SKEMA TUTORIAL

c) 70% // 60% -85% CO2 moves through plasma as bicarbonate ion /
i. HCO3-
CO2 from tissue cells (diffuses into plasma and) enter RBC /
ii. erythrocytes

iii. CO2 combines with water to form carbonic acid / H2CO3
iv. Reaction catalysed by the enzyme carbonic anhydrase
v. Carbonic acid /H2CO3 dissociates, forming proton/ H+ and bicarbonate
ions/ HCO3-
vi. HCO3- diffuses out of the RBC
vii. Transported in solution in the plasma to lungs
viii. As bicarbonate ions diffuse out of the RBC, chloride ions / Cl- replace

ix. them (diffuse inwards from the plasma)
x.
xi. The process also known as chloride shift
xii.
xiii. To maintain electrical neutrality
Haemoglobin combines with hydrogen ions / H+

Preventing a decrease in pH

In the lungs, carbon dioxide concentration is low

xiv. Carbonic anhydrase reaction proceeds in reverse direction

converting carbonic acid /H2CO3 into water and carbon dioxide
xv. Carbon dioxide diffuses out (from RBC and plasma) into the alveoli

CLO 7.1: Gaseous Exchange and Its Control in Mammals
c) Analyse the oxygen dissociation curve of haemoglobin

3.
a)
Hb + 4O2 --------> Hb(O2)4

b)
i. Oxygen dissociation curve is a sigmoid
ii. Haemoglobin has high affinity for oxygen

iii. Blood can be saturated with oxygen (up to 98%)
iv. even when the partial pressure of oxygen is low
v. When oxygen concentration is low (in respiring tissues)
vi. percentage of blood saturation with oxygen will reduce rapidly
vii. Causing oxyhaemoglobin to release more oxygen/causing HbO2

dissociate into O2 and Hb.
viii. This is shown by the steepest part of the curve

ix. In the lungs, partial pressure of oxygen is high
x. oxygen combines with haemoglobin to form oxyhaemoglobin
xi. The more the curve is displaced to the right the less readily the

haemoglobin
xii. pick up oxygen/ the more easily it releases oxygen
xiii. The more the curve is displaced to the left, the more readily the

haemoglobin pick up oxygen and the less readily it releases oxygen.

CLO 7.1: Gaseous Exchange and Its Control in Mammals
d) Compare oxygen dissociation curve of haemoglobin and myoglobin

4.

a)
H:Oxygen dissociation curve for haemoglobin
M:Oxygen dissociation curve for myoglobin

b)
Skeletal muscles

C)
(i)
Myoglobin – 90%
Haemoglobin – 25%

(ii)
Myoglobin has a high affinity for oxygen; hence it retains its oxygen in
the resting cell
It only begins to release oxygen when partial pressure oxygen is below
20mmHg, which is during vigorous muscle activity
It act as a store of oxygen in muscle, only releasing the oxygen when
the supplies of oxyhaemoglobin have been exhausted.

b)
i. Haemoglobin has a high affinity towards oxygen molecule
ii. The higher the partial pressure of oxygen molecules, the higher the
percentage of saturation of haemoglobin with oxygen

iii. At low partial pressure of oxygen, more oxygen will dissociate from
oxyhaemoglobin, less haemoglobin binds with oxygen

iv. Myoglobin has a higher affinity towards oxygen even at a low partial
pressure of oxygen. The percentage of its saturation with oxygen is
still high

v. It is only release oxygen when oxygen level in the blood is very low.

c) Shift to the right.
Because more oxygen released from haemoglobin // Haemoglobin less
affinity to oxygen

d)
Oxygen will be release at only in specific partial pressure / condition when
is needed
Ensures the oxygen is sufficient for the cellular reaction in cell / tissue
takes place

CLO 7.1: Gaseous Exchange and Its Control in Mammals
e) Analyse the effect of the changes in partial pressure of carbon dioxide
towards oxygen dissociation (Bohr effect).

5.
a)
Bohr effect refer to displacement of oxygen-haemoglobin dissociation
curve that shift to the right by a change in pH // increase of
oxyhaemoglobin dissociation due to lower pH / increase in CO2
concentration

b)
because of cooperative binding characteristic

c)
Curve C

d) ]
i. Active tissue has a high amount of CO2
ii. CO2 reacts with water in blood forming carbonic acid

iii. Catalyse by carbonic anhydrase
iv. Carbonic acid dissociates to form bicarbonate ion and hydrogen ion
v. Hydrogen ion lowers the pH
vi. Causing the oxygen dissociation curve to shift to the right
vii. Haemoglobin affinity towards oxygen is reduced because hydrogen

ions affect the shape of haemoglobin
viii. Oxyhaemoglobin dissociates

ix. More oxygen release to the active tissues
x. To prevent blood become more acidic, hydrogen ion reacts with

haemoglobin and form haemoglobinic acid

CLO 7.2: Roles of chemoreceptors in controlling breathing
a) State the types of chemoreceptors.
b) Explain the role of chemoreceptors in controlling the rate of normal
breathing.

