Bio p2 Mega Ques

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BIO MYE 2014

Form 4 Chapter 2

Chloroplast Collects light energy and converts it into chemical energy.

Grana is involved in the light reaction of photosynthesis.

Vacuole Q: How does it provide support for herbaceous plants?A: Turgor pressure of the fluid in the vacuole pushes the cell contents and

plasma membrane against the cell wall. This creates support for the plant.

Q: the role to maintain turgidityA: The vacuole maintains the osmotic concentration of the cell sap. If the cell is

flaccid, more water will enter the cell by osmosis.

Q: components that builds up vacuole

A: Cell sap

Cell Wall Q: Components that builds up cell wall

A: Cellulose

Q: Characteristics of cell wallA: Cell wall is tough and rigid to maintain the shape of the cell and provide

mechanical support.

Nucleus Q: function of chromosome

A: Carry genetic material which is transferred from the parent to the offspring.

Golgi Apparatus Packaging, processing and transport center of carbohydrates, proteins andglycoproteins.

Rough Endoplasmic Reticulum Transports proteins made by ribosomes out of the cell

Soft Endoplasmic Reticulum Synthesize lipids and detoxification of drugs

Lysosomes Contain hydrolytic enzymes which hydrolyze damaged organelles, eliminate

waste and absorb useful nutrients

Q: Cell in a concentrated salt solution

A: The cell is plasmolysed. The concentrated salt solution is hypertonic to the cell. Water diffuses out of the cell

through osmosis. The cytoplasm and vacuole shrink. The plasma membrane pulls away from the cell wall.

Q: Why do sperm cells contain mitochondria

A: Sperm cells need a large amount of energy so that it can swim towards the uterus for fertilization.

Q: How xylem vessel is adapted to transport water in plant

A: A xylem vessel is a continuous, hollow tube which will ensure the continuous upward flow of water and minerals

easily. Its wall is made up of lignin so that it will support the plant.

Q: What process do cells undergo to become specific cells that perform specific functions

A: Differentiation

Q: What is meant by cell specialization:A: When cells undergoes differentiation to carry out a specific function

Q: Function of cardiac muscles

A: Contract and relax to pump blood

Q: Importance of sin as integumentary systemA: Protective barrier against infection of pathogens or mechanical injuries. Regulates body temperature with the

presence of hair or by sweating. Eliminate urea by sweating.



Q(E): How Amoeba can carry out life processes such as Nutrition, Locomotion and Excretion

NUTRITION: via phagocytosis using their pseudopodia

Use pseudopodia to engulf food particle and form food vacuole.

Lysosome bind with food vacuole, release hydrolytic enzyme to digest food particle

Digested food is diffused into the cytoplasm to the used.

Undigested food is left behind

LOCOMOTION: Ameboid movement using pseudopodia Pseudopodia extend and their tips are anchored onto the ground followed by slow of cytoplasm into

pseudopodia (cytoplasmic projection)

EXCRETION: carbon dioxide and ammonia is excreted via simple diffusion across plasma membrane

Water excreted via osmoregulation using contractile vacuole

External environment more hypotonic than cytoplasm of amoeba

Cause water to diffuse into amoeba via osmosis and collected in contractile vacuole

When contractile vacuole reaches maximum size, it will contract and expel the water into the external

environment.

Q(E): Comparison of reproduction process between amoeba and paramecium

Reproduction of amoeba is asexual only by binary fission of parent cell when condition conducive

When condition not conducive, form spores. Reproduction of paramecium can be asexual and sexual. Asexual by binary fission of parent cell when

condition conducive.

Paramecium also reproduces through conjugation.

Q(E): Characteristics of xylem tissues and contribution of function

Consists of tracheid and xylem vessels: long tube joined end to end from root to shoot

Ensure continuous water flow from root to shoot

Cell walls thicken with lignin to prevent food substances from entering the cell

Causing cells to die when mature

No cytoplasm and it is hallow when mature

Good for water and mineral flow

Lignin for mechanical support and strength



Form 4 Chapter 3

SBP 20121. Diagram 1.1 shows a plant cell in normal condition, and two conditions of the plant cell when immersed in different

concentrations of sucrose solutions.

