLYPEPTIDE DETACH FROM MRNA AND

RIBOSOME .

1+1(ANY 2)FROM

( A),(B),(C)

(B) I) SUCCINATE/SUCCINIC ACID, FUMARATE/FUMARIC ACID 1+1

II) THE ACTIVE SITE OF AN ENZYME HAS A SHAPE AND STRUCTURE

(CONFIGURATION) WHICH ONLY COMPLEMENTS(SIMILAR) THE SHAPE

OF ITS SUBSTRATE MOLECULES.ENZYMES AS “LOCK’ AND SUBSTRATE AS ‘KEY’

SUBSTRATE MOLECULES ATTACHES INTO ENZYME ACTIVE SITE, FORMING AN ENZYME-SUBSTRATE COMPLEX.BONDS IN THE SUBSTRATE MOLECULES ARE FORMED OR

BREAKDOWN IN THE E-S COMPLEX, PRODUCTS ARE FORMED.THE PRODUCT MOLECULES LEAVE THE ACTIVE CENTER AS THEY

1+1+1(ANY 3)





2

HAVE ENTIRELY DIFFERENT STRUCTURE FROM SUBSTRATE

MOLECULES .

TOTAL 10


2. ( A) A: RESTING POTENTIAL

B: DEPOLARIZATION

C: REPOLARISATION

D: REFRACTORY PERIOD

1111

(B) - THE ACTIVE TRANSPORT OF SODIUM AND POTASSIUM IONS IN AND OUT

ACROSS THE MEMBRANE/ 3N A+ OUT, 2 K + IN - THE MEMBRANE IS MORE PERMEABLE TO POTASSIUM IONS

COMPARED TO SODIUM IONS ; THUS MORE POTASSIUM IONS DIFFUSE OUT OF THE NEURONE COMPARED TO SODIUM IONS DIFFUSE IN

1

111

( ANY 2)

(C) B-DEPOLARIZATIO VS, C -REPOLARISATION

- SODIUM-GATED CHANNELS OPEN VS POTASSIUM-GATED CHANNESL OPEN - SODIUM IONS ENTER THE NEURONE VS POTASSIUM IONS LEAVE THE NEURONE - THE INSIDE OF A NEURONE VS BECOMES POSITIVE VS THE INSIDE OF A

NEURONE BECOMES NEGATIVE

1

111

( ANY 2)

(D) - THE POTASSIUM-GATED CHANNELS REMAIN OPEN / SLOW TO CLOSE

- THIS ALLOWS FURTHER EFFLUX OF POTASSIUM IONS 11

( ANY 1)

(E) - LIMITS THE FREQUENCY IN WHICH IMPULSES MAY FLOW - ENSURES THAT IMPULSES FLOW IN ONE DIRECTION ALONG THE NERVE

11

( ANY 1)

TOTAL 10

Q3 SUGGESTED ANSWERS

M ARK

3. ( A)

DESCRIPTION HUMORAL IMMUNE

RESPONSE

C ELL‐MEDIATED IMMUNE

RESPONSE

ORIGIN OF CELLS BONE MARROW BONE MARROW

PRIMARY CELLS B CELLS/ B

LYMPHOCYTES T CELLS/ T

LYMPHOCYTES

SITE OF CELL MATURATION BONE MARROW THYMUS GLAND

1+1

1+1

1+1

(B) - AN ANTIGEN EXPOSURE TRIGGERS TO DIVIDE - TO PRODUCE MEMORY B CELLS(M-CELLS) AND PLASMA CELLS.