6.
a)
J : medulla oblongata
K : vagus nerve
L : carotid body
M : aortic body

b)
J : has respiratory centre that contains the centra chemoreceptor which
are stimulated by the high concentration of H+. it then sends the
impulse to the diaphragm and intercostal muscles to increase the rate
of breathing.
M : Specialized cell in chemoreceptor that detect the increase in CO2 or
H+ concentration in blood and send impulse to the respiratory centre.

c)
Concentration of CO2
Concentration of H+

d)

Exercise increases cellular respiration
This increase the concentration of CO2/H+ in blood which is then detected
by the chemoreceptor
Nerve impulse from the chemoreceptor are sent to the respiratory centre
that will increase the rate of breathing
Excess CO2 will be expel.

e) As more CO2 diffuse from the blood into alveoli, the concentration CO2 in
the blood decrease
When concentration CO2 returned to normal and less H+ produced the
chemoreceptor will not be stimulated, hence the breathing rate back to
normal

7. (a) Central chemoreceptor and peripheral chemoreceptor

(b) Chemoreceptors in carotid and aortic bodies are sensitive to small
i. changes in the concentration of CO2 in blood
Increase in CO2 level in blood / decrease in blood pH stimulates the
ii. chemoreceptor
Chemoreceptors send nerve impulse to the inspiratory centre in the
iii. medulla
The inspiratory centre sends out nerve impulses via the phrenic nerve
iv. to diaphragm
And via the thoracic / intercostal nerve to the external intercostal
v. muscles
Causing them to increase the rate of contraction
vi. Inhalation occurs
vii. As the lung expands, stretch receptors in the alveoli walls of lungs are
viii. stimulated
Impulse pass along the vagus nerve
ix. To the expiratory centre in the medulla and inhibits the inspiratory
x. centre / witch off the inspiratory centre
The diaphragm and external intercostal muscle relax
xi. Expiration / exhalation takes place
xii.

xiii. The lungs are no longer stretched and stretch receptors are in no
longer stimulated

xiv. The expiratory centre become inactive and inspiration / inhalation
begins again

xv. The whole cycle is repeated rhythmically throughout the life of the
organism

CLO 7.3 : Gaseous exchange and control in plants.
a) Explain the regulation of stomatal opening and closing based on starch
sugars hypothesis 7.

8.

a)
Cell P: guard cell
Cell Q: epidermal cell / subsidiary cell

Organelle R: chloroplast

b)
P (guard cell) have chloroplast while Q (epidermal cell) do not have
chloroplast

c)
Photosynthesis occur cause the CO2 decrease, H+ decrease/ pH level
increase.
In high pH, enzyme phosphorylase catalyses the transformation of starch
into glucose (sugar) / glucose-1-phosphate in the guard cells.
Water potential guard cell is lower compare to subsidiary cells caused the
water from subsidiary cell enter the guard cell by mitosis.
Guard cell become turgid

d) In the presence of light/ day time photosynthesis occurs Guard cells
i. synthesize sugar
It increases the osmotic pressure of the guard cell// it lowers the
ii. water potential of the guard cell.
Water from surrounding mesophyll tissue/ epidermal cell/
iii. neighbouring cell/ adjacent cell/subsidiary cell enters guard cell.
By osmosis
iv. Guard cells become turgid and stomata open.
v.
vi. In the absence of light/ night, pH low/ accumulation of carbon
dioxide.
vii. Sugar converted into starch
viii. Decrease osmotic pressure of guard cell// increase water potential in
guard cell.
ix. Guard cells release water to the surrounding mesophyll cells/
epidermal cell/ neighbouring cell/ adjacent cell/ subsidiary cell
x. By osmosis
xi. Guard cell becomes flaccid and stomata closed



CHAPTER 8: TRANSPORT SYSTEM

8.1 MAMMALIAN HEART AND ITS REGULATION
(a) Explain the initiation of heartbeat.
(d) Explain the factors affecting heartbeat.

a) : Sinoatrial node
A : Atrioventricular node
B : Bundle of His
C

b)

Initiates the heartbeat.

c)

To ensure the atria empty completely before ventricle contract.

d) (Low blood pH will be) detected by chemoreceptor in the walls of carotid arteries & aorta

/carotid bodies & aortic bodies and medulla oblongata). 1m

Which then send an impulse to the cardiac accelerator centre / CAC in medulla

oblongata. 1m

CAC send output signal via sympathetic nerve to stimulate adrenal gland.1m

to releases epinephrine (which stimulates the heart rate to increase). 1m (any 2)

e)

To supply oxygenated blood to the muscles/ to supply more O2

8.1 MAMMALIAN HEART AND ITS REGULATION
(b) Explain cardiac cycle.

2.

a)

One complete sequence of pumping and filling of heart with bloods // The sequence of
events of a heartbeat.

b)

0.8 seconds

c)

Phase A : Atrial and ventricular diastole - 0.4 seconds
Phase B : Atrial systole and ventricular diastole - 0.1 seconds
Phase C : Ventricular systole and atrial diastole - 0.3 seconds

d)
i.

Increases.

ii.