(a) (i) Name P, Q and R. [3m]

P: Cell Wall

Q: Cytoplasm

R: Vacuole

(a) (ii) State condition Y of the plant cell. [1m]

A: Turgid

(b) Explain how condition X and condition Y of

the plant cell occur. [4m]

Condition X : Cell maintains shape. Solution

outside the cell

is isotonic to the sap cell of plant cell. Water

diffuses in and out of the cell at equal rate.

Condition Y: Solution outside the cell is hypotonic

to cell sap of plant cell. Water diffuses into thecell by osmosis. Plasma membrane is pressed

against the cell.

(c) A chemical substance inhibits the respiration process in the root hair cells of the plant.

Explain the effect to the transport of mineral ions into the root hair cells. (F5C1) [2m]

A: The cell is unable to produce energy. Active transport does not occur. Thus, mineral ions cannot be transportedinto the cell.

(d) Diagram 1.2 shows a wilted plant after supplied with an excess amount of fertilizer.

Draw a labeled diagram to show the condition of a leaf cell of the plant.

A:



Form 4 Chapter 5

SBP 20122. Diagram 2 shows a meiotic division of an animal

cell.

(a) Name phase S and membrane U. [2m]

Phase S: Phrophase IMembrane U: Nuclear membrane

(b) (i) Explain the chromosomal behavior during phase

S. [2m]

A: Crossing Over. Exchange of genetic material between homologous chromosome.

(b) (ii) Explain how the chromosomal behavior in (b) (i)

contributes to the survival of an animal species. [2m]

A: Crossing over causes variation to occur among

animal species.

This will enhance the ability to adapt in different

environment.

(c) Explain the chromosomal behavior during phase T.[2m]

A: Homologous chromosomes are arranged in line at

metaphase plate

(d) Complete the diagram of the daughter cells. Explain the occurrence of daughter

cells V and W. [4m]

A: During Anaphase 2 Sister chromatids (of one chromosome) are not separated.

Spindle fibres are not fully formed.



Q: Explain process of crossing over

A: Crossing over occurs between the non-sister chromatids of the homologous chromosome. At the chiasmata there is

an exchange of genetic material.

Q: Function of crossing over

A: Gives variation and new recombination of genes to the chromosomes

Q: Function of Meosis

A: Variation and maintain diploid number of chromosomes of sexually reproductive organisms from one generation toanother.

Q: Function of Mitosis

A: Repair damaged cell

Q(E): Cloning

Mammary cells from sheep A were extracted and transferred into low nutrient culture medium. (Shortage of

nutrients cause semi starved cells to switch off their genes and stop dividing)

Unfertilized cell was removed from ovary of sheep B.

Nucleus of egg was extracted and discarded. (enucleated)

The two cells were placed together. An electric pulse was given to stimulate fusion of the cells.

The nucleus from sheep A is put into an enucleated egg cell.

The fused cell walls was grown in culture and divided repeatedly by mitosis.

The early embryo was implanted in the uterus of a surrogate mother.

Surrogate mother gives birth to a sheep genetically identical to Sheep A.

Q(E): Advantages and disadvantages of cloning

ADVANTAGES

Multiply copies of useful genes or clones in a shorter time and in larger numbers – Cloning of Escherichia

Coli that has been genetically manipulated to produce important pharmaceutical drugs like insulin and

growth hormones.

Choosing of desirable characteristics – Trans generic crops which are resistant to pests, diseases andherbicides.

Vegetative reproduction – can take place any time, propagate endanger species, Low cost, less space

DISADVANTAGES

No genetic variation – same level of resistance

Shorter life span

Decreasing biodiversity – Offspring which is genetically identical to parent

Disrupts natural equilibrium of system

May not be safe for consumption – unknown effects, long term effects to consumers



Form 4 Chapter 7

SBP 20126. (a) D6.1 shows the surface view of lower epidermis in a leaf of a

plant. D6.2 shows part of cross section of a woody stem. Explain the

gas uptake for respiration through pores M and N in the plant. [6m]

A: During the day, stoma (M) is open. Oxygen from the atmosphere

diffuses through stoma into intercellular air spaces of mesophyllcells. The cctt of oxygen in the cells becomes lower than the cctt of

oxygen in the air spaces. The difference in concentration gradient

allows oxygen to diffuse continuously from the air spaces into the

cells.

At the lenticels (N) oxygen from atmosphere diffuses into the airspaces between cork cells which are loosely arranged. Oxygen then

diffuses into the stem and root tissues.

