ANSWER BIO SCORE SCHEME F4

BIO~SCORE Form4 SCHEME 2020

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BIO~SCORE Form4 SCHEME 2020

CHAPTER 1: INTRODUCTION TO BIOLOGI
PG ANSWER
2 1. ...life...study... 2. ...living things...

Field of Study in Biology
1. Bioinformatic: ...computer technology... 2. Microbiology: ...microorganisms...
3. Biotechnology: ...products... 4. Zoology: ...animals.
5. Botany: ...plants. 6. Ecology: ...environment.
7. Physiology: ...body. 8. Anatomy: ...structures...
9. Genetic: ...genes...heredity... 10. Entomology: ...insects.
11. Taxonomy: Classification... 12. Histology: ...tissues.
3 Medical field: ...clinical practice... Pharmacy: ...synthetic vitamins...medical...
Food technology: ...microorganisms... Agriculture: ...transgenic...
Mind test 1

4 a) Fume hood: ...hazardous gasses...
b) Lamina flow cabinet: ...airflow...
c) Biological safety cabinet: ...contaminated...pathogens...
Mind Test 2
d) Emegency shower station: It is very important when the body is exposed to harmful
chemicals.
e) Eyewash station: Used to wash eyes when exposed to chemicals or harmful substances.

5 2. Safety Glasses: ...eye... Lab Coat: ...body... Shoes: ...glass...chemicals...
Mind Test 3: face mask
3. Substances can disposed: ....low...harmless...pH5...pH9.
Substances cannot disposed: ...Mercury. ...alcohol. ...surgery... ...Pottasium. ...Formaldehyde.
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BIO~SCORE Form4 SCHEME 2020

PG ANSWER
...5....9...

6 Standard Operating Procedure

1. Category A: ...sharp waste...sterilized...

2. Category B: ...autoclave...

3. Category C: ...pack...

4. Category D: ...decontaminated...

Spillage of Chemical Substances Spillage of Mercury

1. ...teacher... 1. ...lab assistant.

2. ...prohibited... 2. ...prohibited...

3. ...sand. 3. ...sulphur...

4. Scoop... 4. ...fire department.

7 Laboratory Safety Procedures
a) …without supervision. b) …eat… c) Wash… e) Clean…disinfectant…

Mind Test 4
1. ...unplug…electrical… 2. …panic. 3. …designated emergency plan…
4. …items… 5. Assemble…. 6. …fire…
…hot… …. report… …. flammable… …smell…
…. gloves… …buried… …. antiseptic…microorganisms. …disinfectant…

Mind Test 4
1. Isolate the victim from accident area.
2. Provide first aid.
3. Immediately contact teacher or lab assistant
4. Declare the accident compound as prohibited area.

8 Manipulated… Responding….
…heartbeat rate…number of laps…

9

10

Characteristic of Histograms: …unit… …frequency… …any gaps…
Biological Drawings: a) …shaded. b) …clear…unbroken.
11 c) …ruler… d) …titled. e) Detail…
Example Biological Drawings: cilia Macronucleus Contractile vacuole
Sagittal Plane
Frontal Plane
Horizontal Plane

12 a) Section: Cross …longitudinal…

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BIO~SCORE Form4 SCHEME 2020

PG ANSWER
Cross
Longitudinal

13 b) Direction:
Anterior: Front…
Posterior: …back…
Lateral: …centre.
Ventral: …body…
Dorsal: …back…
Superior: …above…
Inferior: …under…
Medial: …middle…

Mind Test 7
Anterior…. Dorsal…posterior

14- Problem statement: Is there the present of reducing sugar in rice porridge, honey and bee sugar?
17 Hypothesis: Honey bees contain reducing sugar but rice porridge and sugar cane do not contain

reducing sugar

Variables:
Manipulated variable: food samples(sugarcane, rice porridge and honey bees)
Responding variable : the present or absent of reducing sugar
Fixed variable : volume of food samples
Procedure:
1. Prepare a test tube containing 2ml of sugar cane.
2. 10 drops of Benedict solution are added on the solution.
3. The test tube is carefully heated by suspending in a hot waterbath at about 40ºC - 50ºC for 5
minutes
4. Any colour change on the mixture is recorded and record it in a table. Brick red precipitate
are formed in the presence of a reducing sugar
5. The experiments were repeated using rice porridge and honey bees sample. The colour change
on Benedict solution was recorded.
6. Record all the changes in the table provided.
7. Repeat the experiment to get more accurate result.

Presentation of data : The present of reducing sugar
Food sample Colour change on the food samples
Sugar cane
Rice Porridge
Honey bees

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BIO~SCORE Form4 SCHEME 2020

CHAPTER 2: CELL BIOLOGY AND CELL ORGANISATION

PG ANSWER
19 ....basic

...iodine

20

1. The cellular components observed are nucleus, cytoplasm and cell wall.
2. Plant cells have fixed shape due to the presence of cell wall.

Plant cells have fixed shape

Mind Test 1:

(b) To allow the methylene blue solution to stain the glycogen granules in the cheek cell

hence giving a clearer vision of the cell under the microscope.

(c) Watering technique.

21 Lisosome: hydrolytic enzyme…food…organelles

Plasma membrane: phospholipids…proteins…selectively permeable…waste

*Ribosome : …RNA....proteins

Nucleus : .. cellular activities.

Centrioles : ...microtubules..spindle fibres...

Golgi apparatus: ..proteins...enzymes.... ...secretory vesicles...

Cytoplasm: .... medium...

*Endoplasmic reticulum: ...ribosomes... ....transport vesicles....

Smooth endoplasmic reticulum: ..lipids..glycerol

Mitochondrion...inner.. ...energy...glucose

22 Vacuole : ..tonoplast.. …cell sap…turgid

Chloroplasts : ....chlorophyll ...photosynthesis.

Cell wall: …cellulose. ..Fully permeable ...mechanical support...

Mind Test 2

(a) P - Tonoplast Q – Rough endoplasmic reticulum R-chromatin

S - Nucleolus T- smooth endoplasmic reticulum

U - Mitochondrion V- Golgi apparatus

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PG ANSWER

(b) Vacuole .... Contains sugars, mineral salts and pigments

Smooth endoplasmic reticulum ... Involves in the synthesis of lipids.

Lysosome .... Contains digestive enzyme

Chloroplast ... Absorbs sunlight energy during photosynthesis

23 Similarities: Both cells contain nucleus, plasma membrane, mitochondrion, ribosome,

endoplasmic reticulum, cytoplasm and golgi apparatus

Differences:

Animal cell Structure Plant cell

Shape Have a fixed shape

Cell wall Have a cell wall

Do not have chloroplast Chloroplast

Vacuoles Contain vacuoles

in the form of glycogen Carbohydrate stored in the form of starch

Centrioles Do not have centrioles

Living Proceses in Unicellular Organisms

5. ...only one ...
6. ...... excrete, respond to stimuli, move, reproduce and grow.

24 Living Processes:

Movement : ...cilia.... its axis. ...pseudopodia.... cytoplasm...

Nutrition : cilia...oral groove... Food vacuoles ...digestive enzymes...cytoplasm...

...phagocytosis .... …food vacuole... ...lysozyme…lysosome... Nutrients...

Respiration: ...simple diffusion…

Respond to stimuli: …. moving away

25 Growth: …cytoplasm.
Excretion: ...diffusion. …osmosis ...contractile vacuole. ...maximum...
...contracts... excess water. …osmoregulation.
Reproduction: …fission mitotically... conjugation. ...spores
Mind Test 4: P1: Amoeba sp. Approaches the food particle using its pseudopodia.

P2: Two pseudopodia extend out and enclose the food particle and forming food

vacuoles. P3 Food vacuole fuses with lysosome and the digestive enzyme inside the

lysosome (lysozyme) digest the food into soluble nutrients which will be absorbed

into the cytoplasm. P4: The feeding mechanism is known as phagocytosis.

26 1. .... types...size, shape and structure.
2. .structure...function…
3. ...various…
Red blood cell: …biconcave disc…oxygen.
Nerve cell: .... nerve impulses.
Muscle cells: .... striped…movement.
Sperm cell: ......tail...chromosomes

27 Mind Test 5
Sieve tube: ... Long cylindrical….
Xylem vessel: ...Long continuous….
Palisade mesophyll cell: … Long cylindrical cell…

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PG ANSWER

Spongy mesophyll cell: ... Cells are loosely arranged…

Guard cell: .... Specialised epidermal cells...

Root hair cell: .... have long projection…

2. ...shape...specified….function. ...tissue.

3. Epithelial...Connective...Muscle...Nerve...

28 Epithelial tissues: …diffusion….gasses. …trachea…cilia…mucus…dust
particle…cilia. …skin…oesophagus. …mucus...
Connective tissue: transport .... oxygen...... pathogen. .... Calcium
phosphate…organs
Adipose tissue: fat…dermis….skin…store energy…insulates…. body.