44

AV valves are closed and semilunar valves opened.

iii.
Blood flow out from the ventricles to the pulmonary artery and aorta.

e)
When the pressure in atria is greater/higher than the pressure in ventricles.

3.
- The phase in the cardiac cycle is ventricular systole and atrial diastole (0.3 s)
- The ventricles contract
- Ventricular pressure increases
- and atrial pressure decreases
- When ventricular pressure higher than/exceed atrial pressure
- Tricuspid and bicuspid/mitral close // AV valves close
- Blood pump out of ventricular to the pulmonary artery and aorta
- Recoil of the blood against the closed AV valves
- This produce the “lub” sound/first heart sound
- Preventing blood from flowing back into the atria

8.1 MAMMALIAN HEART AND ITS REGULATION
(b) Explain ECG.

4.

a)

Record the changes in electrical activity during the heart beat.

b)

i. P wave : Atrial depolarization.
ii. QRS wave : Ventricular depolarization

iii. T wave : Ventricular repolarization

c) Atrial contraction(systole).

d) Ventricular contraction (systole).

e)
i. PQ segment : Impulse delayed at AV node.
ii. ST segment : Ventricle are contracting and emptying.
iii. TP segment : Ventricles are relaxing and filling.

45

3.2 HUMAN LYMPHATIC SYSTEM
(b) Describe the transport of lipids from the small intestine into the blood stream.

5.
a)

Transport the digested fats from the villi of the small intestines to heart.

b)

Fatty acids and glycerol.

c)

- Fatty acid and glycerol will combine with protein and
- mixed with cholesterol to form chylomicrons / lipoprotein.

d)

Chyle.

e)

Because digested lipid is not water soluble and do not mix with watery blood // to
prevent digested lipid from deposited (and become plague) in the wall of arteries
which can narrow or completely block the arteries // to prevent arteriosclerosis.

8.2 HUMAN LYMPHATIC SYSTEM

(a) Describe the pathway of lymph from tissue to blood circulatory system.

6.
• Tissues fluid enters the lymphatic capillaries
• Lymph capillaries conduct the lymph to larger vessels called lymphatic veins
• Lymph veins enter lymph nodes (small organized masses of lymph tissues)
• From lymph node, lymph are drain into lymphatic duct
• Lymphatic duct can divide into thoracic duct
• and right lymphatic duct
• Right lymphatic duct is responsible for draining the lymph from upper right quadrant of
body
• The thoracic duct is much larger and drains lymph from the rest of the body
• Lymph from thoracic duct will enter the heart via left subclavian vein
• Lymph from right lymphatic duct will enter the heart via right subclavian vein

8.2 TRANSPORT IN PLANTS
(a) State the pathway of water movement in root.
(d) Explain the lateral pathway of water and minerals is transported from
surrounding soil to the root vascular system.

7.

a)
A. : Epidermis
B. : Cortex
C. : Endodermis

46

D. : Stele
b)

Pathway 1 : Apoplastic (pathway)
Pathway 2 : Symplastic (pathway

c)

- Vacuolar (pathway)
- Transmembrane (pathway

d)

Water and solutes move through/along the cell wall and extracellular spaces.

e)

Water and solutes move along the cytoplasm of the adjacent cell through the
plasmodesmata.

f)
Ensures that no minerals can reach the vascular tissue of the root without crossing a
selectively permeable plasma membrane.
Prevents solutes that have accumulated in the xylem from leaking back into the soil
solution.

8.3 TRANSPORT IN PLANTS
(d) Describe the movement of water via xylem by transpiration-cohesion-tension
mechanism and root pressure.

8.

a)
Transpiration.

b)
As water diffuse out/evaporates through stomata, it creates a negative pressure / tension
on the water (to replace the lost amount of water).
so water is consequently pulled upwards as it replaces the water that is lost.
The pulling force is known as transpirational pull.

c)

47

Cohesion and adhesion

d)

Cohesion – water molecules are attracted to another water molecules by
hydrogen bond.
Adhesion – adhesion of water molecules with the wall of xylem cell by hydrogen
bond.

e)

Root pressure.

f)

- At night, root cell actively pumping mineral ions into the stele/xylem
- (At the same time,) endodermis helps prevent these ions from leaking out
- This causes the accumulation of mineral ion in stele
- Water potential within the stele become low
- Water molecules diffuse into the stele from the root cortex
- As water flows from root cortex, it will push up the xylem sap.

8.3 TRANSPORT IN PLANTS
(d) Explain the Pressure Flow Hypothesis in phloem.

9.
a)

Pressure Flow hypothesis.

b)
Sucrose

c) Provide ATP during actively transport of sucrose from source into sieve tube.

d) Passive and active transport.

e) - Accumulation of sucrose lowers the water potential in the sieve tube.
- Water from xylem will diffuse into the sieve tube
- by osmosis
- The entering of water molecules/ the osmotic flow of water into sieve tube will
generate high hydrostatic pressure in that area.

f)

- High hydrostatic pressure in sieve tube at source area
- and low hydrostatic pressure in sieve tube at sink area
- Due to pressure gradient, X will be translocated from high hydrostatic pressure
area to low hydrostatic pressure area.

g)

Sucrose is consume /used up for cellular respiration

h)

Flower, fruit.

