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PROGRAM PENINGKATAN PRESTASI AKADEMIK PEPERIKSAAN AKHIR TAHUN TINGKATAN 4 2012 PERATURAN PEMARKAHAN BIOLOGI KERTAS 2 1 JABATAN PELAJARAN NEGERI SELANGOR MAJLIS PENGETUA SEKOLAH MENENGAH (MPSM)

SKEMA BIOLOGI K2

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Page 1: SKEMA BIOLOGI K2

PROGRAM PENINGKATAN PRESTASI AKADEMIK

PEPERIKSAAN AKHIR TAHUN TINGKATAN 4 2012

PERATURAN PEMARKAHAN

BIOLOGI

KERTAS 2

1

JABATAN PELAJARAN NEGERI SELANGOR

MAJLIS PENGETUA SEKOLAH MENENGAH (MPSM)

Page 2: SKEMA BIOLOGI K2

ANSWER SCHEME

PEPERIKSAAN AKHIR TAHUN TINGKATAN 4 2012Question 1Ques Criteria Marks

1(a)(i) Able name structure Q

Answer:

Q : Epithelial tissue 1

1

1(a)(ii ) Able to state on adaptation of QSample answer

P1 : consists of many / a group of epithelial cells P2 : has cilia

Any 1

11

1

1( b ) Able to state two functions of structure Q in the trachea.Sample answer:P1 : build the lining wall of the tracheaP2 : Cilia to sweep dustP3 : Secrete mucus that traps dust particles

Any 2

111

2

1( c )(i) Able to state the cell organization of level 4Answer : System 1

1

(c)(ii) Able to explain one function of trachea in respiratory systemSample answer:.P1: Allows oxygen intake / carbon dioxide eliminationP2; during inhalation/exhalation P3 : as trachea does not collapseP4 : made up of rings of cartilage

Any 2

1111

2

1( d ) Able to explain one adaptation of the lung.Sample answer:

P1: has many alveoliP2 : increase total surface areaP3 : efficiency of gases exchangeP4 : surface moist

1111

2

2

Page 3: SKEMA BIOLOGI K2

P5: easy for respiratory gases to dissolveAny 2

1

1( e ) Able to explain one difference in structure of the lung in individual M and NSample answer

P1 : surface area of the lung in M is larger than NP2 : lung M has more smaller ( rounded ) structures/ alveolus /air sacs compare to NP3 : as( more ) alveolus/ (rounded) structures/air sacs in N have been damagedP4:by cigarette smoke in NP5: N alveolus shows symptom of emphysemaP5 : decrease rate of gases exchangeAny three

11

1

11

3

Total 12

Question 2Num Criteria Marks

2 ( a ) ( i ) Able to name ways of transport X and Y and give reasons

Sample answer

X : Active transport t/ Passive transport

Reason : Transportation molecules is against concentration gradient //lower molecules to higher molecules concentration

Y : Facilitated diffusion

Reason : Transportation is following concentration gradient

//higher molecules concentration to lower molecules concentration

1

1

1

1

4

2( a )( ii ) Able to state two differences between way of transport X and Y .

Sample Answer

2

3

Page 4: SKEMA BIOLOGI K2

2( b )

X YNeeds energy ATP Does not need energy

Against concentration gradient

Follows concentration gradient

Occurs in living cells Occurs in living and non living cell

Any 2

Able to explain how amino acids molecules are transported by Y

Sample answer

P1 : Concentration of amino acids molecules is higher in outside cell compare to inside cellP2 : amino acids molecules binds to the binding site of carrier proteinP3 :carrier protein changes shapeP4 :allow molecules to pass through to the other side of plasma membrane P5 : carrier proteins change back to its original shape

Any 3

1

1

1

1

1

11

1

3

2( c ) Able to explain the effect of cyanide in the transport of minerals

Sample answer

P1 : Active transport cannot be carried outP2 : Active transport needs energyP3 : cyanide disrupts respirationP4 : no energy is producedP5 : less/no mineral intake

Any 3

11111

3

Total 12

4

Page 5: SKEMA BIOLOGI K2

Question 3Num Criteria Marks

3(a) Able to name the type of cell division and stage M and N.Answer

Type of cell division: Meiosis Stage M : Metaphase I Type of cell division : Mitosis Stage N : Metaphase

1111

4

3(b) Able to state the chromosomal behaviour of cell N.Sample Answer

P1: the chromosome /sister chromatids line up at the metaphase plate /equatorP2: the spindle fibre holds on the chromosome at the centromere

Any 1

1

1

1

3(c)(i) Able to name structure P. AnswerCentriole 1

1

3(c)(ii) Able to explain what happen to the daughter cell formed if there is no structure P in the cellSample answer