29 Tendon: …muscles…bones.

Ligament: …bones…bones.

Smooth muscles: …small intestines…involuntary action…peristaltic…

Skeletal muscles: …skeletal bones…contract and relax…voluntary…

Cardiac muscles: …heart...involuntary

Nerve tissues: …neurons…muscles…glands.

Tissue Organisation: …Meristematic tissues…Permanent tissues…

30 Parenchyma tissue: …thin-walled…starch…sugar…support…shape…

Collenchyma tissue: …unevenly thickened…support…

Sclerenchyma tissue: …lignin…mechanical…

Density of Certain Organelles and Functions of Specialised Cells
Sperm cells: …energy…

Flight muscle cells in birds and insects: …contract…

Meristematic cells: …mitosis…

Palisade…mesophyll cells: …chlorophyll…energy…

Pancreatic cells: …enzymes…hormones…

31 Sequence of Levels of Organisation in Multicellular Organisms: ...many...system.
Integument…system: …protect…Endocrine…Lymphatic…Muscular… Urinary….

Respiration

32 Mind Test 6:

(i) (a) Epithelial tissue (b) Sweet gland (c) Erector muscle

(d) Blood vessel (e) Sebum gland (f) Hair

(ii) Epithelial tissue, connective tissue, nerve tissue and muscle tissue.

(iii) a. regulates body temperature b. eliminates urea from the body c. as a

protective layer (from ultraviolet rays and pathogen attack)

(iv) Because skin is built up by various types of tissues which are joined together to

perform specific functions.

33 Organs and Systems in Plants:
Shoot : …photosynthesis…pollination.
Root system: …support

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CHAPTER 3: MOVEMENT OF SUBSTANCES ACROSS PLASMA MEMBRANE
PG ANSWER
35 2. into..... plasma membrane..... exits

Mind test 1
Glucose...Mineral salts....Amino acids......Vitamins.....Urea and Uric acid.
3. plasma membrane .... internal environment
Components of A Plasma Membrane

1. Protein... phospolipid
Mind Test 2

36 Mind Test 3
a) The pore and the carrier proteins which float freely in the phospholipid bilayer form
mosaic pattern which changes constantly.
b) The proteins and the phospholipids are free to move sideways within the membrane
which causes the membrane to have a fluid characteristic.
Mind Test 4
Glycerol....2.
2. extracellular environment, cytoplasm, face
Mind Test 5
Lipid.... small
Shapes...big....amino acid....facillitated diffusion...active

37 Water...flexible...carbohydrates....lipid.....reseptor
Mind Test 6

1. Selectively permeable
3. membrane…visking
4. permeable
38 1. size....polarity...ionic charges
39- Mind Test 7
41 Glucose molecules. The size of glucose is smaller than starch molecules causing the
starch to remain in the visking tubing and cannot move across the tubing membrane.

Experiment osmometer

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PG ANSWER

Discussion
1. increases.... into....osmosis
2. movement....down
3. bigger...smaller
4. decreases
Conclusion Water molecules permeate from an area of high water potential to an area of
low water potential. Hypothesis is accepted.

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PG ANSWER
42 Passive transport – high…. Low….down… dynamic equilibrium

Active transport – low…high…. Against…accumulation….excretion
43 Mind test 8

1.purple
2. higher
3. high concentration….. randomly

(b) (i) wáter
(ii) high… low
(iii) sucrose

44 Mind Test 9
i) X: Distilled water Y : 20% sucrose solution
ii) Osmosis

Mind Test 10
1. The sucrose solution of arm M is increased while the level of sucrose solution of N
arm decreases
2. 0.6% sucrose solution is hypotonic compared to 5% sucrose solution
3. Water molecules diffuse across the selectively permeable membrane by osmosis
4. The movement of water molecules is from 0.6% sucrose solution (N arm) to 5%
sucrose solution (M arm)
5. The concentration of sucrose solution on arm M will decrease while the concentration
of sucrose on arm N will increase
45 (c) (i) ion.. amino acid…. Transport protein…..cannot

(ii) energy… down
(iii) same
(iv) specific
(v) specific site
46 Mind Test 11

Mind Test 12
1. against

2. cellular respiration
3. binds

4. site….changes
5. accumulation… excretion

6. Pump
Mechanism of active transport

a) Binding site…..ATP
b) ADP

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BIO~SCORE Form4 SCHEME 2020

PG ANSWER
47 c) energy… outside

d) original
e) into
f) outside
g) sodium-potassium
h) proton, hydrogen ions

48 Mind Test 13
Similarities.
1. bind with the molecule or ion.

2. specific active sites to transport specific molecules

3. allow substances to pass through them.

Mind Test 14
*Refer to mind test 13 above.
49 Mind Test 15

*Y- yellow

50 Facilitated diffusion....diffuse into.... osmosis...higher... energy
51 Mind test 16

(a)
1. Movement of calcium ions against the concentration gradient of moving from low concentration to high
concentration
2. This process requires energy / ATP
3. Carrier proteins combine with calcium ions and change their shape and allow calcium ions to cross the
carrier protein
4. This results in the accumulation of calcium ions in the cell

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PG ANSWER
(b)
1. The oxygen concentration in the outer environment of Paramecium is higher than in the paramecium
cell. Oxygen will diffuse into the cell paramecium through simple diffusion process.
2. The carbon dioxide concentration in the paramecium cell is higher than in the outer cell environment.
Carbon dioxide will diffuse out of the cell paramecium through via simple diffusion process

52 Mind Test 17
Definition: Osmosis is the process of net movement of water molecules randomly from a region of higher
potential of water to a region of lower potential of water across a selectively permeable membrane.
Movement....low.... high.....high......low
Mind test 18

53 Mind Test 19
Hypotonic – osmosis, osmotic pressure, haemolysis

Isotonic – Size, shape
Hypertonic – out, shrink, crenation

54 Mind Test 20
Hypotonic – turgid, support

Isotonic – same
Hypertonic – osmosis, hypotonic,deplasmolysis

55 Mind Test 21
0.24 mol/dm3

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PG ANSWER
56 Mind test 22

5% sucrose solution
5% sucrose solution was isotonic to the cell sap of plant cell
Water molecules will diffuse into and out of cells at the same rate
Plant cell become flaccid.

30% sucrose solution
30% sucrose solution was hypertonic to cell sap of plant cell
Water diffuse out of the cell through osmosis
Cells become flasid and plasmólisis occurs
The cytoplasm and vacuoles shrinks
The plasma membrane is pulled away from the cell wall

Sucrose solution 0.5%
0.5% sucrose solution was hypotonic to the cell sap of plant cell
Water molecules diffuse into the cell sap by osmosis
// cell becomes turgid and créate turgor pressure
The plasma membrane is pushed against to the cell Wall

57- Mind Test 23 Final length of potatoes strip
61 (a) (cm)

Concentration of sukrose 5.6
solution (M) 5.3
4.6
0.2 4.2
0.4
0.6
0.8

(b)(i)
Observation 1
If the concentration of sucrose solution is 0.2M, the lenght of potatoes strip is 5.6 cm.
Observation 2
If the concentration of sucrose solution is 0.8M, the lenght of potatoes strip is 4.2 cm

(ii) Inference 1

0.2M sucrose solution is hipotonic to the sap sel of potatoes, wáter diffuse into the potatoes strip

by osmosis

Inference 2 :

0.8M sucrose solution is hipertonic to the sap sel of potatoes, wáter diffuse out the potatoes strip

by osmosis

(c)

Variables Method to handle
Manipulated variable : Use different concentration of sucrose solution
Concentration of sucrose solution (0.2M,0.4M,0.6M dan 0.8M )

Responding variable : Measure and record the length of potatoes by using
Final length of potato strip /percentage ruler//
different in length of potato strip Calculate the percentage difference in length of
potatoes strip by using formula
Final lenght – initial length x 100

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PG ANSWER

Initial length
Constant variable :
Volume of sucrose solution/initial length Fix the initial length of potatoes strip at 5.0 cm
of potatoes strip

(d) Hypothesis :

As the sucrose concentration increases, the length of potatoes strip decreases

(e) (i)

Concentration of sucrose (M) 0.2 0.4 0.6 0.8

Initial length of potatoes strip (cm) 5.0 5.0 5.0 5.0

Final length of potatoes strip (cm) 5.6 5.3 4.6 4.2

Percentage different in length (%) 12.0 6.0 -8.0 -16.0

62 (f) - the concentration of sucrose solution that is isotonic to the cell sap of potatoes is 0.5M

- At this point, the concentration of sucrose does not cause any change in the length of the

potato strips

- The amount of water diffuse in and out of the cell is the same rate

(g) –the process of wáter molecule diffuse into or diffuse out of the potatoes strip

through the selectively permeable membrane

- it is shown by the final length of potatoes strip / different change in length of potatoes strip

- affected by the concentration of sucrose solution

(h) – the final length of potatoes strip is more than 5.6cm.

