Stpm Trial 2009 Bio Q&a (melaka)

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1. The movement of water through the vascular tissue in plants relies on which property of water? A Good solvent for ions and polar molecules (B- ~i Strong cohesive forces between molecules c' High specific heat capacity D Changes in density with temperature 2. The diagrams show four different molecules.

P ~H.oH H,/H C HO/

\

C

\9'"1 HI

c-c I

I

/H 'OH

H,

I

S

CHzOH

CHzOH

~-Q

H

R

Q .,0

N-C-C H/ I 'OH H

6-0\ H, I~

C HO/ \OH

C C

HI

c-c I

H

OH

o~ /OH

I

OH

I

/OH ' H

CH 2

H,

I

;0

N-C-C H/ I ' OH H

Which of the following below correctly shows the information about the molecules above? Contains a carboxyl Forms 1,6 glycosidic Forms peptide bonds bonds in glycogen group by condensation A P P PandQ B Q R P and R R R C RandQ 0' \ S P Qand S

3. The graph shows rates of simple diffusion and facilitated diffusion, of substance X across a cell surface membrane, as the concentration of substance X increases. simple diffuSion rate of diffusion! albitrary unrts

~"-----

laclillated diffusloo

concentratioo of substance X

Why does the rate of facilitated diffusion level off whereas the rate of simple diffusion does not? A Facilitated diffusion is limited by the number of protein channels in the membrane. :,,!3 !Facilitated diffuston is limited by the number of protein pumps in the membrane. C- Facilitated diffusion requires ATP which will eventually be used up. D Only facilitated diffusion is affected by the kinetic energy of the molecules that are diffusing. --- ~

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4. The diagram shows a drawing of an electron micrograph of a cell.

T

Match the structures in the cell with their functions? Synthesis of lipid

A B

~

C D

Synthesis of protein and carry out intracellular transport

R R S T

Breakdown of aged and damaged organelles

p T T S

Provides energy via aerobic metabolism

s'

Q Q

p

P

Q

R

P

Accumulates raw substances and secretes them from cells T S R

5. The epithelium shown below is part of the tissues of a mammalian urinary bladder.

What is the type of epithelium shown?

----

c~)rransjtional epithelium

B Stratified cuboidal epithelium C Stratified squamous epithelium D Stratified columnar epithelium

3

Q

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6.

According to the induced-fit hypothesis of enzyme action, which of the following statements is true?

A The binding of the substrate molecule depends on the shape of the active site.

B The allosteric site of the substrate molecule is used to bind to the enzyme. (~) The binding of the substrate changes the shape of the enzyme's active site slightly. D The substrate molecule modifies its shape to wrap around the enzyme molecule. 7.

In the presence of a fixed concentration of a competitive inhibitor, which of the following would best characterize an enzyme~ysed reaction when the concentration of the substrate is increased? I

The K.n increases.

II The maximum rate of reaction, V max increases.

III The maximum rate of reaction, V max decreases. IV The inhibition decreases. V The inhibition does not change. A I, II and IV B I, III and V (~ "IV only D Vonly 8.

The bacteria, Escherichia coli, were cultured for many generations in a medium containing a heavy isotope of nitrogen, I~. They were then transferred to a medium containing the light isotope of nitrogen, I~. They were given time to replicate DNA and left to reproduce for another three generations. After that, ultracentrifugation was canied out and the result of ultra violet light absotption was observed. Which one of the following shows the result of the experiments for generation 1, 2 and3?

I

m

n

4

IV

v

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Generation I I I IT IT

A B C

~

Generation IT ill N

Generation ill IT V

ill

N

N

V

9. The table below shows three different green plants and their characteristics. Plant I Tomato IT Maize ill Cactus

Which of the following is the correct match? ,

""

~)

B C D

I

IT

(a) (b) (b) (c)

(b) (a)

ill (c) (c)

(c)

(a) (b)

(a)

10. Which of the following statements are true of the light stage of photosynthesis? J.;, I The first electron acceptor'tsYSI is molecule R II The first electron acceptor of PSIT is molecule Q ill The energy level of PSI is much higher than that ofPSII N The reaction centre of PSI is P700 and the reaction centre ofPSII is P680 V The reaction centre of PSI is P680 and the reaction centre ofPSII is P7()O

Q

A IandN B IT,illandV C I,n,illandN (D': I, II, ill and V

tZ

J

J \ I

"" \

5

/

,)

., ,

Y

(,"

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11. The schematic diagram below shows the link reaction and Krebs cycle of aerobiosis.

By referring to the diagram, at which of the stages are NAD+ reduced?

