STPM Trial 2009 Bio2 Q&A (Johor)

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CONFIDENTIAL*/SULIT*

964/2

BIOLOGY (BIOLOGI)

PAPER 2 (KERTAS 2)

STRUCTURE AND ESSAY (SRTUKTUR DAN ESEI)

Two and a half hours (Dua jam setengah)

JABATAN PENDIDlKAN NEGERI JOHOR

PEPERIKSAAN ·PERCUBAAN STPM

2009

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 answerson the answer sheets provided. Begin each answer on a fresh sheet ofpaper.Answers should be illustrated by large, clearly labelled diagrams whereversuitable.

Answers may be written in either English or Malay.

Arrange your answers in numerical order and tie the answer sheets to this

booklet.

Arahan kepada calon:

For examiner's use(Untuk kegunaan

Pemeriksa)

2

3

4

56

7

Jawab semua soalan dalam Bahagian A dalam ruang yang disediakan. 8Jawab mana-mana empat soalan daripada Bahagian B. Untuk bahagian ini, tulis 9jawapan anda pada helaian jawapan yang dibekalkan. Mulakan setiap jawapan f-----::1-= -O- -+- - - - l

pada helaian kertas yang baru. Jawapan hendaklah disertai gam bar rajah yang Totalbesar dan mempunyai label yang jelas di mana-mana yang sesuai.Jawapan boleh ditulis dalam bahasa Inggeris atau bahas Melayu. (Jumlah)

Susunjawapan anda mengikut tertib berangka dan ikat helaianjawapan bersamadenan buku soalan ini.

964/2

This question paper consists of 10 printed pages.

(Kertas soalan ini terdiri daripada 10 halaman bercetak)

*This question paper is CONFIDENTIAL until the examination is over.*Kertas soalan ini SULIT sehingga peperiksaan kertas ini tam at.

[Turn over (lihat sebelah)CONFIDENTIAL*

SULIT*

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2

Section A [40 marks]

Answerall questions in this section.

1. Diagrams (i) and (ii) show the stages of the life cycle of a green plant. The pictures in

diagram (i) are not in the correct sequence.

c

A

8

...

D E F

Diagram (i)

Asexual generation

Spore

J

ZygoteSexual generation

r---G-a-m-e-te---'I

Diagram (ii)

(a) The life cycle of this plant shows alternation of generations. Define the termalternation ofgenerations

............................... ................................................................ .. .................................. .................. .

[2 marks]

(b) Which of the pictures labelled A - F in diagram (i) is known the young sporophyte?

• • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • ••••••• 1 •• . A ••••••••••••••••••••••••••••••••••••• • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • •• • • • •

[1 mark]

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3

(c) Name the male and female r e p r o d u c ~ i v e structures of the gametophyte

•••• •••••• 1 ....... .. . . . . . . .. . .. . . . . . . . . .. . ............ . . . . . . . . . . . . . . . . .. . , •••••••••• • ••••••••••••••• •• •• •• • ••••••• ••• • • ••••••••••••••••••••••••• •• • •

••••• •• •••••••••••.•••••••••• "0 ......... . . ", •••••••••••• " • ..• •••••• 6, ___ •••• •• ••••••• • •• ••••• • •• ••• •••••••• • ••••••••••••••••••••••••• •• • ••••••• •• ••••

........ ........ .. .. .. .. .... .. ......... .... .......... ... .. ........ ..... .... .. ...... .................... .. ..... .......................................[2 marks]

(d) Mark on Diagram (ii) the positions where mitosis and meiosis occur.

[2 marks]

(e) State three characteristics of the plant that are considered more advanced compared

to bryophytes .

..... ............................ .. ............ .... ............... .. ........ . - ..... .. .... . ......... ............... "...... ................................ .

........ ............... ................................... .... .......................... ...... ..... ............................ ........ ..... ...... .. .

........................ " .. ........ .. ......... .. ............................ ...................................................................... .

