Modul Ka f4 2011 Combine(t)

CHAPTER 2 – CELL STRUCTURE AND CELL ORGANISATION QUESTION 1 : No 1(a)

Marking Criteria

Marks

Able to draw and label a green plant cell

6

Cell wall Chloropla st Cytoplasm vacuole nucleus

2

Diagram – 2 marks [function, neat] Labels : 4

5 correct – 4 marks 4 correct – 3 marks 3 correct – 2 marks 2 correct – 1 mark. (b)

Able to state the functions of three organelles in the plant cell Organelles 1. Nucleus

Functions - controls all activities of the cell // - determine characteristics / cell functions.

2. Vacuole

- store chemicals such as water/ amino acid/ sugar/ mineral / waste products // - regulates water balance // supports

3. Mitochondrion

- the site of energy production // cellular respiration

4. Lysosome

-breaks down complex organic molecules / protein/ lipid / polysaccharides /nucleic acids

5. Chloroplast

-traps sunlight (energy) during photosynthesis

6. Ribosome

- the site of protein / enzyme.synthesis

3 + 3

6

No

Marking Criteria 7. RER

Marks

- Transports protein (made by ribosomes) to Golgi apparatus/ other parts of the cell.

8. SER

- synthesis of lipid / phospholipids / steroids // - detoxification

of drugs /

poisons. 9. Golgi Body/ Apparatus

-

Processing/

packaging of

/

transporting

centre

the

synthesised

proteins (such as

enzyme / hormone / antibody, phospholipids

and

carbohydrates & glycoproteins (such as mucus). (c)

Compare and contrast a plant cell with an animal cell. Similarities :

8

Both have a nucleus/ a plasma membrane/ mitochondria/ ribosomes/

1

endoplasmic reticulums/ cytoplasm/ Golgi apparatus

1 Any 2

Differences : Animal Cell

Structure/ Characteristic Does not have a fixed Shape shape Absent Cell wall Absent Chloroplast Absent // very small (if Vacuole any) Food (carbohydrate) is Food storage stored in the form of glycogen Present Centriole

Plant Cell Has a fixed shape

1

Present Present Present

1

Food (carbohydrate) is stored in the form of starch Absent

1

1

1 1

QUESTION 2 :

No 2(a)

Marking Criteria

Marks

Able to state how the structures are adapted to their functions.

10

Red blood cell : Fact

Explanation

F1 : Biconcave E1 : to increase the TSA/V ratio // E2 : allows rapid diffusion of oxygen into or out of disc shaped the rbc F 2: no nucleus

E1 : thereby having more space to contain the respiratory pigment / haemoglobin // E2 : the haemoglobin pigment has a high affinity for oxygen

F3 : the plasma membrane is elastic

E1 : allowing it to squeeze through the narrow blood capillaries // E2 : to transport oxygen to the whole body

1 1 1 1 1 1 F1/F2/F3 – 1m

Efferent neurone Fact

Explanation

F1: has a long axon

E1 : to conduct /transmit nerve impulses// E2 : to the effector

F2 : has dendrons / dendrite

E1 : to receive impulse

F3 : Has the node of Ranvier

E1 : speed up impulse transmission

F4 : has myelin sheath

E1 : acts as an electrical insulator

F5 : contain a lot of mitochondria

E1 : to generate energy / ATP // E2 : that is required for the transmission of nerve impulses

1 1 1 1 1 1 1

Sperm cell Fact

b(i)

Explanation

F1: has a tail

E1 : to swim towards the ovum in the fallopian tube

1

F2 : contain a lot of mitochondria

E1 : to generate energy / ATP // E2 : needed for locomotion

1

F3 : the head contains hydrolytic enzymes

E1 : can break down the plasma membrane of the ovum E2 : to allow fertilisation

Able to describe how Amoeba can carry out the following life

1

processes : nutrition and reproduction. Answer : Nutrition 1

P1 : holozoic nutrition // feeds on bacteria / small green algae / diatoms found in water

1

P2 : pseudopodia are formed to surround food,

1

P3 : forming a food vacuole

1

P4 : food is digested by hydrolytic enzymes

1

P5 : the soluble products are absorbed and assimilated

1

P6 : the undigested food is egested Any 4 Reproduction P1 : Ameoba reproduce asexually through binary fission

1

P2 : a mature amoeba will divide at the nucleus, followed by the a

1

division of the cytoplasm before forming two daughter cells (in favorable condition) P3 : spore with thick protective cyst walls are produced.

1 Any 2

b(ii)

Able to explain why Amoeba does not burst when its immerse in the

4

distilled water ; P 1 : Have contractile vacuole to regulates amount of water/ osmotic

1

pressure P2 : Water diffuse into amoeba cause by distilled water is hypotonic

1

than amoeba. P3 : Then excess of water in the cell diffuse into// enters the

1

contractile vacuole by osmosis P4 : When the contractile vacuole enlarges / increase in size to

1

maximum size. P5 : its contract to release/ expel / throw out excess water to the external environment

QUESTION 3:

1

No 3(a)

Marking Criteria P1 : Cell is basic units in the organism / man example muscle cell/ epithelium cell P2 : Tissues are groups of similar cells that have a similar structure and performing similar function. P3 : Epithelial tissue consists of one or a few layers of epithelial cell, and is found covering the outside of the body or lining organs and glands. P4 : Epithelial tissue acts as a barrier or protection. It helps to protect organs from injury and fluid loss P5 : It is found lining the oesophagus, stomach, intestines, villi and most parts of the digestive system P6 : The epithelial cells are adapted for absorption or secretion

Marks 1 1 1

1 1 1

P7 : The wall of the upper part of the oesophagus (an organ) contain voluntary muscles

1

P8 : Peristalsis is assisted by the circular muscles and longitudinal muscles (smooth muscle tissue)

1

P9 : All the muscle tissues are formed from groups of muscle cells P10 : The wall of the stomach (an organ) has three layers of smooth muscle tissue

