SKEMA TUTORIAL

CHAPTER 2: ECOLOGY

CLO: 2.1 :a) Define ecosystem
b) Describe lake ecosystem based
i. light penetration (photic and aphotic)
ii.distance from shore and water depth (littoral, limnetic)

1.

a)
All the organisms in a given area as well as the abiotic factors with which they
interact; one or more communities and the physical environment around them

b)
A : littoral zone
B : limnetic zone
C : aphotic zone
D : benthic zone

c)
Organism: Benthos
Example: Nematode/ protozoa/ bacteria/ crustacean

d)
Photic zone because it has the highest light penetration required for
photosynthesis

e)
Littoral zone Because it has the highest number of microhabitats

f)
i. Based on light penetration.
ii. Photic zone.

iii. The upper layer where light is sufficient for photosynthesis.
iv. Aphotic zone
v. The lower layer receives little light/ where light is insufficient for

photosynthesis.
vi. Based on distance from the shore and water depth.
vii. Littoral zone
viii. Shallow/water close to shores
ix. Rooted and floating aquatic plant/insect/water

strider/crustaceans/frogs/fish/weed.
x. Diversity is greatest here.
xi. Limnetic zone.
xii. Well-lit, open surface waters/far away from the shore
xiii. Occupied by tiny organism such as

xiv. plankton/zooplankton/phytoplankton/fish/cyanobacteria/diatoms/green algae.
xv. Benthic
xvi. The sediment at the bottom
xvii. Zone of decomposition
xviii. Dominated by bacteria/ decomposer/benthos

CLO: 2.1 : c) Describe terrestrial ecosystem of tropical rainforest stratification (emergent,
canopy, understory, shrub, ground layer/ forest floor)

2.
a)

A: Emergent
B: Canopy
C: Understory
D: Shrub layer

b) i)
Small leaves/ some species are deciduous during the brief dry season.

ii)
Snake, lizard, butterflies, birds, squirrel, insects

c) i) Canopy / Understory

ii)
To get enough sunlight

d)
Usually warm the whole year and receive heavy rainfall
The consistence sunlight provides essential energy for photosynthesis

e)
Plant in understory strata must be shade-tolerant since they're growing under
the canopy trees. The size of the tree also smaller compare to size of plant in
canopy.

CLO: 2.2 : Energy flow through ecosystem.
a)Explain the energy transfer in ecological pyramids in relation to trophic level.
(CLO 3)
b)Calculate energy loss in each trophic level

3.
a)

Photosynthesis

b)
Feeding process

c)
10 %

d)
10/100 X 440000kJ = 44000kJ (NPP primary consumer)
10/100 X 44000 kJ = 4400kJ (NPP secondary consumer)

c)
Some of the energies are lost as heat by respiration and metabolic activity.
Some of the organisms may have died and are decomposed by
saprophytes.

d)
Large amount of energy is lost to the environment during each transfer as heat
through respiration, excretion and transpiration.
Each trophic level receives less energy than the level below it.
At the 4th or 5th trophic level, only a small amount of energy is left.
So, it is insufficient to support the further trophic levels.

e)
GPP = 4500kJ + 500kJ = 5000kJ

f)
425/ 500 kJ X 100 = 85%

.4.
a)

Grazing food chain

b)
Organism obtain organic molecules by synthesizing them from organic materials.

c)
Producer

d)
Energy is loss through evaporation/ Transpiration/ Respiration. (any two)

e)

Four
f)

i) Herbivores: Insects/ squirrel/ rabbit/ rat/ grasshopper
ii) Carnivore P: Insectivorous bird
iii) Carnivore Q: Hawk/ fox
iv) Detritivores: Rag worm/ earthworm

g)
a.
The Pyramid of numbers
The Pyramid of biomass
The Pyramid of energy

5.
i. Food chain reflect the transfer of energy in a ecosystem.
ii. Original source of energy is from the sun// solar energy.

iii. Primary producers// photosynthetic organisms trap (1-5% of) light energy
received for biomass production.

iv. Primary consumers// herbivores feed on plants // primary producers.
v. Energy is transferred from first trophic level to the second trophic level.
vi. 90% of the energy is lost to the environment // only 10% of the biomass

produced consumed by the herbivores // 10% of the energy is transferred to the
next trophic level.
vii. There is further energy lost through respiration and excretion.
viii. Secondary consumer feed on primary consumers.
ix. Energy is transferred from second to the third trophic level.
x. tertiary consumers feed on secondary consumers.
xi. There are seldom more than five trophic levels.
xii. Large amount of energy is lost during each transfer (within & between
organism) // each trophic level received less energy than the level below it.
xiii. At fourth or fifth level, only a small amount energy left / insufficient to support
further trophic level.
xiv. Limiting the number of members in the higher trophic level

CLO: 2.3 : Biogeochemical cycle.
a) Describe biogeochemical cycle components (cycling pool and reservoir
pool in Carbon and Nitrogen cycles.
b) Illustrate phosphorus cycle.

6. a)
Cycling of chemical elements or nutrients from the abiotic environment to the
organism and then back to the abiotic environment.

b) Biogeochemical cycle can be divided into 2 components:
i. A large reservoir pool of elements.
ii.

iii. Chemical elements in inorganic forms in the abiotic components of
biosphere.

iv. From where the element are absorbed by autotrophic organism and
converted into complex organic form.

v. Example: sediments/ soil, atmosphere, ocean, fossil fuels.
vi. An exchange/ cycling pool.
vii. Chemical elements or nutrients in complex organic forms in the biotic

components of biosphere.
viii. The elements is passed along food chain and ultimately converted back into

inorganic form to be released into the environment.
ix. Example: plants, animals.

c)

(i)

A : CO2 in atmosphere B : Dissolved CO2 C : Fossil fuel (gas, oil and coal)

(ii)
Break down, remove and recycle waste product and dead matter.
Recycle these products and return into the environment as carbon dioxide in
form of respiration.