11

(C) - INTERLEUKIN-1 1

(D) - INDUCES HELPER T CELLS TO SECRETE INTERLEUKINE-2 1

TOTAL 10


4. ( A) CURVE S – ABSOLUTE GROWTH CURVE CURVE T – ABSOLUTE GROWTH RATE CURVE

11

(B) S – ACTUAL INCREASE IN SIZE OVER A PERIOD OF TIME

T – THE INCREASE IN SIZE PERUNIT TIME 11

(C) ANNUAL PLANTS 1

(D) I) - LIMITED GROWTH (OCCURS IN ANNUAL PLANTS)- AS THEY COMPLETE THEIR LIFE CYCLE IN A YEAR - THE GROWTH CURVE IS A SIGMOID CURVE

111

( ANY 2)

II) - WOODY PLANTS ARE PERENNIAL PLANTS - WHICH DISPLAY UNLIMITED GROWTH - THE GROWTH CURVE IS A SERIES OF SIGMOID

111http://edu.joshuatly.com/

http://fb.me/edu.joshuatly




3

( ANY 2)

TOTAL 10





4


Section B (Essay Questions)


5. ( A) SINGER’S MODEL OF MEMBRANE

- LABELS: PHOSPHOLIPID BILAYER, CHANNEL PROTEINS, CARRIER PROTEINS

CHOLESTEROL - DRAWING IN ONE PLANE, , BUT NOT IN 3-DIMENSIONAL

1( ANY 1)

1

M AX = 2

(B) - THE BIOLOGICAL CELL MEMBRANE ACTS AS BARRIER AND ARE SELECTIVELY

PERMEABLE - THE MEMBRANE CONSISTS OF A FLUID BILAYER OF PHOSPHOLIPIDS AND

VARIOUS PROTEIN MOLECULES EMBEDDED IN IT. SOME OF THIS PROTEIN

MOLECULES ACT AS ION CHANNELS, CARRIER PROTEIN OR PUMPS

- THE PHOSPHOLIPIDS BILAYER HAS A HYDROPHOBIC MIDDLE REGION MADE

UP OF HYDROPHOBIC FATTY ACIDS TAILS- THE PHOSPHOLIPIDS BILAYER IS PERMEABLE TO VERY SMALL UNCHARGED

MOLECULES LIKE OXYGEN, AND CARBON DIOXIDE, STEROID BASED

HORMONE, FATTY ACIDS AND ALCOHOL(SIMPLE DIFFUSION) - SIMPLE DIFFUSION OF WATER MOLECULES ACROSS THE SEMIPERMEABLE

CELL MEMBRANE IS CALLED OSMOSIS. - SOME INTEGRAL MEMBRANE PROTEIN FORM HYDROPHILIC ION CHANNELS.

THIS ALLOWS DIFFUSION OF VARIOUS CHARGED IONS E.G. K+, N A+, C A+, AND HCO3-, 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

- OTHER LARGE SIZED HYDROPHILIC MOLECULES SUCH AS GLUCOSE ARETRANSPORTED ACROSS THE CELL MEMBRANE THROUGH FACILITATED

DIFFUSION USING A PROTEIN CARRIER MOLECULES

- IN FACILITATED DIFFUSION, THE BINDING OF SUBSTANCES TO THE SPECIFIC

PROTEIN CARRIER CAUSES THE CARRIER TO CHANGES ITS SHAPE AND THE

SUBSTANCE IS RELEASED INTO THE CELL

- TRANSPORT PROTEIN ON THE CELL MEMBRANE CAN ALSO TRANSPORT

SUBSTANCE ACROSS THE CELL MEMBRANE AGAINST THE CONCENTRATION

GRADIENT THROUGH ACTIVE TRANSPORT. - IN ACTIVE TRANSPORT, THE SHAPE OF PROTEIN CARRIER CHANGES USING

ENERGY (ATP) - EXOCYTOSIS AND ENDOCYTOSIS ARE ACTIVE TRANSPORT PROCESSES THAT

MOVE MATERIAL IN BULK ACROSS THE CELL MEMBRANE

- EXCOCYTOSIS INVOLVE THE TRANSPORTATION OF SUBSTANCES OUT OF THE

CELL IN BULK THROUGH THE FUSSION OF VESICLE MEMBRANE WITH THE

CELL MEMBRANE

- IN ENDOCYTOSIS THE BULK SUBSTANCES IS 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