1 1

P11 :There are connective tissues in the digestive system. The connective tissues include all the blood cells

1

P12 : The digestive system is made up of several organs that work together as one unit. They are the stomach, liver, gall bladder, pancreas, small intestine, large intestine (colon) and rectum

1

10

3(b)

Able to explain how the leaf modified to carry out its various function.

characteristics

2

• Allow gaseous exchange between air space and enviroment • Allow loss of water vapour through transpiration Less stomatal pores • To reduce excessive loss of water on the upper surface vapour than lower surface Palisade mesophyll • To absorb as much light as possible cells are packed closely together under the leaf’s upper epidermis Palisade mesophyll • To increase the rate of cells are also photosynthesis packed with chloroplasts Spongy mesophyll • Allow efficient diffusion of gases cells are loosely • for photosynthesis which uses packed. carbon dioxide and releases oxygen // • respiration which uses oxygen and releases carbon dioxide • transpiration which results in loss of water vapour.

2

2. Thick waterproof cuticle on its upper surface 3. The epidermis has stomatal pores

5.

6.

7.

Function • Allow efficient absorption of sunlight and carbon dioxide for photosynthesis, absorption of oxygen for transpiration and loss of water through transpiration. • • To reduce excessive loss of water during transpiration

1. Flat and broad with a large surface area to volume ratio

4.

10

8. Xylem vessels in the • Supply the leaf cells with water for leaf’s veins photosynthesis and transpiration

2

2 2

2

2

2 2

9. Phloem tissue / sieve tubes in the leaf’s veins

• Transport the organic products of photosynthesis to other parts of the plant.

CHAPTER 3 – MOVEMENT OF SUBSTANCES ACROSS THE PLASMA MEMBRANE Question 1 No 1(a)

Marking Criteria •

Marks

Organelle involve in osmoregulation is contractile

Total Marks

1

vacuole •

Water diffuses into Amoeba sp. by osmosis and fills

1

the contractile vacuole •

When the vacuole is filled with water to its maximum

1

size, • (b)

It will contract to expel the excess water to its

1

4

surroundings Similarities :

1

•

Both occurs in living cells

•

Both occurs through a semi-permeable membrane

•

Both require carrier protein to bind with the substances

1 1

2

ANY 2

Differences : Facilitated diffusion Molecules move down the

Active transport Molecules move against

concentration gradient Molecules move through

the concentration gradient Molecules move through

pore proteins or carrier

carrier proteins only

1

1

proteins Occurs until a dynamic

Results in accumulation of

equilibrium is achieved

substances in the cell or removal of substances from

ATP or energy is not

the cell ATP or energy is required

1

required Not dependent on cellular

Dependent on cellular

1

respiration Not affected by inhibitors

respiration Inhibited by inhibitors such

1

•

as respiratory poisons Vinegar is acidic and has low pH

•

This prevent the growth of microorganisms in

6 1

( c)

mangoes

1

•

The mangoes can be preserved to last longer

1

•

Concentrated salt solutions hypertonic to the tissue of fish

1

•

Water diffuse out of fish cell by osmosis

1

•

Water also diffuse out of bacteria cell to the surroundings

•

The bacteria cells become plasmolysed

•

This prevent the growth of bacteria which cause food spoilage

1 1 1

8

1

20

Question 2 No 2(a)

Marking Criteria

Marks

•

Amino acid is a large water soluble molecule

1

•

It requires carrier protein to move across the

1

Total Marks

membrane •

Amino acid will bind to the carrier protein which is

1

specific to it •

Carrier protein will change its shape to bring the amino acid molecule across the membrane

•

Lastly the carrier protein will release the amino acid

1 1

5

and returns to its original shape

(b) •

Fertilizers which are added to the soil dissolve in soil water

•

The soil water become hypertonic to the cell sap of plant roots

•

1 1 1

Water diffuse from the cell sap into the soil by osmosis

1

•

The cells become plasmolysed

1

•

Wilting occurs and the plants eventually die

5

(c)

1 •

20% salt solution is hypertonic to red blood cell

1

•

Water diffuse out of cell by osmosis

1

•

Cell become shrink

•

Crenation occur

1 1

ANY 3

(d)

3

1 1

•

Distilled water is hypotonic to red blood cell

•

Water diffuse into the cell by osmosis

•

Cell swell up and eventually burst

1

•

* Haemolysis occur

1

ANY 3

1

1

3

1 •

30% sucrose solution is hypertonic to potato strip

1

•

Potato cell become plasmolysed

1

•

Distilled water is hypotonic to potato strip

•

Water diffuse into cell sap of potato strip by osmosis

•

Cell becomes turgid again

•

The cell is said to have undergone deplasmolysis

4

ANY 4

20

CHAPTER 4 – CHEMICAL COMPOSITION OF THE CELL Question 1 No 1(a)(i)

Marking Criteria •

Intracellular enzymes are produced and retained in

Marks

Total Marks

1

the cell •

For the use of the cell itself

1

•

Extracellular enzymes are produced in the cell but

1

secreted from the cell

(a)(ii)

•

To function externally

•

The nucleus contain DNA which carries the information for the synthesis of enzymes

•

The genetic information is transcribed from DNA to RNA in the nucleus

•

1 1 1 1

Proteins depart from RER wrapped in vesicles that bud off from the membrane of the RER

•

1

Proteins that are synthesized at the ribosomes are transported through the space within RER

•

4

The RNA leaves the nucleus and attaches to ribosomes on the rough endoplasmic reticulum

•

1

1

The transport vesicles fuse with the Golgi Apparatus and empty their content into the membranous space

•

1

The protein are further modified during their transport in Golgi Apparatus

•

Secretory vesicles containing enzymes bud off from

1 8

Golgi Apparatus and travel to plasma membrane •

These vesicles fuse with plasma membrane and

1

release the extracellular enzymes ANY 8

(b)

1 •

R is enzyme-substrate complex

•

The substrate molecule is represented by the ‘key’