P1: no spindle fibres formedP2 : chromosomes are not held at the centromere during metaphase P3: (sister) chromatids are not separatedP4 ( sister ) chromatids not able to move to the opposite polesP5 : daughter cell with unequal number of chromosomes formed P5: may lead to genetic disorder disease//any suitable name of the genetic disorder disease

Any 3

11

111

3

3(d) Able to explain the importance of process Q to an organismSample answer

P1: process P is crossing over// exchange of genetic materialP2 : leads to variationP3 : better adaptation to the environment P4 : better chance of survival of the speciesP5 : better adaptation to the environment

Any 3

11111

3

Total 12

5

Page 6: SKEMA BIOLOGI K2

Question 4Num Criteria Mark

s4(a) Able to identify organ R and S

Answer:S : StomachT : Pancreas

11

2

4(b)(i) Able to state the function of S/ stomach in the digestion process.Sample answer

P1 : Gastric glands in S produced / secrete gastric juiceP2 : Consists of pepsin / rennin / hydrochloric acidP3 : Pepsin hydrolyses protein to polypeptidesP4 : Rennin coagulates soluble milk / caseinogens to casein. P5 : Hydrochloric acid provides an acidic medium// kill bacteria//stops the action of amylase

Any 2

1111

1

2

4(b) (ii) Able to state the effect of taking meal at irregular time.Sample answer:

P1 ( Having meals at irregular times ) result in the absence of food in the stomach at certain period.P2 : gastric juice (secreted by gastric gland) act on (the epithelial lining of) the stomach wallP3 : result to gastric ulcer / gastritis// pain in the stomach

Any 2

1

1

1

2

4(c) Able to state the function of pancreas and effect of cancer of pancreas to the process of lipid digestion

Sample answer

P1: Cancer of pancreas prevents the pancreas from releasing the pancreatic juice into duodenumP2 : Lipase not secreated P3 : Lipids not hydrolysed by lipase P4 :to produce fatty acid and glycerol.

Any 2

1

111

2

6

Page 7: SKEMA BIOLOGI K2

4(d) Able to state two adaptations of vilii in absorption process.

Sample answer

P1 : Has dense network of blood capillaries / lactealP2 : Has a very thin wall / one cell thick wallP3 : Has moist surfaceP4 : Has large number of villus P5 : to increase the rate of absorption of nutrient.

Any 2

11111

2

4(e) Able to explain the how amino acids is assimilated by liver

Sample answer:

P1 : Excess amino acid will be transported to liver/RP2 : By deamination processP3 : Liver /R hydrolysed amino acid into glucose and urea.P4 : This will lower the concentration of amino acid in the bloodP5: Liver /R synthesized plasma proteins// forms enzymes/hormones

Any 2

111

1

1

2

Total 12

7

Page 8: SKEMA BIOLOGI K2

Question 5No. Criteria Mark

5(a) ( i ) Able to name species in zone R , S and TAnswer

Zone R : Avicenia / SonneratiaZone S : RhizophoraZone T : Brugeira

111

3

5(a)( ii) Able to explain two adaptations of species in zone R to survive in the ecosystem. Sample answer

P1 : Has (long)cable rootP2 :To give support in muddy soil/ traps the mud/soilP3 : Has viviparity seed /seed germinate while still being attached to parent treeP4 : To float / avoid sink in sea water / stands upright in the mudP5 : Has pneumatophores / breathing rootsP6 : To obtain oxygen during high tideP7 : Leaves have thick cuticle// sunken stomataP8 : Reduce transpiration/water loss

Any 4

111

11111

4

5(b) Able to explain the changes in the ecosystem that will happen to zone S after 10 years Sample answer

P1 : Zone S has prop rootsP2 : trapped mud/sedimentsP3 : hardened the soilP4 : increase the soil levelP5 : soil more fertileP6 : condition no more favorable for species in zone SP7 : Species in zone T replace species in zone S

Any 3

1111111

3

5(c) Able to explain one importance of mangrove swamp to maintain biodiversitySample answer

P1 : Provides habitat to organismsP2 : Provides breeding places for organisms P3 : Avoid from extinctionP4 : Balanced the carbon cycleP5 : Maintain a balanced ecosystem Any 2

11111

2

Total 12

8

Page 9: SKEMA BIOLOGI K2

Question 6Num Criteria Marks

6 ( a ) Able to explain the adaptations of alveolus in gaseous exchange.

Sample answer

P1: the wall of alveolus is only one cell thickP2: enables rapid diffusion of gasesP3: inner surface of each alveolus is lined with a layer of moist epithelial cellsP4: enables respiratory gases to dissolve in the fluid before diffusing across alveolar wallP5: outer surface is covered by a dense network of blood capillariesP6 : which provides a large surface area for rapid diffusion of gases to and from the alveoli.