A I, ill, vn and IX . VI, IV, VI and X C II, V, vn and vm o ill, VI, vn and X

12. What products are fonned when anaerobic respiration takes place in muscles and yeast cells? Muscles

Pyruvate, NAD+, ATP Ethanol, NADH, ADP, '.~ ) Lactate, NAD+ , ATP, o Lactate, NAD+, ATP, C~ A B

~,

Yeast cells

Ethanol, NAD+, ATP Lactate, NADH, ADP, CO2 Ethanol, NAD+, ATP, CO2 Ethanol, NAD+, ATP,

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13. Which one of the foUowing is not a correct match between the organisms and their mode of nutrition? A Butterflies and aphids: liquid feeders B Centipedes and earthworms: detritus feeders C Frogs and lizards: macrophagous feeders ~, Alligators and monkeys: microphagous feeders

14. Which process in erythrocytes is catalysed by carbonic anhydrase enzyme? A Dissociation of oxyhaemoglobin to release oxygen ,.~ Binding of haemoglobin with oxygen to form oxyhaemoglobin ~ Dissociation of carbonic acid to release hydrogen carbonate ions D Reaction between water and carbon dioxide to form carbonic acid 15. The graph shows the oxygen dissociation curve for fish and humans.

.

~

Fish~;~ ' . ~ ",

~

,

,,

~

,

:,

.

/

/ .

I

Z

PartW pr~ssure of oxygen :: . ..

--,."

~. ;

.

The oxygen dissociation curve for fish is more to the left of the oxygen dissociation curve for hwnans because A ',the haemoglobin of fish has a higher affinity for oxygen than that of humans. B at low partial pressure of oxygen. the haemoglobin of fish take up less oxygen compared to the haemoglobin ofhwnans. . C the concentration of oxygen in water is lower than in air. D at low partial pressure of oxygen. the haemoglobin of humans is more saturated than the haemoglobin of fish.

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16. Which one of the following is true of a hypothesis that describes the opening and closing mechanism of the stoma of a plant?

K+ W concentration concentration

A B C D" \1\)\j'\ IJ.-)

Decreased Decreased Unchanged Increased

Water potential in the guard cell

High

Decreased Increased Unchanged Decreased

Low

High Low

Condition of stomatal pore Closed Closed Opened Opened

17. These events occur during the cardiac cycle: I II III IV

Activation of the atrioventricular node, Activation of the sinoatrial node. Closing of the atrioventricular valves. Closing of the semi-lunar valves.

Which one of the following is the correct sequence of events in the cardiac cycle from the beginning to the end? A B C

-OJ

I, II, IV, III I, III, II, IV II, I, IV, III II, III, I, IV

18. Which of the following does not involve active transport across membranes?

A

The movement of mineral ions from apoplast to symplast

\.Ji) The movement of sugar from one sieve tube element to the next C Uptake of potassium ions by guard cells during stomatal opening D The movement of mineral ions into cells of the root cortex 19. Which of the following explain the fate of the excess amino acids in the protein metabolism of mammals? I II III IV

Excreted as albumin in urine. Deamination occurs and amino acid is excreted as ammonia in urine. Deamination occurs and amino group is excreted as urea. Transamination to produce other amino acids .

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

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20. In the kidney, the blood pH is increased by the movement of hydrogen ions from the A distal convoluted tubule into the peritubular capillaries. B peritubular capillaries into the distal convoluted tubule. peritubular capillaries into the proximal convoluted tubule. D proximal convoluted tubule into the peritubular capillaries.

@)

21. Which of the following occurs as a result of increase in insulin production? "\'"A, Increase in glucose absorption in the intestines B Decrease in glucose absorption in the intestines C Decrease in conversion of glycogen to glucose D Increase in conversion of glucose to glycogen 22. The diagram below shows the changes in membrane potential in an axon.

! '".0 d

s

N

40

0

&

~

o

.5 -70r---.

~ I II III IV

S~ulus

.......- ~2-3... , """""""ir---- TIme/ms O t 4

L...-...

Na+ channels open K+ channels open Na+ channels close K+ channels close

What happens at points M, Nand

AI ,--,/

B

Cc D

M I I II II

m 0

N II, III III III I, IV

IV II

I III

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23. Which of the following occur in the myofibrils when a muscle contracts?

~ C D

Sarcomere Shortens Lengthens Shortens Shortens

Hzone Shortens Shortens Lengthens No change in length

A band No change in len~ No change in length Shortens Shortens

I band Shortens Shortens Lengthens Shortens

24. What is the difference in the mechanism of action between steroid honnone and peptide honnone? A Target cells respond faster to steroid honnone than to peptide honnone. B Steroid honnone enters the cytoplasm and then passes to the nucleus whereas . peptide hOWlone remains in the cytoplasm. ~ Steroid honnone binds to a receptor protein whereas peptide honnone binds to a G protein. o Steroid honnone activates the expression of a specific gene whereas peptide hormone activates a protein! enzyme already present in the cell. 25. If a long-day plant has a critical night length of 10 hours, which of the following 24-hour cycles would prevent flowering? A 15 hours light / 9 hours dark .B 14~ hours light / 9~ hours dark (~ ) 12 hours light / 12 hours dark rr 4 hours light / 8 hours dark / 4 hours light / 8 hours dark 26. The table below shows four types of antibodies and their action in the immune system of hwnans. Types of antibody I II III IV V

IgM IgG IgA IgO IgE

Function (a) agglutination of antigen (b) blocks the binding of bacteria to body cells (c) fuses with bacteria to enable phagocytes to engulf it (d) causes allergic response (e) act as the antigen's receptor in order to trigger B cells

Which of the following combination correctly describe the type of antibody and its function?