1

while the enzyme molecule is represented by the 1

‘lock’ •

The substrate has a shape that is complementary to the active site of the enzyme

•

The substrate binds to the active site to form an enzyme-substrate complex , like a key fits into a lock

•

1

Reaction occurs that is the enzyme catalyses the conversion of substrate to products

•

The products are released from the enzymes

•

The enzyme remain unchanged at the end of the

1 1 1 1

8

reaction •

20

The enzyme is now free to bind with another molecule of substrate

Question 2 ITEM NUMBER 2(a)(i)

(a)(ii)

MARKING SCHEME

SUB

TOTAL MARKS

•

Enzyme

MARK 1

•

Antibody

1

•

Hormone

•

Plasma protein

•

P is primary structure

•

It refers to the linear sequence of amino acids in a

1

Any

1

2

1 1 2

polypeptide chain •

Q is secondary structure

•

It refers to the polypeptide chain

1 1 1

•

1

beta pleated sheet

1

•

R is tertiary structure

•

It refers to the way the helix chains or beta1

pleated sheets •

3

are coiled or folded into three dimensional shape of a polypeptide chain

(b)

3

That is coiled to form alpha-helix chain or folded into

1 1

•

Amylase

•

Dissolve and remove starch stains

•

Protease

•

Dissolve and remove protein stains and blood

1 1 1 6 1

•

Lipase

•

Dissolve and remove fats, oils and grease stains

(c)

1 •

Lipase is used in the ripening of cheese in dairy

1

products industry •

Cellulase is used to extracts agar from seaweed in

1

seaweed products •

Cellulase is used to breakdown cellulose and removes seed coats from cereal grains in cereal

1

grain products •

Trypsin is used to remove hair from animal hides in

1

leather industry •

Amylase is used to convert starch flour into sugar in

1

the making of bread in baking industry • •

Zymase is used to convert sugar into ethanol to

1

make alcoholic drink in beer/wine making industry

4

Protease is used to remove the skin of fish in fish

20

product

ANY 4

CHAPTER 5 – CELL DIVISION QUESTION 1 No (a)

Marking Criteria

Marks

Able to state the important of mitosis Sample Answer : P1 : in growth process // to increase the number of cells ( during the

1

growth process) 1

P2 : Cell replacement // To replace dead and damaged P3 : Regeneration // Production of new cells

1

P4 : Asexual reproduction // the production of new individuals ( from

1

parent organism ) (b)

3

Any 3

Able to explain the differences between process of mitosis and

10

meiosis using appropriate diagram mitosis Prophase

meiosis Prophase I

1

location of chromosome is at random

homologous chromosome synapsis

no crossing over/ no chiasmata

crossing over/ chiasmata

Metaphase

Metaphase I

1 1

1

chromosome are arrange at the middle of cell

homologous chromosome are arrange at the middle of cell

1

No

Marking Criteria

Anaphase

Marks

Anaphase I

1

chromatid move to the opposite

homologous chromosome will

pole

move to the opposite

Telophase

Telophase I

1

1

Each daughter cell has the same

Each daughter cell has half number

number of chromosome

1

2 daughter cell that is 2n

4 daughter cell that is n

only one division

two times division

1

does not cause genetic variation

cause genetic variation

1 1

Diagram: 4 (c)

Exp: 6

= Max: 10

Able to explain a cloning in animal to produce large number of animal much faster than through sexual reproduction : P1 : examples: cow/ sheep P2 : remove a somatic cell donor of animal and growth in a culture P3 : unfertilised ovum from donor is obtained P4 : the nucleus ovum is removed P5 : an electric pulse stimulates the fusion between the somatic cell and the ovum

1 1 1 1 1 1

P6 : the cell divide by mitosis P7 : embryo is form P8 : then, the embryo is planted into surrogate mother

1 1

max

No

Marking Criteria

Marks 7

QUESTION 2: No (a)(i)

Marking Criteria

Marks

Able to explain what happens in the cell during phase X. Sample answer : P 1 : Protein are synthesized // Mitochondria / chloroplast are synthesized

1 1

P2 : DNA are synthesized 1

P3 : The cell accumulate energy.

2 Any 2 P

(a)(ii)

Able to describe tissue culture technique. Sample answer: P1 : Tissue culture technique

1

P2 : Tissue culture technique is used to produce (high quality of seedling)oil palm seedlings in vitro/any suitable example.

1

P3 : The leaves/shoot/stem/root tissues are cut out.(These cut out plant tissues are called explants).

1

P4 :The pieces of meristematic tissue (explants) are cultured in sterile nutrient medium, in suitable pH and with addition of plant growth substances.( at least 2 factors)

1

P5 :The flasks containing the tissue are stored in an incubator at 37°C for 2/3 weeks.

1

P6 : The cell divide by mitosis to produce callus.

1 1

P7 : The callus is then cut into small pieces.

(b)

P8 : The small pieces of callus tissues are then cultured in sterile nutrient medium.

1

P9 : When it has grown to a suitable size, the clone is transferred to the nursery. Any 6 P

1

Able to discuss Advantages / strength:

6

No

Marking Criteria P1 : Genetic engineering involves genes manipulation / transfer / modification in organisms to produce certain products.

1

P2 : Example; products in pharmacy such as insulin / antibiotics; food products based on plants / animals; agricultural / agrochemical products.

1

P3 : The products produced are very similar to the original / natural materials because the same genes are used / particularly chosen genes are transfered.

1

P4 : The production of products is faster especially with the use of microorganisms/bacteria.

1

P5 : Microorganisms such as bacteria are suitable to be used as gene vectors / they have free DNA in the form of ring / plasmid.

1

P6 : High reproduction rate of bacteria/microorgansm in optimal culture mediums able to produce a large amount of chosen genes / products / insulin / antibiotics.

1

P7 : Can be used by thousands of people who need them / widespread usage.