Any 4

111

1

11

4

6( b ) Able to explain inhalation process in human

Sample answer

P1: the external intercostal muscles contractP2: internal intercostal muscles relaxP3: rib cage move outwards / upwardsP4: diaphragm contracts and flattensP5: volume of thoracic cavity increasesP6: causing air pressure within alveoli to decreaseP7: higher atmospheric pressure outside causes air to rush in

Any 4

1111111

4

6( c ) ( i ) Able to explain the need for anaerobic respiration.

Sample answer

P1: oxygen in muscle cells is insufficientP2: causes oxygen debtP3: lactic acids accumulates in muscle cellsP4: may lead to muscle crampP5 : panting/deep breathP6 : to pay “oxygen debt ”

Any 2

111111

2

6( c )( ii ) Able to state 5 differences between aerobic respiration and 5

9

Page 10: SKEMA BIOLOGI K2

anaerobic respiration

Sample answer

P1: Oxidation of glucose is complete in aerobic respiration but oxidation of glucose is incomplete in anaerobic respirationP2: Oxygen is used in aerobic respiration but oxygen is not used in anaerobic respiration.P3: Large amount of energy (2898 kJ)is released during aerobic respiration but a small amount of energy (150 kJ in muscle cells, 210 kJ in yeast fermentation)is released during anaerobic respiration.P4: 38 molecules of ATP is released during aerobic respiration but 2 molecules of ATP is released during anaerobic respiration.P5: Anaerobic respiration occurs in the cytoplasm and mitochondria but anaerobic respiration occurs in in the cytoplasm only.P6: The products of aerobic respiration are carbon dioxide, water and energy but the products of anaerobic respiration are ethanol, carbon dioxide and energy in yeasts /lactic acid and energy in muscle cells.

Any 5

1

1

1

1

1

1

6( d ) Able to explain the regulation of oxygen and carbon dioxide by his body.

Sample answer

P1: when the oxygen level in the blood is lowP2: the carbon dioxide level increases due to cellular respirationP3: carbon dioxide reacts with water in blood to form carbonic acidP4: causing blood pH to dropP5: the peripheral chemoreceptors are stimulatedP6: and send nerve impulses to the respiratory centreP7: this stimulates the respiratory muscles to contract and relax fasterP8: breathing and ventilation rates increase more oxygen inhaled and oxygen level in blood returns to normal.

Any 5

111111

1

1

5

Question 7Num Criteria Marks

10

Page 11: SKEMA BIOLOGI K2

7(a) Able to explain how extracellular enzyme is produced and transported out of a cell.Sample Answer

P1: Nucleus contain genetic information (of protein synthesis in) / DNAP2: Ribosome is the site of protein synthesis P3: the synthesized proteins in the ribosomes are transported through (the spaces between the) rough endoplasmic reticulum.P4: Proteins (depart from the rough endoplasmic reticulum) wrapped in transport vesicles.P5: The transport vesicles bud off from the sides of the rough endoplasmic reticulum.P6: The transport vesicles travel to Golgi apparatus.P7:Transport vesicles fuse with the membrane of the Golgi apparatus P8: and empty their contents into (the membranous space of the) Golgi apparatus.P9: The proteins are processed, modified, and repackaged in the Golgi apparatus.P10: Secretory vesicles containing these protein/modified proteins/extracellular enzyme bud off from the Golgi apparatus membraneP11: and transported to the plasma membrane.P12: these vesicles will then fuse with the plasma membraneP13: Before releasing the protein/modified proteins outside the cell as enzymes.

Any 6

1

11

1

1

11

1

1

1

111

6

7(b) Able to explain the mechanism of lock and key hypothesis. Sample Answer

P1: Enzyme has active site P2: the active site has distinctive shapeP3: enzyme highly specificP4: Substrate binds to the active site of the enzymeP5: to form enzyme – substrate complexP6: the enzyme catalyses the substrate to form productP7: the enzyme is now free to bind with another substrate

Any 4

1111111

4

7(c) Able to explain the effect of temperature on the rate of enzyme reaction.Sample Answer

6

11

Page 12: SKEMA BIOLOGI K2

P1: At low temperature rate of enzyme reaction is slowP2: Rate of enzyme reaction is maximum at their optimum temperatureP3: as temperature increase, the chances of substrate to bind to the active site of enzyme increaseP4: for every 10 C rise in temperature, the rate of reaction increaseP5: above the optimum temperature, any increase in temperature will no longer increase the rate of reactionP6:At temperature above 60 C, enzyme is denatured so the rate of enzyme reaction is zero.P7: bond that hold the enzyme molecules begin to breakP8: alter the three dimensional shape of enzymeP9: substrate no longer fit into the active site of enzymeP10: the enzyme lose their activities