A

B' ) C'

D

I (a) (e) (a) (b)

II (c) (b) (c) (a)

III (d) (c) (b) (c)

IV (b) (d) (e) (e)

V (e) (a) (d) (d)

10

\(

(

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27. The table below shows the types of immune response and the cells that are involved. Types of immune response

Cells

I hwnoral immunity II Cell-mediated immunity

(a) T cells (b) suppressors cells (c) B cells (d) helper T cells (e) plasma cells (f) cytotoxic cells

1

I Which of the following correctly match the type of response with the type of cells? A B C D 28. A person infected by HIV will develop AIDS and die. The immune system fails to eradicate the virus because I the body fails to produce anti-HIV antibodies. II the double-stranded viral DNA becomes incorporated as provirus into the DNA of the infected cell and remains dormant. III the unusually high mutation rates give rise to variations in viral antigens. IV the decrease in the T -helper cell level results in the loss of humoral and cellmediated immunity. A I and II

B II and III C I, II and III

IJ>; II, III and IV 29. The table below shows five kinds of animals and their methods of asexual reproduction.

I r

Animal

I II III IV V

Axolotl larva Aphi~

Apis Planaria Fasciola

(a) (b) (c) (d) (e)

Method of asexual reproduction Paedogenesis Fragmentation and regeneration Polyembryony Haploid parthenogenesis Diploid partehogenesis

II

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fA ", \: " B

I (a) (a)

C D

(b) (b)

-

<

II (d) (e) (d) (e)

III (e) (d) (e) (d)

IV

V

(c)

(b)

(b)

(c) (c) (a)

(a) (c)

Y3

)(

30 The table shows a list of sexual reproductive structures in angiosperms and lower plants

Structures in angiosperm

Structures in lower/primitive plants

I Contents in pollen grain II Contents of embryo sac III Stamen IV Carpel "V Anther

(a) female gametophyte (b) microsporangium (c) male gamete (d) microsporophyll (e) megasporophyll

Which of the following shows the correct match?

A B

I (c) (a)

(b)

~"

(b)

(c)

(c)

(b)

D '" J

II (a)

III (d) (d) (e) (a)

IV

V

(e) (c) (d) (d)

(b)

'/

A

(e) (a) (e)

31. The table below shows the type of germinal layers and the tissues/organs which are formed from it.

I II III

fA'. B

C D

Germinal layer Ectoderm Mesoderm Endoderm I (a) (a) (b)

II (b)

(c)

(b)

(c) (c)

(a) (b) (c)

Tissue/Organ Alimentary canal, liver, pancreas Nervous system, skin Connective tissue, muscle, skeleton

III

\j (

(c) (b) (a) (a)

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32.Foetal development in humans involves a series of mitotic divisions which result in an increase in the number of cells. However, there is no corresponding increase in the overall size. This occurs A after the formation of zygote , B_ ----..,after the formation of morula \C/ .after the formation of blastula D after the formation of gastrula

x fA

33. The table below shows patterns of growth and examples of growth patterns in organisms. Growth patterns

Examples of growth patterns

I Allometric II Isometric

(a) (b) (c) (d)

an

Increase in size of a baby to become adult Increase in size of zygote to become a blastula Increase in weight of a child Increase in number of cells in a zygote

Which of the following combinations are correctly matched?

A B C"; D~

I

II

(b) (d) (a) (c)

(a) (a) (b) (b)

34. The diagram below shows the inheritance of albinism in a human family. The allele for albinism is recessive to the nonnal allele.

tr.r

.AIbmo·....

•

o

o

AIbmomaie

NonnIt fIrnaM

/.

Normal male

What is the probability that the next child born to individuals P and Q will be an albino male? . A. 118

Cij) 114 C. 112 D. 3/4

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35. Consider the cross AaBb x aabb that yields

42A B 39 aabb lOA bb

9 aaB What conclusions can be deduced from this cross?

I II III IV

This is a test cross COV is 19% The alleles are linked. The genotype of the heterozygous parent can be represented as Aa/bB

A. B. \CJ D:

II and III I, III and IV I, II and III I, II, III and IV

36. The three diagrams below show the pattern formed by three-pairs of homologous chromosomes during prophase I of meiosis.

==- 0 a pair of

{;==;:;===~nonnaI dHomcJIOmes

I1IUIM8CIchomoflomw

(Ii)

Which of the following is a correct match of the mutation that has occurred in the mutated chromosomes? Deletion (i) (ii) (ii) (iii)

Addition or duplication (ii) (i)

Inversion

(iii)

(i) (i)

(ii)

(iii) (iii)

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37. The table below shows types of genetic abnormalities and their causes.