48

*Note to lecturer : any other cell that not synthesis food (cannot carry out
photosynthesis) but receive and use that food from source, can be accepted.

49

CHAPTER 9
HOMEOSTASIS

9.1 CONCEPT OF HOMEOSTASIS
(a) Explain the concept of homeostasis and describe the homeostatic control
system.

1.

a)
Homeostasis is the process of maintaining biologically stable conditions
inside a living organism through physiological processes.

b)
P: Receptor
Q : Control centre
R : Effector

C) Receptor detects stimulus/ any changes from the set point/ input in
• physical or chemical factors
and sends input (in the form of nerve impulses) to the control center.
• Control centre evaluates the stimulus/ input it receives from receptors
• to trigger/ stimulate/ generates output commands/action that will
• correct the changes.
and send to the effector
• Effector receives output from the control center
•
• produces response/ effect that changes the controlled condition. //
carries out the information of corrective mechanisms action
• through negative feedback mechanism
• to restore condition back to normal.

9.2 NEGATIVE FEEDBACK MECHANISM
(b) Explain the negative feedback mechanism in controlling the blood glucose
level.

2.

a)
Homeostatic mechanism that stops or reduces the intensity of the
original stimulus and consequently causes a change in a variable that is
opposite in direction to the initial change.

b)
Cell P : Beta cells of the islets of Langerhans
Cell Q : Alpha cells of the islets of Langerhans

(source: Biology 12th ed.Campbell. page:992)

c)
Eating carbohydrate rich meal / any other suitable answer

d)
Hormone X : Insulin

e)
i. Stimulate conversion of excess glucose into fat.
ii.
Stimulate conversion of excess glucose to glycogen //
glycogenesis.
iii.
Stimulate the cell to take out more glucose from the
bloodstream into the body cell / by increases the
permeability of plasma membrane to glucose.

f) Hormone Y : Glucagon

g)
• Stimulates conversion of glycogen to glucose / glycogenolysis.
• Stimulates conversion of amino acid and glycerol or fatty acid to
glucose / gluconeogenesis. (any 1)

h)
• Blood glucose level fall from normal set point
• detected by alpha cells of islets of Langerhans in pancreas
• trigger the secretion of glucagon hormone
• Glucagon stimulates conversion of glycogen to glucose /
glycogenolysis in the liver or muscle

• and stimulate conversion of amino acid and glycerol or fatty acid to
glucose / gluconeogenesis in the liver.

• Glucose released into bloodstream
• and causes blood glucose level increase to normal set point //

normal blood glucose level is restored.

• Secretion of glucagon is inhibited (when blood glucose level is
restored to normal set point)

9.3 HUMAN HOMEOSTATIC ORGAN: STRUCTURE AND FUNCTION OF KIDNEY
(a) Describe the structure of nephron.
(b) Analyses the processes in urine formation.
(c) Describe the counter current multiplier mechanism in urine formation.

3.

a)

X : Renal artery E :Thin segment of ascending limb

Y : Renal vein F :Thick segment of ascending

limb

A : Bowman’s capsule G :Vasa recta

B : Glomerulus H :Collecting duct

C : Proximal convoluted tubule I :Peritubular capillaries

D : Loop of Henle J :Distal convoluted tubule

b)
Nephron

c)
(Glomerular) ultrafiltration

d) The high hydrostatic blood pressure in glomerulus.
• due to the diameter of afferent arteriole is larger than efferent
• arteriole.
Highly coiled glomerulus
• provide large surface area for filtration.
• High permeability of the glomerulus
• due to the numbers perforated walls of the capillaries and the
• podocytes
forming a filtration membrane (Any 6)
•

e)
Water // urea // salts // glucose // amino acid // sodium ion // potassium
ion // chloride ion // bicarbonate ion // other small dissolved solute from
blood (Any 1)

f)• Blood cell // platelet // most of plasma protein (Any 1)
▪ Because its large molecule.

g)
(Selective) reabsorption

h)
(Tubular) secretion

i) Loop of Henle / D and vasa recta / G

j)
Maintains high salt concentration / high osmolarity in inner medulla, to
form concentrated urine.

k) Have long loop of Henle and extend deep into medulla
•

• Contribute to the greatest osmotic gradient in inner medulla
• Enable kidney to conserve water
• by producing small volume of very concentrated urine.

4.
• As the filtrate flows from the cortex to the medulla in the descending
limb,
• water leaves the tubule by osmosis because the interstitial fluid is very
hypertonic.
• The water moves straight into the ascending vessel of vasa recta
surrounding the loop of Henle.
• With the loss of water, Na+ and Cl- becomes more concentrates in the
descending limb.
• As the filtrate flows up the loop of Henle the Na+ and Cl- diffuse out from
the thin segment of the ascending limb.
• In the thick segment of the ascending limb, the Na+ and Cl- are actively
transported into the interstitial fluid of renal medulla.
• (As the two limbs are so close to one another), this produces a high
concentration Na+ and Cl- around the descending limb.
• The filtrate entering the distal tubule is hypotonic compare to the
peritubular fluid.
• As the filtrate descends again toward the medulla in the collecting duct,
water is move out by osmosis into the hyperosmotic interstitial fluids
• This concentrates salt, urea and other solutes in the filtrate.
• Some urea leaks out from the lower part of the collecting duct into the
interstitial fluid,
• along with NaCl, urea contributing to the high interstitial osmolarity of
the inner medulla.