- distilled wáter is hipotonic solution

- more wáter is diffuse into the vacoule by osmosis

(i)

Type of solution Sodium chloride solution
Hipotonic 0.10 %
Isotonic 0.89 %
Hypertonic 3.00%

63 Wilting phenomenon : hypertonic… diffuses… flaccid… plasmolysis… wilted
1. wáter loss
2. isotonic… 0.85-0.90 %

64 1. vesicle…surrounded….hydrophobic región…hydrophilic región..protect…gastric juice…targeted
2. wáter loss…dehydration..same…into…. perspiration

65 Purify…salt removal…pressure… filtered out…. pure

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PAGE CHAPTER 4 : CHEMICAL COMPOSITIONS IN THE CELL

67 ANSWER

1. PROPERTIES OF WATER MOLECULE
(a)
a) ...inorganic....... hydrogen.......oxygen
b) ....charge ……… ………. negative........ universal
c) ....soluble …….. biochemical
(b) (i) …… attach
ERATA:
i) Adhesion force is the force that causes water molecules to attach to another surface
(ii) Cohessive force is the force that causes water molecules to attach to each other.

(iii) capillary action ………. narrow
2. Specific Heat Capacity Of Water.

i) 4200
ii) thermal energy ……….. one
iii) …. absorbs ……. temperature.
Mind Test 1:
1. ….. natural …………. molecules
2. ……. Biochemical
3. ………….osmotic
4. ….. solvent ………………. synovial fluid
5. …. transport …………. Urea
Mind Test 2
1. Maintains the vitality of plant cells to support plant cells.

2. High affinity, in which water molecules attach to each other and cause water
to rise continuously in vascular tissue

Elements In Carbohydrates

1. ......hydrogen........ oxygen

68 Type of Carbohydrates

Polysaccharides

.............................. glycogen

.................plant

Animal........

..................cell wall

Disaccharides

...................... maltose

.........................sukrose

.............................laktose

Explanation...........

1. Sweet 1.............tasteless

2.

69 3. ........Water
4. .......sukrose

Mind Test 3:

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PAGE ANSWER BIO~SCORE Form4 SCHEME 2020
Q [Starch]
P[Glycogen] R [Cellulose]

70 Mind Test 4:
a)i) L: Hydrolysis......................... M: Condensation
(ii) 3 molecules of water are released
b) Process X: Hydrolysis ........................ Process Y: Condensation

71 ........... yellow.
The starch is hydrolyzed by amylase to maltose, maltose are small molecules and can diffuse
out of the tube visking into distilled water. The starch molecule is large and not can seep out
of the visking tube..
Importance of Carbohydrates
3.................cellulose
Mind Test 6
Grapes because it contains glucose. Molecular glucose is simple and does not need to be
digested. It is then absorbed and transported to cells to oxidize for energy production. As for
bread, it needs to be cooked first. So it provides energy even though it contains high
carbohydrates. Sausage is a protein, it's a low energy source compared to carbohydrates.

72 4.3 Protein
1. ..................nitrogen
3........................amino acids

Mind Test 7
Answer: D

73 Mind Test 8-
a) Consists of a linear polypeptide chain with amino acid sequences or peptides.
b) Protein X undergoes hydrolysis, binding to polypeptides is expressed to form
peptides.

Mind Test 9:
a) Eggs, chicken and soy milk contain high protein. This protein can help to recover
from surgery with new tissue repair. Proteins can build antibodies to fight or kill
pathogens that may enter the mother's body through injury.

74 4.4 LIPIDS
2. .........Carbon, hydrogen......
3. .....................glyserol........ fatty acids.......

Type of Lipids
1. ............ phospholipids
(I) ......... triglycerides.

75 vi)……. saturated fat …………….. unsaturated fat.
b)…………. water-resistant
d) (ii)…………… Cholesterol

Mind Test 10:
1………. three fatty acids…… triglycerides.
2………condensation

3……………..hydrolisis

Type of fat Saturated fat Unsaturated fat

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PAGE ANSWER

1.Example Butter Oil

2. The presence of a Has double bond
double bond between Can
carbon atoms in fatty
acids Liquid
Low
No double bond

3. Ability to form

chemical bonds with Cannot

additional hydrogen

atoms

4.Condition at room Solid
temperature

5. Cholesterol content High

76 Importance of Lipids in Cells
1..........energy
3.........insulator
4 Solvents

Mind Test 11:
Fat acts as a heat insulator to prevent it’s conduction of heat from the body to the
aquatic environment and thus maintaining its body temperature.

4.5 Nucleic Acids
Elements in nucleic acids.
1………………. polymer ……. nucleotide
2……………..nitrogen
Nucleotide Structure
1…………… Phosphate ….. pentose ……. nitrogen bases.
2.... ribose
3................adenine...........urasil
4.........RNA
77 i) ................two....................helixs.....

ii) ..... histone
iii) …….genetic.....
iv) .... chloroplasts
a) (i).......... single

(ii) .................... tiamine
(iii)......DNA.... protein
iv) ............. messenger................ synthesis.
v)...................... ribosome.
78 Importance of Nucleic Acids in Cells
1...............characteristic....................
2......... genetic code................... polypeptides.............
3....... three...........
Mind Test 13
Answer
B
79 Chromosome Formation of DNA and Protein

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PAGE ANSWER

1.......... polynucleotide........not...
2..................... nucleosomes

Mind Test 14

I.
II.

CHAPTER 5 : METABOLISM AND ENZYME
PG ANSWER

81 Metabolism
1. ......chemical reactions........catabolism........anabolism.
2. ......ATP..........carbohydrate.......

3.

Catabolism Anabolism
a) .........breaking down.......complex....... b) ........absorbs energy.......
b) ........releases energy.......

Mind Test 1
1. A

2. K

82 Enzymes and Its Necessity In Metabolism
1. .........protein…….accelerate……
2. ………metabolism……….

3. ……..substrate…….active sites……..enzyme-substrate complexes.

Naming of Enzymes Catabolism Reaction / Hydrolisis
Substrate Glucose + Galactose

Lactose sucrase

Sucrose Glucose + Glucose
Maltose Glucose + Glucose

Cellulose Glycerol + Fatty acid

Lipid Maltose

Starch
(amylose)

83 General Properties of Enzymes
2. ……..reversible.
3. …….not change……

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4. …….small…..

5. …….active sites…….specific.

6………inhibitors…….cyanide..

Involvement of Organelles in Production of Intracellular and Extracellular Enzyme
EXTRACELLULAR
 Secreted out…..
 ……pepsin, tripsin,…...
84 Extracellular Enzyme Production
Nucleus
 .........RNA.......ribosome.

Ribosome and RER
 ........lumen.......
 .......transport vesicle.

Transport vesicles
 .......Golgi apparatus.....

Golgi apparatus
 .......modified.......
 .......secretory vesicle.

Secretory vesicles
 .......plasma membrane.......extracellular.......

85 The Mechanism Of Enzyme Reaction With ’Lock And Key’ Hypothesis
1. .......polypeptide.......dimensional...
2. ......active sites.....
3. ......complements......specific.
4. .......lock ....key.....
Mind Test 2
1. Sucrase / enzyme is a lock while sucrose / substrate is a key
2. Sucrose bind to active site of sucrase ( lock binds with key )
3. An unstable sucrose-sucrase complex is formed
4. Hyrdolisis reaction occurs, releases product which are fructose and sucrose from
its active site.
5. Sucrase does not change at the end of the reaction
6. Enzyme can be reused to react on other substrates.

86 Activation energy
1. Activation…..bonds…..
2. ……require high…..
3. ……reduce……biochemical…..

Mind Test 3
1. Substrate combine with enzyme to form enzyme-substrate complex
2. To lower the activation energy
3. Reactions on the substrate can occur more easily

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4. Speed up the rate of reactions
5. Product can be produced easily
The Mechanism of Enzyme Action with the change in Factors

b) pH
d) Substrate concentration

87 a) Temperature
 …..low….
 ….kinetic energy…..frequency if effective collision…..
 …..eccelerated.
 …..2 times……optimum temperature.
 …...maximum……
 …denaturation…..hydrogen…..destroyed.

Mind Test 4
A: 370C
B: 450C
C: 650C

88 b) pH
 …..pH….
 …..2.0…..8.5.
 …..charges…..active site……reduce……
 ……function.
 Extreme………denatured.