I II III IV V

A

B

- ~

8

D

Genetic abnormality Sickle-cell anaemia Thalassaemia major Down's syndrome Klinefelter's syndrome Turner's syndrome I P P R R

II R T T

Q

P _Q R S T

Cause An extra X chromosome 44 autosomes + 1 X chromosome Base substitution of haemoglobin gene Trisomy 21 Base deletion of the haemoglobin gene

III Q Q

IV

s s

V T R

S p

p

T

Q S

38. In a human population, one out of25000 babies is born with sickle-cell anaemia. The baby is homozygous recessive (genotype ss). Which of the following statements are true? I II III IV

Frequency for the recessive allele is 0.0063 Frequency for the dominant allele is 0.9937 Frequency for normal babies is 98.74% Number of carriers in the 25000 babies is 313

A I and II only I, II and III only , ~) I, II and IV only U I, II, III and IV

a

39. Which of the following statements are true about the mechanism of gene regulation and expression of lactose operon? I II III IV

Allolactose forms complex with repressor molecule The shape of the repressor molecule is altered allosterically. The repressor molecule is attached to the operator. High concentration of glucose increases the activation of the lactose operon system

. A..

I and II

.J!) I, II and III C I, II and IV D II, Illand IV

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43. The table below shows the taxonomic groups and taxon for the housefly.

Tnonomic group

Taxon

I

Kingdom

(a)

Arthropoda

II

Phylum

(b)

Insecta

III Class

(c)

Diptera

NOrder

(d)

Animalia

Which of the following combinations is correct?

A B C

,,~)

I a b c d

II d c b a

III b a a b

IV c d d c

44. The table shows four groups of plants and their characteristics. Group

I II III N

Characteristics

Pteridophytes Bryophytes Angiosperms Fungi

(a) (b) (c) (d)

Chitinous cell wall; store carbohydrate as glycogen. No vascular tissues; flagellated male gamete. Siphonogamous fertilization; seed development Vascular tissues; zoidogamous ft:;rtilization.

Which of the following combinations are correctly matched? A ( B"""', C D

I (a) (d) (d)

(b)

II (b) (b)

III (c) (c)

(c) (a)

(b) (c)

IV (d) (a) (a) (d)

45. Which of the following would provide the best information for distinguishing phylogenetic relationships between several species that are almost identical in anatomy? .A The fossil record B Homologous structures C Comparative embryology D';Molecular comparisons of DNA and amino acid sequences

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40. A study on 400 mice about their resistance towards a type of poison has been carried out. The resistance characteristic is controlled by the dominant allele R. 36% of the mice population is found to be resistant towards the poison. Calculate the number of mice expected to have Rr genotype.

A 16

1!,, 72

0

128 D 256

41. The diagram below shows the gene regulation and expression of a lac operon when lactose is present

What are represented by P, Q, R and S?

p A

)3",

R Promoter

Structural enes Promoter

s

~ ~R~e--~------~--~--~~--+---~------~~--~~----~

D

Re

42. The following shows the steps taken during a DNA finger printing process. I II III IV

Fragments separated by gel electrophoresis Radioactive probes with complementary base sequences are added Exposed to X-ray film DNA treated with endonuclease

The correct sequence of the steps above is

A I, II, II, IV R III, II, I, IV C · IV, I, II, III D I, II, IV, III

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46. Which of the following shows the characteristics of continuous variation? I II III IV

Controlled by polygene Shown by qualitative characteristics Influenced by environmental factors The classes show a normal distribution

I(

A I, II and III B I, III and IV 'G) II, III and IV D I, II, III and IV

47. Which of the following statements are true of an ecosystem? I II · III IV

Phytoplankton are producers The last consumer obtains the highest energy Ecosystem is an open system with input and output of energies Heterotrophs include herbivores, carnivores, omnivores, decomposers and detri tivores.

A IandII

a III and IV , C :1, III and IV o II ,III and IV 48. Species stTategy refers to the ability to compete b<:tween twQ

sp~~i... O)

in

A

lulbi,,",

Which one of the following is incorrect for r and k strategies?

A

B

.C) D

r Strat~y

K Strat~

Can disperse easily and colonise new habitat Progeny are small in size Large number of progeny Have ability to survive in a crowded habitat

Achieve equilibrium in population size. Progeny are large in size. Small number of progeny. Do not have strong ability to compete.

Y

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49. A student analyses the population of species X in an area S by using 1m x 1m quadrats (as shown in the diagram below). The numbers in the quadrats represent the number of species X. Calculate the density and frequency of species X.

Quadrat

~ @] ~ ~ ~ @] @] §] ~ @] ~

" 19

10m

@J ~ ~ <

5m---:;)

Frequency(%) 100 50 100 50

Density (moL) 22.4 22.4 44.8 44.8

o B C D

50. Which ofthe following are the correct statements for Liebig's law and Shelford's law I

The growth of an organism is dependent on the amount of essential material which is presented to it in minimum quantity. II The growth of an organism is dependent on the amount of essential material which is presented to it in maximum quantity III Organisms have an ecological minimum and maximum with limits of tolerance below the minimum and exceeding the maximum IV Organisms have an ecological minimum and maximum with a range in between which represents the limits of tolerance.