- IN RECEPTOR-MEDIATED ENDOCYTOSIS, LIGAND (CHOLESTEROL

MOLECULES) BIND TO SPECIFIC RECEPTORS IN COATED PITS ON CELL

MEMBRANE. - ALL THESE STRUCTURES AND ITS RELATED PROCESS ENABLE THE CELL

MEMBRANE TO FUNCTION AS SEMIPERMEABLE MEMBRANE AS WELL AS

ENABLE THE CELL MEMBRANE TO REGULATE THE MOVEMENT OF

SUBSTANCES IN AND OUT OF THE CELL

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1( ANY 13)

M AX = 13http://edu.joshuatly.com/ http://fb.me/edu.joshuatly




5

TOTAL = 15

Q6 SUGGESTED ANSWERS M ARK 6. ( A)

ANAEROBIOSIS IN PLANTS ANAEROBIOSIS IN ANIMALS

OCCURS IN ROOTS / SEED / STORAGE

ORGAN/TUBER OCCURS IN SKELETAL MUSCLES

OCCURS WHEN ROOT ARE

SUBMERGED //OXYGEN SUPPLY

DECREASES

OCCURS WHEN MUSCLES ARE

ACTIVELY CONTRACTING

PRODUCTS ARE ETHANOL AND

CARBON DIOXIDE PRODUCT IS LACTIC ACID

PYRUVATE CHANGED / REDUCED TO

ETHANOL FIRST THEN TO ITS FINAL

PRODUCT, ETHANOL

PYRUVATE DIRECTLY CHANGED / REDUCED TO ITS FINAL

PRODUCT,LACTIC ACID ETHANOL IS POISONOUS / CAN KILL

THE PLANT CELLS L ACTIC ACID IS STILL USEFUL / NOT

POISONOUS

ETHANOL CANNOT BE CONVERTED

BACK TO GLUCOSE L ACTIC ACID CAN BE CONVERTED

BACK TO GLUCOSE

ANAEROBIC RESPIRATION IS NOT

NEEDED TO SUSTAIN THE SUPPLY OF

ATP

ANAEROBIC RESPIRATION IS USEFUL

TO SUSTAIN THE SUPPLY OF ATP

1/0

1/0

1/0

1/0

1/0

1/0

1/0

( ANY 6)

M AX = 6

(B) ROLES OF NAD AND NADP IN A MESOPHYLL PALISADE CELL:

- NAD INVOLVED IN CELLULAR RESPIRATION - 2 MOLECULES OF REDUCED NAD / NADH PRODUCED IN GLYCOLYSIS - 1 / 2 MOLECULES OF NADH PRODUCED IN LINK REACTION - 3 / 6 MOLECULES OF NADH PRODUCED IN KREB CYCLE - NAD ACCEPTS HYDROGEN / ELECTRON AND TRANSFERS / CARRY TO THE

INNER MITOCHONDRIAL - MEMBRANES / CRISTAE / CYTOCHROMES / ETC / MATRIX - NADP INVOLVED IN PHOTOSYNTHESIS - VII NADP ACCEPT ELECTRON AND HYDROGEN ION FORMING NADPH IN

NON-CYCLIC - PHOTOPHOSPHORYLATION / LIGHT DEPENDENT REACTION / LIGHT

REACTION

- NADPH USED IN C ALVIN CYCLE / DARK REACTION / LIGHT INDEPENDENTREACTION

- NADPH IS USED TO REDUCE CO2 / IN REDUCTION PHASE - NADPH REDUCES GLYCERATE -1,3-DIPHOSPHATE INTO 1

GLYCERALDEHYDES- 3-PHOSPHATE / PGAL

11111

111

11

11

( ANY 9)

M AX = 9

TOTAL = 15





6

Q7 SUGGESTED ANSWERS M ARK 7. ( A) - THESE TYPES OF PLANTS ARE SHORT DAY PLANTS

- THE PLANTS REQUIRE A PERIOD OF DARKNESS EQUAL TO / LONGER THAN

CRITICAL NIGHT LENGTH - LIGHT / SPOTLIGHT FLASHES FROM VEHICLES INTERRUPT / SHORTEN THE

PERIOD OF DARKNESS - CAUSES THE CONVERSION OF PR INTO PFR - PFR IS THE (BIOLOGICAL) ACTIVE FORM - HIGH CONCENTRATION / LEVEL OF PFR INHIBIT FLOWERING IN SDP’S