Mind Test 5
√ - Keep at 40 ºC temperature

89 c) Enzyme concentration
 ………concentration…….more…….maximum……
 ………limiting………excess of active sites……..
 ……..added.
 ……..doubled…….

d) Substrate concentration
 ……..substrate……..
 ……..increase……..constant.
 ……..occupied…….saturated.
 ……..enzyme concentration………

Mind Test 6
90 √ - Enzyme concentration is a limiting factor at level P

√ - As substrate concentration increases from Q to R, more molecules bind to the
active site of the enzyme

Mind Test 7
P1: Molekul P is an inhibitor
P2: Rate of reaction by enzyme T decrease/ low
P3: Molecule P competes with substrate S to bind to the active site
P4: Substrate S cannot bind with enzyme active site
P5: Enzyme-substrate complex cannot be formed

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P6: Less product is produced

91 Study the Effect of Temperature on the activity of Amylase Enzyme
......effect.....temperature......amylase enzyme.
S1: .......amylase solution.....
S2: .......boiling tube.......
S5: .......different temperatures........

92 Water temperature/ The end observation of Time taken for iodine to
0C experiment remain yellow (minutes)

5 34

28 14
37 10
45 12
55 34
b)i) Observation

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1. The time taken for iodine to remain yellow at 37 0C was 10 minutes.
2. The time taken for iodine to remain yellow at 55 0C was 34 minutes.

93- ii) Inferences

95
1. At 370C, starch hydrolysis takes the fastest time at optimum temperature, more product /

maltose produced
2. At 550C, starch hydrolysis takes the slowest time at low temperature, less product / maltose
produced

.

c)

Variables How to recognize variables

Temperature Using different temperatures such as 50C, 280C, 370C, 450C
dan 550C.

Time taken for iodine to Record (jot down) time taken for iodine to stay yellow using
remain yellow stopwatch

Volume of amylase Keep amylase volume constant at 2ml

d) Hypothesis
The higher the temperature , the higher the rate of starch hidrolysis / fast time taken (to
remain yellow).

e)i)

Temperature Time taken for iodine to Amylase rate activity
( 0C ) remain yellow (minutes) ( minute -1 )
5 0.03
34
28 0.07
14
37 0.10
45 10 0.08
12
55 0.03
34

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96 f)
 The higher the temperature , the higher the rate of amylase activity // At 370C, the rate of
amylase activity is maximum.
 More chances of colision between amylase and starch.

 More product / maltose will be formed.

g)
 Starch hydrolysis is the broken down of /digested of starch molecule by amylase,
 It is shown by the time for iodine to remain yellow .
 Starch hydrolysis is affected by difference temperature.

h)
 Time taken for iodine to remain yellow more than 34 minute.
 65oC is a higher temperature and cause amylase enzyme denatured.
 Therefore, starch hydrolisis become slow / product (maltosa) formed become less.

i)

Materials Apparatus

Starch , amylase , buffer solution /pH, White tiles, thermometer, water boil
iodine

96 Applications of Enzymes in Daily Life
1. .......bacteria......
2. ......inert......insoluble.......separated.......

Enzyme and it uses

Amylase

(a) …....starch….
(b) ……starch…..

Selulase
(a) …….cellulose
(b)

i. ........cellulose......
ii. ……breaks…...
iii. Jelly……
Protease
(a) …….detergent……
 ……..digest…..
 Speed up……
 ……..optimum……
(b) ……soften……
(c) Soften.......remove the fish’s skin.

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97 Lipase

(a) Cheese......

Trypsin :

…….skin.

Ligninase:

.......lignin.......
Lactase:
……...lactose …… lactic acid……
Pectinase
……..pectin bonds…….

CHAPTER 6: CELL DIVISION

PG ANSWER

99 Mind Test 1

Mitosis Reproductive

Diploid chromosomal number Meiosis

Haploid chromosomal number

1. Paternal, female
2. Homologous, similar

3. Chromatin

100 1. √
2. √

3. √
4. Pairs, X

5. √
6. X

7. √

Phases in a Cell Cycle
1. duplication, daughter
2. Interphase, mitosis, M phase

101 Interphase: Plasma
Membran
G1 Phase
e Nucleus
a)The cell increases in size with a bigger
nucleus and the chromosomes are in the Nuclear
form of long and fine threads membrane
(chromatin).
Chromati
b) Cellular components (mitochondria and Centrionles
endoplasmic reticulum) are synthesized.

c)Proteins are synthesized for the use inside
the cell.

d) The cell metabolism rate is high.

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102 Mind Test 3:

PROPHASE
-chromatin

-chromatids
-centromere

-disintegrates
-centrioles, spindle

METAPHASE
-centrioles
-centromeres, centre, equatorial

-attached
-divide

ANAPHASE
-shorten, contract, apposite poles

-poles

103 TELOPHASE
-chromosome

-complete, identical
-chromatin

-disappear
-nuclear membrane

Mind Test 4
a)

i. Stage IV – Stage II –Stage III – Stage I
ii.

Stage Name of stage Stage Name of stage
I Prophase III Anaphase
II Metaphase IV Telophase

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104 b) Metaphase: chromosome are arranged in equatorial plane
i. Anaphase: sister chromatids apart to the opposite poles
ii.
Telophase: sister chromatids arrived at the opposite poles of the cell, identical
iii. chromosomes is formed.

c)

Stages PROPHASE ANAPHASE METAPHASE TELOPHASE

Structure Not divided Not divided Not divided Divided
inside the cell Randomly Moves to the arranged at the at opposite
Cytoplasm arranged opposite poles equatorial plane poles
Chromosome Present Absent
Absent Present Present
Spindle fibre Absent Absent
Nuclear Absent
membrane

d) Identify the different stages of mitosis in the following diagram

105 Mind Test 5
Animal cells have centrioles to produce the spindle fibers during mitosis while plant
cells do not have centrioles however they could still form the spindle fibers
Animal Cells : shrinks, membrane

106 1. The Necessity of Mitosis
a) New cells, embryo, growth
b) dead
c) damaged, injury, regeneration
d) reproducing, unicellular, binary fission
2. Mitosis is applied in many fileds like:
a) Medicine: stem cell, damaged

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107 Mind Test 6
a) i- tissue culture
ii- can produce good trait banana trees in large numbers and shorter period of time.
b) i- No, ii – Yes, iii- No, iv-Yes
c)

Bone Brain Ovum Skin Ovarium Sperm
marrow X
epithelium X

108 Definition of Meiosis Species Chromosome
number
1. Gametes Cats
Prawns 2n n
Types of Cells which carry out Meiosis Humans
38 19
1. Somatic, testes, ovary 254 127
2. Primary spermatocyte, primary oocyte 46 23
109 The Necessity of Meiosis
1. Gametogenesis, diploid, maintained
2. Genetic, genetic

Phase in Meiosis
1. Meiosis I, replication
2. Diploid, two, 4 cells haploid

Mind Test 7 Prophase I Metaphase 1 Anaphase I
Meiosis I Telophase I Metaphase II Anaphase II

Meiosis II Prophase II
Telofasa II

110 Mind Test 8
PROPHASE I
 Shorten, thicken
 centromere
 Bivalent
 Homologous
 Tetrad
 Crossing over, non-sister
 chiasma
 combination
 Nucleus
 centrioles

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111 METAPHASE I
 Homologous
 Did not

ANAPHASE I
 contract, pull
 Separate
 centromere
 four, two

TELOPHASE I
 arrive
 nuclear membrane

112 MEIOSIS II –Mind Test 9

PROPHASE II
 Spindle fibres

METAPHASE II
 Random

ANAPHASE II
 Separate
 Chromosomes

TELOPHASE II
 Cytokinesis
 four
 half
 Gametes

113 Mind Test 10
Similarities

2. interphase
3. cytokinesis
4. DNA undergoes replication only once in both mitosis and meiosis

Mitosis Aspect of differences Meiosis
Somatic cell Place of events
Ovarian/ testicular/
Produces new cells for Function reproductive cells
growth and repair of Produces gametes
damaged tissue.
Once Number of cell 2 times
division Occur during Prophase I
Does not occur
Does not occur Synapsis for occur
2 sets of daughter homologous 4 sets of daughter
chromosomes
Crossing over

Number of daughter

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Diploid (2n) number of cells at the end of cell Haploid (n), half number
chromosomes division of chromosomes from the
parent cell.
Similar to daughter cells Number of non identical to parent cell
chromosomes in each
No variation Have variation genetic
daughter cell

Genetic composition
of daughter cells

Genetic variation

114 Mind Test 11
a) R: mitosis S: meiosis
b) 3 reasons :
i. R produced 2 daughter cells, while S produced 4 daughter cells
ii. The number of chromosomes in cell S, become half.
iii. Cell did not divide once only for R, but for S, parent cells divide 2 times
iv. Content of genetic daughter cells is the same for R, but different for S.

Mind Test 12
a)

114 b)

115 c)1. Mitosis produced 2 dughter cells but meiosis produced 4 daughter cells.
2. Mitosis produced the same genetic of daughter cells but meiosis produced different
daughter cells from parent cells.