1

P8 : Able to produce a variety of proteins / recombinant proteins / enzymes used in food industries / medicine / agriculture.

(c)

Marks

1

P9 : Genetic engineering technique is used to solve criminal cases through DNA finger printing / DNA fragments analysis.

1

P10 : Other uses /examples; metal extraction from oxide/any suitable examples Any 6 P

1

Able to describe the effect of cycle malfunctions to the body. Sample answer P1: The exposure damage the DNA of the cell P2: A cell divides through mitosis repeatedly. P3: Produces cancerous cell P4: Due to (severe ) distruption to the mechanism that controls the cell cycle

1 1 1 1

P5: Cancerous cells divide freely / uncontrollably heeding the cell cycle control

1

6

No

Marking Criteria P6: (these cells ) compete with surrounding normal

Marks

cells to obtain nutrient / energy (for growth)

1

6

P7: Invade / destroy neighbouring cells 1

P8: (they can spread to other organ and) initiate cancers there .

1 Any 6 P TOTAL

20

QUESTION 3 No 3(a)i

Marking Criteria

Marks

Able to explain the principles used in the cloning technique P1 : Cloning is an asexual reproductive process of producing

1

clones//does not involve gamete P2 : A clone is a group of cells//organism//a population of organisms produced from a single ancestral cell. P3 : A clones genetically identical P4 : The technique can be used to produce high quality of organism / 3(a)i

orchids/ oil palm / cocoa plants. Able to state the advantages and disadvantages of this technique.

i

Advantages -

multiply copies of useful genes or clones can be produced in a shorter time and in larger numbers

1 1

3

1

marks

Any 4 advantages and

1 1

-

produce clones through asexual reproduction

1

-

produce clones that are resistant to diseases

1

3 disadvantages or Any 3 advantages and 4 disadvantages

Disadvantages

3(b)

-

clones have the same level of resistance towards certain diseases

1

7

-

may undergo mutations which can disrupt natural equilibrium

1

marks

-

do not show any genetic variations

1

-

unable to adapt to the changes of the environment

1

Able to describe how cytokinesis occurs in plant and animal cells to produce two daughter cells. -

process of cytoplasmic division

1

-

begins before nucleus division is complete / during telophase to form

1

No

Marking Criteria

Marks

two daughter cells -

in animal cells, actin filaments in the cytoplasm contracts

1

-

to pull a ring of the plasma membrane inwards

1

-

forming a groove called a cleavage furrow

1

-

the cleavage furrow pinches at the equator of the cell and deepens

1

progressively until two daughter cells are separated -

in plants cells, the membranous vesicles are formed along the equator between the two nuclei

1

-

the vesicles fuse to form a cell plate

-

the cell plate grows outwards until its edges fuse with the plasma membrane of the parents cell

-

1

at the end of cytokinesis , cellulose fibres are produced by the cells to

1 1 10

strengthen the new cell walls

marks

QUESTION 4 : No 4(a)

Marking Criteria

Marks

Able to explain the process that occurs in stage P, Q, R and R. Stage P : Prophase I -

chromosome becomes shorter/thicker

-

homologous chromosome come together form bivalent through synapsis

1 1

-

non sister chromatids exchange segments of DNA // crossing over

1

-

nucleus membrane disappears

1

-

spindle fibres form

1

Stage Q : Metaphase I - pairs of homologous chromosome arrange at metaphase plate Stage R: Anaphase I -

spindle fibre pull the homologous chromosome away from one

another and move to the opposite poles Stage S: Telophase 1 -

chromosome arrive at the poles

1

1

No -

(b)

Marking Criteria spindle fibre disappears

Marks

1

6

1

marks

Able to state the different between meiosis I and meiosis based on stage

4

P, Q, R and R. Stage

P

Meiosis I Homologous chromosome come

Meiosis II chromosome locate at

together form bivalent through

random no crossing over 1

synapsis.There is crossing over Pairs of homologous

chromosome arrange at

chromosome arrange at

metaphase plate in a

R

metaphase plate Homologous chromosome move

straight line Chromatid move to

S

towards opposite poles Chromosome arrive at the poles

opposite poles Chromatid arrive at poles

Q

1

1 1

(c)

10

Explain briefly how meiosis involved in genetic variation. F1 - crossing over during prophase I of meiosis E1 – non sister chromatids of homologous chromosome break at the

1 1

chiasma E2 – segments of the chromatids exchange places E3 – segment of the marternal chromatids become attached to the

1 1

paternal chromatids E4 – new combinations of genes are produced on these chromatids F2 – independent assortment of chromosome E1 – at metaphase I the homologous pairs of chromosomes arranged on

1 1 1

the metaphase plate at random E2 – each homologous pair of chromosomes is positioned relative to the poles of the cell independent of other pairs

1

No

Marking Criteria E3 – there is independent assortment of maternal and paternal

Marks 1

chromosomes into daughter cells E4 – result in a variety of gametes each with different combinations of maternal and paternal chromosome

:

1

CHAPTER 6 : NUTRITION No Marking Criteria 1(a) Able to state the meaning of malnutrition.

Marks

Sample answers 1-

Malnutrition is a condition due to taking an unbalanced

1+1

2

Max 8

8

diet in which certain nutrients are lacking, 2(b)

in excess or in the wrong proportions

Able to name and explain the disease in Diagram 6.1.1, Diagram 6.1.2 and Diagram 6.1.3 related to malnutrition. Sample answers Diagram 6.1.1 1- Kwashiorkor 2- A child does not receive sufficient protein in his diet. 3- has the characteristic sign of scaly skin, thin muscles , thin hair and a swell of the body Diagram 6.1.2 4 - Rickets 5 - Vitamin D deficiencies 6 - poor teeth and bone formation in children 7 - leads to softening and weakening of the bones. Diagram 6.1.3 8 - Obesity 9 - excessive intake of food rich in fat 10 - body weight exceed by 20% of ideal/ normal weight

No

Marking Criteria

Marks

1(c )

Able to explain the long term effects of his diet to his

Max 10

10

health. Sample answers Bread 1- Bread is rich in carbohydrates. 2- cause high glucose content in blood. 3- leads to Diabetes mellitus. 4- Excess glucose/glycogen is converted to fats. 5- leads to obesity. Meat 6-

Meat is rich in protein.