[3 marks]

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4

2. Cystic fibrosis is a recessive condition that affects about 1 in 2,500 babies in the

Caucasian population of the United States.

(i) Calculate the frequency of the recessive allele in the population.

[2 marks]

(ii) Calculate the frequency of the dominant allele in the population.

[1 mark](iii) Determine the percentage of heterozygous individuals (carriers) in the population.

[1 mark]

(b) In a population of butterflies brown body (8) is dominant to white (b). 40% of butterflies

in the population are white . Using the information given, calculate the following:

(i) The percentage of butterflies in the population that are heterozygous.

[3 mark]

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5

(ii) The frequency of homozygous dominant individuals.

[1 mark](c) Give two conditions for a population to be in Hardy-Weinberg equilibrium.

[2 marks]

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6

3. In lactose metabolism, f3-galactosidase enzyme catalyses the hydrolysis of lactose. In an

E. coli bacterial cell, this enzyme would not be produced if lactose is absent. The

diagram below shows the mechanism involved in the regulation of lactose metabolism

when lactose is absent.

Structural OperatorRegulator

genes gene gene

1Repressor

molecule

(a) The diagram above illustrates the mechanism of an operon concept

(i) Name the operon concept shown.

[1 mark]

(ii) What type of operon is it?

... ............................ ........... ............................... ....... .................. ...................... .........................

I1 mark](iii) Explain your answer to (a) (ii).

................................ ........... ............... ......... ................. .. ........ .. ....- ....... ............ ...........................

[1 mark]

(b) (i) Name the gene that codes for the f 3 - g a l ~ c t o s i d a s e enzyme.

[1 mark]

(ii) What products are formed when f 3 - g a l a c t o s i d ~ s e acts on lactose?

[1 mark]

(c) State the function of the regulator gene in this operon.

[1 mark]

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7

(d) Explain briefly how the absence of lactose prevents the production of the 13-

galactosidase enzyme .

.... .............................. ............................ ..... ........ ..........................................................................

...... ............................. ................................................ .... .... ..... .... ..... ... ... . ...... .. ..... ............. ..... ... .

...................... ..... ..... .... .. .......... .... ..... .... ... .. .......................... ................. ......... .. ... .... .........................

. . . ........ ... .. ,......... .. ......... ... ..... .. ... ..... .............. .. " ...... .' ........ ..............[2 marks]

(e) If a mutation a l t ~ r $ the' promoter in this operon, suggest what would happen to theproduction of the enzyme ~ - g a l a c t o s i d a s e enzyme.

............... ... .................................. ... .............. .. ....... .... ......._............ .. ...... .. .. ...... .. .... .........................•••••••••••••••••••• , ., •••••••• •• 0, ' 00 0 ••• • •••••••• • •• • ·••••• 0 ••••••••• ••••• •• •••••••••••••••• ••••••• • ••••••••••••••••••••••••••••••••••••••• •••• •

[2 marks]

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8

4. Taxonomists organise species into hierarchical systems to show phylogenetic

relationships. They find algae difficult to classify. The photosynthet ic pigments in algae

are used to distinguish among the major groups or phyla . These phyla are the green

algae (Chlorophyta) , the brown algae ( Phaeophyta ) and the red algae ( Rhodophyta ).

The major features of Chlorophyta, Phaeophyta and Rhodophyta are shown in Table 1.

Features Chlorophyta Phaeophyta RhodophytaVegetative Unicellular, Simple or Unicellular,

morphology colonial, thalloid branched, filamentous,

filamentous, thalloid

thalloid

Flagellate stages in I I X

the life cycle

Carbohydrate Starch Laminarin Floridean starch

storage product

Cell wall. Cellulose Cellulose Cellulose

polysaccharides Alginic acid Polysulphate

Fucinic acid esters

Photosynthetic I I Ipigment:

Chlorophyll a

Chlorophyll b I X X

B carotene I I ILutein I X IFucoxanthin X / X

R- phycoerythrin X X IR- phycocyanin X X /

Table 1

(a) With reference to Table 1:

i. Name two features that all three phyla share with the plant kingdom .