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d) 1. In Meiosis, the number of chromosomes is half from parent cells in daughter

cells (garmet)
2. Every garmet cells only carry one set chromosome from parent cells,
haploid (n)
3. When two garmets combine during fertilization, the number of haploid remained and
zygote cells is formed.
e) 1. Zygote cell that are formed, contained 47 of chromosomes.
2. Baby formed might have Down’s syndrome / Turner’s syndrom
Mind Test 13

116 The Effects of Abnormal Mitosis on Human Health
2. uncontrollably, formation
3. benign, through operation
4. radiation, carcinogen, genetics
5. continuously, tumour, spread, destroying, normal cells
6. blood, lymphatic, the formation
7. organ damage, death

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117 Mind Test 14

a) Cell cancer is formed when normal cell undergo uncontrolled mitosis. Cell cancer
will uncontrolly divided and formed tumor, which is abnormal cell. Cell cancer will
spread to other cell from blood vessel or lymph vessel.

b) 1. Avoid from exposed to radioactive radiation
2. Practice healthy lifestyle / avoid pressure in life / avoid smoking and drugs

c) Radiation from radioactive will slow down the growth of cancer cells. Uncontrolled
division of cell and formation of chromosomes will stop due to radiation from
radioactive. When the cells unable to divide, the cells will die and dissaper from
body.

d) Eribulin will stop the function of spindle fibers to pull chromosomes to opposite poles.
When chromosomes unable to move, the cells will stop from division. Uncontrolled
mitosis can be stopped.

118 Mind Test 15
a) Pregnant woman in older age (40 years and above), have reproductive cells that
have a long cease of meiosis. These cells are highly susceptible to complications
during chromosome movement in meiosis. This can lead to the formation of non-
haploid gametes. Gametes with non-haploid chromosomes number can cause
Down’s syndrome baby if they fuse with a normal gamete from the opposite parent.

b) X-rays can disrupt chromosome movement during mitosis. Cells whose chromosomes
are not separated properly cause abnormal organ growth. The baby in the womb
might be deformed.

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CHAPTER 7 – CELLULAR RESPIRATION

MS ANSWERS

1. ..anabolisme... catabolisme… releases
120 2. .. energy

The main substrate in energy production

1. .. oxidation … energy
2. ..glucose

3. … carbohydrates
..photosynthesis … organic … glucose… chlorophyll

(i) Aerobic (ii) Anaerobic (iii) Fermentation
Respiration Respiration ..complete… limited oxygen …..

..glucose.. oxygen. .. breakdown …energy without

121 1. .. breakdown of glucose...oxygen... chemical... glucose.. carbon dioxide ..adenosine triphosphate
(a) .. cytoplasm...glycolysis..glucose..enzymes..one..2..pyruvate
(b)... oxydation.. carbon dioxide.. energy

First stage Second stage
OXYDATION OF PYRUVATE
GLIKOLISIS Carbon dioxide + Water + Energy

Glucose Pyruvate

3. .. energy

Mind Test 1:

i. Glucose..6O2..2898kJ
.. contraction.. relaxation... respiratory.. providing energy.. increase.. generate.. contraction

122 Experiment to Study Aerobic Respiration Process (page and )

- Problem Do living organisms carry out aerobic respiration?
124 Statement

Hypothesis Living organisms use oxygen and release carbon dioxide during aerobic

respiration.

Variables Presence of living organisms
Procedure Increase in the level of coloured liquid
Initial level of coloured liquid

1. Prepare the apparatus as shown in the figure above.
2. Prepare two boiling tubes labelled A and B.
3. Fill both boiling tubes with 10 g soda lime.
4. Put the wire gauze in the middle of boiling tube A.
5. Put a cockroach on the wire gauze in boiling tube A while

the boiling tube B is left empty.
6. Wipe all connections of the apparatus with petroleum jelly.
7. Close the screw clip and mark the height of the initial level

of the coloured liquid in the capillary tube for both boiling
tubes.
8. Leave the apparatus for an hour.
9. Measure and record the final height of the coloured liquid in

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MS ANSWERS
Discussion
both capillary tubes after an hour with a ruler.
10. Record your observations in the following table.

1. .. control ... respiration
2. .. absorb carbon dioxide
3. i... increase... aerobic respiration... oxygen

ii...carbon dioxide... soda lime.
4. .. lower... less energy... move ... Rate of respiration... oxygen

Conclusion Living organisms use oxygen and release carbon dioxide during aerobic

respiration.

125 1. .. complete.. limited oxygen.. without

2. .. second … oxygen.

3. ..alcohol.. yeast.. plant .. human muscle cell

(a) .. complete

(i) .. oxygen.. alcohol fermentation... alcohol fermentation..ethanol..zymase... Carbon dioxide

126 (ii) .. germinate.. oxygen.. alcohol fermentation.. energy...ethanol.. alcohol dehydrogenase..
carbon dioxide

Mind Test 2:
1. .. alcohol fermentation.
2. ..glucose
3. ..ethanol... energy
4. .. limited oxygen... without
5. .. tolerance..ethanol
7. .. oxygen... surface
8. .. air spaces
9. .. oxygen

127 (b)(i) Yogurt.. fermentation... lactic acid fermentation...lactose... yogurt... sour
(ii) .. exceeds.. oxygen debt.... partial.. two..150.. lactic acid.. muscle cramps.. fatigue.
... more..

128 Oxygen debt... Oxydation.. oxygen debt

Mind Test 3:
1. Both involve the breakdown of glucose to produce energy.
2. Both occur without the presence of oxygen.
3. Both produce 2 ATP molecules

Sel otot Sel yis

Undergo alcohol fermentation

Etanol is produced

No carbon dioxide is produced

Mind Test 4:

i. a. … 2 Lactic Acid + …

b. … 2 Carbon dioxide + 2 Ethanol + …

ii. a. Lactic Acid Fermentation

Breakdown of glucose without the presence of oxygen to produce ATP energy. 2
2
molecules of lactic acid produced.

b. Alcohol fermentation

Breakdown of glucose without the presence of oxygen to produce ATP energy.

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MS ANSWERS
molecules of carbon dioxide and ethanol produced.

129 Mind Test 5:

i. To eliminate dissolved oxygen in the solution.
ii. To prevent oxygen from atmosphere dissolved into the solution.
iii. P1. .. cloudy … carbon dioxide produced by yeast fermentation... (…red colour turns yellow)

P2. .. does not change due to lack of yeast.
P3. .. cloudy … bubbly.
P4… alcohol fermentation … carbon dioxide gas (bubbles) … ethanol.
P5… increase in temperature indicating heat released …(ATP produced)

P6... smell of ethanol… detected..
P7... still clear … no bubbles.
P8.. does not increase … cannot be detected.. .. acts as a control.

Glukosa Zimase Tenaga + Etanol + Karbon dioksida
(C6H12O6) Yis
(210 kJ //2ATP) + (2 C2H5OH) + 2CO2

130 Mind Test 6:
1. Both start in the cytoplasm.

2. Both produce chemical energy in the form of ATP.

3. Both involve the breakdown of glucose and coversion to chemical energy.
4. Both begin with glycolysis where glucose breakdown into pyruvate.

5. Both occur in animals, plants, yeast and bacteria.

Respirasi aerob Fermentasi

1. .. oxygen Occur without the presence of oxygen or in condition where
oxygen is limited.
2. .. cytoplasm … mitochondria
3. .. glucose .. cytoplasm

4. One … 2898kJ Incomplete glucose oxidation

5. energy, carbon dioxide and … glucose … 210kJ (alcohol fermentation)) … 150kJ (lactic
water. acid fermentation)

6. … + Oxygen - Energy and lactic acid in human/ animals (lactic acid
Carbon dioxide + Water + …
fermentation)

- Energy, ethanol and carbon dioxide in yeast/plant

(alcohol fermentation)

(Lactic acid fermentation)

Glucose …+…

(Alcohol fermentation)
… carbon dioxide + Ethanol + …

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CHAPTER 8: RESPIRATORY SYSTEM IN HUMAN AND ANIMAL

PG ANSWER
132 Mind Test 1:

(a)......gaseous exchange....respiratory surface.....simple diffusion .

(b)....unicellular...large total surface area to volume ratio.....multicellular ....small...alveolus .

(c)

(i) Small in size and numerous in number, thus total surface area is bigger to increase

the rate of gaseous exchange.

(ii) Thin wall as thin as one cell to ease gaseous diffusion.

(iii) Wall is always moist to allow gas to dissolve in them.

(iv) Numerous network of blood capillary (except insects) to transport respiratory gases.