(iii)
HCO3- / CO3 2-
CaCO3

(iv)
Extra carbon dioxide is contributing to the accelerated greenhouse effect
which is causing global warming

7.

a)
Basic component of proteins in plant and animals tissues.

b)
A: Nitrification
B: Ammonification
C: Denitrification
D: Nitrogen fixation

b)
A: Nitrobacter and Nitrosomonas
D: Rhizobium and Azotobacter

c)
Symbiotic nitrogen fixing bacteria are able to fix nitrogen and converting them into
nitrogenous compounds. The bacterium gives the plant source of nitrogen. In
return the host plant supplies carbohydrates to the bacteria.

d)
Nitrification is a process where nitrogenous compound are oxidized into nitrate
salts the soil while denitrification is a process where nitrates are converted into
nitrogen gas.

Nitrification involves two types of nitrifying bacteria, Nitrosomonas and Nitrobacter
while denitrification involve two types of denitrifying bacteria, Pseudosomonas
and Thiobacillus.

8.

a)
Organism required phosphorus as a major constituent of nucleic acid,
phospholipids ,ATP and other energy-storing molecule and as a mineral
constituent of bone ant teeth.

CLO 2.4 : Conservation and management.
a) Describe sustainable development.
b) Explain threats to biodiversity in Malaysia.
c) Illustrate conservation of diversity in Malaysia.

9. a)
Development which meeting the needs of the present without compromising the
ability of future generations to meet their own needs.

b) Sustainable Forestry, Sustainable Yield in Agriculture and Sustainable Yield in
Fishery

c)
(i) Climate change: we want to emit less carbon into atmosphere.
(ii) Renewable: These sources will not run out.
(iii) Security: Reduced dependence on foreign energy like oil

CLO 2.5 : Population Ecology
a)Explain biotic potential and environmental resistance and their effects on
population growth.
b)Explain carrying capacity and its importance.
c)Describe natality and mortality and their effects on the rate of population growth.
d)Explain population growth curves (state the basic forms of growth curves).
i) Exponential growth curve (human)
ii) Logistic growth curve (Paramecium sp.)
e)Explain the limiting factors affecting the population size:
i) Density dependent factors.
ii) Density independent factors.

1.

a)
Graph I
The number of offspring that would survive to reproductive stage under ideal
conditions.

b)
Graph I: Exponential growth curve
Graph II : Logistic growth curve/ Sigmoid growth curve

c)
A: Lag phase. Population growth is slow. The period where the organism prepare
for growth with very little or no cell multiplication.
B: Log phase.Population growth increase rapidly or exponentially. Growth of the
population is at near their biotic potential. All individuals in the population have
access to plenty of food, water and shelter. Predators are scarce and
reproduction is successful.
D: Stationary phase. Population growth is constant.

:
d)

Predation/ competition /lack of resources (food supply/water etc) / disease/
parasitism/ territorial

2.
a)
The organisms are the beginning to adapt with the new environment.

b)
Phase Q: Natality rate is higher than mortality rate
Phase R: Natality rate is equal to mortality rate.

c)
Environmental resistance is the factors that limit the biotic potentials of the
population

d)
Space / Predation / Toxicity

e)
Prey – Paramecium aurelia Predator – Didinium nasutum

f)
Dilute the medium or ensure no methyl cellulose is added to the medium.
This is to ensure the movement of the prey is not limited.

3.
a) i.

2000

ii.
Prey : Paramecium caudatum
Predator : Didinium nasutum

b) i.
Limited food source

ii.
Logistic growth curve

iii.
Increase
Because less number of predator

c)
Natality, mortality,emigration, immigration

4.
a)

To make sure food sources are sufficient/ as a food source

b) i.
Logistic growth curve.

ii. Competition to get the food / has achieved the carrying capacity.

c)
They cannot escape from Tribolium sp./ being eaten by Tribolium sp.

d) i.
Interspecific competition.

ii.
Density-dependent factor.
Predator/ Parasite Infection/ Disease/ Territorial behavior

e) Space.

5.

Mark given:
(i) Correct Axis: – X and Y - with title –>1 mark
(ii) Correct Phase with labelling ---> 1 mark
(iii) Correct Curve ----> 1 mark

i. Logistic growth curve has 4 phases which are Lag phase, Log phase, Decelerating
phase and Stationary phase.

ii. Lag phase.
iii. Population growth is slow.
iv. Population just tries to adapt to the new environment.
v. Very few organisms in a population at early stage of growth.
vi. Log/ exponential phase.
vii. Population growth increase rapidly/ exponentially.
viii. A period when there are no constraints on population growth due to excessive

nutrients/food, enough space & no accumulation of waste product / diseases.
ix. Decelerating phase.
x. Population growth slow down.
xi. A period when there are limitations on population growth due to limited food

supply & not enough space because of competition between individuals.
xii. Stationary phase.
xiii. Population growth is constant.
xiv. Size of population remain constant (rate of population growth is zero/ no net

growth).
xv. Number of offspring produced is equal to number of offspring died (birth rate =

death rate).
xvi. Example population is Paramecium sp.

6. Density dependent factors is the factors that population growth rates are affected
by population size.
i. Competition for resource/food/growth requirement.
When the population density is high/resources are limited, the competition
ii. increase.
iii. Competition occurs between individual of the same population/ intraspecific
competition //between population of different species / interspecific competition.
iv. Territorial behaviour.
There is little or no overlap between neighbouring territories of the same species.
v. Species without territories often die from predator/starvation.
vi. Predation/ predator-prey interaction relationship.
vii. An interspecific interaction which result in cyclic fluctuation in population size.
viii. An increase in the prey population support an increased in predator population.
ix. Lead to a drastic decrease in number of prey organism.
x. Overcrowding/ limited space.
xi. Occur at high population size.
xii. Cause excessive show a decline in health/hormonal change.
xiii. Which cause infertility /aggressive behaviour.
xiv. Parasitism/disease/ health
xv. Parasitic infection /disease spread more easily in a high-density population.
xvi. Accumulation of toxic waste/waste product.
xvii. As population grows, waste product from metabolic reaction accumulates.
xviii. And become toxic to the member of the population.
xix. Density independent factor is factors which tend to affect the size of a population
xx. independent or regardless to the population density.
xxi. A typical abiotic condition.
Example storm, fires, drought, flood and other natural occurrences.
xxii.
xxiii.

CHAPTER 3: SELECTION AND SPECIATION

CLO 3.1 : Selection
(a) Define natural selection.
(b) State three types of natural selection: stabilizing selection,
disruptive selection and directional selection.
(c) Analyse stabilizing selection, disruptive selection and directional selection
with examples
(d) Define artificial selection.
(e) Describe inbreeding and outbreeding.