7-

Excess protein is converted to urea

8-

Causing liver failure/ kidney failure.

9 - Excess uric acid due to consumption of excess protein cause gout Cheese 9-

Cheese is rich in fats/lipids.

10- Results in high cholesterol level in blood 11- Lipid/cholesterol deposited in inner wall of arteries 12- Lumen of arteries become smaller 13- Cause cardiovascular disease. 14- Excess fat also leads to obesity TOTAL

20

No

Marking Criteria

Marks

2(a)(i)

Able to describe the type of nutrition in organism P and organism

2+2

Total Mark 4

6

6

R. Sample answers Organism P 1 - Autotrophic nutrition 2 - Synthesize its own glucose / starch from carbon dioxide and water with the help of light energy through the process of photosynthesis Organism Q 3 - Heterotrophic nutrition/ holozoic 4 - Obtain its food source/organic substances from the surroundings (eat plant/ producer) Able to compare the process of cellulose digestion in organism Q (ii)

and organism R. Sample answers Similarities 1 - Both have alimentary canal which are made up of the oesophagus, stomach, small intestine and large intestine. 2 - Both are unable to produce cellulase to digest cellulose. Differences Organism Q

Organism R

1 The type of diet is omnivores

The type of diet is herbivores

2 Stomach has one chamber

Stomach has four chambers

3 Microorganisms in the digestive

Bacteria and protozoa in rumen

tract do not play an important

and reticulum secrete the

role in digestion of cellulose/ do

enzyme cellulase to digest

not have enzyme cellulase to

cellulose

digest cellulose. 4 The food from the mouth is

The food from the mouth is

swallowed to the stomach

swallowed to the rumen and

without regurgitation.

reticulum, then it is regurgitated into the mouth to be chewed again before being swallowed into the omasum.

No

Marking Criteria

Marks

3(a)

Able to explain the effect of eating too much of this kind of mangoes

5

Total Mark 5

5

5

Max 5

5

on the digestion of food in Y. Sample answers 1 - mangoes with vinegar contain much acid, so its reduces the pH value/ increases acidity in the duodenum 2 - Acid medium is less suitable for the action of enzyme lipase, amylase and trypsin 3 - less/ no lipid is digested/hydrolysed to fatty acid and glycerol by lipase 4 - less/ no starch is digested/hydrolysed to maltose by amylase 5 - less/ no polypeptide is digested/hydrolysed to peptides by trypsin

(b)

Able to state what he should do to handle health problems that may arise from the removal of organ Z. Sample answers 1 - Reduce the intake of high carbohydrate food / protein /fatty

food.

2 - Get insuline injection when needed / if glucose level too high. 3 - Get glucagons injection when needed / if glucose level too low. 4 - Pancreas transplant. 5 - Eat more vegetables / fruits. Able to state the functions of X. (c)(i)

Sample answers Functions of Y/liver 1 - Maintenance of blood glucose level under the influence of insulin and glucagons. 2 - Synthesis plasma protein such as fibrinogen / prothrombin from amino acids.

3 - Synthesis bile. 4 - Storage of nutrients such as fat-soluble vitamins (A & D)/ B12/ ferum/ copper/ potassium. 5 - Detoxification of poisonous substances such as

alcohols/drugs/

toxins/ pesticides/carcinogens /poisons. 6 - Deamination of amino acids. 7 - Produce heat. Able to explain the process of assimilation of amino acids and (c)

glucose in X.

(ii)

Sample answers

Max 5

5

The process of assimilation amino acids and glucose in Y Amino acids 1 - Excess amino acids cannot be stored and are broken down into urea by a process called deamination before being excreted by the kidneys Excess amino acid

urea

2 - When there is a short supply of glucose and glycogen, the liver converts amino acids into glucose. 3 - Plasma protein can be synthesized from amino acids and is used for blood clotting and osmoregulation. Glucose 4 -

Excess glucose is converted into glycogen and stored in the liver. Excess glucose

glycogen

5 - When the blood sugar level falls, glycogen is converted back to glucose. 6-

Glucose in the liver is used for cellular respiration. Glucose + Oxygen

Energy + Carbon dioxide + water TOTAL

No

Marking Criteria

20 Marks

Total

4(a)

Able to explain the consequences to his health.

Max 10

Mark 10

2

2

Marks

Total Mark

Sample answers 1

- Fried chicken and meat in the hamburger are high in fat and cholesterol

2

- Excessive intake of fats can be lead to obesity

3

- Excessive cholesterol and fats are deposited in the lumen of arterial wall.

4

- This cause the lumen becomes narrow and leads to arteriosclerosis and hypertension

5

- this will cause the decreasing of blood supply to tissues, leads to heart attack and stroke

6

- Bread in hamburger, soft drink and ice cream are sweet and high in sugar

7

- Sweetened food can cause tooth decay

8

- and obesity

9

- Excessive intake of sugars may eventually lead to diabetes mellitus

10 - Vegetables and fruits are good source of vitamins, minerals and fibres 11 - Insufficient intake may lead to deficiency diseases, dry and scaly skin ,scurvy (other suitable examples) 12 - Insufficient in fibres causes constipation

4(b)(i)

Able to state the meaning of photosynthesis. Sample answers 1 - Photoynthesis is a process whereby a green plant produces glucose / starch from carbon dioxide and water 2 - in the presence of chlorophyll and sunlight.

No

Marking Criteria

4(b)(ii) Able to explain the main stages in photosynthesis.