133 Respiratory Structure of Insects and Its Adaptations:

(a) ……tracheal system. ………tracheoles

(b) ......spiracle......abdomen (c) ......chitin.....collapse

2. Adaptations in Respiratory Structure:
(a) ......tracheoles....
(b) ......body or muscle cells......
(c) ......moist...thin......dissolve.....

4. Diagram of cockroach tracheal structure:
...chitin......spiracle....tracheoles

Mind Test 2:
(a) Grasshoppers are active insects. Air sacs contain more oxygen supply to oxidize glucose
producing more energy for the contraction of flight muscles.
(b)

One cell thick √ Big total surface area √

Always moist √ Has a network of many blood
capillaries

134 Respiratory Structure of Fish and Its Adaptations:

1. (a)...gills... (b) .....filament.....archs

2. (a) ......lamellae......increase total surface area....
(b) .......increase rate of gaseous diffusion.
(c) .......transport respiratory gases efficiently.

Mind Test 3:
The fish will die, because its gill filaments will stick to each other, reducing the total surface area
for gaseous exchange. The fish will experience lack of oxygen.

135 Respiratory Structure of Frogs and Its Adaptations:
(a) ......lungs....moist skin.

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2. (a) (ii) .........diffusion. (iii) ......dissolve. (iv) ......transport......

(b) (i) ......permeable....... (ii) ........mucus........dissolve. (iii) ........skin...transport......

Mind Test 4:
There are mucus glands to ease oxygen gas in the air to dissolve and diffuse into the blood
capillaries on the skin for gaseous exchange.

136 Respiratory Structure of Humans and Its Adaptations:
1. (a) ......alveolus. .......thoracic cavity......rib cage.
(b) Labels:Trachea ... Bronchus ... Bronchioles ... Alveolus

Mind Test 5:
Bronchus....bronchiole....nasal cavity....mouth cavity.....trachea....intercostal muscles...diaphragm

2. (i) Small in size......large in number........ (ii) Moist wall......... (iii) Thin wall ...... one cell

thick ....... (iv) .......blood capillaries .......

137 Comparisons and Contrasts of Respiratory Structures in Humans and Animals

Similarities:

(a)..moist (b).....diffusion (c)....big (d)....transport...

Differences:

Humans Insects Fish Frogs

..tracheole.. ...blood capillary.

...moist ...environment ...moistened..

Mind Test 6:

Alveolus √

138 Mind Test 7: Insect
Differences 2. ……..spiracle.
3. …..valves…..
Human 4. ……..chitin.

2. ……nostril.

4. ……cartilage.
5. …….alveolus.

Mind Test 8: √
shortness of breath

139 Breathing Mechanisms of Humans and Insects
3. (a)INSECTS:
Inhaling: (i) Abdominal..... Exhaling: (i) .......contract. (ii).......decreases, ........increases.

140 3. (b) FROGS:
Inhaling:
P: ..........close.........

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Q: .........enters.........
R: ..............raised, ........
S: ..................inflate. .......
Exhaling:T: ................lungs...........
141 3. (c)FISH:
Inhale
1. .........lowered.......closed.....
2. ............increases, ..... decreases.
3. .............enter........
Exhale
2. ........pressure......
3. ---------respiratory gases.......
142 3. (d) HUMANS:
(i) ........diaphragm.
(ii) .........external intercostal muscles....internal intercostal muscles.

Exhalation
1. External intercostal muscle relax, internal intercostal muscles contract.

2. Rib cage moves downwards and inwards.

3. Diaphragm muscles expand and diaphragm curves upwards.

4. Volume in thoracic cavity decreases, increasing the internal pressure in

thoracic cavity (higher than the atmospheric pressure).

5. Causes air to be forced out of the lungs.

143 Mind Test 9:
1.Inhalation 2.external intercostal muscle contracts , internal intercostal muscle relaxes
3.rib cage moves upwards and outwards 4.diaphrag muscle contract,diaphragm become flat
5.volume:increase, Pressure:decrease 6.air move in / entering lungs

Mind Test 10: √
Pressure in lungs increase √
Volume of thoracic cavity decreases

Comparing and Contrasting the Breathing Mechanisms of Humans and Animals
Similarities:
......inhale.....exhale.....volume......pressure

144 Differences:
Humans: .....nostrils.......intercostal...diaphragm.....rib cage......
Insects: .......spiracle.....abdominal
Fish: Water.....mouth....mouth....operculum.......floor......opercular mucles.......
Frogs: .......nostrils.........
Mind Test 11:
1. Heat from cigarette smoke increases the temperature in respiratory tract.

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2. Tracheal walls and alveolus become dry.

3. Cilia at the trachea will be destroyed.

4. Foreign particles/toxins cannot be filtered and causes more mucus to be produced.

5. Coughs will occur as a reflex action to eliminate mucus and pathogens.

6. Respiratory tract / trachea and bronchi will narrow, causing bronchitis.

7. Structure of alveolus is destroyed, causing emphysema.

8. Total surface area for gaseous exchange decreases, causing difficulty to breathe.
145 External and Internal Respiration

1. .........alveolus....
2. ......body tissues,.....
3. .......simple diffusion. ......partial pressure.......

At X :(a) .........higher......blood capillaries.
(b) .............low.........alveolus.
(c) ..........high........ pCO2.
(d) ………..oxyhaemoglobin………blood…….

146 At Y:(a) ........low........dissociates.......diffuses.......
(b) ......cellular respiration.......
(c) ......carbon dioxide. .......higher........
(d) (i) ......bicarbonate......
(ii) ............carbaminohaemoglobin
(iii).......... plasma

Mind Test 12:
(a) The main method is carbon dioxide is transported in the form of bicarbonate ions
(HCO3-). Carbon dioxide diffuses into erythrocytes and combines with water to form
carbonic acid (H2CO3). Carbonic acid is not stable, thus it dissociates into bicarbonate
and hydrogen ions (H+). This reaction is catalysed by carbonic anhydrase enzyme.

(b) ……carbonic acid……hydrogen ion + bicarbonate ion

147 Mind Test 13:
1. Alveolus
2. pO2 in alveolus is higher than in blood capillaries
3. from alveolus into blood capillaries
4. pCO2 in alveolus is lower than in blood capillaries
5. from blood capillaries into alveolus
6. after the exchange, oxygen in blood combines with haemoglobin pigments in
erythrocytes to form oxyhaemoglobin and is transported by blood to the heart to be
pumped to the entire body as oxygenated blood

CoPD towards the Humans Respiratory System
1. ........asthma, chronic bronchitis....emphysema.
148 (a).........trachea.......inflammation.......
(b) ........coughing..........mucus.

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(c) ......alveolus.......walls......decreases. ..........difficulty.

149 Mind Test 14:
Carbon monoxide: ......carboxyhaemoglobin........
3,4 Benzo-(α)-pyrene: …..carcinogenic…….
Nicotine: ……cancer……constricts……
Heat and Dryness: Corrodes the lungs……..
Nitrogen dioxide: …….acidic…….

CHAPTER 9 : NUTRITION AND HUMAN DIGESTIVE SYSTEM

PG ANSWER
151 1 ....alimentary canal...

2....salivary gland., gastric...,.... intestinal ...

Mind Test 1:oesophagus stomach duodenum ileum large intestine

Digestion:

.....breaking ..complex....absorbed......cells

....surface area....enzyme..

Physical..

 Mechanical....surface area..  ...enzyme....water.

 ..mouth...  ....bloodstream.

 .....peristalsis..

152 Carbohydrate Digestion in the Mouth:
1. ...stimulates....
2. (a)Amylase.... (b)Mucus..... (c)Antimikrobial.... (d)Buffer....
3. .....food bolus...
4....mucus
5. ... rhythmic..... sphincter...

Protein Digestion in the Stomach:
1. ......chief..,parietal..goblet..
153 2.....gastric juices 4....pepsin... 5 (a)....acidic medium (d)...bacteria
7....chyme..pyloric....
Mind Test 2:

mashed potatoes ileum
Roasted chicken Stomach
Cheese mouth

Mind Test 3:
1. Hydrochloric acids inside gastric juices from stomach will flow back into oesophagus.
2. Individual will experience heart burn, nausea and regurgitation, difficulty to swallow

and excessive belching.
3. Known as acid reflux or ‘Gastroesophageal Reflux Disease’ (GERD).