• ••••••••••• ••••••••••••••••••••••••••••••••• t ••• •• • ••••• • •••••• • •••••••• • ••••••••••••••••••• to .......................... •••••••••••• ••••••••••••• •

•••••••• ...................................... o ........................................................... ••••••••••••••••••••••••••••••••••••••••••

[2 marks]

ii. List two characteristics of brown alaae which are not found in the other two phyla

......... . .......................................................... .................................................................................•••••••••••••••••••••••• • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • ••• •• ••••• •••••••••••• •••••••••••• •••••• _ • • • • • • • • • • • • • • • • • • •• • •••••••••••••••••••••••••••••

........................................................ .... ................................................................-....................... .

[2 marks]

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9

iii. Suggest two reasons why taxonomists think that the three phyla of algae may notbe closely related.

.... ... , .............................. 0. 0 . . . .. . .. . .. . .... . ••• • ••• ••••••• • •••••••• ••••• • •• ••• • ••• 0 •• ••• • 0 • • • • 0 • ••• •• •••••••• • •••••• • ••••••• ••••••••

•••••• • • • •••••••••••••••• •• •••• •• " " ' " • •••••• •• • 0 •••• 0 .0 .. . .. . 0 •• 0 •••• • • ••• ••• •• • • •• • •• •••• ••••••• ,. , •• •• , •• • 0 ....... ..........................

................. .... ..... .. .. .. ....... ... ............. .. .. ..... .... ... ... .. ...... .... ..... ... ... ... .... ... .. .. ................................. ..

[2 marks](b) State one function of flagella in unicellular algae.

. ...... ,......... ..... .... ... ... ............ ........ ..... .... ....... .... ......... .... ......... ... .... ............... .. ................... .

[1 mark]i. Define the term taxon.

.... .................. ... ............... .. ......................... ............... .. .. ........................ ............ ... ..... .... ... ...,.......... .. .

[1 mark]ii. State one major function of a natural classification.

................................................................................ ............_...... ... .............. ................ .. ...... ... ...

[1 mark]iii. State one major function of an artificial classification .

•• ••• •••• • •• •••••••• • •• • • •••••••• •••••• •• •••• •• • •••• • • • • ••••• •••• • ••••• ••• •• ••• ••• • •• ••••• •• • • " •• • ••••••••• ••••• ••••••••• •••• •• ow ••••••• ••••• •• • •••

[1 mark]

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10

Section B [60 marks]

Answer any four questions in this section.

5. a) Polysaccharides play an important role as structural'and storage compounds in plants.By giving one example for each of these compounds, explain how their molecularstructures are related to their functions.

[8 marks]

6. C h e m o a u t o t r o p h and photoautotroph are two groups of autotrophic organisms.

7. (a) Explain what is meant by the Bohr effect. [5 marks]

(b) Explain the regulation of breathing in humans. [10 marks]

. \

8. (a) Describe how an action potential is transmitted along a non-myelinated neurone.

[7 marks]

(b) Explain how a nerve impulse is transmitted across a synapse.

[8 marks}

9. (a) Differentiate between gene mutation and chromosomal mutation.

[4 marks]

(b) Explain the different types of gene mutation, giving specific examples where

relevant.

[8 marks

(c) Down's Syndrome is caused by chromosomal mutation. Explain how this may occur

. [4 marks]10. (a) In recombinant DNA technology, a desired gene is obtained and inserted into a host

cell to be cloned.

(i) Describe the ways of obtaining a desired gene.

. [7 marks]

(ii) Explain how the desired gene can be inserted into a host cell.

[5 marks]

(b) List three applications of recombinant DNA technology in the medical field.

[3 marks]

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No

1(a)

(b)

(c)

(d)

(e)

2(a)

(i)

. ii)

(iii)

(b)(i)

(ii)

(iii)

BIOLOGY SCHEME

TRIAL STPM 2009

PAPER 2

Answer

The alternation between a haploid gametophyte generation that produceshaploid gametes and a diploid sporophyte generation that produces diploid

sporeswhere one of the generations is a dominant generation.