Max 8

8

Sample answers Light reaction 1 - Chlorophyll absorbs/ traps light energy to produce ATP and electrons/ chemical energy 2 - Photolysis of water produces H+ and OH- ions 3 - H+ ion combines with electron to form hydrogen atom 4 - Hydrogen/ ATP/ NADPH will be used in the dark reaction 5 - Light reaction occurs in grana Dark reaction 6 - 6 - The process takes place in the absence of light/ does not need light 7 - CO2combines with hydrogen/ hydrogen and is reduced to form glucose and water 8 - Glucose molecules undergo condensation/ converted/ stored to starch 9 - Formation of glucose and starch occurs in chemical reaction chain requires ATP/ chemical energy 10- Chemical reactions need ATP energy 11 - Dark reaction occurs in stroma TOTAL

No

Marking Criteria

Marks

20

Total Mark

5(a) Able to explain the necessity for food processing.

Max 10

10

Marks

Total Mark 10

Sample answers 1 – prevent food spoilage 2 – (food spoilage) causes by the action of microorganism 3 – decomposing bacteria/fungi on carbohydrate/protein 4 – produced carbon dioxide, water, ammonia hydrogen 5 – make food become toxic 6 – Oxidation of food when cut/expose to air 7 – oxygen react with enzymes/chemicals released by cell 8 – Increase it commercial value 9 – food additives is added in preserving the freshness of food 10 – Improve the taste/appearance/texture 11 – Intention of diversifying the uses of food 12 – increased the variety of products

No

Marking Criteria

5(b) Able to describe how each method can preserve food for a

Max 10

long period of time Sample answers Pasteurisation 1-

milk is treated to 63oC for 30 minutes//72oC for 15 seconds

2-

followed by rapid cooling to below 10oC

3-

destroy bacteria but not the spores

4-

retains the natural flavour of milk//nutrients//vitamin B

5-

must refrigerated to avoid the growth of sperms

Canning 6-

use heat sterilisation

7-

kill microorganisms and spores

8-

steamed at high temperature and pressure to drive out air

9-

sealed while the food is being cooled

10 - vacuum in the can prevent growth of microorganism Refrigeration 11 - stored at temperature below 0oC 12 - prevent the growth of microorganisms/the germination of spores

TOTAL

CHAPTER 7

20

No 1(a)

Marking Criteria

Marks

Able to name organ X and organ Y

1

X: Lung (b)

1

Y: Gills Able to draw and label the respiratory surfaces of organ X and Y

(c)

2

D-2 L-2

Organ X

Total

4

Organ Y

Able to compare and complain the respiratory system in human and fish.

adaptations

of

Similarities and explanation 1. both lungs and gills have large surface area for gas

1

exchange. 2. Human lungs have numerous alveoli while the gills have

1

numerous filaments. 3. Both respiratory systems are thin / one cell thick.

1

4. This ensures more efficient respiratory gases diffusion.

1

5. Both respiratory systems have many network of blood

1

capillaries 6. dissolve more respiratory gases.

1

7. Both systems use muscles to change the pressure of

1

respiratory cavity / thoracic and mouth cavities. 8. Human has diaphragm and intercostals muscles while fish

1

has muscles in the mouth and operculum. 9. Enable/ allowing oxygen to be transported to the body cells

1

through blood vessels. 10. Both human and fish have a closed blood circulatory

1

system. Differences and explanation 1. To ensure the supply of oxygen is continuous and sufficient

1

2. The trachea are reinforced with chitin to prevent them from

1

collapsing whereas the density/water pressure prevent the gills from sticking together 3. The position of respiratory organs ensures the respiratory

1

surface is always moist. 4.The surface of gas exchange/the alveoli is located inside the body to prevent from dehydrating whereas the gills are located

1 Max10

No 2(a)

Marking Criteria

Marks

Total

1. Inhalation/ breathing in

1

6

2. External intercostal muscles contract, internal intercostal

1

Able to explain the state of breathing shown in Diagram 7.2.

mucles relax. 3. The rib cage move upwards and outwards.

1

4. The diaphragm contracts and flattens.

1

5. Volume of the thoracic cavity increase resulting in

1

reduced air pressure in alveoli. 6. Higher atmospheric pressure outside causes the air to 1

rush in. (b)

Able to explain the effects of the breathing mechanism if structure R is unable to function.

1

1. Structure R is diaphragm. 2. Less/no change in volume in the thoracic cavity/ lung

1

3. Less/ no change in air pressure in the thoracic cavity/

1

lung 4. Less/ no air exchange/ less/no intake of O2/ less/no CO2

Max4

1

expelled 5. Resulting difficulty in breathing in and out (c)

1

Able to describe the regulation of the carbon dioxide concentration in body fluid. 1. During vigorous exercise, the partial pressure of carbon

1

dioxide increase. 2. Carbon dioxide dissolve in blood to form carbonic acid

1

3. and decrease the blood pH value

1

+

4. The drop in pH / increases in H ions detected by central

1

chemoreceptors (in medulla oblongata) and peripheral chemoreceptors at the aortic bodies and carotid bodies. 5. Nerve impuls is then send to respiration centre in m-Ob

1

6. The respiratory centre send nerve impulse to the

1

diaphragm and the intercostals muscles 7. Causing the respiratory muscles to contract and relax

1

faster. 8.

results the breathing and ventilation rates increase.