9.3 HUMAN HOMEOSTATIC ORGAN : STRUCTURE AND FUNCTION OF KIDNEY
(d) Relate the regulation of blood water content with ADH.

4. : Posterior pituitary gland
: ADH / antidiuretic hormone
a) : Collecting duct

B
C
D

b)
Eating salty food // sweating profusely / losing water through sweating
// any other suitable answer

c)
Osmoreceptor

d)
• Hormone C / ADH bind and activate the membrane receptors on the
surface of collecting duct.
• Activated receptor initiates a signal transduction cascade
• that direct the insertion of aquaporin into the membrane lining
collecting duct.
• More aquaporin channel resulting more water recapture, reducing
urine volume.

(source: Biology 12th ed.Campbell. page:1046)

e)
• Alcohol inhibits posterior pituitary gland
• from releasing ADH/ antidiuretic hormone / hormone C
• Collecting duct not permeable to water
• Leading to excess urinary water loss
• and dehydration (which may cause some symptoms of hangover)

(source: Biology 12th ed.Campbell. page:1047)

CHAPTER 10: COORDINATION

CLO10.1 Nervous system. (b) Explain formation of resting and action potential.

1.

a.
Rapid change in the membrane potential that occurs when nerve cell membrane is
stimulated. [2 marks]

b.
1: Resting potential
2: Rising phase of action potential
3: Falling phase of action potential
4: Undershoot [4 marks]

c
The voltage-gated sodium ion channels close. Voltage-gated potassium ion channels
open. Potassium ion leave/pump out of the cell causes the inside becomes more negative
than the outside. [4 marks]

CLO 10.1 Nervous system. (b) Explain formation of resting and action potential.

2. a.
The membrane potential of a neuron when it is not transmitting signals // the membrane not
be stimulated or conducting impulse.
[1 mark]

b.
By facilitated diffusion of sodium ion and potassium ions down the concentration gradient.
By active transport of sodium ion and potassium ions by the sodium potassium pump.
[2 marks]

CLO 10.1 Nervous system. (d) Describe the structure of synapse. (e) Explain the mechanism of
synaptic transmission across synapses.

3. a.

P = Synaptic vesicles containing neurotransmitter [4 marks]
Q = vesicle releasing neurotransmitter
R = Neurotransmitter
S = Synaptic cleft

b.
Ca2+ causes the synaptic vesicles to fuse with the presynaptic membrane and then rupture.
[1 mark]

c. The arrival of an action potential at axon terminal, depolarizes the presynaptic membrane.
The depolarization opens voltage-gated Ca2+ channels, triggering an influx of Ca2+ into synaptic
knob.

Rise in Ca2+ causes synaptic vesicles to fuse with the presynaptic membrane.
Synaptic vesicle releasing the neurotransmitter into the synaptic cleft by exocytosis. [2 marks]
d.
Synaptic delay occurs due to the neurotransmitter need to diffuse across the synaptic cleft
because the impulse is chemically transmitted. [2 marks]

e.
To prevent repeated action potential as the neurotransmitter will be hydrolysed
and returned to the presynaptic membrane. [1 mark]

f.
To prevent the postsynaptic cell from being continuously stimulated.
To allow another synaptic potential to occur.

CLO 10.2 Mechanism of muscle contraction. (c) Describe the structure of sarcomere. (d) Explain
the mechanism of muscle contraction based on Sliding Filament Theory.

4.
a.
A =Myosin / thick filaments
B = Actin / Thin filaments
C = H zone
D = A band [4 marks]
b.
C = becomes shorter
D = No changes [2 marks]
c. Sliding – Filament Theory [1 mark]

d.
When muscles contract, actin and myosin would slide past each other. [1 mark]

e.
To bind to ATP by hydrolyzing ATP into ADP + Pi. [1 mark]

f.
To bind to troponin so troponin will change its shape. [1 mark]

CLO 10.2 Mechanism of muscle contraction. (c) Describe the structure of sarcomere. (d) Explain
the mechanism of muscle contraction based on Sliding Filament Theory.

5.

a.
i.
I band [1 mark]

ii.
A band [1 mark]

b.
i.
Binds myosin by the change of configuration to release actin filament and back to the low
energy configuration [2 marks]

ii.
Calcium binds to troponin, and rearranges the troponin-tropomyosin complexes exposing
the binding sites on actin [2 marks]

iii.
Defines the border between one sarcomere to another/ joins the sarcomeres [1 mark]

c.
i. Sarcomere : becomes closely packed / striated / compact

ii. A band : Myosin head pull actin filament. The length remain unchanged [2 marks]

d.
hand, arms, thigh, lower arms etc [1 mark]

CLO 10.3 Hormones in mammals. (a) State types of hormone. (b) State the types of mechanism
of hormone action. (c) Explain the mechanism of hormone action:
(ii) Second messenger(cAMP): non-steroid hormone (adrenaline and glucagon).