A

B

rC~

D

Shelford's Law II IV I III

Liebig's Law IV I III II

END OF PAPER ONE

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2 Section A [40 marks]

Answer gil questions in this section. I.

The diagram below shows the flow of electrons in non-cyclic and cyclic photophosphorylation. High

:-.10

\...

Electron energy I~vel

v:.::

~

1electron transport chain

~)

O~ t...-S

~

FJectron energy level

transport chain

l-"")~ ~~

Photo system I

Photosystem II

Light energy

Light energy L ow

Low'--'---- - - ' - - - - - - - - - - - - -

(a)

State the precise location of photo phospho

tion in a chloroplast.

...... Y':~.lqf~;~ ..~ ....~~................. .................................. ... .... • •• • •• • • • •• • • • • • • • • • • •• •• • • •••• • 0 •• •• • • • • • • • • • • • • • • •

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

of light in photophosphorylation. . Y:'~ ... :I~

....phd-:>~Wf'?m .. .!... }~.. ...r!1~~ .J:t!It1. ... .
••

••

•

••

(c)

0

••

0

•••••

•

••••

••••

••••

•

• • • •• • ••

[2 marks]

... .~:~~.... . .. ~~? ... ~ .... .P..~ .. •••

r... .

(2 marks]

••••••••••••

•

•

Explain how non-cyclic photoposphorylation.

••

••••

•

•

•

••

•

••

•

•••••

•••

•

•

••••

•

photophosphorylation

0

1... .

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

••••••••••••••••

differs

•

••

•

••••

••

•

••

from cyclic [4 marks]

~C!j.c;V.," .... r·~.:f,· f~ ~ fK;L"J l,¢.l~!:"! .. ..:~ ,.~ .f.~... ~~. 4· ...I'.~. X .I{¥>! .(\~ .. ... . .~~.te.t~f1~ ..?~. ~
.. 4M ....r:. ..?.. ~ .... Md....lh~~1JiF....((P:... ~4: .... ~(~«~.I... Jr.,.. .... f£1.ftj~ ... " .. f.-.... ~c}1... ... ~~~~ .. ~.~.. . P't)A~ ... ~.~.....~.~..~~! .. .~.~~. :.D.O.~·

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3

Ir-rp

. ~~.~.~~~.V~~~~~~~...~.... 9.~./.~PX .. 0~~···~~·~·~:@·r·~...~.~~ . . .~t!\ .. ~~.'=-

... ~~~~.\~.~~~~.~ ................................................... .

(d)

"-

State two ways in which oxidative phosphorylation in mitochondria resembles [2 marks] photophosphorylation in chloroplasts.

.. .~11 .. :\\ .. ~ ~ '!1~~7'1t' . • . .\\ ...Vl'!~"""Jlt ..................................... . .... .. ...... .. ...................... .. ........ ......... ......... ~ .. ................................. .

,

..

,.

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4

2

Different plants require different daylengths / periods of daylight for flowering. These plants can be classified into three groups. (a) (i) Using a suitable example, explain what is meant by day-neutral plants. [2 marks]

...p..\0J:~. ...~d... .5\?~.....':01~.....Q:.~~.~~....~..... " W:V.....

.. ......... ~ .... y~ ........

0

. . . Q~... , .. . . .::~;). =~~::;:::~~~~;;~;~.:~~;~~;:~~;;~~=~~::~. . . . .. period of daylight and flowering on a particular plant. The results ofthe experiment is as shown below. '-<

C!J

~

100

0 !i:::: '-'

'" -5 ...., V>

c: '" 50 0.

'0 C!J

bO

!S

c: CI) ~

C!J

0..

0 8

10

Which photoperiod group doe your answer.

12

4

16

is particular plant belong? Give a reason for [2 marks]

........ ..~hvt ... ~.q~-: . ~ ... ~iw.-0.: .. Jt. ...~~.~....~~~ .. ~ ...~)q~.Il.~ .. .~.~ .... . ...........~~ .... f.~;: ,.~~...~ ... !;.I. ~~: .....~PY.t~... ~~ ....~~!: ........ .. .. ... ...... . (b)

The diagram below shows three different treatm plant. No flOWellng

B

Flowering

C~,-_--, t=

I

o

6

12

Timelhow's

18

24

No floweIing

Keys

D

Li,bt

•

Dark

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5 (i)

From the data given, state w' period or dark period.

Jo.v

\()

.

.

is crucial in determining flowering: light [1 mark]

. ...

.'

.'

, , '.

. (

"

...... .........~ ... ~ .. ~ ........................... : ............................ . ................... .. "

.