1

9. So excess co2 is eliminated from the body,

1

10. the carbon dioxide

1

return to normal

concentration and pH of the blood

10

3(a)

Able to differentiate the cellular respiration process that occurs in cell X and tissue Y. Cell X Anaerobic respiration

Tissue Y Aerobic respiration

(Because oxygen is absent) (Because oxygen is present_ Glucose is not completely Glucose is completely oxidized// Glucose

Fermented/ oxidized/ convert

to

ethanol and energy. The quantity of

breakdown//

CO2, Glucose is oxidize to CO2, water and energy. energy The quantity of

2

8

2

energy

produced is lower/ 2 ATP/ 210 produced is higher/ 38 ATP/ kJ (per molecule of glucose)

2898 kJ (per molecule of

Respiration equation:

glucose) Respiration equation:

Glucose → Carbon dioxide +

Glucose + Oxygen → Carbon

Ethanol + Energy

(b)

dioxide + Water + Energy

2

2

Able to state what is process Q and molecule X. Process Q

- Anaerobic respiration

Molecule X - Lactic acid

Max 6 1 1

Able to explain how molecule X can be remove from muscle cell. 1. Inhale more oxygen by doing fast and deep breathing. 2. Excess oxygen taken in during inhalation is used to oxidise lactic acid to carbon dioxide and water. 3. This oxidation process takes place in the liver. 4. Thus the oxygen debt is the amount of oxygen needed to

1 1 1 1

remove the lactic acid from the muscle cells. Lactic acid + oxygen (c)

carbon dioxide + water +

1

energy Able to explain the exchange of respiratory gases. 1. Respiratory surfaces in human is alveoli.

1

2. The concentration of oxygen in the alveoli is higher than its

1

concentration in the blood capillaries. 3. Oxygen in the alveoli diffuses into the blood capillaries.

1

4. The concentration of carbon dioxide in the blood capillaries is

1

higher than its concentration in the alveoli. 5. Carbon dioxide diffuses from the blood capillaries of the lungs

1

6

4(a)

Able to explain how gaseous exchange occurs in the alveoli and blood capillaries P1: Gas exchange is driven by diffusion //

1

max 10

Diffusion of a gas depends on differences in partial pressure between the two regions P2: thus does not require energy (for exchange). P3: The molecules move down a concentration

1 1

gradient. P4: Oxygen moves from the alveoli which is high oxygen concentration P5: to the blood which has lower oxygen concentration

1 1 1

P6: due to the continuous consumption of oxygen in the body. P7: Conversely, carbon dioxide is produced by

1

metabolism P8: has a higher concentration in the blood than in the air of alveoli P9: carbon dioxide diffuses out of the blood capillaries

1 1

into the alveoli P10: Oxygen in the lungs first diffuses through the alveolar wall and dissolves in the blood plasma. P11: then diffuse into red blood cells P12 (Oxygen) bind to hemoglobin. P13: allows a greater amount of oxygen to be

1 1 1 1

transported in the blood (b)

Able to explain the process of energy production by the athlete during and after the race. 1. The muscle cells of the athlete undergoes anaerobic

1

respiration to produce energy 2. During intensive physical activity / running / sprinting// when the athlete start running (t = 0), oxygen requirement increase

1

immediately to produce large amount of energy 3. The athlete holds his breath for a short period of time // the athlete breath is shallow during running 4. The oxygen supplied by breathing between t = 0 minute to 6

1 1

minute is insufficient for cellular respiration 5. The muscle cells are now in the state of oxygen debt // undergo oxygen deficit

1

max 10

CHAPTER 8: DYNAMIC ECOSYSTEM Question 1 No 1(a)

Marking Criteria - Colonisation is a process in which organism starts to inhibit an

Marks 1

unhabitat area such as bare ground and forms a colony. - Succession is the gradual process where one community changes its

1

environment so that it is replaced by another community. - Diused pond, pioneer species are Phytoplankton (microscopic algae)

1

and submerged plant (Hydrilla sp., Cabomba sp., elodea sp.) begin to grow and carry out photosynthesis. - When they die and decompose, organic matter converted into humus at the pond base, the pond become shallow.

1

Total

-The condition becomes favourable for floating plants such as water

1

hyacinths (Eichornia sp.) and duckweeds (Lemna sp.) -They spread covering water surface and prevent sunlight from

1

reaching the submerged plant causing these plants to die since they cannot photosynthesise. - The decomposed plants add more organic matter and the pond

1

becomes more shallow. - The emergent plants (sedges, cattails) replace the floating plants.

1

- They grow from the edge of the pond towards the middle of the pond

1

as the pond becomes more shallow. - When these plants die, their decomposed remains are added as

1

sediments to the base of the pond thus reduces the depth of the pond. - The condition becomes suitable for land plants like small herbaceous weeds.

1

- Gradually, the land becomes much drier and more land plants (shrubs, herbs and large woody plant) start to grow.

1 ANY 10

1. b)

- A jungle emerges and turns into a tropical rainforest which form a

1

climax community. – It is a balanced ecosystem which involves the interaction

1

between the abiotic and biotic factors. - Ensures conversation of biodiversity, preservation of flora, fauna and

1

organism. - Preventing the extinction of flora and fauna.

1

- Maintaining major sources of human food such as ulam, meat, honey

1

and sources of traditional medicinal herbs. - Sustains the food webs in the ecosystem.

1

- Preventing disruption of the natural cycles of water/carbon and also

1

balancing photosynthesis and respiration.

- Provides natural water catchment areas.

1

- Preserves natural resources for recreational activities and

1

eco- tourism. - Reduces stress and promotes a healthy lifestyle.

1

- A balanced ecosystem prevents the loss of plants which will cause a

1

reduction in food resources and food chains. - It will also cause soil erosion and flash floods.

1

- And the extinction of some animal and plant species.

1 ANY 10

Question 2 No Marking Criteria 2. a) - The ground is too soft and muddy soil, unable to support plants.

Marks 1

i)

1

- Waterlogged conditions of the soil decrease the amount of oxygen for root respiration. - The swamp water has a high concentration of salt and is hypertonic

1

compared to the cell sap of the root cells. - The plants growing in swamp will have the problem of dehydration.

1

- Seeds that fall into the muddy swamp will die of dehydration/

1

insufficiency of oxygen. - The swamp is exposed to the sun leads to a high rate of transpiration.

1

Total

- As a result, the plants growing there will lose water very fast by transpiration.

1

ANY 5

2.