1.

(a)
A: directional
selection B:
disruptive selection

(b) Selection B acts against intermediate phenotype and favours the two
extreme phenotypes. The intermediate phenotypes are gradually
decrease in number and split the population into two subpopulations.

(c)
Similarity
Both selection A and stabilising selection reduce genetic
variability. Differences
Selection A favours the extreme phenotype while stabilising
selection favours Intermediate phenotype // Selection A act
against one extreme phenotype while stabilising selection act
against both extreme phenotypes.
Selection A occur when there Is a gradual change in the environment
while stabilising selection occurs when there is no change in the
environment.

The ability to adapt with Its environment // have
greatest fitness.

i. Stabilising selection is natural selection that favours intermediate
phenotype.

ii. Acts against both extreme phenotypes.
iii. Intermediate phenotypes have greater survival and reproduce

successfully.
iv. This selection reduces genetic variability.
v. It is occurring in most population when there Is no change in

the environment ..Example: wing length of hawk / eagle

i. Disruptive selection is natural selection that favours both extreme
phenotypes.

ii. Both extreme phenotypes have greater survival and reproductive success.
iii. Increases the chances of the advantageous allele to be passed on to next

generation.
iv. Selection against the intermediate phenotype.
v. The intermediate group decrease from generation to generation.
vi. This can split a population into two subpopulations.
vii. Promotes genetic variability.
viii. Occurs when environmental condition is varied

Example: variety of beak size and shape of Galapagos / Cameroon finches

i. Directional selection is natural selection that favours one extreme
phenotype.

ii. Act against the other extreme phenotypes.
iii. Shifts the curve/ mean towards the favoured phenotypes.
iv. Reduces genetic variability.
v. Occurs when there is a gradual change in the environment make

population inhabit area with extreme condition / migrate to new habitat
with different environmental conditions. Example: industrial melanism /
Biston betularia/ beak size of Galapagos finches

2.

(a)

Large beak size.

Because their food is still available / Only large seeds Is available.

(b)

(c)
Stabilising selection: Wing length of hawk / eagle
Disruptive selection: Variety of beak size and shape of Galapagos/
Cameroon fiches Directional selection: Industrial melanism / Biston
betularia // Beak size of Galapagos finches

4..
(a) Human Intervention in animals or plants reproduction to ensure certain
desirable traits are represented in the next generation.
(b)
Inbreeding and outbreeding

(c)
More meat and more milk.

(d)
Increase homozygosity that led to a loss of vigour / reduction in fertility.
Reduce the variability of the genome / Eliminate heterozygous genotype
from the inbred population.

(e) Inbreeding is the mating between closely related individuals while
outbreeding is mating between unrelated / distantly related
individuals.
Inbreeding increase homozygosity while outbreeding Increase
heterozygosity. Inbreeding shows less variation while
outbreeding shows more variation.
Inbreeding produce offspring with less vigour while outbreeding produce
hybrid more vigour.

(f)
Have greater strength/ resistant to disease/ longer lifespan

(g)
i. decreased hybrid vigour; decreased biological fitness of a
population (inbreeding depression), which is its ability to
survive and reproduce.
ii. increased incidence of metabolic disorders, structural
abnormalities and Inherited disease conditions caused by
harmful recessive genes.
iii. loss of genetic variation; Increased in homozygosity.
iv. decreased fertility; poor breeding success.
v. greater risk of spontaneous abortions; still birth and early
infant deaths, high juvenile mortality.
vi. shorter lifespan of animals and decreased yield in crops.

CLO 3.2 : Speciation
a) Define the biological species concept.
b) Describe modes of speciation: sympatric and allopatric.
c) State the processes that leads to speciation.
d) Relate these processes to speciation.
i. Reproductive isolation.
ii. Genetic drift.
iii.Hybridization.
iv. Adaptive radiation.

5.
(a) A species is one or more populations of individuals that are

interbreeding under natural conditions and producing fertile offspring

and are reproductively isolated from other such populations.

(b)
R: allopatric speciation
S: sympatric speciation

(b)

Modes R / Allopatric speciation gene flow is interrupted / cause by geographical barrier

while modes S / sympatric speciation takes place in geographically overlapping

population/ cause by sexual selection / non-random mating // formation of polyploidy//

new ecological niches. - 1 mark

Modes R / Allopatric speciation, population forms a new species while geographically

isolated but modes S / sympatric speciation (involved a small population) becomes a new

species without geographic speciation. – 1 mark

Mode R / Allopatric speciation, population undergo evolutionary change and reproduction

isolation established while modes S / sympatric speciation involved a reproductive barrier

isolates subset of a population without geographic separation from the parent species.

- 1 mark

Any 1 answer - 1 mark

(c)
A population forms a new species without geographical barrier.
Results from the new ecological niches/ non-random mating due to
sexual selection. A reproductive barrier / isolation has arisen / evolve
between a subset of a population.

6.

(a)
Bottleneck effect

(ii) At P

(b) : A small colony of between 20 and 100 Individuals survived
Q on Guadalupe Island of Baja California.
: From that small colony have come the approximately 160,000
R elephant seals found on the Pacific coast today.

(c) Reduce the adaptability of the species to environmental change/ the
species may not be able to adapt to new selection pressures such as
climatic changed / a shift in available source.

(d)
Extinct

7.

(a) Founder effect occur when small population is isolated/ migrated out
from original large population and form new colony with different gene
pool.

(b)
Because they marry within their own community / Inbreeding occur which
prevents new genetic variation from entering the population.
Many of them are homozygous for harmful recessive allele. Children are
therefore more likely to inherit two copies of the particular recessive
genes that lead to genetic disease.

(c)
Finnish people / Afrikaner

8.

(a)
Radiation

(b)
Competition
A species may encounter a wide variety of unoccupied habitats

(c) Enables the animals to occupy a specific niche each even though they
are living together.
Enables the organism to be isolated physically then
reproductively. One species diversify to many related
species.
Formation of new species.