6.
a.
the cAMP acts as the second messenger and initiates a chain of reaction inside the cell
[2 marks]
b. i)
to increase blood glucose level by glucagons [1 mark]

ii)
X: adenylate cyclase
Y: G protein [2 marks]

c.
The hormone reacts with protein receptors on the target cell’s plasma membrane causing a
change in the shape of receptor protein [2 marks]

d.
The binding of receptor activates an enzyme that catalyzes the conversion of ATP
to cAMP. cAMP act as second messenger and initiates chain of reactions inside
the cell thus amplifies the molecule [2 marks]

e.
Gene Activation. Steroid hormone. [2 marks]

CLO 10.1 Nervous system. (c) Describe the characteristics of nerve impulse.

7.

1. The presence of myelin sheath
- act as electrical insulator
- The effect of action potential ‘jumps’ from node to node and passes along the
myelinated axon faster.

2.Axon
- The bigger the diameter, the higher velocity/ the resistance is reduced as the diameter is
big [4 marks]

CLO 10.1 Nervous system. (f) Compare the transmission of impulse across synapse and along the axon.

8. TRANSMISSION OF IMPULSE ALONG THE AXON
TRANSMISSION OF IMPULSE AT THE SYNAPSE

Impulse is chemically transmitted. Impulse is electrically transmitted.

The conduction of impulse involves two The impulse travels along one neuron
adjacent neurons

Calcium ions are required for triggering the Calcium ions are not required
release of neurotransmitters into synaptic

cleft

Can be excitatory or inhibitory Only excitatory

Depolarization is due to influx of sodium ions into the neuron than an outflow of potassium ions

There is synaptic delay because of the time Action potential occurs immediately.
taken for the release of neurotransmitter & for

the diffusion of this substance across the
synaptic cleft.

[3 marks]

CLO 10.1 Nervous system. (a) State the organization of the nervous system.

9.

a.
Y : Peripheral Nervous System (PNS)
Z: Central Nervous System (CNS) [2 marks]

b.
Y : Sensory/ afferent neuron and motor/efferent neuron
Z : Interneuron [2 marks]

c.
Voluntary responses [1 mark]

CLO 10.2 Mechanism of muscle contraction. (a) Describe the structure of neuromuscular junction.
(b) Explain impulse transmission at the neuromuscular junction.

10.

a.
The synapse between motor neuron and muscle fiber [2 marks]

b
C : Synaptic vesicle containing acetylcholine
F : Acetylcholine [2 marks]

c.
Calcium ion channel [1 mark]

d.
- When A arrives at the end of axon:
∙ Ca2+ channels open / Ca2+ diffuses into axon terminal
∙ Ca2+ stimulates synaptic vesicles to fuse to presynaptic membrane

∙ Acetylcholine released into synaptic cleft by exocytosis

∙ Acetylcholine binds to Na+ channel / receptor on the postsynaptic membrane

∙ Na+ diffuses into post synaptic cell (K+ diffuses out)

∙ Depolarisation / excitatory post synaptic potential (EPSP) occurs [6 marks]

CLO 10.4 Photoperiodism. (a) Explain the role of phytochrome in the regulation of flowering.

11.
a.
X: Short Day Plant
Y: Long Day Plant [2 marks]

b.
Plant Y [1 mark]

c.
Not flower because a flash of red light during the night converts PR to PFR, so flowering is
inhibited. [2marks]

d
chrysanthemum [1mark]

CLO 10.1 Nervous system. (c) Describe the characteristics of nerve impulse.

12.
1. All- or – None Law
- Action potential occurs maximally or not at all.
- If the stimulus is too low, there is no action potential.
2. Refractory period
- A short time immediately after an action potential in which the neuron cannot
respond to another stimulus.
- This interval sets a limit on the maximum frequency at which action potential can be
generated.
- Ensure that the impulses move only in one direction, from cell body to the axon terminals.
[6 marks]

CLO 10.1 Nervous system. (b) Explain formation of resting and action potential.

13.
Generation of action potential along an axon occur in 5 stages:
1. Resting state
- both the sodium & potassium gated channels and voltage-gated ion channels are closed /
- The membrane’s resting potential is maintained by the sodium
- potassium pump & the sodium and potassium channels of the membrane which permits
facilitated diffusion.
2. Depolarization
– a stimulus opens some gated sodium channels
– sodium inflow through those channels and depolarizes the membrane (axon) /
– if the depolarization reaches the threshold (-55mV), it triggers an action potential.
3. Rising phase of action potential
– depolarization over threshold level which stimulate voltage
– gated sodium channels to open.
– more sodium influx into the axon results in the further depolarization.
– in increase depolarization cause more gated sodium channels to open.

4. Falling phase of the action potential
– voltage – gated sodium channels closed. Blocking sodium inflow.
– after impulse transmitted in the axon, voltage-gated potassium channels open, rapid
outflow potassium. Make the inside of the axon membrane negative again.