• ... • • • • • • • • • • • • • • • "•••••••• • ••••.••••••••••••••• i ••••••••••••.•• • •••••••••••••.••••• . •••••••••••••••••••••• ••

(ii)

Use the results from t experiment to deduce the photoperiodic group t o ' which this plant bong. [ 1 mark] (

..... 5~~~~ .. ~ .. .~\tM-t .... ~ t ...itv~ ... ~....~~~~..1~ ... .l~~ ... ~. ~.~~:. (c) In a further experiment, the 15 hours of dark period is interrupted by flashes of red Iight(R) and / or far-red light (F).

D

no flowering

R

F

E

flowering

R F R F

III o

6

12

no flowering

24

18

Time I hours

What is your conclusion about the effects of the two types of 'ght on flowering? [4 marks]

ts,............... ~ J~O'lI.lI.\ :-!-.~ .........•••••

INs-

..A~, ..... p.te~;......~ ~ ..... ..I.~:]b.!. .... dlJ.~t:.... k .... .r.~.\J. ~.~

W'Y'~r'

.. @.y\<; .... r~~ ...~~······f{?~.~.~ ... .r!.~.~...!!:5~:.!!./X.Hk.....~ .. ~.~....

)

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[1 mark]

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[2 marks]

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9 4 The diagram below shows a DNA fragment containing a desired gene and a bacterial plasmid.

DNA ATGGATCCA ATGGATCCA TACCTAGGT gene TACCTAGGT plasmid

TACCTAGGTACT ATGGATCCATGA

(a)

(i) Identify a restriction site found on the DNA fragment and plasmid. [1 mark]

0

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(ii) State and explain the characteristics of the restriction site named in (a) (i).

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10

(c)

In some genetic engineering processes, a synthetic gene is inserted into a bacterial host. This process is shown below.

marlier lenc:

bacterial cen

marlier ,cnc

(i) What is the term

US~d to scribe

the function of the plasmid in this process? [1 mark]

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[1 mark]

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(d) Human insulin is now genetically engineered to treat diabetes. .~ advantages in using this type of insulin. [3]~

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11

Section B [60 marks] Answer any four questions in this section. 5. (a) By drawing a clear labelled diagram, show the structure ofa DNA molecule. [4 marks] (b) Elaborate on the importance of the DNA structure in relation to (i) gene replication and (ii) protein synthesis 6. (a) Describe briefly the structure of the mammalian liver.

[11 marks] . [5 marks] .

(b) Describe the functions of (i) (ii)

the liver and the kidney

as homeostatic organs.

[10 marks]

7. (a) By giving suitable examples, explain briefly what is meant by the terms, oviparity. ovoviviparity and viviparity. [6 marks] (b) Describe the parturition process in humans. ~.

[9 marks]

(a) Escherichia coli is a bacterium that will only produce the enzymes necessary for utilizing lactose when it is placed in lactose solution. Explain how this enzyme induction is controlled. [9 marks] (b) Explain how a zygote with the genotype XO is formed in human. What is the effect on an individual carrying the chromosome XO? [6 marks]

9. By giving a named example, describe the taxonomical characteristics of the following taxonomical groups of organisms: (i) (ii) (iii) (iv) (v)

Arthropoda Echinodermata Fungi Mollusca Spermatophyta

•

10. (a) Describe the different types of ecological pyramids. (b) Discuss the factors which influence the size of a population.

[15 marks]

[6 marks] [9 marks]

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(d) (i) Similar to human insulin! less probability of side effects; 0 (ii)Cheaper than extracts from animal sources 0 (iii)No need to kill animal! ethical reasonsO (iv)Can be produced in large quantity to meet increasing demandO

Section B Sea)

Structure of DNA/Deoxyribonucleic Acid o Diagram CD o Labels: Double helix! spiral double chain; phosphate molecule; deoxyribose; correct base pairing: A=T, G==C; correct number of hydrogen bonds; distance between 2 base pairs = 3.4nm1lO base pairs or 1 complete turn)=14nm Any 6 = Max 4 0" 3!f.:J "'f-IIW\, (Note: Explanation not required)

@

(b) (i) Importance of DNA Structure Relating to Gene Repli~ation. o DNA contains the code/information for semi-conservative replication CD o The DNA double helix is able to separate when hydrogen bonds break between bases Q) o Free nucleotides in the nucleoplasm are able to pair complementarily on the exposed polynucleotides Q) " o The number of bases and pairing is fixed by the length/size of the DNA molecule o Pairing is constant: A=T, G == C Q) o The bases on the original/parental strand ensures the corre-et pairing of base sequence Q) o The content of the DNA does not change and its characteristics/codes/information are passed down''to the daughter cellsQ) Any 5 (ii) Importance of DNA Structure Relating to Protein synthesis o The DNA double helix opens/separates at a specific point/part to expose its sequence of specific genes/template CD o The weak hydrogen bonds enable the DNA double helix to open easily and this is catalysed by DNA polymerase Q) o Free nucleotides then arrange accurately in a complementary manner on the template/one of the exposed DNAQ)