(b) Diagram 6.3 shows a rice plant in a paddy field. Explain

the respiration that occurs in the root cells. [6m]

A: The plant carries out anaerobic respiration. Glucose is

broken down in the absence of oxygen to release energy.Ethanol and carbon dioxide are also produced. Cells in the

roots of rice plants are extremely tolerant of ethanol. Many of

the roots are very shallow. This is because the roots use the

oxygen which diffuses into the water surface. Rice stem

contain a large number of air spaces The air spaces allow

oxygen to penetrate through to the cells of roots as it is

growing in the absence of oxygen.

(c) D6.4 shows the changes in the volume of CO2 absorbed or

released by a plant in different light intensity.

Explain the relationship between the rate of photosynthesis &

the rate of respiration in the plant at points P, Q, R & S. [8m]

At P, in low light intensity, only respiration occurs. Hence large

quantity of CO2 is released. As light intensity increases the

quantity of CO2 released decreases because part of CO2 produced during respiration is used for photosynthesis. Sugar is

used in respiration more than it is produced in photosynthesis

At Q, all the CO2 released from respiration is equivalent to CO2

used up during photosynthesis. Rate of photosynthesis is equal

to the rate of respiration.

This point is called compensation point where net gaseous

exchange is zero

At R, as light intensity increases, the rate of photosynthesis become faster than the rate of respiration. The CO2

needed is obtained from the atmosphere while excess O2 is released into the atmosphere.

At S, it is the light saturation point. An increase in light intensity does not increase the rate of photosynthesis.



Q(E): Respiratory structure and breathing mechanism of an insect

Q(E):Aerobic respiration in plant cells – describe intake of oxygen by plants into the cells for respiration

Q(E): Differences between the processes that take place in organelle P and organelle Q

Q(E): Concentration of carbon dioxide in the blood is regulated during vigorous activity (Carotid bodies, carotidartery, aortic bodies, aorta, chemoreceptor in medulla oblongata, respiratory center, heart, diaphragm, peripheral

nerve)

During vigorous activity, the concentration / the partial pressure of carbon dioxide increases as a result of

active cellular respiration

the carbon dioxide react with water to form carbonic acid which results in a drop in the pH level of the blood

and tissue fluid that bathing the brain

The drop in pH is detected by the central chemoreceptors in the medulla oblongata

and detected by peripheral chemoreceptors ( carotid bodies and aortic bodies )

The central chemoreceptors and pheripheral receptors send nerve impulses to the respiratory centre in the

medulla oblongata

The respiratory centre sends nerve impulses to the diaphragm and the intercostal muscles, causing the

respiratory muscle to contract and relax faster

As a result, the breathing and ventilation rate increase causes more oxygen inhaled and the oxygen

concentration return to the normal level

As excess carbon dioxide is eliminated from the body, the carbon dioxide concentration and pH value of the

blood return to normal level

Q: Explain why the pulse rate takes several minutes to return to normal after a vigorous activity.

After vigorous activity, the pulse rate takes several minutes to return to normal because during the activity

the oxygen intake is not able to meet the oxygen demand of the body.

Respiration has to take place anaerobically /anaerobic respiration occur

As a result, lactic acid accumulates in the muscle.

So more oxygen is needed to oxidize the lactic acid and to provide the energy for the recovery of the muscle.

Q: Explain the process that occurs in theroots of both plants which result in the

condition shown in diagram 6.2(b) and

6.3(b).

Anaerobic respiration process

involves the breaks down of glucose in the absence of oxygen or in a limited supply of oxygen to release

energy

Oxidation of glucose is incomplete which 2 molecules of ATP are produced

The by-products are ethanol and energy

Take place in the cytoplasm

Terrestrial Plant respires aerobically.

Aerobic respiration process involves the oxidation of glucose in the presence of oxygen to release energy

In situation of waterlogged condition // less amount of dissolve oxygen





Form 5 Chapter 2

SBP 2012

3. Diagram 3.1 shows flight muscles of a bird.

(a) Name X and Y. [2m]

X: Pectoralis major muscleY: Tendon

(b) Explain why structure X has an abundance of

mitochondria. [2m]

A: Mitochondria produce energy. Structure X

needs a lot of energy for contraction.

(b) (ii) Explain the function of X in locomotion of bird. [2m]

A: When X contracts, it will create a pulling force. Thus the wings are pulled down.

(c) Explain the effect to the locomotion of bird if structure Y is torn. [2m]A: Pulling force created by contraction of muscle cannot be transferred to the bone. Bone cannot be pulled downward.