(d)
Reproductive isolation // hybridisation // genetic drift

9.
i. the allopatric speciation occurs when a population is geographically
isolated into two subpopulations.
ii. mutation may occur Independently in one or both subpopulations
giving rise to new alleles.
iii. different natural selection pressures will act on both subpopulations.
iv. gene flow from one subpopulation to the other subpopulation is prevented.
v. both subpopulations will eventually give rise to the formation of two new
species.
vi. the sympatric speciation occurs when two or more new species are
formed from a single species in the same geographical location.
vii. the reproductive isolation prevents members of two subpopulations
from producing fertile offspring.
viii. reproductive isolating mechanism refers to any structural, functional or
behavioural characteristics that prevents successful reproduction from
occurring.
ix. Isolation can be brought about by prezygotic barriers.
x. Such as temporal isolation, habitat isolation, behavioural isolation,
mechanical and genetic isolation.
xi. That prevent mating.
xii. Isolation can be brought about by postzygotic barriers.
xiii. Such as hybrid inviability, hybrid sterility or hybrid breakdown.
xiv. That reduce the viability or fertility of the zygotes after mating has occurred.

10.
i. Habitat/ ecological Isolation
ii. Two species live in different habitat within same area and do not meet /
breed
iii. Seasonal/ temporal Isolation
iv. The two species mating/ breed/ flowering at different times/ season.
v. Behavioural Isolation
vi. The two species exhibit different courtship behavioural pattern
vii. Mechanical Isolation

viii. Two related species have structural differences in their
reproductive organ/incompatible to each other.

ix. Gametic Isolation
x. Gametes of difference species are chemically incompatible so

they fall to fuse to form a zygote.
[Any 4]

11. Post-zygotic isolating mechanism Is a mechanism that prevents
i. the zygotes from developing into normally functioning fertile
Individuals.
ii. There are 3 categories to describe post-zygotic Isolating mechanism:
iii. Hybrid inviability
iv. hybrid zygotes fall to develop / mature.
v. Hybrid sterility.
vi. F1 hybrid fall to produce functional / viable gametes.
vii. Hybrid breakdown.
viii. F2 or backcross hybrid have reduced viability/ Infertility.

12. Hybridisation is the process of Interbreeding between individuals of
i. different species.
The F1 hybrids are sterile.
ii. During meiosis in the F1 hybrids, chromosomes from each parent
iii. cannot pair together to form homologous chromosome.
Meiosis cannot occur; therefore, gamete cannot be produced.
iv. Occasionally chromosome doubling may occur in F1 hybrids.
v. Fertile F2 hybrids is produced.
vi. Because homologous chromosome can pair up during meiosis as two
vii. sets of parental chromosomes are present.
Thus, they represent a new biological species.
viii.

13.
i. adaptive radiation refers to the formation of diverse species from a
single ancestral species in a short period of time to fill many
ecological niches.
ii. this Is an evolutionary process driven by mutation and natural selection.
iii. adaptive radiation often occurs as a result of an organism arriving in an

environment with unoccupied niches, such as a newly formed lake or
isolated Island chain.

iv. the colonising population may diversify rapidly to take advantage of all
possible niches.

v. an example of adaptive radiation Is shown by Darwin's finches.

vi. these finches are closely related but have different beak shapes and sizes.

vii. The beaks are adapted to the type of food that the finches eat.











CHAPTER 4
BIOCATALYSIS

4.1 PROPERTIES OF ENZYME AND MECHANISM OF ACTIONS
(a) State the properties of enzymes.

1.
A biological catalyst that are capable of speeding up the rate of biological
reactions by lowering the activation energy and remain unchanged at the end of
the reactions.
All enzyme are globular protein.
Function to lowering the activation energy.
Reaction is extremely specific.
Not altered at the end of reaction.
Required in small amount to catalyse the reaction.
Reaction is extremely efficient.
May catalyze the reversible reaction.
Enzyme can be denatured by extreme pH/high temperature

4.1 PROPERTIES OF ENZYME AND MECHANISM OF ACTIONS
(c) Explain how enzyme lowering the activation energy
(e) State the factors that affect the enzymatic reaction.

2.

FIGURE 1
a)

b) No catalyse reaction occur. So the rate of reaction become slow and
activation energy become high.

c)
pH
Substrate concentration

d) Enzymes consist of polypeptide chains held in particular position by
several bonds such as hydrogen bonds, disulphide bonds, ionic bonds.
When high temperature the enzymes molecules vibrate and twist so rapidly
that some of their non-covalent bonds break.
The polypeptide chain open up and randomly arranged.
The protein loses its three dimensional globular shape.
The shape of active site no longer complementary with the shape of its
substrate.
The enzyme become denatured, so the rate of reaction will decrease/slow
down.

4.1 PROPERTIES OF ENZYME AND MECHANISM OF ACTIONS
(b) State the six classes of enzyme according to International Union of
Biochemistry and Molecular Biology (IUBMB) classification.
(d) Illustrate to explain the mechanism of enzyme action based on Induced Fit
Model.

3.

a)
Induced Fit mechanism

b)

I : Reactant/substrate III : Enzyme
II : Active site of enzyme IV : Products

c)
Bring the molecule close to each other (correct orientation)
Stress on bond in the susbtrate (making it easier to reach transition state)
Providing a favourable microenvironment

d)
. Because enzyme have a specific active site in which only substrate that

have complementary shape with the active site of an enzyme can bind to
form an enzyme-substrate complex.

e)

Enzyme classes Type of reaction catalysed
Oxidoreductases Catalyse all redox/oxidation-reduction reaction. By
the transfer of hydrogen , oxygen or electrons from
one molecule to another.

Transferases Catalyse the transfer of functional group from one
Hydrolases molecule to another
Lyases Catalyse the hyrdrolysis of substrate//breakdown of
Isomerases substrate by addition of water
Ligases Catalyse the breaking of chemical bond without the
addition of water
Catalyse isomerization//the arrangement of atoms
within molecule by converting one isomer to another
Catalyse reactions in which new chemical bonds are
formed and uses ATP as energy source

4.2 COFACTOR
(b) Explain the three types of cofactors and functions of metal ion activator,
coenzyme and prosthetic group.

4.

a) Y is not a protein and it will not denature

b)
Y is prosthetic group. Example is haem.