- Root system which is highly branched and widespreads underground

a) ii)

cable roots to give good support to the plants. - Pneumatophores (breathing root) which grow protruding upwards

1 1

above the ground to allow gaseous exchange. - The cell sap of the roots cells has a higher osmotic pressure than the

1

soil water that surrounds them to enable water enter the root by osmosis. - Hence, the cells are able to withstand the high salt content of the

1

swamp. - Excess salt is eliminated through hydatodes found at the lower

1

epidermis of leaves. - Viviparous seeds which germinate while still attached to the mother

1

plants. - The long radical produced will let the seedling stick into the ground,

1

not submerged into the soft and waterlogged soil or drift away. - Thick layer of cuticle covers the leaves and succulent which help to

1

reduce the rate of transpiration and store water. ANY 5

10

Question 3 No 3.a)

Marking Criteria -The diagram shows a saprophytic fungus.

Marks 1

- It obtain its nutrient by secreting digestive enzymes onto the

1

Total

substrate.

3. b)

- The enzymes digest the complex substances into simple forms.

1

- The simple forms are then absorbed by hypha.

1

- Owls are predators and rats are preys.

1

- In the month of January until April, the increase in the prey’s is

1

followed by an increase in predator population. - Due to abundance of food.

1

- However from April to August, when the number of predator

1

4

increases, the number of preys will then decrease. - This is because the high number of predators will easily

consume

1

the prey. - When the prey reduces, the predators will have less to eat.

1

- There is intraspecific competition.

1

- The number of predators also reduces in the following months from

1

August to December. -The prey-predator relationship takes place in cycle.

1

- This keep the population of both organisms in a dynamic equilibrium.

1

10

Marks 1

Total

ANY 6

Question 4: No 4.

Marking Criteria - Rhizobium bacteria inside the root nodules of legumes and Nostoc sp. found freely in the soil fixed the nitrogen in air. - Decaying bacteria/ fungi decompose plant/ animal/ dead organism/

1

waste product - To form ammonium compound.

1

- Nitrosomonas sp. / nitrifying bacteria converted ammonium

1

compound to nitrite. - Nitrobacter sp. /nitrifying bacteria convert nitrite to nitrate.

1

- Nitrate is absorbed by plant to form plant protein.

1

- ( Plant protein ) eaten by an animal to form animal protein.

1

- Denitrifying bacteria reduce the nitrate content in the soil.

1

- by converting the nitrate into the nitric oxide and nitrogen gas.

1

- Nitrogen gas goes back into the atmospheric to complete the

1

nitrogen cycle.

10

CHAPTER 9: ENDANGERED ECOSYSTEM Question 1 No 1

Marking Criteria - Industries/ factories/ vehicle contribute to air pollution.

Marks

- Smoke/ fine solid particles can cause respiratory problem.

1

- oxides of nitrogen/ sulphur dioxide dissolve in rain to produce acid

1

Total

rain. - (acid rain) causing the soil become acidic/ unsuitable for cultivation

1

of crops/ leaching of mineral/ corrosion of metal. - Increase Carbon dioxide in atmosphere causes the greenhouse /

1

global warming. - Industrial/ domestic/ agricultural activities produce waste to contribute water pollution. - Agrochemical / pesticides/ insecticides used by farmer flow into the river/ lead to the poisoning of aquatic organism.

1

4

- Agricultural run-offs contain excess nitrates/ phosphates lead to

1

eutrophication. - (eutrophication) causes the BOD value will increase thus may harm

1

the aquatic organisms. - Effluents from electronics factories contain heavy metals/ mercury/ cadmium kill the aquatic organism/ disturb food chain.

1

4

- Discharged of hot water from industries / glass building cause thermal pollution. - Increase the water temperature in the river causing died aquatic

1

organisms - Increase the atmosphere temperature.

1

2

Marks 1

Total

Question 2: No 2

Marking Criteria - The phenomenon is acid rain. - Factories / motor vehicles released large amount of oxides of nitrogen to the atmosphere.

1

- and sulphur dioxide.

1

- oxides of nitrogen combines with water vapour (in the atmosphere) to

1

form nitric acid. - Sulphur dioxide combines with water vapour (in the atmosphere) to

1

form sulphuric acid. - The rain falls as acid rain.

1

Effects: - May corrode buildings.

1

- Aquatic lives may die due to acidic water/ low pH

1

- Minerals in soil will be dissolved/ washed into rivers.

1

- Soil becomes infertile / not suitable for agriculture.

1

6

- Plants may die due to infertile soil/ acidic soil.

1

- The ecological balance of ecosystem disrupted.

1 ANY 4

10

Question 3: No 3

Marking Criteria - The destruction of ozone layer is due to the increasing levels of

Marks 1

Total

chlorofluorocarbons (CFCs) in the atmosphere. - CFCs are a group of chemical compounds that contain chlorine,

1

fluorine and carbon. - CFCs are widely used as coolants in air conditioners and refrigerators, as propellants in aerosol cans and as foaming agents in

1

the making of polystyrene packaging, pillow, cushions etc. - The CFCs in the atmosphere are struck by UV light forming chlorine

1

atoms. - Chlorine atom then breaks the ozone molecule into chlorine

1

monoxide and oxygen gas. - Chlorine monoxide then reacts with the free oxygen atom in the

1

atmosphere to form chlorine atom and oxygen molecule. - The chlorine atoms repeat the breaking of the ozone molecules

1

causing the depletion of the ozone layer continuously. ANY 4

4

The effects of excessive ultraviolet radiation on human: - reduction of the body’s immune system.

1

- Skin cancer - Cataract of the eyes.

1

2

1

1

1

1

ANY 2 The effect on plants: - reduction of the growth therefore reducing crop yields. Effect on aquatic organism - Death of plankton, reduce food supply to aquatic organism, fisherman’s catch is reduced. Steps to overcome this problem: - Reduce or stop using CFC or chlorine based products.

1

- Replace CFC with HCFC.

1

- Use wrapping papers instead of polystyrene boxes.

1

- Patch up the holes in the ozone layer by firing frozen ozone balls into

1

2

the atmosphere. ANY 2

10