(d) Diagram 3.2 shows two types of vertebrae in human

backbone.

State two differences in structure between the vertebra P andvertebra Q. [2m]

(e) A man has swollen ankle and is painful during movement

after having a habit of taking high protein diet. Explain the

relationship between the diet and the disease he

suffers. [2m]

A: He suffers gout arthritis due to deposition of uric acid in his joint. His diet contains a lot of protein. (Gout is a type

of arthritis. It occurs when uric acid builds up in blood and causes inflammation in the joints.)

Structure P Structure Q

Smaller centrum Larger centrumHas transverse foramen Has no transverse foramen

Has short transverse process Has long transverse process

Has large spinous process Has small spinous process



Form 5 Chapter 3

SBP 2012

7. (a) Diagram 7.1 shows the negative feedback mechanism in regulating the water balance in human blood. Explain

the corrective mechanism when the osmotic pressure of the blood increases. [6m]

A: Osmotic pressure of the blood increases whenwater content in the blood is low. This is detected

by the osmoreceptor cells in the hypothalamus and

are stimulated. Hypothalamus send nerve impulse

to pituitary gland to secrete more ADH into the

blood. The adrenal gland is not stimulated to

release aldosterone so less amount of salt isreabsorbed. ADH is transported by blood to

kidney. Distal convoluted tubule and collecting

duct become more permeable to water. More water

reabsorbed by blood. Less urine which is more

concentrated and darker in colour is produced.Blood osmotic pressure returns to normal. The

decrease in blood osmotic pressure creates a

negative feedback mechanism so that more ADH is

not secreted.

(b) Diagram 7.2 shows the sequence of

organs and tissue that responded when a

man was attacked by a robber.

Explain the involvement of nervous

system and endocrine system in this

situation. [8m]

The receptors in the eyes detect stimulus. The receptors trigger nerve impulses to be sent to the brain. The brain

integrates and interprets the information. Nerve impulses is sent to adrenal gland. Adrenaline is secreted and

transported by blood to the liver, lungs and heart.In the liver, glycogen converted into glucose. There is an increase in

breathing rate, ventilation rate and heartbeat. Thus there will be more blood, glucose and oxygen in skeletal muscles.

Skeletal muscles become energized. This enables a person to fight off an attacker or flee immediately from danger.

** the brain needs to be highly alert to mobilize various parts of the body into immediate action

(c) Diagram 7.3 shows transmission of impulse through a synapse.

Explain the effect of a pain killer drug on the transmission of nerveimpulses. [6m]

Nerve impulses reach axon terminal/synaptic knob. This triggers the

synaptic vesicles to move towards the presynaptic membrane and

fuse with the membrane. Neurotransmitters are released into the

synapse/synaptic cleft. Drugs disintegrate the neurotransmitters. Less

neurotransmitters reach post synaptic membrane. Hence less new

nerve impulses transmitted.



Q: Function of spinal cord

A: Controls reflex action

Q: Know how to draw afferent and efferent neurone

Q: Difference between Afferent and Efferent NeuroneA: Both neurone has myelin sheaths, axons, cell bodies and dentrites. Afferent neurone has a cell body at the side of

the cell. The efferent neurone has a cell body at the end of the cell. The Affarent neurone has a short axon while the

efferent neurone has a long axon.

Q: Define synapse

A: The spaces between the pre-synaptic membrane of a neurone and the post-synaptic membrane of a neurone.Transmission of newve impulses take place in the synapse.

Q: Transmission of chemical signals across the synapse

Electrical impulse reaches presynaptic membrane

It triggers synaptic vesicles to move towards the presynaptic membrance and fuse with the membrane Neurotransmitters are released into the synaptic cleft

Neurotransmitters bind to specific receptors on the postsynaptic membrane

This leads to the generation of new electrical signals

Impulse moves along the post-synaptic neurone

*the release of neurotransmitters is in one direction, from the synaptic knob to the postsynaptic neurone.

Q: explain knee jerk reflex

When hammer strikes knee, sensory receptors in quadriceps muscles are stimulated.

Nerve impulses are triggered

Afferent neurone transmits nerve impulses to efferent neurone in the spinal cord

Neurotransmitters are released and diffuses across the synapse.