5. Undershoot
– when the charge inside of the axon lower than threshold level (-55mV), gated sodium
channels are slowly closed.
– the continued potassium ion outflow makes the membrane more negative than resting
potential.
– within one or two million second, the voltage - potassium channels close.
– the membrane potential returns to the resting potential.
[8 marks]

CLO 10.1 Nervous system. (e) Explain the mechanism of synaptic transmission across synapses.

14.
- The arrival of axon potential at the axon terminal, depolarizes the presynaptic membrane.
- The depolarization opens voltage gated Ca2+ channels, triggering an influx of Ca2+ into synaptic
knob.
- Rise in ca2+ ion causes synaptic vesicles to fuse with the presynaptic membrane.
- Synaptic vesicle releasing the neurotransmitter into the synaptic cleft by exocytosis.
- the neurotransmitter diffuses across the synaptic cleft & attaches to a specific receptor site on
the ligand voltage gated ion channel in the postsynaptic membrane.
- Triggers opening ligand voltage gated ion channel, allowing Na+inflow into the postsynaptic
neurons.
- This leads to a depolarization of the postsynaptic membrane.
- The depolarization response is known as an excitatory postsynaptic potential (EPSP).
- If the EPSP is large enough to reach the threshold level, an action potential is generated and is
transmitted along the postsynaptic neuron.
- Enzymes present in the synaptic cleft that break down the neurotransmitter immediately
e.g: Acetylcholinesterase.
- Acetylcholinesterase splits acetylcholine into acetate & choline.
- The choline is taken up by the presynaptic cell & combine with the acetyl coenzyme A to reform
acetylcholine. [8 marks]

CLO 10.1 Nervous system. (g) Explain the mechanism of action of drugs (e.g. cocaine) on the
nervous system.

15.
- Cocaine effect neurons in the brain’s pleasure pathway that located in limbic system.
- The cells used neurotransmitter dopamine
- Cocaine prevent the reuptake dopamine by the presynaptic membrane
- Excess dopamine in synaptic cleft and continually binds to receptors in the postsynaptic
membrane.
- Therefore, depolarization occurs repeatedly which results in the continuous impulse transmission.
- When the receptors within postsynaptic are exposed to high level of dopamine, for prolonged
period, postsynaptic membrane often responds by decreasing the number of receptors.

- When cocaine is no longer taken and dopamine level return to their normal concentration, the
smaller number of dopamine receptors are available for the neurotransmitter to bind to is
insufficient to fully activate nerve cell

- To produce the same pleasure high level of dopamine is required.
- So more cocaine has to be taken to increase dopamine level.

[7 marks]

CLO 10.2 Mechanism of muscle contraction. (d) Explain the mechanism of muscle contraction
based on Sliding Filament Theory.

16.

- Action potential is propagated along plasma membrane and down T tubules.
- Action potential triggers Ca2+ channel to open, Ca2+ release from the sarcoplasmic reticulum into
the sarcoplasm.
- The Ca2+ bind to the troponin-tropomyosin complex, changing its conformation.
- Troponin displaces tropomyosin, exposing the myosin binding sites on the actin filament.
- At thick filament, ATP is bound to the myosin when the muscle fiber is at rest.
- Each myosin molecule has a head containing enzyme ATPase which catalyses hydrolysis of ATP
to ADP and inorganic phosphate (Pi).
- The ADP and Pi initially remain attached to the myosin head and is in its high-energy
configuration.
- The myosin head binds to actin (expose myosin binding site at actin), cross bridge are formed. -
ADP + Pi are then release from the myosin head, the myosin head return to its low-energy
configuration.
- The head bends by angle 450 in a flexing motion (the power stroke) that pulls the actin filament
closer to the center of the sarcomere, propelling the actin filament to slide past the myosin
filament.
- Muscle contract
- After the process is completed, the myosin heads, bound to another ATP molecule, allowing the
myosin to detach from the actin, thereby breaking the cross bridge and beginning a new cycle.
[10 marks]

CLO 10.3 Hormones in mammals. (c) Explain the mechanism of hormone action: (i) Gene activation:
steroid hormone (ii) Second messenger(cAMP): non-steroid hormone (adrenaline and glucagon).

17.
i.
- Steroid hormones pass through the cell surface membrane.
- Steroid hormone bind to a receptor protein in the cytoplasm forming hormone-receptor
complex.
- The complex formed enters the nucleus.
- The hormone exert a direct effect on the chromosomes by switching on genes and
stimulating transcription (mRNA formation).
- The mRNA enters the cytoplasm
- mRNA is translated into new proteins such as enzyme and other proteins that can carry out a
response to the hormonal signal.

ii.

- Non steroid hormones combine with receptors in target cell membranes; the receptors have a
binding site and an activity site.

- The hormone-receptor complex (as first messenger) triggers a drop of biological activity. -
The hormone-receptor complex generally activates a G protein, which then activates the
enzyme adenylate cyclase that is bound to the inner cell membrane.

- This enzyme removes two phosphates from ATP to produce cyclic AMP (the second
messenger), which in turn activates protein kinase that activate proteins.

- These activated proteins induce changes in the cell.