11

1

111

( ANY 5)

M AX = 5

(B) - ESTROGEN LEVEL HIGH IN THE END / LAST WEEK OF PREGNANCY

- STIMULATE THE DEVELOPMENT OF OXYTOCIN RECEPTOR IN UTERINE WALL

// INCREASE THE SENSITIVITY OF UTERUS MUSCLE/MYOMETRIUM TO

OXYTOCIN - BABY’S HEAD / GROWING FETUS PRESS AGAINST THE MOTHER CERVIX - CAUSING THE CERVIX AND UTERINE WALL TO STRETCH - THIS STIMULATES MATERNAL POSTERIOR PITUITARY TO RELEASE OXYTOCIN - AND ALSO STIMULATES FETAL PITUITARY GLAND SECRETES ACTH- ACTH TRIGGERS FETAL ADRENAL GLAND TO SECRETE CORTICOSTEROIDS

HORMONE - CORTICOSTEROIDS HORMONE TRIGGERS PLACENTA TO SECRETE

PROSTAGLANDIN - OXYTOCIN ALSO STIMULATES PLACENTA / FETAL TO PRODUCE

PROSTAGLANDIN - BOTH PROSTAGLANDIN AND OXYTOCIN STIMULATES A POWERFUL

CONTRACTION OF MYOMETRIUM / UTERUS WALL - THE MORE FREQUENT THE MYOMETRIUM CONTRACTS, THE MORE

OXYTOCIN/ PROSTAGLANDIN PRODUCED - THE UTERUS CONTRACTION ALSO STIMULATES STRETCH RECEPTORS IN

THE UTERINE WALL AND THE CERVIX TO TRIGGER FURTHER RELEASE OF

OXYTOCIN - THE CERVIX DILATES / WIDEN, ITS TISSUE SOFTENS AND BECOME MORE

FLEXIBLE AND AMNIOTIC BAG BURST - CONTRACTIONS BECOME STRONGER AND STRONGER / MORE FREQUENT

HELP TO PUSH BABY DOWN/OUT

1

1//1

11111111

1

1

1

1

( ANY 10)

M AX = 10

TOTAL= 15





7


8. ( A) OVIPARITY:

- WHEN THE EGGS ARE LAID AND HATCHED OUTSIDE THE FEMALE BODY

- EXAMPLE: REPTILES / BIRDS/ FROGS / MOST FISH / INSECTS I SOME

SPECIES OF SHARKS OVOVIVIPARITY:

- THE EGG IS FERTILIZED AND STORED I RETAINED WITHIN THE OVIDUCT

- THE EMBRYO IS NOURISHED BY THE YOLK IN THE EGG II OBTAINS ITS

NUTRIENTS FROM THE YOLK - THE YOUNG ARE BORN AFTER BEING HATCHED WITHIN THE MOTHER'S BODY - EXAMPLE: SOME SPECIES OF SHARKS

VIVIPARITY:

- THE EGG IS FERTILISED WITHIN THE FEMALE BODY AND THE EMBRYO

FORMED IS NOURISHED BY NUTRIENTS OBTAINED FROM THE MOTHER'SBODY / THROUGH THE PLACENTA. THE BABY IS BORN ALIVE

- EXAMPLE: ALL MAMMALS AND SOME SPECIES OF SHARKS.