They function as carrier and transfer groups of atoms, single atom or
electron from one molecule to another.

c)
Coenzyme and metal ions

d)
Metal ions

4.2 INHIBITORS
(b) Explain the roles and types of inhibitors: competitive inhibitors and non –
competitive inhibitors
(c) Analyse graph related to competitive and non competitive inhibition.

5.
a) Competitive inhibitor. Because the maximum rate of reaction still can be
achieved when increase in substrate concentration. But it take longer time
to form the products.

b)
Inhibitor 1: Malonate/malonic acid
Inhibitor 2: Isoleucine

c)
Inhibitor I /competitive inhibitor and substrate compete for the same active
site of enzyme while Inhibitor II /non-competitive inhibitor and substrate
bind to different site on enzymes/allosteric sites

In the presence of of inhibitor I, maximum rate of reaction can be reached
when more substrate concentration was added while In the presence of of
inhibitor II rate of reaction decrease /cannot reached to maximum rate of
reaction

d)
An enzyme active site has regions, which form chemical bonds to hold the
substrates.
The shape of the enzyme’s active site exactly complementary to the shape
of the substrate molecule, which combines with it.
This configuration has led to the name of the lock and key model of
enzyme action.
A competitive inhibitor which has similar shape to the substrate binds
to the active site of an enzyme, thereby preventing the binding of the
substrate.
However it can be displaced by a substrate molecule.
So the substrate and inhibitor compete for the active site.

CHAPTER 5: CELLULAR RESPIRATION & FERMENTATION

5.1: Aerobic respiration
(b) Outline the complete oxidation of glucose which
involves
glycolysis, Krebs cycle and oxidative phosphorylation.

1.

a)
A. Glucose
B: Pyruvate
C: Acetyl CoA
D: FADH2
E: NADH
F: Water
G: Oxygen
H: CO2

b)
i) Glycolysis : Cytoplasm / cytosol
ii) Formation of Acetyl CoA : Mitochondria
iii) Krebs Cycle : Matrix of mitochondria
iv) ETC and chemiosmosis : Inner membrane of mitochondria

c)

Process Net ATP via NADH FADH2 Total ATP produce via
Substrate level Oxidative
phosphorylation
phosphorylation

A→B 2 ATP 2 0 6 ATP
NADH

B→C 0 20 6 ATP
NADH

Krebs 2 ATP 62 22 ATP
cycle NADH FADH2

TOTAL : 4 ATP TOTAL : 34 ATP

5.1.1: Glycolysis.
5.2: (a) Illustrate to explain glycolysis pathway (from glucose to pyruvate)
Fermentation and its application

2.

a)
Process glycolysis.
Location in cytosol / cytoplasm

b)
Step P & Q.
To increase the energy level of molecule // to supply energy to the
molecule // so that it will become chemically reactive and will be easily break
down.

c)
Step P: Hexokinase
Step Q: Phosphofructokinase

d)
Step S.
As reducing agent // transfer its electron to Electron Transport Chain

e)
0 (zero) NADH and 1 pyruvate

f)
2 pyruvates, 2 net ATP, 2 NADH + 2 H+ and 2 H2O

g)
Pyruvate will undergo lactate fermentation // Pyruvate will be reduced into
lactate (by oxidation of NADH → NAD+)

5.1.2 Krebs cycle.
(a) Illustrate to explain Krebs cycle.

3.

a)
Further oxidation of acetyl CoA into CO2 and produces NADH,
FADH2 & ATP // Metabolic reaction that oxidizes acetyl CoA,
organic fuel derived from pyruvate to CO2

b)
Matrix of mitochondria

c)
P : Oxaloacetate

Q : Citrate
R : Succinate

d)

1. Condensation/ entry of acetyl group
2. Isomerization
3. Oxidative decarboxylation
4. Oxidative decarboxylation
5. Substrate level phosphorylation
6. Dehydrogenation / oxidation
7. Hydration
8. Dehydrogenation / oxidation

5.1.3 Oxidative phosphorylation: Electron Transport Chain and Chemiosmosis
4.
(a) Illustrate to explain electron transport chain: The pathway of
electron

transport is NADH dehydrogenase, Ubiquinone/CoQ, cyt c
reductase,

cyt c, cyt c oxidase.
(b) Explain chemiosmosis: proton motive force.
(c) Explain complete oxidation of one molecule of glucose in

active cells to produce 38 ATP.

a)
Protein X : ATP synthase
Protein complex I : NADH dehydrogenase
Protein complex II : Succinate dehydrogenase
Protein complex III : Cytochrome c reductase
Protein complex IV : Cytochrome c oxidase
Molecule Z : Water

b)
Inner membrane of mitochondria

c)
NADH : Protein complex I / NADH dehydrogenase

FADH2 : Protein complex II / Succinate dehydrogenase

d)
Energy that was released from oxidation of NADH and FADH2 at protein
complex / Energy that was released from exergonic flow of electron from
NADH and FADH2.

e) Because protein complex II is not a proton pump.

f)
Chemiosmosis is an energy-coupling mechanism that uses energy stored in
the form of H+ gradient across a membrane to drive cellular work
(such as ATP synthesis)

g)
Proton-motive force is the force used to drive H+ back down their gradient,
across the membrane through by ATP synthase/ protein X.
The ATP synthase harness the proton-motive force/the flowing of H+ to
phosphorylate ADP into ATP.

h)
Y will combine with 2H+ and then reduced to form water/H2O

i)
NADH : 3ATP
Reason : Oxidation of one NADH will totally pumped out 3H+(which then
flow through ATP synthase to synthesize 3 ATP)
FADH2 : 2 ATP
Reason : Oxidation of one FADH2 will totally pumped out 2H+(which then
flow through ATP synthase to synthesize 2 ATP)

5.1: Aerobic respiration
(b) Outline the complete oxidation of glucose which involves glycolysis,
Krebs cycle and oxidative phosphorylation.

5.2: Fermentation and its application.
(a) Explain lactate and alcohol fermentation.