This triggers the efferent nerurone to send new nerve impulses to the effector which is the quadriceps muscle Quadriceps muscle contracts, causing leg to swing forward

Q: Is the brain needed for knee jerk reflex action to occur

A: No. This type of reflex only involve the spinal cord. The brain is reserved for more complex tasks.

8A

P1 - Blood from the renal artery enters the glomerulus through the afferent arteriole and out through the efferent

arteriole.

P2 - The diameter of afferent arteriole is bigger than the diameter of the efferent arteriole causing a high hydrostatic

pressure in the glomerulus.P3 - The fluid and solute in the glomerulus is filtered out through the capillary wall into the lumen of the Bowman’s

capsule.

P4 - The glomerular filtrate has the same composition as the plasma except that it does not contain any of the largercomponent such as red blood cell and plasma proteins.

P5 - The process of filtration produced as a result of this pressure is known as ultrafiltration.

8B

P1- When the osmotic pressure of blood is lower than usual, the osmoreceptor cells in the hypothalamus are less

stimulated/ the pituitary gland is less stimulated, less ADH is secreted.

P2- (A lower level of ADH) causes the structure P/ distal convoluted tubule and structure Q/ collecting duct to

become impermeable to water .

P3 - Less water is reabsorbed( from the filtrate into the blood).

P4 - The blood osmotic pressure rises / return to normal range

P5 - When the osmotic pressure of blood is higher than usual the osmoreceptor cells in the hypothalamus are

stimulated /the pituitary gland is stimulated, more ADH is secreted.

P6 - (A higher level of ADH) causes the distal convoluted tubule and collecting duct to become more permeable towater .



P7 - More water is reabsorbed( from the filtrate into the blood).

P8 - The blood osmotic pressure drops / return to normal range.

8Ci

-The regulation of the physical and chemical factors in the internal environment to maintain a constant internal

environment for the survival of human.

-Physiological processes in the body can proceed at optimum rates in order to promote harmonious growth anddevelopment OR Enzymatic and other metabolic processes and reactions in the body will not be able to function

normally without homeostasis.

8Cii

P1 - When the level of glucose in blood increases, beta cell from islet of langerhans/ langerhans cells in the pancrease

are stimulated.

P2 - Insulin / Hormone X is produced and secreted into the bloodstream.

P3 - Hormone X is carried in the blood by hepatic portal vein to the liver.P4 - In the liver, hormone X /insulin converts the excess glucose in the blood to glycogen.

P5 - In liver cells, the excess glucose in the blood will be converted to lipids

P6 - This causes the level of glucose to decrease and return to normal.

P1 – When the level of glucose in blood decreases, alpha cell from islet of langerhans/ langerhans cells in the pancrease are stimulated.

P2 – Glucagon/ Hormone Y is produced and secreted into the bloodstream.

P3 – Hormone Y is carried in the blood by hepatic portal vein to the liver.

P4 – In the liver, hormone Y / glucagon converts the stored glycogen in the liver to glucose.

P5 - Glucagon also increases the conversion of amino acids and fatty acids in the liver cells to glucose

P6 - This causes the level of glucose to rise and return to normal



Form 5 Chapter 4

SBP 20125. Diagram 5 shows the changes and regulation of hormones during a menstrual cycle.

(a) Explain the effect of follicle stimulating

hormone (FSH) on structure P. [2m]

A: (FSH) stimulates the growth / development ofthe primary follicle. Structure P will grow /

develop / becomes secondary follicle / Grafiaan

follicle.

(b) Explain the effect if the level of hormone Y

is low.

A: Ovulation will not occur. Grafiaan follicle

will not release the secondary oocyte (into the

Fallopian tube).

(c) Explain one difference between primary

oocyte and secondary oocyte.

A: Primary oocyte is diploid while secondary

oocyte is haploid.

(d) (i) Complete the graph starting from point X

in Diagram 5 to show the changes in the

thickness of endometrium wall if fertilization

does not occur. [1m]

A: Able to complete the graph that shows the

thickness of endometrium wall is decreasing /

thinning.

(d) (ii) Explain the graph that you drawn in d(i). [2m]

A: Corpus luteum degenerate. No progesterone is secreted (to thicken the endometrium wall).

(e) Explain the importance of structure Q during foetal development. [3m]

A: Q secrete progesterone. Progesterone will thicken and maintain the endometrium wall. Endometrium wall ready for

implantation of embryo // prevent miscarriage / abortion.

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Bio p2 Mega Ques