11

1

1111

1

1( ANY 7)

M AX = 7

(B) M ARCHANTIA SP.(BRYOPHYTE) VS DRYOPTERIS SP.(FILICINOPHYTE)

SIMILARITIES :- EUKARYOTIC - AUTOTROPHIC/PHOTOSYNTHETIC - NO LOCOMOTION - CELLULOSE CELL WALL - SHOW ALTERNATION OF GENERATION: HAPLOID GAMETOPHYTE AND

DIPLOID SPOROPHYTE - HOMOSPOROUS

DIFFERENCES :

M ARCHANTIA SP.(BRYOPHYTE) DRYOPTERIS SP.(FILICINOPHYTE) ALTERNATION OF GENERATIONS IN

WHICH THE GAMETOPHYTE

GENERATION IS DOMINANT

ALTERNATION OF GENERATIONS IN

WHICH THE SPOROPHYTE IS

DOMINANT

SPOROPHYTE IS ATTACHED TO, AND

IS DEPENDENT UPON, THE

GAMETOPHYTE FOR ITS NUTRITION

G AMETOPHYTE IS REDUCED TO A

SMALL, SIMPLE PROTHALLUS

NO VASCULAR TISSUE, THAT IS NO

XYLEM OR PHLOEM V ASCULAR TISSUE PRESENT (XYLEM

AND PHLOEM) IN SPOROPHYTE:

BODY IS A THALLUS, OR

DIFFERENTIATED INTO SIMPLE

‘LEAVES’ AND ‘STEMS’

NO TRUE ROOTS, STEMS OR LEAVES: THE GAMETOPHYTE IS ANCHORED BY

FILAMENTOUS RHIZOIDS

LEAVES RELATIVELY LARGE AND

CALLED FRONDS

SPOROPHYTE IS THEREFORE HAS

TRUE ROOTS, STEMS AND LEAVES

SPORES ARE PRODUCED BY THE SPORES PRODUCED IN SPORANGIA

11111

1

1/0

1/0

1/0

1/0

1/0

1/0http://edu.joshuatly.com/ http://fb.me/edu.joshuatly




8

SPOROPHYTE IN A SPORE CAPSULE

ON THE END OF A SLENDER STALK

ABOVE THE GAMETOPHYTE

WHICH ARE USUALLY IN CLUSTERS

CALLED SORI ( ANY 8)

M AX = 8


9. ( A) - CLEAVAGE IS A SUCCESSION OF RAPID MITOTIC DIVISIONS THAT FOLLOW

FERTILIZATION.- THE CYTOPLASM OF THE ZYGOTE IS PARTITIONED INTO BLASTOMERAS

WITHOUT GROWTH.- 4 DAYS AFTER FERTILIZATION, THE ZYGOTE DEVELOPS INTO SOLID MASS OF

CELLS CALLED A MORULA.- THE MORULA CONTINUES TO DIVIDED TO FORM A HOLLOW FLUID – FILLED

BALL OF 100 CELL CALLED A BLASTULA.- 7 DAYS AFTER FERTILIZATION, THE BLASTULAR BECOMES A BLASTOCYST

AND IMPLANTS INTO THE UTERINE LINING.- 9 DAYS AFTER FERTILIZATION, A BLASTOCYST MIGRATE INWARDS/ MOVE TO

- INTERIOR LOCATION / INVAGINATE.- G ASTRULA HAS THREE EMBRYONIC GERM LAYERS- ECTODERM,MESODERM

AND ENDODERM.- THE GERM LAYERS DEVELOP INTO REDIMENTS OF ORGAN DURING

ORGANOGENESIS.

1

1

1

1

1

111

1

( ANY 8)

M AX = 8(B) - ALLOWS THE EXCHANGE OF SUBSTANCES SUCH AS OXYGEN, CARBON

DIOXIDE,NUTRIENTS, HORMONES AND UREA.- DOES NOT ALLOW BLOOD FROM THE MOTHER AND FETUS TO MIX. - THIS PREVENTS INCOMPATIBLE BLOOD FROM MIXING AND CLOTTING /

AGGLUTINATING.- ACTS AS A PARTIAL BARRIER TO THE PASSAGE OF HARMFUL PATHOGENS/

MICROORGANISMS /HARMFUL SUBSTANCES.- HOWEVER, VIRUSES(HIV),NICOTINE, ALCOHOL, CERTAIN DRUGS CAN

CROSS THE PLACENTA. - PERMITS THE MATERNAL AND FETAL BLOOD SYSTEMS TO FUNCTION AT

DIFFERING PRESSURES WITH HARMING EITHER THE MOTHER OR THE FETUS.

