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 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 English or Malay. Arrange your answers in numerical order and tie the answer sheets to this booklet.

Arahan kepada cal on:

For examiner's use (Untuk kegunaan

Pemeriksa)

2

3

4

5

6

7

Jawab semua soalan dalam Bahagian A dalam ruang yang disediakan. 8 Jawab mana-mana em pat soalan daripada Bahagian B. Untuk bahagian ini, tulis 9 jawapan 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 Total besar 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 bersama denan buku soalan ini.

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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]

Answer all 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 Zygote

Sexual generation

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

Diagram (ii)

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

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

[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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(c) Name the male and female reproduc~ive structures of the gametophyte . •••• •••••• • 1 ....... ... ..... , ... ... ....... .. . ............ ..... . ......... . .. . , •••••••••• • ••••••••••••••• 0 •• •• ••• ••••••• ••• • • •••••••••••••••••••••••••

0 ••• •

••••• •• •••••••••••.•••••••••• "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 Operator Regulator

genes gene gene

1 Repressor

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 f3-gal~ctosidase enzyme.

[1 mark]

(ii) What products are formed when f3-galactosid~se 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 alt~r$ the' promoter in this operon, suggest what would happen to the production of the enzyme ~-galactosidase enzyme .

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

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

[2 marks]

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4. Taxonomists organise species into hierarchical systems to show phylogenetic relationships. They find algae difficult to classify. The photosynthetic 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 Rhodophyta Vegetative 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 I pigment: Chlorophyll a

Chlorophyll b I X X B carotene I I I Lutein I X I Fucoxanthin X / X R- phycoerythrin X X I R- phycocyanin X X /

Table 1

(a) With reference to Table 1:

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

• ••••••••••• ••••••••••••••••••••••••••••••••• It ••• , •• • ••••• t • •••••• • •••••••• • ••••••••••••••••••• to .......................... t •••••••••••• t ••••••••••••• •

•••••••• t ...................................... t oo ........................................................... i ••••••••••••••••••••••••••••••••••••••••••

[2 marks]

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

......... '. ........................................................... / ................................................................................ .

•••••••••••••••••••••••• • •••••••••••••••••••••••••••••• -• ••• •• ••••• ! •••••••••••••••••••••••••••••• _ •••••••••••••••••••• • •••••••••••••••••••••••••••••

..................................................... ,... .................................................................... -........................ .

[2 marks]

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iii. Suggest two reasons why taxonomists think that the three phyla of algae may not be closely related .

.... ... !', .............................. 0. 0 . . . .. ~ . ... ... ..... 0 ••• • ••• ••••••• • •••••••• 0 ••••• • •• ••• • ••• 0 •• ••• • 0 • • • • 0 • ••• •• •••••••• • •••••• • ••••••• ; ••••••••

•••••• • • • •••••••••••••••••• ~ •••• A •• '~. ""'" • •••••• 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 .

•• ••• •••• • •• •••••••• • •• • • •••••••• •••••• •• •••• ••• •••• • • • • ••••• ", •••• " • ••••• ••• •• ••• ••• • •• ••••• ••• • " •• • ••••••••• ••••• ••••••••• •••• •• row ••••••• ••••• •• • •••

[1 mark]

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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 molecular structures are related to their functions.

[8 marks]

6. Chemoautotr~oph 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 produces haploid gametes and a diploid sporophyte generation that produces diploid spores where one of the generations is a dominant generation .

F

male : antheridum Female: archegonium

Zygote

,-------~Sexual 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

. 1 1 1 1

1 1

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 the population is 0.02 (or 2%). ~ 5'0 The 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, then the equation will be as follows: 2pq = (2)(.98)(.02) = 0.04 or 1 in 25 are 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

1 1

1

1

1 1

1

1 1

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3 a(i)

a(ii)

a(iii)

No Migration Random mating must occur Large population -No selection

- lac operon

- inducible operon

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

(Any Two) Total

1

1

1

1 1 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 ~-galactosidase would be 1

continuously 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 1

Total 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 compound mixture of amylase and amylopectin amylase - unbranched chain of a-glucose forms helix structure amylopectin - branched chains of a-glucose compound stabilized by countless hydrogen bond compact and insoluble readily 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 triglyceride

and three molecules of water

gJ~/ce-t-ol

DiAGRAM-2m ()

II ( CE2O:~ ~- H(j)-C-:C:E-·'.-C""~ i I . ---- () . -.... --.

II i H'O:B:-j-.---"~.,gJ-.~---cc ;:-" ; ,,--(: Eo

~., . C=-::oV_=-.J.-~~P-~-(CE:) ::--C(-.:r:: r -'

1 o

C.1·H 2-0-,~I _;CH:::" '---CH' o

" rE:-'-O-~-:'CE~; "--C :0,

C2::; ·· · -· -o-c-:~c-t-:::; :~CE:-

- 3F :0

ii. importance of lecithin in cell membrane structure

fatty a cid

2

Structure of lecithin - Lecithin is a type of phospholipids molecule 1 consisting of a hydrophilic head and two hydrophobic tails Structure 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

. Photoautotr~ph

Done by green plants or organisms

which has the chlorophyll pigment 1

Synthesise organic

compounds from carbon

dioxide and water

Synthesise organic compounds from J-i\63."~ S\l-J:),fl-'

i~or9~rll~_~QDl(?.9unds 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\()JJ"~

\efv~ -~~~, -

Saprophytic organisms can be defined as: organisms that obtain their nutritional needs from dead and decaying organic materials 1 Cannot synthesise their own food Secrete enzymes such as amylase, proteases, lipase which digest 1 their food extracellulary 1 Absorb the digested products through the cell surfaces 1 Give example: Mucor, Rhizopus, mushroom 1 Ecologically important because they act as a decomposer 1 Break down the dead organism and waste product 1 The decomposed material which contains chemical elements can 1 be 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 4 15

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

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

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

- thus increasing the carbon dioxide concentration reduces the affinity of 1 haemoglobin 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 results in an increase in the thoracic cavity volume, bringing about 1 inspiration

- alveolus and bronchioles expands during inspiration stimulating the 1 stretch receptors within the walls of the alveoli and bronchioles to send impulses 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 outer intercostal 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 move

to 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+ ions diffuse quickly from synaptic cleft or extracellular fluid into the 1 synaptic knob

- this influx of Ca2+ causes the synaptic vesicle to fuse with the presynaptic 1 membrane

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

- 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 1 DNA 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 1 the 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 an amino acid but it is a different amino acid.

- an example of genetic disorder caused by substitution is sickle-cell 1 anaemia, where the base thymine in the code for glutamic acid is substituted by the base adenine in the gene that codes for the (3- 1 polypeptide chain.

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

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

- in insertion, an extra nucleotide base pair is inserted into the DNA base 1 sequence of a gene causing the whole base sequence to be shifted one place 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 one place forward .

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

- insertions and deletions are usually more harmful than substitution and 1 inversion 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 1 base in the j3-globin allele and this results in a lack of j3-polypeptide chains of the haemoglobin molecule.

(c) max 8 - Down syndrome is an example of aneuploidy that is instead of 46 1 chromosomes 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 1 anaphase II of meiosis.

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

- ~hen a sperm containing 23 chromosomes fuses with an ovum containing 24 chromosomes and the zygote formed contains three 1 chromosome 21, trisomy.

- the individual may be a male or female usually with flat, broad faces, 1 slanted 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.

1

1

1

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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

1

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 1

allowing 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.

1

(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

1

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