5.

a)
Y : Pyruvate / Pyruvic acid Z : Acetyl CoA

b)
When oxygen is not present (substance Y will undergo process P
and Q) in order for NADH to be reoxidized into NAD+ for reuse as an
electron carrier for glycolysis.

b)
P process.
Because does not release CO2 / not undergo decarboxylation

c) P process not produce intermediate substance but Q process produce

intermediate substance (acetyldehyde / ethanal)
P process does not release CO2 but Q process release CO2 through
decarboxylation process.

d)
Inner membrane of mitochondria

e)
4 ATP

f)
Because process R and S involve both substrate level phosphorylation and
oxidative phosphorylation where the production of ATP is from the oxidized
NADH and FADH whereas in process P and Q, only involved substrate level
phosphorylation which is a direct phosphorylation of ADP with a phosphate
group by using the energy obtained from a coupled reaction

5.2: Fermentation and its application.
(a) Explain lactate and alcohol fermentation.

6.

a)
A catabolism process of converting glucose to lactate (animal cell) or ethanol
and CO2 (plant cell) in an absence of oxygen //
A catabolic process that makes a limited amount of ATP from
glucose (or other organic molecules) without an electron transport
chain and produces alcohol or lactic acid.

b)
L : Pyruvate
M: Ethanal / Acetaldehyde
N: Krebs cycle / Citric acid cycle / Tricarboxylic acid cycle (TCA)

c)
Hexokinase and Phosphofructokinase

d)
Cytosol / Cytoplasm

e)

L to M : Decarboxylation

M to ethanol: Reduction

5.2: Fermentation and its application.
(a) Explain lactate and alcohol fermentation.
(b) State the importance of fermentation in industry

7.

a)
A : Pyruvate
B : Lactate / Lactic acid
C : Acetaldehyde / ethanal
D : CO2

b)
Glycolysis

c)
Alcohol fermentation / ethanol fermentation

d)
Fermentation occur in order for NADH to be reoxidized into NAD+ for reuse as
an electron carrier for glycolysis / to keep supplying the NAD+ so that
glycolysis process can occur continuously even in absent of oxygen //
Fermentation is alternative pathway to produce ATP in absent of oxygen

e)
Act as reducing agents, that oxidize into NAD+.
Electrons released are used to reduce the acetaldehyde/ethanal to ethanol

f)
Wine, bread making, dairy industry to make yogurt and cheese

5.1: Aerobic respiration.
(a) State the need of energy and the role of respiration in living organisms.

5.1.1: Glycolysis.
(a) Illustrate to explain the glycolysis pathway (from glucose to pyruvate).

8. a)

i. Adenosine triphosphate
ii. Product of cellular respiration
iii. ATP consists of an adenine molecule, a ribose molecules & 3

phosphate groups bonded to each other.
iv. Phosphate bond are unstable and contains energy.
v. As a main source of energy in respiration.
vi. When the terminal phosphate group is split off of ATP, ADP

(adenosine diphosphate) & a phosphate (Pi) are formed.
vii. Energy released is used to drive anabolic reactions.
viii. Act as energy carrier.

b)
Step 6: oxidation & phosphorylation
• Glyceraldehyde-3-phosphate (G3P) is oxidized & NAD+ is reduced
To NAD+ + H+
• For each glucose molecule, 2 NADH are produced Glyceraldehyde-3-
phosphate (G3P) is then phosphorylated on C1
• The phospate source is inorganic phosphate (not ATP), which is
present in the cytosol
• The new phospate bond is a high energy bond with even more
potential energy
• Produce 1,3-bisphosphoglycerate
Step 7: Substrate level phosphorylation
• ATP is produced by substrate-level phosphorylation
• For each glucose molecule, 2 ATP are produced
• Produce 3-phosphoglycerate
Step 10: Substrate level phosphorylation
• ATP is produced by substrate level phosphorylation
• A phosphate group is transferred from PEP to ADP
• For each glucose molecule, 2 ATP are produced
• Produce pyruvate

5.1.1: Glycolysis.
(b) Describe link reaction (conversion of pyruvate to Acetyl CoA)

9.

i. Link reaction is a process to convert 3C pyruvate to 2C acetyl CoA
ii. Occurs in intermembrane space of mitochondria
iii. Involved three main steps
iv. Step 1: Pyruvate undergo decarboxylation by releasing one molecule of

CO2
v. Step 2: The remaining 2C fragment is oxidized to acetate in which then,

reduced NAD+ into NADH
vi. Step 3: Coenzyme A (CoA) is attached to the acetate by an unstable bond that

makes the acetyl group very reactive. This will produce acetyl CoA.
vii. Acetyl CoA then enters the Krebs cycle.

5.1.2: Krebs cycle. (a) Illustrate to explain Krebs cycle.
5.1: Aerobic respiration
5.2: Fermentation and its application.

10. a) Explain the steps involved in Krebs cycle. (CLO3, C4) [8 marks]

Step 1: (condensation/ entry of acetyl group)
• the unstable 2C acetyl CoA breaks
• 2C acetyl groups binds with 4C oxaloacetate to form 6C citrate

Step 2: (isomerization)

• Citrate is isomerized to isocitrate

Step 3: (oxidative decarboxylation) 2 major events:
• Isocitrate loses CO2 leaving a 5C compound
• 5C compound is oxidized and NAD+ is reduced to NADH + H+
• producing α-Ketoglutarate

Step 4: (oxidative decarboxylation)
• Removal of CO2
• α-Ketoglutarate is oxidized and NAD+ is reduced to NADH + H+
• attachment of CoA to form succinyl CoA (high energy bond)

Step 5: (substrate-level phosphorylation)
• The high energy bond of succinyl CoA breaks
• CoA is displaced by a phosphate group, which is then transferred to
guanosine diphosphate (GDP) to form guanosine triphosphate (GTP)
• GTP donates a phosphate group to ADP to form ATP
• These reaction converts succinyl CoA to succinate

Step 6: (oxidation/ dehydrogenation)
• Succinate is oxidized to fumarate and FAD is reduced
• 2H are transferred to FAD to form FADH2

Step 7: (hydration)
• Water is added to fumarate which rearranges the chemical bonds to form
malate

Step 8: (oxidation/ dehydrogenation)
• Malate is oxidized and NAD+ is reduced to NADH + H+
• Oxaloacetate is regenerated
• 2 turns of Krebs cycle produce 2 ATP by substrate-level phosphorylation
• Reduce cofactors (6 NADH + H+ & 2 FADH2 per glucose) carry high
energy electrons to the ETC

b)

Similarities : Marks
Both processes produced ATP 1
Both processes involves glycolysis which produce pyruvate 1

Differences: 2/2
2/0
Aerobic respiration Fermentation

Involves glycolysis, Involves only glycolysis
Krebs cycle and
oxidative

phosphorylation. Occurs in cytoplasm 2/0
only
Occurs in cytoplasm 2/0
and Glucose molecule is 2/0
mitochondria incompletely broken
down. 2/0
Glucose molecule is 2/0
completely broken Product could be ethanol
down. and carbon dioxide // lactate

Products are water and or lactic acid
carbon dioxide.
Oxygen is not required
Oxygen is required
Less ATP is produced (2
More ATP is produced ATP)
(38 ATP)

CLO 5.1.3: Oxidative phosphorylation: ETC and chemiosmosis. (c) Explain complete
oxidation of one molecule of glucose in active cell to produce 38 ATP.