- ACTS AS A TEMPORARY ENDORINE ORGAN, PRODUCING HCG, ESTROGEN

AND PROGESTERONE. THIS PREVENTS OVULATION AND MENSTRUATION.- PERMITS CERTAIN MATERNAL ANTIBODIES TO BE TRANSFERRED TO THE

FETUS, PROVIDING FETUS WITH IMMUNITY TO CERTAIN DISEASES.

1

11

111

1

1

1

( ANY 7)

M AX = 7

TOTAL= 15





9


10. A) (I) SPECIATION :- A PROCESS OF FORMATION OF ONE OR MORE SPECIES

- FROM PREEXISTING SPECIES / AS A RESULT OF A RESULT OF ISOLATION, NATURAL SELECTION, GENETIC DRIFT OR HYBRIDIZATION.

(II) GENE POOL:- THE GENE POOL IS THE TOTAL OF ALL THE GENES AND ALLELES WHICH ARE

PRESENT IN A SEXUALLY REPRODUCING POPULATION.

11

1

M AX = 3

I) GEOGRAPHICAL ISOLATION:- EXAMPLE GEOGRAPHICAL ISOLATION / EXTRINSIC ISOLATION ARE

- GEOGRAPHICAL BARRIERS LIKE DESERT, MOUNTAINS, OCEANS AND RIVER.

- THESE BARRIERS SEPARATE THE DEMES AND PREVENT THEM FROM

MEETING AND BREEDING WITH EACH OTHER.

- THIS MODE OF SPECIATION IS CALLED ALLOPATRIC SPECIATION / POPULATION SEPRATION BY GEOGRAPHICAL BARRIER.

- (IF ISOLATION IS COMPLETE ) THERE IS NO LONGER A GENE FLOW BETWEEN

THE DEMES.- MUTATION, SELECTION PRESSURES, GENETIC DRIFT AND ADAPTIVE

DISPERSION RESULT IN THE PRODUCTION OF DIFFERENT GENOTYPES FOR

EACH DEMES.- THE 2 ISOLATED DEMES ARE NOT ABLE TO INTERBREED

111

1

1

1

1( ANY 4)

M AX = 4

II) PRE-ZYGOTIC

- PREVENT MATING / FERTILIZATION BETWEEN SPECIES.- ECOLOGICAL ISOLATION / HABITAT ISOLATION-SPECIES LIVE IN DIFFERENT

HABITATS.- TEMPORAL ISOLATION- SPECIES HAVE DIFFERENT BREEDING SEASON TIME.

- BEHAVIOURAL ISOLATION – SPECIES HAVE DIFFERENT MATING 9EHAVIOR

- MECHANICAL ISOLATION-MORPHOLOGICAL OR STRUCTURAL DIFFERENCES

IN THE REPRODUCTIVE ORGAN / PHYSIOLOGICAL DIFFERENCES PREVENT

NORMAL MATING FOR ANIMAL OR POLLINATION FOR PLANTS.- GAMETIC ISOLATION – FEMALE AND MALE GAMETES NOT COMPATIBLE /

FERTILIZATION DOES NOT OCCUR.

POST-ZYGOTIC

- B ARRIERS OPERATE AFTER HYBRID ZYGOTES ARE FORMED.- HYBRID INVIABILITY – GENE OF TWO PARENT SPECIES ARE NOT

COMPLITABLE AND THE HYBRIDS DO NOT SURVIVE.- HYBRID STERILITY – HYBRIDS ARE STERILE

11

11

1/1

1/1

111

1

1http://edu.joshuatly.com/ http://fb.me/edu.joshuatly




- (THERE ARE UNABLE TO BRING ABOUT GENE FLOW BETWEEN THE PARENT

SPECIES.)- HYBRID BREAKDOWN-HYBRIDS ARE FERTILE BUT THE OFFSPRING ARE

INFERTILE

( ANY 8)

M AX= 8

TOTAL= 15

gerak gempur perak stpm 2012 biology

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