F

male : antheridumFemale: archegonium

Zygote

,-------S e x u a l organism

. Gamete

Body is differentiated into stem, leaves and fibrous roots.Vascular tissue consists of tracheids and sieve tubes.Dominant sporophyte.Free gametophyte. ( Any 3)

Marks

. 111

1

11

1

1

1

1

1

Total ·1

q2 = 1/2,500 or 0.0004, q =0.02.

The frequency of the cystic fibrosis (recessive) allele in thepopulation is 0.02 (or 2%). 5'0The frequency of the dominant (normal) allele in the population (p)is simply 1 - 0.02 =0.98 (or 98%).Since 2pq equals the frequency of heterozygotes or carriers, thenthe equation will be as follows: 2pq = (2)(.98)(.02) = 0.04 or 1 in 25are carriers.

bb = q2 = 0.4, q = 0.63,

Since p + q = 1( then p must be 1 - 0.63 = 0.37,2pq = 2 (0.37HO.63) = 0.47.

p2 or (0.37)2 = 0.14.

No mutations

11

1

1

11

1

1

1

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3

a(i)

a(ii)

a(iii)

No MigrationRandom mating must occurLarge population -No selection

- lac operon

- inducible operon

- it is stimulated to be switched on when lactose is present

(Any Two)Total

1

1

1

11

1

1

10

b(i) - lac z 1

b(ii) - glucose and galactose 1

c - regular gene codes for repressor molecule/protein

d * - the active repressor molecule binds to the operator gene and blocks the 1

attachment of RNA polymerase to the promoter- this prevents the transcription of genes of lac Z, lac Y and lac A so no 1

mRNA can be made

e - the inactivated repressor molecule can no longer bind to the operator 1

4 (a)(i)

(ii)

(iii)

(b)

(c) (i)

(ii)

(iii)

gene

- the operon remains switched on and ~ - g a l a c t o s i d a s e would be 1continuously produced

- The presence of chlorophyll a- The presence of cellulose

- The presence of fucoxanthin- The presence of alginic acid

- Polysaccharides in the cell wall are different- The storage compounds are different.

- Difference in pigment

1

1

1

(Any Two) 2

- Movement/motility

- A taxon is a group that contains organisms that share some basic

features that indicate they share a common ancestry.

- Natural classification reflects the evolutionary or phylogenetic

1

1

relationship based on homologous characteristics. 1

- In artificial classification, the analogous characters are used

for utilitarian / medical/economical purpose 1Total 10

3 d) Note that:

- promoter increases the rate of attachment for RNA polymerase to start transcription

of the structural gene.

- galactosidase is prevented from being produced if the operator gene is blocked by

the repressor molecule.

- In the absence of lactose, the operon is switched off, meaning repressor molecule

binds to the operator gene and prevents gene transcription

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5 a. important role as structural and storage.. cellulose-structural compound- made of long chain of f3-glucose- unbranched chain run parallel to each other- have cross-linkage that gives stability and strength- insoluble- fibers laid in layers in different directions adding further strength

starch-storaQe compoundmixture of amylase and amylopectinamylase - unbranched chain of a-glucose forms helix structureamylopectin - branched chains of a-glucosecompound stabilized by countless hydrogen bondcompact and insolublereadily hydrolysed to form sugar when required

b i.' the esterification process involves condensation reaction

[max 4]

[max 4]

- between one molecule of glycerol and three molecules of fatty acids 1

- three ester bonds are formed to produce a molecule of triglycerideand three molecules of water

g J c e - t - o l

DiAGRAM-2m()

II( C E 2 O : ~ H ( j ) - C - : C : E - · ' . - C " " ~ i I .--- () . -.... --.