11.

- 38 ATP in aerobic respiration
Glycolysis
- 2 ATP used
- Glucose Glucose-6-phosphate
- Fructose-6-phosphate Fructose-1,6-bisphosphate
- 4 ATP formed via substrate-level phosphorylation.
- 2 ATP during conversion 1,3- bisphosphoglycerate to 3-phosphoglycerate
- 2 ATP during conversion phosphoenolpyruvate to pyruvate
- Net production: 2 ATP from glycolysis

Link reaction
- Pyruvate → Acetyl CoA to form 2 NADH

Krebs cycle
- 2 ATP via substrate level phosphorylation
- During conversion succinyl CoA to succinate

Electron Transport Chain (ETC)
- 3ATP produced from 1NADH
- 2ATP produced from 1FADH2
- 2 NADH from glycolysis are transported
- from cytoplasm to the mitochondrion
- via shuttle glycerol phosphate (GP)
- 2 NADH form glycolysis → 2 x 3 ATP = 6 ATP
- 2 NADH from link reaction → 2 x 3 ATP = 6 ATP

- 6 NADH from Krebs → 6 x 3 ATP= 18 ATP
- 2 FADH2 from Krebs → 2 x 2 ATP = 4 ATP
- Overall 34 ATP via oxidative phosphorylation and 4 ATP directly

from substrate level phosphorylation produce 38 ATP

CHAPTER 6
PHOTOSYNTHESIS

6.1 OVERVIEW OF PHOTOSYNTHESIS
(a) Outline the complete process of photosynthesis: Light dependent and light
independent reactions

1.

a)
Light dependent reaction

b)
Thylakoid membrane /granum/ grana

c) : Water molecules
P : Primary Electron Acceptor
Q : ATP
R : Oxygen
S

d)

ATP and NADPH

e)

Photolysis of water

f)

PSI: Reaction centre P700 & antenna molecules (light harvesting complex)
PSII: Reaction centre P680 & antenna molecules (light harvesting complex)
g)

Oxidizing agent/electron carrier

2.
a)
F : Light dependent reaction
G : Light independent reaction/Calvin cycle

b)
F: Occur in thylakoid membrane
G: Occur in the stroma

c)
A is light energy/photon.
The function is to excite the electron/photoactivation of the reaction centre.

d)
18 molecules

As reducing agent

e)
not enough light energy to carry out light dependent reactions. So, the rate of
photosynthesis will be reduced.

6.2 ABSORPTION SPECTRUM OF PHOTOSYNTHESIS
(a) State the photosynthetic pigments involved in photosynthesis

3.
● Photosynthetic pigment is a substance that absorbs / traps visible light
(narrow band between 380 nm to 750 nm.
● TWO main photosynthetic pigments are chlorophyll a (absorb best in red and
blue wavelength) chlorophyll b (has different absorption spectrum and funnel
the energy from these wavelengths to chlorophyll a)

6.3 LIGHT DEPENDENT REACTION
(a) Explain the cyclic and non cyclic photophosphorylation.

4.

Cyclic photophosphorylation
● Photoexcited electrons are released from reaction center PSI (P700) will pass

through electron transport chain (Ferredoxin, Cytochrome complex and
Plastocyanin) and then back to the reaction center PSI
● Energy released is used to produce ATP by chemiosmosis
● no NADPH produced

Non cyclic photophosphorylation
● The light energy/photon is absorbed by the antenna chlorophyll molecules/

antenna complex and is transferred to the PSII reaction center (P680)
● Electrons in the reaction center PSII get excited/move to the higher energy

level//PSII become photoactivated
● Photoexcited electrons then accepted by primary electron acceptor
● Photoexcited electrons from PSII pass along through a series of electron

carrier/electron transport chain in the thylakoid membrane (Plastoquinone,
Cytochrome complex and Plastocyanin) to the reaction center PSI
● Energy was released as the electrons pass through the electron transport
chain and is used to produce ATP by chemiosmosis
● At the same time, light energy is absorbed by antenna pigments in reaction
centre PSI (P700) and the photoexcited electron is moved to a higher energy
level and accepted by the primary electron acceptor.
● The photoexcited electron from PSI passes through electron transport chain
(Ferredoxin and NADP reductase) and accepted by NADP+ to produce NADPH.
● Electrons which loss from PSI are replaced by electrons from PSII
● Electrons which are lost from in PSII are replaced by electrons from photolysis
of water.

6.4 LIGHT INDEPENDENT REACTION / CALVIN CYCLE
(a) Explain Calvin cycle involving carbon fixation, reduction and regeneration
of RuBP

5.

a) : 3-phosphoglycerate
J : Glyceraldehyde-3-phosphate
K : Ribulose-1,5-bisphosphate
L

b)
Stroma

c)
3-phosphoglycerate is phosphorylated by ATP to form 1,3
bisphosphoglycerate
1,3 bisphosphoglycerate is reduced by NADPH
And the phosphate group is removed to form Glyceraldehyde-3-phosphate
(G3P).

d)

● Regeneration of CO2 acceptor/RuBP
● K can be reconverted to L by rearranging the carbon skeleton of five

molecules of glyceraldehydes-3-phosphate into three molecules of
RuBP.
● These reaction require three ATP molecules

e)
Stage 1 :Carbon dioxide fixation
● Ribulose-1,5-bisphosphate (RuBP) combines / fix with CO2
(carboxylation process).
● Catalyzed by Ribulose-1,5-bisphosphate carboxylase-oxygenase
/Rubisco
● to form the six-carbon intermediate which is unstable and split to form

two molecules of 3-phosphoglycerate (PGA).
Stage 2 : Carbon dioxide reduction

● 3-phosphoglycerate(PGA) is phosphorylated by ATP to form
1,3-bisphosphoglycerate.