Any 6

11

1

1

1

1

1111

7(d) Able to explain the uses of enzyme technology in wine industrySample Answer

P1 : Zymase is use as a catalystP2 : to convert glucoseP3 : to ethanolP4 : a type of alchohol / wineP5 : during fermentation

Any 4

11111

4

Total 20

Question 8Num Criteria Marks

12

Page 13: SKEMA BIOLOGI K2

8(a) Able to explain the good and bad effects of spraying fertilizers on agriculture and River S.Sample Answer

Good ImpactP1: Crops grow healthy P2: Increase crop yieldP3: Supplying minerals such nitrates and sulphates

Any 2Bad ImpactP3 :nitrates/sulfates dissolved in rain waterP4: washed away / leached into River SP5: river water rich in nitrates/sulphates P6: resulting in the rapid growth of algae on the surface of the riverP7: eutrophication /algal bloomP8 : block/prevent sunlight to penetrate into the river waterP9: alga used oxygen for decomposition / higher rate of decompositionP10: decrease the oxygen content (of the River S)P11 : increased BOD level of waterP12: cause the death of aquatic organisms ( in the River S )

Any 8

111

1111

111

111

10

8(b) Able to explain the bad effects of the presence of industries to of the environment and health of the residents in the town areaSample Answer

P1: (Burning of coal and oil )in factories releases harmful gases /sulphur dioxide/ nitrogen dioxide/carbon dioxide/smoke/ dust /dirt.P2 : increase excess particles/pollutants to the environmentP3 : leads to air pollutionP4: sulphur dioxide/ nitrogen dioxide cause acid rainP5: which destroys buildings/ sculpture of stone /vehiclesP6: acid rain reduces the pH of soil / waterP7: causes death of organismsP8 : (excess) carbon dioxide cause Green House effectP9: increase temperature of the environmentP10 : cause disease to the residents / difficulty in breathing/coughing/skin cancer /any suitable disease P11 : disaster / droughtP12: Smoke/ dust / soot cause hazeP13 :irritates the eyes/ sight difficulty

Any 10

1

111111111

111

10

Total 20Question 9

No Criteria Mark

13

Page 14: SKEMA BIOLOGI K2

9(a) Able to explain the mechanism of photosynthesis.Sample Answer:

Light reaction

P1: Light reaction occurs in the presence of lightP2: Light reaction occurs in the grana in the chloroplastP3 :During light reaction chlorophyll captures light energy which excites the electron of the chlorophyllP4: The electron leaves the chlorophyll moleculeP5 : Energy from the excited electron is used to form ATP P6: Light energy is also used to split water molecules into hydrogen ions and hydroxyl ions/photolysis of waterP7: The hydrogen ions combine with the electron to form hydrogen atomsP8 : The hydroxyl ions loses an electron to form a hydroxyl groupP9 : The hydroxyl groups then combine to form water and oxygen

Dark reaction

P7 : Dark reaction occurs in the absence of lightP8 : Dark reaction takes place in the stroma in the chloroplastP9 : Hydrogen atoms are used to fix carbon dioxide during reduction of carbon dioxideP10: Carbon dioxide is reduced into glucose and water P11 : The glucose undergoes condensation to form starchP12 : Starch is stored in chloroplast as starch grains Any 10

1111

1

11

11

111111

1 1

10

(b) Able to explain the effect of the lacking of micronutrients and macronutrients in plants. Sample Answer:

P1: Ferum is a micronutrientP2 : Lacking/deficiency of ferum cause yellowing of young leaves P3 : Copper is a micronutrientP3 : Lacking/deficiency of copper cause death of tips of young shoots and stunted growthP4 : Zinc is a micronutrientP5 : Lacking/deficiency of zinc cause stunted growthP6 : Boron is a micronutrientP7 : Lacking/deficiency of boron cause leaves become thick/curledP8 : Nitrogen is a macronutrientP9 : Plant cannot synthesise chlorophyll/enzymes for photosynthesisP10 : Lacking/deficiency of nitrogen cause stunted growth/chlorosis/yellow leafP11 : Phosphorus is a macronutrientP12 : Plant cannot synthesise ATP needed for photosynthesisP13 : Lacking/deficiency of phosphorus cause poor root growthP14 : Potassium is a macronutrient

11

1

11

111

11111

10

14

Page 15: SKEMA BIOLOGI K2

P15 : Plant cannot synthesise protein and cofactor of enzymesP16 : Lacking/deficiency of potassium cause yellow leaves

Any 10

11

Total 20

END OF MARK SCHEME

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