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o Complementary bases precisely pair with each other : !\=U . G == C to ensure accurate transcription CD o On the mRNA, 3 bases/triplet base forms a codon CD o Each specific codon is a code for a specific amino acid CD o After transcription, pair accurately mRNA sloughs off/separates from the template and move through the nuclear pore to the cytoplasm to continue the process of translation on the ribosome CD o Four different bases on the DNA are sufficient to supply all the codons for all the different types of amino acids needed CD Any 6 6 (a) Liver Structure

• • •

Consists of many lobules CD Each lobule contains many liver cells/ hepatocytes CD Arranged in rows/liver cords radiating from the centre to the side! arranged radially around a central vein (a branch of hepatic vein) CD Between liver cords/ rows of liver cells, there are minute channels/ canaliculi which contains/ channels bile! liver cells are surrounded by minute channels called canaliculi CD Sinusoid, minute blood-filled space/channel in the liver, ' connects arteriole branches and hepatic portal vein with the central veil1. CD Blood supplies come from hepatic artery and hepatic portal vein CD Branches of the hepatic artery and hepatic portal vein are found between the liver lobules CD Each liver cell is not differentiated. It has distinct nucleus and Golgi body, many mitochondria and lysosomes and richly contains glycogen granules and fat dropletsCD (Any S:Max 5 marks)

•

• • • •

(i) Liver as a homeostatic organ

i.

In the regulation of blood glucose/ sugar level (Ji (b)

To remove excess glucose from the blood ifin excess or to raise the blood glucose level if it decreases CD (c) Wh~n blood glucose is in excess/level is high: \ • more glucose is absorbed into the liver cells and respired/ break down to ~ CO2 and water/ more sugar is respired
0t

(d)

Q

•

When blood glucose level is low/lacking: Glycogen stored in liver cells is broken down to glucose under the stimulation/effect of glucagon
.,) ...

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In the regulation of amino acid Amino acid molecules are taken up by the liver cells to synthesise protein CD Transminated to other amino acids CD Are deaminated CD (Any I)

". • /.

In the regulation of lipid

/

.

J •

L.

Fat is converted by liver cells to glycogen CD Producing cholesterol when the uptake through inadequate/insufficient/lacking CD Excess blood cholesterol is remove via the bileCD (any 1)

the

diet

IS

(ii) Kidney as a homeostatic organ

· l

(e) In the regulation of water in body fluid! blood osmotic pressure: CD 0 High blood osmotic pressure causes pituitary gland to secrete ADHCD o ADH increases the permeability of distal tubule and collecting duct of /IN!lX 3 renal nephron CD o More water reabsorbed in the distal tubule and collecting ductCD o Vice-versa ~ (Any 2) (f) In the regulation of blood pH CD wiK ~>'f' 2. Excess hydrogen ions in the blood are excreted' actively through the distal \Y tubule of the nephron CD (Any 1)

rr ,

(g) In the regulation of mineral ion concentration CD ~c4'~ 0'" """"j o~ ".,... tA£.~-cJ~ 2. Excess mineral ions are not reabsorbed . They are eliminated! removed in (0 the urine CD 3. Inadequate sodium ions in blood, sodium ions rue reabsorbed via the distal tubule and collecting duct under the influence of aldosteron~. CD (Any I) ~I 7 I .\-

I

! +- I

Oviparity: ··.· • when the eggs are laid and hatched outside the female body CD • Example Reptiles / B~rds / Frogs / most fish / insects I some species of sharks CD Max2 Ovoviviparity • The egg is fertilized and stored I retained within the oviduct CD The embryo is nourished by the yolk in the egg II obtains its nutrients fro~ the yolk CD The young are born after being hatched within the mother's body CD Any 1 • Example Some species of sharks CD Max 2

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Viviparity • The egg is fertili zed within the female body and the embryo formed is no uri shed by nutrients obtained from the mother's body / through the placenta. The baby is born alive CD • Example All mammals and some species of sharks. CD Max 2 (b)

• • • • • • • • • • • • •

8 (a)

Parturition High level of oestrogen produced by the placenta initiates parturition. CD Triggers the release of prostaglandin from the uterine wall CD This enable the uterine muscle to contract CD The strong myometrium! uterine contraction sends positive feedback to mother's pituitary gland and the foetus to release oxytoxin CD Both prostaglandin and oxytoxin stimulate powerful uterine contraction CD This increase in frequency as the labour progresses CD The contraction stimulates stretch receptors in the uterine wall and the cervix CD stimulate more oxytoxin to be released CD cervix dilates and soften its tissues and becoming more flexible CD The first stage of labour ends with the cervix dilating/opening to about 10 cm CD In the second stage of labour, strong contractions, aided by the mother' s pushing, expel the foetus through the vagina. CD The uterus contracts again and separate the placenta from the uterine wall CD The placenta is passed out and the umbilical cord is tied and cut This is the final/third stage of labour CD Max 9 . . Total: 15 marks ",,_"')t. '1 Regulation of Enzyme Induction

o A well-labelled diagram (Refer to appropriate diagran1 in any reference text) <.D o To utilize lactose, E. coli requires j3-galactosidase~ lactose permease ( or gala~to~e . permease) and galactosIde transacetylase. CDI'