IIi H'O:B:-j-.-- - " . , g J - . ~ - c c ;:-" ; ,,- -(:Eo

C = - : : o V _ . J . - ~ ~ P - ~ ( C E : ) : : - - C ( - . : r : : r

-'

1o

C . - 0 - , ~ _ ; C H : : ' - - - C H ' o

"O - ~ - : ' C E " - - C :0 ,

C 2 : : ; o - c - : ~ c t - : : : : C : -

- 3F :0

ii. importance of lecithin in cell membrane structure

fatty a ci d

2

Structure of lecithin - Lecithin is a type of phospholipids molecule 1

consisting of a hydrophilic head and two hydrophobic tailsStructure of cell membrane - The cell membrane is made up of two 1

phospholipids layers with the hydrophilic head on the outside of the

bilayer

Importance of cell membrane - The lecithin bilayer forms a boundary 1

separating the cell contents from the external environment

Importance of cell membrane - Being hydrophobic, it is selectively 1

permeable and regulates the movement of substances across the

membrane

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6(a)

6(b)

Chemoautotroph

Done by bacteria

.P h o t o a u t o t r ~ p h Done by green plants or organisms

which has the chlorophyll pigment 1

Synthesise organic

compounds from carbon

dioxide and water

Synthesise organic compounds from\ 6 3 . " ~ S\l-J:),fl-'i ~ o r 9 ~ ( ? . 9 u n d s such as carbon

dioxide, water

1

Energy - from oxidation of

inorganic substances such

as H2S, ammonia and iron

Energy supply - from the (sun)light 1

\.1 ( J \ ( ) J J " ~

f v ~ --

Saprophytic organisms can be defined as: organisms that obtain

their nutritional needs from dead and decaying organic materials 1Cannot synthesise their own foodSecrete enzymes such as amylase, proteases, lipase which digest 1their food extracellulary 1Absorb the digested products through the cell surfaces 1Give example: Mucor, Rhizopus, mushroom 1Ecologically important because they act as a decomposer 1Break down the dead organism and waste product 1The decomposed material which contains chemical elements can 1be reused (absorbed) by the saprophytes and other autotrophs.

Obligate parasite

Unable to live independently

without the presence of a host for

supply of nutrient

Unable to reproduce independently

e.g. Tapeworm (Taenia solium)

Always exist as an obligate

parasite

(Any 2) Max 2 mark

Facultative parasite

Able to live independently

without the presence of,a host

for supply of nutrient

Able to reproduce

independently

e.g. bootlace fungus

(Armillaria mellea)

When under stressful

condition; it can be an

obligate parasite.

(Any 2) Max 2 marks

1

1

1

1

3

8

Total 415

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7(a) - dissociation of carbonic acid in the erythrocytes causes an increase in 1the concentration of hydrogen ions resulting in reduction of the pH

- this results in the oxyhaemoglobin dissociating to release haemoglobin 1which combines with the excess hydrogen ions to form haemoglobinicacid ( HHb ), as a buffering effect

- increasing the carbon dioxide concentration increases the rate of 1oxyhaemoglobin dissociation

- thus increasing the carbon dioxide concentration reduces the affinityof

1haemoglobin towards oxygen, a process called Bohr's effect- Bohr's effect results in a shift of the oxygen dissociation curve of 1

haemoglobin to the right 5

(b) - the breathing cycle is controlled by the breathing centre located in the 1

medulla oblongata- this breathing centre consist of the inspiratory centre and the expiratory 1

centre- the inspiratory centre sends impulses to the outer intercostal muscles 1

and diaphragm bringing about contraction while the inner intercostal

.muscle relaxes

- this resultsin

an increasein

the thoracic cavity volume, bringing about 1inspiration- alveolus and bronchioles expands during inspiration stimulating the 1

stretch receptors within the walls of the alveoli and bronchioles to sendimpulses to the expiratory centre

- the expiratory centre sends inhibitory impulses to the inspiratory centre 1- the inspiratory centre then stops sending impulses to the diaphragm

and outer intercostals muscle causing them to relax. 1- this brings about a decrease in thoracic cavity volume resulting in

expiration 1- when the volume in the alveolus and bronchioles are reduced, the

stretch receptors are no longer stimulated to fire inhibitory impulses to

the expiratory centre 1

- inspiratory centre once again sends impulse to the diaphragm and outerintercostal muscle bringing about contraction and inspiration 1