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PG ANSWER
Carbohydrate, protein & lipid Digestion in small intestine:

2.......lipid 3. ......pancreas 4. ......liver,............bile duct

154 Liver:
(c)(i)....alkaline.. (ii) ........chyme... (iii) Emulsify lipid....
Duodenum:
(a)Lipid.....fatty acid...glycerol (b)..amylase....maltose (c).....polypeptides
Pancreas
(a)....pancreatic juice....pancreatic duct.... (b) ... trypsin
Ileum:.
(a)....intestinal glands......juice
 sucrose,
 maltase.....maltose....glucose
 erepsin....peptides into amino acid
 lipid....lipid ..into fatty acid and glycerol
 lactase.....lactose into glucose and galactose

155 Uji Minda 4:
Mouth: salivary gland: saliva,salivary amylase,...maltose
Stomach: gastric gland: gastric juice , pepsin,....polypeptides @ pepton
Duodenum: Liver: Bile,emulsify......
Pancreas: pancreatic juice,.....glycerol
pancraetic amylase
ileum: intestinal juices : sucrase: ....fructose....
maltase: ....glucose + glucose
Erepsin: ....amino acid
Lactase ....galactose

156 Mind Test 5:
1......protein...lipid 2.....pepsin....polypeptides @ pepton
3......trypsin....pancreas...alkaline 4....emulsified...bile salt.....lipase....glycerol
5. amino acid....alkaline 6....fatty acid.

Mind Test 6:
1. Orange juices make the pH medium inside the duodenum decrease/ become more acidic
2. Acidic condition reduce the amylase reaction
3. Starch in duodenum less / cannot be hydrolyzed

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PG ANSWER
157 Mind Test 7:

(a) pH & temperature
(b) amylase speed up the hydrolysis of starch into maltose
(c)

-the mixture is dropped into iodine solution every two minutes

-time taken for iodine remain yellow is taken using stopwatch

-the rate of hydrolysis / digestion is calculated using formulae :1/time

Mind Test 8: Result +ve Result -ve
Test Name (Reagent) blue black remains
Iodine
Benedict's Reagent yellow

Biuret Test brick red remains
precipitate blue
Millon test
Grease Spot Test purple
Sudan III Test solution
Emulsion Test red
precipitate

no
trace

tiny
droplets

158- Absorption:
159 1.Villi... 2. ........fructose,....glycerol
epithelium cell: (a) ...microvilli.... (b) ......mucus....

(c)Glucose,galactose,amino acid, mineral salt...water soluble vitamin
(d)Fructose
(e)Fatty acid ...glycerol.....
Blood capillary: (a)..... hepatic portal vein (b)....lacteal.....blood capillary
Lacteal: (b)......laceal. (c).......thoracix duct (d)....thoracix ducts

Mind Test 9:
(i)....nutrient absorption.. (ii).......increase ....surface area (iii)...Goblet cells (vi)....secretes..
(v)...transport (vi)... absorb....lipid digestion
160 Mind Test 10:
The shortage of mitochondria causes less energy generated; less active transport occurs. This
causes less glucose and amino acids can be absorbed into the blood capillaries.
Assimilation:
1.......lymphatic....assimilation 2.......complex 3.....hepatic portal vein......liver
4....thoracix duct
161 The Function of Liver:

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PG ANSWER
1.....content
2.LIVER: (a).....glycogen (b) .......decreases....glucagon
(a)....protoplasma....enzyme...hormone (c)....deamination...
BODY CELLS: (a)....energy... (b) ...protein.. ..lipid.
(a) ... protein plasma.... (b)......glucose.

(a) ...adipose (b)....cholesterol (c) .......oxidised..
162 3 ...detoxification 5 .....storage......fat soluble vitamin.....minerals

Mind Test 12:

(1) Heart cirrhosis patients will face digestion problems. Lipid digestion becomes slow.
(2) Less bile salts can be produced, fats is difficult to emulsify into tiny droplets.
(3) Glucose content in blood is difficult to be regulated. Glycogen cannot be stored in

liver.
(4) Detoxification of toxin and drug doesn't happen. There is a risk that the person will be

diagnosed with a cancer.
(5) The production of enzymes, hormones, antibodies decrease, causing the immunity level

to decrease as well.
(6) Protein is difficult to be metabolised, thus the blood osmotic pressure is hard to be

regulated / will cause oedema at hands, legs, and abdomens.
(7) The presence of scar tissues increase the blood pressure in abdomens. Next, lymphs will be

swollen, causing the quantity of red blood cells and white blood cells to decrease

Mind Test 13:

Name: Hepatic Portal Vein

Reason:.....absorption.....assimilation.

163 Defaecation:

1. ....colon,

Colon:......peristalsis......K......reabsorbed....faeces.....fibres...toxic substances...

Rectum:.....defaecation.

164 Energy Value:

2 the quantity of heat released by one gram of food to increase the water temperature by 1 degree

Celsius

5...twice.....

Mind Test 14;

Label: thermometer, boiling tube, water, food sample

Answer/Calculation: 1050 Jg-1 @ 1.05 KJg-1

165 Mind Test 15:

Types of food Quantity Energy Energy
taken ( g ) ( kJ per 100 g) ( kJ per g )

White rice 400 1500 1500 X 400/100 = 6000

Fat 40 3000 40 X 3000/100 = 12003000

X40/100 = 1200 1200

Fish 150 300 300 X 150/100 = 450300 X
150/ 100 = 450

Papaya 40 40 40 X 40/100 = 16 40 /100
= 16

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BIO~SCORE Form4 SCHEME 2020

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Amount of energy obtained 766676567656

Mind Test 16: Volume of solution or juice needed to Percentage of Concentration
Solution Vitamin C (%) of Vitamin C
bleach DCPIP (ml)
/ Fruit Juices (mg ml-3)

12 purata

Ascorbic acid 1.0 1.0 1.0

Lime juice 3.0 3.0 3.0 0.03 0.33

Orange juice 5.2 5.3 5.2 0.02 0.19

Pineapple 4.8 5.0 4.9 0.02 0.20
juice

166 Mind Test 17:

The process to produce the juice involves heat or hight temperature to make it more long lasting/

durable. However, high temperature will also destroy most ascorbic acid. Therefore the

concentration of vitamin C decreases.

Modification Of Diet: (b)......energy....
1.(a).....calcium & phosphorus...bone & teeth...
(c)..health...repair....osteoporosis

167 Mind Test 18:
(1) His BMI level is 30 and above. He is categorized as obese.
(2) He is at risk to suffer from cardiovascular diseases, diabetes, kidney failure, heart failure,
exhaustion and easily stressed.
(3) To overcome this he needs to reduce his intake of carbohydrate food so less starch is
coverted into sugar and less sugar is converted into fat
(4) He also needs to avoid fatty food to prevent the deposition of cholesterol in the artery and
also reduce the risk of having cardiovascular diseases.
(5) In contrast he should practice 'Pinggan Sihat Malaysia' which is 'suku, suku, separuh'
(quarter, quarter, half of a plate) while increasing intake of vegetables and fruits.
(6) Practice healthy lifestyle such as exercising frequently to burn fat
(7) Avoid sleeping late, junk food and fast food
(8) He can also get hospital treatment such as the bariatric surgery
(9) to shorten intestines or to reduce the size of stomach
(10) so that the individual will only eat a small amount of food, thus the stored fat and protein

will also burnt to produce energy

168 Health Issue Related to modification or organ and system:

1(b) ......full..., (c)...fat,.....

2(a) ...pancreatic juices ... (b)....decreases.... (c)....insulin...glucagon..

3.(a)...bile... (b) ...emulsified.. (c) ....lipids..

4.(a)...indeigestion (b) .....hydrolysis...

169 Mind Test 19:

surgery to remove gall bladder prevents the storage of bile

-bile deficiency decreases the rate of fat hydrolysis because fat could not be emulsified and the

optimum pH could not be readied.

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BIO~SCORE Form4 SCHEME 2020

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-individual has to reduce intake of fatty food

Health issue related to Defaecation:

1. ....high-fiber....softens.... 2.(d)...good bacteria

constipation:....stool /faeces....colon....slowly...water

haemorrhoid:....high pressure...swelling..

colon cancer:.....constipation....toxicants...out of control..

170 Eat food containing Take a balanced and Avoid food that are

various nutrients nutritious diet. high in fats and

and increase intake sugar.

of food with fiber. Good eating Chew on food until

Avoid from eating too habit: it becomes small
much (overeating) or particles.
not eating at all Consumption of
(undereating). the right amount Drink 2 to 3 litres of
Avoid eating junk water per day and
food and food with of food at the reduce consumption
low nutritional value. right time of carbonated
drinks.
GERD:...close Eat meals at
regular times.

Bulimia:...vomit out ...cardiovascular disease

171 Gastritis: .1...inflammation.... 2.late....time...gastric juice... .3. .....ulcer.... 4. alcohol. 5....antacid
Anorexia:1....psychological...unattractive...
2. ..refrain .... eating...excessively...3...underweight...mental,emotional,..menstrual
4. ....therapy...counselling..
Dysmorphia:.1...muscular... 2. ... intensive...

172 Diabetes:1...glucose... 2......glyogen... 3.....insulin...
Pica:.1..disorder
Obesity:.1..excessive... 2. .....converted.. 3. .cardiovascular...carbohydrate...exercise..