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o Thus, when the bacterium, E. coli is present in lactose solution, some of the lactose is converted to allolactose, which acts as an inducer. CD o This inducer binds with the repressor and changes its comformation/shapc so that the repressor molecule is unabled to attach to the operator. CD o So, the operator is turned on. This leads to the expression of the structural genes/transcription occurs followed by translation and enzymes are produced. CD o On tpe other hand, in the absence of lactose, there is no inducer to bind to the repressor. Thus, the repressor will bind to the operator to inactivate it/to block it and prevents the RNA polymerase from attaching/moving across it. Hence, transcription cannot occur and no enzymes are produced. CD-Max 9 marks (b) Explanation On The Formation of an XO Zygote

o o o o o o o o

This is due to no-disjunction during oogenesis CD The egg/ovum formed may not have an X-chromosome CD / .lr...-e_;, lftNY- i/0j'~ -c<: )~. d!",,~c,_>.,,~,;>t, If such egg/ovum is fertilized by a sperm carrying a normal X-chromosome, the offspring will have the genotype XOCD XO individuals will suffer from Turner's Syndrome CD It is a barren!sterile/infertile female CD Ovaries and breasts do not develop CD Do not menstruate/no menstrual cycle and do not ovulate 'CD Dwarf, deaf and with low 10 CD Any 6

* 1i1lL-

. lSra) (i) Arthropoda

QJ o o o o o o

r

V- ~. + ::2.- ~ ff . Eukaryotic; coelomate; triploblastic CD Bilateral symmetry CD Segmented body/ shows segmentation CD Exoskeleton consists of hard chitin CD Posseses jointed limbs/appendages CD Open blood circulatory system CD Any 3

(ii) Echinodermata

o Eukaryotic; coelomate; triploblastic CD o Pentarneric radial symmetry at adulthood but larvae show bilateral symmetry CD o Without tpetarneric segmentation! no head CD o Presence ·ofwater vascular system with tube feet for gaseous exchange/ excretion! locomotion CD o Simple guts CD o Spine/calcium plates below skin CD o Sexual reproduction via binary/sexual fission!CD Any 3

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(iii) Fungi o o o o

o o o

EukaryoticCD Chitinous wall CD Absence of chlorophyll CD Heterotrophic/saprotrophic or parasitic
(iv) Mollusca

o o o o o o

o

Eukaryotic; coelomate (reduced)CD Soft body and without metameric segmentation CD Body with a head, a muscular foot and visceral mass CD Mantel might secrete a shell and forms cavity for respiratory organs involve in gaseous exchange CD Radula is present to grind food CD Sexual reproduction CD Larval stage is a trocophore larvaCD Any 3

(v) Spermatophyta o o o o o

Produces seeds (protectedlunp.rotected in ovar.y). CD Dominant sporophyte generation with the gametophyte generation greatly reduced CD Well-developed vascular bundles CD Heterosporous
10 (a) 1 Number pyramid CD /\I~·c.eQ..y' • shows the number of individual organisms that occupy the trophic levels in a food chain CD '

2

Biomass pyramid
•

based on the total dry weight, calorific value or other measurements for the amount of liVing substance in each subsequent trophic level. ())

3 Energy pyramid CD

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•

shows the rate of energy flow or production at each subsequent trophic level. The best ecological pyramid to illustrate as a whole the functional condition of a community.
(b)

Factors affecting population size/ growth - Fluctuation! increase or decrease of population size is due to interaction between two intrinsic factors: birth (natality) rate and death (mortality) rate of a population.
I Population age: - a population with a high number of aged members/individuals results in low biotic potential and this limits/reduce the size of a population. // Conversely a population with a high number of young individuals with high biotic potential increases population size. 2 Food,water, oxygen, etc: - Inadequate/lack of essential needs such as food, oxygen, etc.inhibit growth/reproduction of organisms in a population. - This also increases death rate and the size of population decreases. ( and vice-versa)

3 Predation: - When number of predators increases, population will decrease. 4 Diseases outbreak/epidemic /parasitism/environmental disaster. - These will increase death rate and will decrease the size of 2opulation 5 Limited space/lack of space/Qvercrowding. - The lack of space or overcrowding resultsvin behavioral dis.tnrbahces such as ,~. mating failure, aggressive behaviour, cannabalism. 6 Accumulation of toxic excretory products! contamination/pollution - These factors sl(,?w down population growth thus population size decrease. / '1

8 Migration/emigration and immigration - Emigtation causes populauon to decrease whereas immigration causes population to incrcase . 9 Territorial behaviour: - decreases overcrowding and competition. - It ensure food supplies and space are adequate for offspring! mating. This causes population size to increase.

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