● 1,3 bisphosphoglycerate is then reduced by NADPH followed by
removal of inorganic phosphate to form Glyceraldehyde-3-phosphate
/G3P / PGAL.

● For every three CO2 that enter the Calvin cycle, six molecule G3P will
be produced

● Only one molecule G3P will exit the cycle to be used to synthesis
sugar/glucose molecule

● and the other five molecule G3P remain in the cycle
Stage 3 : Regeneration of Ribulose-1,5-bisphosphate/RuBP

● Five molecules G3P will undergo rearrangement to regenerate three
molecules RuBP

● These reaction require three molecules ATP from light dependent
reaction

6.5 ALTERNATIVE MECHANISM OF CARBON FIXATION: C4 AND CRASSULACEAN
ACID METABOLISM (CAM) PATHWAYS
(a) Explain photorespiration and state the alternative mechanism of carbon
fixation (C4 and Crassulacean Acid Metabolism (CAM) pathways)

6.

a)
● Photorespiration is a metabolic pathway that consumes O2 and ATP
and releases CO2. It can reduce photosynthesis output
● The closing of stomata causes gaseous exchange does not occur (CO2
in and O2 out)
● The effect is CO2 concentration in the plant will decrease while O2
concentration will increase
● As the O2 concentration increases, Ribulose-1,5-bisphosphate
carboxylase-oxygenase will catalyze the fixation of
Ribulose-1,5-bisphosphate with O2 rather than CO2
● Producing unstable 5C compound
● Unstable 5C compound then split into two molecule which are
3-Phosphoglycerate(3C) and 2-Phosphoglycolate (2C)
● As a result, it reduces the production of 3-phosphoglycolate molecules
which are used to synthesize organic molecules such as glucose.
● The rate of photosynthesis also will be decrease

b)
● Phosphoenolpyruvate carboxylase/ PEP Carboxylase
● Ribulose-1,5-bisphosphate carboxylase- oxygenase/ Rubisco

c)

Mesophyll cell and bundle sheath cell

d)
1: PEP carboxylase adds CO2 to phosphoenolpyruvate
2: By carboxylation to form oxaloacetate

e)
Oxidation and decarboxylation

f)
The product/ 3C compound(pyruvate) is phosphorylated by ATP

g)
To reduce photorespiration

h)
Corn / sugarcane

6.5 ALTERNATIVE MECHANISM OF CARBON FIXATION: C4 AND CRASSULACEAN
ACID METABOLISM (CAM) PATHWAYS
(b) Compare carbon fixation in C4 and CAM plants

7.
a)
Crassulacean Acid Metabolism plant/ CAM plant

b)
G: malic acid / malate H: vacuole

c)
The lost of water by evaporation is less compared to C3 and C4 plant// water
lost is minimized

d)
Use more ATP and NADPH// Undergo slow growth

e)
● Both use PEP carboxylase and RuBP carboxylase oxygenase to fix
carbon dioxide
● Both add CO2 into organic intermediates before entering the Calvin
cycle.
● Both minimize photorespiration

f)

● For pathway in FIGURE 5 (CAM pathway), carbon dioxide fixation only
occurs in mesophyll cell while in C4 pathway, carbon dioxide fixation
occurs in mesophyll cell and bundle sheath cell /

● In the CAM pathway, carbon dioxide fixation involves spatial separation

while in the C4 pathway, carbon dioxide fixation involves temporal
separation.

g)
The cycle required a supply of NADPH and ATP
which was the product of light dependent reactions that require light energy
from the sun.

8.

C4 plants CAM plants.
CO2 fixation occurs twice. First in CO2 fixation occurs twice. Both occur in
mesophyll cell and then in bundle mesophyll cells.
sheath cell
CO2 acceptor is PEP in mesophyll cell and CO2 acceptor is PEP and RuBP in
RuBP in bundle sheath cell mesophyll cell

CHAPTER 7: GASEOUS EXCHANGE AND ITS CONTROL

CLO 7.1: Gaseous Exchange and Its Control in Mammals
a) Describe the structure of haemoglobin and its characteristics as respiratory

pigments.

1.
a)
i. Conjugated protein
ii. Consists of globin and iron/Fe2+/ ferum iron containing molecules/
haem
iii. Haemoglobin (Hb) is a tetramer// consisting of 4 polypeptide chains.
iv. Consists of 2 alpha polypeptide chains and 2 beta polypeptide chains
(2α and 2β)
v. Forms quaternary structure
vi. Each haemoglobin molecule contains four iron molecules/ haem
groups.

b) Has high affinity for oxygen molecules when oxygen molecules
i. concentration is high (affinity decrease when oxygen molecules
concentration is low).
ii. Carry oxygen in the form of oxyhaemoglobin
iii. The reaction of oxygen with haemoglobin is temporary/ completely
iv. reversible//Hb+O2 ↔ HbO2
v. Carry oxygen from lungs to tissues where partial pressure of oxygen,
vi. PO2 is high// Hb
Can release oxygen in tissue where partial pressure oxygen is low
vii. PO2
viii. Each haemoglobin molecule is capable of carrying four oxygen
molecules// each iron molecules bind to oxygen molecule//
ix. cooperative binding.
Haemoglobin loads// unloads one oxygen molecules at a time.
Carry carbon dioxide in the form of carbaminohaemoglobin (HbCO2/
HbNHCOOH) from tissues to lungs where partial pressure PO2 is low
Acts as a buffer for H+.

CLO 7.1: Gaseous Exchange and Its Control in Mammals
b) Describe three ways of carbon dioxide transport from respiring tissue to
lungs.

2.

a) J: Glucose//C6H12O6
K: Carbonic anhydrase
L: Haemoglobinic acid // HHb
M: Chloride ion //Cl-

b) II: To maintain electrochemical balance // charge balance
III: To prevent the blood from becoming acidic (by H+) // acts as a buffer