- the cycle is repeated 10

Total .1§

8 1\0'1-

(a) - when a p1yelinated neurone is sufficiently stimulated, an action potential

is generated. 1- this sets up a local current which de polarizes the adjacent region 1- the influx of sOdium ions from the extracellular fluid into one region of the

axon creates a local circuit in that region 1

- the increase in sodium ions in the axoplasm repels the cations to moveto the adjacent region which is more negatively charged 1

- this increases the membrane potential in the adjacent region and opens

up sodium voltage gated channels 1- sodium ions diffuse into the neurone and the membrane is depolarized 1

- when the threshold level is exceeded, a new action potential is generated 1- the local current at one region, therefore, induces a new action potential

in the adjacent region which keeps moving in a forward direction 1

Max:7

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( b) - when a nerve impulse arrives at a synaptic knob, calcium gated channels 1

in the presynaptic membrane opens

9 (a)

(b)

- Ca2+ ons diffuse quickly from synaptic cleft or extracellular fluid into the 1

synaptic knob- this influx of Ca

2+ causes the synaptic vesicle to fuse with the presynaptic 1

membrane

- vesicles release neurotransmitter molecules into the synaptic cleft by 1exocytosis

- neurotransmitter molecules diffuse across the cleft and bind to the 1

receptors on the postsynaptic membrane 1- this binding triggers the opening of sodium channels- Na+ ions diffuse into the postsynaptic neurone, depolarising the 1

postsynaptic membrane

- a new potential, known as eXCitatory postsynaptic potential (EPSP) is 1

generated- if the EPSP is large enough to reach the threshold level, an action 1

potential is generated and is transmitted along the postsynaptic neurone.

Max:8

- Gene mutation is the change in the sequence of nucleotide bases of the 1DNA that corresponds to a particular gene in an organism

- also known as point mutation. 1- frameshift mutation and missense mutation are different forms of gene 1

mutaiion.- Chromosomal mutation is the change in the structure of the chromosome

also known as chromosomal aberration 1

- or the change in the number of the chromosomes in an organism. 1- aneuploidy and euploidy which consists of allopolyploidy and

autopolyploidy are different forms of chromosomal mutation.

max 4

- The four possible ways that gene mutation can occur are through 1

substitution, inversion, insertion, and deletion.

- in substitution, a nucleotide base pair is replaced by another base pair in 1the DNA nucleotide sequence of the gene.

- and they are usually missence mutations as the new nucleotide base 1

alters one genetic code to a different code which may still code for anamino acid but it is a different amino acid.

- an example of genetic disorder caused by substitution is sickle-cell 1anaemia, where the base thymine in the code for glutamic acid is

substituted by the base adenine in the gene that codes for the (3- 1polypeptide chain.

- in inversion, two or more nucleotide base pairs have been reversed in the 1DNA base sequence within the gene.

- the altered genetic code may result in a different amino acid in the 1polypeptide chain and the formation of a non-functional protein .

- in insertion, an extra nucleotide base pair is inserted into the DNA base 1sequence of a gene causing the whole base sequence to be shifted oneplace backward. (frameshift mutation)

- in deletion, a nucleotide base pair is deleted from the DNA base 1

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sequence of a gene causing the whole base sequence to be shifted oneplace forward .

- both insertion and deletion are frameshift mutation and every single 1triplet code after the insertion or deletion point is altered.

- insertions and deletions are usually more harmful than substitution and 1inversion because of the frameshift mutations which often lead to

production of non-functional proteins.- 13- Talassaemia major is a genetic disorder caused by the deletion of a 1base in the j3-globin allele and this results in a lack of j3-polypeptide chainsof the haemoglobin molecule.

(c) max 8- Down syndrome is an example of aneuploidy that is instead of 46 1chromosomes there are 47 chromosomes in the individual.