CLO 10.4 Photoperiodism. (a) Explain the role of phytochrome in the regulation of flowering.

18.
- Phytochrome are pigment in plant
- PR absorbs red light
- PFR absorbs far red light
- PR is converted to PFR when receiving red light
- PFR is converted to PR when receiving far red light
- During dark, PFR rapidly revert to PR
- The hormone for flowering plant is called florigen
- For long day plant PR absorb red light and converted to PFR
- PFR induce flowering
- For short day plant PFR absorb far red light and converted to PR
- Hence long dark period is required to induces flowering in these short day plant [10 marks]

CHAPTER 11
IMMUNITY

11.1 IMMUNE RESPONSE
(a) Define immunity and state the types of immunity
(b) Describe the general structure of antibodies and state the classes based on
its structure.

1.

a)

Ability of an organism to recognize and defend itself against specific
pathogens and antigens

b)
Antibody / immunoglobulin

c)
Adaptive immunity

d) : Antigen binding site
A : Light chain
B : Heavy chain
C

e)

Disulphide bond / bridge / linkage

f)

IgA - Immunoglobulin A
IgG - Immunoglobulin G
IgM - Immunoglobulin M
IgD - Immunoglobulin D
IgE - Immunoglobulin E

11.1 IMMUNE RESPONSE
(c) State the roles of lymphoid organs in immunity such as:
• Thymus
• Spleen
• Tonsil
• Lymph nodes
• Bone marrow
(d) Explain the various type of antigen and antibody interactions:

• Neutralisation
• Opsonization
• Activation of complement system and pore formation

2.

a)
B lymphocytes. Produced in bone marrow

b)
A: Neutralization
D: Activation of complement system and pore formation

c)
● Antibodies bind to the antigen forming antigen-antibody complex.
● The binding of antibodies to antigen activates the complement protein.
● Activated complement protein generates a membrane attack complex.
● Formation of pores on the plasma membrane.
● Water and ions enter the cell, and the cell of bacteria lysed.

3.
(i) Thymus

Site of development of T lymphocytes / The thymus secretes a hormone,
thymosin, that causes pre-T-cells to mature (in the thymus) into T-cells

(ii) Spleen

To remove worn out blood cells and bacteria from the blood ( blood purifier)

(iii) Tonsils

Defense against bacteria and other foreign agents or bodies

(iv) Lymph nodes

Filters damaged cells, traps foreign molecules and particles such as pathogens /
Macrophages and phagocytes in lymph nodes destroy pathogens and foreign
bodies / lymphocytes in lymph nodes produce antibodies to destroy antigen / are
areas of concentrated lymphocytes and macrophages along the lymphatic veins

(v) Bone marrow

Site for lymphocytes production. B-lymphocytes (B-cells) mature in the bone
marrow. T lymphocytes (T-cells) mature in the thymus gland.

4.
Neutralization

● The antibody blocks viral attachment sites or coats bacterial toxins,
making them ineffective - Phagocytic cells eventually destroy the
antibody-bacteria complex

● In a process called Opsonization, the bound antibodies enhance
macrophage attachment to, and thus phagocytosis of the microbes.

Activation of complement system and pore formation

● The antigen-antibody system activates the complement system
● In an infection, the first in a series of complement fixation proteins is activated,

triggering a cascade of activation steps with each component activating the
next in series.
● Upon completion, it results in the lysis of many types of viruses and
pathogenic cells.

11.2 DEVELOPMENT OF IMMUNITY
(a) State the two types of immune response.
(b) Explain humoral and cell mediated immune response against infection.

5.

a)
Third line of defense

b)
Humoral immune response and cell mediated immune response

c)
B lymphocyte / B cell , T lymphocyte / T cell

d)
● A fragment of foreign protein (antigen) inside the cell associated with
an MHC/MHC I/MHC II molecule and transported to the cell surface.
● The combination of MHC molecule and antigen is recognized by T
cell/cytotoxic T cell/ Helper T cell, alerting it to the infection.

11.2 DEVELOPMENT OF IMMUNITY
(c) Explain the primary and secondary immune response

6.

a)

Bacterial infection occurs and an antigen is presented for the first time. This
is taken for the B cells and T cells to proliferate. B cells differentiate into
plasma cells and specific antibodies are secreted.

b)

Between 0 and 7 days.

c)

Both between 0 and 7 days and between 28 and 35 days

d)

Immunisation/Vaccination

e)
Primary response : IgM
Secondary response : IgG

f)

Generating memory cells ensures that a receptor specific for a particular
epitope will be present. There will be more lymphocytes with this specificity
than in a host that had never encountered the antigen.

7.
● The humoral immune response is an immune response that is mediated
by antibodies produced by the B lymphocytes (B cell)
● When a B lymphocytes recognises a specific antigen,
● it multiplies rapidly through mitosis and produces large number of clones
● Most of clones differentiate into plasma cells while a small number
differentiate into memory cells
● The plasma cells begin to produce large quantities of antibodies specific
to the antigen. The antibodies will bind with and cause the destruction of
antigen
● The memory cells remain in the body even after the infection is over. They
enable a faster and greater immune response to occur if the same
antigen is encountered in the future