CHAPTER 10 TRANSPORT IN HUMANS AND ANIMALS

MS ANSWER
174 The necessity of transport systems in complex multicellular organisms

175 1. ……. essential substances …… cellular waste products ……..

2. ……. diffusion …….. large …..

3. …….large……small…….complex…….diffusion……internal transportation system.

4.

TSA/V 6 3 2

Types of Circulatory Systems in Multicellular Organisms

Closed circulatory System
a) Continuous

b) across

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176
Insect: Open Circulatory System
177
a) .......haemocoel.....
178 b) .......diffusion.......
c) .......ostium.......

Fish: Single Closed Circulatory System
a) .......atrium.....ventricel.

b) The deoxygenated blood........
c) .......the gill capillaries.....

d) –

e) .......carbon dioksida......
f) .......heart.

g) .......each.......

Amphibians:
a) ......three..... Pulmocutaneous Circulation ........... Systemic Circulation.

b) ...... lungs..... skin.....veins.
c) ......mixed.....ventricel.
e) ......low.......

Human:
a) .......atria......ventricle.
b) ......pulmonary......systematic.
c) ......right ventricle......lungs.....oxygenated blood...... left atrium
d) ...... heart.......lungs.......
e) ......twice......
Similarities and differences between circulatory systems in complex multicellular
organisms
Similarities

(a) …………multicellular………….
(b) ………...heart…………………..
(c) ………..nutrients …………….
(d) ……….valves ………………

Differences:

Organism Insects Fish Amphibian Human
One s

Types of circulatory Open blood Double
system circulatory
Four
Number of heart system
chambers - One
Number of atrium
Number of ventricle -
-

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BIO~SCORE Form4 SCHEME 2020

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Separation of

oxygenated blood - - Incomplete
and deoxygenated

blood

Mind Test 2: Q – Human / bird
i) P – Fish

179 ii) Similarities:

P1: Both have closed circulatory system
P2: The blood flows in blood vessels
P3: Both have heart / atria / ventricle
P4: The heart acts as pumping organ / pumping blood

Differences : Organism Q / Human

N Organism P/ Fish
o

1 Single circulatory which is bood Double circulatory which is the blood

flows through heart once in each flows through the heart 2 times in

complete circulation. each complete circulation.

2 The heart have 2 chambers // The ......4 ...// .........2 atrium and 2
heart consists of 1 atrium and 1 ventricle
ventricle
Have septum
3 No septum Oxygenated blood flows from lungs
to heart
4 Oxygenated blood flows from gill
to body cells / tissues

5 .........gill

180 Mind Test 3
1. Atrium
2. Ventricle
3. Bicuspid and tricuspid valve
4. Corda tendinea
5. Semilunar valves
6. Aorta
7. Pulmonary artery
8. Vena cava
9. Pulmonary veins
10. Septum

181 Composition of Human Blood

Human Blood: Plasma
Platletes: ...... bone marrow.....nucleus.

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BIO~SCORE Form4 SCHEME 2020

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Soluble materials: ....... Na+, Mg2+, Cl-......
........ carbon dioxide……..
…….. globulin-...............
Leukosit (Granulosit): ...... Granulated.....berlobus.
Eusinofil: ........ inflammation....
Neutrofil: ....... phagocytosis.....
Basofil: .......heparin....
Monosit: ..... phagocytosis.
Limfosit: ....... antibodies......

182 Mind Test 4:

a) P: Erythrocyte Q: Leukocytes
Reason: Cell P do not has nucleus while cell Q has nucleus and size cell P is

smaller than cell Q.

b) It secretes the thrombokinase enzyme in the mechanism of blood clotting.

c) P1: It does not contain nucleus to allow more haemoglobin to be contained.
P2: Duplicate discs to increase surface area to accelerate gas exchange.
P3: It can be shaped like an ellipse to increase its surface in contact with the blood

capillary wall to increase respiratory gas absorption into the cell.

d) In liver and spleen P cells are destroyed by macrophage cells through phagocytosiss

e) Spinach contains many ferum / iron. The iron atoms are necessary for the formation of
haemoglobin pigments. So more erythrocytes can be built and more oxygen can be
transported to the cells especially muscle cells / tissues. More glucose oxidation occurs.
More energy can be produced for muscle contraction. This makes Popeye stronger.

183 Types of Blood Vessels

Artery
a) ....... oxygenated..... pulmonary artery.
b) ....... leaves...... breaking……………
c) ....... arterioles………..capillaries.

Capillaries
a) ....... thinnest.....
b) ....... respiratory gaseous …….. body cell........
simple diffusion .
c) .......smallest........

Vein
a) ....... rejoin....... venules..

b) ....... veins…………..
c) ........ deoxygenated blood...... back.......

DIAGRAM: Connective tissue , Smooth muscle
184 Mind Test 5

Arteri Kapilari Vena

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BIO~SCORE Form4 SCHEME 2020

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........ thick....... ....... thin,......

..... heart..... Smallest

No valve ...... backflow......

...... artery......

High High
......... pulmonary artery.. ........ pulmonary veins.

185- Mind Test 6: DIAGRAM
186 P:Artery Q:Vein R: Arteriol es S:Venul
T: Connective tissue U:smooth muscle V:Endothelium W:valve
186
MECHANISM OF HEART BEAT
187
Mechanism of Heartbeat of Human’s Hearts
2. ...... cardiac muscles...... electric impulses...... contract......
3. ....... myogenic....... impulse..............
4. ....... forces.......

Pumping of the heart
1. .......... pacemaker.
2. ......... specific....... atrium.......
3. ......... electrical impulses........ walls........ rhythmically........ sinoatrial node.
4. ......... the bundle of His......

1. ......... electrical......
2. ........ pressure......
3 ......... spread......
4 ........ ventricles....... lungs....... aorta.
6. ....... closing.........

Contraction of skeletal muscles around the veins
1. ......... force....... vein,...... insufficient.......

2 ......... against.........
3. ……. valves ……. One...............
4. a) ……. smooth ……. venule

b) ……. skeletal …… open ……. flowing back

DIAGRAM: a)...closed b)....opens c).....relaxes

MECHANISM OF BLOOD CLOTTING
The Necessity for Blood Clotting mechanism:
Mind Test 7

 Stop or minimise the loss of blood on the injured blood vessel.
 Prevents microorganisms such as bacteria from entering the bloodstream through the

damaged blood vessel.

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BIO~SCORE Form4 SCHEME 2020

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 Maintain the blood pressure because excessive blood loss will lower blood pressure
to a dangerous level.
 Maintains the blood circulation in order to transport cells requirement (nutrients and
oxygen).

188 Mechanism of Blood Clotting:

1. …….. Platelet……. thrombokinase….. Prothrombin…..

Mind Test 8
1. . ……… constricts …….. blood loss …….
2. ………. Collagen fibers …….. platelets ……. plaques ……
3. …………Thrombokinase ……. liver.
4. ……… enzyme ….. fibrin
5. ……… scab.
6. ……… healed ……..

189 Health Issues related to Blood Clotting

Haemophilia:

(a) ….hereditary…………clotting……….
(b) Excessive bleeding …………………..

Thrombosis
(a)……………………………… thrombosis.
(b)………………. platelet …………………
(c) ……………….damage…………….. slow.

Embolism
(a) …………removed ………………………..

(b) ………trapped……..block……………….embolism.

Blood Grouping Of Humans
ABO Blood Group

Blood Group Antigen on Antibody in
Red Blood Cells the blood serum

Antigen A Anti-A

B Anti-A and Anti-B

Antigen A and Antigen B

O

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BIO~SCORE Form4 SCHEME 2020

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190 ABO Blood Group with Blood Donation

1. …………. blood group type ……………. antibodies ……………antigen ……

2. …………………… agglutination …………

Blood Group Can donate blood to Can receive blood from
blood groups blood groups

B

O

Mind Test 10: Blood group recipient
Donor A, B, O and AB

AP
B
C

D

Rhesus Factor (Rh)

1. …………antigen ……………..red blood cells,……………….
2. …………aggregates ………..antibodies ………………………
3. …………Rh-positive ………..Rh-negative.
4. …………recipient,…………antibody.
5. ………….agglutination……….recipient.

191 Pregnancy and the Rhesus Factor

……………. anti ….. cross……
D antibodies.

Mind Test 11:
1. ………………………………….anti-D antibodies.
2.
3. ………….second……….Rh-positive………………. cross through……….erupt.
4. …………miscarriage……..erythroblastosis fetalis…………………………
5. ………...anti-Rhesus globulin……………………….

192 Health Issues Related To The Human Circulatory System

1. ………………. atherosclerosis, arteriosclerosis, angina, hypertension, myocardial
infarction (heart attack), thrombosis coronary and stroke.

2. Atherosclerosis ……………*High Blood Pressure/Hypertension

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