- it is a result of non-disjunction during meiosis. 1- the two chromosomes number 21 fail to separate during anaphase I or 1anaphase II of meiosis.

- the gametes produced contain 24 chromosomes (2 copies of 1chromosome 21) and 22 chromosomes (no chromosome 21)

- ~ h e n a sperm containing 23 chromosomes fuses with an ovumcontaining 24 chromosomes and the zygote formed contains three 1chromosome 21, trisomy.

- the individual may be a male or female usually with flat, broad faces, 1slanted eyes, short palms and are mentally retarded .

max 4

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10(a) There are three ways to obtain a desired gene:

(i) (1) Producing the gene from mRNA by using reverse transcriptase

- When a gene is active/expressed, it can produce a few thousand

molecules of mRNA which are complementary to the gene. A probe

is used to identify the required mRNA.

- From the mRNA, a copy of the original gene/DNA can be produced

by using retrovirus. The enzyme involved is reverse transcriptase.

- DNA produced this way is known as complementary DNA or cDNA.

(2) Synthesising the desired gene artificially

- The sequence of bases in a gene can be determined from the

sequence of amino acids in the protein that it codes for.

- Based on that knowledge, a gene can be synthesised by using

nucleotides and joining them in the right order.

(3) Cutting the desired gene from the donor's DNA by using restriction

endonucleases.

- Restriction endonucleases are enzymes produced by bacteria to cut

up the DNA of viruses which attack them . Restriction endonucleases

are used to cut a donor's DNA to obtain the desired gene.

- Restriction enzymes cut DNA at specific base sequences known as

restriction sites. More than 2000 restriction enzymes have been

discovered, each with its specific restriction sites.

- Restriction sites are polindromes. This means the base sequence of

one strand reads the same as its complementary strand in the

1

1

1

1

1

1

1

opposite direction. 1

- Restriction enzymes make staggered cuts, producing single

stranded sticky ends which can be used to join up DNA fragments by

hydrogen bonding.

- By using restriction endonucleases, the DNA of donor organism is cut

into many fragments of various lengths. The fragments are then

separated by means of gel electrophoresis.

- The DNA fragment which contains the desired gene is identified by

using a gene probe. It is called the target DNA.

Total =11 (maximum =7)

- The desired gene is joined to a fragment of DNA known as a vector.

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(a)(ii)

(b)

Two commonly used vectors are bacterial plasmids and

bacteriophage lambda ( " ) DNA.

- Bacterial plasmids are cut by using the same restriction

endonuclease as those used to cut the donor DNA so as to produce

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complementary sticky ends. 1

- The target DNA is joined to the bacterial plasmid or phage" DNA by

means of their sticky ends. The deoxyribose sugars and the

phosphate groups are ligated by using DNA ligase.

- The resulting recombinant DNA molecules are then transferred into 1

host cells, usually E coli bacteria. This is done by adding

recombinant DNA molecules to a culture flask containing E coli.

Calcium ions are added and the flask is warmed. Such a treatment

gives rise to pores in the cell surface membrane of Ecoli, thus 1allowing the recombinant DNA molecules to enter. The process is

called transformation/transduction.

- If bacteriophage" is used as a vector, insertion of recombinant DNA

is done by infecting E coli with the phage.

Total 5

(1) Recombinant DNA technology has been used to make bacteria

produce humulin (human insulin) for use by diabetics.

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(2) Farm animals have been engineered to be "pharmaceutical .1

factories", i.e. made to produce rare human proteins such as 0-1 -

antitrypsin enzyme and human growth hormone for treating diseases

like emphysema and dwarfism.

(3) Diseases such as haemophilia, cystic fibrosis, muscular dystrophy

and cancer are caused by defective genes. Recombinant DNA

technology is used in gene therapy for treating such diseases by

replacing defective genes with normal genes.Total 3

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STPM Trial 2009 Bio2 Q&A (Johor)

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Transcript of STPM Trial 2009 Bio2 Q&A (Johor)